Introduction to Real Analysis (Instructor Solution Manual) [1 ed.] 3030269019, 9783030269012

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Christopher Heil

Introduction to Real Analysis: Solutions Manual

Last Updated: April 21, 2019

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2019 by Christopher Heil

Solutions to Exercises and Problems

These are my solutions to most of the exercises and problems from my test “An Introduction to Real Analysis.” Additionally, there are detailed verifications of claims stated in the text without proof in the course of various lemmas, theorems, and so forth. Of course, many problems have solutions other than the ones I sketch here In particular, there could very well be easier solutions than the ones I give (and over the years, my students have often shown me better solutions, many of which have been incorporated here). These solutions have not been proofread as carefully as has the text proper, so the probability of errors is correspondingly higher. Please send comments and corrections to “[email protected]”. Non-Distribution Notice. This solutions file is meant for the use of course instructors only. Please do not repost or further distribute this file without the permission of the author. It can be helpful to share a few selected solutions with students after they have worked the problems themselves, but even then it is best to keep the distribution limited. In order to learn, students have to struggle to find their own solutions, there simply is no substitute for this effort, and looking at a solution too early takes this opportunity away from the student. Christopher Heil.

Solutions to Exercises and Problems from Chapter 1 1.1.6 “⇒.” Assume that E is closed and there exist points xn ∈ E such that xn → x ∈ X. If xn = x for some n then x = xn ∈ E, so we may assume that xn 6= x for every n. In this case x is an accumulation point of E. Suppose that it was the case that x ∈ / E. Then E C is an open set that contains x, so there must exist some r > 0 such that Br (x) ⊆ E C . But then every point of E is at least a distance r from x. In particular, d(xn , x) > r 1

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for every, which contradicts the fact that xn → x. Therefore we must have x ∈ E. “⇐.” Suppose that the limit of every convergent sequence of points from E belongs to E. If E is not closed then E C is not open, so there must exist some point x ∈ E C such that no open ball Br (x) centered at f is entirely contained in E C . Considering r = n1 in particular, this tells us that there must exist a point xn ∈ B1/n (x) that is not in E C . But then xn ∈ E, and we have d(xn , x) < n1 , and therefore xn → x. By hypothesis, it follows that x ∈ E, which is a contradiction since x belongs to E C . Consequently E must be closed. 1.1.7 (a) Let F be the set of all possible limits of elements of E:  F = y ∈ X : there exist xn ∈ E such that xn → y .

We must show that F = E. Choose any point y ∈ F. Then, by definition, there exist points xn ∈ E such that xn → y. Since E ⊆ E, the points xn all belong to E. Hence y is a limit of elements of E. But E is a closed set, so it must contain all of these limits. Therefore y belongs to E, so we have shown that F ⊆ E. In order to prove that E is a subset of F, we will first prove that F C is an open set. To do this, choose any point y ∈ F C . We must show that there is a ball centered at y that is entirely contained in F C . That is, we must show that there is some r > 0 such that Br (y) contains no limits of elements of E. Suppose that for each k ∈ N, the ball B1/k (y) contained some point from E, say xk ∈ B1/k (y) ∩ E. Then these xk are points of E that converge to y (why?). Hence y is a limit of points of E, which contradicts the fact that y ∈ / F. Hence there must be at least one k such that B1/k (y) contains no points of E. We will show that r = 1/k is the radius that we seek. That is, we will show that the ball Br (y), where r = 1/k, not only contains no elements of E but furthermore contains no limits of elements of E. Suppose that Br (y) did contain some point z that was a limit of elements of E, i.e., suppose that there did exist some xn ∈ E such that xn → z ∈ Br (y). Then, since d(y, z) < r and since d(z, xn ) becomes arbitrarily small, by choosing n large enough we will have d(y, xn ) ≤ d(y, z) + d(z, xn ) < r. But then this point xn belongs to Br (y), which contradicts the fact that Br (y) contains no points of E. Thus, Br (y) contains no limits of elements of E. Since F is the set of all limits of elements of E, this means that Br (y) contains no points of F. That is, Br (y) ⊆ F C . In summary, we have shown that each point y ∈ F C has some ball Br (y) that is entirely contained in F C . Therefore F C is an open set. Hence, by definition, F is a closed set. We also know that E ⊆ F (why?), so F is one of the closed sets that contains E. But E is the smallest closed set that contains E, so we conclude that E ⊆ F.

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(b) “⇒.” Assume that E is dense in X. Then E = X, so part (a) implies that every point x ∈ X is a limit of elements of E. “⇐.” Suppose that every point x ∈ X is a limit of elements of E. Then part (a) implies that E = X, so E is dense in X. 1.1.14 Let {Vi }i∈J be any open cover of f (K). Each set Ui = f −1 (Vi ) is open, and {Ui }i∈J is an open cover of K. Since K is compact, this cover must have a finite subcover {Ui1 , . . . , UiN }. But then {Vi1 , . . . , ViN } is a finite subcover of f (K), so f (K) is compact. 1.1.15 For this proof we let BrX (x) and BsY (y) denote open balls in X and Y, respectively. (a) ⇒ (b). Suppose that f is continuous, and
choose any point x ∈ X and any ε > 0. Then the ball V = BεY f (x) is an open subset of Y, so U = f −1 (V ) must be an open subset of X. As x ∈ U, there exists some δ > 0 such that BδX (x) ⊆ U. If y ∈ X is any point that satisfies dX (x, y) < δ, then we have y ∈ BδX (x) ⊆ U, and therefore
f (y) ∈ f (U ) ⊆ V = BεY f (x) .
Consequently dY f (x), f (y) < ε.

(b) ⇒ (c). Assume that statement (b) holds, choose any point x ∈ X, and suppose that xn ∈ X are such that xn → x. Fix any ε > 0, and let δ > 0 be the number whose existence is given by statement (b). Since xn → x, there must exist some N > 0 such that dX (x,
xn ) < δ for all n > N. Statement (b) therefore implies that dY f (x), f (xn ) < ε for all n > N, so we conclude that f (xn ) → f (x) in Y.

(c) ⇒ (a). Suppose that statement (c) holds, and let V be any open subset of Y. Suppose that f −1 (V ) were not open in X. Then there is some point x ∈ f −1 (V ) such that there is no radius r > 0 for which the open ball Br (x) is a subset of f −1 (V ). In particular, we have for each n ∈ N that the ball B1/n (x) is not contained in f −1 (V ), and therefore some point xn ∈ B1/n (x) such that xn ∈ / f −1 (V ). As a consequence, d(x, xn ) < 1/n for every n, but f (xn ) ∈ / V for any n. Now, x ∈ f −1 (V ), so f (x) does belong to V. Since V is open, there is some open ball centered at f (x) that is entirely contained in V. That is, there is some radius ε > 0 such that Bε (f (x)) ⊆ V. On the other hand, we have xn → x, so by applying statement (c) we must have f (xn ) → f (x). Consequently, there is some N > 0 such that d(f (x), f (xn )) < ε for all n ≥ N. But then f (xN ) ∈ Bε (f (x)) ⊆ V, which contradicts the fact that f (xN ) ∈ / V. This is a contradiction, so f −1 (V ) must be open, and therefore f is continuous.

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1.1.19 (a) Suppose that x 6= y, and let r = d(x, y)/2. By the definition of a metric, we have r > 0. If z ∈ Br (x) ∩ Br (y), then d(x, y) ≤ d(x, z) + d(z, y) < 2r = d(x, y), which is a contradiction. Therefore Br (x) ∩ Br (y) = ∅, so we can simply take U = Br (x) and V = Br (y). (b) Suppose that xn → x and xn → y. Then 0 ≤ d(x, y) ≤ d(x, xn ) + d(xn , y) → 0

as n → ∞.

Consequently d(x, y) = 0, so x = y. 1.1.20 Suppose (xn )n∈N is a Cauchy sequence and there exists a subsequence (xnk )k∈N that converges to x. Then given ε > 0, there exists some K > 0 such that d(xnk , x) < ε for all nk > K. Also, there exists an N such that d(xm , xn ) < ε for all m, n > N. Suppose that n > N. Then since the nk are strictly increasing, there exists some nk that is greater than both K and N. For this nk we have d(xn , x) ≤ d(xn , xnk ) + d(xnk , xn ) < ε + ε = 2ε. This is true for all n > N, so xn → x.

1.1.21 (a) We are given that d(xn , xn+1 ) < 2−n for every n. Choose any ε > 0, and let N be large enough that 2−N +1 < ε. If n > m > N, then d(xm , xn ) ≤

n−1 X

k=m

d(xk , xk+1 ) ≤

∞ X 1 1 1 = m−1 < N −1 < ε. 2k 2 2

k=m

Hence {xn }n∈N is Cauchy. (b) Let {xn }n∈N be a Cauchy sequence in X. Then there exists an integer N1 such that 1 m, n ≥ N1 =⇒ d(xm , xn ) < . 2 Set n1 = N1 . Then there exists an N2 such that m, n ≥ N2 =⇒ d(xm , xn )
N1 . Let n2 = N2 . Then we have n1 , n2 ≥ N1 , so 1 d(xn2 , xn1 ) < . 2 Continuing in this way, we can construct n1 < n2 < · · · such that ∀ k ∈ N,

d(xnk+1 , xnk ) < 2−k .

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1.1.22 “⇒.” This direction is immediate. “⇐.” Suppose that every subsequence of (xn )n∈N has a subsequence that converges to x, but the full sequence (xn )n∈N does not converge to x. Then there exists an ε > 0 such that given any N we can find an n > N such that d(xn , x) > ε. Iterating this, we can find a subsequence (xnk )k∈N such that d(xnk , x) > ε for every k. But then no subsequence of (xnk )k∈N can converge to x, which is a contradiction. 1.1.23 “⇒.” Suppose that xt → x as t → 0+ , and let (tk )k∈N be any sequence of real numbers in (0, c) such that tk → 0. Choose any ε > 0. Then since xt → x, there exists a δ with 0 < δ < c such that d(xt , x) < ε whenever 0 < t < δ. Since tk → 0, there exists an N > 0 such that 0 < tk < δ for all k ≥ N. Hence for every k ≥ N we have d(xtk , x) < ε, and therefore xtk → x. “⇐.” Suppose that xt 6→ x as t → 0+ . Then there exists an ε > 0 such that for each δ = 1/k we can find a real number tk ∈ (0, c) with 0 < tk < 1/k but d(xtk , x) ≥ ε. Hence (tk )k∈N is a sequence of real numbers in (0, c) such that tk → 0, but xtk 6→ x as k → ∞. Remark: By passing to a subsequence of {tk }k∈N we can obtain a subsequence {sj }j∈N such that sj decreases monotonically to zero yet xsj 6→ x as j → ∞. 1.1.24 (a) Fix x ∈ Rd and ε > 0. We are given that  h(x) = inf g(y) : y ∈ Br (x)

and h(x) 6= −∞. Since h(x) is a real number, by definition of inf there must be some z such that |z − x| < r

and

g(z) ≤ h(x) + ε.

Let δ = r − |z − x|, and suppose that |y − x| < δ. Then |y − z| ≤ |y − x| + |x − z| < δ + |x − z| = r, so z ∈ B(y). Therefore

 h(y) = inf g(t) : t ∈ B(y) ≤ g(z) ≤ h(x) + ε.

Since this is true for all y with |y − x| < δ, we conclude that h is usc at x. The result can fail if h(x) = −∞ at some point. For example, if ( 0, x ≤ 0, g(x) = −1/x, x > 0, then

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  x ≤ −1, 0, h(x) = −∞, −1 < x ≤ 1,   −1/(x − 1), x > 1,

and h is not usc at x = 1.

(b) “⇒.” Suppose that f is continuous at x. Then given ε > 0, there is a δ > 0 such that |x − y| < δ =⇒ f (x) − ε ≤ f (y) ≤ f (x) + ε. Hence f is both usc and lsc at x. “⇐.” Suppose f is both usc and lsc at x and fix ε > 0. By definition of usc, there is a δ1 > 0 such that |x − y| < δ1 =⇒ f (y) ≤ f (x) + ε. Similarly, by definition of lsc, there is a δ2 > 0 such that |x − y| < δ2 =⇒ f (y) ≥ f (x) − ε. Taking δ = min{δ1 , δ2 }, we see that if |x − y| < δ, then f (x) − ε ≤ f (y) ≤ f (x) + ε, which implies that |f (x) − f (y)| ≤ ε. Hence f is continuous at x. (c) Fix ε > 0. Since g(x) = inf fα (x), α∈J

the definition of infimum implies that there must exist an α ∈ J such that fα (x) ≤ g(x) + ε. Since fα is usc at x, there exists a δ > 0 such that |x − y| < δ =⇒ fα (y) ≤ fα (x) + ε. Hence, if y satisfies |x − y| < δ, then g(y) ≤ fα (y) ≤ fα (x) + ε ≤ g(x) + 2ε. Therefore g is usc at x. (d) “⇒.” Assume that f is usc at every point x ∈ Rd . To show that {f ≥ a} is closed, suppose that xn ∈ {f ≥ a} and xn → x as n → ∞. Given ε > 0, we have by definition that there is some δ > 0 such that |x − y| < δ =⇒ f (y) ≤ f (x) + ε. Since xn converges to x, there is some N ∈ N such that

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|x − xn | < δ,

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n ≥ N.

Therefore a ≤ f (xn ) ≤ f (x) + ε,

n ≥ N.

Hence a ≤ f (x) + ε, and since ε is arbitrary, it follows that a ≤ f (x). Thus x ∈ {f ≥ a}, which implies that {f ≥ a} is closed.

“⇐.” Now suppose that {f ≥ a} is closed for each a ∈ R. Then {f < a} is open for each a. Fix x ∈ Rd , and choose ε > 0. If we set a = f (x) + ε then we have f (x) < a. Therefore x ∈ {f < a}, which is an open set. Consequently, there exists some δ > 0 such that the open ball Bδ (x) is contained in {f < a}. This implies that |x − y| < δ =⇒ f (y) < a = f (x) + ε. Hence f is usc at x. (e) Suppose that f was not bounded above on K. Then for each integer n there would exist a point xn ∈ K such that f (xn ) > n. Since K is compact, the sequence {xn }n∈N must have a convergent subsequence, i.e., there exist integers nk and a point x0 ∈ K such that xnk → x0 . But f is usc and f (xnk ) > nk , so f (x) ≥

lim sup f (x) ≥ lim sup f (xnk ) = ∞.

x→x0 ,x∈K

k→∞

This contradicts the fact that f was assumed to be finite at every point. 1.2.4 (a) Given x, y ∈ X, kxk = k(x − y) + yk ≤ kx − yk + kyk, so kxk − kyk ≤ kx − yk. By reversing the roles of x and y we obtain the inequality kyk − kxk ≤ kx − yk, so we conclude that kxk − kyk ≤ kx − yk. (b) Assume xn → x and fix ε > 0. Then there is an N > 0 such that kx − xn k < ε for all n > N. Therefore, if m, n > N then kxm − xn k ≤ kxm − xk + kx − xn k < 2ε, so {xn }n∈N is Cauchy.

(c) Suppose that {xn }n∈N is Cauchy. Then there exists an N > 0 such that kxm − xn k < 1 for all m, n ≥ N. Therefore, for n ≥ N we have kxn k = kxn − xN + xN k ≤ kxn − xN k + kxN k ≤ 1 + kxN k. Hence, for an arbitrary n we have  kxn k ≤ max kx1 k, . . . , kxN −1 k, kxN k + 1 .

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(d) By the Reverse Triangle Inequality, if xn → x, then kxk − kxn k ≤ kx − xn k → 0.

(e) If xn → x and yn → y, then

k(x + y) − (xn + yn )k = k(x − xn ) + (y − yn )k ≤ kx − xn k + ky − yn k → 0. (f) Suppose xn → x and cn → c. Then C = sup |cn | < ∞, so kcx − cn xn k ≤ kcx − cn xk + kcn x − cn xn k = |c − cn | kxk + |cn | kx − xn k ≤ |c − cn | kxk + C kx − xn k → 0. 1.2.11 If y, z are any two points in Br (x) and 0 ≤ t ≤ 1, then




x − (1 − t)y + tz = (1 − t)(x − y) + t(x − z)

≤ (1 − t) k(x − y)k + t kx − zk

< (1 − t) r + tr = r. Therefore (1 − t)y + tz ∈ Br (x), so Br (x) is convex. 1.2.12 “⇐.” Suppose that Y is a closed subspace of X, and let {xn }n∈N be a Cauchy sequence in Y. Then {xn }n∈N is a Cauchy sequence in X, so there exists some vector x ∈ X such that xn → x. Since Y is closed, this vector x must belong to Y. Therefore every Cauchy sequence in Y converges to an element of Y, so Y is complete. “⇒.” Suppose that Y is a Banach space with respect to the norm of X. Suppose that xn ∈ Y and xn → x ∈ X. Then {xn }n∈N is a Cauchy sequence in Y and hence must converge to some vector y ∈ Y. Thus, {xn }n∈N is a sequence in X such that xn → x and xn → y. By uniqueness of limits, we conclude that x = y ∈ Y. Exercise 1.1.6 therefore implies that Y is closed. PN 1.2.13 The partial sums sN = n=1 xn of the series converge to x in norm, so by the continuity of the norm and the Triangle Inequality (which by induction applies to any finite sum), we see that

∞

X

xn

= kxk =

n=1

lim ksN k

N →∞

X

N

xn = lim

N →∞ n=1

(continuity of the norm) (definition of sN )

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≤ =

lim

N →∞ ∞ X

n=1

9 N X

n=1

kxn k

kxn k

(Triangle Inequality) (definition of infinite series).

1.2.14 Suppose that f, g ∈ span(S) are given. Then there exist functions fn , gn ∈ S such that fn → f and gn → g in norm. Therefore fn + gn → f + g in norm. As fn + gn ∈ S for every n, it follows that f + g belongs to the closure of S, which is span(S). Therefore span(S) is closed under vector addition, and a similar argument shows that it is closed under scalar multiplication. Therefore span(S) is a subspace of X. By definition span(S) is a closed set, so it is a closed subspace. (b) Suppose that M is a closed subspace of X and S ⊆ M. Since M is closed under vector addition and scalar multiplication, it follows that S ⊆ M. Since M is closed under limits, it follows that M contains every limit of elements of S. The set of all such limits is the closure of the span, so we have shown that span(S) ⊆ M. 1.3.2 Note first that if f (x) = ∞ for some x, then kf − fn ku = ∞ since fn is bounded. This contradicts the assumption that kf − fn ku → 0, so f (x) must be a scalar for every x. Let x be any fixed point in X, and choose ε > 0. Then, by definition of uniform convergence, there exists some integer n > 0 such that kf − fn ku < ε (in fact, this will be true for all large enough n, but we need only one particular n for this proof). Since fn is continuous, there is a δ > 0 such that for all y ∈ X we have d(x, y) < δ

=⇒

|fn (x) − fn (y)| < ε.

Consequently, if y ∈ I is any point that satisfies d(x, y) < δ, then |f (x) − f (y)| ≤ |f (x) − fn (x)| + |fn (x) − fn (y)| + |fn (y) − f (y)| ≤ kf − fn ku + ε + kfn − f ku < ε + ε + ε = 3ε. Therefore f is continuous. To see that f is bounded, observe that fn is bounded because it belongs to Cb (X), and f − fn is bounded because kf − fn ku < ε. As the sum of two bounded functions is bounded, we conclude that f = fn + (f − fn ) is bounded, and therefore f ∈ Cb (X). 1.3.5 Suppose that c1 , . . . , cN are scalars that are not all zero, and p(x) =

N X

k=0

ck pk (x) =

N X

k=0

ck xk

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is such that p = 0 on I. Without loss of generality, we may assume that cN 6= 0. Then p is a nonzero polynomial of degree N and every x ∈ I is a root of p. Therefore p has uncountably many roots. Yet the Fundamental Theorem of Algebra states that p has only finitely many roots, so this is a contradiction. Therefore every ck must be zero. 1.3.6 “⇒.” Suppose that f : Rd → C is uniformly continuous, and fix ε > 0. Then there exists some δ > 0 such that kx − yk < δ =⇒ |f (x) − f (y)| < ε. Consequently, if kak < δ then kx − (x − a)k = kak < δ for every x, so kf − Ta f ku = sup |f (x) − f (x − a)| ≤ ε. x∈R

This says that kf − Ta f ku → 0 as a → 0. “⇐.” Suppose that kf − Ta f ku → 0, and fix ε > 0. Then there exists some δ > 0 such that kf − Ta f ku < ε whenever kak < δ. Consequently, if kx − yk < δ and we set a = x − y, then |f (x) − f (y)| = |f (x) − f (x − a)| ≤ kf − Ta f ku < ε. Hence f is uniformly continuous. 1.3.7 Suppose that {fn }n∈N is Cauchy in C0 (Rd ) (with respect to the uniform norm). Then it is Cauchy in Cb (Rd ), which is complete, so there exists a function f ∈ Cb (Rd ) such that fn → f uniformly. We need only show that f ∈ C0 (Rd ). To do this, fix ε > 0. Then there exists an integer N > 0 such that kf − fn ku < ε for all n ≥ N. In particular, kf − fN ku < ε. Since fN ∈ C0 (Rd ), there exists an R > 0 such that |fN (x)| < ε for all kxk > R. Hence for |x| > R we have |f (x)| ≤ |f (x) − fN (x)| + |fN (x)| ≤ ε + ε = 2ε. Thus f (x) → 0 as |x| → ∞, so f ∈ C0 (Rd ), and therefore C0 (Rd ) is complete. Let f be any function in C0 (R). If we fix ε > 0, then there exists some R > 0 such that |f (x)| < ε/2 for all kxk > R. Let B be the closed unit ball of radius 2R centered at the origin. This is a compact set, so f is uniformly continuous on B. Hence there exists a δ < R such that if a, b ∈ B and ka − bk < δ, then |f (a) − f (b)| < ε. Now choose any points x, y ∈ Rd such that kx−yk < δ. If x and y both belong to B, then |f (x)−f (y)| < ε. Suppose that x ∈ / B. In this case kxk > 2R. Hence, by the Reverse Triangle Inequality, kyk ≥ kxk − ky − xk ≥ 2R − δ > R. Therefore kxk, kyk > R, so

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|f (x) − f (y)| ≤ |f (x)| + |f (y)|
N + 1.

Each function gN belongs to Cc (R). If N ≤ |x| ≤ N +1 then |gN (x)| ≤ |g(N )|. Since g(x) → 0 as |x| → ∞, it follows that lim kg − gN ku =

N →∞

≤

lim

sup |g(x) − gN (x)|

N →∞ |x|≥N

lim


sup |g(x)| + |g(N )|

N →∞ |x|≥N

= 0.

Thus gN → g uniformly. As a consequence, {gN }N ∈N is a Cauchy sequence with respect to the uniform norm. Since uniform convergence implies pointwise convergence, the function g is the only function that the sequence {gN }N ∈N can converge to uniformly. But g does not belong to Cc (R), so although {gN }N ∈N is a Cauchy sequence in Cc (R), it does not converge uniformly to any element of Cc (R). Therefore Cc (R) is not complete. 1.3.11 Each function gk belongs to Cc (R), because gk is continuous and it is supported on the compact interval [−2k , 2k ]. The uniform norm of gk is kgk ku = 2−k . Consequently ∞ X

k=1

P

kgk ku =

∞ X

k=1

2−k = 1 < ∞,

so the series gk converges absolutely in Cc (R). It also converges P absolutely in the larger space C0 (R). The series converges in C0 (R), but g = gk is an element of C (R) that is not compactly supported, and as a consequence the 0 P series gk does not converge in the space Cc (R).

1.4.2 Case 1: Real-Valued Functions Assume that f is a differentiable realvalued function on I whose derivative is bounded, and set

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K = kf ′ ku = sup |f ′ (t)|. t∈I

is finite. Choose any two points x < y in I. Then f is differentiable everywhere on the interval (x, y) and is continuous on [x, y]. Because f is real-valued, the Mean Value Theorem therefore implies that there exists a point c between x and y such that f (y) − f (x) f ′ (c) = . y−x Rearranging, we see that

|f (y) − f (x)| = |f ′ (c)| |y − x| ≤ K |y − x|. Case 2: Complex-Valued Functions Suppose that f is a differentiable complex-valued function on I whose derivative is bounded, write f = g + ih where g and h are real-valued. Since f ′ is bounded and f ′ = g ′ +ih′ , the functions g ′ and h′ are bounded. Since g is real-valued, Case 1 implies that g is Lipschitz and kg ′ ku is a Lipschitz constant for g. Similarly h is Lipschitz with Lipschitz constant kh′ ku . Therefore, given any points x, y ∈ I, we compute that 1/2  |f (x) − f (y)| = |g(x) − g(y)|2 + |h(x) − h(y)|2 1/2  ≤ kg ′ k2u |x − y|2 + kh′ k2u |x − y|2 = K |x − y|,

where K = kg ′ ku + kh′ ku .

1.4.3 Since h(x) = x2 sin(1/x), we have for x 6= 0 that h′ (x) = 2x sin For x = 0,

 1 1  1 1 1 + x2 cos − 2 = 2x sin − cos . x x x x x

h(0 + t) − h(0) t2 sin(1/t) = lim = lim t sin(1/t) = 0. t→0 t→0 t→0 t−0 t

h′ (0) = lim

Hence h is differentiable everywhere and h′ is bounded on [−1, 1], although h′ is not continuous at x = 0. Therefore Problem 1.4.2 implies that h is Lipschitz on [−1, 1]. 1.4.4 (a) Uniform continuity follows immediately from the definition of H¨ older continuity, just as it does for Lipschitz continuity. (b) If f is H¨ older continuous with α > 1 then

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α f (x) − f (y) ≤ lim C |x − y| = lim C |x − y|1−α = 0. lim y→x y→x y→x x−y |x − y|

Therefore f is differentiable and f ′ (x) = 0 for every x, so f is constant. (c) Let f (x) = x1/2 . If 0 ≤ x < y < ∞ then |f (x) − f (y)| = (y 1/2 − x1/2 ) =

y 1/2 + x1/2 y 1/2 + x1/2

y−x y 1/2 + x1/2

= (y − x)1/2

(y + x)1/2 y 1/2 + x1/2

≤ |y − x|1/2

(2y)1/2 y 1/2

= 21/2 |y − x|1/2 , so f is H¨ older continuous on [0, ∞) with exponent α = 1/2. The sum of two functions that are H¨ older continuous with exponent α also is H¨ older continuous with the same exponent, so f is H¨ older continuous on R with exponent α = 1/2. Suppose f was H¨ older continuous for some exponent α > 1/2. Then for y > 0 we would have y 1/2 = |f (0) − f (y)| ≤ C |0 − y|α = Cy α . 1

Hence y 2 −α ≤ C for all y > 0. Letting y → 0 gives a contradiction.

(d) The function g is continuous on the closed interval [0, 21 ], so it is uniformly continuous on this interval. As g is constant outside of [0, 12 ], it is uniformly continuous on the real line. Fix any constant α > 0. Then lim

x→0+

1 |f (x) − f (0)| = − lim+ α |x − 0|α x→0 x ln x = − lim

x→0+

= lim

x→0+

x−α ln x

αx−α−1 x−1

= lim αx−α x→0+

= ∞. Therefore g is not H¨ older continuous at the origin for any exponent α > 0.

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1.4.5 (a) By definition, 0 ≤ kf kC α < ∞ for each f ∈ C α (I). Suppose that kf kC α = 0. Then kf ku = 0, so f = 0 (note f is continuous). If f ∈ C α (I) and c ∈ C, then

|cf (x) − cf (y)| = |c| kf kC α . |x − y|α

kcf kC α = kcf ku + sup x6=y

Suppose f, g ∈ C α (I). Then kf + gkC α = kf + gku + sup x6=y

|(f + g)(x) − (f + g)(y)| |x − y|α

≤ kf ku + kgku + sup x6=y

|f (x) − f (y)| + |g(x) − g(y)| |x − y|α

≤ kf kC α + kgkC α . Therefore k · kC α satisfies the Triangle Inequality, therefore is a norm.

We will use the criterion of Theorem 1.2.8 to show that C α (I) is complete, i.e., we will show that every absolutely convergent series in C α (I) is convergent. Suppose that {fn }n∈N is absolutely convergent in C α (I). For each n let Cn = sup x6=y

Then we have both X kfn ku < ∞

|fn (x) − fn (y)| . |x − y|α

and

C =

n

X n

Cn < ∞.

P Consequently, the series f = fn converges absolutely with respect to the uniform norm k · k. Since Cb (I) is complete, it follows that f ∈ Cb (I). We must show that the series converges to f in the norm of C α (I). Since |f (x) − f (y)| ≤

∞ X

n=1

|fn (x) − fn (y)| ≤

we have f ∈ Cα (I). Further,

∞ X

n=1

Cn |x − y|α = C |x − y|α ,

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N X

f − fn

Cα

n=1

N X

fn = f −

+ sup n=1

≤

∞ X

15

n=N +1

u

kfn ku +

x6=y

∞ X

    N N X X f (x) − fn (x) − f (y) − fn (y) n=1

n=1

|x − y|α

Cn

n=N +1

→ 0 as N → ∞. P Hence the series f = fn converges in the norm of C α (I), so we conclude α that C (I) is complete. (b) If we set α = 1 then only notational changes are needed in the proof given in part (a). For example, we should write k · kLip instead of k · kC α , and so forth.

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Solutions to Exercises and Problems from Chapter 2 2.1.7 Remark: There is a typo in the statement of this exercise in the text. The word “nonoverlapping” should be removed from the hypotheses. The proof below does not require that the boxes in question be nonoverlapping. Let Rk = Q ∩ Qk . Each Rk is a (possibly degenerate) box, and we have Q = R1 ∪ · · · ∪ Rn . We will show that vol(Q) ≤

n X

vol(Rk ).

n X

vol(Rk ).

k=1

Since degenerate boxes have zero volume, we can ignore those boxes and assume that each Rk is nondegenerate. In this case, Q is a union of the finitely many boxes R1 , . . . , Rn . The only difference between this case and that of Lemma 2.1.6 is that these boxes need not be nonoverlapping. Therefore, if we extend the sides, then we obtain smaller boxes whose union is Q, but possibly with duplications in the smaller boxes. Applying Lemma 2.1.6 the volume of Q equals the sum of the volumes of a subset of these smaller boxes (selected to cover Q without duplicates). Including the duplicates gives a possibly larger sum. This gives us the inequality vol(Q) ≤

k=1

Since vol(Rk ) ≤ vol(Qk ), we conclude that vol(Q) ≤

n X

vol(Qk ).

k=1

Alternative approach. We proceed by induction. The conclusion holds trivially if n = 1. Suppose that the conclusion holds for any collection of n ≥ 1 boxes that cover Q. Let Q1 , . . . , Qn , Qn+1 be boxes such that Q ⊆ Q1 ∪ · · · ∪ Qn ∪ Qn+1 . Allowing degenerate boxes, Q ∩ Qk is a box and vol(Q ∩ Qk ) ≤ vol(Qk ). Therefore we can replace each Qk by Q∩Qk , i.e., we can assume that Qk ⊆ Q for each k. Further, since a degenerate box has volume zero, we can assume that Qk is nondegenerate for each k. Although Q\Qn+1 is not a box in general, we can write it as a union of finitely many nonoverlapping boxes K1 , . . . , Km . Further, {Kj ∩ Qk }nk=1 is a cover of Kj by boxes, so by the inductive hypothesis we have that

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vol(Kj ) ≤

n X

k=1

vol(Kj ∩ Qk ).

However, Q = K1 ∪ · · · ∪ Km ∪ Qn+1 , and the boxes on the right-hand side intersect only along their boundaries. Hence m S Qk ⊇ (Kj ∩ Qk ), j=1

where the boxes on the right-hand side intersect only along their boundaries. Therefore vol(Q) =

m X

vol(Kj ) + vol(QN +1 )

j=1

≤ =

m X N X

vol(Kj ∩ Qk ) + vol(QN +1 )

N X m X

vol(Kj ∩ Qk ) + vol(QN +1 )

j=1 k=1

k=1 j=1

≤

N X

vol(Qk ) + vol(QN +1 ),

k=1

so the result follows by induction. S∞ 2.1.15 (a) If x ∈ lim sup Ek , then x belongs to k=j Ek for every j ∈ N. Therefore, for each j, there must exist some k ≥ j such that x ∈ Ek . If x belonged to only finitely many Ek , then this couldn’t happen, so x must belong to infinitely many of the set Ek . This reasoning is reversible, so we conclude that lim sup Ek precisely consists of the points that lie in infinitely many Ek . T∞ If x ∈ lim inf Ek , then there is some j such that x belongs to k=j Ek . Hence for that j we have x ∈ Ek for every k ≥ j. Again, this reasoning is reversible. S∞ (b) Set Fj = k=j Ek and F = lim sup k → ∞Ek =

∞ T

Fj .

j=1

Then F ⊆ Fj , so |F |e ≤ |Fj |e for every j. Applying countable subadditivity,

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∞ ∞ X S Ek ≤ lim 0 ≤ |F |e ≤ lim |Fj |e = lim |Ek |e = 0, j→∞ j→∞ j→∞ k=j

e

k=j

P∞

the final equality following from the fact that k=1 |Ek |e < ∞. Since lim inf Ek ⊆ lim sup Ek , it must have exterior measure zero as well. S∞ 2.1.16 Set Fj = k=j Ek and F = lim sup Ek = k→∞

∞ T

Fj .

j=1

Then F ⊆ Fj , so |F |e ≤ |Fj |e for every j. Applying countable subadditivity, ∞ ∞ X S 0 ≤ |F |e ≤ lim |Fj |e = lim Ek ≤ lim |Ek |e = 0, j→∞ j→∞ j→∞ k=j

e

k=j

P∞

the final equality following from the fact that k=1 |Ek |e < ∞. Since lim inf Ek ⊆ lim sup Ek , it must have exterior measure zero as well. 2.1.20 Let E = R1 ∪ · · · ∪ Rn where R1 , . . . , Rn are nonoverlapping boxes. By definition of exterior measure, or by applying monotonicity, we have |E|e ≤

n X j=1

|Rj |e =

n X

vol(Rj ).

j=1

Let {Qk } be any countable covering of E by countably many boxes, and fix ε > 0. Given k ∈ N, let Q∗k be a box that contains Qk in its interior but is only slightly larger in the sense that vol(Q∗k ) ≤ (1 + ε) vol(Qk ). Since Qk ⊆ (Q∗k )◦ , the interiors of the boxes Q∗k form an open covering of E: As E is compact, this covering must have a finite subcovering. Hence there exists some integer N > 0 such that N S

j=1

Rj = E ⊆

We wish to show that n X j=1

N S

(Q∗k )◦ ⊆

k=1

vol(Rj ) ≤

N X

N S

k=1

Q∗k .

vol(Q∗k ).

k=1

We apply the same idea as in the proof of Lemma 2.1.6. That is, we extend the sides of the boxes Q∗k to obtain a grid-like covering of E by smaller boxes. There can be duplicates in this covering. We can ignore any smaller boxes

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whose interiors lie completely outside of E. Hence we have a collection of smaller boxes that cover E. Recall that E is the union of the finitely many nonoverlapping boxes R1 , . . . , Rn . Each box Rj is covered by a distinct subset of these smaller boxes, and those smaller boxes make a grid-like cover of Rj , possibly with overlaps. Hence the sum of the volumes of the boxes Rj is bounded by the sum of all the volumes of the smaller boxes. That sum is itself bounded by the sum of the volumes of the boxes Q∗k . This gives us the desired inequality N n X X vol(Rj ) ≤ vol(Q∗k ). j=1

k=1

Putting this all together, we obtain n X j=1

vol(Rj ) ≤

N X

k=1

vol(Q∗k ) ≤ (1 + ε)

N X

k=1

vol(Qk ) ≤ (1 + ε)

X

vol(Qk ).

k

Taking the infimum over all such coverings by boxes, we see that n X j=1

vol(Rj ) ≤ (1 + ε) |E|e .

Finally, since ε is arbitrary, this yields n X j=1

vol(Rj ) ≤ |E|e .

Alternative proof. Let Q1 , . . . , QN be nonoverlapping boxes, and set E = Q1 ∪ · · · ∪ QN . By subadditivity, N N X X N S |Qk |e = vol(Qk ), |E|e = Qk ≤ k=1

k=1

k=1

so our task is to prove the opposite inequality. Let {Rℓ } be a cover of E = Q1 ∪ · · · ∪ QN by countably many boxes. For each fixed k, the collection {Rℓ ∩ Qk }ℓ is a covering of Qk by boxes, so X vol(Qk ) = |Qk |e ≤ vol(Rℓ ∩ Qk ), k = 1, . . . , N. ℓ

Also, {Rℓ ∩ Qk }N k=1 is a finite collection of nonoverlapping boxes contained in Rℓ . A variation on the ideas in Lemma 2.1.6 or Exercise 2.1.7 shows that

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20 N X

k=1

vol(Rℓ ∩ Qk ) ≤ vol(Rℓ ).

Therefore N X

k=1

vol(Qk ) ≤

N X X

k=1

ℓ

X

vol(Rℓ ∩ Qk ) ≤

vol(Rℓ ).

ℓ

Since this is true for every covering, we conclude that N X

k=1

vol(Qk ) ≤ inf

nX ℓ

o vol(Rℓ ) =

N S Qk = |E|e , k=1

where the infimum is taken over all possible coverings of E = Q1 ∪ · · · ∪ QN by countably many boxes Rℓ . 2.1.24 Fix x ∈ C. Then x has a ternary expansion x = .d1 d2 d3 · · · =

∞ X dk k=1

3k

where each di is either 0, 1, or 2. This ternary expansion is unique, except for points that are ternary rationals. Any point that lies in an open interval removed at the kth stage of the construction of C satisfies dk = 1 (more precisely, one of its ternary expansions has dk = 1). For example, at stage 1 the open interval ( 13 , 23 ) is removed, and d1 = 1 for all points in this interval. No point in that open interval has an alternative ternary expansion that begins either with d1 = 0 or d1 = 2. Continuing in this way, we see that x has a ternary expansion that contains only the digits 0 and 2. The set of sequences {dk }k∈N where every dk = 0 or dk = 2 is uncountable. Only a countable number of such sequences correspond to ternary expansions P dk that converge to the same number x. Hence the set of numbers x = ∞ k=1 3k with dk either 0 or 2 is uncountable. Finally, the point x = 14 does belong to C, because every digit in its ternary representation is either 0 or 2: ∞ X 1 0 2 0 2 2 = + 2 + 3 + 4 + ··· = . 4 3 3 3 3 9k k=1

2.1.25 The Cantor set is closed because it is the intersection of the closed sets Fn . In order to show that C contains no intervals, fix any point x ∈ C. Then x has a ternary expansion

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x = .d1 d2 d3 · · · =

∞ X

k=1

dk 3k

where each di is either 0 or 2. Define xn

n X dk = .d1 · · · dn 000 · · · · · · = 3k k=1

and yn = .d1 · · · dn 111 · · · =

n X dk

k=1

3k

+

∞ X

k=n+1

1 . 3k

Then xn ∈ C and yn ∈ R\C (note that yn has a unique ternary expansion). Moreover, lim |x − xn | = lim

n→∞

n→∞

∞ ∞ X X 2 dk ≤ lim = 0. k n→∞ 3 3k

k=n+1

k=n+1

Hence x is an accumulation point of C (and therefore C is a perfect set). Similarly yn → x, so x is a accumulation point of [0, 1]\C as well. Consequently C contains no interior points, and every point in C is a boundary point of C. 2.1.29 This follows from countable subadditivity. 2.1.30 Suppose that Rd \Z is not dense in Rd . Then its closure E = Rd \Z is a closed set that is not equal to Rd . Therefore Rd \E is a nonempty open set, so it contains some open ball Br (x). Hence Br (x) ⊆ Rd \E = Rd \ (Rd \Z) ⊆ Z, which implies that the set Z has positive measure. 2.1.31 Let Zk = Z ∩ [k, k + 1] and let Zk2 = {x2 : x ∈ Zk }. By subadditivity, it suffices to show that each set Zk2 has zero measure. For simplicity, consider the case k > 0. Since P |Zk | = 0, there exists a covering Zk ⊆ ∪[aj , bj ] by intervals such that (bj − aj ) ≤ |Zk | + ε = ε. By replacing [aj , bj ] by [aj , bj ] ∩ [k, k + 1] we can assume that aj ≤ bj ≤ k + 1 for every j. Since Zk2 ⊆ ∪[a2j , b2j ], we therefore have |Zk2 | ≤

X j

(b2j − a2j ) =

X j

(bj − aj ) (bj + aj ) ≤ (2k + 2) ε.

This is true for every ε > 0, so |Zk | = 0. 2.1.32 Fix ε > 0. For each k ∈ Z, let 
Γk = y, f (y) : y ∈ [k, k + 1] .

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If we show that |Γk |e = 0 for every k, then it will follow from subadditivity that |Γf |e = 0. Since the argument is similar for any particular k, consider k = 0. Since f is uniformly continuous on [0, 1], there exists some δ > 0 such that ∀ x, y ∈ [0, 1],

|x − y| < δ =⇒ |f (x) − f (y)| < ε.

Choose N large enough that δ <  k−1

Ik =

N

1 N.

(A)

Set

 k ,N ,

k = 1, . . . , N.

For each k, choose any point xk ∈ Ik . If y ∈ Ik , then we have |y − xk | ≤

1 < δ, N

and therefore |f (y) − f (xk )| < ε by equation (A). Hence, if we set   Jk = f (xk ) − ε, f (xk ) + ε and Qk = Ik × Jk ,

then

y ∈ Ik =⇒ (y, f (y)) ∈ Qk . Since every point y ∈ [0, 1] belongs to some Ik , it follows that Γ0 = Therefore



N
S Qk . y, f (y) : y ∈ [0, 1] ⊆ k=1

|Γ0 | ≤

N X

k=1

|Qk |e =

N X 2ε

k=1

N

= 2ε.

Since ε is arbitrary, it follows that |Γ0 |e = 0.

2.1.33 Since A ⊆ B ∪ (A△B), subadditivity implies that |A|e ≤ |B|e + |A△B|e . Since the measures are finite, we can subtract to obtain |A|e − |B|e ≤ |A△B|e . A symmetric argument shows that |B|e − |A|e ≤ |A△B|e . 2.1.34 Let I = inf

X

 vol(Qk ) ,

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where the infimum is taken over all countable collections of boxes {Qk } such S that E ⊆ Q◦k . S S It E ⊆ Q◦k , then E ⊆ Qk , so by definition of exterior measure we have X |E|e ≤ vol(Qk ). k

Taking the infimum over all such coverings, it follows that |E|e ≤ I. If |E|e = ∞, then the converse inequality I ≤ |E|e is immediate, so assume S that |E|e < ∞. Fix any ε > 0. Then there exist boxes Qk such that E ⊆ Qk and X vol(Qk ) ≤ |E|e + ε. k

Let

Q∗k

be a box such that Qk ⊆ (Q∗k )◦ and

vol(Q∗k ) ≤ (1 + ε) vol(Qk ). S

(Q∗k )◦ , so by definition of I we have X X X
I ≤ vol(Q∗k ) = (1 + ε) vol(Qk ) ≤ (1 + ε) |E|e + ε .

Then E ⊆

k

k

k

Since |E|e is finite and ε is arbitrary, we obtain I ≤ |E|e . 2.1.35 (a) Fix n ∈ N. For each integer k = 0, . . . , n − 1, let Qk be the square with sidelengths n1 whose bottom left corner sits at ( nk , nk ). The n squares Q0 , . . . , Qn−1 cover L, so |L|e ≤

n−1 X k=0

vol(Qk ) =

n−1 X k=0

1 n 1 = 2 = . 2 n n n

Since n is arbitrary, it follows that |L|e = 0.

(b) Every ray or line in R2 can be written as a countable union of line segments. Therefore, if we show that every line segment has exterior measure zero, then it follows from subadditivity that every ray or line has exterior measure zero as well. Further, by monotonicity, if we show that every closed line segment has measure zero, then any line segment that is missing one or both endpoints also will have measure zero. Let L be a closed line segment in R2 of length ℓ. Exterior Lebesgue measure is translation-invariant by Lemma 2.1.11, so we may assume that one endpoint of L sits at the origin. Suppose L is a vertical line segment, say the segment from (0, 0) to (0, b). Given ε > 0, the box Q = [−ε, ε] × [0, b] covers L, so |L|e ≤ vol(Q) = 2bε. Since ε is arbitrary, it follows that |L|e = 0. So, suppose that L is not vertical. Then L is the line segment connecting (0, 0) to (ℓ, ℓm) for some real numbers ℓ and m. For simplicity assume that

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both ℓ and m are positive. Fix n > 0. For each integer k = 0, . . . , n− 1, let Qk be the square with sidelengths n1 whose bottom left corner has coordinates ( nk , km n ). That is, the bottom left corner of Qk sits on L. The n squares Q0 , . . . , Qn−1 cover L, so n−1 X

|L|e ≤

vol(Qk ) =

k=0

n−1 X k=0

n 1 1 = 2 = . 2 n n n

Since n is arbitrary, it follows that |L|e = 0.

2.1.36 Fix ε > 0. Let R be the unit cube in Rd−1 , i.e., R =

d−1 Y

[0, 1] =

j=1

 (x1 , . . . , xd−1 ) : 0 ≤ xj ≤ 1 .

Let (ck )k∈Zd−1 be any sequence of positive numbers such that X ck ≤ 1. k∈Zd−1

For each k = (k1 , . . . , kd−1 ) ∈ Zd−1 , set Qk = (R + k) × [−ck ε, ck ε], where R + k is the translation of R by the vector k, i.e.,  R + k = = (x1 , . . . , xd−1 ) : kj ≤ xj ≤ kj + 1 .

Since R + k is a cube in Rd−1 whose sides each have length 1, the set Qk is a box in Rd−1 with volume vol(Qk ) = 1 · · · 1 · 2ck ε = 2ck ε. Further, {Qk }k∈Zd−1 is a cover of S by boxes, so X X |S|e ≤ vol(Qk ) = 2ck ε ≤ 2ε. k∈Zd−1

k∈Zd−1

Since ε is arbitrary, it follows that |S|e = 0. 2.1.37 Recall that if U is a d × d real matrix, then the following statements are equivalent: •

• •

U is an orthogonal matrix. U is invertible and such that U −1 = U T . The columns of U are an orthonormal basis for Rd .

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•

25

U is a linear “rigid motion” (a composition of rotations and flips) that maps Rd onto itself. kU xk = kxk for all x ∈ Rd .

•

U x · U y = x · y for all x, y ∈ Rd .

•

Suppose that U is an orthogonal matrix and B = Br (x) is a ball. Then U (B) = {U y : kx − yk < r}

= {U y : kU x − U yk < r} = {z : kU x − zk < r}

= Br (U x).

Thus U maps balls to balls of the same radius. Now let S be an arbitrary proper subspace of Rd , and let S ⊥ be the orthogonal complement of S with respect to the usual dot product on Rd . Choose any nonzero unit vector z ∈ S ⊥ . Then S ⊆ Z = z ⊥ . Let U be an orthogonal matrix such that U z = ed = (0, . . . , 0, 1). For example, to do this we can extend z to an orthonormal basis z1 , z2 , . . . , zd−1 , z for Rd , let T be the orthogonal matrix that has z1 , z2 , . . . , zd−1 , z as columns, and then let U = T −1 . Since T ed = z, we have U z = T −1 z = ed , and U is orthogonal because the inverse of an orthogonal matrix is orthogonal. We claim that U (S) ⊆ U (Z) ⊆ {ed }⊥ = Rd−1 × {0}. To see this, choose any x ∈ S. Then x ∈ Z, so x · z = 0. Hence U x · ed = U x · U z = x · z = 0, so U x ∈ {ed }⊥ . This shows that U (S) ⊆ {ed }⊥ . Applying Problem 2.1.36, it follows that |U (S)|e ≤ |Rd−1 × {0}|e = 0. Fix ε > 0. Then there exists some open set V ⊇ U (S) such that |V |e < ε. By Lemma 2.1.5, there exist nonoverlapping cubes Qk such that V = ∪Qk . Note that S Qk = V U (S) ⊆ k

and

|U (S)|e ≤

X k

|Qk |e = |V |e < ε.

Let 2rk be the sidelength of Qk , and let xk be the center of Qk . Any cube of sidelength 2r is contained in a ball of radius d1/2 r. Therefore, if we set

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sk = d1/2 rk

and

Bk = Bsk (xk ),

then Qk ⊆ Bk . Since U (S) ⊆ ∪Qk , it follows that S −1 S −1 S ′ S ⊆ U (Qk ) ⊆ U (Bk ) ⊆ Bk . k

k

k

Now, since U is an orthogonal matrix, U −1 is also an orthogonal matrix, so Bk′ = U −1 Bk is a ball with the same radius sk as Bk . Let Q′k be the cube with the same center as Bk′ and with sidelengths 2sk , i.e., Q′k is the smallest cube with sides parallel to the coordinate axes that contains Bk′ . Then we have S ′ S −1 S −1 S ′ Qk . S ⊆ U (Qk ) ⊆ U (Bk ) ⊆ Bk ⊆ k

The volume of

Q′k

k

k

k

is

|Q′k |e = (2sk )d = (2d1/2 rk )d = dd/2 (2rk )d = dd/2 |Qk |e . Hence |S|e ≤

X k

|Q′k |e ≤ dd/2

X k

|Qk |e < dd/2 ε.

Since ε > 0 is arbitrary, we conclude that |S|e = 0. 2.1.38 (a) Case 1. If δk = 0 for some k then D(E) is contained in a proper subspace of Rd . In fact, this subspace is “parallel to the coordinate axes.” For simplicity of presentation, suppose that it is δ1 that is zero. Then D(E) ⊆ S = {0} × Rd−1 = {(0, x2 , . . . , xd ) : x2 , . . . , xd ∈ R}. For each k = (k2 , . . . , kd ) ∈ Zd−1 , fix an εk > 0 and let Qk be the box Qk = [−εk , εk ] × [k2 , k2 + 1] × · · · × [kd , kd + 1]. Then D(E) ⊆ S ⊆ so |D(E)|e ≤ |S|e ≤

X

k∈Zd−1

S

Qk ,

k∈Zd−1

vol(Qk ) =

X

2εk .

k∈Zd−1

By choosing εk appropriately, we can make the final sum on the line above as small as we like. Therefore |D(E)|e = 0, and hence we have the desired equality |D(E)|e = 0 = |δ1 · · · δd | |E|e . Case 2. Assume that δk 6= 0 for every k. Let {Qk } be a countable covering of E by boxes. Note that D(Qk ) is a box in Rd , and

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vol(D(Qk )) = |δ1 · · · δd | vol(Qk ). Therefore {D(Qk )} is a countable covering of D(E) by boxes, so X X |D(E)|e ≤ vol(D(Qk )) = |δ1 · · · δd | vol(Qk ). k

k

Taking the infimum over all such coverings, we see that |D(E)|e ≤ |δ1 · · · δd | |E|e . The converse inequality follows similarly, by covering D(E) by boxes. (b) Because Lebesgue measure is invariant under translations, we have |Br (x)| = |Br (0) + x| = |Br (0)|. Applying part (a), we therefore find that |Br (x)| = |Br (0)| = |rB1 (0)| = rd |B1 (0)|, so we can take Cd = |B1 (0)|. 2.1.39 “⇒.” Suppose that |E|e = 0. Then for each integer n ∈ N, there exists a collection of boxes {Qnk }k∈N such that E ⊆

S k

Qnk

∞ X

and

vol(Qnk ) < 2−n .

k=1

The collection of boxes {Qnk }k,n∈N (where we allow duplicates in the sequence) satisfies ∞ X ∞ ∞ X X vol(Qnk ) ≤ 2−n = 1 < ∞. n=1 k=1

n=1

Further, if x ∈ E then for each n there exists some integer kn such that x ∈ Qnkn . Hence x belongs to infinitely many of the boxes Qnk . “⇐.” Suppose {Qk }k∈N is a collection of boxes such that each x ∈ E P belongs to infinitely many Qk and vol(Qk ) < ∞. Then E is contained in lim sup Qk , so the Borel–Cantelli Lemma implies that |E|e = 0. S r∈Q

2.1.40 Since Z has measure zero and Q is countable, | Therefore there must exist some point S (r − Z). x 6∈ r∈Q

For this x we have

x∈ / r−Z

for any r ∈ Q,

and therefore x 6= r − z

for any r ∈ Q or z ∈ Z.

(r − Z)|e = 0.

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Therefore, if r ∈ Q, then r 6= z + x for any z ∈ Z, and consequently r ∈ / Z + x. That is, Q ∩ (Z + x) = ∅, which tells us that Z + x contains no rational points. 2.1.41 (a) We have either N S

U =

(ak , bk )

or

U =

∞ S

(ak , bk ),

k=1

k=1

where the (ak , bk ) are disjoint open intervals. For simplicity of notation, we will consider the second possibility only; the case of a finite union is similar. By subadditivity, |U |e ≤

∞ X

k=1

∞ X

|(ak , bk )|e =

k=1

(bk − ak ).

We must prove the opposite inequality. Fix any integer N, and let M be large enough that ak +

1 M

1 M,

< bk −

k = 1, . . . , N.

Then fix any m ≥ M, and observe that N  S

ak +

k=1

1 m , bk

−

1 m



N S

⊆

k=1

(ak , bk ) ⊆ U.

  1 1 Since ak + m , bk − m for k = 1, . . . , N are finitely many disjoint boxes, we compute that N  S  1 1 (monotonicity) |U |e ≥ ak + m , b k − m k=1

=

N X

k=1

=

N X

k=1

 vol ak +

b k − ak −

1 m , bk

2 m

−

1 m



(Exercise 2.1.20)


.

This is true for every m ≥ M, so by letting m → ∞ we see that |U |e ≥

lim

m→∞

N X

k=1

b k − ak −

Finally, letting N → ∞, it follows that

2 m




=

N X

k=1


b k − ak .

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|U |e ≥

lim

N →∞

29 N X

k=1

b k − ak




=

∞ X

k=1

(b) The complement of the Cantor set is

U = [0, 1]\C = 13 , 23 ∪ 91 , 29 ∪


b k − ak . 7 8 9, 9

Applying part (a), it follows that




∪ ··· .

1 1 1 + 2· + 4· + · · · = 1. 3 9 27 P∞ P∞ be any two points in D. 2.1.42 Let x = k=1 ck 3−k and y = k=1 dk 3−kP ∞ Then each ck and dk is either 0 or 1, so x + y = k=1 (ck + dk )3−k where ck + dk = 0, 1, 2 for each k. Hence x + y ∈ [0, 1], so D + D ⊆ [0, 1]. Now let x be an arbitrary point in [0, 1]. Then we can write x = P∞ −k where each ak is either 0, 1, or 2. Define k=1 ak 3     0, ak = 0, 0, ak = 0, ck = 0, ak = 1, and dk = 1, ak = 1,     1, ak = 2, 1, ak = 2. |U |e =

Then the points

y =

∞ X ck 3k

and

z =

k=1

∞ X dk k=1

3k

belong to D, and x = y + z ∈ D + D. Therefore D + D = [0, 1]. 2.1.43 We have |Fn+1 |e ≤ (1 − α) |Fn |e , so |Fn |e → 0 as n → ∞. Consequently |Cα |e = 0. If Cα contained an interior point then, by monotonicity, |Cα |e would be positive. Hence Cα contains no interior points. As Cα is closed by construction, we have Cα = Cα◦ ∪ ∂Cα , so this also implies that Cα = ∂Cα . It remains to show that Cα is perfect. Since Cα is closed it contains all of its accumulation points. Therefore, given x ∈ Cα we must show that x is an accumulation point of Cα . For each n ∈ N, the point x must belong to one of the closed intervals whose union is the set Fn . Let xn be one of the endpoints of the interval that contains x (if x is one of the endpoints, then let xn be the opposite endpoint). Then xn ∈ Cα \{x}, and |x − xn | decreases to zero as n → ∞. Therefore x is an accumulation point of Cα . 2.1.44 This problem is similar to, but slightly different from, Problem 2.1.43.

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Let F0 = [0, 1]. The exterior measure of F0 is |F0 |e = 1. 1 Divide F0 into 10 nonoverlapping intervals of length 10 . Remove the interior of the fourth interval to form F1 , i.e., 5 4 ] ∪ [ 10 , 1]. F1 = [0, 10

The set F is contained in F1 . By monotonicity and subadditivity, |F |e ≤ |F1 |e ≤

9 . 10

1 . Now write F1 as a union of 90 nonoverlapping intervals of length 100 Considering these in groups of 10, remove the interior of the fourth interval from each group of 10 to form the set F2 . As there are 9 groups, we have removed 9 intervals. This leaves us with 90 − 9 = 81 nonoverlapping intervals 1 . By monotonicity and subadditivity, we therefore have of length 100

81 |F |e ≤ |F2 |e ≤ = 100



9 10

2

.

Continuing in this way, we see that for every n ∈ N we have  n 9 |F |e ≤ |Fn |e ≤ . 10 Therefore |F |e = 0. 2.1.45 Suppose that S = {x1 , x2 , . . . } is a countably infinite perfect subset of Rd . Let n1 = 1 and r1 = 1. Set U1 = Br1 (xn1 ), the open ball of radius r1 centered at xn1 . Since xn1 is an accumulation point of S, there exist infinitely many elements of S that belong to U1 . Let n2 be the first integer greater than n1 such that xn2 ∈ U1 . Let U2 = Br2 (xn2 ) be a nonempty open ball centered at xn2 such that and xn1 ∈ / U2 . U2 ⊆ U2 ⊆ U1 Let n3 be the first integer greater than n2 such that xn3 ∈ U2 . Let U3 = Br3 (xn3 ) be a nonempty open ball centered at xn3 such that U3 ⊆ U3 ⊆ U2

and

xn2 ∈ / U3 .

Continuing in this way we construct nested decreasing open balls Un . Set K =

∞ T

n=1


Un ∩ S .

Since S is closed and Un is compact, the sets Un ∩ S are compact and nested decreasing. The Cantor Intersection Theorem therefore implies that K is

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nonempty. However, K ⊆ S by definition. Yet x1 = xn1 ∈ / U2 , so x1 ∈ / K. Also, x2 , . . . , xn2 −1 are not contained in U1 , so they do not belong to K. The point xn2 does not belong to K because it does not belong to U3 , and so forth. In fact, no element of S can belong to K, which is a contradiction. 2.2.22 (a) Suppose that K ⊆ Rn is compact, and let {Vi }i∈J be any open cover of f (K). Since f is continuous, each inverse image Ui = f −1 (Vi ) is open. Consequently {Ui }i∈J is an open cover of K. As K is compact, there must exist some finite subcover of K, say {Ui1 , . . . , UiN }. But then {Vi1 , . . . , ViN } is a finite cover of f (K), so f (K) is compact. (b) Suppose that H is an Fσ -set in Rn . Then, by definition, we can write H = ∪Fk where each set Fk is closed. Setting Fkj = Fk ∩ [−j, j]n , we have H = ∪j,k Fkj where each set Fkj is compact. Part (a) shows that a continuous function maps compact sets to compact sets, so   S S f (Fkj ). f (H) = f Fkj = j,k

j,k

Each f (Fkj ) is compact and therefore closed, so f (H) is an Fσ -set. 2.2.21 We write out the details of the equivalent of statements (a) and (c) in the proof of Lemma 2.2.21. (a) ⇒ (c). Suppose that E is measurable. Then by Lemma 2.2.15, for each S k ∈ N we can find a closed set Fk ⊆ E such that |E \Fk | < 1/k. Set H = Fk and let Z = E \H. Then H is an Fσ -set, H ⊆ E, and Z = E \H ⊆ E \Fk for every k. Hence |Z|e ≤ |E \Fk | < 1/k for every k, so |Z| = 0. (c) ⇒ (a). If E = H ∪ Z where H is an Fσ -set and |Z| = 0, then E is measurable since both H and Z are measurable. 2.2.30 Suppose first that f is continuous and bounded above, and fix M ∈ R. We claim that f ≤ M everywhere

⇐⇒

f ≤ M a.e.

One direction is easy, for if f ≤ M everywhere, then certainly f ≤ M a.e. For the converse, choose any M > 0, and suppose that there is a point x ∈ U where f (x) > M. Then since f is continuous and U is open, there must be an open ball B containing x such that f (y) > M for all y ∈ B. But then f > M on a set with positive measure, i.e., it is not true that f ≤ M a.e. Hence this shows by contrapositive argument that if f ≤ M a.e., then f ≤ M everywhere. Consequently,

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32

esssup f (x) = inf{M : f (x) ≤ M a.e.} x∈U

= inf{M : f (x) ≤ M for every x} = sup f (x). x∈U

Hence the supremum and essential supremum of f coincide in this case. If f is continuous but not bounded above, then its supremum is ∞. However, if we fix M then f (x) > M for some x, and hence f exceeds M on some open ball since it is continuous. Therefore the essential supremum of f exceeds M, and since M is arbitrary it follows that the essential supremum of f is ∞ as well. 2.2.31 Suppose that F, K are nonempty, disjoint subsets of Rd such that F is closed and K is compact. Suppose that dist(F, K) = 0. Then there exist points xk ∈ F and yk ∈ K such that kxk − yk k < k1 . Since K is compact, the sequence {yk }k∈N contains a convergent subsequence, say ynk → z. The point z must belong to K since K is closed. Then kxnk − zk ≤ kxnk − ynk k + kynk − zk → 0

as k → ∞.

Hence z is a limit point of F. As F is closed, this implies that z ∈ F, which contradicts the fact that F and K are disjoint. However, we can create nonempty disjoint closed sets whose distance is zero. For example, in R we can take S S [2k − k2 , 2k − k1 ]. [2k, 2k + 1] and B = A = k∈N

k∈N

Embedding these sets into Rd gives examples in any dimension. Another example in R2 is A = {(x, 0) : x ≥ 1}

and

B = {(x, x1 ) : x ≥ 1}.

2.2.32 If either A and B has infinite measure then both sides of the desired equality are infinite, and so the result follows in this case. Therefore we can assume that |A|, |B| < ∞. Write A and B as disjoint unions: A = (A\B) ∪ (A ∩ B)

and

B = (B \A) ∪ (A ∩ B).

Since Lebesgue measure is countably additive, |A| = |A\B| + |A ∩ B|

and

|B| = |B \A| + |A ∩ B|.

On the other hand, we can write A ∪ B as the following disjoint union: A ∪ B = (A\B) ∪ (A ∩ B) ∪ (B \A)

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33

and therefore |A ∪ B| = |A\B| + |A ∩ B| + |B \A|. Since all quantities are finite, we can sum and rearrange to obtain |A| + |B| = |A\B| + 2 |A ∩ B| + |B \A| = |A ∪ B| + |A ∩ B|. 2.2.33 Let Z =

S

(Em ∩ En ).

m6=n

Since there are only countably many pairs of integers m 6= n, the set Z has zero measure. Let Fn = En \Z. Then {Fn }n∈N is a sequence of disjoint measurable sets and |Fn | = |En | for every n, so ∞ ∞ S S  S  X X ∞ ∞ ∞ Fn = |Fn | = |En |. En = Fn ∪ Z = n=1

n=1

n=1

n=1

n=1

2.2.34 Each sphere is measurable since it is a closed set. Let m = |S1 |, the measure of the sphere of radius 1. Since Sr = {x : kxk = r} = {rx : kxk = 1} = rS1 , it follows from Problem 2.1.38 that |Sr | = rd |S1 | = rd m. Fix 0 < α < 1, and set rn = αn/d r and E =

∞ S

n=1

Sr n .

The spheres on the right-hand side of the preceding equation are disjoint, so |E| =

∞ X

n=1

|Srn | =

∞ X

rnd m =

∞ X

αn rd m =

n=1

n=1

α rd m. 1−α

On the other hand, E ⊆ Br (0), so this implies that α rd m = |E| ≤ |Br (0)| = Cd rd , 1−α where Cd is a fixed constant that depends only on the dimension d. If m > 0, then by letting α ր 1 we see that Cd ≥

lim

α→1−

α m = ∞, 1−α

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34

which is a contradiction. Therefore we must have m = 0. Another approach is to note that Sr is contained in the annulus Br+ε (0)\Br−ε (0), and again use Problem 2.1.38 to compute the measure of these two balls. 2.2.35 First proof. We are given that |E ∩ (E + t)| = 0 for every t 6= 0. Suppose that |E| > 0. Write E = ∪k∈Z Ek , where Ek = E ∩ [k, k + 1). The sets Ek are disjoint and their union is E, so at least one of them must have positive measure. By monotonicity we still have |Ek ∩ (Ek + t)| = 0 for every t 6= 0, so we can reduce to the case where Ek is a bounded set. For simplicity (translate Ek if necessary), we can further assume that k = 0. That is, if we set F = E0 then we have |F | > 0 and F ⊆ [0, 1] and |F ∩ (F + t)| = 0 for every t 6= 0. Let S = {rk }k∈N be a list of all the rational numbers in [0, 1]. Then S (F + rk ) ⊆ [0, 2], F +S = k∈N

so F + S has finite measure. However,

j 6= k =⇒ |(F + rj ) ∩ (F + rk )| = 0, so it follows from Problem 2.2.33 that ∞ ∞ S X X |F + S| = (F + rk ) = |F + rk | = |F | = ∞. k∈N

k=1

k=1

This contradicts the fact that F + S is a bounded set. Hence we must have |E| = 0. Second proof. The proof of the Steinhaus Theorem (which is not covered until Section 2.4) can be adapted to this problem. 2.2.36 No. Example 1. Let E = F = R, and let P(x, y) be the statement x − y 6∈ Q. That is, P(x, y) is true if x − y is irrational. Then for every x ∈ R, we have that x − y ∈ / Q for a.e. y. Thus it IS true that for every x ∈ E, P(x, y) is true for a.e. y ∈ F. However, there are NO y such that P(x, y) is true for every x, because there is no real number y such that x − y is irrational for every x. Therefore it is NOT true that

c Solutions 2019 Christopher Heil

for a.e. y ∈ Y,

35

P(x, y) is true for every x ∈ E.

Example 2. Take E = F = [0, 1], and let P(x, y) be the statement “x 6= y.” For every fixed x ∈ [0, 1] we have x 6= y for a.e. y since there is only one y such that x = y. But there is no y such that P(x, y) is true (x 6= y) for every x. 2.2.37 (a) ⇒ (b). Suppose that E is measurable. Then, by definition of measurability, there exists an open set U ⊇ E such that |U \E|e < ε. By Lemma 2.2.15, there exist an closed set F ⊆ E such that |E \F |e < ε. Subadditivity therefore implies that |U \F | ≤ |U \E|e + |E \F |e < 2ε. (b) ⇒ (c). Assume that statement (b) holds. Then for each k ∈ N, there exists an open set Uk and a closed set Fk such that Fk ⊆ Ek ⊆ Uk and T S |Uk \Fk | < k1 . Let U = Uk and F = Fk . Then U is a Gδ -set and F is an Fσ -set and F ⊆ E ⊆ U. Further, for each k we have |U \F | ≤ |Uk \Fk |
0. Then there exists an open set U ⊇ E such that |U \E|e < ε. Since U is open, there exist nonoverlapping boxes Qk such that U = ∪∞ k=1 Qk . Since ∞ X

k=1

|Qk | = |U | < ∞,

we can choose M large enough that S =

M S

k=1

Qk ,

P∞

k=M+1

A = E \S,

|Qk | < ε. Let B = S \E.

Note that S is a finite union of nonoverlapping boxes and E = (S ∪ A) \ B. Since

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36

A = E \S ⊆ U \S ⊆

∞ S

we have |A|e ≤ |U \S| ≤

∞ S

k=M+1

Finally, B = S \E ⊆ U \E, so

Qk ,

k=M+1

Qk ≤

∞ X

k=M+1

|Qk | < ε.

|B|e ≤ |U \E|e < ε. Second proof (better). Suppose that E is measurable, and fix ε > 0. Then there exists an open set U ⊇ E such that |U \E|e < ε. Since U is open, there exist nonoverlapping boxes Qk such that U = ∪∞ k=1 Qk . Since ∞ X

k=1

|Qk | = |U | < ∞,

we can choose M large enough that S =

M S

k=1

Qk ,

A =

P∞

k=M+1

∞ S

|Qk | < ε. Let

Qk ,

k=M+1

B = U \E.

Then |A|e < ε by countable subadditivity, |B|e < ε by construction, S is a finite union of nonoverlapping boxes by definition, and (S ∪ A) \ B = U \ B = E. (b) ⇒ (a). First Proof. Fix ε > 0. By hypothesis, E = (S ∪ A)\B, where S is a finite union of nonoverlapping boxes and |A|e , |B|e < ε. Since S is measurable, let U ⊇ S be an open set such that |U \S| < ε. Although we don’t know that A is measurable, we can find an open set V ⊇ A such that |V | ≤ |A|e + ε. Consequently, |V | ≤ |A|e + ε < 2ε. Let G = U ∪ V. Then G is open, and since U ⊇ S and V ⊇ A, we have that G ⊇ S ∪ A ⊇ E. After some tedious set-theoretic calculations, we can see that G\E ⊆ (U \S) ∪ V ∪ B. Therefore |G\E|e ≤ |U \S| + |V | + |B|e ≤ ε + 2ε + ε = 4ε,

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so E is measurable. Second Proof. We will use Carath´eodory’s Criterion to show that E is measurable. Let X be any subset of Rd . By subaddivity, we have |X|e ≤ |X ∩ E|e + |X \E|e . Therefore we just have to prove the opposite inequality. Given ε > 0, we have by statement (b) that E = (S ∪ A)\B where S is a union of finitely many boxes and |A|e , |B|e < ε. Note that E ⊆ S∪A

and

S \B ⊆ E.

A set-theoretic calculation shows that X \(S \B) = (X \S) ∪ (X ∩ B). Therefore, since S is measurable, |X ∩ E|e + |X \E|e ≤ |X ∩ (S ∪ A)|e + |X \(S \B)|e

≤ |X ∩ S|e + |X ∩ A|e + |X \S|e + |X ∩ B|e = |X ∩ S|e + |A|e + |X \S|e + |B|e = |X ∩ S|e + ε + |X \S|e + ε

= |X|e + 2ε

(S is measurable).

But ε is arbitrary, so this shows that |X ∩ E|e + |X \E|e ≤ |X|e . Since X is any subset of Rd , Carath´eodory’s Criterion therefore implies that E is measurable. 2.2.39 We are given that 0 < |E|e < ∞. Since 0 < α < 1 and |E|e 6= 0, we have α1 |E|e > |E|e . Therefore, by Theorem 2.1.27 there exists an open set U ⊇ E such that 1 |E|e ≤ |U | ≤ |E|e . α Since U is an open subset of Rd , we can write it as a countable union of nonoverlapping cubes Qk . By Corollary 2.2.17, we have |U | = Suppose that we had

∞ X

k=1

|Qk |.

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38

|E ∩ Qk |e < α |Qk | for every k. Then |E|e = |E ∩ U |e X ≤ |E ∩ Qk |e k

= |E ∩ Q1 | + < α |Q1 | + = α |U | ≤ α

X

k>1

X

k>1

|E ∩ Qk |e

α |Qk | (we separated terms to emphasize 0. For example, if we let an = 2−2n , then

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s =

∞ X

39

2n−1 2−2n =

n=1

∞ X

1 1 2−n = , 2 n=1 2

so |P | = 1/2 in this case. If we take an = 2−3n then s =

∞ X

2n−1 2−3n =

n=1

∞ 1 X −2n 1 2 = , 2 n=1 6

so |P | = 5/6 in this case. If 0 < ε < 1 and we take an = 2−2n+1 ε, then s =

∞ X

2n−1 an =

n=1

∞ X

2−n ε = ε,

n=1

so |P | = 1 − s = 1 − ε in this case. The construction of P ensures that P contains no open intervals, and therefore P ◦ = ∅. This also implies that its complement U = [0, 1]\P is dense in [0, 1]. To see why, suppose that U was not dense. Then its closure U would be a proper closed subset of [0, 1]. Consequently [0, 1]\U would be a nonempty open set, so it must contain some open interval (a, b). But then (a, b) ⊆ [0, 1]\U ⊆ [0, 1]\U = P, which contradicts the fact that P contains no open intervals. 2.2.43 (a) If A is measurable, then Lemma 2.2.15 implies that |A| = |A|i .

(b) First Proof. We are given a set A ⊆ Rd such that |A|e < ∞ and |A|e = |A|i . For each k ∈ N, there exists an open set Uk ⊇ A such that |A|e ≤ |Uk | ≤ |A|e +

1 . k

By definition of |A|i , there exist closed sets Fk ⊆ A such that |A|i −

1 ≤ |Fk | ≤ |A|i . k

Now set G = ∩Uk and H = ∪Fk . Then G is a Gδ -set and H is an Fσ -set, so they are measurable. Since |A|e ≤ |G| ≤ |Uk | ≤ |A|e + and

1 k

1 ≤ |Fk | ≤ |H| ≤ |A|i , k by letting k → ∞ we see that |G| = |A|e and |H| = |A|i . Since these numbers are finite, it follows that |A|i −

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40

|G\A|e ≤ |G\H| = |G| − |H| = |A|e − |A|i = 0. Lemma 2.2.21 therefore implies that A is measurable. Second Proof. This solution uses Carath´eodory’s Criterion, which is not proved until Section 2.3. Assume A ⊆ Rd satisfies |A|e < ∞ and |A|e = |A|i . Fix ε > 0. Then by definition of |A|i , there exists a closed F ⊆ A such that |F | ≥ |A|i − ε = |A|e − ε. Since F is measurable, Carath´eodory’s Criterion implies that |A|e = |A ∩ F |e + |A \ F |e = |F | + |A \ F |e . Since all quantities are finite, we therefore have |A \ F |e = |A|e − |F | ≤ ε. Hence A is measurable. (c) Let N be a nonmeasurable subset of [0, 1], and let A = (−∞, 0) ∪ N . Then |A|e = |A|i = ∞, but A is not measurable. (d) Assume that A ⊆ E where E is a measurable set. Suppose that F is a closed subset of A. Then E \A ⊆ E \F, we have |E \A|e ≤ |E \F |. Since E is measurable, countable additivity therefore implies that |F | + |E \A|e ≤ |F | + |E \F | = |E|. Taking the supremum over all closed sets F ⊆ A, we obtain |A|i + |E \A|e ≤ |E|. For the converse inequality, let U be any open set that contains E \A. If x ∈ E \U, then x ∈ E but x ∈ / U. Therefore, if x ∈ / A then x ∈ E \A ⊆ U, which is a contradiction. Hence we must have x ∈ A, which shows that E \U ⊆ A. As E \U is measurable, its inner and outer measures are equal, so |E \U | = |E \U |i ≤ |A|i . Using this, countable additivity, and the fact that E ∩ U ⊆ U, it follows that |E| = |E \U | + |E ∩ U | ≤ |A|i + |U |. Taking the infimum over all open sets U ⊇ E \A, we obtain

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41

|E| ≤ |A|i + |E \A|e . 2.2.44 “⇒.” If A and B are both measurable, then we can apply countable additivity to conclude that |E| = |A| + |B|. “⇐.” First Proof. Suppose that |E| = |A|e + |B|e . By Theorem 2.1.27, there exist Gδ -sets G ⊇ A and H ⊇ B such that |G| = |A|e

and

|H| = |B|e .

By replacing G with G ∩ E and H with H ∩ E, we can assume that G and H are measurable subsets of E that satisfy |G| = |A|e and |H| = |B|e . Note that E \H ⊆ A ⊆ G. Therefore |G\A|e ≤ |G\(E \H)| = |G| − |E \H| = |G| − (|E| − |H|) = 0. Therefore A is measurable. A similar argument shows that B is measurable, or we can simply observe that since B = E \A and both E and A are measurable, B must be measurable as well. Alternatively, we can note that |E| ≤ |G ∩ H| + |G\H| + |H \G| ≤ |G ∩ H| + |G| + |H| = |G ∩ H| + |E|. Therefore |G ∩ H| = 0. But G\A ⊆ G ∩ B ⊆ G ∩ H, so |G\A| ≤ |G ∩ H| = 0. Therefore A is measurable. Second Proof. Suppose that |E| = |A|e + |B|e . Since E is measurable, Problem 2.2.43 implies that |E| = |A|i + |E \A|e . Therefore |A|e + |B|e = |E| = |A|i + |E \A|e = |A|i + |B|e , which implies that |A|e = |A|i . Since A has finite exterior measure, Problem 2.2.43 implies that A is measurable. Therefore B = E \A is measurable as well.

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2.2.45 Let E = Z, and define f : E → R by f (n) = n for n ∈ Z. Then f is continuous, but its essential supremum is zero while its supremum is ∞. 2.2.46 (a) Let H = ∩Uk be a Gδ -set. Then T 
S d R \Uk . Uk = H C = Rd \ k

k

Each set Rd \Uk is closed, so H C is an Fσ -set.

(b) If E = {xk }k∈N is countable, then we can write E = {x1 } ∪ {x2 } ∪ · · · . Each singleton {xk } is closed, so E is an Fσ -set.

(c) Given x ∈ Rd , let Qk be the cube centered at x with sidelengths Then ∞ T Qk , {x} =

1 k.

k=1

so {x} is a Gδ -set. Thus every singleton is a Gδ -set, so at least some countable sets are Gδ -sets. Not every countable set is Gδ ; for example, Example 2.2.19 shows that the set of rationals is not a Gδ -set. Some countably infinite sets are Gδ -sets, including { n1 }n∈N . To prove that this particular set is a Gδ -set, for all integers k, n ∈ N let Unk be the interval Unk = Then

S k n Un

1

n

−

 1 1 1 , + . kn(n + 1) n kn(n + 1)

is open for each k, and  ∞ ∞  T S k Un = n1 n∈N . k=1

n=1

(d) Define A = Q ∩ [0, 1]

and

B = (R\Q) ∩ [2, 3].

Since A is countable, it is an Fσ -set. However, it is not a Gδ -set, for the same reason that Q is not a Gδ -set (see Example 2.2.19). Similarly, the set B is Gδ , but it is not Fσ . We claim that E = A ∪ B is neither Gδ nor Fσ . If E was Gδ , then we could write E = ∩Un where each Un is open. Then Unk contains the point n1 , 1 for any m 6= n. But then we would have but does not contain m
T A = E ∩ (−1, 2) = Un ∩ (−1, 2) , n

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which is a contradiction since A is not a Gδ -set. On the other hand, if E was Fσ , then we could write E = ∪Fn where each Fn is closed. But then we would have
S B = E ∩ [2, 3] = Fn ∩ [2, 3] , n

which is a contradiction since B is not an Fσ -set. Although A is not a Gδ -set, it is countable, so is a countable union of singletons. Each singleton is Gδ by part (b), and the set B is also a Gδ -set, so by writing   S E = {x} ∪ B, x∈A

we have expressed E as a countable union of Gδ -sets. Therefore E is a Gδσ set.

2.2.47 (a) “⇒.” Assume that f is continuous at x, and fix ε > 0. Then there exists some δ > 0 such that |f (x) − f (y)| < ε whenever y ∈ Bδ (x). Consequently, if y, z ∈ Bδ (x) then we have |f (y) − f (z)| < 2ε, and therefore oscf (x) < 2ε. Since ε is arbitrary, we conclude that oscf (x) = 0. “⇐.” Suppose that oscf (x) = 0, and fix ε > 0. Then, by definition of infimum, there exists some δ > 0 such that  sup |f (y) − f (z)| : y, z ∈ Bδ (x) < ε.

Taking z = x, it follows that for all y ∈ Bδ (x) we have |f (y) − f (x)| < ε. Therefore f is continuous at x.

(b) We will show that the set E = {x ∈ Rd : oscf (x) ≥ ε} contains all of its limit points and therefore is closed. Suppose that oscf (xn ) ≥ ε and xn → x as n → ∞. We must show that oscf (x) ≥ ε. Fix δ > 0. Then there exists some integer n ∈ N such that |x − xn | < δ2 . Since oscf (xn ) ≥ ε, we must have  sup |f (y) − f (z)| : y, z ∈ Bδ/2 (x) ≥ ε. Fix 0 < η < ε. Then there exist some points y, z ∈ Bδ/2 (x) such that |f (y) − f (z)| ≥ ε − η. Now, |x − y| ≤ |x − xn | + |xn − y|
0

As η > 0 is arbitrary, it follows that oscf (x) ≥ ε. (c) By part (a), we can write D as D =

 S n 1o x ∈ Rd : oscf (x) > 0 = . oscf ≥ k k∈N

Part (b) tells us that {oscf ≥ ε} is closed, so we conclude that D is an Fσ -set. 2.2.48 (a) Case 1 : |A|e < ∞. Let A be any subset of Rd whose exterior measure is finite. By Lemma 2.2.20, there exists a Gδ -set H ⊇ A such that |A|e = |H|. Now let E be any measurable subset of Rd . Applying the Carath´eodory Criterion and monotonicity, we see that |H ∩ E| + |H \E| = |H| = |A|e = |A ∩ E|e + |A\E|e ≤ |H ∩ E| + |H \E|. Therefore we have the equality |A ∩ E|e + |A\E|e = |H ∩ E| + |H \E|.

(A)

However, since monotonicity implies that |A ∩ E|e ≤ |H ∩ E|

|A\E|e ≤ |H \E|,

and

and since all of the quantities involved are finite, the only way that equation (A) can hold is if |A ∩ E|e = |H ∩ E|

|A\E|e = |H \E|.

and

This completes the proof for the case where A has finite measure. Case 2: Arbitrary sets. Let A be any subset of Rd . For each k ∈ Z, set Ak = A ∩ [−k, k]d . S

Each set Ak has finite measure, and A = Ak . By Case 1, for each k we can find a Gδ -set Hk ⊇ Ak such that |Ak ∩ E|e = |Hk ∩ E| For each j ∈ Z, set

for every measurable E.

c Solutions 2019 Christopher Heil

45 ∞ T

Gj =

Hk .

k=j

As each Hk is a countable intersection of open sets, Gj is likewise a countable intersection of open sets and therefore is a Gδ -set. Now set ∞ S

H =

Gj .

j=1

Note the following facts. •

G1 ⊆ G2 ⊆ · · · .

•

If j ≤ k, then Aj ⊆ Ak ⊆ Hk . Therefore Aj ⊆ S

S

∞ T

Hk = Gj .

k=j

•

A = Aj ⊆ Gj = H.

•

Unfortunately, H need not be a Gδ -set. Now let E be any measurable subset of Rd . Then H ∩E =

∞ S

j=1

(Gj ∩ E),

and this is a countable union of nested increasing sets. Consequently, ∞ S (Gj ∩ E) |H ∩ E| = j=1

= lim |Gj ∩ E| j→∞

≤ lim sup |Hj ∩ E| j→∞

= lim sup |Aj ∩ E|e j→∞

(by continuity from below) (since Gj ⊆ Hj ) (by definition of Hj )

≤ |A ∩ E|e

(since Aj ⊆ A)

≤ |H ∩ E|e

(since A ⊆ H).

Hence |H ∩ E| = |A ∩ E|e . (b) Although the set H in part (a) need not be a Gδ -set, we know that there exists a Gδ -set S ⊇ H such that |S| = |H|. Let Z = S \H, so Z is a set of measure zero. Then

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|S ∩ E| = |(H ∪ Z) ∩ E| = |(H ∩ E) ∪ (Z ∩ E)| ≤ |H ∩ E| + |Z ∩ E| = |A ∩ E|e + 0 ≤ |S ∩ E|. Consequently |S ∩ E| = |A ∩ E|e , so S is the Gδ -set that we seek. S

(c) Let E = Ek , and let H be the set found in part (a). Then |A ∩ E| = |H ∩ E| ∞ S = (H ∩ Ek ) k=1

=

∞ X

k=1

=

∞ X

k=1

(by part (a))

e

|H ∩ Ek |e

(countable additivity)

|A ∩ Ek |e

(by part (a)).

2.2.49 (a) Since ∅ ∈ L(Rd ), we have ∅ = ∅ ∩ A ∈ L(A). Fix F ∈ L(A). Then F = E ∩ A where E ∈ L(Rd ). Therefore E C = d R \E ∈ L(Rd ), so A\F = A ∩ F C = A ∩ (E C ∪ AC ) = A ∩ E C ∈ L(A). Therefore L(A) is closed under complements in A. Suppose Ak ∈ L(A) for k ∈ N. Then Ak = Ek ∩ A where Ek ∈ L(Rd ). Since ∪Ek ∈ L(Rd ), we have   S S S (Ek ∩ A) = Ek ∩ A ∈ L(A). Ak = k

k

k

Therefore L(A) is closed under countable unions of subsets of A.

(b) Let M = {E ⊆ A : E ∈ L(Rd )}. If F ∈ M, then F ⊆ A and F ∈ LRd . Hence F = F ∩ A ∈ L(A). Conversely, suppose that F ∈ L(A). Then F = E ∩ A where E ∈ L(Rd ). Since A is measurable and L(Rd ) is closed under finite intersections, we conclude that F ∈ L(Rd ). Therefore f ∈ M. 2.2.50 Since the empty set is countable, it belongs to Σ and therefore Σ is nonempty.

Choose any E ∈ Σ. If E C = X\E is countable then it belongs to Σ. On the other hand, if E C is not countable then E must be countable since E

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belongs to Σ. Hence (E C )C is countable, so E C ∈ Σ. Therefore Σ is closed under complements. Now suppose that Ek ∈ Σ for k ∈ N. If every Ek is countable then so is ∪Ek , so ∪Ek ∈ Σ in this case. Otherwise some particular Ej must be uncountable. But since Ej ∈ Σ we then know that EjC is countable. Since S

Ek

k


C

C

=

T k

EkC ⊆ EjC ,

we conclude that ∪Ek is countable, and therefore ∪Ek belongs to Σ in this case. Hence Σ is closed under countable unions. 2.2.51 (a) This is a special case of part (b). (b) Let {Σi }i∈J be a collection of σ-algebras on X. Since ∅ and X belong to every σ-algebra on X, they belong to the intersection of any collection of σ-algebras. Suppose that Ek ∈ Σ for all k in some countable index set K. Then given any fixed i, we have Ek ∈ Σi for every k. Therefore E = ∪k Ek ∈ Σi since Σi is closed under countable unions. Hence E belongs to every Σi , so E belongs to Σ by definition of intersection. Closure under complements is similar. (c) This follows from part (b). 2.3.2 Here is an alternative proof of Theorem 2.3.2. Suppose that E1 ⊆ E2 ⊆ · · · are measurable. If we set E0 = ∅, then ∞ S

k=1

Ek =

∞ S

j=1

(Ej \Ej−1 ),

and the sets on the right-hand side above are disjoint. Therefore, by countable additivity, ∞ ∞ ∞ X S S |Ej \Ej−1 | (E \E ) = E = j j−1 k k=1

j=1

j=1

=

=

lim

N →∞

N X j=1

|Ej \Ej−1 |

N S (Ej \Ej−1 ) lim N →∞ j=1

=

lim |EN |.

N →∞

2.3.6 If either E or F is empty then the result is immediate, so we may assume that E and F are both nonempty.

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(a) This follows from the definition of volume and the fact that the measure of a box equals its volume. (b) Suppose that U, V are open and nonempty. If (x, y) ∈ U × V, then x ∈ U and y ∈ V. Therefore exists a ball Brn (x) in Rm and a ball Bsn (y) in Rn such that Brm (x) ⊆ U and Bsn (y) ⊆ V. Let t = min{r, s}, and let Btm+n (x, y) be the ball in Rm+n centered at (x, y). The reader should check that Btm+n (x, y) ⊆ Brm (x) × Bsn (y) ⊆ U × V. This shows that U × V is open, and hence measurable. We can find countably many nonoverlapping boxes Qk ⊆ Rm and Rℓ ⊆ Rn such that S S U = Qk and V = Rℓ . k

ℓ

By Corollary 2.2.17,

|U | =

∞ X

k=1

|Qk |

and

Further,

S

U ×V =

k,ℓ

|V | =

∞ X ℓ=1

|Rℓ |.

(Qk × Rℓ ),

and this is a union of nonoverlapping boxes. Therefore, ∞ X ∞ S X |U × V | = (Qk × Rℓ ) = |Qk × Rℓ | k,ℓ

=

=

k=1 ℓ=1 ∞ X ∞ X

ℓ=1

=

|Qk | |Rℓ |

ℓ=1 k=1 ∞ X ∞ X ∞ X ℓ=1

(by Corollary 2.2.17)

k=1

(by part (a))

 |Qk | |Rℓ |

|U | |Rℓ | = |U | |V |.

Hence the result holds for open sets. (c) Suppose that G, H are two bounded Gδ -sets. By definition of a Gδ -set, T T Vk , Uk and H = G = k

k

and by intersecting with a sufficiently large open ball if necessary, we can assume that the Uk , Vk are bounded open sets. Appealing to Lemma 2.2.20(b), we can also assume that the Uk are nested decreasing, and likewise the Vk are nested decreasing.

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Since all of the sets involved have finite measure, it follows from continuity from above that |G| = lim |Uk | k→∞

|H| = lim |Vk |.

and

k→∞

We claim that, because the Uk and Vk are nested, G×H =

∞ T

k=1

(Uk × Vk ).

To see this, choose (x, y) ∈ G × H. Then x ∈ G so x ∈ Uj for some j, and similarly y ∈ H so y ∈ Vℓ for some ℓ. Let k be the smaller of j and ℓ. Then x ∈ Uj ⊆ Uk and y ∈ Vℓ ⊆ Vk , so (x, y) ∈ Uk × Vk . This shows that G × H ⊆ ∩k (Uk × Vk ), and the converse inclusion follows easily. Consequently G × H is a Gδ -set and hence is measurable. Since the sets Uk × Vk are bounded and nested decreasing, continuity from above implies that |G × H| = lim |Uk × Vk | = lim |Uk | |Vk | k→∞ k→∞    = lim |Uk | lim |Vk | = |G| |H|. k→∞

k→∞

Hence the result holds for bounded Gδ -sets.

(d) Next, suppose that E is any bounded measurable set and Z is any set with measure zero. Choose ε > 0, and let {Qk }k be a covering of E by boxes and {Rℓ }ℓ a covering of Z by boxes such that X X |Qk | ≤ |E| + ε and |Rℓ | ≤ |Z| + ε = ε. k

ℓ

Then {Qk × Rℓ }k,ℓ is a covering of E × Z by boxes, so XX XX |E × Z|e ≤ |Qk × Rℓ | = |Qk | |Rℓ | k

ℓ

k

=

ℓ

X k

 X  |Qk | |Rℓ | ℓ

≤ (|E| + ε) ε.

Since this is true for every ε, we conclude that |E × Z|e = 0 = |E| |Z| (and therefore E × Z is measurable). An alternative approach is to cover E and Z by open sets and apply part (b). S If E is an arbitrary measurable set and |Z| = 0, then we can write E = Ek where each Ek is bounded. The above work shows that |Ek × Z| = 0. Since E × Z is the union of the sets Ek × Z, it follows that |E × Z| = 0 as well.

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(e) Let E1 , E2 be arbitrary bounded measurable sets. Then we can write E1 = H1 \ Z1 and E2 = H2 \ Z2 where H1 , H2 are bounded Gδ -sets and |Z1 | = |Z2 | = 0. By replacing Z1 by Z1 ∩ H1 , we can assume that Z1 ⊆ H1 , and likewise Z2 ⊆ H2 . After some set-theoretic calculations, we see that
E1 × E2 = (H1 × H2 ) \ (Z1 × H2 ) ∪ (H1 × Z2 ) ,

and therefore E1 × E2 is measurable since all of the sets on the line above are measurable. |Z1 × H2 | = |H1 × Z2 | = 0, and therefore |(H1 × H2 ) \ (E1 × E2 )| = (Z1 × H2 ) ∪ (H1 × Z2 ) = 0.

Since H1 × H2 is a Gδ -set, this shows that E1 × E2 is measurable. Further, since all the sets involved have finite measure, we have that |E1 × E2 | = |H1 × H2 | − (Z1 × H2 ) ∪ (H1 × Z2 ) = |H1 × H2 |

= |H1 | |H2 | = |E1 | |E2 |. Hence we have proved the result for bounded measurable sets. Finally, suppose that E, F are arbitrary measurable sets. For k, ℓ ∈ N, define Ek = {x ∈ Rm : k − 1 ≤ |x| < k} and Fℓ = {x ∈ Rn : ℓ − 1 ≤ |x| < ℓ}. Then E×F =

S

k,ℓ

(Ek × Fℓ ),

and this is a disjoint union. Therefore, by countable additivity, even if the sums are infinite we have |E × F | =

∞ X ∞ X k=1 ℓ=1

|Ek × Fℓ | = =

∞ X ∞ X

k=1 ℓ=1

X ∞ k=1

|Ek | |Fℓ |

 X  ∞ |Ek | |Fℓ | = |E| |F |. ℓ=1

Hence the result holds for arbitrary measurable sets. 2.3.12 Let Q be a cube in Rd with sides of length s. By the Pythagorean Theorem, the diameter of this cube is d1/2 s. That is, x and y are any two

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points in Q, then kx − yk ≤ d1/2 s. Since f is Lipschitz, it follows that kf (x) − f (y)k ≤ K kx − yk ≤ Kd1/2 s. Thus, the diameter of the set f (Q) is at most Kd1/2 s. Consequently f (Q) is contained in a closed ball of diameter at most Kd1/2 s, and hence is contained in a cube with sidelengths Kd1/2 s. Therefore the measure of f (Q) is at most |f (Q)|e ≤ (Kd1/2 s)d = K d dd/2 sd = C |Q|, where C = 2d K d dd/2 is a fixed constant that does not depend on the box Q. 2.3.16 (a) Let Q be any cube in Rd . Then Q = rQ0 + x for some r > 0 and x ∈ Rd . Using the translation-invariance of Lebesgue measure, the dilation property of Lebesgue measure proved in Problem 2.1.38, and the fact that L is linear, it follows that |L(Q)| = |L(rQ0 + x)| = |rL(Q0 ) + L(x)|

linearity

= |rL(Q0 )|

translation-invariance

= rd |L(Q0 )|

dilation property

= |Q| dL . Note that this step holds for every matrix L, singular or nonsingular. (b) If U is an open set, then we can write U as a countable union of nonoverlapping cubes Qk . Therefore   ∞ S |L(U )| = L Qk k=1

∞ S L(Qk ) = k=1

≤

=

∞ X

k=1

∞ X

k=1

|L(Qk )|

(subadditivity)

dL |Qk |

= dL |U |. We will show that if L is nonsingular, then we actually have the equality |L(U )| = dL |U |. Each box Qk is the disjoint union of its interior Q◦k and its boundary Zk = ∂Qk . We know that the boundary of a box has measure zero, so

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52 ∞ S

Z =

Zk

k=1

has measure zero as well. Then we can write U as the disjoint union U = Z ∪

∞ S

k=1

A set-theoretic calculation shows that L(U ) = L(Z) ∪

Q◦k .

∞ S

k=1

L(Q◦k ).

Moreover, since L is invertible it is a bijection, and therefore this is a disjoint union of sets. Since L is Lipschitz, Theorem 2.3.13 implies that L(Z) has measure zero. Therefore, by countable additivity, ∞ X ∞ S ◦ L(Qk ) = |L(Z)| + |L(Q◦k )|. |L(U )| = L(Z) ∪ k=1

k=1

But |L(Z)| = 0, and by applying additivity again we see that |L(Qk )| = |L(Q◦k ∪ Zk )| = |L(Q◦k ) ∪ L(Zk )|

= |L(Q◦k )| + |L(Zk )|

=

(additivity)

|L(Q◦k )|.

Therefore |L(U )| = |L(Z)| + = 0 +

∞ X

k=1

=

∞ X

k=1

∞ X

k=1

|L(Q◦k )|

|L(Qk )|

dL |Qk |

(work above) (work above) (by part (a))

= dL |U |. Note that we used countable additivity in these computations, which requires disjoint sets, and disjointness follows from our assumption that L is nonsingular. (c) Assume that L is nonsingular. Let H be a bounded Gδ -set. Then there exist bounded nested decreasing T open sets Uk such that H = Uk . Since L is nonsingular, it is a bijection. Therefore

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L



∞ T

53

Uk

k=1

Hence



=

∞ T

L(Uk ).

k=1

  ∞ ∞ T T L(Uk ) Uk = |L(H)| = L k=1

k=1

= lim |L(Uk )|

(continuity from above)

= lim dL |Uk |

(by part (b))

= dL |H|

(continuity from above).

k→∞

k→∞

Now let E be any bounded measurable set. Then there exists a bounded Gδ -set H and a set Z that has measure zero such that E = H \Z. As a consequence, |E| = |H|. Since L : Rd → Rd is linear and therefore Lipschitz, it maps sets with measure zero to sets with measure zero. Hence |L(Z)| = 0. We also know that L maps measurable sets to measurable sets, so L(E) is measurable, and we have |L(E)| = |L(H \Z)| ≤ |L(H)| = dL |H| = dL |E|. To obtain the opposite inequality, we observe after some set-theoretic calculations (see the Lemma below) that L(H \Z) ⊇ L(H)\L(Z), although equality need not hold on the line above. Since H is bounded and L is linear, L(H) is a bounded set. Therefore L(H) and L(E) have finite measures, so since L(Z) has zero measure it follows that |L(E)| = |L(H \Z)| ≥ |L(H)\L(Z)| = |L(H)| − |L(Z)| = dL |H| − 0

(as H is Gδ )

= dL |E|. Combining the above work, we see that |L(E)| = dL |E|. If E is an arbitrary measurable set, then we can write E as a disjoint union of countably many bounded measurable sets Ek . Then L(E) is the disjoint union of the sets L(Ek ), so countable additivity implies that X X |L(E)| = |L(Ek )| = dL |Ek | = dL |E|. k

k

Lemma. f (A)\f (B) ⊆ f (A\B). (In general, equality need not hold.)

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Proof. If y ∈ f (A)\f (B), then y = f (x) where x ∈ A, but y 6= f (z) for any z ∈ B. If we had x ∈ B, then we would have y = f (x) with x ∈ B, which is a contradiction. Therefore x ∈ / B, so we have y = f (x) where x ∈ A\B. Thus y ∈ f (A\B). ⊓ ⊔ (d) Let δ1 , . . . , δd be the diagonal entries of ∆. Then, using Problem 2.1.38, we have d∆ = |∆(Q0 )| = |δ1 · · · δd | |Q| = | det(∆)|. Note that this calculation holds for every diagonal matrix, singular or nonsingular. (e) Let B = B1 (0) be the open unit ball in Rd . If V is an orthogonal matrix then V (B) = B. An orthogonal matrix is nonsingular, so we can apply part (c) to compute that |B| = |V (B)| = dV |B|. As |B| = 6 0, it follows that dV = 1. (f) If A and B are nonsingular, then we can apply part (c) to obtain dAB = |(AB)(Q0 )| = |A(B(Q0 ))| = dA |B(Q0 )| = dA dB . (g) Given a nonsingular d × d matrix L, write L = W ∆V T where V, W are orthogonal matrices and ∆ is a diagonal matrix. Combining the preceding steps and noting that V T is also an orthogonal matrix, we see that dL = dW d∆ dV T = | det(∆)|. On the other hand, since the determinant is multiplicative, we have det(L) = det(W ) det(∆) det(V T ) = det(∆). Therefore dL = | det(L)|. Singular matrices. If L is singular, then range(L) is a proper subspace of Rd . If Problem 2.1.35(c) has been worked, then we can appeal to that problem and conclude that range(L) has measure zero. Consequently, dL = |L(Q0 )| ≤ |range(L)| = 0 = | det(L)|, and so the problem is done. However, if Problem 2.1.35(c) has not been worked, then we do not yet know that arbitrary proper subspaces have measure zero. One way to handle this is to observe that rotations are nonsingular, and therefore our work for nonsingular matrices shows that Lebesgue measure is invariant under rotations. Every proper subspace can be rotated to a proper subspace that is

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parallel to the coordinate axes, and we can easily show that those proper subspaces have measure zero. Hence an arbitrary proper subspace has measure zero, and again we are done. Yet another approach is to modify the steps used for nonsingular matrices to show that dL = 0 when L is singular. We show how to do this below. We already observed in part (b) that if U is an open set, then |L(U )| ≤ dL |U |.

(A)

Suppose that H is a bounded Gδ -set. As in part (c), let Uk be bounded, T nested decreasing open sets such that H = Uk . Since ∞  ∞ T T L L(Uk ), Uk ⊆ k=1

k=1

we have

∞ ∞  T T L(Uk ) |L(H)| = L Uk ≤ k=1

(subadditivity).

k=1

= lim |L(Uk )|

(continuity from above)

≤ lim dL |Uk |

(by equation (A))

= dL |H|

(continuity from above).

k→∞

k→∞

If E is a bounded measurable set, then E = H \Z where H is a bounded Gδ -set |Z| = 0. Hence |L(E)| = |L(H \Z)| ≤ |L(H)| ≤ dL |H| = dL |E|. Let A and B be any d × d matrices. Then B(Q0 ) is a bounded measurable set, so dAB = |(AB)(Q0 )| = |A(B(Q0 ))| ≤ dA |B(Q0 )| = dA dB . Consequently, if L is a singular matrix and we let L = W ∆V T be the SVD for L, then we have dL ≤ dW d∆ dV T = dW | det(∆)| dV T = 0. (In fact, we proved before that dW = 1 = dV T ), but we do not even need that here.) Hence dL = | det(L)| even when L is singular. 2.3.17 Since |An | → |E|, we can choose n1 < n2 < · · · so that |E \Ank | = |E| − |Ank | < 2−k |E|,

k ∈ N.

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Note that the first equality on the preceding line holds because E has finite measure (see Lemma 2.3.1). Let A = ∩Ank . Applying Lemma 2.3.1 again, we compute that ∞ T Ank |E| − |A| = |E \A| = E \ k=1

S ∞ = (E \Ank ) k=1

∞ X

≤

k=1 ∞ X


0. To see that this can fail if the measure of E is infinite, set E = R and An = [2n , 2n+1 ]. Then |An | = 2n → ∞ = |E|, but ∩Ank = ∅ for every choice of indices n1 < n2 < · · · . 2.3.18 “⇒.” If E is measurable, then Carath´eodory’s Criterion implies that |A|e = |A ∩ E|e + |A\E|e for every set A. “⇐.” First Proof. Assume that |Q| = |Q ∩ E|e + |Q\E|e holds for every box Q. Let U be any open subset of Rd . Then there exist nonoverlapping cubes Qk such that U = ∪Qk . Therefore |U | = |(U ∩ E) ∪ (U \E)| ≤ |U ∩ E|e + |U \E|e S  S  Qk \E = Qk ∩ E + e

k

S = (Qk ∩ E) + k

≤

=

X k

e

|Qk ∩ E|e +

e

k

k

|Qk \E|e

 X |Qk ∩ E|e + |Qk \E|e k

e

k

S (Qk \E)

X

(subadditivity)

(subadditivity) (all terms nonnegative)

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=

X k

57

|Qk |

(hypothesis)

= |U |

(nonoverlapping cubes).

Consequently |U | = |U ∩ E|e + |U \E|e . Now let H be any Gδ -set in Rd with |H| < ∞. Then there exist open sets U1 ⊇ U2 ⊇ · · · with finite measures such that H = ∩Uk . Hence |H| ≤ |H ∩ E|e + |H \E|e   ≤ inf |Uk ∩ E|e + |Uk \E|e k

(subadditivity) (monotonicity)

= inf |Uk |

(previous case)

= |H|

(continuity from above).

k

Consequently |H| = |H ∩ E|e + |H \E|e in this case as well. Finally, let A be an arbitrary subset of Rd . If |A|e = ∞, then subadditivity implies that ∞ = |A|e ≤ |A ∩ E|e + |A\E|e ≤ ∞, and therefore |A| = |A ∩ E|e + |A\E|e in this case. Otherwise, if |A|e < ∞, then there exists some Gδ -set H ⊇ A such that |H| = |A|e < ∞. Therefore |A|e ≤ |A ∩ E|e + |A\E|e

(subadditivity)

≤ |H ∩ E|e + |H \E|e

(monotonicity)

= |H|

(previous case)

= |A|e . Hence |A|e = |A ∩ E|e + |A\E|e . Thus the equality |A|e = |A ∩ E|e + |A\E|e holds for every set A, so Carath´eodory’s Criterion implies that E is measurable. Second Proof, for sets with finite measure. Assume that |E|e < ∞, and suppose that |Q| = |Q ∩ E|e + |Q\E|e holds for every box Q. If we fix ε > 0, then there is an open set U ⊇ E such that |E|e ≤ |U | ≤ |E|e + ε. Since U is open, there exist nonoverlapping cubes Qk such that U = ∪Qk . Therefore S  S  Qk \E |E|e + |U \E|e = Qk ∩ E + k

e

k

e

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S = (Qk ∩ E) + e

k

≤ =

X k

X k

|Qk ∩ E|e +

S (Qk \E)

e

k

X k

|Qk \E|e

|Qk |

(subadditivity) (hypothesis)

= |U |

(nonoverlapping cubes)

≤ |E|e + ε. Since E has finite measure, we can subtract |E|e from both sides to obtain |U \E|e ≤ ε. Therefore E is measurable. Q. Can this proof be extended to cover sets with infinite measure? Third Proof. Let A be any subset of R. By subadditivity we always have |A|e ≤ |A ∩ E|e + |A\E|e , so we only need to prove the opposite inequality. If |A|e = ∞, then this is immediate: |A ∩ E|e + |A\E|e ≤ ∞ = |A|e . Therefore we can assume that |A|e < ∞. Fix ε > 0. Then there exist boxes Qk that cover A and satisfy X |A|e ≤ |Qk | ≤ |A|e + ε. k

Therefore S  S  Qk \E |A ∩ E|e + |A\E|e ≤ Qk ∩ E + e

k

e

k

S S = (Qk ∩ E) + (Qk \E) e

k

≤ =

X k

X k

|Qk ∩ E|e +

e

k

X k

(monotonicity)

|Qk \E|e

|Qk |

≤ |A|e + ε. This is true for every ε > 0, so we conclude that

(subadditivity) (hypothesis)

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|A ∩ E|e + |A\E|e ≤ |A|e . Therefore E is measurable. 2.3.19 (a), (b), (c) If 0 < s < t < ∞ then E ∩ Bs (0) ⊆ E ∩ Bt (0), Hence f is monotone increasing. For each t > 0, define Et = E ∩ Bt (0). Each set Et is measurable and has finite measure. The sets Et are nested, i.e., s < t =⇒ Es ⊆ Et . Monotonicity therefore implies that f (s) = |Es | ≤ |Et | = f (t). Since

S

t>0 Et

= E, continuity from below implies that lim |Et | = |E|.

t→∞ T

Since t>0 Et = E ∩{0} and since |Et | < ∞ for every t, continuity from above implies that lim+ |Et | = 0. t→0

To be more precise, there is a technicality that we should address. Continuity from above and below is stated for sequences of sets indexed by the natural numbers, not for sequences indexed by a continuous parameter. To circumvent this, we use Problem 1.1.23. Let tk → 0 be any discrete sequence of positive real numbers that decreases monotonically to zero as k → ∞. Then we can apply continuity from above to conclude that lim |Etk | = 0.

k→∞

Because |Etk | converges to the same value (zero) for every sequence tk → 0, Problem 1.1.23 implies that limt→0+ |Et | = 0. A similar approach justifies the conclusion that limt→∞ |Et | = |E|. Next, set  St = x ∈ Rd : kxk = t . Then

S s 0. Since f (t) increases to |E| as t → ∞, there is some R > 0 such that f (R) = |Et | > |E| − ε. Consequently, |E \ER | < ε.

Let C be the constant such that |Br (x)| = Crd (note that C depends only on the dimension d). If s ≤ t ≤ R + 1, then |f (t) − f (s)| = |Et | − |Es | = |Et \Es | ≤ |Bt (0) \ Bs (0)| = Ctd − Csd = C (t − s) (td−1 + td−2 s + · · · + tsd−2 + sd−1 ) ≤ C (t − s) d(R + 1)d . Therefore, if we set δ = min



 1 ε , , 2 Cd(R + 1)d

then t − s < δ =⇒ |f (t) − f (s)| < ε. On the other hand, if R < s < t then Et \Es ⊆ E \ER , so |f (t) − f (s)| = |Et \Es | ≤ |E \ER | < ε. Since δ < 1, given any s < t that satisfy t − s < δ, we must have either s, t < R + 1 or s, t > R (or both). In either case, |f (t) − f (s)| < ε, so f is uniformly continuous. Second proof. For t 6= 0, set

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g(t) = |E| − f (|t|), and let g(0) = |E|. By part (a), g ∈ C0 (R). All functions in C0 (R) are uniformly continuous, so g is uniformly continuous on R, and hence on any interval contained in R. For t > 0 we have f (t) = |E| − g(t). Since constant functions are uniformly continuous, so is f. 2.3.20 (a) For each t > 0, define At = E ∩ Bt (0). Each set At is measurable and has finite measure. The sets At are nested, S i.e., if s < t then As ⊆ At . Since t>0 At = E, continuity from below implies that lim |At | = |E| > 0. t→∞

Since

T t>0 At

= E ∩ {0}, continuity from above implies that lim |At | = 0.

t→0

As 0 < |At | < ∞ for every t, it follows that there must exist some t > 0 such that 0 < |At | < |E|. Therefore, the set A = At is measurable, and both A and E \A have positive measure. (b) By part (a), there exists a set E1 ⊆ E such that we have both |E1 | > 0 and |E \E1 | > 0. Applying part (a) to the set E \E1 , we obtain a set E2 that is disjoint from E1 and satisfies |E2 | > 0 and |E \ (E1 ∪ E2 )| > 0. Continuing in this way we create disjoint sets E1 , E2 , . . . contained in E that each have positive measure. (c) Since E has finite measure, Problem 2.3.19 implies that the function f (t) = |E ∩ Bt (0)| is uniformly continuous on (0, ∞). Further, lim f (t) = 0

t→0+

lim f (t) = |E|.

and

t→∞

Therefore, there must exist a t such that |Et | =

|E| . 2

Set E1 = Et and F1 = E \E1 . By countable additivity, |F1 | = |E| − |E1 | =

|E| . 2

Applying the previous argument with E replaced by F1 , there exists some set E2 ⊆ F1 such that

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|E| |F1 | = . 2 4 By construction, E1 and E2 are disjoint. Repeating this procedure, we obtain the desired sets Ek . |E2 | =

(d) Case 1. Assume that |E| < ∞, and let n be large enough that n1 < |E|. 1 By Lemma 2.2.15, there exists a closed set F ⊆ E such that |E \F | ≤ 2n . Since E and F are measurable, we therefore have |F | ≥ |E| −

1 . 2n

For each m ∈ N, set Fm = F ∩ Bm (0). Then Fm is compact, and it follows from continuity from below that |Fm | → |F | as m → ∞. Therefore, there 1 . Hence exists some m such that |Fm | ≥ |F | − 2n |Fm | ≥ |F | −

1 1 ≥ |E| − . 2n n

Therefore we can take Kn = Fm for that m. Case 2. Assume that |E| = ∞. By Lemma 2.2.15, there exists a closed set F ⊆ E such that |E \F | ≤ 1. Since E and F are measurable, we must have |F | = ∞. For each n ∈ N, set Kn = F ∩ Bn (0). Then Kn is compact. Further, by continuity from below, |Kn | → |F | = |E|. (e) As in Problem 2.3.19, let f (t) = |E ∩ Bt (0)|,

t > 0.

That problem shows that f is continuous and monotone increasing, f (t) → 0 as t → 0+ , and f (t) → |E| = ∞ as t → ∞. Therefore, there is a t such that f (t) = 1. Let A1 = E ∩ Bt (0), so |A1 | = f (t) = 1. Since A1 has finite measure, the set E \A1 has infinite measure. Applying the same argument, there must exists a set A2 ⊆ E \A1 such that |A2 | = 1. We can again repeat the argument to find A3 , A4 , . . . . 2.3.21 Suppose that no such point x exists. Then for each x ∈ E there is some δx > 0 such that |E ∩ Bδx (x)| = 0. By Problem 2.3.20(d), there exists a compact set K ⊆ E such that |K| > 0. Then {Bδx (x)}x∈E is an open cover of K, so there must exist finitely many points x1 , . . . , xN ∈ E such that

c Solutions 2019 Christopher Heil

K ⊆ But then

N S

63

Bδk (xk ),

N S

K = K ∩E ⊆ so |K| ≤ which is a contradiction.

where δk = δxk .

k=1

k=1


Bδk (xk ) ∩ E ,

N N X X Bδ (xk ) ∩ E = 0 = 0, k

k=1

k=1

2.3.22 Let K be a Lipschitz constant for f. Given a generic cube Q in Rn with sides of length s, we will derive an estimate for the measure of f (Q) in terms of |Q| and s. By the Pythagorean Theorem, the diameter of Q is n1/2 s. That is, x and y are any two points in Q, then kx − yk ≤ n1/2 s. Since f is Lipschitz, it follows that kf (x) − f (y)k ≤ K kx − yk ≤ Kn1/2 s. Thus, the diameter of the set f (Q) is at most Kn1/2 s. Consequently f (Q) is contained in a closed ball of radius at most Kn1/2 s, and hence is contained in a cube with sidelengths 2Kn1/2 s. Therefore the measure of f (Q), which is a subset of Rm , is at most |f (Q)|e ≤ (2Kn1/2 s)m = 2m K m nm/2 sm = C sm−n |Q|,

(A)

where C = 2m K m nm/2 is a fixed constant that does not depend on the box Q. Given k ∈ N, subdivide the unit cube Q0 = [0, 1]n into k n nonoverlapping cubes Q 1 , . . . , Qk n . Each box Qj has sidelengths that

1 k.

Applying equation (A) with s =

1 k,

we see

C |Qj | . k m−n Since Q0 is the union of the nonoverlapping cubes Qj , j = 1, . . . , k n , we therefore obtain |f (Qj )|e ≤ C ( k1 )m−n |Qj | =

n kn X kS |f (Qj )|e f (Qj ) ≤ |f (Q0 )|e = j=1

e

j=1 n

≤

k X C |Qj | j=1

k m−n

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n

= =

C k m−n C k m−n

k X j=1

|Qj |

|Q0 | =

C k m−n

.

Since this is true for every k ∈ N and since m − n > 0, it follows that |f (Q0 )|e = 0. The same argument shows that |f (Q0 + ℓ)|e = 0 for every ℓ ∈ Zn . Hence S n |range(f )|e = |f (R )|e = f (Q0 + ℓ) = 0. ℓ∈Zn

e

2.3.23 Define T : R2 → R by T (x, y) = x − y. We seek to prove that T −1 (E) = {(x, y) ∈ R2 : T (x, y) ∈ E} is measurable. Let A =



 11 . 10

Then we can rewrite T −1 (E) as follows: T −1 (E) = {(x, y) ∈ R2 : x − y ∈ E} = {(x, y) ∈ R2 : x − y = z for some z ∈ E} = {(x, y) ∈ R2 : x = y + z for some z ∈ E} = {(y + z, y) ∈ R2 : y ∈ R, z ∈ E}     11 y = : y ∈ R, z ∈ E 10 z = {A(y, z) : y ∈ R, z ∈ E} = A(R × E). Now, R × E is a measurable subset of R2 , and A is a linear mapping of R2 to itself, so A maps measurable sets to measurable sets. Therefore A(R × E) = T −1 (E) is measurable. 2.3.24 (a) Fix x ∈ Rd , and suppose that xn → x. Fix ε > 0. By definition of the distance function, for each n there exists a point yn ∈ E such that dE (xn ) = dist(xn , F ) ≤ |xn − yn | ≤ dist(xn , F ) + ε = dE (xn ) + ε. Hence for each n we have

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dE (x) = dist(x, F ) ≤ |x − xn | + |xn − yn | ≤ |x − xn | + dE (xn ) + ε. Consequently,   dE (x) ≤ lim inf |x − xn | + dE (xn ) + ε n→∞     ≤ lim sup |x − xn | + lim inf dE (xn ) + ε n→∞

n→∞

= 0 + lim inf dE (xn ) + ε. n→∞

Since ε is arbitrary, it follows that dE (x) ≤ lim inf dE (xn ). n→∞

On the other hand, given any y ∈ E we have dE (xn ) ≤ |xn − y| ≤ |xn − x| + |x − y|, so lim sup dE (xn ) ≤ n→∞



 lim sup |xn − x| + |x − y| = 0 + |x − y|. n→∞

This is true for each y ∈ E, so lim sup dE (xn ) ≤ inf |x − y| = dist(x, E) = dE (x). n→∞

y∈E

Combining the above estimates, we see that dE (x) = lim dE (xn ), n→∞

so dE is continuous at x. (b) We have Er = d−1 E (−∞, r), the inverse image of the open interval (−∞, r) under dE . Since dE is continuous, the set Er is open. (c) Suppose E is closed. If dE (x) = dist(x, E) = 0 then there exist points yn ∈ E such that yn → x. Therefore, since E is closed, it must contain x. Conversely, if x ∈ E then |x − x| = 0, so dE (x) = dist(x, E) = 0.

(d) Let E be a closed subset of Rd . By part (b), E1/k is open for each T k ∈ N. By definition, E ⊆ E1/k . Suppose that x ∈ E1/k for each k ∈ N. Then dist(x, E) < 1/k for every k, so dist(x, E) = 0. Part (c) therefore T implies that x ∈ E. Hence E = E1/k , so E is a Gδ -set. (e) This follows from part (d) and the fact that the complement of a Gδ -set is an Fσ -set.

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(f) Suppose E is compact, and let (rn )n∈N be any sequence of positive real numbers such that rn ց 0. The sets Ern are nested decreasing with n. Since E is compact, it is bounded. Therefore E ⊆ Br (0) for some r > 0. Consequently Er1 ⊆ Br+r1 (0), so Er1 is bounded and therefore has finite measure. Each set Ern contains E, T and if x ∈ Ern for every n then dist(x, E) = 0. Since E is closed, it thereT fore follows from part (c) that x ∈ E. Thus E = Ern . Continuity from above therefore implies that |Ern | → |E| as n → ∞. As this is true for every sequence rn ց 0, it follows that |Er | → |E| as r → 0+ . Counterexample for generic closed sets. Consider the set E = {(x, 0) ∈ R2 : x ∈ R}. This is the x-axis in R2 , which is a closed but unbounded set. We have Er = R × (−r, r), which has measure |Er | = ∞ for every r > 0. Therefore |Er | does not converge to |E| = 0 as r ց 0. Counterexample for bounded open sets. Finally, let {rk }k∈N be an enumeration of Q ∩ [0, 1]. Fix 0 < ε < 1. For each k, let Ik be an interval of length 2−k ε that contains rk , and let U = ∪Ik . Then U is an open set that contains Q ∩ [0, 1], and U is bounded and has measure at most |U | ≤ ε. However, since U contains a dense subset of [0, 1], we have Ur ⊇ [0, 1] for every r > 0. Hence |Ur | ≥ 1 for every n, yet |U | ≤ ε < 1, so |Ur | does not converge to |U | as r ց 0. 2.3.25 (a) By definition, B is the intersection of every σ-algebra Σ that contains U. Since each such σ-algebra contains U, the intersection contains U by definition. Problem 2.2.51 shows that B is a σ-algebra. Therefore is it closed under complements, so it contains all the complements of the open sets, which are the closed sets. Since B is also closed under countable unions, it contains every countable union of closed sets, which means it contains all of the Fσ -sets. Taking complements, B is closed under countable intersections, and therefore contains the Gδ -sets. Repeating this process, we see that B contains the Fσδ -sets and so forth. (b) By definition, B is the intersection of every σ-algebra Σ that contains U. One of these σ-algebras is the Lebesgue σ-algebra L, simply because every open set is Lebesgue measurable and therefore belongs to L. Therefore B is a subset of L by the definition of intersection. (c) If E is measurable, then there exists a Gδ -set B such that B ⊇ E and Z = B \E has measure zero. Therefore E = B \Z where B is Borel and |Z| = 0. 2.4.5 We fill in the details of the claim made in the proof of Theorem 2.4.5 that

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[0, 1) ⊆

∞ S

k=1

(N + rk ) ⊆ [−1, 2].

The second inclusion is easy because N ⊆ [0, 1) and each scalar rk belongs to [−1, 1]. To prove the first inclusion, choose any point x ∈ [0, 1). Then x belongs to some equivalence class of the relation ∼, so there exists some point y ∈ N such that x ∼ y. Hence x = y + r where r is rational, and since both x and y belong to [0, 1), we must have r ∈ [−1, 1]. Hence r = rk for some k, and therefore x ∈ N + rk for that k. 2.4.8 (a) Suppose that E1 ⊆ E2 ⊆ · · · is a sequence of nested increasing subsets of Rd , and set E = ∪En . There exists a Gδ -set G ⊇ E with |G| = |E|e , and for each n there exists a Gδ -set Gn ⊇ En with |Gn | = |En |e . Define Hn = G ∩

∞ T

Gm

and

H =

m=n

∞ S

Hn .

n=1

Then Hn , H are Lebesgue measurable, and we have by nestedness that En =

∞ T

Em ⊆ G ∩

m=n

Hence

∞ T

Gm = Hn ⊆ Gn .

m=n

|En |e ≤ |Hn | ≤ |Gn | = |En |e , so |Hn | = |En |e . Further, E =

∞ S

n=1

so

En ⊆

∞ S

n=1

Hn = H ⊆ G,

|E|e ≤ |H| ≤ |G| = |E|e . Therefore |H| = |E|e . Finally, H1 ⊆ H2 ⊆ · · · , so continuity from below for Lebesgue measurable sets implies that lim |En |e = lim |Hn | = |H| = |E|e .

n→∞

n→∞

(b) Let N be the nonmeasurable subset of R constructed in Theorem 2.4.4. This set N has the property that rational translates of N are disjoint. Set Nk = N ∩ [k, k + 1]. Since N = ∪Nk and measurability is preserved under countable unions, at least one of these sets Nk must be nonmeasurable. By translating, we can assume that N0 is a nonmeasurable subset of [0, 1]. Then translates of N0 by rationals r ∈ [0, 1] are all disjoint and N0 + r ⊆ [0, 2] for each r ∈ Q ∩ [0, 1]. Let {rk }k∈N be an enumeration of Q ∩ [0, 1], and set Ek = N0 + rk . Define ∞ S Ek . FN = k=N

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Then FN ⊆ [0, 2] for every N, so |F1 |e < ∞. Also, the sets FN are nested decreasing, so by monotonicity their exterior measures form a decreasing sequence of nonnegative real numbers. However, |FN |e ≥ |EN |e = |N0 |e > 0, so |FN |e converges to some strictly positive number. On the other hand, we have ∩FN = ∅, so ∞ T FN = |∅|e = 0 < |N0 |e ≤ lim |FN |e . N =1

N →∞

e

2.4.9 The proof of Theorem 2.4.4 shows that there exists a nonmeasurable set N such that {N + r}r∈Q is a partition of R into disjoint sets. Suppose that A ⊆ R satisfies |A|e > 0. Let Ar = A ∩ (N + r). Then S A = Ar . r∈Q

If Ar is measurable and has positive measure, then Theorem 2.4.3 implies that Ar − Ar contains an interval centered at 0, which contradicts the definition of N. Therefore, if Ar is measurable then |Ar | = 0. Consequently, if every Ar is measurable then we have |A| = 0, which is a contradiction. Hence some Ar must be nonmeasurable. 2.4.10 We will take the longer approach of generalizing the Steinhaus Theorem to higher dimensions. Step 1. Let Q = [0, s]d . We claim that if t = (t1 , . . . , td ) ∈ Rd , then |Q ∩ (Q + t)| ≤

d   X d

k=0

k

sk ktkd−k .

First assume that 0 ≤ tk for every k. In this case we have 0 ≤ tk ≤ ktk for every k, so Q ∪ (Q + t) ⊆ [0, s + ktk]d , and therefore |Q ∪ (Q + t)| ≤ (s + ktk)d =

d   X d k s ktkd−k . k

k=0

A similar argument applies if any tk is negative, so the claim follows. Step 2. Now we generalize the Steinhaus Theorem. We claim that if E ⊆ Rd is Lebesgue measurable and |E| > 0, then the set of differences  E − E = x − y : x, y ∈ E

contains an open ball Br (0) for some r > 0.

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To see this, we apply Problem 2.2.39 and conclude that there exists a cube Q such that the measure of the set F = E ∩ Q satisfies |F | = |E ∩ Q| >

3 |Q|. 4

(A)

The statement of Steinhaus’ Theorem is invariant under translations, so by translating E, F, and Q we can assume that Q = [0, s]d where s > 0. Choose any t ∈ Rd . If F and F + t are disjoint, then we must have 2sd = 2 |Q| < 2 ·

4 |F | 3

(by equation (A))

4 |F ∪ (F + t)| 3 4 ≤ |Q ∪ (Q + t)| 3 d   4 X d k s ktkd−k ≤ 3 k =

(since F and F + t are disjoint) (by monotonicity) (by the Lemma).

(B)

k=0

However,

d   4 X d k 4 lim s ktkd−k = sd < 2sd . 3 k ktk→0 3 k=0

Therefore if ktk is small enough then equation (B) cannot hold. Hence there is some r > 0 such that ktk < r =⇒ F and F + t are not disjoint. Therefore, if ktk < r then there is some point x ∈ F ∩ (F + t). So, x = y + t for some y ∈ F, which implies that t = x − y ∈ F − F. This shows that F − F contains the open ball Br (0), and therefore E − E must contain this ball as well. Step 3. Define a relation on Rd by declaring that x ∼ y if and only if every component of x − y is rational. This is an equivalence relation, so by the Axiom of Choice there exists a set N that contains exactly one element of each distinct equivalence class of this relation. The distinct equivalence classes partition Rd , so their union is Rd . Therefore S S S S (N + r). {r + x} = (Qd + x) = Rd = x∈N

x∈N r∈Qd

r∈Qd

Since exterior Lebesgue measure is translation-invariant, the exterior measure of N + r is exactly the same as the exterior measure of N. Combining this fact with countable subadditivity, we see that

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X X S ∞ = |R|e = |N |e . (N + r) ≤ |N + r|e = d r∈Q

e

r∈Qd

r∈Qd

Consequently, we must have |N |e > 0. However, any two distinct points x 6= y in N belong to distinct equivalence classes of the relation ∼, so some component of x and y must differ by an irrational amount. Therefore N − N contains no open balls, so the Steinhaus Theorem implies that N cannot be Lebesgue measurable. 2.4.11 No. Let N be a nonmeasurable subset of R. Then A = {0} × N has measure zero as a subset of R2 , so it is measurable. However, A0 = {y ∈ R : (0, y) ∈ {0} × N } = N, which is not measurable. 2.4.12 Property (a) fails because µ([0, 1]) = ∞. Suppose that E1 , E2 , . . . are disjoint subsets of R. If the cardinality any Ej is infinite then so is E = ∪Ek , so in this case we have X µ(E) = ∞ = µ(Ej ) = µ(Ek ). k

Otherwise every setP Ek is a finite set, so the cardinalities are additive and we again have µ(E) = k µ(Ek ). Thus µ is countably additive, so property (b) holds. The cardinality of set is invariant under translations, so property (c) holds. 2.4.13 Property (a) holds because δ([0, 1]) = 1. Suppose that E1 , E2 , . . . are disjoint subsets of R, set E = ∪Ek . If 0 ∈ E then 0 ∈ Ej for some unique j, so in this case we have δ(E) = 1 = δ(Ej ) =

∞ X

δ(Ek ).

k=1

On the other hand, if 0 ∈ / E then 0 ∈ / Ek for any k, so in this case we have δ(E) = 0 =

∞ X

δ(Ek ).

k=1

Thus δ is countably additive, so property (b) holds. Property (c) fails. For example, if E = {0} then δ(E) = 1 but δ(E +1) = 0. 2.4.14 (a) We must show that lim f (x) = f (0) = |E|.

x→0

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By subadditivity, for every x we have f (x) = |E ∩ (E − x)| ≤ |E|. Hence lim sup f (x) ≤ f (0) = |E|. x→0

As the limits from the left and right are entirely similar, it therefore suffices to show that lim inf f (x) ≥ f (0) = |E|. + x→0

Step 1 : E = (a, b) is a finite open interval. Note that E − x = (a − x, b − x). For x > 0 we have ( (a, b − x), E ∩ (E − x) = (a, b) ∩ (a − x, b − x) = ∅,

0 < x < b − a, x ≥ b − a,

Therefore, for x > 0, ( b − a − x, 0 < x < b − a, f (x) = 0, x ≥ b − a. Hence lim f (x) = lim+ (b − a − x) = b − a = |E|.

x→0+

x→0

Step 2 : E = U is a union of finitely many disjoint open intervals. In this case we can write U =

N S

(ak , bk ).

k=1

Since we have finitely many disjoint intervals (ak , bk ), we may order them as a1 < b 1 < a2 < b 2 · · · < aN < b N . Let If 0 < x < δ, then

o nε , b 1 − a1 , . . . , b N − aN . δ = min N

f (x) = |U ∩ (U − x)| N N S S (ak − x, bk − x) = (aj , bj ) ∩ j=1

k=1

N S (aj , bj ) ∩ (ak − x, bk − x) = j,k=1

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N S ≥ (ak , bk ) ∩ (ak − x, bk − x) k=1 N S = (ak , bk − x) k=1

=

N X

|(ak , bk − x)|

N X

(bk − ak − x)

N X

(bk − ak ) − N x

k=1

=

k=1

=

k=1

(disjoint intervals)

= |U | − N x > |U | − ε. This shows that lim inf f (x) ≥ |U |. x→0+

Step 3 : E = U is a union of infinitely many disjoint open intervals. In this case we can write ∞ S

U =

(ak , bk ).

k=1

For each N ∈ N, set UN =

N S

(ak , bk ).

k=1

Using monotonicity and Step 2, we compute that lim inf f (x) = lim inf |U ∩ (U − x)| + + x→0

x→0

≥ lim inf |UN ∩ (UN − x)| + x→0

≥ |UN |. As continuity from below implies that |UN | → |U |, it follows that lim inf f (x) ≥ |U |. + x→0

Step 4 : E is an arbitrary bounded measurable set. Fix x > 0 and ε > 0. Then there exists a bounded open set U ⊇ E such that

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|U \E| < ε. Set A = U \E. We claim that E ∩ (E − x) ⊇

    U ∩ (U − x) \ A ∪ (A − x) .

To see why, suppose that y is any element of the right-hand side set. Then y ∈ U,

y ∈ U − x,

y∈ / A,

y∈ / A − x.

Therefore y ∈ U \E = A, and y ∈ (U − x)\(A − x) = E − x. This shows that y ∈ E ∩ (E − x) and proves the claim. Note that A ∪ (A − x) ⊆ U ∩ (U − x). As all sets are measurable and have finite measure, we can therefore compute that     f (x) = |E ∩ (E − x)| ≥ U ∩ (U − x) \ A ∪ (A − x) = U ∩ (U − x) − A ∪ (A − x)
≥ U ∩ (U − x) − |A| + |A − x| = U ∩ (U − x) − 2|A| ≥ U ∩ (U − x) − 2ε. Applying Step 3, we obtain U ∩ (U − x) − 2ε ≥ |U | − 2ε ≥ |E| − 2ε. lim inf f (x) ≥ lim inf + + x→0

x→0

As ε is arbitrary, it follows that

lim inf f (x) ≥ |E|. + x→0

(b) We will prove the Steinhaus Theorem. Assume that E is measurable and |E| > 0. Let F be a subset of E that has positive and finite measure. If we show that F − F contains an interval, then the larger set E − E also contains an interval. Therefore it suffices to consider the case that E has finite measure. Fix ε > 0. Since f (x) = |E ∩ (E − x)| is continuous at x = 0 and since f (0) = |E| > 0, there exists a δ > 0 such that f (x) ≥ ε for all |x| < δ. Thus,

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|x| < δ =⇒ |E ∩ (E − x)| = f (x) ≥ ε > 0. Consequently, if |x| < δ then E ∩ (E − x) is not empty, so there exists some point y ∈ E ∩ (E − x). That is, y ∈ E and y ∈ E − x. Hence y = z − x for some z ∈ E. But then x = z − y ∈ E − E. This proves that (−δ, δ) ⊆ E − E.

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Solutions to Exercises and Problems from Chapter 3 3.1.14 We are given a monotone increasing function f : E → R whose domain E is a measurable subset of R. Fix any a ∈ R. If f (x) < a for every x ∈ E, then {f < a} = E, which is measurable. Otherwise there exists at least one x ∈ E such that f (x) ≥ a. Hence {f ≥ a} is not empty. Therefore s = inf{f ≥ a} = inf{x ∈ E : f (x) ≥ a} is a finite real number. Case 1. Suppose that s ∈ E. If x ∈ E and x < s, then f (x) < a by definition of infimum. Hence x ∈ / {f ≥ a} in this case. If x ∈ E and x ≥ s, then f (x) ≥ f (s) ≥ a. Hence x ∈ {f ≥ a} in this case. Therefore {f ≥ a} = E ∩ [s, ∞), and this is measurable. Case 2. Suppose that s ∈ / E. If x ∈ E and x < s, then f (x) < a by definition of infimum. Hence x ∈ / {f ≥ a} in this case. If x ∈ E and x ≥ s, then x > s (since s ∈ / E) and f (x) ≥ a by definition of infimum. Hence x ∈ {f ≥ a} in this case. Therefore {f ≥ a} = E ∩ (s, ∞), which is measurable. Thus, in any case the set {f ≥ a} is measurable, so f is a measurable function. 3.1.15 “⇒.” If f is measurable then {f > a} is measurable for every real number a. “⇐.” Assume that {f > r} is measurable for every rational r. Fix a ∈ R and set E = {f > a}. The set Er = {f > r} is measurable for every rational r > a, and S Er . E = r∈Q,r>a

Therefore E is measurable since there are only countably many rationals greater than a. 3.1.16 Assume f is defined on all of E. If x ∈ E and f (x) 6= −∞ then f (x) ∈ (−∞, ∞], and therefore f (x) > k for some k ∈ Z. Hence   S E = {f = −∞} ∪ {f > k} . k∈Z

Each of the sets on the right is measurable by hypothesis, so f is measurable.

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If f is only defined on E \Z where |Z| = 0, then the same argument shows that E \Z is measurable. Since Z has measure zero, it follows that E is measurable as well. 3.1.17 (a) If f is measurable, then {f = a} = {f ≥ a} ∩ {f ≤ a} is measurable. (b) Let N be a nonmeasurable subset of (0, ∞). Define ( x, if x ∈ R, f (x) = −x, if x ∈ / R. If a ≤ 0 then {f = a} = {0}, which is measurable. If a > 0 then {f = a} is either {a} or ∅, so is measurable. Thus {f = a} is measurable for every a ∈ R. However, {f > 0} = N, which is not measurable, so f is not a measurable function. 3.1.18 (a) “⇐.” Suppose that f −1 (U ) is measurable for each open set U ⊆ R. Then for each a ∈ R we have that {f > a} = {x ∈ Rd : a < f (x)} = f −1 (a, ∞) is measurable, so f is measurable. “⇒.” Suppose that f : Rd → R is measurable, and let U ⊆ R be any open set. Then we can write U as a countable disjoint union of open intervals (possibly including infinite open intervals), say U = ∪(aj , bj ). Since f −1 (aj , bj ) = {aj < f < bj } = {aj < f } ∩ {f < bj }, we conclude that f −1 (aj , bj ) is measurable for each j, and hence f −1 (U ) = ∪f −1 (aj , bj ) is measurable.

(b) “⇒.” Suppose that f : Rd → C is measurable. Then its real part fr and its imaginary part fi are both measurable. For simplicity let us identify C with R2 . In particular, with this identification we write f (x) = (fr (x), fi (x)). Given an open strip (a, b) × R in C, we have
f −1 (a, b) × R = fr−1 (a, b), which is measurable since fr is measurable.. Similarly,
f −1 R × (c, d) = fi−1 (c, d)

is measurable. Consequently the inverse image of the open rectangle

(a, b) × (c, d) = (a, b) × R ∩ R × (c, d)

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is measurable. Every open subset of C can be written as a countable union of open rectangles, so it follows that f −1 (U ) is measurable for every open set U ⊆ C. “⇐.” Suppose that the inverse image of any open subset of C is measurable. Again identifying C with R2 , if we fix a ∈ R then the set (a, ∞) × R is open in C. Hence
{fr > a} = fr−1 (a, ∞) = f −1 (a, ∞) × R

is measurable. Therefore fr is a measurable function, and similarly fi is measurable, so we conclude that f is measurable. 3.1.19 (a) Let En = {|f | ≤ n}. Each set En is measurable since |f | is measurable. We have E1 ⊆ E2 ⊆ · · ·

and

B =

S

En = {|f | < ∞}.

Continuity from below therefore implies that |En | → |B|. As {f = ∞} has measure zero, we have |B| = |E| > 0, so there must be some n such that |En | > 0. By definition, f is bounded on En , so we can take A = En . (b) Let En = {|f | ≥ We have

1 n }.

Each set En is measurable since |f | is measurable.

E1 ⊆ E2 ⊆ · · ·

and

B =

S

En = {f 6= 0}.

Continuity from below therefore implies that |En | → |B|. As B = {f = 0} has positive measure, there must be some n such that |En | > 0. By definition, |f | ≥ n1 on E, so we can take A = En . 3.2.4 We verify the claim made in the proof of Lemma 3.2.4 that if a > 0 then {1/g > a} = {0 < g < 1/a}. Suppose that x ∈ 1/g > a. This means that 1/g(x) > a. Since a > 0, we cannot have g(x) = ∞, since 1/∞ = 0. Likewise g(x) 6= −∞. Therefore g(x) is a finite real number. Since a > 0, we must have g(x) > 0. Therefore the inequality 1/g(x) > a implies that 1/a > g(x), so x ∈ {0 < g < 1/a}. Now assume that x ∈ {0 < g < 1/a}. Then 0 < g(x) < 1/a. In particular, g(x) is a positive real number. Consequently, the normal rules of arithmetic apply and we have 1/g(x) > a, and x ∈ 1/g > a. 3.2.7 We verify the claim made in the proof of Lemma 3.2.7 that {f > a} =

∞ S

{fn > a}.

n=1

Suppose that f (x) > a, i.e., sup fn (x) > a. By the definition of a supremum, there must exist elements of {fn (x)}n∈N that lie as close to the supre-

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mum as we like. Hence, there must be at least one m ∈ N such that fm (x) > a. S Therefore x belongs to {fn > a}. S Now suppose that x ∈ {fn > a}. Then there is some m ∈ N such that fm (x) > a. But then f (x) = sup fn (x) ≥ fm (x) > a, so x ∈ {f > a}. 3.2.9 (a) For k ∈ N and j ∈ Z define half-open intervals Ij,k = Define a step function φk =

hj j + 1  , . k k

X

f ( kj ) χIj,k .

j∈Z

The function φk is measurable because {φk > a} is a union of countably many half-open intervals. Fix x ∈ R. For each k ∈ N, let jk be the unique integer such that x ∈ Ijk ,k and let xk be the left-hand endpoint of that interval, i.e., xk = We have φk (x) = f

jk k

jk . k


= f (xk ).

Since x, xk both belong to Ijk ,k , we have |x − xk | ≤

1 → 0 as k → ∞. k

Thus xk → x as k → ∞. If f is continuous at x, then this implies that lim φk (x) = lim f (xk ) = f (x).

k→∞

k→∞

(b) If f is continuous at a.e. x, then the measurable functions φk constructed in part (a) converge pointwise a.e. to f. Consequently f is measurable. (c) For each k ∈ N and j ∈ Zd , define Ij,k =

d h Y ji − 1 ji  . , k k i=1

Each “half-open cube” Ij,k is measurable, and each point x ∈ Rd belongs to a unique set Ij,k . The proofs of part (a) and (b) carry over with only minor changes.

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3.2.10 Write f and g in real and imaginary form as f = fr + ifi and g = gr + igi . Then fr , fi , gr , and gi are all measurable real-valued functions. (a) Since f + g = (fr + ifi ) + (gr + igi ) = (fr + gr ) + i (fi + gi ), we see that f + g is measurable. (b) Writing f g = (fr + ifi ) (gr + igi ) = (fr gr − fi gi ) + i (fr gi + fg gr ), we see that f g is measurable. (c) Since 1 1 gr − igi gr gi = = 2 = 2 −i 2 , g gr + igi gr + gi2 gr + gi2 gr + gi2 and since gr2 −gi2 is a real-valued function that is nonzero a.e., we see that 1/g is measurable. Combining this with part(b), it follows that f /g is measurable as well. (d) If h(x) = limn→∞ fn (x) exists for a.e. x, then the real and imaginary parts of fn must converge pointwise for almost every x. Hence the real and imaginary parts of h are measurable, so h is measurable. (e) Follows from parts (a) and (d). (h) Since L−1 is a linear bijection, Exercise 2.2.22 implies that L−1 maps measurable sets to measurable sets. Therefore the domain L−1 (E) is a measurable set. Suppose that U is any open subset of C. Then (f ◦ L)−1 (U ) = L−1 (f −1 (U ). Since f is measurable and U is open, the set f −1 (U ) is measurable (see Problem 3.1.18). Exercise 2.2.22 shows that linear bijections map measurable sets to measurable sets. Since L−1 is a linear bijection, it follows that (f ◦ L)−1 (U ) is a measurable set. Applying Problem 3.1.18 again, it follows that f ◦ L is measurable. 3.2.15 We give the details of the proof of Corollary 3.2.15 for complex-valued functions. Assume f : R → C is complex-valued, and write f = fr + ifi where fr , fi are real-valued. By Case 1, there exist simple functions φrn , φin such that φrn → fr and φin → fi , with |φrn | ≤ |fr | and |φin | ≤ |fi |. Set φn = φrn + iφin . Then φn is a complex-valued simple function, φn → f pointwise, and |φn |2 = |φrn |2 + |φin |2 → |fr |2 + |fi |2 = |f |2 .

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As before, the convergence is uniform on any set on which f is bounded. 3.2.16 Case 1: c ∈ R. Without loss of generality, consider c = 0. Define Z1 = {f = ∞} ∩ {g = −∞}, Z2 = {f = −∞} ∩ {g = ∞}. The sets Z1 and Z2 are measurable since f and g are measurable. Therefore Z = Z1 ∪ Z2 is a measurable set as well. Define F (x) = (f · χZ C )(x) =

(

f (x), x ∈ / Z, 0, x ∈ Z,

G(x) = (g · χZ C )(x) =

(

g(x), x ∈ / Z, 0, x ∈ Z.

and

The function h defined in the problem statement is h = F + G. Fix a ∈ R. If a > 0 then {F > a} = {f > a} \ Z. If a ≤ 0 then

{F > a} = {f > a} ∪ Z.

In any case, {F > a} is measurable, so F is a measurable function. Similarly, G is measurable. Hence a − G = a + (−1)G is measurable, and therefore {h > a} = {F + G > a} = F > a − G is measurable. Consequently h is a measurable function. Case 2: c = ∞. Define

( f (x), F (x) = ∞,

and G(x) =

(

g(x), ∞,

x∈ / Z, x ∈ Z, x∈ / Z, x ∈ Z,

so h = F + G. If a ∈ R then {F > a} = {f > a} ∪ Z,

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which is measurable. Hence F is measurable and likewise G is measurable. A lemma from the text therefore implies as before that {F > a − G} is measurable, and this is the same set as {h > a}. Case 3: c = −∞. This is similar to Case 2. 3.2.17 Step 1. Suppose first that f, g are nonnegative, extended real-valued, measurable functions on E. In this case we have 0 ≤ f (x) g(x) ≤ ∞ for every x ∈ E. If a < 0 then {f g > a} = E, which is measurable. If a = 0 then {f g > 0} = {f > 0} ∩ {g > 0}, which is measurable. Fix a > 0. Let Q+ = Q ∩ (0, ∞), and set   S  E = {f > r} ∩ g > ar . r∈Q+

The set E is measurable since Q+ is countable and both f and g are measurable. We will show that {f g > a} = E. If x ∈ E then there is some positive rational r such that f (x) > r and g(x) > a/r. Therefore f (x) g(x) > a, so x ∈ {f g > a}. We will prove the opposite inclusion by a contrapositive method, i.e., we suppose that x ∈ / E and show that x ∈ / {f g > a}. So, we are given that x ∈ / E. We have assumed that f and g are nonnegative. If either f (x) = 0 or g(x) = 0 then f (x) g(x) = 0, so x ∈ / {f g > a} in this case. Therefore, it suffices to concentrate on the case where f (x) and g(x) are both strictly positive. Let rk be rationals such that 0 < rk ≤ f (x) and rk → f (x) as k → ∞. Since x ∈ / E, the point x cannot belong to  {f > rk } ∩ g > rak for any k. Since f (x) > rk , this implies that we must have g(x) ≤ a/rk for every k. Hence f (x) g(x) ≤ af (x)/rk , so f (x) g(x) ≤ lim

k→∞

a f (x) = a. rk

Thus x ∈ / {f g > a}, which completes the proof that {f g > a} = E. Therefore, we have shown that {f g > a} is measurable for each a ∈ R, so the product function f g is measurable. Step 2. Now let f, g be arbitrary extended real-valued, measurable functions on E. Splitting into positive and negative parts, we can write f = f + − f − where f + , f − are nonnegative. By definition, if f (x) = ∞ then

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f + (x) = ∞ and f − (x) = 0, so we never encounter the indeterminate form ∞ − ∞ in the representation f = f + − f − . Similarly writing g = g + − g − , we have f (x) g(x) = f + (x) g + (x) − f + (x) g − (x) − f − (x) g + (x) − f − (x) g − (x).

(A)

By Step 1 we know that the four products f ± g ± are each measurable. Suppose f (x) = ∞, which implies that f + (x) = ∞ and f − (x) = 0. If g(x) ≥ 0 then g − (x) = 0, and therefore out of the four terms on the righthand side above, only one is nonzero. Similarly, if g(x) ≤ 0 then g + (x) = 0, and again only one of the four terms is nonzero. Hence if f (x) = ∞ then there are no indeterminate forms on the right-hand side of equation (A) above. Similarly considering the cases f (x) = −∞, g(x) = ∞, and g(x) = −∞, we see that even though f ± and g ± are extended real-valued, there are no x’s for which the right-hand side of equation (A) is indeterminate. Consequently, Lemma 3.2.1 implies that f + g + − f + g − − f − g + − f − g − is measurable, and therefore f g is measurable. 3.2.18 Assume first that each fk is extended real-valued. By Lemma 3.2.7, g(x) = lim sup fk (x)

and

h(x) = lim inf fk (x) k→∞

k→∞

are measurable functions. Therefore n o L = x ∈ X : lim fk (x) exists = {g ≤ h} k→∞

is a measurable set. Now suppose that each fk is complex-valued. Write fk = gk + ihk , where gk and hk are real-valued. Then, by the argument for extended real-valued functions, n o Lg = x ∈ X : lim gk (x) exists k→∞

and

Lh =

n o x ∈ X : lim hk (x) exists k→∞

are measurable. Since n o L = x ∈ X : lim fk (x) exists = Lg ∩ Lh , k→∞

it follows that L is measurable as well. To show that the set S is measurable, set sN (x) =

N X

n=1

Then sN is measurable, and

|fn (x)|.

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n o S = x ∈ X : lim sN (x) exists . N →∞

Therefore S is measurable by the arguments given above. 3.2.19 Since we are interested in an almost everywhere property and since the boundary of I contains at most two points, it suffices to assume that I is an open interval. Extend f to all of R by setting f (x) = 0 for x ∈ / I. Define fn (x) =

f (x + n1 ) − f (x) 1 n

,

x ∈ R.

Each function fn is measurable on R, and therefore its restriction to E is measurable because E is a measurable set. Since fn converges pointwise to f ′ (x) at each point of E, it follows that f ′ is measurable on E. 3.2.20 Extended real-valued functions. Suppose that f : Rd → [−∞, ∞] is measurable and ϕ : Rd → Rd is a bijection such that ϕ−1 is Lipschitz. Given a ∈ R, we have {f ◦ ϕ > a} = (f ◦ ϕ)−1 (a, ∞] = ϕ−1 (f −1 (a, ∞]). Since f is measurable, the set f −1 (a, ∞] = {f > a} is measurable. Exercise 2.2.22 shows that a Lipschitz function maps measurable sets to measurable sets. Since ϕ−1 is Lipschitz, it follows that {f ◦ L > a} is a measurable set. Complex-valued functions. Suppose that f : Rd → C is measurable and ϕ : Rd → Rd is a bijection such that ϕ−1 is Lipschitz. Write f = g + ih where g and h are real-valued. Then g ◦ ϕ and h ◦ ϕ are both measurable by the preceding case. Since (f ◦ ϕ)(x) = f (ϕ(x)) = g(ϕ(x)) + ih(ϕ(x)) = (g ◦ ϕ)(x) + i(h ◦ ϕ)(x), we have f ◦ ϕ = (g ◦ ϕ) + i(h ◦ ϕ), so f is measurable. 3.2.21 (a) Define An = {|f | > n}

and

A = {|f | = ∞}.

Each set An is measurable since |f | is measurable, and |A| = 0 by hypothesis so A is measurable as well. Since E has finite measure, so does A1 . By construction, A1 ⊇ A2 ⊇ · · · and A = ∩An . Continuity from above therefore implies that |An | → |A| = 0. Consequently, if we choose n large enough then we will have |An | < ε. Note that f is bounded on E \An , in fact, |f | ≤ n on this set. Since E \An is measurable, it contains a closed set F such that

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|F | > |E \An | −

ε ε ε ≥ |E| − − = |E| − ε. 2 2 2

Therefore |E \F | < ε, and f is bounded on the closed set F. (b) Fix ε > 0, and define f (x) = sup |fn (x)|

Ek = {f ≤ k} = {x ∈ E : f (x) ≤ k}.

and

n

Then f is measurable, and, by hypothesis, f (x) = sup |fn (x)| ≤ Mx < ∞, n

so f is finite at every point. It therefore follows from part (a) that there exists a closed set F ⊆ E such that |E \F | < ε and f is bounded on F, say f ≤ M on F. But then |fn (x)| ≤ f (x) ≤ M for every n and every x ∈ F.

3.2.22 (a) The empty set is measurable, and f −1 (∅) = ∅, which is measurable. Therefore ∅ ∈ Σ. Suppose that B ∈ Σ, i.e., B and f −1 (B) are measurable. Then B C is measurable. Since f −1 (B C ) = f −1 (B)C and f −1 (B) is measurable, its complement is measurable as well, and therefore f −1 (B C ) is measurable. Hence B C ∈ Σ. Now suppose that B1 , B2 , . . . all belong to Σ. Then they are measurable, S so Bk is measurable as well. Further, f −1 (Bk ) is measurable for every k. Therefore S S f −1 ( Bk ) = f −1 (Bk ) S

is measurable. Therefore Bk ∈ Σ. Thus Σ contains the empty set, is closed under complements, and is closed under countable unions. Therefore Σ is a σ-algebra. (b) Suppose that U ⊆ R is open. Then U is a Borel set, and since f is measurable, Problem 3.1.18 implies that f −1 (U ) is measurable. Hence U ∈ Σ. Now, by definition, B is the smallest σ-algebra that contains the open sets. Since Σ contains every open set, it follows that B ⊆ Σ.

(c) If B is any Borel set then B belongs to Σ, and therefore f −1 (B) is measurable.

3.3.4 (a) This is an immediate consequence of the definition of the L∞ -norm and the space L∞ (E). (b) Given f ∈ L∞ (E) and c ∈ C, we have kcf k∞ = esssup |cf (x)| x∈E

 = inf M ∈ [0, ∞] : |cf (x)| ≤ M for a.e. x ∈ E

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 = inf M ∈ [0, ∞] : |f (x)| ≤ M/|c| for a.e. x ∈ E  = |c| inf K ∈ [0, ∞] : |f (x)| ≤ K for a.e. x ∈ E = |c| esssup |f (x)| x∈E

= |c| kf k∞. (c) Fix f, g ∈ L∞ (E). Let Zf be the set of x ∈ E such that |f (x)| > kf k∞ , and let Zg be the set of x ∈ E such that |g(x)| > kf k∞ . Both Zf and Zg have measure zero, so Z = Zf ∪ Zg has measure zero. If x ∈ / Z, then |f (x) + g(x)| ≤ |f (x)| + |g(x)| ≤ kf k∞ + kgk∞ . Since |Z| = 0, it follows that |f + g| ≤ kf k∞ + kgk∞ a.e. Applying Lemma 2.2.28(b), it follows that kf + gk∞ = esssup |f (x) + g(x)| ≤ kf k∞ + kgk∞ . x∈E

(d) “⇒.” Suppose that kf k∞ = esssupx∈E |f (x)| = 0. By Lemma 2.2.28(a), this implies that |f (x)| ≤ 0 a.e., which is equivalent to saying that f = 0 a.e. “⇐.” If f = 0 a.e. then |f (x)| ≤ 0 a.e. Consequently, kf k∞ = esssup |f (x)| ≤ 0. x∈E

Since the essential supremum of |f | must be nonnegative, we conclude that kf k∞ = 0. 3.3.8 “⇒.” Suppose that fn → f in L∞ -norm. For each n, let Zn = {|f − fn | > kf − fn k∞ }. By definition of the L∞ -norm, each set Zn has measure zero. Therefore the set ∞ S Zn Z = n=1

also has measure zero. If x ∈ / Z, then x ∈ / Zn for any n, so |f (x) − fn (x)| ≤ kf − fn k∞ for every n. Letting kf − fn ku denote the uniform norm on the set E \Z, we therefore have kf − fn ku =

sup |f (x) − fn (x)| ≤ kf − fn k∞ → 0

x∈E \ Z

as n → ∞.

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Thus fn converges uniformly to f on E \Z. “⇐.” Suppose that Z is a set of measure zero such that fn → f uniformly on E \Z. Letting kf − fn ku denote the uniform norm on the set E \Z, we have |f (x) − fn (x)| ≤ kf − fn ku , all x ∈ E \Z. As Z has measure zero, we therefore have |f (x) − fn (x)| ≤ kf − fn ku ,

a.e. x ∈ E.

Hence kf − fn k∞ ≤ kf − fn ku . The opposite inequality follows by definition. Therefore kf − fn k∞ = kf − fn ku → 0 as n → ∞. Thus fn converges to f in L∞ -norm on E. 3.4.4 (a) Assume that fn → f in L∞ -norm. For each n, the set  Zn = |f − fn | > kf − fn k∞ S

has measure zero. Hence Z = Zn has measure zero, and

sup |f (x) − fn (x)| ≤ kf − fn k∞ → 0 as n → ∞.

x∈Z /

Therefore fn → f uniformly on E \Z. This implies that fn → f almost uniformly. (b) Suppose that fn → f almost uniformly. Then for each k ∈ Z, there exists a measurable set Ak ⊆ E such that |A| < 2−k and fn → f uniformly on E \Ak . Let Z = ∩Ak . Then |Z| = 0, and if x ∈ / E \Z then x ∈ E \Ak for some k, which implies that fn (x) → f (x) as n → ∞. Hence fn → f a.e.

3.4.5 (a) fn = χ[−n,n] converges pointwise to the constant function f = 1, but the convergence is not uniform on any unbounded subset of R. Another example is fn (x) = x/n, which converges pointwise to the zero function, but the convergence is not uniform on any unbounded subset of R.

(b) Suppose that |E| > 0. Even if each fn is finite a.e., we must require f to be finite a.e. For example, if fn (x) = n for x ∈ E, then fn is finite everywhere and fn converges pointwise to the function f = ∞, but the convergence is not uniform on any subset of E. 3.4.6 (a) Let {fn }n∈N be the sequence of Shrinking Triangles from Example 3.4.1. If we fix ε > 0, then fn converges uniformly to the zero function on the interval [ε, 1]. As the set A = [0, ε] has measure ε, it follows that fn converges almost uniformly to the zero function.

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However, we have k0 − fn k∞ = 1 for every n, so fn does not converge to the zero function in L∞ -norm. (b) Set fn (x) = x/n for x ∈ R. Then fn converges pointwise to the zero function on R, but the sequence does not converge almost uniformly. 3.4.7 Set E1 = E. Applying Egorov’s Theorem with ε = |E|/2, there is a measurable set A1 ⊆ E1 such that fn → f uniformly on A1 and |E1 \A1 | < |E|/2. Set E2 = E1 \A1 . Applying Egorov’s Theorem to the set E2 with ε = |E|/4, there is a measurable set A2 ⊆ E2 such that fn → f uniformly on A2 and |E2 \A2 | < |E|/4. Note that E2 \A2 = E \(A1 ∪ A2 ), and A2 is disjoint from A1 . We continue inductively in this way. Assuming the construction at stage k is complete, we set Ek+1 = Ek \Ak = E \ (A1 ∪ · · · ∪ Ak ). Applying Egorov’s Theorem to the set Ek+1 with ε = |E|/2k+1 , there is a measurable set Ak+1 ⊆ Ek+1 such that fn → f uniformly on Ak+1 and |E \ (A1 ∪ · · · ∪ Ak ∪ Ak+1 )| = |Ek+1 \Ak+1 |
1} → 0.

Therefore there exists some n1 such that

Likewise, we have

1 n ≥ n1 =⇒ {|f − fn | > 1} ≤ . 2 {|f − fn | > 1 } → 0. 2

so there exists some n2 such that

1 n ≥ n2 =⇒ {|f − fn | > 12 } ≤ . 4

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In fact, this holds for all sufficiently large n2 , so we can choose n2 so that n2 > n1 . Continuing in this way, we obtain the desired indices n1 < n2 < · · · . 3.5.7 Suppose that fn converges to f almost uniformly, and choose any ε > 0 and η > 0. By definition of almost uniform convergence, there exists a measurable set A ⊆ E such that |A| < η and fk → f uniformly on E \A. Therefore, there is some N > 0 such that sup |f (x) − fn (x)| < ε,

n > N.

x∈A /

Hence {|f (x) − fn (x)| ≥ ε} ⊆ A for all n > N. Therefore, for all n > N we have {|f − fn | ≥ ε} ≤ |A| ≤ η. This says that

lim {|f − fn | > ε} = 0.

n→∞

m

Since this is true for every ε > 0, we have shown that fn → f. m

3.5.11 Pointwise convergence is clear. If fn → f, then there is a subsequence that converges pointwise a.e., and therefore f = 0 a.e. However, since we have |{fn 6= 0}| = ∞ for every n, the functions fn do not converge in measure to the zero function. 3.5.12 For each x ∈ R, the sequence {fn (x)}n∈N limit:   1, n 1 − |x| = 0, f (x) = lim n→∞ 1 + |x|n   −1,

converges to the following |x| < 1, x = ±1, |x| > 1.

That is, fn converges pointwise to this function f. Since each fn is continuous and bounded but f is not continuous, the convergence cannot be uniform. See Figure 3.4 for a plot of f25 . 1.0

0.5

-2

1

-1

-0.5

-1.0

Fig. 3.4 Graph of f25 .

2

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Fix ε > 0. Note that −1 ≤ fn (x) ≤ 1 for every x. If f (x) ≥ 1 − ε, then 1 − |x|n ≥ (1 − ε) (1 + |x|n ) ≥ 1 − ε. Hence |x|n ≤ ε, so |x| ≤ ε1/n . If f (x) ≤ −1 + ε, then 1 − |x|n ≤ (−1 + ε) (1 + |x|n ), and therefore |x|n − 1 ≥ (1 − ε) (1 + |x|n ) ≥ 1 − ε.

Hence |x|n ≥ 2 − ε, so |x| ≥ (2 − ε)1/n . In summary,

{|f − fn | ≤ ε} ⊆ {|x| ≤ ε1/n } ∪ {|x| ≥ (2 − ε)1/n }. Therefore {|f − fn | > ε} ⊆ (ε1/n , (2 − ε)1/n ) ∪ (−(2 − ε)1/n , −ε1/n ), so
lim {|f − fn | > ε} ≤ lim 2 (2 − ε)1/n − ε1/n = 0.

n→∞

n→∞

m

Hence fn → f.

3.5.13 The argument for this problem is the same for either the complexvalued or the extended real-valued finite a.e. cases. m

m

(a) Suppose fk → f and fk → g, and fix ε > 0. If |f (x) − g(x)| > 2ε then 2ε < |f (x) − g(x)| ≤ |f (x) − fk (x)| + |fk (x) − g(x)|. Therefore either |f (x) − fk (x)| > ε or |fk (x) − g(x)| > ε, so    |f − g| > 2ε ⊆ |f − fk | > ε ∪ |g − fk | > ε .

This is true for every k, so by monotonicity and subadditivity,      |f − g| > 2ε ≤ lim |f − fk | > ε + |g − fk | > ε = 0. k→∞

Consequently,

 ∞   |f − g| > 0 = S |f − g| > n=1

1 n

 ∞ X  |f − g| > ≤ n=1

1 n

= 0.

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That is, the set where f differs from g has measure zero, so f = g a.e. (b) Fix ε > 0. Then we know that   lim |f − fk | > ε = 0 = lim |g − gk | > ε . k→∞

k→∞

Suppose that x is such that (f (x) + g(x)) − (fk (x) + gk (x)) > 2ε.

Then

2ε < (f (x) + g(x)) − (fk (x) + gk (x)) ≤ |f (x) − fk (x)| + |g(x) − gk (x)|.

Therefore we must have either |f (x) − fk (x)| > ε or |g(x) − gk (x)| > ε, which tells us that    |(f + g) − (fk + gk )| > 2ε ⊆ |f − fk | > ε ∪ |g − gk | > ε . Therefore, by monotonicity and subadditivity,  lim |(f + g) − (fk + gk )| > 2ε k→∞    ≤ lim |f − fk | > ε ∪ |g − gk | > ε k→∞

≤ lim

k→∞

= 0.

    |f − fk | > ε + |g − gk | > ε m

This is true for every ε > 0, so we conclude that fk + gk → f + g. m

m

(c) Assume that |E| < ∞, fk → f, and gk → g. Case 1 : f = g = 0 a.e. In this case we have that   |fk gk − f g| > ε = |fk gk | > ε .

Hence it suffices to assume that f and g are the zero function. Since {|fk gk | > ε} ⊆ {|fk | > ε1/2 } ∪ {|gk | > ε1/2 }, it follows that    |fk gk | > ε ≤ |fk | > ε1/2 + |gk | > ε1/2 → 0. m

Hence fk gk → f g in this case.

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Case 2 : gk = g for every n. Define En = {|g| > n}. Then E1 ⊇ E2 ⊇ · · · and E = ∩Ek has measure zero since g is finite a.e. Since |E| < ∞, we have by continuity from above that |En | → 0. Choose ε > 0, and fix any η > 0. Then we can find a n such that |En |
N then n η ε o < . |f − fk | > n 2

Note that if x ∈ EkC then |g(x)| ≤ n. Therefore, for k > N we have 
|f g − fk g| > ε ≤ {|f g − fk g| > ε} ∩ EnC + |En | n  η ε o ∩ EnC + ≤ |f − fk | > |g| 2 n o ε η ≤ |f − fk | > + n 2 η η < + = η. 2 2

This shows that

 lim |f g − fk g| > ε = 0,

k→∞ m

so fk g → f g.

Case 3 : Arbitrary fk , gk . m By Case 1, we have that (f − fk )(g − gk ) → 0, and by Case 2, we have that m m f gk → f g and fk g → f g. Therefore, fk gk − f g = (fk gk − f gk − fk g + f g) + (f gk − f g) + (fk g − f g) = (fk − f )(gk − g) + (f gk − f g) + (fk g − f g) m

→ 0. m

m

Alternative Proof. Fix any n1 < n2 < · · · . Then fnk → f and gnk → g, so there exist k1 < k2 < · · · such that fnkj → f a.e. Hence

and

gnkj → g a.e.

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fnkj gnkj → f g a.e. But E has finite measure, so Corollary 3.5.8 implies that m

fnkj gnkj → f g. Thus every subsequence of fn gn n∈N has itself a subsequence that converges m to f g in measure. Problem 3.5.14 therefore implies that fn gn → f g.

(d) Let g(x) = 1/x and set gk = g · χ[−k,k] . Then g − gk = 0 on the interval [−k, k], and for |x| > k we have |g(x) − gk (x)| ≤ 1/k. Therefore, given ε > 0, m for all k large enough we will have |g(x) − gk (x)| < ε for all x, so gk → g. m Now set fk (x) = f (x) = x. We certainly have that fk → f. However, f (x)g(x) = 1 for all x and fk gk = χ[−n,n] , so if 0 < ε < 1 then we have 

|f g − fk gk | > ε



= [−n, n]C ,

which has infinite measure. Therefore fk gk does not converge to f g in this case. m

(e) Since fn → f, there exists a subsequence {fnk }k∈N such that fnk → f pointwise a.e. Since |fnk | ≥ δ a.e. for each k, it follows that |f | ≥ δ a.e. Let Z be the set of all x such that either |fn (x)| > δ for some n or |f (x)| > δ. This set Z has measure zero by subadditivity. Suppose that x ∈ /Z is such that 1 1 f (x) − fn (x) ≥ ε. Then

1 1 |fn (x) − f (x)| = − ≥ ε, |f (x) fn (x)| f (x) fn (x)

so

|fn (x) − f (x)| ≥ ε |f (x) fn (x)| ≥ εδ 2 .

Since Z has measure zero, it follows that   1  1 2 → 0 f − fn ≥ ε ≤ |fn − f | ≥ εδ Therefore

as n → ∞.

1 m 1 fn → f .

3.5.14 (a) ⇒ (b). This direction is immediate. (b) ⇒ (a). Suppose that every subsequence of {fn }n∈N has a subsequence that converges in measure to f, but the full sequence {fn }n∈N does not converge in measure to f. Then there must exist an ε > 0 such that  |f − fn | > ε does not converge to 0.

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Therefore, there must exist some η > 0 such that for every N there exists some n > N such that  |f − fn | > ε ≥ η.

In particular, there is some n1 > 1 such that  |f − fn1 | > ε ≥ η.

Then there must exist some n2 > n1 such that  |f − fn2 | > ε ≥ η.

Continuing in this way, we find n1 < n2 < · · · such that  |f − fn | > ε ≥ η, all k ∈ N. k

But then no subsequence of {fnk }k∈N can converge in measure to f, which is a contradiction. m

3.5.15 (a) Assume fn → f and ϕ is uniformly continuous. Note that ϕ ◦ fn and ϕ ◦ f are measurable because ϕ is continuous. Fix ε > 0. Since ϕ is uniformly continuous, there exists a δ > 0 such that |x − y| ≤ δ =⇒ |ϕ(x) − ϕ(y)| ≤ ε. For each n ∈ N, set

An =

m



|f − fn | > δ .

Since fn → f, we know that |An | → 0. Now, if x ∈ / An , then |f (x) − fn (x)| ≤ δ, and therefore |ϕ(f (x)) − ϕ(fn (x))| ≤ ε. Hence

 |ϕ ◦ f − ϕ ◦ fn | > ε ⊆ An ,

and therefore

 lim |ϕ ◦ f − ϕ ◦ fn | > ε ≤ lim |An | = 0.

n→∞ m

n→∞

Thus ϕ ◦ fn → ϕ ◦ f.

To show that convergence in measure can fail if ϕ is not uniformly continuous, let fn (x) = x − n1 , f (x) = x, ϕ(x) = x2 . Then |f (x) − fn (x)| = However,

1 n

for every x, so fn → f uniformly and in measure.

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|(ϕ ◦ f )(x) − (ϕ ◦ fn )(x)| = |x − (x −

1 2 n )|

2x 1 = − 2 . n n

This will exceed ε on a set of infinite measure. Hence ϕ◦ fn does not converge in measure to ϕ ◦ f. m

(b) Assume fn → f and |E| < ∞. Let {gn }n∈N be any subsequence of m {fn }n∈N . Then gn → f, so there exists a subsequence {hn }n∈N of {gn }n∈N such that hn → f pointwise a.e. Let Z be the set of points where convergence does not occur. If x ∈ / Z then hn (x) → f (x), and therefore ϕ(hn (x)) → ϕ(f (x)) since ϕ is continuous. Hence ϕ ◦ hn → ϕ ◦ f pointwise a.e. Since E has m finite measure, this implies that ϕ ◦ hn → ϕ ◦ f. Applying Problem 3.5.14, we m conclude that ϕ ◦ fn → ϕ ◦ f. The same counterexample as before shows that the assumption |E| < ∞ is necessary here. 3.5.16 The proof is similar to that of Problem 1.1.20. Fix ε > 0 and η > 0, and let γ = min{ε, η}. m

Since fnk → f, there exists some K > 0 such that {|f − fn | > γ} < γ. nk > K =⇒ k

Since {fn }n∈N is Cauchy in measure, there exists an N such that {|fm − fn | > γ} < γ. m, n > N =⇒

Suppose that n > N. Then since the nk are strictly increasing, there exists some nk that is greater than both K and N. Since {|f − fn | > 2ε} ⊆ {|f − fn | > 2γ} ⊆ {|f − fnk | > γ} ∪ {|fnk − fn | > γ}, we have {|f − fn | > 2ε} ≤ {|f − fn | > γ} + {|fn − fn | > γ} < γ + γ ≤ 2η. k k This is true for all n > N, so

lim {|f − fn | > 2ε} = 0.

n→∞ m

That is, fn → f.

3.5.17 The argument for this problem is the same for either the complexvalued or the extended real-valued finite a.e. cases. First we state a little lemma.

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Lemma. If {fn }n∈N is a sequence in a normed space and kfn+1 − fn k < 2−n for every n, then {fn }n∈N is Cauchy. Proof. Choose any ε > 0, and let N be large enough that 2−N +1 < ε. If n > m > N, then kfn − fm k ≤

n−1 X

k=m

kfk+1 − fk k ≤

Hence {fn }n∈N is Cauchy.

∞ X 1 1 1 = m−1 < N −1 < ε. 2k 2 2

k=m

⊓ ⊔

Now we give the solution to the problem. (a) ⇒ (b), (c) ⇒ (d). These implications are immediate. (b) ⇒ (a). Assume that statement (b) holds, and fix ε, η > 0. Let δ = min{ε, η}. Then there exists an N > 0 such that n > N =⇒ {|f − fn | > δ} < δ.

Consequently, for n > N we have {|f − fn | > ε} ≤ {|f − fn | > δ} < δ ≤ η. Therefore statement (a) holds.

m

(a) ⇒ (d). Suppose that fn → f, and fix ε > 0. Then there exists an N > 0 such that  ∀ n ≥ N, |f − fn | > ε2 < ε2 . By the Triangle Inequality,

|fm (x) − fn (x)| ≤ |fm (x) − f (x)| + |f (x) − fn (x)|. It follows from this that   S |fm − fn | > ε ⊆ |f − fm | > 2ε |f − fm | > 2ε . Consequently, if m, n ≥ N, then    |fm − fn | > ε ≤ |f − fm | > ε + |f − fn | > ε < 2 2

ε 2

+

ε 2

= ε.

(d) ⇒ (a). This is Theorem 3.5.10.

(d) ⇒ (c). Assume that statement (d) holds, and fix ε, η > 0. Let δ = min{ε, η}. Then there exists an N > 0 such that m, n > N =⇒ {|fm − fn | > δ} < δ.

Consequently, for m, n > N we have

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{|fm − fn | > ε} ≤ {|fm − fn | > δ} < δ ≤ η.

Therefore statement (c) holds.

3.6.2 (a) ⇒ (b). Assume that f is measurable, and fix ε > 0. For each k ∈ N, set Ek = E ∩ {x ∈ Rd : k − 1 ≤ |x| < k}. By Theorem 3.6.1, there exists a compact set Fk ⊆ Ek such that |Ek \Fk |
N, we have xn ∈ Fk ∪ Fk+1 . Case 1. Suppose Fk+1 contains only finitely many xn with n > N. In this case there is some M such that xn ∈ Fk for all n > M. Since Fk is closed and xn → x, it follows that x ∈ Fk ⊆ F. Furthermore, we have f (xn ) → f (x) because f |Fk is continuous. Case 2. If Fk contains only finitely many xn , then a symmetric argument shows that x ∈ F and f (xn ) → f (x). Case 3. Suppose Fk and Fk+1 each contain infinitely many xn with n > N. Let {yn }n∈N be the subsequence of {xn }n>N consisting of points in Fk , and let {zn }n∈N be the subsequence of points in Fk+1 . Since Fk is closed and yn → x, we have x ∈ Fk ⊆ F and f (yn ) → f (x). Similarly, we obtain x ∈ Fk+1 ⊆ F and f (zn ) → f (x). Since Fk and Fk+1 are disjoint, the subsequences {yn }n∈N and {zn }n∈N “partition” the original sequence {xn }n>N . It follows that f (xn ) → f (x). The combination of these three cases proves that F is closed and f is continuous on F. (b) ⇒ (c). Assume that statement (b) holds. Given ε > 0, let F be the set whose existence is given by statement (b). Since F is a closed subset of Rd and f |F is continuous, the Tietze Extension Theorem implies that there exists a continuous function g : Rd → C that equals f on the set F. (c) ⇒ (a). Assume that statement (c) holds.

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Set E1 = E. Applying statement (c) with ε1 = min{|E|/2, 1}, there is a closed set F1 ⊆ E1 and a continuous function g1 such that g1 = f on F1 and |E \F1 | < |E|/2. Set E2 = E1 \F1 . Applying statement (c) to the set E2 with ε2 = ε1 /2, there is a closed set F2 ⊆ E2 and a continuous function g2 such that g2 = f on F2 and |E2 \F2 | < ε1 /2. Note that E2 \F2 = E \(F1 ∪ F2 ), and F2 is disjoint from F1 . We continue inductively in this way. Assuming the construction at stage n is complete, we set En+1 = En \Fn = E \ (F1 ∪ · · · ∪ Fn ). Applying statement (c) to En+1 with ε = |E|/2n , there exists a closed set Fn+1 ⊆ En+1 and a continuous function gn+1 such that gn+1 = f on Fn+1 and ε1 |E \ (F1 ∪ · · · ∪ Fn ∪ Fn+1 )| = |En+1 \Fn+1 | < n . 2 Let F = ∪Fn . By construction, Z = E \F has measure zero. Moreover, we can write E as a countable union of disjoint sets: E = Z ∪ F1 ∪ F2 ∪ · · · . The function g =

∞ X

n=1

gn · χFn

is measurable (though not necessarily continuous), and g = f on the set F. Consequently g = f a.e., so f is measurable.

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Solutions to Exercises and Problems from Chapter 4 PN PM 4.1.3 Let φ = j=1 aj χEj and ψ = k=1 ak χEk be the standard representations of φ and ψ. (a) If Ej ∩ Fk 6= ∅, then there is some point x ∈ Ej ∩ Fk , and therefore aj = φ(x) ≤ ψ(x) ≤ bk . On the other hand, if Ej ∩ Fk = ∅, then |Ej ∩ Fk | = 0. Applying these facts and equations (4.2) and (4.3), we conclude that Z

φ =

E

M X N X j=1 k=1

aj |Ej ∩ Fk | ≤

M X N X

j=1 k=1

bk |Ej ∩ Fk | =

Z

ψ. E

R (b) If φ = 0 a.e., then either ak = 0 or |Ek | = 0 for each k, and hence E φ = 0. R Conversely, E φ = 0 then ak |Ek | = 0 for each k, and therefore either ak = 0 or |Ek | = 0 for each k. Consequently, φ = 0 a.e. (c) φ χA is simple because φ χA =

M X

aj χEj χA =

M X

aj χEj ∩A .

j=1

j=1

Applying Lemma 4.1.2, it follows that Z

φ χA =

E

M X j=1

aj |Ej ∩ A| =

Z

φ.

A

(d) We have φ χA =

M X

ak χEk ∩A

and

k=1

Therefore Z

φ =

A

=

Z

M X

ak χEk ∩An .

k=1

φ χA

E N X

k=1

=

φ χA n =

N X

k=1

(by part (c))

ak |Ek ∩ A| ak

∞ X

n=1

|Ek ∩ An |

(by Lemma 4.1.2)

(by countable additivity)

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= = =

∞ X N X

n=1

R

ak |Ek ∩ An |

(nonnegativity)

φ χA n

(by Lemma 4.1.2)

n=1 k=1 ∞ Z X

n=1 E ∞ Z X

99

φ

(by part (c))

An

φ = ∞ for some k then there is nothing to prove, so we may R assume that Ak φ < ∞ for every k. If we set A0 = ∅, then (e) If

Ak

∞ S

∞ S

Ak =

j=1

k=1

(Aj \Aj−1 ),

and the sets on the right-hand side above are disjoint. Further, Aj+1 = Aj ∪ (Aj+1 \Aj ), and the sets on the right are disjoint. Lemma 4.1.2 therefore implies that Z Z Z φ, φ + φ = Aj+1 \ Aj

Aj

Aj+1

where all of these integrals are finite. Applying Lemma 4.1.2 again, we see that Z ∞ Z X φ φ = A

j=1

= = =

lim

N →∞

lim

N →∞

lim

N →∞

Aj \ Aj−1 N Z X j=1

Z

AN

Z

Aj

φ −

φ − Z

Z

Aj−1

 φ

φ

A0

φ.

AN

4.1.8 (a) We fill in the details on the converse inequality in the proof of part (a) of Lemma 4.1.8. and the proofs of parts (b) and (c). For the converse inequality, note that Z  Z f χA = sup φ : 0 ≤ φ ≤ f χA , φ simple on E . E

E

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Let φ be any particular simple function on E such that φ ≤ f χA . Since φ is identically zero off of A, we have φ = φ χA . Therefore Exercise 4.1.3 implies that Z Z Z χ φ = φ A = φ. E

E

A

On the other hand, if we restrict our attention to the domain A then φ is a simple function on A and φ ≤ f on A, so Z Z φ ≤ f. A

A

Combining the preceding equations, we see that Z Z f χA ≤ f. E

A

(b) If φ is any simple on E such that φ ≤ f then we also have R R function φ ≤ g, and therefore E φ ≤ E g by the R definition of the integral of g. Since the integral of f is the supremum of E φ over all such simple functions φ, it R R follows that E f ≤ E g. (c) This follows directly from the fact that if c ≥ 0, then sup cx : x ∈ S = c sup x : x ∈ S.

4.1.10 “⇐.” Suppose that f = 0 a.e.R If φ is a simple function such that 0 ≤ φ ≤ f, then φ = 0R a.e., and hence E φ = 0. Taking the supremum over all such φ, we obtain E f = 0. R “⇒.” Suppose that E f = 0. Then for each n ∈ N, Tchebyshev’s Inequality implies Z n 1 o f = 0. 0 ≤ x ∈ E : f (x) > ≤ n n E Hence {f >

1 n}

has measure zero for each n > 0, and therefore {f > 0} =

has measure zero. Thus f = 0 a.e.

∞  S

f>

n=1

1 n



4.1.11 Let Ek be any collection of closed disjoint intervals such that |Ek | = 2−2k ,

k ∈ N.

S

Let E = Ek , and define f =

∞ X

k=1

For each n ∈ N, set

2 k χE k .

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101

ϕn =

n X

2 k χE k .

k=1

Then ϕn is a nonnegative simple function, and Z

ϕn = E

n X

k=1

2k |Ek | =

n X

n X

2k 2−2k =

k=1

2−k .

k=1

Further, ϕn ր f, so the Monotone Convergence Theorem implies that Z Z ϕn = 1. f = lim n→∞

E

E

Thus f is a nonnegative function whose integral is finite, and f is finite at every point. 4.1.12 Since the integral of f is finite, Problem 4.1.11 implies that f (x) is finite for a.e. x. Since 0 ≤ g(x) ≤ ∞ for every x while f is finite almost everywhere, g(x) − f (x) can take an indeterminate form on at most a set of measure zero. Lemma 3.2.2 therefore implies that g − f is measurable. Since g − f ≥ 0, the integral of g − f exists as a nonnegative, extended real number. The additivity of the integrals of nonnegative functions given in Theorem 4.2.3 implies that, in the extended real sense, Z Z Z Z g. (A) (g − f ) + f = f = (g − f ) + E

Case 1:

R

E

E

E

E

g < ∞. In this case both f and g have finite integrals. Also, 0 ≤ g − f ≤ g,

so the integral of g − f exists and Z Z g < ∞. (g − f ) ≤ 0 ≤ E

E

Consequently, all of the integrals in equation (A) are finite, so we can rearrange to obtain Z Z Z f. g − (g − f ) = E

E

E

R If Case 2: E g = ∞. Now we suppose that the integral of g is infinite. R the integral of g − f was finite, then equation (A) would imply that E g is infinite. This is a contradiction, so we must have Z (g − f ) = ∞. E

Since f has finite integral, it follows that

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102

Z

E

Z

(g − f ) = ∞ =

Z

g −

E

f. E

4.2.5 (a) Observe that f χA = f χ∪An =

∞ X

f χA n .

n=1

Therefore, the result follows by applying Corollary 4.2.4 to the functions f n = f χA n . R (b) If Ak = ∞ for some k then there is nothing to prove, so we may R assume that Ak f < ∞ for every k. If we set A0 = ∅, then ∞ S

∞ S

Ak =

j=1

k=1

(Aj \Aj−1 ),

and the sets on the right-hand side above are disjoint. Further, Aj+1 = Aj ∪ (Aj+1 \Aj ), and the sets on the right are disjoint. Part (a) therefore implies that Z Z Z f, f + f = Aj+1 \ Aj

Aj

Aj+1

where all of these integrals are finite. Applying part (a) again, we see that Z

f =

A

∞ Z X

= = =

lim

N →∞

lim

N →∞

lim

f

Aj \ Aj−1

j=1

N →∞

N Z X j=1

Z

AN

Z

Aj

f −

f − Z

Z

Aj−1

f



f

A0

f.

AN

4.2.8 Assume that Fatou’s Lemma holds, and suppose 0 ≤ fn ր f. Since R R fn ≤ f we have fn ≤ f for each n. Combining this with Fatou’s Lemma, we conclude that Z Z Z Z Z f. fn ≤ fn ≤ lim sup lim inf fn ≤ lim inf f = E

E n→∞

n→∞

E

n→∞

E

E

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Therefore equality holds in the line above. 4.2.9 By monotonicity of the integral, we have Z Z 0 ≤ fn ≤ f ≤ ∞, E

E

n ∈ N.

These integrals could be infinite, but we at least have the given inequalities. Combining this with Fatou’s Lemma, we see that Z Z lim inf fn f = E n→∞

E

≤ lim inf n→∞

Z

≤ lim sup n→∞

≤

Z

fn

(Fatou)

E

Z

fn

E

f.

E

Consequently equality holds throughout the lines above. 4.2.10 For each n, k ≥ 0 set En = {f ≥ n}

Fk = {k ≤ f < k + 1}.

and

These are measurable sets. Further, the Fn are disjoint, and ∞ S

En =

Fk .

k=n

In particular, since f is nonnegative we have E = E0 =

∞ S

Fk ,

k=0

and therefore

Z

f =

E

Also, by disjointness, |En | = Therefore

∞ Z X

k=0

∞ X

k=n

f.

(A)

Fk

|Fk |.

(B)

c Solutions 2019 Christopher Heil

104 ∞ X

n=1

|En | =

∞ X ∞ X

|Fk |

by equation (B)

∞ X k X

|Fk |

all terms nonnegative

n=1 k=n

=

k=1 n=1

=

∞ X

k |Fk |

k=1

Fk

k=1

∞ Z X

=

∞ Z X

≤ ≤

since k ≤ f on Fk

f

Fk

k=1

Z

k

f

by equation (A)

E

< ∞

since f is integrable.

4.2.11 Let En = E ∩ [−n, n]d , and set fn = f χEn . Then 0 ≤ fn ≤ f and fn → f pointwise, so Problem 4.2.9 implies that Z Z Z f = lim fn = f. lim n→∞

En

n→∞

E

E

The result therefore follows by taking A = En with n large enough. R 4.2.12 Since E f is finite, f must be finite a.e. Let g(x) = lim inf f (x)n . n→∞

Since products of measurable functions are measurable, we know that f (x)n is measurable for each n, and hence g is measurable as well. By Fatou’s Lemma, Z Z Z 0 ≤ g(x) dx ≤ lim inf f (x)n dx = f (x) dx < ∞. n→∞

E

E

E

This implies that 0 ≤ g(x) < ∞ a.e. Therefore {f > 1} must have measure zero, since g(x) = ∞ for all x such that f (x) > 1. Thus 0 ≤ f (x) ≤ 1 a.e. Consequently, f (x)2 ≤ f (x) a.e. Therefore f (x) − f (x)2 ≥ 0 a.e., but we have Z

0

1


f (x) − f (x)2 dx = 0.

It therefore follows from Exercise 4.1.10 that f (x) = f (x)2 a.e. Hence, except for a set of measure zero, f (x) must be either 0 or 1. Setting A = {x ∈ E :

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105

f (x) = 1}, it follows that f = χA a.e. Since f is measurable, A must be a measurable set. 4.2.13 Note R first that the function f is measurable and is nonnegative a.e. Its integral E f therefore exists and is a nonnegative extended real number (if we like, by redefining f on a set of measure zero, we can assume that f is nonnegative at all points, not just a.e.). R Proof in the case that E f < ∞. R R We are given that E fn → E , so the integral of fn must be finite for all R n large enough. Without loss of generality, we may therefore assume that f is finite for every n. By Fatou’s Lemma, E n Z Z Z fn . (lim inf fn ) ≤ lim inf f = A

A

n→∞

n→∞

A

A similar inequality holds for integrals on E, and for integrals on E \A. Since all of the integrals involved are finite, we therefore compute that Z Z Z fn ≤ lim sup fn f ≤ lim inf A

n→∞

n→∞

A

A

Z Z fn − ≤ lim sup n→∞

≤ lim sup n→∞

= ≤

Z

E

Z

 Z + lim sup −

fn

n→∞

E

E

f − lim inf

E

f −

Z

n→∞

Z

fn E \A



Z

fn

E \A



fn

E \A

f =

E \A

Z

f.

A

Consequently equality holds on each line above, which implies that Z Z Z fn = lim sup fn . f = lim inf n→∞

A

n→∞

A

A

R

R

fn exists and equals A f. R Counterexample when E f = ∞. Consider fn = χ(−∞,0] + n1 χ[0,n] .

Therefore the limit of

A

We have fn → f = χ(−∞,0] pointwise on R, yet on the subset A = [0, ∞) we have Z ∞ Z 1 fn = 1, f = 0. 0

0

c Solutions 2019 Christopher Heil

106

Note that the integral of f on R is ∞. 4.2.14 For simplicity of presentation, and without loss of generality, we will take the domain of f to be [0, 1]. Since f is continuous on the closed interval [0, 1], we know that it is both Riemann integrable and Lebesgue integral on [0, 1]. Let I be the Riemann R1 integral of f on [0, 1], and let 0 f be the Lebesgue integral of f. For each integer n > 0, set ∆ = n1 . Also set xk =

k , n

k = 0, 1, . . . , n,

noting that xk implicitly depends on n. Since f is continuous, for each k = 1, . . . , n there is a point x∗k ∈ [xk−1 , xk ] where f achieves its minimum on that interval (again note that x∗k implicitly depends on n). Then Ln =

n X

f (x∗k ) ∆n

k=1

is a lower Riemann sum for f. Since f is continuous, these lower Riemann sums converge to I as n → ∞. On the other hand, Ln is the Lebesgue integral of the step function φn =

n X

f (x∗k ) χ[xk−1 ,xk ) .

k=1

If we restrict to dyadic integers of the form n = 2j , then we obtain a monotonically increasing sequence: φ2j ≤ φ2j+1 . Furthermore, if we set φn (1) = f (1), then φ2j (x) converges pointwise to f (x) for each x ∈ [0, 1]. The Monotone Convergence Theorem therefore implies that the Lebesgue integral of φ2j increases to the Lebesgue integral of f. Therefore Z 1 Z 1 f = lim φ2j = lim L2j = I. 0

j→∞

0

j→∞

Hence the Lebesgue integral of f equals the Riemann integral of f.

4.2.15 This is an immediate consequence of the Dominated Convergence Theorem, but we will give a direct proof based on Rthe MCT. Without loss of generality, we can assume that E f1 < ∞. Since fn ց f, we have 0 ≤ f1 − fn ր f1 − f. Applying the Monotone Convergence Theorem and part (a), it follows that

c Solutions 2019 Christopher Heil

Z

E

f1 −

Z

Z

fn =

E

E

107

(f1 − fn ) →

Z

E

(f1 − f ) =

Z

E

f1 −

Z

f.

E

Since all quantities are finite, we can rearrange to obtain Z Z fn → f. E

Counterexample if

R

E

fk = ∞ for every k. Let E = [0, ∞) and define fn (x) =

x n,

x ≥ 0.

We have fn ց f = 0, but for every n we have Z ∞ fn = ∞. 0

Therefore, the integral of fn does not converge to the integral of f = 0. Another counterexample fn = n1 on any domain E ⊆ Rd with |E| = ∞. 4.2.16 “⇒.” Assume that f is measurable. Since f is nonnegative, the definition of the Lebesgue integral tells us that Z  Z sup φ : 0 ≤ φ ≤ f, φ simple = f. E

E

R

R If ψ is any simple function such that f ≤ ψ, then E f ≤ E ψ, so Z  Z f ≤ inf ψ : f ≤ ψ, ψ simple . (B). E

E

We must show that equality holds on the preceding line. For each n ∈ N, let ψn be the function obtained by rounding up f to the nearest integer multiple of 2−n . Explicitly, if we fix x ∈ E, then since f is bounded there is a unique integer j ≥ 0 such that j−1 j < f (x) ≤ n . 2n 2 We then set

j . 2n Since f is bounded, ψn is defined at every x ∈ E and takes only finitely many distinct values. The fact that f is measurable implies that the set on which ψn takes a particular value is measurable. Therefore ψn is measurable, so it is a simple function. Letting M be an upper bound for f, we Rhave f ≤ ψn ≤ M +1 for every n. Since E has finite measure, it follows that E ψn < ∞. By construction, ψn ց f. ψn (x) =

c Solutions 2019 Christopher Heil

108

Since ψ1 has finite integral, Problem 4.2.15 therefore implies that Z Z f = lim ψn . n→∞

E

E

Consequently, equality holds in equation (B). “⇐.” Assume that the equality of sup and inf holds, and let Z  Z  I = sup φ : 0 ≤ φ ≤ f, φ simple = inf ψ : f ≤ ψ, ψ simple . E

E

Since f is bounded, there is a constant M such that f ≤ M. Therefore, if φ is a simple function and φ ≤ f, then φ ≤ M. Consequently Z Z M = M |E|, φ ≤ E

E

and therefore I ≤ M |E| < ∞. Now, by the definition of sup and inf, there exist simple functions 0 ≤ φn ≤ f ≤ ψn such that lim

n→∞

Z

φn = I = lim

n→∞

E

Z

ψn .

(A)

E

From here we give two ways to finish the proof. Method 1. Set φ = sup φn

and

ψ = inf ψn . n

n

Then φ and ψ are each measurable and nonnegative, and for every n ∈ N we have 0 ≤ φn ≤ φ ≤ f ≤ ψ ≤ ψn . Although we do not know whether f is measurable, both φ and ψ are measurable, so Z Z Z Z ψn ψ ≤ φ ≤ φn ≤ E

E

E

E

for every n ∈ N. Consequently, Z Z Z Z φn = I. ψ ≤ lim φ ≤ φn ≤ I = lim n→∞

E

n→∞

E

E

E

As ψ − φ ≥ 0, we can apply Problem 4.1.12 and compute that Z Z Z φ = 0. ψ − (ψ − φ) = E

E

E

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109

As ψ − φ is nonnegative, this implies that ψ − φ = 0 a.e. But φ ≤ f ≤ ψ, so we conclude that φ = f = ψ a.e. Therefore f is measurable. Method 2. Since φn ≤ ψn , equation (A) implies that Z (ψn − φn ) kψn − φn k1 = E

=

Z

E

ψn −

Z

E

φn → I − I = 0

as n → ∞.

That is, ψn − φn → 0 in L1 -norm. Consequently, there exists a subsequence such that ψnk − φnk → 0 pointwise a.e. But φnk ≤ f ≤ ψnk , so for a.e. x we have |f (x) − φnk (x)| ≤ |ψnk (x) − φnk (x)| → 0 as n → ∞. Therefore φnk → f a.e., and hence f is measurable. 4.2.17 (a) Assume first that E has finite Lebesgue measure. Fix ε > 0, and for n ≥ 0 define En = {εn ≤ f < ε(n + 1)}. These sets are measurable and disjoint subsets of E, and Γf ⊆

∞ S

n=0


En × [εn, ε(n + 1) .

Hence the exterior Lebesgue measure of Γf satisfies |Γf |e ≤ =

∞ X En × [εn, ε(n + 1)

n=0 ∞ X

n=0

ε |En |

S ∞ = ε En n=0

≤ ε |E|.

Since this is true for every ε > 0, we conclude that |Γf |e = 0. If |E| = ∞, then we can write E = ∪Ek where each set Ek has finite measure. Applying the preceding case and subadditivity, we again obtain |Γf |e = 0.

(b) Step 1. If f = χF is a characteristic function, then the region under the graph is simply Rf = F × [0, 1], which is measurable. Further, we have Z Z Z χF (x) dx = f (x) dx. dx = |Rf | = |F | = F

E

E

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110

Hence the result holds when f is a characteristic function. PN Step 2. Now suppose that φ = k=1 ck χEk is the standard representation of a nonnegative simple function. Then the sets Ek are disjoint, so Rφ =

N S

k=1

which is measurable, and

Ek × [0, ck ],

Z Z N N X X Ek × [0, ck ] = χE k = ck φ.

|Rφ | =

k=1

k=1

E

E

Hence the result holds for simple functions.

Step 3. Now let f be an arbitrary measurable, nonnegative function on E. Then we can find simple functions φn such that φn ր f. The union of the regions under the graphs of φn is “almost” the region under the graph of f, but it may not include all the points on the graph of f. In fact,   S Rf = Rφn ∪ Γf . n

Since Γf and Rφn are measurable sets, the region under the graph of f is measurable. We compute that S |Rf | = Rφn (since |Γf | = 0) n

= lim |Rφn | n→∞ Z = lim φn

(continuity from below)

=

(Monotone Convergence Theorem),

n→∞

Z

(by Step 2)

E

f

E

so the result holds for f. 4.2.18 (a) We give a simple direct proof of Fatou’s Lemma for series. Note that all of the terms akn are nonnegative. Therefore all of the quantities that appear in the problem statement do exist in the extended real sense. A basic fact for liminf’s is that lim inf xn + lim inf yn ≤ lim inf (xn + yn ). n→∞

n→∞

n→∞

Therefore, for each fixed positive integer N,

c Solutions 2019 Christopher Heil N X

k=1

111 N X

lim inf akn ≤ lim inf n→∞

n→∞

akn ≤ lim inf n→∞

k=1

∞ X

akn

k=1

The final quantity on the right-hand side of the line above is independent of N. Therefore we can take the limit as N → ∞ to obtain ∞ X

k=1

lim inf akn = n→∞

lim

X N

N →∞

lim inf akn

k=1

n→∞



∞ X

≤ lim inf n→∞

akn .

k=1

(b) The statement and proof of the MCT for series is as follows. Monotone Convergence Theorem Suppose that for each k ∈ N we have 0 ≤ akn ր bk as n → ∞. Then lim

n→∞

∞ X

akn =

k=1

∞ X

bk .

k=1

Proof. Using part (a) and the fact that for a fixed k we have akn ≤ bk for every n, it follows that ∞ X

bk =

k=1

∞ X

k=1

lim akn

n→∞

≤ lim inf n→∞

≤ lim sup n→∞

=

∞ X

∞ X

akn

(by part (a))

k=1 ∞ X

bk

k=1

bk .

k=1

4.3.6 (a) By hypothesis,Rf and g Rare extended real-valued functions such that f ≤ g a.e., and both E f and E g exist. Suppose first that f and g are Rnonnegative,R so 0 ≤ f ≤ g a.e. Then Z = {g > Rf } has measure zero, so Z f = 0 = Z g by Exercise 4.1.10, and R E \ Z f ≤ E \ Z g by Lemma 4.1.8 Consequently, Exercise 4.2.5 implies that Z

E

f =

Z

E \Z

f+

Z

Z

f ≤

Z

E \Z

g+

Z

Z

g =

Z

g.

E

Now assume that f andR g are arbitrary extended real-valued functions such R + + that f ≤ g a.e. and both E f and E g exist. We must have 0 ≤ R f +≤ gR a.e. − − and 0R ≤ g ≤R f a.e., so it follows from the above work that E f ≤ E g + and E g − ≤ E f − . If all of these quantities are finite, then we can subtract

c Solutions 2019 Christopher Heil

112

to obtain Z

f =

E

Z

E

f+ −

Z

E

f− ≤

Z

E

g+ −

Z

g− =

E

Z

g.

E

R R Suppose that E f + = ∞. Then we must have E f − < ∞ since R we+know that R g = ∞ and f exists. Also, it follows from the assumption f ≤ g that E RE − g < ∞. Consequently, the subtractions Rperformed E R above are still valid in the extended real sense, so we still obtain E f ≤ E g. A similar argument works if any of the other integrals is infinite. (b) Assume f = g a.e. If f and g are extended real-valued, then we have both f ≤ g a.e. and g ≤ f a.e. Hence this case follows from part (a). If f and g are complex-valued, then after splitting into real and imaginary parts, we see that fr = gr a.e. and fi = gi a.e. Hence Z Z Z Z Z Z g. gi = gr + i fi = fr + i f = E

E

E

E

E

E

R (c) Suppose E f exists and A is a measurable subset of E. R Consider R first the case where f is extended real-valued. R At least one of + or E f − must be finite, let us say that we have E f + < ∞. Since Ef f + χA ≤ f + , it follows from part (a) that Z Z Z f + < ∞. f + χA ≤ f+ = E

E

A

R

R

Hence A f + exists and is finite. Since A f − exists as a nonnegative extended real number, it follows that Z Z Z + f− f − f = A

A

A

exists. If f is complex-valued, then we can reduce to the real case by breaking into real and imaginary parts. (d) If f is extended real-valued a.e., then f + = 0 a.e. and R 0 − R + and f = f = 0 a.e.R Consequently E f = 0 = E f by Exercise 4.1.10, so it follows that E f = 0. If f is complex-valued and f = 0 a.e., we argue similarly by breaking into real and imaginary parts. −

(e) Suppose that f : E → [−∞, ∞] is integrable and c ∈ R. If c ≥ 0 then (cf )+ = cf + and (cf )− = cf − . Applying part (c) of Lemma 4.1.8, it follows that

c Solutions 2019 Christopher Heil

Z

113

Z

cf =

E

E

(cf )+ −

E

cf + −

Z

=

= c

Z

Z

Z

cf −

E

+

E

(cf )−

E

f −c

Z

−

f

= c

E

Z

f.

E

Next consider c = −1. We have (−f )+ = −f − and (−f )− = −f + , so  Z Z Z − + f = − f − f − E

= − Z

=

E

Z

f+ +

E

f

−

E

Z

=

E

Z

−

Z

E

f− E

f+

E

(−f )+ −

Z

(−f )− =

E

Z

(−f ).

E

Finally, if c ≤ 0, then −c ≥ 0, so by combining the above steps we find that Z Z Z Z Z f. f = c (−c)(−f ) = (−c) (−f ) = −(−c) cf = E

E

E

E

E

This completes the proof for extended real-valued functions. Now we turn to complex-valued functions. Suppose that f : E → C is integrable and c ∈ C. Given c = a + ib ∈ C, we have from the previous case and the definition of the integral of a complex-valued function that Z Z (a + ib) (fr + ifi ) (cf ) = E

E

= =

Z

E

(afr − b fi ) + i (b fr + afi )

E

(afr − b fi ) + i

Z

= a

Z

E

fr − b

= (a + ib) = c

Z

E

Z

Z

E

f.

Z

(b fr + afi )

E

fi + ib

E

fr + i

Z

Z

E

E

fi

fr + ia 

Z

E

fi

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114

R R (f) Part (c) implies that An f exists for each n, as does A f. Assume f is extended real-valued. Exercise 4.2.5 then implies that Z Z XZ XZ + + − f = f and f = f −. E

At least one of is justified:

An

n

R

E

f + or Z

R

E

n

An

f − must be finite. Hence the following calculation

E

f =

E

Z

E

=

XZ n

=

An

f−

E

f

+

An

XZ n

Z

f+ −

X Z n

=

f+ −

XZ n

−

f−

An

Z

f

An

−



fn .

An

If f is complex-valued, we argue similarly by breaking into real and imaginary parts. R R (g) Part (c) implies that An f exists for each n, as does A f. R If Ak f = ∞ for some k then there is nothing to prove, so we may assume R that Ak f < ∞ for every k. Since Aj+1 = Aj ∪ (Aj+1 \Aj )

and the sets on the right-hand side are disjoint, part (f) implies that Z Z Z f. f + f = Aj+1 \ Aj

Aj

Aj+1

Further, all of these integrals are finite. If we set A0 = ∅, then A =

∞ S

k=1

Ak =

∞ S

j=1

(Aj \Aj−1 ),

and the sets on the right-hand side above are disjoint. Applying part (f), we therefore compute that

c Solutions 2019 Christopher Heil

Z

f =

A

115

Z

∞ X

= = =

f

(by part (f))

Aj \ Aj−1

j=1

lim

N →∞

lim

N →∞

lim

N →∞

N Z X

Aj

j=1

Z

AN

Z

f −

Z

f − Z

f

Aj−1



f A0

f.

AN

4.3.9 Given a measurable set E ⊆ Rd , we have Z Z Z χE (x − a) dx = χE+a (x) dx = |E + a| = |E| = Rd

Rd

χE (x) dx.

Rd

Hence the integral of a characteristic function is invariant under translations. Taking linear combinations, this fact extends to simple functions. Given a nonnegative function f : Rd → [0, ∞], there exist simple functions φn that increase pointwise to f. The functions φn (x−a) increase pointwise to f (x−a), so by applying the Monotone Convergence Theorem we see that Z Z f (x − a) dx = lim φn (x − a) dx n→∞

Rd

= lim

n→∞

=

Z

Z

Rd

φn (x) dx Rd

f (x) dx.

Rd

Now suppose that f : Rd → [−∞, ∞] is an arbitrary extended real-valued function whose integral exists. Then the integrals of f + and f − both exist, with at most one of these being infinite. Applying the translation-invariance proved for nonnegative functions, it follows that Z Z Z f (x − a) dx = f + (x − a) dx − f − (x − a) dx Rd

Rd

=

Z

Rd

=

Z

f + (x) dx −

Z

Rd

f − (x) dx

Rd

f (x) dx.

Rd

Finally, if f is complex-valued then we write f = fr + ifi and use the fact that the integrals of fr and fi are invariant under translations.

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The proof for invariance under reflection is similar, starting with the calculation Z Z Z χE (x) dx. χ−E (x) dx = | − E| = |E| = χE (−x) dx = Rd

Rd

Rd

This equality then extends by cases to generic functions. An alternative approach is to use Problem 4.2.17, i.e., compare the regions under the graph of f (x − a) or f (−x) to the region under the graph of f (x). An argument in that spirit is written out in the solution to Problem 4.3.10 4.3.10 We could proceed as in the solution to Problem 4.3.9, but instead we will write out a proof based on Problem 4.2.17. Note that since f is defined on the set E, the composition f ◦L is defined on L−1 (E). Wince L−1 is an invertible linear transformation, it maps measurable sets to measurable sets. Therefore L−1 (E) is measurable. Moreover, Theorem 2.3.15 tells us that |L−1 (E)| = | det(L−1 )| |E| = where C = | det(L−1 )| =

|E| = C |E|, | det(L)| 1 . | det(L)|

Step 1. Suppose that f is nonnegative and finite everywhere on E. We will relate the regions under the graphs of f and f ◦ L. Writing components vertically and using block matrix notation, we have    x −1 Rf ◦L = : x ∈ L (E), 0 ≤ y ≤ f (Lx) y  −1   L (w) = : w ∈ E, 0 ≤ z ≤ f (w) z  −1     L 0 w = : w ∈ E, 0 ≤ z ≤ f (w) 0 1 z  −1     L 0 w = : w ∈ E, 0 ≤ z ≤ f (w) 0 1 z  −1  L 0 = (Rf ). 0 1 Since



L−1 0 det 0 1



= det(L−1 ) = C,

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Theorem 2.3.15 therefore implies that the measure of Rf ◦L is C times the measure of Rf : |Rf ◦L | = C |Rf |. Consequently, by using Problem 4.2.17 to relate the integrals to the measures of the regions under the graph, we have Z Z f. (f ◦ L) = |Rf ◦L | = C |Rf | = C E

L−1 (E)

The remainder of the proof simply consists of proceeding through cases to extend this equality to general functions. Step 2. Assume f is nonnegative and finite a.e. Let Z = {f = ∞}, and define g(x) = 0 for x ∈ Z and g(x) = f (x) otherwise. Then g = f a.e. and g is nonnegative and finite everywhere. Since L−1 is an invertible linear transformation, it maps sets of measure zero to sets of measure zero. Therefore L−1 (Z) has measure zero, so g ◦ L = f ◦ L a.e. Applying Step 1, it follows that Z Z Z Z (f ◦ L) = (g ◦ L) = C g = C f L−1 (E)

L−1 (E)

E

E

Step 3. Assume that f is an extended real-valued function such that exists. Writing f = f + − f − , it follows from Step 2 that Z Z ± f ±. (f ◦ L) = C

R

E

f

E

L−1 (E)

R

R

R

+ − Since E f Rexists, at most R −one of E f or E f can be infinite. + If both E f and E f are finite, then we can subtract to obtain Z Z Z f (Lx) dx = f + (Lx) dx − f − (Lx) dx L−1 (E)

L−1 (E)

= C

Z

E

= C

Z

f + (x) dx − C

Z

L−1 (E)

f − (x) dx

E

f (x) dx.

E

R R Even if one of E f + or E f − is infinite, the above calculation is still valid in the extended real sense. R Step 4. Now assume that f is a complex-valued function such R that RE f exists. Write f = g + ih where g and h are real-valued. Then E g and E h both exist and are finite, and by Step 3 we have

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118

Z

L−1 (E)

(g ◦ L) = C

Z

g

Z

and

E

L−1 (E)

(h ◦ L) = C

Z

h.

E

Since all of these quantities are finite, we therefore have Z Z Z (f ◦ L) = (g ◦ L) + i (h ◦ L) L−1 (E)

L−1 (E)

= C

Z

E

= C

Z

g − iC

Z

L−1 (E)

h

E

f.

E

4.4.5 (a) This is an immediate consequence of the definition of the L1 -norm and the space L1 (E). (b) Given f ∈ L1 (E) and c ∈ C, we have Z kcf k1 = |c| |f | E

= |c|

Z

E

|f |

(by Lemma 4.1.8)

= |c| kf k1 . (c) Fix f, g ∈ L1 (E). Then |f + g| ≤ |f | + |g|, so Z kf + gk + 1 = |f + g| (by Lemma 4.1.8) ≤ =

Z

Z

E

E

E


|f | + |g|

|f | +

Z

E

|g|

(by Theorem 4.2.3)

= kf k1 + kgk1 . R (d) “⇒.” If kf k1 = E |f | = 0, then Exercise 4.1.10 implies that |f | = 0 a.e., which is equivalent to f = 0 a.e. “⇐.”R If f = 0 a.e. then |f | = 0 a.e., so Exercise 4.1.10 implies that kf k1 = E |f | = 0.

4.4.10 We fill in some of the missing details in the proof of Theorem 4.4.10. R R R (a) The proof given in the text shows that Ek (f + g) = Ek f + Ek g when k = 2. We will give the details for k = 1, 3, and 6. Sets E1 and E6 . These two cases follow immediately because f and g are both nonnegative on E1 , and strictly negative on E6 .

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Set E3 . On this set we have f ≥ 0, g < 0, and f + g < 0 (instead of f + g ≥ 0 as it was on E2 ). Since −f − g and f are both nonnegative on E3 , we compute that Z Z Z Z − (f + g) + f = (−f − g) + f (by Exercise 4.3.6(e)) E3

E3

E3

= =

Z

E3

E3

(−f − g) + f

E3

(−g) = −

Z

(by Theorem 4.2.3)

Z

g.

E3

Since each integral is finite, we can rearrange to obtain Z Z Z (f + g) = f + g. E3

E3

E2

Sets E4 and E5 . These are entirely symmetrical to E2 and E3 , with the roles of f and g interchanged. (b) We finish the proof of Theorem 4.4.10 for the complex-valued case. It follows from equation (4.15) that f + g is integral. Splitting into real and imaginary parts, and applying part (a) and the definition of the integral of a complex-valued function, we compute that Z Z (f + g) = (fr + gr ) + i(fi + gi ) E

E

=

Z

(fr + gr ) + i

E

=

Z

fr +

E

=

Z

f+

α

kx χ[0,1] k1 =

Z

(fi + gi )

E

gr + i

E

Z

E

fi + i

Z

gi

E

g.

E

E

4.4.15 If α 6= −1, then

Z

Z

Z

0

1

xα dx = lim

t→0+

1 − tα+1 . α+1

This is finite if and only if α + 1 > 0, i.e., α > −1. A similar calculation shows that the Lp -norm is infinite if α = −1. Hence xα χ[0,1] (x) belongs to L1 (R) if and only if −1 < α < ∞. Similarly, if α 6= −1 then Z ∞ tα+1 − 1 xα dx = lim kxα χ[1,∞) k1 = , t→∞ α + 1 1

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120

which is finite if and only if α + 1 < 1, i.e., α < −1. The case α = −1/p leads to an infinite Lp -norm. Hence xα χ[0,1] (x) belongs to L1 (R) if and only if −∞ < α < −1. 4.4.16 (a) The function f (x) = belongs to C0 (R) but is not integrable.

1 |x| + 1

(b) For each integer n > 0, choose an > 0 and 0 < bn < 1/2, and let the graph of g be a triangle of height an on the base [n, n + 2bn ]. Precisely, we define

g(n) = 0 = g n + 2bn and g n + b n = an ,

then we let g be linear on the intervals [n, n + bn ] and [n + bn , n + 2bn ], and everywhere not yet defined we set g(x) = 0. This function g is continuous, and ∞ X kgk1 = an b n . n=1

By choosing an and bn , we can make this quantity finite. For example, we can let an increase if bn decreases correspondingly, such as an = n and bn = n−3 . In this case g is continuous and integrable, but it does not belong to C0 (R), and in fact g is not even bounded.

(c) Assume that f is uniformly continuous, but f (x) does not converge to zero as x → ∞. Note that this does not say that f must converge to some other value as x → ∞. On the other hand, it does tell us that there exists some ε > 0 such that for each R > 0 there is a point x > R such that |f (x)| > 2ε. Since f is uniformly continuous, there is a δ > 0 such that |x − y| < δ

=⇒

|f (x) − f (y)| < ε.

Now, there must exist some point x1 > 1 such that |f (x1 )| > 2ε. Hence if x ∈ (x1 − δ, x1 + δ) then 2ε < |f (x1 )| ≤ |f (x1 ) − f (x)| + |f (x)| < ε + |f (x)|. Thus |f (x)| > ε on the interval (x1 −δ, x1 +δ). Then we repeat this argument. There must exist some x2 > x1 + δ such that |f (x2 )| > 2ε. As before we find that |f (x)| > ε on the interval (x2 − δ, x2 + δ). Continuing in this way, |f | is bounded below by ε on infinitely many disjoint intervals of length 2δ, which implies that f is not integrable. (d) Suppose that f is integrable and a = limx→∞ f (x) exists. By breaking into real and imaginary parts, it suffices to assume that f is real-valued.

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121

Suppose that a > 0. Taking ε = a/2, there is some R > 0 such that f (x) > ε for all x > R. But then f is not integrable, which is a contradiction. Similarly, we cannot have a < 0, so it follows that a = 0. 4.4.17 (a) Since f is integrable, it is finite a.e. Let h(x) = f (x) whenever f (x) is finite, and set h(x) = 0 when f (x) = ±∞. Then h is measurable and finite at every point. Therefore g(x) − h(x) never takes an indeterminate form. As h and g are both measurable, it follows from Lemma 3.2.1 that g − h is measurable. As g − f = g − h a.e., it follows that g − f is also measurable. Since g ≥ f a.e., we have g − ≤ f − a.e. As f is integrable, it follows that Z Z Z 0 ≤ g− ≤ f− ≤ |f | < ∞. E

E

E

R

Further, E g + exists as a nonnegative, extended real number. Therefore the integral of g on E exists, and Z −∞ < g ≤ ∞. E

Also, g − f ≥ 0 a.e., so the integral of g − f exists and is a nonnegative, extended real number. If g is integrable, then by the linearity of the integral for integrable functions, we immediately obtain Z Z Z (g − f ) = g− f, E

E

E

so in this case Rwe are finished. On the other hand, if g is not integrable, then we must have E g + = ∞ and therefore Z Z Z + g = g − g − = ∞. E

E

E

Since f is integrable, it follows that Z Z g − f = ∞. E

E

As g is not integrable, f is integrable, and the sum of integrable functions is integrable, the function g − f cannot be integrable. Since g − f ≥ 0 a.e., we therefore have Z (g − f ) = ∞. E

Consequently,

Z

E

(g − f ) = ∞ =

Z

E

g−

Z

f. E

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122

(b) First we consider the extension of the MCT. Suppose that fn ≥ g a.e., where g is integrable, and fn ր f on E. Then fn − g ≥ 0 for every n, and fn − g ր f − g a.e. Applying part (a) and the Monotone Convergence Theorem (or more precisely the MCT variation derived in Theorem 4.3.7), we therefore obtain Z Z Z (f − g) g = f − E

E

E

= lim

n→∞

= lim

n→∞

= As

R

E

g is finite, it follows that Z



Z

E

Z

fn −

Z

Z

fn

−

Z

fn .

E

lim

n→∞

f = lim

n→∞

E

(fn − g)

E



g

E

Z

 g.

E

E

Now we consider the extension of Fatou’s Lemma. The argument is similar. Assume that fn ≥ g a.e., where g is integrable, and let f = lim inf fn . Then fn − g ≥ 0 a.e., so by Fatou’s Lemma and part (a) we have Z Z Z (f − g) g = f − E

E

E

=

Z

E

lim inf (fn − g) n→∞

≤ lim inf n→∞

= lim inf n→∞

Z

E

Z

(fn − g)

E

fn −

Z

E

g



  Z Z g. fn − = lim inf n→∞

E

E

As g is integrable, it follows that Z Z fn . f ≤ lim inf E

n→∞

E

To see that the assumption that g is integrable is necessary, let E = R and consider fn = − n1 . Then fn ≥ −1 for every n and fn ր 0, but

c Solutions 2019 Christopher Heil

Z

fn = −∞ → / 0 =

R

Also,

123

Z

lim inf fn =

R n→∞

n→∞

0. R

0 = 0

R

which is strictly greater than lim inf

Z

Z

Z

R

fn = −∞.

4.4.18 Set E = R and fn = n1 χ[−n,n] . Then fn converges uniformly to the zero function, but fn does not converge to the zero function in L1 -norm because Z Z ∞ 1 n 1 = 1. k0 − fn k1 = |fn | = n −n −∞ 4.4.19 Since f (0) = 0 and f is differentiable at x = 0, the limit lim

x→0

f (x) − f (0) f (x) = lim = f ′ (0) x→0 x x−0

exists and is finite. Consequently, f (x)/x is bounded on some neighborhood of 0. Precisely, there is a δ > 0 such that f (x) < ∞. M = sup x 0δ

Combining these two inequalities, we see that f (x)/x is integrable on R, hence its integral on R exists. R 4.4.20 Since f ∈ L1 (Rd ), we know that E |f | exists and is finite. Consequently, Problem 4.3.10 implies that Z Z 1 |f | < ∞. |f ◦ L| = | det(L)| Rd Rd

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124

Therefore f ◦ L ∈ L1 (Rd ).

4.4.21 (a) If f ∈ L1 (E) and g ∈ L∞ (E), then for almost every x we have |f (x) g(x)| ≤ |f (x)| kgk∞ . Therefore Z

E

|f g| ≤

Z

E

Z

|f | kgk∞ = kgk∞

|f | = kgk∞ kf k1 < ∞,

E

so f g ∈ L1 (E). (b) By Problem 2.3.20(a), there exists a measurable set A ⊆ E such that 0 < |A| < ∞. By part (c) of that same problem, there exist disjoint, measurable subsets Ak of A such that |Ak | = 2−k |A|. Set f =

∞ X

2k/2 χAk .

k=1

Then f ∈ L1 (E), because Z

|f | =

E

∞ X

k=1

2k/2 |Ak | =

∞ X

k=1

2k/2 2−k |A| < ∞.

However, f 2 ∈ / L1 (E), because Z

E

|f |2 =

∞ X

k=1

2k |Ak | =

∞ X

k=1

2k 2−k |A| = ∞.

(c) For simplicity of notation, assume first that f, g ≥ 0 a.e. Since f 2 and g 2 are integrable, they must be finite a.e. The function (f +g)2 is integrable, because Z Z   2 (f + g)2 ≤ 2 max f, g E

E

= 4

Z

E

 max f 2 , g 2

Z   f 2 + g2 ≤ 4 E

< ∞.

The same reasoning shows that (f − g)2 is integrable. Since the sum of integrable functions is integrable, it follows that

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125

(f + g)2 − (f − g)2 4

fg =

is integrable. For the general case, if f and g are arbitrary measurable functions such that |f |2 , |g|2 ∈ L1 (E), then |f | and |g| are nonnegative functions that satisfy |f |2 , |g|2 ∈ L1 (E). Therefore, the previous case implies that |f g| is integrable. However, f g is integrable if and only if |f g| is integrable, so it follows that f g ∈ L1 (E). 4.4.22 We are given that f is integrable on [a, b] and Z x f (t) dt = 0, all x ∈ [a, b]. a

If f is complex-valued, then we can write f = g + ih where both g and h are real-valued. The hypotheses on f imply that Z x Z x h g = 0 = a

a

for every x ∈ [a, b]. Therefore, it suffices to assume that f is real-valued. Suppose that there exists a measurable set E ⊆ [a, b] with positive measure such that f (x) > 0 for every x ∈ E. Then there exists a Rclosed set F ⊆ E such that 0 < |F | < |E|. Since f is nonnegative on F, if F f = 0 then we would have f = 0 a.e., which is a contradiction. Therefore Z f > 0. F

Since U = (a, b)\F is an bounded open subset of the real line, U is a union S of at most countably many disjoint open intervals. Write U = (ak , bk ) where the intervals (ak , bk ) are disjoint. Since 0 =

Z

b

f =

a

it follows that

XZ k

bk

ak

f =

Z

U

Z

Z

f +

f = −

Z

f < 0.

F

Therefore there must be at least one k such that Z bk f 6= 0. ak

But then

f,

U

F

c Solutions 2019 Christopher Heil

126

0 =

Z

bk

f =

Z

ak

f +

bk

f = 0 +

Z

bk

ak

ak

a

a

Z

f 6= 0,

which is a contradiction. Consequently, f cannot be strictly positive on any set with positive measure. That is, f ≤ 0 a.e. A symmetric argument shows that f ≥ 0 a.e., so we conclude that f = 0 a.e. 4.4.23 (a) First proof. Let M = sup kfn k1 < ∞. n

Note that the function f is measurable because it is the pointwise a.e. limit of measurable functions. It is integrable because we can use Fatou’s Lemma to compute that Z Z Z |f | ≤ lim inf |fn | ≤ sup |fn | = M < ∞. kf k1 = n→∞

E

Note that Z

E

|f − fn | ≤

Z

E

n

E

|f | +

Z

E

E

 |fn | ≤ M + M = 2M.

Therefore, for every n ∈ N we have Z Z Z Z |fn | ≤ M. |f − fn | ≤ |fn | − |f − fn | ≤ −2M ≤ − E

E

E

E

Thus, the sequence of real numbers Z Z |f − fn | |fn | − an = E

E

is bounded both above and below, and therefore hasR a finite liminf and limsup. Our goal is to prove that these are both equal to E |f |. Since |fn | = |fn − f + f | ≤ |fn − f | + |f |, we have |fn | − |fn − f | ≤ |f |, and therefore Z  Z Z |f |. |fn − f | ≤ |fn | − lim sup n→∞

E

E

On the other hand, |f − fn | ≤ |f | + |fn |,

E

(A)

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127

so |f | + |fn | − |f − fn | ≥ 0. Applying Fatou’s Lemma, we therefore compute that Z Z   2 |f | = lim |f | + |fn | − |f − fn | E n→∞

E

≤ lim inf n→∞

=

Z

E

Z

|f | +

E

|f | + lim inf n→∞

Z

E

|fn | −

E

|fn | −

Z

Z

Z

E

E

 |f − fn |

 |f − fn | .

Therefore Z

E

|f | ≤ lim inf n→∞

Z

E

|fn | −

Z

E

 |f − fn | .

(B)

Combining equations (A) and (B) therefore gives the result. Second proof. This proof is from Stroock, “Essentials of Integration Theory”; but it uses the DCT, which is not proved until Section 4.5. Let M = sup kfn k1 < ∞. n

The function f is measurable because it is the pointwise a.e. limit of measurable functions. It is integrable because we can use Fatou’s Lemma to compute that Z Z Z |f | ≤ lim inf |fn | ≤ sup |fn | = M < ∞. kf k1 = n→∞

E

Note that

E

n

E

|fn | − |f | − |f − fn | → 0 a.e.

Also, |fn | − |f | − |f − fn | ≤ |fn | − |f − fn | + |f | ≤ |fn − (f − fn )| + |f |

(Reverse Triangle)

= 2|f | ∈ L1 (E). Therefore we can apply the Z lim n→∞

Since

Dominated Convergence Theorem to obtain

E

|fn | − |f | − |f − fn | = 0.

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128

Z Z Z Z |fn | − ≤ |fn | − |f − fn | − |f | , |f − f | − |f | n E

E

E

E

the result follows. (b) Let

fn = χ[−n,n]

and

f = 1.

Then fn is integrable and fn → f pointwise. However, we have Z ∞ |f | = 1 ∞

while lim

Z

n→∞

E

|fn | −

Z

E


|f − fn | = lim 2n − ∞ = −∞. n→∞

4.4.24 “⇒.” This follows from the Reverse Triangle Inequality, which tells us that kf k1 − kfn k1 ≤ kf − fn k1 .

“⇐.” Assume that kfn k1 → kf k1 . In this case we have sup kfn k1 < ∞, so Problem 4.4.23 implies that Z  Z Z |f | = lim |f − fn | kf k1 = |fn | − n→∞

E

E

E

  = lim kfn k1 − kf − fn k1 . n→∞

Since kfn k1 converges, it follows that kf − fn k1 converges as well, and lim kf − fn k1 = kf k1 − lim kfn k1 = kf k1 − kf k1 = 0.

n→∞

n→∞

4.5.3 For almost every x we have |f (x)| = lim |fn (x)| ≤ g(x) ∈ L1 (E), n→∞

so f is integrable on E. Also, |f (x) − fn (x)| ≤ |f (x)| + |fn (x)| ≤ 2g(x) a.e., so we have 2g − |f − fn | ≥ 0 a.e. Applying Fatou’s Lemma, we therefore have Z Z lim inf (2g − |f − fn |) g = 2 E

E n→∞

≤ lim inf n→∞

Z

E

(2g − |f − fn |)

c Solutions 2019 Christopher Heil

= 2

129

Z

g + lim inf n→∞

E

= 2

Z

Z

E

g − lim sup n→∞

E

Z

(−|f − fn |)

E

|f − fn |.

Rearranging, we see that 0 ≤ lim sup n→∞

Consequently

Z

lim

n→∞

E

4.5.5 (a) Define En =

Z

E

|f − fn | ≤ 0.

|f − fn | = 0. 

|f | ≤ n ,

and set fn = f · χEn . Since f is integrable, it must be finite a.e. Therefore fn → f pointwise a.e., and furthermore we have |fn | ≤ |f |. Since f is integrable, the Dominated Convergence Theorem implies that fn → f in L1 -norm. (b) Choose any ε > 0. Then there exists an N such that kf − fn k1 < ε/2 for all n ≥ N. Note that f − fn = f · χEnC . Let δ = ε/(2N ), and suppose that |A| < δ. Then Z Z Z |f | |f | + |f | = A∩EN

A

≤

Z

N +

A∩EN

Z

A \ EN

A

|f − fN |

≤ N |A| + kf − fN k1
0. By definition of the distance to a set, there exists a point z ∈ A such that dist(y, A) ≤ d(y, z) ≤ dist(y, A) + ε. Since z ∈ A, we therefore have dist(x, A) ≤ d(x, z) ≤ d(x, y) + d(y, z) ≤ d(x, y) + dist(y, A) + ε. Consequently dist(x, A) − dist(y, A) ≤ d(x, y) + ε. A symmetric argument shows that dist(y, A) − dist(x, A) ≤ d(y, x) + ε. Hence |dist(x, A) − dist(y, A)| ≤ d(x, y) + ε, and since ε > 0 is arbitrary, the result follows. (d) Part (c) shows that f (x) = dist(x, A) is Lipschitz. All Lipschitz functions are uniformly continuous. Explicitly, if we fix ε > 0, then we can take δ = ε. If x, y ∈ X satisfy d(x, y) < δ, then part (c) implies that |f (x) − f (y)| = |dist(x, A) − dist(y, A)| ≤ d(x, y) ≤ δ = ε. Therefore f is uniformly continuous. 4.5.9 Choose f ∈ L1 (Rd ). Given ε > 0, we can find g ∈ Cc (Rd ) such that kf − gk1 < ε. Fix R > 0 such that supp(g) ⊆ [−R, R]d . Since g is uniformly continuous, there exists a δ > 0 such that

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|a| < δ =⇒ kg − Ta gk∞
0 such that for each δ > 0 there is a point t ∈ (x − δ, x + δ) such that |f (x) − f (t)| ≥ ε. Fix any particular k ∈ N, and for convenient of notation let Γ stand for Γk , and write  Γ = a = x0 < x1 < · · · xn = b .

Since x ∈ / S, we know that x is not equal to any xj . Hence x ∈ (xj−1 , xj ) for some j. Hence there is some δ > 0 such that (x − δ, x + δ) ⊆ (xj−1 , xj ). Let t be the point specified above. Either f (x) ≥ f (t) or f (t) ≤ f (x). If f (t) ≥ f (x), then f (t) − f (x) ≥ ε, Then  ψk (x) = Mj = sup f (u) : u ∈ [xj−1 , xj ] ≥ f (t) ≥ f (x) + ε  ≥ inf f (u) : u ∈ [xj−1 , xj ] + ε = mj + ε = φk (x) + ε.

Hence ψk (x) − φk (x) ≥ ε. On the other hand, if f (x) ≥ f (t), then f (x) − f (t) ≥ ε, In this case,  ψk (x) = Mj = sup f (u) : u ∈ [xj−1 , xj ] ≥ f (x) ≥ f (t) + ε  ≥ inf f (u) : u ∈ [xj−1 , xj ] + ε = mj + ε = φk (x) + ε.

Hence ψk (x) − φk (x) ≥ ε in this case as well. 4.5.15 (a) For 1 ≤ x ≤ 2 we have

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Further,

2 n sin x/n n2 |x|/n 1 = ∈ L1 [1, 2]. 1 + nx2 ≤ nx2 |x|

xn(n/x) sin x/n n2 sin x/n = lim n→∞ n→∞ 1 + nx2 1 + nx2    xn 1 sin x/n 1 = lim = lim ·1 = . n→∞ 1 + nx2 n→∞ x/n x x lim

Therefore we can apply the Dominated Convergence Theorem (or, since the domain has finite measure, the Bounded Convergence Theorem) and compute that 2 Z 2 2 Z 2 n sin x/n 1 lim dx = dx = ln x = ln 2. 2 n→∞ 1 1 + nx 1 x 1 (b) For x ∈ [0, 1] we have sin xn = 1 lim n→∞ xn

and

sin xn 1 xn ≤ 1 ∈ L [0, 1].

The DCT therefore implies that Z 1 Z 1 sin xn 1 dx = 1. lim dx = n→∞ 0 xn 0

For x > 1, Z ∞ Z ∞ Z ∞ sin xn 1 sin xn x−n dx = dx ≤ → 0. xn dx ≤ n x n − 1 1 1 1 Therefore the limit in question is 1.

4.5.16 (a) “⇒.” This direction is trivial. R “⇐.” Assume that A f = 0 for every measurable set A ⊆ E. By breaking into real and imaginary parts, it suffices to consider the case that f is extended real-valued. Let A = {f > 0}. This is a measurable set, so by hypothesis we R have A f = 0. However, f is nonnegative on A, so Exercise 4.1.10 implies that f is zero almost everywhere on A. Since f is strictly positive on A, it follows that |A| = 0. Similarly, B = {f < 0} has measure zero. Consequently f = 0 almost everywhere on E. (b) Let En = {|f | ≤ n}. By Exercise 4.5.5(a), f χEn → f in L1 -norm. Therefore, if n is large enough then we will have Z |f | = kf − f χEn k1 < ε. E \ En

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Since f is bounded on En , it therefore suffices to take A = En . 4.5.17 First proof. Fix ε > 0. Since f is integrable, Exercise 4.5.5 implies R that there exists a constant δ > 0 such that E |g| < ε for every measurable set E satisfying |E| < δ. If x ∈ R and 0 ≤ h < δ, then the measure of the interval [x, x + h] is less than δ, so Z x+h Z x+h Z x Z x+h |F (x + h) − F (x)| = f− f = f ≤ |f | ≤ ε. 0

0

x

x

A similar argument applies if −δ < h ≤ 0, so we conclude that F is uniformly continuous. Second proof. Given x, a ∈ R, we compute that Z x−a Z |F (x − a) − F (x)| = f (t) dt − −∞

x

−∞

f (t) dt

Z x Z x f (t) dt = f (t − a) dt − −∞ −∞ Z x Ta f (t) − f (t) dt ≤ −∞

≤ kTa f − f k1 .

Therefore kTa F − F ku = sup |F (x − a) − F (x)| ≤ kTa f − f k1 → 0, x∈R

so F is uniformly continuous by Problem 1.3.6. 4.5.18 We will closely follow the proof of the DCT for functions that is given in Exercise 4.5.3. For each k ∈ N, set ck = sup |akn |. n∈N

P Then ck ≥ 0, and by hypothesis we have ck < ∞. Since |bk | ≤ ck , it follows P that we also have |bk | < ∞. For all positive integers k and n we have the estimate |bk − akn | ≤ |bk | + |akn | ≤ 2ck . Therefore 2ck − |bk − akn | ≥ 0,

k, n ∈ N.

Applying Fatou’s Lemma (Problem 4.2.18), it follows that

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2

∞ X

∞ X

ck =

k=1

lim inf (2ck − |bk − akn |) n→∞

k=1

≤ lim inf n→∞

= 2

∞ X

∞ X

k=1

(2ck − |bk − akn |)

n→∞

k=1

= 2

∞ X

k=1

∞ X

ck + lim inf

k=1

ck − lim sup n→∞

(−|bk − akn |)

∞ X

k=1

|bk − akn |.

Rearranging, we see that 0 ≤ lim sup n→∞

Consequently lim

n→∞

Therefore

∞ X

k=1

∞ X

k=1

|bk − akn | ≤ 0.

|bk − akn | = 0.

X X ∞ X ∞ ∞ bk − akn = (bk − akn ) k=1

k=1

k=1

∞ X

≤

k=1

|bk − akn |

→ 0 as n → ∞. 4.5.19 Since f is Riemann integrable on [a + δ, b], it is Lebesgue integrable on that interval, and its Riemann and Lebesgue integrals coincide. Thus, we have Z b

f,

Iδ =

δ > 0.

a+δ

Let δn be any sequence of positive numbers such that δn ց 0. Let fn = f χ[a+δn ,b] . Then fn is nonnegative and Lebesgue integrable on [a, b], and 0 ≤ fn ր f. The Monotone Convergence Theorem therefore implies that Z

a

b

f = lim

n→∞

Z

a

b

fn = lim

n→∞

Z

b

a+δn

f = lim Iδn = I. n→∞

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In particular, since I is finite we conclude that f is integrable on [a, b]. 4.5.20 Set E = R and fn = n1 χ[−n,n] . Then fn converges pointwise to the zero function and |fn | ≤ 1 for every n. However, fn does not converge to the zero function in L1 -norm. 4.5.21 We may assume |E| > 0, as otherwise there is nothing to prove. We are given measurable functions fn on |E| < ∞ such that fn → f a.e. and |fn | ≤ M a.e. for every n. This implies that |f | ≤ M a.e., and therefore |f − fn | ≤ |f | + |fn | ≤ 2M a.e. Since fn and f are bounded and E has finite measure, we know that fn and f are integrable, and hence their integrals on E exist. Fix ε > 0. By Egorov’s Theorem, there exists a set measurable set A ⊆ E with ε |A| < 4M such that fn → f uniformly on E \A. Therefore there exists some N > 0 such that

ε n > N =⇒ (f − fn )χE \ A u < . 2|E| Consequently, for all n > N we have Z Z |f − fn | + kf − fn k1 = ≤

Z

A

2M + A

Z

E \A

E \A

|f − fn |

ε 2|E|

ε |E \A| 2|E| ε ε + |E| ≤ 2M 4M 2|E| ε ε = + 2 2

≤ 2M |A| +

Therefore fn → f in L1 -norm. 4.5.22 First we prove a little lemma. Lemma. Let {xn }n∈N be a sequence in a metric space X, and let x ∈ X be fixed. Suppose that every subsequence {yn }n∈N of {xn }n∈N has a subsequence {zn }n∈N of {yn }n∈N such that zn → x. Then xn → x. Proof. Suppose that every subsequence of {xn }n∈N has a subsequence that converges to x, but the full sequence {xn }n∈N does not converge to x. Then there exists an ε > 0 such that given any N we can find an n > N such that d(x, xn ) > ε. Iterating this, we can find a subsequence {xnk }k∈N

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such that d(x, xnk ) > ε for every k. But then no subsequence of {xnk }k∈N can converge to x, which is a contradiction. ⊓ ⊔ Now we return to the problem at hand. m We are given that |fn | ≤ g ∈ L1 (E) and fn → f. Let {gk }k∈N be any m subsequence of {fn }n∈N . Then gk → f, so there is a subsequence {hj }j∈N of {gk }k∈N such that hj (x) → f (x) a.e.R Since |hRj | ≤ g for all j, the Dominated Convergence Theorem implies that hj → f. That is, every subsequence has a subsequence we have convergence. It therefore follows from R for which R the lemma that fn → f as n → ∞. m Next, we apply this fact to the functions |f −fn |, which satisfy |f −fn| → 0. m Since fn → f, we know there is a subsequence fnj that converges pointwise a.e. to f. Since |fnj | ≤ g for all j, we conclude that |f | ≤ g a.e., and hence |f − fn | ≤ 2g a.e. Therefore we can apply our previous work to the sequence {|f − fn |}n∈N and obtain Z Z kf − fn k1 = |f − fn | → 0 = 0. E

E

4.5.23 Nothing changes in the problem if we allow f to be extended realvalued instead of just nonnegative. Therefore, we assume that f : E → R [−∞, ∞] is an integrable function, and I = E f > 0. Interpreting the ball of radius zero as the empty set, for each r ≥ 0 set At = E ∩ Br (0) and It =

Z

f =

Ar

Z

f χA r .

E

If we fix r ≥ 0, then lim f χAs = f χAr a.e.

s→r +

Furthermore, for every s > r we have |f χAs | ≤ |f | ∈ L1 (E). The Dominated Convergence Theorem therefore implies that Z Z χ lim Is = lim f χAr = Ir . f As = s→r +

s→r +

E

E

Thus, Ir is continuous from the right (as a function of r). A similar calculation shows that Ir is continuous from the left. Applying the DCT again, we similarly compute that

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lim Ir = I0 = 0

r→0+

and lim Ir =

r→∞

Z

f = I.

E

The Intermediate Value Theorem therefore implies that Ir takes every value strictly between 0 and I as r ranges over (0, ∞). Further, if we take A = ∅ then we have Z f = 0, A

while if A = E then

Z

f = A

Z

f = I.

E

R Hence A can be chosen so that RA f takes value between 0 and I, inclusive. Remark: If f ≥ 0 a.e., then A f cannot take any value outside of [0, I]. R However, if f is negative on a set with positive measure, then A f could lie outside of this range for some subsets A. 4.5.24 Since ln (1 + t) and t agree at t = 0 and since their derivatives satisfy ∀ t ≥ 0,

d 1 d ln (1 + t) = ≤ 1 = t, dt 1+t dt

we have ln (1 + t) ≤ t for all t ≥ 0. Therefore, if we set

then we have

 f (x)  fn (x) = n ln 1 + n

f (x) = f (x). n Thus each fn is dominated by the integrable function f. Since f is integrable, it must be finite a.e. For each x such that f (x) < ∞, we have  f (x) n lim fn (x) = lim ln 1 + = ln ef (x) = f (x). n→∞ n→∞ n 0 ≤ fn (x) ≤ n

Hence fn (x) → f (x) for a.e. x. Consequently, the Dominated Convergence Theorem implies that fn → f in L1 -norm, and therefore the integral of fn converges to the integral of f.

4.5.25 We have f (x) = 0 for all x ∈ K, and f (x) > 0 for all x ∈ / K. Let K1 be the set of points that are a distance at most 1 from K, i.e., K1 = {x ∈ Rd : f (x) ≤ 1}.

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Then K ⊆ K1 , and

g = (1 − f ) χK1 .

Since g is identically zero outside of K1 , we have g n → 0 on Rd \ K1 . On K1 \ K we have 0 ≤ 1 − f (x) < 1, so g n → 0 on this set. On K we have g n = 1 for every n. Therefore g n → χK pointwise. Additionally, g n ≤ χK1 ∈ L1 (Rd ), so we can apply the Dominated Convergence Theorem to compute that Z Z χK = |K|. lim gn = n→∞

Rd

Rd

4.5.26 First Proof. Since E has finite measure, the function χE is integrable on Rd . For simplicity of notation, let Eh = E + h. First, we compute that 2 |E| = |E| + |Eh |

(translation invariance)

= |E ∪ Eh | + |E ∩ Eh |

(Problem 2.2.32)

= |E ∩ Eh | + |E \ Eh | + |Eh \ E| + |E ∩ Eh |. = 2 |E ∩ Eh | + |E \ Eh | + |Eh \ E|. Second, by disjointness and the strong continuity of translation, Z Z χEh \E χ |E \ Eh | + |Eh \ E| = E\Eh + Z = |χE − χEh | Z = |χE − χE+h | = k χE − T h χE k 1 → 0

as h → 0.

Therefore 2 |E ∩ Eh | = 2 |E| − |E ∩ Eh | − |E \ Eh | → 2 |E|

as h → 0.

Second proof. Z ∞  Z ∞    χE∩E+h − χE χE χE+h − χE = −∞ −∞ Z ∞ χ χ ≤ E E+h − χE −∞

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≤ =

Z

∞

−∞

Z

E

|χE+h − χE |

|Th χE − χE |

= kTh χE − χE k1 → 0

as h → 0.

Therefore Z

|E ∩ (E + h)| =

∞ −∞

χE∩E+h →

Z

∞ −∞

χE = |E|.

4.5.27 We have for almost every x that |f (x)| = lim |fn (x)| ≤ lim gn (x) = g(x) ∈ L1 (E), n→∞

n→∞

so we conclude that f ∈ L1 (E). Since g + gn − |f − fn | ≥ 0 a.e., Fatou’s Lemma implies that Z Z lim inf (g + gn − |f − fn |) g = 2 E n→∞

E

≤ lim inf n→∞

≤ lim inf n→∞

=

Z

Z

E

Z

≤

= 2

g +

n→∞

Z

g + lim sup n→∞

E

Z

E

Z

E

g + lim inf

E

Z

(g + gn − |f − fn |)

g − lim sup n→∞

E

E

Z

gn − gn −

gn

E

Z

E

Z

E

Z

E

 |f − fn |

 |f − fn |

 Z  |f − fn | + lim inf − n→∞

|f − fn |.

Rearranging, we find that 0 ≤ lim sup n→∞

and therefore

Z

E

|f − fn | ≤ 0,

E

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lim

n→∞

4.5.28 Define

Z

E

|f − fn | = 0.

fn (x) =

1+

gn (x) =

1+

g(x) = e−x .


x −n n


x −n , n

sin nx ,

Clearly, |fn (x)| ≤ gn (x),

x ≥ 0.

We have gn → g pointwise on [0, ∞), since lim gn (x) = lim 1 +

n→∞

n→∞

This also shows that


x −n n

= e−x = g(x).

lim fn (x) = lim gn (x) sin nx = e−x · 0 = 0,

n→∞

n→∞

so fn → 0 pointwise on [0, ∞). Also, gn ∈ L1 [0, ∞) for n ≥ 2 because Z ∞ Z ∞
x+n −n dx gn (x) = n 0

0

= nn

Z

∞

(x + n)−n dx

0

∞ x + n1−n = n 1 − n 0 n

t nn x + n1−n lim 1 − n t→∞ 1 − n 0   nn = lim (t + n)1−n − n1−n 1 − n t→∞
nn 0 − n1−n = 1−n =

nn n1−n n−1 n < ∞. = n−1 =

This also shows that

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lim

n→∞

Z

141

∞

gn = 1 =

0

Z

∞

e−x dx =

0

Z

∞

g.

0

The conditions for the Generalized DCT are therefore met, so Z ∞ Z ∞ 0 = 0. fn = lim n→∞

4.5.29 Assume we see that

R1 0

0

0

xn f (x) dx = 0 for every n. Taking complex conjugates, Z

1

f (x) xn dx = 0,

for every n.

0

Forming linear combinations, we therefore have that Z

1

f (x) p(x) dx = 0,

for every polynomial p.

0

Fix ϕ ∈ C[0, 1]. By the Weierstrass Approximation Theorem, there exist polynomials pn such that kpn − ϕk∞ → 0 as n → ∞. Therefore Z 1 Z 1 Z 1 ≤ f (x) ϕ(x) dx f (x) (ϕ(x) − p (x)) dx f (x) p (x) dx + n n 0

0

0

≤ kf k1 kϕ − pn k∞ + 0 → 0 as n → ∞.

Since f is bounded and [0, 1] has finite measure, f is integrable. Therefore we conclude that Z 1 f (x) ϕ(x) dx = 0 0

for every continuous ϕ. Since C[0, 1] is dense in L1 [0, 1] we can select continuous ϕn such that kf − ϕn k1 → 0. Hence Z 1 Z 1 Z 1 2 f (x) (f (x) − ϕn (x)) dx + f (x) ϕn (x) dx |f (x)| dx ≤ 0

0

0

≤ kf k∞ kϕ − pn k1 + 0

Therefore

R1 0

→ 0

as n → ∞.

|f (x)|2 dx = 0, so f (x) = 0 a.e.

4.5.30 Assume that the functions ft satisfy the given hypothesis. Choose any sequence {tk }k∈N in (0, c) such that tk → 0. Then the DCT implies that lim kf − ftk k1 = 0.

k→∞

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Applying the Lemma 4.4.9, it follows that lim kf − ft k1 = 0.

t→0+

4.5.31 (a) We are given that f is integrable, and hence f is measurable by definition. Fix ω ∈ R. The function hω (x) = sin ωx is measurable because it is continuous. The product of measurable functions is measurable, so f hω is measurable, and since f is integrable and hω is measurable their product is integrable. Therefore F (ω) exists at every point ω. Now we show that F is continuous at ω = 0. Since sin 0x = 0, we have F (0) = 0. Also, for every x where f is defined and f (x) is finite (which is a.e. x), we have lim f (x) sin ωx = f (x) sin 0x = 0.

ω→0

Further, |f (x) sin ωx| ≤ |f (x)| ∈ L1 (R). The Dominated Convergence Theorem therefore implies that Z ∞ Z ∞ 0 dx = 0 = F (0). lim g(ω) = lim f (x) sin ωx dx = ω→0

ω→0

−∞

−∞

Therefore g is continuous at ω = 0. Part (c) asks for continuity at arbitrary points ω, and this follows by using a similar argument. (b) Now assume that xf (x) ∈ L1 (R). Recall that lim

y→0

sin y = 1. y

Since F (0) = 0, if we are allowed to interchange the limit and the integral then Z ∞ F (ω) − F (0) sin ωx F ′ (0) = lim = lim f (x) dx ω→0 ω→0 −∞ ω−0 ω Z ∞ sin ωx = lim xf (x) dx ω→0 −∞ ωx Z ∞ xf (x) dx. = −∞

Since we are given that xf (x) is integrable, this shows that F ′ (0) exists if the interchange is allowed. We justify the interchange by verifying that the hypotheses of the Dominated Convergence Theorem are satisfied. First, the integrand converges pointwise a.e., because for every x where f is defined and finite we have

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lim f (x)

ω→0

143

sin ωx sin ωx = lim xf (x) = xf (x). ω→0 ω ωx

Second, the integrand is bounded by an integrable function, because | sin y| ≤ |y| and therefore f (x) sin ωx ≤ f (x) ωx = xf (x) ∈ L1 (R). ω ω

Thus, the DCT is applicable. Part (c) asks for differentiability at arbitrary points ω, and this follows by using a similar argument. (b) This part is not a consequence of part (a), because f (x)/x need not be integrable. We will prove that G is differentiable at any point ω. Set gω (x) = f (x)

sin ωx . x

Since | sin θ| ≤ |θ|, if we fix ω then we have sin ωx ≤ |ωf (x)| ∈ L1 (R). |gω (x)| ≤ f (x) x Hence gω is integrable, so F (ω) =

R

gω (x) dx is well-defined for every ω.

First proof. We use trig identities to write     (ωx + hx) − ωx (ωx + hx) − ωx sin(ωx + hx) − sin ωx cos = 2 sin hx 2 2   hx sin(hx/2) . cos ωx + = hx/2 2 Therefore lim

h→0

gω+h (x) − gω (x) sin(ωx + hx) − sin ωx = lim f (x) h→0 h hx   sin(hx/2) hx = lim f (x) cos ωx + h→0 hx/2 2 = f (x) cos ωx.

Further, for every h we have   gω+h (x) − gω (x) = |f (x)| sin(hx/2) cos ωx + hx ≤ |f (x)| ∈ L1 (R). hx/2 h 2

Applying the DCT, we see that G′ (ω) exists, because

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G(ω + h) − G(ω) h→0 h Z ∞ Z ∞ gω+h (x) − gω (x) = lim dx = f (x) cos ωx dx. h→0 −∞ h −∞

G′ (ω) = lim

Second proof. Define S(x) = sin x. Let ω be fixed. Since S is differentiable, the Mean Value Theorem implies that given t and h, there exists some point ξ (depending on both x and h) such that S(ωx + hx) − S(ωx) = S ′ (ξ). hx Hence gω+h (x) − gω (x) = f (x) S(ωx + hx) − S(ωx) h hx = |f (x) S ′ (ξ)|

= |f (x) cos ξ| ≤ |f (x)| ∈ L1 (R). Also, since S is differentiable, lim

h→0

S(ωx + hx) − S(ωx) = S ′ (ωx) = cos ωx. hx

Therefore gω+h (x) − gω (x) S(ωx + hx) − S(ωx) = lim f (x) = f (x) cos ωx. h→0 h→0 h hx lim

Applying the DCT as before, we see that G′ (ω) exists and has the value G(ω + h) − G(ω) h Z ∞ Z ∞ gω+h (x) − gω (x) dx = f (x) cos ωx dt. = lim h→0 −∞ h −∞

G′ (ω) = lim

h→0

Third proof. Use convolution; see the solution to Problem 4.6.29(c). 4.5.32 Let

Fix x and set

∂f M = sup (x, y) . x,y ∂x

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fk (x, y) =

145




f x + k1 , y − f (x, y) 1 k

,

k ∈ N.

Since f (x, y) is a measurable function of y, so is each function fk . Since we know that ∂f ∂x exists, it follows that ∂f (x, y) = lim fk (x, y) k→∞ ∂x is a measurable function of y. Since we also know that follows that the integral Z 1 ∂f (x, y) dy 0 ∂x

∂f ∂x

is bounded, it

does exist for this x. By hypothesis, f (x, y) is an integrable function of y, so we can define F (x) =

Z

0

1

f (x, y) dy,

x ∈ [0, 1].

Our goal is to show that for every x the following limit exists and has the indicated limit: Z 1 ∂f F (x + h) − F (x) d F (x) = lim = (x, y) dy. (A) h→0 dx h 0 ∂x We will apply the Bounded Convergence Theorem. However, there is a technical issue—that theorem applies to limits of sequences, not to limits of a real parameter. But this is not a major issue, as we can just apply Problem 1.1.23, which tells us that the limit in equation (A) holds if and only if for each sequence hk → 0 we have Z 1 F (x + hk ) − F (x) ∂f lim = (x, y) dy. k→∞ hk 0 ∂x Fix the point x ∈ [0, 1]. We must show that the integrand converges pointwise for each y and is bounded. To show convergence, and let hk be any sequence of real numbers that converges to zero. Note that Z 1 F (x + hk ) − F (x) f (x + hk , y) − f (x, y) = dy. hk hk 0 Since

∂f ∂x

exists, the integrand converges pointwise for every y: f (x + hk , y) − f (x, y) ∂f = (x, y), k→∞ hk ∂x lim

all y ∈ [0, 1].

Now we will show that the integrand is uniformly bounded. Write

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f = fr + ifi , where fr and fi are real-valued. Since ∂f = ∂fr + i∂fi , it follows that for every y, 2 2 1/2  ∂fr ∂fi ∂fr ≤ M. ∂x (x, y) = ∂x (x, y) + ∂x (x, y)

A similar inequality holds for the imaginary part. Now fix any particular y ∈ [0, 1]. Since fr is real-valued, we can apply the Mean Value Theorem. There exists a point ξk between x and x + hk such that ∂f fr (x + hk , y) − fr (x, y) = (ξk , y). hk ∂x Therefore fr (x + hk , y) − fr (x, y) = ∂fr (ξk , y) ≤ M. hk ∂x

A similar inequality holds for the imaginary part. Since |f | ≤ |fr | + |fi |, we conclude that f (x + hk , y) − f (x, y) hk fr (x + hk , y) − fr (x, y) fi (x + hk , y) − fi (x, y) ≤ + hk hk ≤ 2M.

This bound is independent of y and k (and also x, but that point has been fixed anyway). Thus the integrand converges pointwise for every y and is uniformly bounded. Since we are integrating over the finite domain [0, 1], the Bounded Convergence Theorem therefore implies that Z 1 f (x + hk , y) − f (x, y) F (x + hk ) − F (x) = lim dy lim k→∞ 0 k→∞ hk hk Z 1 ∂f (x, y) dy. = 0 ∂x 4.5.33 (a) By definition, we have µ(∅) = 0. Since µ(A) ≥ 0 for every set A, we know that µ(A) is never −∞. Problem 2.4.12 shows that counting measure is countably additive. This shows that counting measure is a signed measure on (Rd , L). Therefore, it only remains to show that µ is not absolutely continuous with re-

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spect to Lebesgue measure. This follows from the fact that |{0}| = 0 but µ({0}) = 1. (b) By definition, we have δ(∅) = 0. Since δ(A) ≥ 0 for every set A, we know that δ(A) is never −∞. Problem 2.4.13 shows that the δ measure is countably additive. This shows that the δ measure is a signed measure on (Rd , L). Therefore, it only remains to show that δ is not absolutely continuous with respect to Lebesgue measure. This follows from the fact that |{0}| = 0 but |δ0| = 1. R R (c) By hypothesis, at least one on Rf + or f − must be finite. Without loss of generality, let us assume that f − < ∞. In this case, νf (A) is a well-defined extended real number for every measurable set A ⊆ Rd . We have Z νf (∅) = f = 0. ∅

d

For every measurable set A ⊆ R , Z Z Z f −. f+ − f = νf (A) = Since

Z

0 ≤

f

A

A

A

−

Z

≤

A

Rd

f − < ∞,

if follows that νf (A) can never be −∞. Suppose that A1 , A2R, . . . are disjoint measurable subsets of Rd , and let A = ∪Ak . Then, since Rd f exists, Exercise 4.3.6(f) tells us that νf (A) =

Z

f =

A

∞ Z X

∞ X

f =

Ak

k=1

νf (Ak ).

k=1

Therefore νf is countably additive. This shows that νf is a signed measure on (Rd , L). Therefore, it only remains to determine whether νf is absolutely continuous with respect to Lebesgue measure. If f is integrable, then Exercise 4.5.5(b) tells us that νf is indeed absolutely continuous with respect to Lebesgue measure. However, this is still true even if f is not integrable. To see why, write f = f + − f − . If |A| = 0, then f + and f − are both zero a.e. on A. Since they are also nonnegative, we can apply Exercise 4.1.10 to obtain Z Z + f − = 0. f = A

A

Consequently νf (A) =

Z

A

f =

Z

A

f+ −

Z

A

f − = 0.

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Hence νf is absolutely continuous with respect to Lebesgue measure. 4.6.4 We give the details of the proof of part (b) of Lemma 4.6.4. (b) Assume that fk ց f, and set gk = f1 − fk

and

g = f1 − f.

Since F is closed under linear combinations, we have gk ∈ F for every k. Further, g is integrable and 0 ≤ gk ր g. Applying part (a), it follows that g ∈ F. Therefore f = f1 − g ∈ F as well. 4.6.12 For every x we have Z

1

f (x, y) dy = 0,

0

and for every y we have Z

1

f (x, y) dx = 0.

0

Also, since |f | = 1/|Qk | everywhere on the square Qk , we have ZZ

Q

|f (x, y)| (dx dy) =

RR

RR

∞ ZZ X

k=1

Qk

|f | =

∞ X

k=1

Since Q f + and Q f − are equal, it follows that RR Consequently, Q f is undefined.

|Qk | RR

Q

1 = ∞. |Qk | f+ = ∞ =

4.6.13 Since the function x is odd, we have Z 1  Z 1 Z 1 Z 1 x 1 dx dy = x dx dy 2 2 −1 −1 1 − y −1 1 − y −1 Z 1 1 = · 0 dy = 0. 2 −1 1 − y

However,

Z

1

−1

and therefore x

Z

1

−1

Consequently,

x dy = ∞, 1 − y2

  x > 0, ∞, x dy = 0 x = 0,  1 − y2  −∞, x < 0.

RR

Q

f −.

c Solutions 2019 Christopher Heil

Z

1

x

0

while

Z

0

−1

x

149

Z

Z

1 −1 1

−1

1 dy dx = ∞ 1 − y2

1 dy dx = −∞. 1 − y2

Therefore the iterated integral Z 1 Z x −1

1

−1

1 dy dx 1 − y2

has the form ∞ − ∞, so it is undefined. 4.6.14 We have d y (x2 − y 2 ) − y (2y) x2 − y 2 = = . 2 2 2 2 2 dy x + y (x + y ) (x2 + y 2 )2 Fix 1 < x < ∞. The function fx (y) =

x2 − y 2 (x2 + y 2 )2

is strictly positive on the intervals (1, x) and (x, ∞). On these two intervals, ∞ Z ∞ y x2 − y 2 = 0− x = − 1 , dy = 2 2 2 2 2 (x + y ) x + y x 2x2 2x x and

Z

1

x

x x2 − y 2 x y 1 1 1 = dy = − 2 = − . (x2 + y 2 )2 x2 + y 2 1 2x2 x +1 2x x2 + 1

Both of these quantities are finite, so fx is integrable on (1, ∞), and Z ∞ x2 − y 2 1 dy = − 2 . 2 + y 2 )2 (x x +1 1 Hence Z

∞

1

Z

∞

x2 − y 2 dy dx = − (x2 + y 2 )2

1

Z

∞

π 1 dx = − . x2 + 1 4

∞

π 1 dy = . y2 + 1 4

1

Similarly, Z

1

∞

Z

∞

1

Note, however, that

x2 − y 2 dx dy = (x2 + y 2 )2

Z

1

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Z

∞

1

so

Z

∞

1

2 x − y2 1 1 1 1 1 (x2 + y 2 )2 dy = 2x + 2x − x2 + 1 = x − x2 + 1 ,

Z

∞

1

2 Z x − y2 dy dx = (x2 + y 2 )2

1

∞

1 1 − x x2 + 1



dx = ∞.

4.6.15 For simplicity of presentation, assume that c > 0 (a symmetric proof covers the case c < 0). Note that χ[x,x+c] (t) = χ[t−c,t] (x). As a function of two variables, F (x, t) = f (t) χ[t−c,t] (x) is measurable, and it is integrable as a function of two variables because Z ∞Z ∞ Z ∞Z ∞ |F (x, t)| dx dt = |f (t) χ[t−c,t] (x)| dx dt −∞

−∞

−∞

=

Z

−∞

= c

−∞

∞

Z

|f (t)|

∞

−∞

Now, since f is integrable we have Z g(x + c) − g(x) = = =

∞

−∞

 χ[t−c,t] (x) dx dt

|f (t)| dt < ∞.

x+c

f −

−∞

Z

Z

Z

x

f

−∞

x+c

f

x Z ∞

f (t) χ[x,x+c](t) dt

−∞

=

Z

∞

f (t) χ[t−c,t] (x) dt

−∞

=

Z

∞

F (x, t) dt.

−∞

Thus g(x + c) − g(x) is defined for every x ∈ R, and it is integrable because Z ∞ Z ∞ Z ∞ |g(x + c) − g(x)| dx ≤ |F (x, t)| dt dx < ∞. −∞

Further, by Fubini’s Theorem,

−∞

−∞

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Z

∞

−∞



151

Z  g(x + c) − g(x) dx =

∞

−∞

Z

=

∞

−∞

Z

=

∞

−∞

Z

=

∞

Z

∞

F (x, t) dt dx

−∞

Z

∞

F (x, t) dx dt

−∞

Z

∞

f (t) χ[t−c,t] (x) dx dt

−∞

f (t) Z

∞

χ[t−c,t] (x) dx dt

−∞

−∞

= c

Z

∞

f (t) dt.

−∞

4.6.16 First Proof. (a) ⇒ (b). Suppose that f = 0 a.e. Then |f | = 0 a.e., so by replacing f with |f | it suffices to assume that f is nonnegative a.e. By Tonelli’s Theorem, fx is measurable, and we have ZZ Z Z f = 0. fx (y) dy dx = E

E×F

F

R

R

Since F fx (y) dy ≥ 0, we must have F fx (y) dy = 0 for almost every x ∈ E. Since f ≥ 0 a.e., it follows that for almost every x ∈ E we have fx = 0 a.e. on F. (b) ⇒ (a). Suppose that for almost every x ∈ E we have fx = 0 a.e. on F. Then we can apply Tonelli’s Theorem to compute that Z Z ZZ |fx (y)| dy dx = 0. |f | = E

E×F

F

Since |f | is nonnegative, this implies that |f | = 0 a.e., and therefore f = 0 a.e. Second Proof. Let Z = {|f | > 0}. Note that ZZ χZ . |Z| = E×F

On the other hand, if we set Zx = {y ∈ F : (x, y) ∈ Z} = {y ∈ F : |f (x, y)| > 0}, then χZx (y) = χZ (x, y). Therefore

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Z

E

Z Z

|Zx | dx =

E

F

E

F

Z Z

=

ZZ

=

χZx (y) dy dx χZ (x, y) dy dx χZ

E×F

= |Z|. Consequently, |Z| = 0 (statement (a)) if and only if |Zx | = 0 for a.e. x (statement (b)). 4.6.17 We use the same techniques as in Problem 4.2.17 to show that Γf and Rf are measurable. Then we compute their measures as follows. For each x ∈ E, set ( [0, f (x)], f (x) < ∞, Ix = [0, ∞) f (x) = ∞. Note that |Ix | = f (x) for each x ∈ E, and χIx (y) = χRf (x, y). Using Tonelli’s Theorem, we compute that Z Z ∞ Z χIx (y) dy dx f (x) dx = E

E

=

ZZ

0

χRf (x, y) (dy dx)

E×[0,∞)

= |Rf |. Note that if x ∈ E and 0 ≤ y < ∞, then χΓf (x, y) = χ{f (x)} (y), even if f (x) = ∞. Therefore we use Tonelli’s Theorem to compute that ZZ χΓf (x, y) (dy dx) |Γf | = E×[0,∞)

Z Z

=

E

Z

=

∞

χ{f (x)} (y) dy dx

0

0 dx = 0.

E

4.6.18 Note that f is continuous and therefore measurable on (0, ∞)2 . Integrating first with respect to x followed by integration with respect to y, we compute that  Z ∞Z ∞ Z ∞ Z n 2 2 f (x, y) dx dy = lim x e−x (1+y ) dx dy (by MCT) 0

0

0

n→∞

0

c Solutions 2019 Christopher Heil

=

153

Z

1 2

0

Z

1 = 2

0

Z

∞



2

lim −

n→∞

e−x (1+y 1 + y2

2

2

2



∞

e−n (1+y lim − n→∞ 1 + y2

∞

  ) x=n

dy

x=0

)

1 + 1 + y2



dy

1 dy 1 + y2 0 Z n 1 1 = lim dy (by MCT) 2 n→∞ 0 1 + y 2  n 1 −1 = lim tan y 2 n→∞ 0   1 −1 −1 lim tan n − tan 0 = 2 n→∞ 1 = 2

=

1 π π = . 2 2 4

On the other hand, integrating first with respect to y followed by integration with respect to x gives Z ∞Z ∞ f (x, y) dy dx 0

0

=

Z

0

=

Z

0

=

Z

0

=

Z

∞

lim

n→∞

∞

lim

n→∞

∞

lim

n→∞

∞

e

−x2

0

=

Z

∞

e−x

0

=

Z

0

∞

e

2

Z

Z

Z

 Z

−x2

n

xe

−x2 (1+y 2 )

dy dx

0 n 0 nx 0

lim

n→∞ ∞

2

x e−x e−x

2 2

y



 dy dx

 2 2 e−x e−t dt dx Z

nx

e

−t2

0

(by MCT)

(t = xy, dt = x dy)



dt dx

2

e−t dt dx

0

2 dx .

Consequently, Tonelli’s Theorem implies that Z

0

∞

e

−x2

2 π dx = . 4

4.6.19 Let a > 0 be fixed. Using integration by parts, we compute that

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154

I =

Z

a

e−tx sin x dx

0

= −e =

−tx

Z a cos x − t 0

a

e−tx cos x dx

0

 Z a
1 − e−at cos a − t e−tx sin x + t 0

= 1−e

−at

cos a − te

Therefore, I =

−at

0

 e−tx cos x dx

2

sin a − t I.

1 − e−at cos a − te−at sin a . 1 + t2

Now, since | sin x| ≤ |x|, Z Z aZ ∞ −tx |e sin x| dt dx ≤ 0

a

a

x

=

Z

∞

e−tx dt dx

0

0

0

Z

a

x

0

1 dx = a < ∞. x

Hence we can apply Fubini’s Theorem to compute that Z a Z aZ ∞ sin x sin x e−tx dx = dt dx x x 0 0 0 Z ∞Z a sin x dx dt = e−tx x 0 0 Z ∞ 1 − e−at cos a − te−at sin a = dt. 1 + t2 0 Since a > 0, sup te−at = t>0

1 . ea

Hence, for each a > 1, 1 − e−at cos a − te−at sin a 3 ≤ ∈ L1 (0, ∞). 1 + t2 1 + t2 Further, we have pointwise that

1 1 − e−at cos a − te−at sin a = . a→∞ 1 + t2 1 + t2 lim

The Dominated Convergence Theorem therefore implies that Z a Z ∞ a sin x 1 π −1 lim dx = lim dt = lim tan t = . a→∞ 0 a→∞ 0 a→∞ x 1 + t2 2 0

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χ[x,1] (t) is a measurable function of t because 4.6.20 Fix 0 < x ≤ 1. Then f (t) t f is measurable, 1/t is continuous a.e., and χ[1,x] is measurable. Also, Z

0

1

Z 1 f (t) f (t) χ[x,1] (t) dt = t dt t x Z 1 1 ≤ |f (t)| dt x x Z 1 1 |f (t)| dt ≤ x 0 =

Therefore g(x) =

Z

0

1

kf k1 < ∞. x

f (t) χ[x,1] (t) dt = t

Z

1

x

f (t) dt t

exists and is a finite scalar for each 0 < x ≤ 1. This shows that g is welldefined (i.e., g(x) actually exists), although we do not know yet whether g is measurable. Define F (x, t) =

f (t) χ[x,1] (t), t

(x, t) ∈ [0, 1]2 .

This is a measurable function on [0, 1]2 , and it is integrable on [0, 1]2 because Z 1Z 1 Z 1Z 1 f (t) χ |F (x, t)| dx dt = t [x,1] (t) dx dt 0 0 0 0 Z 1  Z 1 f (t) χ[0,t] (x) dx dt = t 0 0 Z 1 f (t) = t t dt 0 Z 1 = |f (t)| dt = kf k1 < ∞. 0

Fubini’s Theorem therefore implies that either of the inner integrands in are measurable and integrable. In particular, Z 1 Z 1 f (t) g(x) = dt = F (x, t) dt t x 0

RR

F

is integrable on [0, 1]. Further, Fubini’s Theorem allows us to interchange integrals as follows:

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156

Z

1

g(x) dx =

0

Z

1

0

=

Z

=

1

0

=

Z

1

x

1

0

Z

Z

1

Z

f (t) dt dx t

t

f (t) dx dt t 0 Z t  f (t) dx dt t 0

f (t) dt.

0

4.6.21 For simplicity of notation, by replacing f with |f | we may assume throughout this solution that f is nonnegative. (a) The fact that ω is decreasing follows from the monotonicity of Lebesgue measure. (b) Given t ≥ 0, choose any real numbers sn such that sn ց t. Then the sets {f > sn } are nested increasing and their union is {f > t}. Continuity from below therefore implies that ω(sn ) = {f > sn } → {f > t} = ω(t). This is true for every sequence that decreases to t, so lims→t+ ω(s) = ω(t).

(c) Assume that f is integrable on E. Given t > 0, choose any real numbers sn such that sn ր t. Then the sets {f > sn } are nested decreasing and their intersection is {f ≥ t}. These sets have finite measure because f is integrable. Continuity from above therefore implies that ω(sn ) = {f > sn } → {f ≥ t} .

This is true for every sequence that increases to t, so it follows that lims→t− ω(s) = |{f ≥ t}|. (d) Let A = {(x, t) ∈ E × [0, ∞) : f (x) − t > 0}.

Since f (x) − t is a measurable function of two variables, χA is a measurable function. Given t ≥ 0 and x ∈ E, we have ) ( 1, if f (x) > t χ{f >t} (x) = χA (x, t) = = χ[0,f (x)) (t). 0, if f (x) ≤ t Applying Tonelli’s Theorem, we compute that

c Solutions 2019 Christopher Heil

Z

157

∞

ω(t) dt =

0

Z

0

=

Z

∞ ∞

{f > t} dt

0

=

Z Z E

=

Z

Z

χ{f >t} (x) dx dt

E ∞

χ[0,f (x)) (t) dt dx 0

f (x) dx.

E

(e) This follows from part (d). (f) Since f is integrable, part (e) implies that ω is integrable on [0, ∞). Since ω is also nonnegative and monotone decreasing, it follows that Z

2n ω(2n) = 2

2n

ω(2n) dt ≤ 2

n

Z

2n

n

ω(t) dt → 0 as n → ∞.

This computation does not require n to be integer; it still holds if we let n be real. Therefore, we can replace n by n/2 to obtain n ω(n) → 0

as n → ∞.

Similarly, we compute that 2 n

ω

2 n




= 2

Z

2/n

1/n

ω

2 n




dt ≤ 2

Replacing n by 2n, it follows that
1 1 2 ω = 2n n ω n

2 2n

Z




2/n

1/n

ω(t) dt → 0 as n → ∞.

→ 0 as n → ∞.

4.6.22 (This is one approach to a direct proof, there are others.) We are given that R =

∞ ∞ X X

m=1 n=1

|cmn | < ∞.

Fix an ordering of N × N, i.e., choose a bijection σ : N × N → N. Given any integer K we have K X |cσ(k) | ≤ R. k=1

We have σ(k) = (mk , nk ) for some integers mk and nk . Let M = max{m1 , . . . , mK }

and

N = max{n1 , . . . , nK }.

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Then, since every term in the sum that appears first is a term in the sum that appears second, K X

k=1

|cσ(k) | ≤

N M X X

 |cmn |

m=1 n=1 ∞ ∞ X X

 |cmn |

m=1

n=1

∞ M X X

≤ ≤

m=1

n=1

 |cmn |

= R.

Taking the limit as K → ∞, it follows that ∞ X

k=1

|cσ(k) | ≤ R < ∞,

P Therefore the series x = k cσ(k) converges absolutely. Since the scalar field (R or C) is a Banach space, absolute convergence implies unconditional conP vergence. Consequently, the series x = k cσ(k) converges unconditionally in X. That is, this series converges no matter what bijection σ that we choose, and the vector x is independent of the choice of bijection. Now fix any integer m ∈ N. Then X |cmn | ≤ R < ∞, n

so the series ym =

X

cmn

X

ym

n

converges absolutely. Moreover, X XX |ym | ≤ |cmn | = R < ∞, m

m

n

so the series y =

m

also converges absolutely. Now we compute that N ∞ N X N N X ∞ N X N X X X X X y − + c − c ≤ y − c c mn mn mn mn m=1 n=1

m=1 n=1

m=1 n=1

m=1 n=1

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159

N N ∞ X X X ym + |cmn | ≤ y − m=1

m=1 n=N +1

N X ym + ≤ y − m=1

∞ X ∞ X

n=N +1

m=1

 |cmn |

→ 0 as N → ∞.

This shows that zN =

N X N X

m=1 n=1

cmn → y

as N → ∞.

Let σ be a permutation of N×N obtained by enumerating the “square annuli”     {1, . . . , N + 1} × {1, . . . , N + 1} \ {1, . . . , N } × {1, . . . , N }

P in turn. Then zN is L k=1 cσ(k) for some appropriate integer L. Therefore, the fact that z converges to y says that a subsequence of the partial sums N PL c converge to y. On the other hand, by our work at the beginning k=1 σ(k) of this problem we know that L X

k=1

cσ(k) → x

as L → ∞.

Consequently y = x. Therefore, we have shown that  XX cmn = y = x. n

m

4.6.23 We are given scalars cmn ≥ 0. Hence any infinite series involving these scalars either converges or diverges to infinity. If ∞ X ∞ X cmn < ∞, m=1 n=1

then Problem 4.6.22 implies that

∞ X ∞ X

n=1 m=1

and the two series are equal. Likewise, if

cmn < ∞

c Solutions 2019 Christopher Heil

160 ∞ X ∞ X

n=1 m=1

cmn < ∞,

then Problem 4.6.22 implies that

∞ X ∞ X

m=1 n=1

cmn < ∞

and the two series are equal. Combining the above facts, we have ∞ X ∞ X

cmn < ∞

m=1 n=1

⇐⇒

∞ X ∞ X

cmn < ∞.

∞ X ∞ X

cmn = ∞.

n=1 m=1

The contrapositive statement is that ∞ X ∞ X

cmn = ∞

m=1 n=1

⇐⇒

n=1 m=1

4.6.24 We are given that M =

∞ X

n=1

kfn k1 < ∞,

P i.e., the series fn converges absolutely in L1 -norm. Since each fn is integrable, it follows that ∞ ∞ Z ∞ Z X X X fn ≤ |f | = kfn k1 < ∞. n n=1

E

n=1

Therefore the series

E

n=1

∞ Z X

n=1

fn

E

is an absolutely convergent series of scalars, so it converges to some finite scalar. P To show that the series ∞ n=1 fn (x) converges a.e., set gN (x) =

N X

n=1

|fn (x)|

and

g(x) =

∞ X

n=1

|fn (x)|.

These are series of nonnegative extended real numbers, so they converge pointwise a.e. in the extended real sense. By the Triangle Inequality, for each N we have

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161

kgN k1 ≤

N X

n=1

kfn k1 ≤ M.

Since gN ր g, the Monotone Convergence Theorem implies that Z Z kgk1 = g = lim gN N →∞

E

E

lim kgN k1

=

N →∞

≤

lim

N →∞

N X

n=1

kfn k1

= M < ∞. Therefore g ∈ L1 (E), i.e., g is integrable on E. Hence g is finite a.e. At any point x where g(x) < ∞, we have g(x) =

∞ X

n=1

Consequently, the series f (x) =

|fn (x)| < ∞. ∞ X

fn (x)

n=1

converges absolutely at almost every point x. Since |f | ≤ g, we have f ∈ L1 (E). Also, if we set N X fn (x), hN (x) = n=1

then hN → f pointwise a.e. For every N we have

|hN | ≤ gN ≤ g ∈ L1 (R), so we can apply the Dominated Convergence Theorem. The DCT tells us that hN → f in L1 -norm, and the integral of hN converges to the integral of f. Thus Z X ∞

E n=1

fn =

Z

f =

E

=

=

lim

N →∞

lim

N →∞

lim

N →∞

Z

hN

E

Z X N

fn

E n=1

N Z X

n=1

E

fn =

∞ Z X

n=1

E

fn .

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4.6.25 (a) Let f (x) = e−|x| . If x > 0, then Z (f ∗ f )(x) = e−|x−y| e−|y| dy =

Z

0

e

y−x y

e dy +

−∞

=

Z

0

= e

Z

x

e

y−x −y

e

dy +

0

e2y e−x dy +

−∞

−x

Z

Z

x

e−x dy +

0

Z

∞

∞

ex−y e−y dy x

ex e−2y dy

x

0 −2y ∞ e2y −x x −e + xe + e 2 −∞ 2 x

1−0 0 + e−2x + xe−x + ex 2 2 −x −x e e + xe−x + = 2 2 = e−x

= (1 + x) e−x . A similar computation shows that (f ∗ f )(x) = (1 − x) ex when x < 0. Hence (f ∗ f )(x) = (1 + |x|) e−|x| . 2

(b) Let g(x) = e−x . Then (g ∗ g)(x) = = =

2

Z

Z

Z

2

e−(x

2

−2xy+2y 2 )

e−2(y−xy+x

= e

−x2 /2

= e

−x2 /2

= e

−x2 /2

(c) Let h(x) = xe−x . Then

2

e−(x−y) e−y dy

Z

Z

2

dy

/4+x2 /4)

2

dy

e−2(y−x/2) dy 2

e−2y dy

 1/2 π . 2

c Solutions 2019 Christopher Heil

(h ∗ h)(x) = = =

Z

Z

Z

= e

163 2

2

(x − y) e−(x−y) y e−y dy (x − y) y e−(x

2

−2xy+2y 2 )

(x − y) y e−2(y−xy+x

−x2 /2

= e−x

2

/2

Z

2

dy

/4+x2 /4)

dy

2

(x − y) y e−2(y−x/2) dy

Z 

x −y 2

Z 



 2 x + y e−2y dy 2

 2 x2 − y 2 e−2y dy 4  2   1/2  1/2 2 2 x π 1 π = e−x /2 − e−x /2 4 2 4 2  1/2 2 1 π (x2 − 1) e−x /2 . = 4 2 = e−x

2

/2

4.6.26 This problem carries over to Rd without any changes, so we will give the solution for that setting. (a) By Problem 4.3.10, we can make the change of variable z = x − y to compute that Z Z (f ∗ g)(x) = f (y) g(x − y) dx = f (x − z) g(z) dz = (g ∗ f )(x). Rd

Rd

Hence convolution is commutative. (b) Since f, g, and h are all integrable, we know that f ∗ g and g ∗ h are integrable, and therefore f ∗ (g ∗ h) and (f ∗ g) ∗ h are integrable. We must show that these last two functions are equal a.e. Fubini’s Theorem allows us to interchange integrals as follows: Z (f ∗ g)(y) h(x − y) dy ((f ∗ g) ∗ h)(x) = Rd

=

Z

Rd

=

Z

Z

Rd

f (z) g(y − z) dz h(x − y) dy

f (z)

Rd

=

Z

Rd

f (z)

Z

Rd

g(y − z) h(x − y) dy dz

Rd

g(y) h(x − y − z) dy dz

Z

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=

Z

Rd

f (z) (g ∗ h)(x − z) dz

= (f ∗ (g ∗ h))(x). Therefore convolution is associative. (c) This follows from the linearity of the integral. (d) We have (f ∗ Ta g)(x) = = = =

Z

Z

Z

Z

Rd

f (y) (Ta g)(x − y) dy

Rd

f (y) g(x − y − a) dy

Rd

f (y − a) g(x − (y − a) − a) dy

Rd

(Ta f )(y) g(x − y) dy

= (Ta f ∗ g)(x), and similarly (f ∗ Ta g)(x) =

Z

Rd

f (y) g(x − y − a) dy

= (f ∗ g)(x − a) = (Ta (f ∗ g))(x). 4.6.27 This problem carries over to Rd without any changes, so we will give the solution for that setting. (a) Fix x ∈ Rd . Since g is bounded, so is g(x − y) as a function of y. Since f is integrable, it follows that f (y) g(x − y) is integrable as a function of y, because Z Z |f (y)| kgk∞ dy = kf k1 kgk∞ . |f (y) g(x − y)| dy ≤ Rd

Rd

Therefore (f ∗ g)(x) exists, and by making a linear change of variable we have Z Z (f ∗ g)(x) = f (y) g(x − y) dy = f (x − y) g(y) dy. Rd

(b) We compute that

Rd

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|(f ∗ g)(x + h) − (f ∗ g)(x)| Z Z = f (x + h − y) g(y) dy − ≤

Z

Rd

Rd

R

f (x − y) g(y) dy d

|f (x + h − y) − f (x − y)| |g(y)| dy

≤ kgk∞ = kgk∞ = kgk∞

Z

Rd

|f (x + h − y) − f (x − y)| dy

Rd

|f (z + h) − f (z)| dy

Rd

|T−h f (z) − f (z)| dy

Z

Z

(change of variables z = x − y)

= kgk∞ kT−h f − f k1 → 0 as h → 0,

(strong continuity of translation)

where at the last step we have used the strong continuity of translation on L1 (Rd ) that is established in Exercise 4.5.9. This shows that f ∗g is continuous at x. Since x is arbitrary, we have shown that f ∗ g is continuous on Rd . (c) Using the same computations from part (a), we see that Z |(f ∗ g)(x)| = f (y) g(x − y) dy ≤

≤

Z

Rd

Rd

|f (y) g(x − y)| dy

Rd

|f (y)| kgk∞ dy

Z

= kf k1 kgk∞ . Therefore f ∗ g is bounded and kf ∗ gk∞ = sup |(f ∗ g)(x)| ≤ kf k1 kgk∞ . x∈Rd

4.6.28 This problem carries over to Rd without any changes, so we will give the solution for that setting. (a) Since Cc (Rd ) is contained in both L1 (Rd ) and L∞ (Rd ), Problem 4.6.27 implies that that f ∗ g is continuous and bounded. Therefore, we need only show that f ∗ g is compactly supported. Since f and g are each compactly supported, there is some R > 0 such that f (x) = 0 and g(x) = 0 for all kxk ≥ R. Suppose that kxk > 2R. If kyk ≥ R then f (y) = 0. If kyk < R, then kx−yk > R, and therfore g(x−y) = 0. Hence

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f (y) g(x − y) = 0 for every y ∈ Rd . Thus (f ∗ g)(x) = 0 for all kxk > 2R, so f ∗ g is compactly supported. We can estimate the support of f ∗ g more precisely. Since f and g are both continuous and compactly supported, K = supp(f ) + supp(g) R is a compact set. Suppose that (f ∗ g)(x) = f (y) g(x − y) dy was nonzero. Then there must be a y such that y ∈ supp(f ) and x − y ∈ supp(g) simultaneously, for otherwise f (y) g(x − y) would be zero for every y and therefore (f ∗ g)(x) would be zero. Hence, x = y + (x − y) ∈ supp(f ) + supp(g). This shows that S = {x ∈ R : (f ∗ g)(x) 6= 0} ⊆ supp(f ) + supp(g) ⊆ K. Since the support of f ∗ g is the closure of the set S and since K is closed, it follows that supp(f ∗ g) = S ⊆ K.

(b) Choose any functions f, g ∈ Cc1 (R). Since Cc1 (R) ⊆ Cc (R), part (a) implies that f ∗ g ∈ Cc (R). Therefore, we need only prove that f ∗ g is differentiable and (f ∗ g)′ is continuous. Assume first that f and g are real-valued. Fix x ∈ R. Then for any h ∈ R we have (f ∗ g)(x + h) − (f ∗ g)(x) h Z Z 1 1 = f (y) g(x + h − y) dy − f (y) g(x − y) dy h h Z g(x + h − y) − g(x − y) . = f (y) h

Keep the point x fixed. Let Fh (y) denote the integrand in the final integral, i.e., g(x + h − y) − g(x − y) . Fh (y) = f (y) h For each y ∈ R, we have lim Fh (y) = f (y) lim

h→0

h→0

g(x + h − y) − g(x − y) = f (y) g ′ (x − y). h

Thus the integrand converges pointwise. Further, given any y and h the Mean Value Theorem implies that there exists a point ξ (depending on y and h) such that g(x + h − y) − g(x − y) = g ′ (ξ). h

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Therefore |Fh (y)| = |f (y)| |g ′ (ξ)| ≤ kg ′ k∞ |f (y)| = c |f (y)|. Thus Fh is bounded, independently of h, by the integrable function c|f |. Therefore we can apply the Dominated Convergence Theorem to obtain (f ∗ g)(x + h) − (f ∗ g)(x) h Z g(x + h − y) − g(x − y) = lim dx f (y) h→∞ h Z = f (y) g ′ (x − y) dx

(f ∗ g)′ (x) = lim

h→∞

= (f ∗ g ′ )(x). Hence f ∗ g is differentiable, and its derivative is equal to f ∗ g ′ . Since f and g ′ both belong to Cc (R), part (a) implies that (f ∗ g)′ = f ∗ g ′ belongs to Cc (R). In particular, (f ∗ g)′ is differentiable, so we conclude that f ∗ g ∈ Cc1 (R). Therefore Cc1 (R) is closed under convolution. 4.6.29 (a) Since χE is both integrable and bounded, Problem 4.6.27 implies that χE ∗ χ−E is continuous and bounded. (b) Evaluating at the origin, we see that Z ∞ χE (y) χ−E (0 − y) dy (χE ∗ χ−E )(0) = −∞

=

Z

∞

−∞

=

Z

χE (y) χ−E (0 − y) dy

∞

χE (y) χE (y) dy

−∞

= |E| > 0. Since χE ∗ χ−E is continuous, we conclude that it is nonzero on some open ball B that contains 0. We claim that χE ∗ χ−E (x) can only be nonzero for x ∈ E − E. To see this, fix x ∈ / E − E. Suppose there is some y ∈ E such that z = y − x ∈ E. Then x = y − z ∈ E − E, which is a contradiction. Hence y ∈ E =⇒ y − x ∈ / E, and therefore χE (y) χE (y − x) = 0 for every y ∈ Rd . Consequently

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(χE ∗ χF )(x) =

Z

∞ −∞

χE (y) χ−E (x − y) dy =

Z

∞

−∞

χE (y) χE (y − x) dy = 0.

Finally, since χE ∗ χ−E is nonzero on the open ball B, it follows that B ⊆ E − E. (c) Note that χ−E (x − y) = 1

⇐⇒

x − y ∈ −E

⇐⇒

x − y = −z,

⇐⇒

y = x + z,

⇐⇒

y ∈x+E

⇐⇒

some z ∈ E some z ∈ E

χE+x (y) = 1.

Therefore χ−E (x − y) = χE+x (y). Hence Z χE (y) χ−E (x − y) dy (χE ∗ χ−E )(x) = = =

Z Z

χE (y) χE+x (y) dy χE∩(E+x) (y) dy = |E ∩ (E + x)|.

Since we proved in part (a) that the convolution is continuous, we compute that lim |E ∩ (E + x)| = lim (χE ∗ χ−E )(x) = (χE ∗ χ−E )(0) = |E|,

x→0

x→0

the final equality following from part (b). The easy way to prove that the second limit is zero is to note that both χE and χ−E belong to L2 (R), and then apply Theorem 9.1.5 to show that χE ∗ χ−E ∈ C0 (R). However, we will give a direct proof. Fix ε > 0, and let K ⊆ E be a compact set such that |E \K| < ε. Since K is compactly supported, we know from Problem 4.6.28 that χK ∗ χ−K is continuous and compactly supported. Hence there is some R > 0 such that (χK ∗ χ−K )(x) = 0 for all |x| > R. Consequently, if |x| > R then |E ∩ (E + x)| = (χE ∗ χ−E )(x) − χK ∗ χ−K )(x) Z Z χ χ χ χ = K (y) −K (x − y) dy E (y) −E (x − y) dy −

c Solutions 2019 Christopher Heil

=

Z

≤

Z

≤

Z

169

Z

χE (y) χ−E (x − y) dy −

χK (y) χ−E (x − y) dy +  Z Z χK (y) χ−E (x − y) dy − χK (y) χ−K (x − y) dy |χE (y) − χK (y)| χ−E (x − y) dy + Z χK (y) |χ−E (x − y) − χ−K (x − y)| dy χE \ K (y) · 1 dy +

Z

1 · χ−E \ −K (x − y) dy

= |E \K| + |E \K| < 2ε. This is true for all |x| > R, so we conclude that |E ∩(E + x)| → 0 as |x| → ∞. 4.6.30 (a) Since f is integrable and g is bounded, we know from Problem 4.6.27 that f ∗ g is continuous. We use the following lemma to evaluate the limit as x → ∞. Lemma A. If f ∈ L1 (R), g ∈ Cb (R), and limx→∞ g(x) = 0, then f ∗ g is continuous and limx→∞ (f ∗ g)(x) = 0.

Proof. We already know that f ∗ g is continuous, so fix ε > 0. Then there exists some R > 0 such that |g(x)| ≤ ε for all x > R. Since f is integrable, an application of the DCT shows that there is some M such that Z ∞ |f (x)| dx < ε. M

If y < x − R then x − y > R and therefore |g(x − y)| < ε. Hence if x > M + R then we compute that |(f ∗ g)(x)| ≤

Z

∞

x−R

≤ kgk∞ ≤ kgk∞

|f (y)| |g(x − y)| dy + Z

∞

x−R

Z

|f (y)| dy + ε

∞

M

|f (y)| dy + ε

Z

Z

Z

x−R

−∞

|f (y)| |g(x − y)| dy

x−R

−∞

|f (y)| dy

∞

−∞

|f (y)| dy

≤ ε kgk∞ + ε kf k1 . This shows that (f ∗ g)(x) → 0 as x → ∞. ⊓ ⊔ A similar lemma applies if g(x) → 0 as x → −∞. Combining these two facts together, we see that if g ∈ C0 (R) then f ∗ g ∈ C0 (R).

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(b) We prove an extension of Lemma A from part (a). Lemma B. If f ∈ L1 (R), g ∈ Cb (R),R and limx→∞ g(x) = r, then f ∗ g is ∞ continuous and limx→∞ (f ∗ g)(x) = r −∞ f.

Proof. Let h(x) = g(x) − r. Then h ∈ Cb (R) and h(x) → 0 as x → ∞, so the lemma from part (a) implies that (f ∗ h)(x) → 0 as x → ∞. Since f is integrable, the convolution of f with a constant function is Z ∞ Z ∞ (f ∗ r)(x) = f (y) r dy = r f. −∞

−∞

Since g = h + r we therefore have (f ∗ g)(x) = (f ∗ h)(x) + (f ∗ r)(x) → 0 + r

Z

∞

−∞

f.

⊓ ⊔

Now we return to the problem at hand, in which we have g(x) =

x . 1 + |x|

This is a continuous function, and g(x) → 1 as x → ∞. Consequently Lemma B implies that Z ∞

lim (f ∗ g)(x) =

x→∞

f.

−∞

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Solutions to Exercises and Problems from Chapter 5 5.1.5 If |x − y| ≤ 3−k , then we have |ϕ(x) − ϕ(y)| ≤ 2−k . Let k ≥ 0 be the unique integer such that 1 1 < |x − y| ≤ k . 3k+1 3 Then |ϕ(x) − ϕ(y)| ≤

2 2k+1

= 2

 1 log3 2 ≤ 2 |x − y|log3 2 . 3k+1

Hence ϕ is H¨ older continuous with exponent α = log3 2. Fix 0 < β < α = log3 2. If x, y ∈ [0, 1] then |x − y| ≤ 1, so

|ϕ(x) − ϕ(y)| ≤ 2 |x − y|α = 2 |x − y|α−β |x − y|β ≤ 2 |x − y|β . Hence ϕ is H¨ older continuous with exponent β. On the other hand, |ϕ(3−k ) − ϕ(0)| = |2−k − 0| = 2−k = (3−k )log3 2 . It follows ϕ cannot be H¨ older continuous for any exponent α > − log3 2.

Fig. 5.5 f (x) = x − ϕ(x).

5.1.6 One example is −ϕ(x)! Another is f (x) = x−ϕ(x), which is pictured in Figure 5.5. We f ′ (x) = 1− 0 = 1 a.e., but f is not monotone increasing. In particular, f (2/3) = 23 − 21 = 61 yet f (1) = 1 − 1 = 0. 5.1.7 First we prove a little lemma, which can be easily modified for functions whose domain is [a, b] instead of R.

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Lemma. If f : R → R is a strictly increasing bijection, then f is continuous. Proof. Let (a, b) be an open interval in R, and let c = g −1 (a) and d = −1 g (b). Then x ∈ f −1 (a, b) ⇐⇒ f (x) ∈ (a, b) ⇐⇒ f (c) = a < f (x) < b = g(d) ⇐⇒ c < x < d. Hence f −1 (a, b) is the open interval (c, d). Since every open subset of R is a union of open intervals, we conclude that f −1 (U ) is open whenever U is open, so f is continuous. ⊓ ⊔

Fig. 5.6 g(x) = ϕ(x) + x.

(a) The function g(x) = ϕ(x) + x is continuous and monotone increasing since it is a sum of two continuous, monotone increasing functions (see Figure 5.6). If a < b then g(a) = a + ϕ(a) < b + ϕ(b) = g(b), so g is strictly increasing (hence injective). Since g(0) = ϕ(0) + 0 = 0 and g(1) = ϕ(1) + 1 = 2, we conclude that the range of g is [0, 2]. Therefore g : [0, 1] → [0, 2] is a bijection, so its inverse h = g −1 : [0, 2] → [0, 1] is also a bijection. Given points x < y in [0, 2], let g(a) = x and g(b) = y. Then we must have a < b since g is strictly increasing, so h(x) = a < b < h(y). Therefore h is strictly increasing, and consequently is continuous by the lemma. (b) The Cantor set C is compact, and continuous functions maps compact sets to compact sets, so g(C) is compact. Alternatively, U = [0, 1] \ C is open

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and h is continuous, so the inverse image of U under h, which is g(U ) = (g −1 )−1 (U ) = h−1 (U ), must be open. Since g is a bijection, g(U ) = [0, 2] \ g(C), and therefore g(C) is closed. We can write U = [0, 1] \ C as a union of countably many disjoint open intervals In . The function ϕ maps the interval In to a constant cn , and the function h(x) = x maps In to itself. Therefore g(x) = ϕ(x) + x maps In to In + cn . Hence S S g(U ) = g(In ) = (In + cn ). n∈N

n∈N

Then intervals In are disjoint, and g is a bijection, so the intervals In + cn are also disjoint. Therefore |g(U )| =

∞ X

n=1

|g(In )| =

∞ X

n=1

|In | = 1.

(c) Problem 2.4.9 implies that g(C) contains a nonmeasurable subset N. As h is a bijection, we have A = h(N ) ⊆ C. Therefore A has measure zero, so it is a Lebesgue measurable set. (d) Since A is a Lebesgue measurable set, f = χA is a Lebesgue measurable function. Since N is contained in [0, 2] and g is a bijection, g(A) = g(h(N )) = N. Since the interval (0, ∞) contains 1 but not 0, we have {f ◦ h > 0} = (f ◦ h)−1 (0, ∞) = g(f −1 (0, ∞)) = g(A) = N. This set is not Lebesgue measurable, so f ◦ h is not a Lebesgue measurable function. 5.2.4 (a) Since f (x) = x sin(1/x), we have for h 6= 0 that h sin h1 1 f (0 + h) − f (0) = = sin . h−0 h h Since this quantity does not converge as h → 0, we see that f is not differentiable at x = 0. For n ∈ N, we have (  2  2 2  nπ  , n odd, f sin = = nπ nπ nπ 2 0, n even. Therefore,

c Solutions 2019 Christopher Heil

174 N  X f k=1

2 (k + 1)π



−f

Hence, if we set

ΓN =

n

−1, 0,

 2  ≥ kπ

X

j=1,...,N j odd

2 . jπ

2 2 2 o , , ..., ,1 , N π (N − 1)π π

then SΓN ≥

X

j=1,...,N j odd

2 . jπ

Since sup SΓN = ∞, we see that f does not have bounded variation. (b) Since g(x) = x2 sin(1/x2 ), we have for x 6= 0 that g ′ (x) = 2x sin

1 1 1 − cos . x2 x x

Hence g is differentiable at any point x 6= 0. At the point x = 0, given h 6= 0 we have h2 sin h12 g(0 + h) − g(0) 1 = = h sin 2 → 0 as h → 0. h−0 h h Hence g is differentiable at x = 0, and g ′ (0) = 0. However, g ′ is not continuous at x = 0. In fact, g ′ is unbounded on [−1, 1]. However, this does not imply that g has unbounded variation (for example, f (x) = x1/2 has an unbounded derivative on [0, 1], yet f has bounded variation). However, an argument very similar to the one used in part (a) can be used to show that g ∈ / BV[−1, 1]. Specifically, if we take s r r n 2 2 2 o , , ..., ,1 , ΓN = −1, 0, Nπ (N − 1)π π then SΓN ≥

X

j=1,...,N j odd

2 . jπ

This shows that sup SΓN = ∞, which tells us that g does not have bounded variation. Now we will show that g ′ is not integrable on [−1, 1]. For each integer n ∈ N, set αn =



2 4nπ

1/2

=



1 2nπ

1/2

,

βn =



2 (4n − 1)π

1/2

.

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We have αn < βn , and the intervals [αn , βn ] are disjoint and are contained in (0, 1). Since g is differentiable and g ′ is continuous on the interval [αn , βn ], we can apply the Fundamental Theorem of Calculus on that interval to obtain Z

βn

αn

g ′ (x) dx = g(βn ) − g(αn ).

Consequently, Z

1

0

|g ′ (x)| dx ≥ ≥ =

∞ Z X

n=1

βn αn

|g ′ (x)| dx

∞ Z βn X

n=1 ∞ X

n=1



αn

g ′ (x) dx

g(βn ) − g(αn )

∞ X 2 (4n − 1)π 1 = − sin 2nπ (4n − 1)π sin 2 2nπ n=1 =

∞ X

n=1

2 (4n − 1)π

= ∞, so g ′ is not integrable. (c) Since h(x) = x2 sin(1/x), we have for x 6= 0 that h′ (x) = 2x sin For x = 0,

 1 1  1 1 1 − 2 = 2x sin − cos . + x2 cos x x x x x

h(0 + t) − h(0) t2 sin(1/t) = lim = lim t sin(1/t) = 0. t→0 t→0 t→0 t−0 t

h′ (0) = lim

Hence h is differentiable everywhere and h′ is bounded on [−1, 1], although h′ is not continuous at x = 0. Therefore h is Lipschitz on [−1, 1] (we can argue directly, or appeal to Lemma 5.2.5). Now that we know that h is Lipschitz, we can either argue directly or appeal to Lemma 5.2.7 to conclude that h has bounded variation. 5.2.8 Since g is uniformly continuous on [a, b], if we fix ε > 0 then there exists a δ > 0 such that |g(x) − g(y)| < ε whenever |x − y| < δ. Therefore, if x ∈ [a, b] and |h| < δ is such that x + h ∈ [a, b], then

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176

Z x+h Z x+h G(x + h) − G(x) 1 = 1 − g(x) g(t) dt − g(x) dt h h x h x Z x+h 1 ≤ |g(t) − g(x)|dt h x Z 1 x+h ≤ ε dt h x = ε. This shows that G is differentiable at x (one-sided differentiability at the endpoints) and G′ (x) = g(x). Since g is continuous, we therefore have G ∈ C 1 [a, b]. 5.2.11 (a) Taking Γ = {a, b}, we have |f (b) − f (a)| = SΓ ≤ V [f ; a, b].

(b) By induction, it suffices to consider the case where Γ ′ contains one more point than Γ. If Γ = {a = x0 < · · · < xn = b} and Γ ′ = {a = x0 < · · · < xj−1 < x′ < xj < · · · < xn = b}, then the inequality SΓ ≤ SΓ ′ follows from the fact that |f (xj ) − f (xj−1 )| ≤ |f (xj ) − f (x′ )| + |f (x′ ) − f (xj−1 )|. (c) If we choose a partition Γ = {c = x0 < · · · < xn = d} of [c, d], then Γ ′ = {a < x0 < · · · < xn < b} is a partition of [a, b]. 5.2.14 We give a slightly different alternative proof of Lemma 5.2.14 Even if the following quantities are infinite, we always have V + [f ; a, b] = sup SΓ+ ≤ sup SΓ = V [f ; a, b]. Γ

Γ

Similarly, we have V − [f ; a, b] ≤ V [f ; a, b]. Therefore, if V [f ; a, b] is finite, then so are V + [f ; a, b] and V − [f ; a, b]. On the other hand, as nonnegative extended real values, V [f ; a, b] = sup SΓ = sup (SΓ+ + SΓ− ) Γ

Γ

≤ sup SΓ+ + sup SΓ− = V + [f ; a, b] + V − [f ; a, b]. Γ

Γ

In particular, if both V + [f ; a, b] and V − [f ; a, b] are finite, then so is V [f ; a, b]. Further, for every partition Γ, SΓ+ = SΓ− + C,

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where C is the fixed, finite constant C = f (b) − f (a). Hence, even if they are infinite, V + [f ; a, b] = sup SΓ+ = sup (SΓ− + C) = V − [f ; a, b] + C. Γ

Γ

In particular, V + [f ; a, b] is finite if and only if V − [f ; a, b] is finite. Now, just as in Problem 5.2.18, we can find partitions Γk , where each Γk+1 is a refinement of Γk , such that lim SΓ−k = V − [f ; a, b].

k→∞

Since SΓ+k = SΓ−k + C where C = f (b) − f (a) is a finite constant, it follows that lim SΓ+k = lim (SΓ−k + C) = V − [f ; a, b] + C = V + [f ; a, b].

k→∞

k→∞

Further, since SΓ−k + SΓ+k = SΓk , even if these quantities are infinite we have V + [f ; a, b] + V − [f ; a, b] = lim SΓ+k + lim SΓ−k k→∞

k→∞

= lim (SΓ+k + SΓ−k ) k→∞

≤ V [f ; a, b] ≤ V + [f ; a, b] + V − [f ; a, b]. Hence, in the extended real sense, V + [f ; a, b] + V − [f ; a, b] = V [f ; a, b]. If V [f ; a, b] is finite, then we have finite limits in the calculations above, so in this case we can rearrange to obtain V + [f ; a, b] − V − [f ; a, b] = lim SΓ+k − lim SΓ−k k→∞

=

k→∞

lim (SΓ+k k→∞

− SΓ−k ) = f (b) − f (a).

5.2.17 “⇒.” Assume that f ∈ BV[a, b] and let Γ = {a = x0 < · · · < xn = b} be any partition of [a, b]. For any complex number z = zr + izi we have |zr | ≤ Therefore



|zr |2 + |zi |2

1/2

= |z|.

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n n X X f (xj ) − f (xj−1 ) = SΓ [f ]. fr (xj ) − fr (xj−1 ) ≤ SΓ [fr ] = j=1

j=1

This is true for every partition, so it follows that V [fr ; a, b] ≤ V [f ; a, b], and a similar inequality holds for the imaginary part. “⇐.” Assume that fr and fi have bounded variation. For any complex number z = zr + izi we have |z| = |zr + izi | ≤ |zr | + |zi |. Therefore, given any partition Γ = {a = x0 < · · · < xn = b} of [a, b], SΓ [f ] =

n X f (xj ) − f (xj−1 ) j=1

≤ =

n  X  fr (xj ) − fr (xj−1 ) + fi (xj ) − fi (xj−1 ) j=1


SΓ [fr ] + SΓ [fi ] .

Taking the supremum over all partitions, we see that
V [fr ; a, b] + V [fi ; a, b] .

V [f ; a, b] ≤

5.2.18 Case 1: V [f ; a, b] < ∞. Recall that

V [f ; a, b] = sup SΓ . Γ

By definition of the supremum, there must exist some partition Γ1 such that V [f ; a, b] − 1 ≤ SΓ1 ≤ SΓ . Also by definition of the supremum, there must exist a partition Γ2′ such that V [f ; a, b] − 21 ≤ SΓ2′ ≤ SΓ . Let Γ2 = Γ1 ∪ Γ2′ .

Then Γ2 is a partition of [a, b], and it is a refinement of Γ2′ , so we must have V [f ; a, b] −

1 2

≤ SΓ2′ ≤ §Γ2 ≤ SΓ .

Continuing in this way, we obtain partitions Γk , with Γk+1 a refinement of Γk , such that V [f ; a, b] −

1 k

≤ SΓk ≤ SΓ ,

k ∈ N.

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Further, Γk+1 is a refinement of Γk , so SΓk ≤ SΓk+1 ,

k ∈ N.

Thus SΓk ր V [f ; a, b]. Case 2: V [f ; a, b] = ∞. The argument is similar in this case. There must exist some Γ1 such that SΓ1 > 1. Then there exists some Γ2′ such that SΓ2′ > 2. Let Γ2 = Γ1 ∪ Γ2′ . Then 2 < Γ2′ ≤ Γ2 . Continuing in this way, we obtain Γk such that SΓk ր ∞. 5.2.19 (a), (b) These follow by adding and subtracting the two equations V + [f ; a, b] + V − [f ; a, b] = V [f ; a, b], V + [f ; a, b] − V − [f ; a, b] = f (b) − f (a). (c) Suppose f ∈ BV[a, b], and let Γ = {a = x0 < · · · < xn = b} be any partition of [a, b]. Then, by the Reverse Triangle Inequality, SΓ (|f |) =

n n X X f (xj ) − f (xj−1 ) = SΓ (f ). |f (xj )| − |f (xj−1 )| ≤ j=1

j=1

It follows that |f | ∈ BV[a, b].

(d) Let Γ = {a = x0 < · · · < xn = b} be a partition of [a, b]. Then SΓ (αf + βg) = ≤

n X

|(αf + βg)(xk ) − (αf + βg)(xk−1 )|

k=1 n  X k=1

 |α| |f (xk ) − f (xk−1 )| + |β| |g(xk ) − g(xk−1 )

≤ |α| Sγ (f ) + |β| Sγ (g). Taking the supremum over all partitions Γ, it follows that V [αf + βg; a, b] ≤ |α| V [f ; a, b] + |β| V [g; a, b] < ∞,

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so αf + βg ∈ BV[a, b]. (e) If Γ = {a = x0 < · · · < xn = b} is a partition of [a, b], then SΓ (f g) =

n X

k=1

≤

|(f g)(xk ) − (f g)(xk−1 )|

n   X |f (xk ) g(xk ) − f (xk−1 ) g(xk )| + |f (xk−1 ) g(xk ) − f (xk−1 ) g(xk−1 )

k=1

≤ kgk∞

n X

k=1

|f (xk ) − f (xk−1 )| + kf k∞

n X

k=1

|g(xk ) − g(xk−1 )|

≤ kgk∞ SΓ (f ) + kf k∞ SΓ (g). Taking the supremum over all partitions Γ, it follows that V [f g; a, b] ≤ kgk∞ V [f ; a, b] + kf k∞ V [g; a, b] < ∞, so f g ∈ BV[a, b]. (f) Suppose g ∈ BV[a, b] and |g(x)| ≥ δ > 0 for all x ∈ [a, b]. Then given a partition Γ = {a = x0 < · · · < xn = b}, we have SΓ ( g1 )

n X 1 1 = g(xk ) − g(xk−1 ) k=1

n X g(xk−1 ) − g(xk ) = g(xk ) g(xk−1 ) k=1

≤

n 1 X |g(xk−1 ) − g(xk )| δ2 k=1

1 = 2 SΓ (g). δ

Taking the suprema over all partitions Γ, we see that V [ g1 ; a, b] ≤

1 V [g; a, b]. δ2

Hence 1/g ∈ BV[a, b]. Combining this with part (c), it follows that if f has bounded variation then the function f /g belongs to BV[a, b] as well. 5.2.20 (a) Suppose that f is Lipschitz on [a, b], and let K be a Lipschitz constant for f. Then given a partition Γ = {a = x0 < · · · < xn = b}, we have

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SΓ (f ◦ g) = ≤

181 n X

k=1 n X

k=1

|f (g(xk )) − f (g(xk−1 ))| K |g(xk ) − g(xk−1 )|

= K SΓ (g) ≤ K V [g; a, b]. Therefore f ◦ g has bounded variation, and V [f ◦ g; a, b] ≤ K V [g; a, b]. √ Now consider the functions f (x) = x and g(x) = x2 sin2 x1 on the interval [0, 1]. The function f is continuous; in fact it is H¨ older continuous with exponent α = 12 . However, f is not Lipschitz on [0, 1]. To see that g has bounded variation, we compute that if x > 0 then


g ′ (x) = 2x sin2 x1 + 2x2 sin x1 cos x1 − x12 = 2x sin2 x1 − 2 sin x1 cos x1 . For x = 0,

h2 sin2 (1/h) g(0 + h) − g(0) = lim = lim h sin2 (1/h) = 0. h→0 h→0 h→0 h−0 h

g ′ (0) = lim

Hence g is differentiable everywhere and g ′ is bounded on [−1, 1]. Therefore g is Lipschitz on [−1, 1], so it has bounded variation. However, q (f ◦ g)(x) =

x2 sin2 (1/x) = x | sin x1 |,

and this function does not have bounded variation on [0, 1].

(b) Choose any partition Γ = {a = x0 < · · · < xn = b}. Since g is monotone increasing, we have c ≤ g(a) = g(x0 ) ≤ g(x1 ) ≤ · · · ≤ g(xn ) = g(b) ≤ d. Therefore Λ =

 c ≤ g(x0 ) ≤ · · · ≤ g(xn ) ≤ d

is “almost” a partition of [c, d], except that it may contain duplicate points. However, if g(xj−1 ) = g(xj ) for some point, then F (g(xj )) − F (g(xj−1 )) = 0. Hence such terms can simply be omitted from the following calculations, i.e., we can treat Λ just as if it was a partition. We compute that

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SΓ [f ◦ g; a, b] = =

n X j=1

n X j=1

|(f ◦ g)(xj ) − (f ◦ g)(xj−1 )| |f (g(xj )) − f (g(xj−1 ))|

≤ |f (g(x0 )) − f (c)| +

n X j=1

|f (g(xj )) − f (g(xj−1 ))|

+ |f (d) − f (g(xn ))| = SΛ [f ; c, d] ≤ V [f ; c, d]. Taking the supremum over all such partitions Γ, it follows that V [f ◦ g; a, b] ≤ V [f ; c, d]. Therefore f ◦ g has bounded variation. 5.2.21 We are assuming that f is Lipschitz on E, with Lipschitz constant K, but we are not assuming that E is an interval in R. Fix A ⊆ E. If |A|e = ∞ then there is nothing to prove, so we may assume that |A|e < ∞. Given ε > 0, there exists an open set U ⊇ A such that |U | < |A|e + ε. Since U is an open subset of R that has finite measure, we can write U = ∪(an , bn ) as a union of countably many disjoint open intervals, each with finite length. Set En = E ∩ (an , bn ). If x, y ∈ En ⊆ (an , bn ), then |f (x) − f (y)| ≤ K |x − y| ≤ K (bn − an ), so f (En ) is contained in an interval in R of length at most K (bn − an ). Since U ∩ E = ∪En , we therefore have |f (A)|e ≤ |f (U ∩ E)|e X ≤ |f (En )|e n

≤

X n

K (bn − an )

= K |U | < K (|A|e + ε). Since ε is arbitrary, we conclude that |f (A)|e ≤ K |A|e .

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5.2.22 (a) Since f is even, it suffices to prove that f ∈ BV[0, 1] if and only if a > b. “⇒.” We will prove the contrapositive statement, which is that if a ≤ b, then f ∈ / BV[0, 1]. Assume that 0 < a ≤ b. For each even positive integer N, set ΓN

1/b  1/b  1/b    2 2 2 , , ..., ,1 . = 0, Nπ (N − 1)π π

That is, ΓN = {0, xN , . . . , x1 , 1} where xj =



2 jπ

1/b

.

Note that if j is odd, then |f (xj )| =



2 jπ

a/b

sin

jπ 2

=



2 jπ

a/b

,

and also, since j + 1 is even, we have f (xj+1 ) =



2 (j + 1)π

a/b

sin (j+1)π = 0. 2

Therefore SΓN ≥

≥

=

X

|f (xj+1 ) − f (xj )|

X



j=1,...,N −1 j odd

j=1,...,N −1 j odd

 a/b 2 π

2 jπ

a/b

X

j −a/b .

j=1,...,N −1 j odd

Since 0 < a/b ≤ 1, it follows that supN SΓN = ∞, and therefore f does not have bounded variation. “⇐.” Assume that a > b. We will prove that f ∈ BV[0, 1]. For each odd integer j ∈ N, set xj =



2 jπ

1/b

.

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Since f is smooth and hence Lipschitz on the [x1 , 1], it suffices to show that f has bounded variation on the interval [0, x1 ]. Given any partition Γ of [0, x1 ], let Γ ′ be a refinement of Γ that includes the points x1 , x3 , . . . , xN , with N an odd integer chosen large enough so that xN is the smallest positive point in Γ ′ . Note that f is either monotone increasing or monotone decreasing on the interval [xj+2 , xj ]. Therefore X SΓ ≤ SΓ ′ = |f (xN ) − f (0)| + |f (xj ) − f (xj+2 )| j=1,...,N −2 j odd

≤ 1 +

≤ 1 +

X

j=1,...,N −2 j odd

X

j=1,...,N −2 j odd

≤ 1 + 2

 a/b a/b  2 2 − jπ (j + 2)π



2 jπ

a/b

+



2 (j + 2)π

a/b 

a/b ∞  X 2 jπ j=1

= C. The number C is a constant that is independent of N, and it is finite since a/b > 1. Taking the supremum over all partitions Γ, we see that V [f ; 0, x1 ] ≤ C < ∞. (b) Recall that α =

b . b+1

Note that 0 < α < 1, and also 0 < α =

b < b. b+1

Since f is even, it suffices to prove that f is H¨ older continuous with exponent α on [0, 1]. This is, we must show that there is a constant C > 0 such that for all points 0 ≤ x < y ≤ 1 we have |f (y) − f (x)| ≤ C |y − x|α . We will show that we can take C = max{2b, 2b + 1}. For x > 0, note that f (x) = xb sin x−b , and therefore |f (x)| ≤ xb .

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Case 1: 0 = x < y ≤ 1. Observe that 0 < α < b and y ≤ 1, so y a ≤ y α . Therefore |f (y) − f (0)| = |f (y)| ≤ y b ≤ y α ≤ C |y − 0|α . Case 2: 0 < x < y ≤ 1 and xb+1 < h where h = y − x. In this case we have 1 x < h b+1 and therefore

b

xb < h b+1 = hα . Also, since h < 1 and α < b, we have hb ≤ hα , and therefore y b = (x + h)b ≤

2 max{x, h}

Consequently,


b

= 2b max{xb , hb } ≤ 2b hα .

|f (y) − f (x)| ≤ |f (y)| + |f (x)| ≤ y b + xb ≤ 2 b hα + hα = (2b + 1) hα ≤ C |y − x|α , Case 3: 0 < x < y ≤ 1 and xb+1 ≥ h where h = y − x. For any 0 < t ≤ 1, we have |f ′ (t)| = btb−1 sin t−1 + tb (cos t−b )(−bt−b−1 ) = btb−1 sin t−b − bt−1 cos t−b ≤ btb−1 + bt−1

=

btb + b 2b ≤ . t t

By the Mean Value Theorem, there exists a point x < ξ < y such that f (y) − f (x) = f ′ (ξ). y−x Recalling that h = y − x, it follows that |f (y) − f (x)| = h |f ′ (ξ)| ≤ h

2bh 2b ≤ . ξ x

(A)

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Now, h ≤ xb+1 , so

1

h b+1 ≤ x,

and therefore

1 1 ≤ h− b+1 . x

Consequently,

b 1 h ≤ h h− b+1 = h b+1 = hα . x Therefore we can continue equation (A) as follows:

|f (y) − f (x)| ≤

2bh ≤ 2bhα ≤ C |x − y|α . x

Combining Cases 1, 2, and 3, we have covered all possible choices of points 0 ≤ x < y ≤ 1. Therefore we have shown that f is H¨ older continuous on [0, 1] with exponent α. (c) This follows from the fact that

b b+1

can take any value between 0 and 1.

5.2.23 (a) We are given complex-valued functions fn on [a, b] that converge pointwise to a limit f. Let Γ = {a = x0 < · · · < xn = b} be any partition of [a, b]. Then, using the discrete version of Fatou’s Lemma, we compute that SΓ [f ; a, b] =

n X j=1

=

n X j=1

|f (xj ) − f (xj−1 )| lim inf |fn (xj ) − fn (xj−1 )| n→∞

≤ lim inf n→∞

n X j=1

|fn (xj ) − fn (xj−1 )|

= lim inf SΓ [fn ; a, b] n→∞

≤ lim inf V [fn ; a, b]. n→∞

Taking the supremum over all partitions Γ, it follows that V [f ; a, b] ≤ lim inf V [fn ; a, b]. n→∞

(b) For an example of a pointwise convergent sequence of functions of bounded variation whose limit is not of bounded variation, consider [a, b] = [0, 1] and ( 1 x sin x1 , kπ ≤ x ≤ 1, fk (x) = 1 . 0, 0 ≤ x ≤ kπ

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5.2.24 First proof. Define x ≥ a.

V (x) = V [f ; a, x],

We have V (a) = 0, and since we defined f (t) = f (b) for t > b, we have V (x) = V (b) for all x > b. Since f (t) = f (a) for t < a, we can set V (x) = 0 for x < a. By Exercise 5.2.11(a), given any x < y we have |f (y) − f (x)| ≤ V [f ; x, y] = V [f ; y, a] − V [f ; x, a] = V (y) − V (x). Assume first that t > 0. Since V is monotone increasing, we compute that kTt f − f k1 = ≤ =

Z

|f (x − t) − f (x)| dt

a

Z

b

a

Z

a

=

b

Z

b


V (x) − V (x − t) dt V −

b+t

V −

b

=

Z

b

Z

b−t

V

a−t

Z

a

V

a−t

b+t

V (b) −

Z

a

0 = t V (b) = t V [f ; a, b].

a−t

A similar argument applies if t < 0. Second proof. This argument is similar to the first proof, except that we split f into monotone increasing parts. Assume first that f ∈ BV[a, b] is real-valued. Since f has bounded variation, we can write f = g − h where g and h are bounded and monotone increasing on [a, b]. Since f is bounded, the same is true of g and h. Therefore, by adding a constant to both g and h, we may assume that g and h are each nonnegative. We extend these functions by declaring that g(x) = g(a), x < a,

g(x) = g(b), x > b

h(x) = h(a), x < a,

h(x) = h(b), x > b.

and Then g and h are monotone increasing on R, and f = g − h on all of R. If t > 0 then, using the fact that g and h are increasing and recalling that we are computing the L1 -norms on the interval [a, b], we see that

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kTt f − f k1 = kTt g − Tt h − g + hk1 ≤ kTt g − gk1 + kTt h − hk1 Z b Z = |g(x − t) − g(x)| dx + a

Z

=

b

Z

Z

b−t

g +

a−t

Z

b

b−t

≤

Z

g −

a

=

Z


g(x) − g(x − t) dx +

a

=

|h(x − t) − h(x)| dx

a

b

Z

b

g −

g +

a−t

b−t

|g(b)| +

Z

h −

Z

|g(a)| +

Z

b−t

h

a−t

h −

a

a−t

Z

b

b−t


h(x) − h(x − t) dx

a

b

a

a

b

Z

b

Z

a

h

a−t

b

b−t

|h(b)| +

Z

a

a−t

|h(a)|

= Ct, where C = |g(b)| + |g(a)| + |h(b)| + |h(a)|. Similarly, kT−t f − f k1 = kT−t g − T−t h − g + hk1 ≤ kT−t g − gk1 + kT−t h − hk1 Z b Z b |h(x + t) − h(x)| dx |g(x + t) − g(x)| dx + = a

a

=

Z

b

a

=

Z


g(x + t) − g(x) dx +

b+t

a+t

=

Z

b+t

g −

b

≤

Z

g −

b+t

b

Z

b

g +

a

Z

Z

a+t

a

|g(b)| +

g + Z

a

b+t

a+t

Z

b

a+t

Z

b

a

h −


h(x + t) − h(x) dx

Z

h

a

b+t

h −

|g(a)| +

b

Z

b

Z

a+t

h

a

b+t

|h(b)| +

Z

a

a+t

|h(a)|

= Ct, with the same C. Hence for all t ∈ R we have kTt f − f k1 ≤ C|t|. Now consider the case of a complex-valued function f ∈ BV[a, b]. Write f = fr + ifi where fr and fi are real-valued. By the above work, there exist constants Cr and Ci such that kTt fr − fr k1 ≤ Cr |t|

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189

and kTt fi − fi k1 ≤ Ci |t|. Hence kTt f − f k1 = kTt (fr + ifi ) − (fr + ifi )k1 = kTt fr + iTt fi − fr − ifi k1 ≤ kTt fr − fr k1 + kTt fi − fi k1 ≤ Cr |t| + Ci |t|. Third proof. For simplicity, assume that t > 0, and let N be large enough that [a, b + t] ⊆ [−N t, N t]. Then Z ∞ kTt f − f k1 = |f (x − t) − f (x)| dx −∞

=

Z

b+t

|f (x − t) − f (x)| dx

a

=

Z

since f (x − t) = f (x) outside [a, b + t] Nt

−N t

=

N −1 X

k=−N

≤ ≤

=

N −1 X

k=−N N −1 X

k=−N

N −1 X

k=−N

|f (x − t) − f (x)| dx Z

(k+1)t

|f (x − t) − f (x)| dx

kt

Z

(k+1)t

V [f ; x − t, x] dx

kt

Z

(k+1)t

kt

V [f ; (k − 1)t, (k + 1)t] dx since (k − 1)t ≤ x − t ≤ x ≤ (k + 1)t

  t V [f ; (k − 1)t, kt] + V [f ; kt, (k + 1)t]

= t V [f ; (−N − 1)t, (N − 1)t] + t V [f ; −N t, N t] additivity of V on disjoint intervals

≤ t V [f ; a, b] + t V [f ; a, b]

since f is constant outside [a, b]

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= 2t V [f ; a, b]. 5.2.25 We can apply previous problems, or argue directly as follows. Let Γ = {a = x0 < x1 < · · · < xn = b} be any finite partition of [a, b]. Using the series version of Tonelli’s Theorem to interchange the sums, we compute that SΓ (f ) =

n X j=1

=

|f (xj ) − f (xj−1 )|

n X X ∞
f (x ) − f (x ) k j k j−1 j=1 k=1

≤ =

n X ∞ X fk (xj ) − fk (xj−1 ) j=1 k=1

∞ X n X fk (xj ) − fk (xj−1 ) k=1 j=1

=

∞ X

SΓ (fk )

k=1

≤

∞ X

V [fk ; a, b].

k=1

Taking the supremum over all partitions of [a, b], we see that V [f ; a, b] ≤

∞ X

k=1

V [fk ; a, b] < ∞.

Therefore f ∈ BV[a, b], and we have the desired inequality relating the variation of f to the sum of the variations of the functions fk . 5.2.26 (a) By definition we have 0 ≤ V [f ; a, b] < ∞ for all f ∈ BV[a, b]. A direction calculation shows that V [cf ; a, b] = |c|V [f ; a, b]. By Problem 5.2.19(d), V [f + g; a, b] ≤ V [f ; a, b] + V [g; a, b]. Consequently kf k = V [f ; a, b] defines a seminorm on BV[a, b]. For simplicity of notation in this proof, let

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kf ku =

sup |f (x)|

x∈[a,b]

be the uniform norm (not the L∞ -norm). All functions in BV[a, b] are bounded, so kf kBV = kf ku + V [f ; a, b] is nonnegative and finite for all f ∈ BV[a, b]. If kf kBV = 0, then we have kf ku = 0, and therefore sup |f (x)| = 0. This implies that f (x) = 0 for every x. Given f ∈ BV[a, b] and c ∈ C, we have kcf kBV = kcf ku + V [cf ; a, b] = |c| kf ku + |c| V [f ; a, b] = |c| kf kBV . Given f, g ∈ BV[a, b], it follows from Problem 5.2.19(d) that kf + gkBV = kf + gku + V [f + g; a, b] ≤ kf ku + kgku + V [f ; a, b] + V [g; a, b] = kf kBV + kgkBV . Therefore k · kBV is a norm on BV[a, b]. (b) Now we must show that BV[a, b] is complete with respect to this norm. We will use the equivalent characterization of completeness given in Theorem 1.2.8. That is, we will show that every absolutely convergent series in BV[a, b] converges in the norm of that space. Suppose that fk ∈ BV[a, b], and ∞ X

k=1

Then we have

kfk kBV < ∞.

∞ X

k=1

kfk ku < ∞,

P so we know that the series fk converges absolutely with respect to the uniform norm. In particular, it converges pointwise, so we can set f (x) =

∞ X

fk (x),

k=1

x ∈ [a, b].

By hypothesis, we have ∞ X

k=1

V [fk ; a, b] ≤

∞ X

k=1

kfk kBV < ∞.

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It therefore follows from Problem 5.2.25Pthat f ∈ BV[a, b]. It remains to show that the series fk converges to f in the norm of BV[a, b]. Given N ∈ N, let gN (x) = f (x) −

N X

∞ X

fk (x) =

k=1

fk (x).

k=N +1

This series converges pointwise, and it follows from part (a) that ∞ X

V [gN ; a, b] ≤

V [fk ; a, b].

k=N +1

Assuming that we really can identify gN with f −

N X

f − fk

k=1

BV

PN

k=1

fk , we have

= kgN kBV = kgN ku + V [gN ; a, b]

∞ ∞ X

X

≤ fk + u

k=N +1

=

∞ X

k=N +1

→ 0

V [fk ; a, b]

k=N +1

kfk kBV

as N → ∞.

PN To avoid the issue of whether gN equals f − k=1 fk in the space BV[a, b], we can argue directly, similarly to how we did in Problem 5.2.25. Choose a partition Γ = {a = x0 < x1 < · · · < xn = b}. Then we have 

SΓ f −

N X

k=1

fk



N N n X X X fk (xj ) + fk (xj−1 ) = f (xj ) − f (xj−1 ) − j=1

=

j=1

=

k=1

n X

f (xj ) −

n X



∞ X

j=1 k=N +1

N X

k=1

fk (xj ) − f (xj−1 ) +

fk (xj ) −

∞ X

k=N +1

fk (xj−1 )

k=1

N X

k=1

fk (xj−1 )

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∞ X

193

n X

k=N +1 j=1 ∞ X

=

|fk (xj ) − fk (xj−1 )|

SΓ (fk )

k=N +1

≤

∞ X

V [fk ; a, b].

k=N +1

The important fact in thePfirst steps of the preceding calculation is that we can work with the series fk in a pointwise fashion. Taking the supremum over all partitions Γ, we see that 

V f−

N X

fk ; a, b

k=1



≤

∞ X

V [fk ; a, b].

k=N +1

Hence

N X

f −

f k

BV

k=1

  N N X X

+ V f− f ; a, b = f − f k k

k=N +1

=

∞ X

k=N +1

Thus the series f =

P

u

k=1

∞

X

≤ f k +

→ 0

u

k=1

∞ X

V [fk ; a, b]

k=N +1

kfk kBV

as N → ∞.

fk converges with respect to the norm of BV[a, b].

(c) The proof that ||| · |||BV is a seminorm is similar to that for k · kBV . To prove the uniqueness requirement, assume that |||f |||BV = 0. Then V [f ; a, b] = 0

and

f (a) = 0.

The fact that the total variation is zero implies that f is constant. Since f (a) = 0, this constant must be zero. Therefore f = 0. This shows that ||| · |||BV is a norm on BV[a, b]. By definition, |||f |||BV = V [f ; a, b] + |f (a)| ≤ V [f ; a, b] + sup |f (x)| ≤ kf kBV . x∈[a,b]

This gives one of the desired inequalities for the norm equivalence. Given f ∈ BV[a, b], by the definition of the total variation we have

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|f (x) − f (a)| ≤ V [f ; a, b],

x ∈ [a, b].

Therefore sup |f (x)| ≤

x∈[a,b]

sup



x∈[a,b]

 |f (x) − f (a)| + |f (a)|

≤ V [f ; a, b] + |f (a)| = |||f |||BV . Hence kf kBV = V [f ; a, g] + sup |f (x)| ≤ |||f |||BV + |||f |||BV = 2 |||f |||BV . x∈[a,b]

Therefore the two norms are equivalent. 5.2.27 (a) Choose any ε > 0. Then, by definition of V [f ; a, b], there exists a partition Γ ′ = {a = x′0 < · · · < x′n = b} such that V [f ; a, b] −

ε ≤ SΓ ′ ≤ V [f ; a, b]. 2

Since f is continuous on the compact domain [a, b], it is uniformly continuous, and therefore we can find an η > 0 such that for x, y ∈ [a, b] we have |x − y| < η =⇒ |f (x) − f (y)|
0 such that |Γ | < δ2 =⇒ V [f ; a, b] − ε ≤ SΓ [f ; a, b] ≤ V [f ; a, b]. Let δ = min{δ1 , δ2 }.

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Fix x ∈ [a, b], and assume that x < y < x + δ. Let Γ0 = {x = x0 < x1 < · · · < xn = y} be a partition of [x, y]. Note that |Γ0 | < δ. Let Γ1 be any partition that extends Γ0 to a partition of [a, b] and satisfies |Γ1 | < δ. That is, Γ1 should have the form Γ1 = {a = y0 < · · · < yj < x0 < · · · < xn < z1 < · · · < zk = b}, and we should have yi − yi−1 < δ and zi − zi−1 < δ for each i, as well as x0 − yj < δ and z1 − xn < δ. Now let Γ2 = {a = y0 < · · · < yj < x0 < xn < z1 < · · · < zk = b}. This is a partition of [a, b], and |Γ2 | < δ. SΓ0 [f ; x, y] =

n X j=1

|f (xj ) − f (xj−1 )|

= SΓ1 [f ; a, b] − SΓ2 [f ; a, b] + |f (xn ) − f (x0 )| ≤ V [f ; a, b] − (V [f ; a, b] − ε) + ε = 2ε. Taking the supremum over all such partitions Γ0 , it follows that whenever x < y < x + δ, we have V (y) − V (x) = V [f ; a, y] − V [f ; a, x] = V [f ; x, y] ≤ 2ε. A similar argument applies from the left, so we conclude that V is continuous at x. (c) Case 1: Real-Valued Functions. Assume that f ∈ C 1 [a, b] is real-valued on [a, b]. Fix any particular partition Γ = {a = x0 < · · · < xm = b}. By the Mean Value Theorem, there exist points ξk ∈ (xk−1 , xk ) such that f (xk ) − f (xk−1 ) = f ′ (ξk ) (xk − xk−1 ). Therefore SΓ =

m X

k=1

|f (xk ) − f (xk−1 | =

m X

k=1

|f ′ (ξk )| (xk − xk−1 ).

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The right-most sum on the preceding line is a Riemann sum for f ′ is continuous, we therefore have lim SΓ = lim

|Γ |→0

|Γ |→0

m X

k=1

|f ′ (ξk )| (xk − xk−1 ) =

Z

b

a

Rb a

|f ′ |. Since

|f ′ |.

On the other hand, part (a) tells us that lim SΓ = V [f ; a, b].

|Γ |→0

Therefore V [f ; a, b] =

Rb a

|f ′ |.

Case 2: Complex-Valued Functions. The argument is similar for complexvalued functions, but we must be careful because the MVT only holds for real-valued functions. Assume that f ∈ C 1 [a, b] is complex-valued, and write f = g + ih where g and h are real-valued. Fix any ε > 0. By part (a), SΓ → V [f ; a, b], so there exists a δ1 > 0 such that |Γ | < δ1 =⇒ V [f ; a, b] − SΓ < ε.

Since h′ is uniformly continuous, there exists a δ2 > 0 such that |x − y| < δ =⇒ |h′ (x) − h′ (y)|
0 Z |Γ | < δ3 =⇒

a

b

|f ′ | − RΓ < ε,

where RΓ denotes any Riemann sum corresponding to the partition Γ. Let δ = min{δ1 , δ2 , δ3 }, and consider any partition Γ = {a = x0 < · · · < xm = b} whose mesh size satisfies |Γ | < δ. By the Mean Value Theorem, there exist points ξk , ηk ∈ (xk−1 , xk ) such that g(xk ) − g( xk−1 ) = g ′ (ξk ) (xk − xk−1 ) and h(xk ) − h( xk−1 ) = h′ (ξk ) (xk − xk−1 ). Therefore

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SΓ =

m X

k=1

=

m X

k=1

=

|f (xk ) − f (xk−1 | |g(xk ) − g(xk−1 + ih(xk ) − ih(xk−1 )|

m X ′ g (ξk ) (x − xk−1 ) + ih′ (ηk ) (xk − xk−1 ) k=1

=

m X ′ g (ξk ) + ih′ (ηk ) (xk − xk−1 ) k=1

=

m  X

g ′ (ξk )2 + h′ (ηk )2

k=1

1/2

(xk − xk−1 ).

This is close to RΓ

m  1/2 X (xk − xk−1 ) = g ′ (ξk )2 + h′ (ξk )2 k=1

=

m X ′ g (ξk ) + ih′ (ξk ) (xk − xk−1 )

k=1

=

m X

k=1

|f ′ (ξk )| (xk − xk−1 ),

R which is a Riemann sum for |f ′ |. In fact, because the meshsize of Γ is less than δ, we have Z b ′ |f | − R Γ < ε. a

We need to determine how close SΓ is to RΓ . Since ξk and ηk both belong to [xk−1 , xk ], we have |ξk − ηk | ≤ xk − xk−1 ≤ |Γ | < δ. Therefore

ε . b−a Given any numbers u, v, w ∈ R, it follows from the Reverse Triangle Inequality for the ℓ2 -norm for points in R2 that 2 (u + v 2 )1/2 − (u2 + w2 )1/2 = k(u, v)k2 − k(u, w)k2 |h′ (ξk ) − h′ (ηk )|
0 such that D+ f ≥ δ on (a, b). Fix a < x < y < b. Since f is continuous on the closed bounded interval [x, y], it has a max at some point in that interval, say at x0 . Suppose that x ≤ x0 < y. Then for all x0 < t < y we have f (t) − f (x0 ) ≤ 0, t − x0 and therefore D+ f (x0 ) = lim sup t→x+ 0

f (t) − f (x0 ) ≤ 0. t − x0

This is a contradiction. Therefore f must achieve its maximum on [x, y] at the point y. Consequently f (x) ≤ f (y). This shows that f is monotone increasing on (a, b). Since f is continuous, it follows that f is monotone increasing on all of [a, b]. Now suppose that we just have D+ f ≥ 0 on (a, b). Fix δ > 0, and let g(x) = f (x) + δx, which is continuous. The limsup of a sum is not the sum of limsups in general, but it is if one of the limsups is a limit. That is the case here: D+ g(x) = lim sup t→x+

g(t) − g(x) t−x

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= lim sup t→x+

f (t) − f (x) δt − δx) + lim+ t−x t−x t→x

= D+ f (x) + δ ≥ δ.

Our work above therefore implies that g is monotone increasing on [a, b]. Hence, given any a ≤ x < y ≤ b we have f (x) + δx = g(x) ≤ g(y) = f (y) + δy. Rearranging, f (y) − f (x) ≥ δ(y − x). Letting δ → 0, we obtain f (y)−f (x) ≥ 0. Thus f (x) ≤ f (y), so f is monotone increasing on [a, b]. 5.4.7 Since 0 ≤ χ[rn ,1] (x) ≤ 1, the series defining f converges at every point, and we have 0 ≤ f ≤ 1. Since each term 2−n χ[rn ,1] is monotone increasing for every n, the function f is monotone increasing. Right-continuity follows from the fact that the uniform limit of rightcontinuous functions is right-continuous. To prove this directly, fix any x ∈ [0, 1]. If x = 1 then there is nothing to prove, so we may assume that 0 ≤ x < 1. Choose ε > 0, and let N ∈ N be large enough that ∞ X

2−n < ε.

n=N +1

For each n = 1, . . . , N, the function 2−n χ[rn ,1] is right-continuous at x. Therefore, there exists some δn > 0 such that x + δn < 1 and x < y < x + δn

=⇒

|2−n χ[rn ,1] (x) − 2−n χ[rn ,1] (y)|
rn then χ[rn ,1] = 1 on an open interval around x, and if x < rn then χ[rn ,1] = 0 on an open interval around x. Since the uniform limit of continuous functions is continuous, it follows that f is continuous at x. Remark: Since every term 2−n χ[rn ,1] is left-continuous at the point x = 1, the function f is left-continuous at that particular rational point. 5.4.8 (a) Since

P

∞ X

n=1

kfn ku =

∞ X

n=1

2−n < ∞,

the series f = fn converges absolutely with respect to the uniform norm. Since C[0, 1] is a Banach space with respect to the uniform norm, it follows P that the series f = fn must converge with respect to the uniform norm.

(b) The function f is continuous by part (a). Each function fn is monotone increasing on [0, 1]. Since uniform convergence implies pointwise convergence, if we fix 0 ≤ x < y ≤ 1 then we have f (x) =

∞ X

n=1

fn (x) ≤

∞ X

fn (y) = f (y).

n=1

Therefore f is monotone increasing on [0, 1]. (c) Fix 0 ≤ x < y ≤ 1. Then there exists some n such that x ≤ an < bn ≤ y. Therefore

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f (x) ≤ f (an ) =

∞ X

(f is monotone increasing)

fk (an )

k=1

= fn (an ) +

X

fk (an )

k6=n

= 0 +

X

fk (an )

k6=n

< 2−n +

X

fk (bn )

(f is monotone increasing)

k6=n

= fn (bn ) +

X

fk (bn )

k6=n

= f (bn ) ≤ f (y)

(f is monotone increasing).

Therefore f is strictly increasing on [0, 1]. (d) Lemma 5.4.4 implies that f is differentiable at almost every point and f ′ (x) =

∞ X

fn′ (x) a.e.

n=1

P ′ Let Z be the set of points where f ′ (x) is not equal to fn (x). Then Z has measure zero. Further, for each individual n, the set Zn = {fn′ 6= 0} has measure zero. Consequently, A = Z ∪ Z1 ∪ Z2 ∪ · · · has measure zero, and if x ∈ / A then f ′ (x) =

∞ X

fn′ (x) =

n=1

∞ X

0 = 0.

n=1

Thus f ′ = 0 a.e. 5.4.9 (a) If x ∈ A, then we have D+ f (x) = lim sup h→0+

f (x + h) − f (x) = ∞. h

Consequently there must exist hn > 0 such that hn → 0 and

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f (x + hn ) − f (x) → ∞. hn Therefore, for all n large enough we have f (x + hn ) − f (x) > M. hn This implies that the interval [x, x + hn ] belongs to B for all large enough n. Hence B is a Vitali cover of A. (b) The Vitali Covering Lemma, combined with the extra conclusion given in equation (5.15), implies that there exist finitely many disjoint intervals in B, say [xk , yk ] for k = 1, . . . , N, such that N A ∩ S [xk , yk ] > |A|e − ε. k=1

e

Consequently N X

k=1

N N S S (yk − xk ) = [xk , yk ] > A ∩ [xk , yk ] > |A|e − ε. k=1

k=1

e

e

(c) Since [xk , yk ] belongs to B, we have

N N   X X
(yk − xk ) > M |A|e − ε . f (yk ) − f (xk ) > M k=1

k=1

(d) Since f is monotone increasing, the intervals [f (xk ), f (yk )] are nonoverlapping. The sum of their lengths cannot be any more than the length of the interval [f (a), f (b)] in which they are contained. Hence M |A|e − ε





0. Since f is uniformly continuous, there exists a δ > 0 such that |f (x) − f (y)| < ε whenever kx − yk < δ. Consequently, if |h| < δ, then for every x ∈ Rd ,

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Z Z 1 1 e |f (x) − fh (x)| = f (x) dt − f (t) dt |Bh (x)| Bh (x) |Bh (x)| Bh (x) Z 1 |f (x) − f (t)| dt ≤ |Bh (x)| Bh (x) Z 1 ≤ ε dt = ε. |Bh (x)| Bh (x) Taking the supremum over all x, it follows that kf − feh ku ≤ ε. Rx 5.5.12 We are given f ∈ L1 [a, b] such that a f = 0 for every a ≤ x ≤ b. Extend f by zero outside of [a, b]. The Lebesgue Differentation Theorem implies that for a.e. x ∈ [a, b] we have 1 h→0 h

f (x) = lim

Z

x+h

f.

x

However, if x ∈ [a, b] and h is small enough that x + h ∈ [a, b], then we have Z

x+h

x

f =

Z

x+h

a

f −

Z

a

x

f = 0 − 0 = 0.

Therefore f (x) = 0 for a.e. x. 5.5.13 Set kh =

1 χB (0) . |Bh (0)| h

Then feh is the convolution of f with kh , since Z Z 1 feh (x) = f (x − t) dt = f (x − t) kh (t) dt = (f ∗ kh )(x). |Bh (0)| Bh (0) Rd

The integral on the preceding line is well-defined because f is locally integrable and kh has compact support. The proof that f ∗ kh is continuous is similar to other problems on convolution (for example, the convolution of an integrable function with a bounded function is continuous). However, we will give a simple direct proof. Suppose that xk → x. Since the measure of the ball Bh (xk ) does not depend on k, it follows that for a.e. t we have 1 1 χBh (xk ) (t) f (t) → χB (x) (t) f (t). |Bh (xk )| |Bh (x)| h Set

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g(t) =

1 χB (x) (t) |f (t)|, |Bh (0)| h+1

and note that g is integrable. If k is large enough then kx − xk k < 1. Hence for all large enough k we have Bh (xk ) ⊆ Bh+1 (x), and therefore 1 1 χB (x ) (t) |f (t)| ≤ χB (x) (t) |f (t)| = g(t) ∈ L1 (Rd ). |Bh (xk )| h k |Bh (0)| h+1 Hence the Dominated Convergence Theorem implies that Z 1 e lim fh (xk ) = lim f (t) dt k→∞ k→∞ |Bh (xk )| B (x ) h k Z 1 f (t) dt = |Bh (x)| Bh (x) = feh (x).

Therefore feh is continuous at x. Finally, if f is integrable, then the submultiplicative property of the L1 norm with respect to convolution implies that kfeh k1 = kf ∗ kh k1 ≤ kf k1 kkh k1 = kf k1 .

Again, we can prove this directly. Note that t ∈ Bh (x)

⇐⇒

kx − tk < h

⇐⇒

x ∈ Bh (t).

Therefore χBh (x) (t) = χBh (t) (x). Applying Tonelli’s Theorem, we compute that Z kfeh k1 = |feh (x)| dx Rd

=

≤ = = =

Z 1 f (t) dt dx Rd |Bh (x)| Bh (x) Z Z 1 |f (t)| χBh (x) (t) dt dx |Bh (0)| Rd Rd Z Z 1 |f (t)| χBh (t) (x) dx dt |Bh (0)| Rd Rd Z Z 1 χBh (t) (x) dx dt |f (t)| |Bh (0)| Rd Rd Z 1 |f (t)| |Bh (t)| dt |Bh (0)| Rd

Z

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Z

=

Rd

207

|f (t)| dt

= kf k1 . 5.5.14 Since the integral of g is nonzero, g cannot be zero a.e. Therefore |g| is not zero a.e., and consequently kgk1 > 0. Given h > 0, the function gh is integrable since it is a dilation of g and g is integrable. In fact, by making a linear change of variables and applying Problem 4.3.10 we have Z Z Z −d |g(t)| dt = kgk1 . |gh (t)| dt = h |g(t/h)| dt = kgh k1 = Rd

Similarly, Z

Rd

Rd

gh (t) dt = h

−d

Rd

Z

g(t/h) dt =

Rd

Z

g(t) dt = 1. Rd

Since f and gh are both integrable, the results of Section 4.6.3 imply that f ∗ gh is defined a.e. and is integrable. We proceed similarly to the proof of Theorem 5.5.3. By hypothesis, g is zero outside of some ball. By taking R large enough, it follows that g is identically zero outside of the ball BR (0). Consequently gh is zero outside of BRh (0). Using Tonelli’s Theorem to interchange the order of integration, we can estimate the L1 -norm of f − f ∗ gh as follows: Z kf − f ∗ gh k1 = |f (x) − (f ∗ gh )(x)| dx Rd

=

Z

Rd

≤ =

Z

Rd

Z

Z f (x)

Z

=

Z

gh (t) dt −

R

f (x − t) gh (t) dt dx d

|f (x) − f (x − t)| |gh (t)| dt dx

Rd

BRh (0)

Rd

Z

Z

ktk 0, then there is some δ > 0 such that

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ktk < δ

kf − Tt f k1
0



1 |Bh (x)|

Z

Bh (x)

|g(t)| dt



= Mf (x) + M g(x), Similarly, M (cf )(x) = sup h>0

1 |Bh (x)|

Z

Bh (x)

1 = |c| sup |B h (x)| h>0

Z

|cf (t)| dt

Bh (x)

|f (t)| dt

= |c| Mf (x). 5.5.16 Given x ∈ Rd , since 0 ≤ fn ≤ fn+1 a.e., we have Z Z 1 1 fn ≤ sup fn+1 = Mfn+1 (x). Mfn (x) = sup h>0 |Bh (x)| Bh (x) h>0 |Bh (x)| Bh (x)

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Consequently, {Mfn (x)}n∈N is a monotonically increasing sequence of scalars for each x. By the same reasoning, Mf (x) is an upper bound to this sequence, since Z Z 1 1 fn ≤ sup f = Mf (x). Mfn (x) = sup h>0 |Bh (x)| Bh (x) h>0 |Bh (x)| Bh (x) An increasing sequence of scalars has a limit in the extended real sense, so we have shown that lim Mfn (x) = sup Mfn (x) ≤ Mf (x).

n→∞

n

It remains to show that equality holds. Applying the Monotone Convergence Theorem (Fatou’s Lemma will also work), we see that Z 1 Mf (x) = sup f h>0 |Bh (x)| Bh (x)   Z 1 fn = sup lim h>0 n→∞ |Bh (x)| Bh (x)   ≤ sup sup Mfn (x) h>0

n

= sup Mfn (x) n

= lim Mfn (x). n→∞

5.5.17 Since Bh (x) is one of the open balls that contain x, we immediately obtain Mf (x) ≤ M ∗ f (x). Suppose that B is any open ball that contains x. Then B = Br (y) for some point y and some radius r > 0, and kx − yk < r. Let z be any point in B. Then ky − zk < r, so kx − zk ≤ kx − yk + ky − zk < r + r = 2r. Hence z ∈ B2r (x). This shows that B ⊆ B2r (x). Note that |B| = Cd rd

and

|B2r (x)| = Cd (2r)d = Cd 2d rd

where Cd is a constant that depends only on the dimension. Therefore |B| = 2−d |B2r (x)|, so Z Z 1 1 1 |f | ≤ −d |f | ≤ −d Mf (x) = 2d Mf (x). |B| B 2 |B2r (x)| B2r (x) 2 Taking the supremum over all open balls B that contain x, it follows that

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M ∗ f (x) ≤ 2d Mf (x). 5.5.18 (a) If f ∈ L1 (Rd ), then by Tchebyshev’s Inequality we have Z {|f | > α} ≤ 1 |f |. α Rd Therefore f belongs to Weak-L1 . Now consider the function f (x) = |x|−d . We have {|f | > α} = {x ∈ Rd : |x|d < 1 } α = {x ∈ Rd : |x| < α−d } = Bα−d (0) = Cd (α−d )d =

Cd , α

where Cd is a constant that depends only on the dimension d. Therefore f belongs to Weak-L1 , but f is not integrable on Rd . (b) The Maximal Theorem implies that if f ∈ L1 (Rd ), then d {M f > α} ≤ 3 α

Z

Rd

|f |.

Therefore M f belongs to Weak-L1 .

5.5.19 (a) Suppose first that E is a measurable set. Then, by the Lebesgue differentiation theorem, for a.e. x we have |E ∩ Br (x)| |Br (x)| Z 1 χE (t) dt = lim r→0 |Br (x)| B (x) r ( 1, x ∈ E, = χE (x) = 0, x ∈ / E.

DE (x) = lim

r→0

Now let A be an arbitrary subset of Rd . By Problem 2.2.48, there exists a Gδ -set H ⊇ A such that |A ∩ E|e = |H ∩ E| for every measurable E ⊆ Rd . Consequently, for each x ∈ H we have DA (x) = lim

r→0

|H ∩ Br (x)| |A ∩ Br (x)|e = lim = DH (x). r→0 |Br (x)| |Br (x)|

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In particular, DA (x) = DH (x) = 1 for almost every x ∈ H. Since A ⊆ H, this implies that DA (x) = 1 for almost every x ∈ A. (b) “⇒.” We showed in part (a) that if E is measurable, then DE (x) = 0 for a.e. x ∈ / E.

“⇐.” Suppose that A ⊆ Rd is such that DA (x) = 0 for a.e. x ∈ / A. As in part (a), there is a Gδ -set H ⊇ A such that |A ∩ E|e = |H ∩ E| for every measurable E ⊆ Rd , and this implies that DA (x) = DH (x) for every x. Let B = H \A. Since B ⊆ H, we have DH (x) = 1 for a.e. x ∈ B. And since B ⊆ AC , we also have DA (x) = 0 for a.e. x ∈ B. Hence 1 = DH (x) = DA (x) = 0,

a.e. x ∈ B.

The only way this is possible is if B has measure zero. Therefore |H \A| = |B| = 0, which implies that A is measurable by Lemma 2.2.21.

Examples. Given 0 < α < 1, let E ⊂ R2 be a sector of angle 2πα in the unit circle, and let x = 0. Then |E ∩ Br (0)| = απr2 for every r ≤ 1, so DE (0) = α. For an example where the limit does not exist, we can use the same idea, but let the angle of the sector oscillate as r shrinks. Or, in one dimension, if we set −2k−1 −2k E = ∪∞ ,2 ], k=1 [2 then

|E∩Br (0)| |Br (0)|

oscillates without limit as r shrinks.

5.5.20 Let DE (x) be the density function defined in Problem 5.5.19. If x ∈ (0, 1), then the problem hypotheses imply that DE (x) = lim

r→0

δ 2r |E ∩ (x − r, x + r)| ≥ lim inf = δ. r→0 2r 2r

Hence DE (x) > 0 for a.e. x ∈ [0, 1]. However, Problem 5.5.19 also implies that DE (x) = 0 for a.e. x ∈ / E. Consequently, B = [0, 1]\E must have measure zero. Applying subadditivity, it follows that 1 = |[0, 1]| = |E ∪ B| ≤ |E|e + |B|e = |E|e + 0. On the other hand, we have |E|e ≤ 1 by monotonicity, so it follows that |E|e = 1. 5.5.21 Assume that 0 ≤ x < y < 1 are Lebesgue points of f. Let 0 < h < 1/2 be small enough that x + h < y,

y + h < 1,

h < λ,

Note that 0 < λ − h < 1 − h < 1.

h < 1 − λ.

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Since [x, x + h] and [y, y + h] are disjoint subsets of [0, 1], the measure of their union is 2h. The complement of their union therefore has measure 1 − 2h. Because h < 1 − λ, we have 1 − 2h > λ − h. which is strictly greater than λ − h. Therefore the complement of [x, x + h] ∪ [y, y + h] contains a set A with measure λ − h. Let E1 = [x, x + h] ∪ A

E2 = [y, y + h] ∪ A.

and

Then |E1 | = |E2 | = λ. The hypotheses on f therefore imply that Z

x+h

x

f +

Z

A

f =

Z

f = 0 =

Z

f =

y+h

f +

y

E2

E1

Z

Z

f.

A

All of the integrals on the line above are finite, so this implies that Z

x+h

f =

Z

y+h

f.

y

x

This holds for all small enough h. Since x and y are Lebesgue points of f, it follows that Z Z 1 x+h 1 y+h f (x) = lim+ f = lim+ f = f (y). h→0 h x h→0 h y Setting c = f (x), we see that f (y) = c for every Lebesgue point y ∈ [0, 1). Since almost every point in [0, 1) is a Lebesgue point, it follows that f is constant a.e. Finally, let E be any subset of [0, 1] such that |E| = λ. Then Z Z 0 = f = c = c |E| = cλ. E

E

Since λ 6= 0, we conclude that c = 0, and therefore f = 0 a.e. 5.5.22 (a) Since f is not zero a.e., there must exist some set F with positive measure and some ε > 0 such that |f | ≥ ε on F. By continuity from below, the measure of F ∩BR (0) converges to |F | as R → ∞. Therefore, if we take R large enough then the set E = F ∩ BR (0) has positive measure and |f | ≥ ε on E. Let cd be the constant such that |Br (0)| = cd rd .

(A)

Fix any |x| > R, and let B = B2|x| (x), the ball of radius 2|x| centered at x. Then E ⊆ BR (0) ⊆ B2|x| (x) = B, and |B| = cd (2|x|)d = cd 2d |x|d .

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Therefore Mf (x) ≥

1 |B|

Z

B

|f | ≥

1 |B|

Z

E

|f | ≥

1 |B|

Z

ε =

E

ε |E| = C |x|−d . cd 2d |x|d

(b) Since the function |x|−d is not integrable on |x| > R, it follows that Mf is not integrable on Rd . (c) Let C and R be as given by part (a), let cd be the constant from equation (A), and set α0 =

C C −d = R 2Rd 2

and

C′ =

cd C . 2

If α < α0 , then α < CR−d and therefore R < (C/α)1/d . Suppose that x is any point such that R < |x| < (C/α)1/d . Then Rd < |x|d


α , C

so part (a) implies that Mf (x) ≥ C |x|−d > C Thus {Mf (x) > α} ⊇

α = α. C

 x ∈ Rd : R < |x| < (C/α)1/d .

We therefore have  {Mf (x) > α} ≥ x ∈ Rd : R < |x| < (C/α)1/d = |B(C/α)1/d (0)| − |BR (0)|

= cd

C − cd R d . α

(B)

Now, because α < α0 , we have α
α} is on the order of 1/α, we cannot give a better estimate in terms of order (though we might be able to improve the constants in the estimates).

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Solutions to Exercises and Problems from Chapter 6 6.1.7 “⇒.” Assume that f ∈ AC[a, b] and fix ε > 0. Let δ be the number whose existence is given in definition of absolute continuity. Assume that {[aj , bj ]} is any countable collection of nonoverlapping subintervals of [a, b] such that X (bj − aj ) < δ. j

Then, by definition of absolute continuity, we have X |f (bj ) − f (aj )| < ε. j

For any complex number z = zr + izi we have |zr | ≤ Therefore X j



|zr |2 + |zi |2

|fr (bj ) − fr (aj )| ≤

X j

1/2

= |z|.

|f (bj ) − f (aj )| < ε.

Therefore fr is absolutely continuous, and a similar argument shows that fi is absolutely continuous. “⇐.” Assume that fr and fi are each absolutely continuous. Fix ε > 0, and let δr and δi be the numbers whose existence is given by applying the definition of absolute continuity to fr and fi , respectively. Let δ = min{δr , δi }. Assume that {[aj , bj ]} is any countable collection of nonoverlapping subintervals of [a, b] such that X (bj − aj ) < δ. j

For any complex number z = zr + izi we have |z| = |zr + izi | ≤ |zr | + |zi |. Therefore X j

|f (bj ) − f (aj )| ≤

X j

 |fr (bj ) − fr (aj )| + |fi (bj ) − fi (aj )|

< ε + ε = 2ε.

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Therefore f is absolutely continuous. 6.1.8 (a) Suppose that f ∈ AC[a, b]. Fix ε > 0, and let δ > 0 be the corresponding number whose existence is implied by the definition of absolute continuity. Let {[aj , bj ]} be any countable collection of nonoverlapping subintervals of [a, b]. Using the Reverse Triangle Inequality, we comcpute that X X f (bj ) − f (aj ) < ε. |f (bj )| − |f (aj )| ≤ j

j

Therefore |f | is absolutely continuous.

(b) Suppose that f, g ∈ AC[a, b] and α, β ∈ C. Fix ε > 0. Let δf and δg be the corresponding numbers whose existence is implied by the definition of absolute continuity. Let δ = min{δf , δg }. If {[aj , bj ]} is a countable collection of nonoverlapping subintervals of [a, b], then X (αf + βg)(bj ) − (αf + βg)(aj ) j

=

X

α f (bj ) − f (aj ) + β g(bj ) − g(aj ) j

≤ |α|

X X g(bj ) − g(aj ) f (bj ) − f (aj ) + |β| j

j

≤ |α| ε + |β| ε. Therefore αf + βg ∈ AC[a, b].

(c) Suppose that f, g ∈ AC[a, b]. Let δf and δg be the corresponding numbers whose existence is implied by the definition of absolute continuity. Let δ = min{δf , δg }. If {[aj , bj ]} is a countable collection of nonoverlapping subintervals of [a, b], then X (f g)(bj ) − (f g)(aj ) j

≤

X   f (bj ) g(bj ) − f (aj ) g(bj ) + f (aj ) g(bj ) − f (aj ) g(aj ) j

≤ kgk∞

X X g(bj ) − g(aj ) f (bj ) − f (aj ) + kf k∞ j

j

≤ ε kgk∞ + ε kf k∞ . Therefore f g ∈ AC[a, b].

Alternative Proof using the FTC. Suppose that f, g ∈ AC[a, b]. By Corollary 6.1.5, f, g are differentiable almost everywhere, so f g is differentiable

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almost everywhere as well. Further, (f g)′ (x) = f (x) g ′ (x) + f ′ (x) g(x) a.e. Since f and g are continuous on [a, b], they are bounded on that interval. Further, f ′ and g ′ are integrable, so f g ′ and f ′ g belong to L1 [a, b], and therefore (f g)′ ∈ L1 [a, b]. Since integration by parts is valid for absolutely continuous functions, we therefore have for x ∈ [a, b] that Z x (f g)′ (y) dy a

=

Z

a

x

f (y) g ′ (y) dy +

Z

x

f ′ (y) g(y) dy

a

= f (x) g(x) − f (a) g(a) −

Z

x

′

f (y) g(y) dy +

a

Z

x

f ′ (y) g(y) dy

a

= f (x) g(x) − f (a) g(a). It therefore follows from Theorem 6.4.2 that f g ∈ AC[a, b]. (d) In light of part (c), it suffices to show that 1/g is absolutely continuous. Assume that |g(x)| ≥ η for all x ∈ [a, b]. Given ε > 0, let δ > 0 be the constant whose existence is implied by the assumption that g is absolutely continuous. Let {[aj , bj ]} be Pany countable collection of nonoverlapping subintervals of [a, b] such that |bj − aj | < δ. Then we have X 1 X g(aj ) − g(aj ) 1 g(bj ) − g(aj ) = g(aj ) g(bj ) j j ≤

1 X ε g(aj ) − g(aj ) ≤ 2 . η j η

Since η is a fixed constant, it follows that 1/g is absolutely continuous. 6.1.9 “⇒.” This follows from the definition of absolute continuity. “⇐.” Assume that the given statement holds. Choose ε > 0, and let δ > 0 be the number whose existence is given by the assumed hypothesis. Let {[aj , bj ]}j∈N be any countably infinite collection of nonoverlapping subintervals of [a, b] such that ∞ X j=1

Then for every finite N we have

(bj − aj ) < δ.

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218 N X j=1

(bj − aj ) < δ,

so by hypothesis, N X j=1

|f (bj ) − f (aj )| < ε.

Taking the limit as N → ∞, it follows that ∞ X j=1

|f (bj ) − f (aj )| ≤ ε.

Consequently f is absolutely continuous on [a, b]. 6.1.10 (a) First Proof. Recall from Problem 5.2.26 that BV[a, b] is a Banach space with respect to the norm kf kBV = V [f ; a, b] + kf k∞ . Suppose that fn ∈ AC[a, b], f ∈ BV[a, b], and fn → f with respect to the norm k · kBV . We will use Problem 6.1.9 to show that f is absolutely continuous. Given ε > 0, there must exist some n such that kf − fn kBV < ε. The function fn is absolutely continuous. Let δ > 0 be the constant given by the definition of absolute continuity corresponding to fn and ε. Let {(aj , bj )}N j=1 be any finite collection of nonoverlapping subintervals of [a, b] such that P (bj − aj ) < δ. By definition of δ,, N X j=1

|fn (bj ) − fn (aj )| < ε.

Because there are only finitely many intervals, we can assume that they are ordered. For simplicity, let us consider the case a < a1 < b1 < a2 < b2 · · · < aN < bN < b. In any other case (e.g., a = a1 or bj = aj+1 for some j), the argument requires only a minor modification. Define a partition  Γ = a < a1 < b 1 · · · < a N < b N < b . Then

c Solutions 2019 Christopher Heil N X j=1

=

219

|f (bj ) − f (aj )| N X (f − fn )(bj ) + fn (bj ) − (f − fn )(aj ) − fn (aj ) j=1

≤

N N X X (f − fn )(bj ) − (f − fn )(aj ) + |fn (bj ) − fn (aj )| j=1

j=1

≤ SΓ [f − fn ; a, b] + ε ≤ V [f − fn ; a, b] + ε

≤ kf − fn kBV + ε < 2ε. Thus, we have shown that if {(aj , bj )}N j=1 is any finite collection of disjoint P P subintervals of [a, b] that satisfy (bj − aj ) < δ, then |f (bj ) − f (aj )| < ε. It therefore follows from Problem 6.1.9 that f ∈ AC[a, b]. Hence AC[a, b] is a closed subset of BV[a, b]. Second Proof. Recall from Problem 5.2.26 that BV[a, b] is a Banach space with respect to the norm kf kBV = V [f ; a, b] + kf k∞ . Suppose that fn ∈ AC[a, b], f ∈ BV[a, b], and fn → f with respect to the norm k · kBV . Then V [f − fn ; a, b] → 0

and

kf − fn k∞ → 0.

In particular, fn converges to f uniformly. Also, Z x Z x Z x ′ ′ f |f ′ − fn′ | ≤ V [f − fn ; a, x] (by Corollary 5.4.3) f − ≤ n a

a

a

≤ V [f − fn ; a, b] → 0.

Hence
f (x) − f (a) = lim fn (x) − fn (a) = lim n→∞

n→∞

Z

a

x

fn′

=

Z

x

f ′.

a

Therefore f is absolutely continuous. Hence f ∈ AC[a, b], so AC[a, b] is a closed subset of BV[a, b]. (b) Let ϕ be the Cantor–Lebesgue function, and let ϕn be the functions used to construct the Cantor–Lebesgue function (see Section 5.1). Each ϕn is continuous and piecewise linear, and hence is Lipschitz and therefore belongs to AC[0, 1]. The functions ϕn converge uniformly to ϕ. Since ϕ is continuous

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and monotone increasing on [0, 1], it belongs to BV[0, 1]. However, ϕ is not absolutely continuous on [0, 1]. 6.1.11 Recall that the Reverse Triangle Inequality implies that for any functions u, v ∈ L1 (E) we have kuk1 − kvk1 ≤ ku − vk1 .

Therefore

|g(x) − g(y)| = kf − xk1 − kf − yk1

(A)

= k(f − x) − (f − y)k1 Z = |x − y| dt = |E| |x − y|, E

where on the right-hand side of equation (A) and in the following integrals we think of x and y as being constant functions. This shows that g is Lipschitz on R, and therefore is absolutely continuous on every finite interval. Since f is integrable it is finite a.e. Therefore there must be some n such that A = {|f | ≤ n} has positive measure. On this set we have −n ≤ f (t) ≤ n, so Z Z |f (t) − x| dt |f (t) − x| dt ≥ g(x) = A

E

≥ =

Z

A


|x| − n dt


|x| − n |A| → ∞ as x → ±∞.

(b) Let ω(t) = {f > t} be the distribution function of f. This function is right-continuous on R (compare Problem 4.6.21). Given a fixed x ∈ R, first consider h > 0. We compute that g(x + h) − g(x) h Z Z 1 1 |f (t) − (x + h)| dt − |f (t) − x| dt = h E h E Z Z

1 1 = f (t) − (x + h) dt − f (t) − (x + h) dt h {f >x+h} h {f x} h {f x+h}

f (t) − (x + h) dt − h

Z

Z

{f >x}

f (t) − x dt h

Z

f (t) − (x + h) f (t) − x dt − dt h h {f >x+h} {f >x+h} Z f (t) − x dt − h {xx+h} {x x} .

u(t) dt

Ah



A similar argument shows that   lim+ −I2 (h) + I4 (h) = − {f < x} . h→0

Hence

lim+

h→0

g(x + h) − g(x) = {f > x} − {f < x} . h

The same limit holds from the left, so g is differentiable at x, and g ′ (x) = {f > x} − {f < x} .

In particular, g is differentiable at every point, although g ′ need not be continuous. Now suppose that x is a local minimum of g. Then we must have g ′ (x) = 0, and hence {f > x} = {f < x} . Conversely, suppose that {f > x} = {f < x} for some x. Then g ′ (x) = 0. Since g is continuous on R and converges to infinity as x → ±∞,

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6.2.2 We fill in some of the details in the proof of Corollary 6.2.3. First we demonstrate that D = ∪Dn . We have Dn ⊆ D by definition, so certainly ∪Dn ⊆ D. On the other hand, if we choose x ∈ D, then f ′ (x) exists and is nonzero. Set ε = |f ′ (x)|. Since the derivative exists at x, there is a δ > 0 such that f (y) − f (x) > ε. 0 < |y − x| < δ =⇒ y−x 2 Choose n large enough that

1 2 > n.

Therefore x ∈ Dn . This shows that D ⊆ ∪Dn . S Second, we verify that Dn ∩J = k Ak . By definition, we have Ak ⊆ Dn ∩J for every k. Conversely, choose any point x ∈ Dn ∩ J. Since Dn ⊆ D ⊆ E, we S have f (x) ∈ f (E) ⊆ Qk , so f (x) ∈ Qk for some k and therefore x ∈ f −1 (Qk ) for that k. Therefore x ∈ f −1 (Qk ) ∩ Dn ∩ J = Ak . Finally, we check that f (Ak ) ⊆ Qk . If we choose y ∈ f (Ak ), then y = f (x) for some x ∈ Ak . By definition of Ak , this implies that x ∈ f −1 (Qk ), and this tells us that f (x) ∈ Qk . Hence y = f (x) ∈ Qk . 6.2.5 Suppose that f is differentiable a.e. on E ⊆ [a, b], and A ⊆ E is such that f (x) = c for every x ∈ A. Write f = g + ih where g and h are real-valued. Then g is differentiable a.e. on A and g(x) = Re(c) for x ∈ A. Therefore g(A) = {Re(c)}, which has measure zero. Corollary 6.2.3 therefore implies that g ′ = 0 a.e. on A. A similar argument shows that h′ = 0 a.e. on A, so it follows that f ′ = g ′ + ih′ = 0 a.e. on A. 6.2.6 Let A = {x ∈ E : f ′ (x) exists}. Then Z = E \A has measure zero. RSince′ f is differentiable everywhere on A, Lemma 6.2.4 implies that |f (A)|e ≤ A |f |. Now, since f is absolutely continuous, f (E) is measurable and |f (Z)| = 0. Therefore Z Z ′ |f ′ |. |f | + 0 ≤ |f (E)| = |f (A) ∪ f (Z)| ≤ |f (A)| + |f (Z)| ≤ A

E

To see that absolute continuity is necessary, let ϕ be the Cantor–Lebesgue function. Then ϕ is differentiable a.e. on E = [0, 1], and ϕ′ = 0 a.e. However, ϕ(E) = [0, 1], so Z |ϕ(E)| = 1 > 0 =

E

|ϕ′ |.

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6.3.5 The measure of f (X) = {f (x) : x ∈ X} = {fr (x) + ifi (x) : x ∈ X} as a subset of C is defined to be the measure of the set n o
A = fr (x), fi (x) : x ∈ X

as a subset of R2 . Note that n o
A ⊆ fr (x), fi (y) : x, y ∈ X = fr (X) × fi (X). Therefore

|f (X)| = |A| ≤ |fr (X)| |fi (X)|. Consequently, if fr (X) and fi (X) each have measure zero, then f (X) has measure zero. To show that the converse implication can fail, define f : [0, 1] → C by x ∈ [0, 1].

f (x) = x = x + 0i,

Then f [0, 1] is contained in the real axis in C, so it has measure zero. Yet |fr [0, 1]| = |[0, 1]| = 1. Another example is f : [0, 1] → C defined by f (x) = x + ix. In this case we have |f [0, 1]| = 0 while |fr [0, 1]| = |fi [0, 1]| = 1. 6.3.6 (a) Given ε > 0, let δ be the corresponding constant for g whose existence is given by the definition of absolute continuity. We are given that f : [c, d] → C is Lipschitz, where the interval [c, d] contains the range of g. Let K be a Lipschitz constant for f, and let {[a Pj , bj ]} be a countable collection of nonoverlapping subsets of [a, b] such that (bj − aj ) < δ. Then X X |g(bj ) − g(aj )| ≤ Kε. |f (g(bj )) − f (g(aj ))| ≤ K j

j

Therefore f ◦ g is absolutely continuous. (b) We are given that f and g are both absolutely continuous, and g is monotone increasing on [a, b]. Fix ε > 0. Since f is absolutely continuous, there exists a δ > 0 such that if {[ck , dk ]} is any countable collection of nonoverlapping intervals in [c, d], then X X (dk − ck ) < δ =⇒ |f (dk ) − f (ck )| < ε. (A) j

k

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Since g is absolutely continuous, there exists an η > 0 such that if {[aj , bj ]} is any countable collection of nonoverlapping intervals in [a, b], then X X |g(bj ) − g(aj )| < δ. (bj − aj ) < η =⇒ j

j

Suppose that {[aj , bjP ]} is any countable collection of nonoverlapping intervals in [a, b] such that (bj − aj ) < η. Then we have X |g(bj ) − g(aj )| < δ. j

Now, since g is continuous and monotone increasing, the Intermediate Value Theorem implies that g[aj , bj ] = [g(aj ), g(bj )], although this “interval” could be a single point. Monotonicity also implies  that [g(aj ), g(bj )] is a collection of nonoverlapping “intervals.” If for some j we do have g(aj ) = g(bj ), then for that j we have |f (g(bj )) − f (g(aj ))| = 0.

 Hence we can still apply equation (A) to the family [g(aj ), g(bj )] even though it may contain some degenerate intervals, and we obtain X |f (g(bj )) − f (g(aj ))| < ε. j

Therefore f ◦ g is absolutely continuous. (c) Since AC[a, b] ⊆ BV[a, b], one implication is trivial. To prove the other direction, suppose that f ◦ g has bounded variation, and write f = fr + ifi , where fr and fi are real-valued. Then fr and fi each have bounded variation. Also, since g is real-valued, (f ◦ g)(x) = f (g(x)) = fr (g(x)) + ifi (g(x)) = (fr ◦ g)(x) + i(fi ◦ g)(x). That is, f ◦ g = (fr ◦ g) + i (fi ◦ g). Therefore it suffices to prove that fr ◦ g and fi ◦ g are individually absolutely continuous. In other words, it suffices to assume that the function f is realvalued. Now, since f and g are each continuous, the composition f ◦ g is also continuous. Suppose that A ⊆ [a, b] and |A| = 0. Then |g(A)| = 0 since g is absolutely continuous. Therefore |f (g(A))| = 0 since f is absolutely contin-

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uous. That is, |(f ◦ g)(A)| = 0. Since we are given that f ◦ g ∈ BV[a, b], the Banach–Zaretsky Theorem therefore implies that f ◦ g ∈ AC[a, b].

6.3.7 (a) We are given g(t) = t2 sin2 1t . The solution to Problem 5.2.20 shows that g is differentiable everywhere and g ′ is bounded, and therefore g is Lipschitz on [0, 1]. Consequently, g ∈ AC[0, 1]. The range of g is contained in [0, 1], and the function f (x) = x1/2 is absolutely continuous and monotone increasing on [0, 1]. However, (f ◦ g)(x) = x | sin x1 | does not have bounded variation on [0, 1], and therefore is not absolutely continuous on this interval. (b) (f ◦ g)(t) = t2 sin2

1 t

is absolutely continuous.

6.3.8 By splitting f into real and imaginary parts, it suffices to assume that f is real-valued. Let Z = {x ∈ [a, b] : f ′ (x) does not exist}. By hypothesis the set Z is countable. Suppose that A ⊆ [0, 1] satisfies |A| = 0. Then |A\Z| = 0. Since f is measurable (it is continuous) and f is differentiable at every point of A\Z, Lemma 6.2.4 implies that Z |f ′ | = 0. |f (A\Z)|e ≤ A\Z

On the other hand, the set A ∩ Z is countable, so f (A ∩ Z) is also countable, and therefore |f (A ∩ Z)|e = 0. Applying countable subadditivity for exterior measure, we see that |f (A)|e ≤ |f (A\Z)|e + |f (A ∩ Z)|e = 0. The Banach–Zaretsky Theorem therefore implies that f ∈ AC[a, b]. First proof. Since g is continuous, its indefinite integral F (x) = R6.3.9 x g(t) dt is absolutely continuous. Further, since g is continuous, Exercise a 5.2.8 tells us that F is differentiable everywhere and F ′ (x) = g(x) for every x ∈ [a, b]. Hence (F − f )′ = F ′ − f ′ = g − g = 0 a.e. Corollary 6.3.4 therefore implies that F − f is constant. Hence f = F + c, so f is differentiable at all points and f ′ (x) = g(x) for all x ∈ [a, b]. Second proof. This proof uses the Fundamental Theorem of Calculus for absolutely continuous functions. Since g is continuous, every point is a Lebesgue point of g. Suppose that x ∈ (a, b). We have x + h ∈ (a, b) for all |h| is small enough, so g(x) = lim

h→0

= lim

h→0

1 h 1 h

Z

x+h

g(t) dt

Fund. Thm. Calculus

f ′ (t) dt

since f ′ = g a.e.

x

Z

x+h

x

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= lim

h→0

f (x + h) − f (x) h

since f ∈ AC[a, b].

This shows that f is differentiable at all points in (a, b), and it also shows that f ′ (x) = g(x) for all x ∈ (a, b). If x = a then the same calculation is valid if we take limits from the right: 1 g(a) = lim h→0+ h 1 = lim+ h→0 h = lim

h→0+

Z Z

a+h

g(t) dt

Fund. Thm. Calculus

f ′ (t) dt

since f ′ = g a.e.

a a+h a

f (a + h) − f (a) h

since f ∈ AC[a, b].

Therefore f is differentiable from the right at the point a and we have f ′ (a) = g(a). A similar argument shows that f ′ (b) = g(b). 6.3.10 (a) If f ∈ AC[a, b] then f ∈ BV[a, b] by Lemma 6.1.3.

For the converse, suppose that f ∈ BV[a, b]. Suppose first that f is realvalued. Since we are told that f is differentiable everywhere on [a, b], we know that f is continuous. Further, if |A| = 0 then Corollary 6.2.4 implies that |f (A)| = 0. Since f has bounded variation, the Banach-Zaretsky Theorem implies that f is absolutely continuous. If f is complex-valued, then by splitting into real and imaginary parts the same argument shows that the real and imaginary parts of f are both absolutely continuous. Therefore f itself is absolutely continuous.

(b) If f is constant on [a, b], then f ′ (x) = 0 for every x. For the converse, assume that f is differentiable everywhere and f ′ = 0 a.e. Then f ′ ∈ L1 [a, b], so Corollary 6.3.3 implies that f ∈ AC[a, b]. Hence f is both absolutely continuous and singular, so it is constant. 6.3.11 (a) Suppose first that f is real-valued. Since f is absolutely continuous on [a + δ, b], it is continuous on that interval. As this is true for all δ > 0, it follows that f is continuous on the interval (a, b]. Since f is assumed to be right-continuous at x = a, we conclude that f is continuous on the interval [a, b]. We are also given that f has bounded variation on [a, b]. Suppose that A ⊆ [a, b] and |A| = 0. Set   An = A ∩ a + n1 , b , n ∈ N.

Then |An | = 0, so |f (An )| = 0 since f is absolutely continuous on [a + n1 , b]. Since A is contained in the union of the singleton {a} and the sets An , we conclude that   ∞ ∞ S S f (An ) f (A) ⊆ f {a} ∪ An ⊆ {f (a)} ∪ n=1

n=1

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has measure zero. Therefore f ∈ AC[a, b] by the Banach–Zaretsky Theorem. If f is complex-valued, then we can apply the same argument to the real and imaginary parts of f separately. (b) The function f (x) = x sin x1 is absolutely continuous on [δ, 1] for all 0 < δ < 1, but it is not absolutely continuous on [0, 1], and it does not have bounded variation on [0, 1]. 6.3.12 Set g(x) = x2 sin(1/x2 ). Since |g(x)| ≤ x2 , we have g ∈ L1 [−1, 1]. Exercise 5.2.4 shows that g does not have bounded variation on [−1, 1], so g cannot be absolutely continuous. Alternatively, the fact that the function g ′ (x) = 2x sin

1 1 1 − cos 2 x x x

is not integrable implies that g ∈ / AC[−1, 1]. 6.3.13 Since a > 0, we know that f is continuous at every point. Also, f (0) = 0, and f (x) = xa sin x−b , x > 0. Consequently f is differentiable everywhere on (0, ∞), and f ′ (x) = axa−1 sin x−b − bxa−b−1 cos x−b ,

x > 0.

Since f is even, f ′ (x) exists for all x 6= 0, and in fact f ′ is continuous on R\{0}. “⇒.” If f ∈ AC[−1, 1] then f ∈ BV[−1, 1], and therefore a > b by Problem 5.2.22. “⇐.” Assume that a > b. By Problem 5.2.22, we know that f ∈ BV[−1, 1]. We will prove that f ′ is integrable on [−1, 1]. For x > 0, |f ′ (x)| ≤ |axa−1 sin x−b | + |bxa−b−1 cos x−b | ≤ axa−1 + bxa−b−1 . Since a > b > 0, we have both a − 1 > −1 and a − b − 1 > −1. Therefore Z 1 Z 1 Z 1 xa−b−1 dx xa−1 dx + b |f ′ (x)| dx ≤ a 0

0

= a =

0

1 1 + b a a−b

a < ∞, a−b

so f ′ is integrable. Now fix any 0 < δ < 1. Since f is differentiable everywhere on [δ, 1] and f ′ is integrable on [δ, 1], Corollary 6.3.3 implies that f is absolutely continuous on the interval [δ, 1]. Since we also know that f has bounded variation, we

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can apply Problem 6.3.11 and conclude that f ∈ AC[0, 1]. Since f is even, it follows that f ∈ AC[−1, 1].

6.4.8 If a > 0 then xα is infinitely differentiable on [a, b], and therefore is absolutely continuous on [a, b]. Hence it suffices to consider xα on the interval [0, 1]. If α ≥ 1 then xα ∈ C 1 [0, 1], so is Lipschitz and therefore absolutely continuous. Therefore we can consider 0 < α < 1. The function f (x) = xα is differentiable at every point in (0, 1], and we have f ′ (x) = αxα−1 ∈ L1 [0, 1]. Given any points 0 ≤ c < d ≤ 1, we can compute (as a proper Riemann integral) that Z

d

′

f (x) dx = α

Z

d

α−1

x

c

c

d α

dx = x

c

= f (d) − f (c).

Taking the limit as c → 0+ , we obtain (since f ′ is integrable and we can apply the DCT): Z

d

f ′ (x) dx = f (d) − f (0),

0

d ∈ [0, 1].

Therefore f satisfies the FTC, so it is absolutely continuous on [0, 1]. 6.4.9 Let f be the reflected Cantor–Lebesgue function on [0, 2], as shown in Figure 5.2. Let g(x) = x. Then f ∈ BV[0, 2] and g ∈ C ∞ [0, 2]. Since f is nonnegative and strictly positive on (0, 2), we have Z

2

f (x) g ′ (x) dx =

0

Z

2

f (x) dx > 0.

0

On the other hand, f ′ = 0 a.e. and f (0) = f (2) = 0, so f (2)g(2) − f (0)g(0) −

Z

0

2

f ′ (x) g(x) dx = 0 − 0 − 0 = 0.

Therefore integration by parts is not valid for this pair of functions. 6.4.10 “⇒.” Suppose that f is Lipschitz. Then there exists a constant K > 0 such that |f (x) − f (y)| ≤ K |x − y| for all x, y ∈ [a, b]. Lemma 6.1.3 shows that f ∈ AC[a, b]. As a consequence, f ′ exists a.e. and is integrable. Further, for each x where the derivative exists, f (x + h) − f (x) ≤ lim K (x + h − x) = K. |f ′ (x)| = lim h→0 h→0 h h Hence |f ′ | ≤ K a.e., so f ′ ∈ L∞ [a, b].

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“⇐.” Assume that f is absolutely continuous and there is a constant M such that |f ′ | ≤ M a.e. Absolutely continuous functions satisfy the Fundamental Theorem of Calculus, so if a ≤ x < y ≤ b then Z y Z y Z y ′ ′ |f (t)| dt ≤ M dt = M (y − x). |f (x) − f (y)| = f (t) dt ≤ x

x

x

Hence f is Lipschitz.

6.4.11 The function f is the indefinite integral of an integrable function, so it is absolutely continuous. The Fundamental Theorem of Calculus therefore implies that f ′ = χU a.e. Since U is the complement of P in [0, 1], it follows that f = 0 on the set P, which is a set with positive measure. It remains to show that f is strictly increasing on [0, 1]. Since χU ≥ 0, we know that f is monotone increasing. Suppose that 0 ≤ x < y ≤ 1. We must show that f (x) < f (y). Suppose that we had f (x) = f (y). Then Z y Z y Z x χU = χU − χU = f (y) − f (x) = 0. x

0

0

But χU ≥ 0, so this implies that χU = 0 a.e. on (x, y). Therefore U ∩ (x, y) has measure zero. However, U ∩ (x, y) is an open set, so this is only possible if U ∩ (x, y) = ∅. Thus the open interval (x, y) is contained in the complement of U, which is the set P. However, P contains no open intervals, so this is a contradiction. Therefore we must have f (x) < f (y). 6.4.12 Define f : [0, 2] → R by f (x) =

(

x, 0 ≤ x < 1, x − 1, 1 ≤ x ≤ x.

Then f is differentiable everywhere on [0, 2] except at x = 1. Hence f is differentiable a.e., yet f is not monotone increasing. Suppose now that f : [a, b] → R is absolutely continuous, differentiable a.e., and f ′ ≥ 0 a.e. Then given a ≤ x < y < b, we have by the FTC that Z y f ′ ≥ 0. f (y) − f (x) = x

Therefore f is monotone increasing on [a, b]. 6.4.13 First proof that f (x) → 0. We will show that f is uniformly continuous on R. To see this, fix any ε > 0. Then since f ′ is integrable, Problem 4.5.5 implies that there exists a δ > 0 such that for any measurable set A ⊆ R, we have Z |A| < δ =⇒ |f ′ | < ε. A

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Fix any points x < y such that y − x < δ. Then f ∈ AC[x, y], so Z y Z y |f (y) − f (x)| = f ′ ≤ |f ′ | < ε. x

x

This shows that f is uniformly continuous. It therefore follows from Problem 4.4.16(c) that the limit of f (x) as x → ∞ is zero, and a symmetric argument applies to x → −∞. Second (direct) proof that f (x) → 0. Since f is integrable, if the limit of f (x) as x → ∞ exists then it must be zero. And this limit does exist, because f ′ is integrable and therefore Z x Z ∞ lim f (x) = lim f′ = f′ x→∞

x→∞

0

0

exists. A similar argument applies to x → −∞. R Proof that f ′ = 0. The previous work establishes that f (x) → 0 as |x| → ∞. Since f ′ is integrable on R and f is absolutely continuous on every finite interval, we therefore have by the DCT and FTC that Z

∞

f ′ = lim

b→∞

0

Z

b


f ′ = lim f (b) − f (0) = −f (0). b→∞

0

Similarly, Z

0

f′ =

−∞

lim

a→−∞

Z

0

f′ =

lim

a→−∞

a

Therefore Z

∞ ′

f =

−∞

Z

0 ′

f +

Z

0

−∞


f (0) − f (a) = f (0).

∞

f ′ = f (0) − f (0) = 0.

6.4.14 Each fn is absolutely continuous, so the Fundamental Theorem of Calculus implies that Z x fn′ = fn (x) − fn (0) = fn (x), x ∈ [0, 1]. 0

Since x−1/2 is integrable on [0, 1], we have h ∈ L1 [0, 1]. Therefore, the function Z x f (x) = h, x ∈ [0, 1], 0

is well-defined and is absolutely continuous on [0, 1]. Further, h − fn′ → 0 and |h − fn′ | ≤ |h| + |fn′ | ≤ 2x−1/2 ∈ L1 [0, 1].

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It therefore follows from the Dominated Convergence Theorem that lim

n→∞

Z

0

1

|h − fn′ | = 0.

Hence, Z x ′ sup |f (x) − fn (x)| = sup (h − fn ) 0≤x≤1 0≤x≤1 0 Z x ≤ sup |h − fn′ | 0≤x≤1

≤

Z

1

0

→ 0

0

|h − fn′ | as n → ∞.

Thus we have shown that fn → f uniformly.

6.4.15 (a) ⇒ (b). Assume that statement (a) holds, and let A = {f ′ = 1}. Then statement (a) tells us that f ′ = χA . Since f is absolutely continuous, f ′ is measurable, and therefore A is measurable. Using the fact that f is absolutely continuous and f (0) = 0, we apply the FTC to compute that Z x Z x χA = |A ∩ [0, x]|. f′ = f (x) = f (x) − f (0) = 0

0

(b) ⇒ (a). Assume that statement (b) holds. Then Z x χA . f (x) = |A ∩ [0, x]| = 0

Since χA is integrable, the Fundamental Theorem of Calculus implies that f is absolutely continuous and f ′ = χA a.e. Therefore f ′ (x) is either 0 or 1 for a.e. x. Finally f (0) = |A ∩ [0, 0]| = 0. 6.4.16 First proof. Let V (x) = V [f ; a, x], the total variation of f on [a, x]. Since f (a) = 0, we have |f (x)| = |f (x) − f (a)| ≤ V [f ; a, x] = V (x). Also, f is absolutely continuous, so V ′ = |f ′ | a.e. by Corollary 6.4.5. Therefore |f f ′ | ≤ V V ′ a.e. That same corollary also implies that V ∈ AC[a, b]. Therefore we can use integration by parts to compute that

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Z

b

a

Hence

Z

b

Z

a

b

|f f ′ | ≤

Z

b

b

V ′ V.

a

V (a)2 V (b)2 V (b)2 − = − 0. 2 2 2

VV′ =

a

Consequently,

Z

V V ′ = V (b)V (b) − V (a)V (a) −

Z

1 1 V (b)2 = 2 2

VV′ =

a

b

V′

a

2

1 2

=

Z

a

b

2 |f ′ | .

Second proof. Since f is absolutely continuous and f (a) = 0, we have Z x = f (x) − f (a) = f (x). a

Therefore Z b a

Z

′

|f (x) f (x)| dx =

Z

≤

Z

≤

Z

=

Z

=

b ′

|f (x)

a b a

x

a

b a

Z

b

t

b

′

b

Z

b

x

b

f ′ (t) dt| dx

a

|f ′ (x) f ′ (t)| dt dx

(A)

|f ′ (x) f ′ (t)| dx dt

(by Tonelli)

|f (t)|

a

a

Z

Z

Z

t

b

|f ′ (x)| dx dt

|f ′ (x) f ′ (t)| dt dx

Therefore Z b 2 |f (x) f ′ (x)| dx ≤ (A) + (B) a

=

Z

b

a

=

Z

b

a

= =

Z

Z

Z

Z b

a b a

x

a b a

|f ′ (x) f ′ (t)| dt dx +

(B).

Z

b

a

|f ′ (x) f ′ (t)| dt dx

Z |f ′ (x)| dx

a

2 |f ′ (x)| dx .

b

|f ′ (t)| dt



Z

b

x

|f ′ (x) f ′ (t)|dt dx

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Third Proof. Let F = f 2 . Since F is absolutely continuous, F is as well, and we have F ′ = (f 2 )′ = 2f f ′ . Therefore Z b Z 2 |f f ′ | = a

a

b

|F ′ |

= V [F ; a, b]

= V [f 2 ; a, b] ≤ V [f ; a, b] kf k∞ + V [f ; a, b] kf k∞ Z b 2 ≤ 2V [f ; a, b]2 = |f ′ (x)| dx .

(by Problem 6.1.8)

a

6.4.17 (a) We are given that f is continuous, f has bounded variation on [a, b], f ′ is integrable on [a, b], and Z

b

a

f ′ = f (b) − f (a).

If we knew that Z

x a

f ′ = f (x) − f (a),

x ∈ [a, b],

then f would satisfy the Fundamental Theorem of Calculus, and this would imply that f is absolutely continuous. However, assuming that this holds only for x = b is not enough to imply that f is absolutely continuous. For example, let f be the reflected Cantor–Lebesgue function on [0, 2] that is pictured in the lecture notes. Then f is continuous and has bounded variation on [0, 2] and f ′ = 0 a.e. on [0, 2] so f ′ is integrable. Further f (0) = f (2) = 0 and therefore Z 2

0

f ′ = 0 = f (2) − f (0).

Yet f is not absolutely continuous.

Now assume that we also require that f be monotone increasing on [a, b]. If x ∈ [a, b] then f is monotone increasing on [a, x], so Theorem 5.4.2 tells us that Z x f ′ ≤ f (x) − f (a).

a

Similarly, f is monotone increasing on [x, b], so Z

b

x

f ′ ≤ f (b) − f (x).

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Combining this with the given hypothesis that Z

b

a

f ′ = f (b) − f (a),

we see that Z

a

x

f ′ ≤ f (x) − f (a) = ≤ =

 Z

   f (b) − f (a) − f (b) − f (x) b

a

Z

x

f′ −

Z

b

f′

x

f ′.

a

Therefore for every x ∈ [a, b] we have Z x f ′ = f (x) − f (a). a

Hence f is absolutely continuous. (b) ⇒. Assume that g ∈ AC[a, b]. Then g ′ exists almost everywhere, so |A| = 0. The Banach–Zaretsky Theorem tells us that absolutely continuous functions map null sets to null sets, so we conclude that |g(A)| = 0. Note that the proof of this direction does not require strict monotonicity, or even that g be monotone increasing. “⇐.” Assume that |g(A)| = 0. Since g is monotone increasing, it is differentiable a.e. and has at most countably many discontinuities, all of which are jump discontinuities. However, since g maps [a, b] onto [c, d], there can be no jump discontinuities. Indeed, if for some a ≤ u < v ≤ b we have g(u+) =

lim g(x) < lim− g(x) = g(v−),

x→u+

x→v

then the open interval (g(u+), g(v−)) will not be contained in the range of g. Therefore g has no discontinuities, and hence is continuous on [a, b]. Even so, g need not be differentiable at every point, so the set A need not be empty. Let E = [a, b]\A, i.e., E is the set of points where g is differentiable. The set E and A are measurable, and [a, b] = E ∪A disjointly. Since g is surjective, it follows that [c, d] = = g(E ∪ A) = g(E) ∪ g(A). Applying subadditivity and monotonicity, it follows that d − c = |[c, d]| ≤ |g(E)|e + |g(A)| ≤ |[c, d]| + 0 = d − c.

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Hence |g(E)|e = d − c. Consequently g(E) differs from [c, d] by a set of measure zero, so it follows that g(E) is measurable. Since g is differentiable everywhere on E, we therefore compute that Z d − c = |g(E)| ≤ g′ (by Lemma 6.2.4) E

≤

Z

b

g′

(as g ′ ≥ 0 a.e.)

a

≤ g(b) − g(a)

(since g is monotone increasing)

= d−c

(since g is surjective).

Rb Therefore equality must hold on each line above, so a g ′ = g(b) − g(a). As g is monotone increasing, part (a) implies that g is absolutely continuous. 6.4.18 (a) ⇒ (b). This follows from Corollary 6.4.5. (b) ⇒ (a). Assume that V ∈ AC[a, b], and fix ε > 0. Then there exists a δ > 0 such that for every countable collection of nonoverlapping subintervals {[aj , bj ]} of [a, b] we have X X |V (bj ) − V (aj )| < ε. (bj − aj ) < δ =⇒ j

j

By definition of the total variation, we have |f (bj ) − f (aj )| ≤ V [f ; aj , bj ] = V (bj ) − V (aj ) = |V (bj ) − V (aj )|. Therefore Pfor any collection of nonoverlapping subintervals {[aj , bj ]} that satisfies (bj − aj ) < δ, we have X X |V (bj ) − V (aj )| < ε. |f (bj ) − f (aj )| ≤ j

j

Therefore f is absolutely continuous on [a, b]. (a) ⇒ (c). This follows from Corollary 6.4.5. (c) ⇒ (b). Let V (x) = V [f ; a, x]. Since V is monotone increasing, we have Z

a

x

V ′ ≤ V (x) − V (a)

and

Z

b

x

V ′ ≤ V (b) − V (x).

(A)

Also, |f ′ | ≤ V ′ a.e. by Corollary 5.4.3. Consequently, if x ∈ [a, b] then

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Z

a

b

Z

|f ′ | =

x

|f ′ | +

a

Z

≤

x

V′ +

a

Z

Z

b

x b

|f ′ |

V′

x

≤ V (x) − V (a) + V (b) − V (x) = V (b) − V (a) Z b |f ′ | (by statement (c)). = a

Hence equality holds throughout the preceding calculation. In light of equation (A), this forces Z

x

V

′

= V (x) − V (a)

a

Z

and

b x

V ′ = V (x) − V (a).

Therefore V is absolutely continuous. Remark: We could also appeal to Problem 6.4.17 since V is monotone increasing. Additional conclusions. Assume now that f ∈ AC[a, b]. By the previous results, we therefore have V ∈ AC[a, b] as well. Consequently, parts (a) and (b) of Problem 5.2.19 imply that
V + [f ; a, b] = 21 V [f ; a, b] + f (b) − f (a) ,
V − [f ; a, b] = 21 V [f ; a, b] − f (b) + f (a) ,

so these functions are absolutely continuous as well. Since (f ′ )+ and (f ′ )− exist and are nonnegative a.e., their Lebesgue integrals are well-defined. These integrals are finite since Z x Z x Z x |f ′ | = V (x), (f ′ )− ≤ (f ′ )+ , 0 ≤ a

a

a

where the final equality is a consequence of Corollary 6.4.5. Therefore we can compute that Z x Z x Z x ′ + ′ − (f ) + (f ) = |f ′ | = V (x) a

and

Z

a

x

(f ′ )+ −

a

Z

a

x

(f ′ )− =

a

Z

a

x

f ′ = f (x) − f (a).

Adding the two preceding equations and applying Problem 5.2.19(b) again, we see that

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Z

a

x

237

(f ′ )+ = V (x) + f (x) − f (a) = 2V + (x).

Similarly, by subtracting we obtain Z x (f ′ )− = V (x) − f (x) + f (a) = 2V − (x). 2 a

6.4.19 (a) At x = 0, for h > 0 we have h3/2 sin h1 f (h) − f (0) = = h1/2 sin h1 → 0 as h → 0+ . h−0 h Since f is even, it follows that f is differentiable at x = 0 and f ′ (0) = 0. Since f is differentiable away from the origin, it follows that f is differentiable everywhere. (b) For x > 0 we have f ′ (x) =

3 2

x1/2 sin x1 − x3/2 x−2 cos x1 =

3 2

x1/2 sin x1 − x−1/2 cos x1 .

Hence, while f ′ is continuous on (0, ∞), it is not bounded on any neighborhood of the origin. Therefore f ′ ∈ / L∞ [−1, 1]. On the other hand, Z

0

1

3 |f (x)| dx ≤ 2 ′

Z

0

1

1/2

x

dx +

Z

1

x−1/2 dx =

0

3 2 + 2 = 3. 2 3

A similar calculation holds from the left, so f ′ is integrable on [−1, 1]. (c) Since f is everywhere differentiable and f ′ is integrable, Corollary 6.3.3 implies that f is absolutely continuous. However, since f ′ is not bounded, Problem 6.4.10 implies that f is not Lipschitz. 6.4.20 Since f is even, for most statements it suffices to consider x ≥ 0. Since a > 0, we know that f is continuous at every point. Also, f (0) = 0, and f (x) = xa sin x−b , x > 0. Consequently f is differentiable everywhere on (0, ∞), and f ′ (x) = axa−1 sin x−b − bxa−b−1 cos x−b ,

x > 0.

Since f is even, f ′ (x) exists for all x 6= 0, and in fact f ′ is continuous on R\{0}. We establish a basic fact about f ′ . Fact: f ′ (0) exists if and only if a > 1. In this case, f ′ (0) = 0. Proof. Since sin h−b does not converge as h → 0+ , the limit

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lim+

h→0

ha sin h−b f (h) − f (0) = lim+ = lim+ ha−1 sin h−b h−0 h h→0 h→0

exists if and only if a > 1. A similar argument applies from the left. Furthermore, if a > 1 then the limit is zero, so when a > 1 we have f ′ (0) = lim

h→0

f (h) − f (0) = lim ha−1 sin h−b = 0. ⊓ ⊔ h→0 h−0

Now we continue with the solution to the problem. (a) We know that f ′ (x) exists for all x 6= 0. Therefore f is differentiable everywhere on [−1, 1] if and only if f ′ (0) exists. By the Fact, this occurs if and only if a > 1. (b) “⇐.” Assume that a ≥ b + 1. Then a ≥ b + 1 > 0 + 1 = 1, so f is differentiable everywhere by part (a), and f ′ (0) = 0. For x > 0 we have f ′ (x) = axa−1 sin x−b − bxa−b−1 cos x−b . Since a − 1 and a − b − 1 are both positive, it follows that for all 0 < x ≤ 1, |f ′ (x)| ≤ |axa−1 sin x−b | + |bxa−b−1 cos x−b | ≤ axa−1 + bxa−b−1 ≤ a + b. A similar bound applies from the left, so sup |f ′ (x)| ≤ a + b.

x∈[−1,1]

Thus f is differentiable everywhere and f ′ is bounded and therefore integrable on [−1, 1]. Consequently, Corollary 6.3.3 implies that f is absolutely continuous. Applying Problem 6.4.10, it follows that f is Lipschitz on [−1, 1]. “⇒.” Assume that f ∈ Lip[−1, 1]. In this case, Problem 6.4.10 implies that f is absolutely continuous and f ′ is essentially bounded on [−1, 1]. Now, f ′ is continuous on (0, ∞), so it is essentially bounded if and only if it is bounded. Recall that f ′ (x) = axa−1 sin x−b − bxa−b−1 cos x−b . As in the solution to Problem 5.2.22, let xj =



2 jπ

1/b

.

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Note that lim xj = 0.

j→∞

If j is even, then sin x−b = sin jπ j 2 = 0

cos x−b = cos jπ j 2 = ±1.

and

Hence for even j, a−b−1 −b a−b−1 |f ′ (xj )| = axja−1 sin x−b − bx cos x . j j j = bxj

Since f ′ is bounded and xj converges to zero, we must have a − b − 1 ≥ 0, and therefore a ≥ b + 1. (c) “⇐.” Assume that a > b+1. By part (a), f is differentiable everywhere and f ′ is continuous on R\{0}. Hence it only remains to prove that f ′ is continuous at x = 0. Since a > 1, we know that f ′ (0) = 0. For x > 0 we have f ′ (x) = axa−1 sin x−b − bxa−b−1 cos x−b . As both a − 1 > 0 and a − b − 1 > 0, it follows that   lim+ f ′ (x) = lim+ axa−1 sin x−b − bxa−b−1 cos x−b = 0 = f ′ (0). x→0

x→0

Therefore f ′ is continuous everywhere, so f ∈ C 1 [−1, 1].

“⇒.” Suppose that f ∈ C 1 [−1, 1]. Then f is Lipschitz, so part (b) implies that a ≥ b + 1. If a = b + 1, then f ′ is not be continuous at x = 0, since in this case we have f ′ (x) = axa−1 sin x−b − bxa x−b−1 cos x−b = axa−1 sin x−b − b cos x−b , and cos x−b does not converge as x → 0+ . Therefore we must have a > b + 1.

6.4.21 (a) Choose f ∈ L1 [a, b] and fix ε > 0. Define g(x) = f (x) for x ∈ [a, b] and g(x) = 0 for x ∈ / [a, b]. Theorem 4.5.8 implies that there exists some continuous function φ on R such that kg − φk1 < ε. Let θ = φ|[a,b] , i.e., θ is φ restricted to the interval [a, b]. Then θ is continuous on [a, b], and kf − θk1 =

Z

b a

|f − θ| =

Z

∞

−∞

|g − φ| = kg − φk1 < ε.

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By the Weierstrass Approximation Theorem, there exists a polynomial p such that ε , kθ − pku < (b − a) where the uniform norm is taken over the interval [a, b]. Consequently, kf − pk1 ≤ kf − θk1 + kθ − pk1 Z b < ε + |θ(x) − p(x)| dx a

Z

≤ ε +

a

b

kθ − pk∞ dx

= ε + (b − a) kθ − pk∞ < ε + ε = 2ε. Rb (b) Suppose that f ∈ L1 [a, b] is such that a f (x) xk dx = 0 for every Rb integer k ≥ 0. By linearity, this implies that a f (x) p(x) dx = 0 for every polynomial p. Let g be any continuous function on [a, b]. By the Weierstrass Approximation Theorem, there exist polynomials pn that converge uniformly to g on Rb [a, b]. Since a f pn = 0 for every n, we therefore compute that Z

a

b

Z f g =

b

a

≤

≤

Z

a

Z

b

a

b

|f | |g − pn |

a

Z

fg −

f pn

b

|f | kg − pn k∞

= kf k1 kg − pn k∞ → 0 as n → ∞.

Rb Hence a f g = 0 for every g ∈ C[a, b]. Theorem 6.4.7 therefore implies that f = 0 a.e. R1 (c) Suppose that f ∈ L1 [0, 1] and 0 f (x) x2k dx = 0 for every integer k ≥ 0. Define g on [−1, 1] by ( f (x), x ≥ 0, g(x) = f (−x), x < 0.

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That is, g is obtained by extending f evenly to the interval [−1, 1]. Given any integer k ≥ 0, Z 1 Z 1 Z 0 g(x) x2k dx = g(x) x2k dx + g(x) x2k dx −1

0

=

−1

Z

1

f (x) x2k dx +

0

=

Z

0

g(−y) (−y)2k (−dy)

(y = −x)

1

Z

1

f (x) x2k dx +

0

Z

1

f (y) y 2k dy

0

= 0. On the other hand, g is even and x2k+1 is odd, so Z 1 g(x) x2k+1 dx = 0, all k ≥ 0. −1

Therefore,

Z

1

g(x) xk dx = 0,

all k ≥ 0.

−1

Consequently, part (a) implies that g = 0 a.e. on [−1, 1]. This shows that f = 0 a.e. on [0, 1]. 6.4.22 (a) Since f is a monotone increasing function on the finite closed interval [a, b], it is bounded, with f (a) ≤ f (x) ≤ f (b) forRall x ∈ [a, b]. y Since f is monotone increasing, if a ≤ x < y ≤ b then x f ′ ≤ f (y) − f (x). ′ As f is integrable, the Dominated Convergence Theorem therefore implies that if we keep x fixed, then Z   f (b−) − f (x) = lim− f (y) − f (x) ≥ lim− y→b

y→b

y

f′ =

Z

Z   f (b−) − f (a+) = lim+ f (b−) − f (x) ≥ lim+ x→a

′

f ′.

x

x

Now letting x → a, we again apply the DCT to obtain x→a

b

b

x

f′ =

Z

b

f ′.

a

′

(b) We know that f is integrable and f ≥ 0 a.e. since f is monotone increasing. Set Z x f ′, x ∈ [a, b], g(x) = f (a) + a

and let h = f − g. Then g is absolutely continuous, and g is monotone increasing because f ′ ≥ 0 a.e. Further, the Fundamental Theorem of Calculus tell us that g ′ = f ′ a.e. Therefore h is singular. So, it only remains to show that h is monotone increasing.

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First, we claim that g ≤ f. To see why, fix x ∈ [a, b]. Then Z x g(x) − f (a) = f′ (definition of g) a

≤ f (x) − f (a)

(f is monotone).

Therefore g(x) ≤ f (x) for every x. Now suppose that a ≤ x < y ≤ b. Since g is absolutely continuous and f ′ = g ′ a.e., Z Z y

f′ =

x

y

x

g ′ = g(y) − g(x).

On the other hand, f is monotone increasing, so Z y f ′ ≤ f (y) − f (x).

(A)

(B)

x

Therefore    f (y) − g(y) − f (x) − g(x)     = f (y) − f (x) − g(y) − g(x) Z y Z y ′ f′ by equations (A) and (B) f − ≥

h(y) − h(x) =



x

x

= 0. Thus, h is monotone increasing. (c) Let S = f −1 (I) = {x ∈ [a, b] : f (x) ∈ I}.

If S = ∅ then there is nothing to prove since g(f −1 (I)) = ∅ in this case, and the empty set has measure zero. So, we may assume that S is not empty. Suppose that x ≤ y both belong to S. If x ≤ z ≤ y, then f (x) ≤ f (z) ≤ f (y). As f (x) and f (y) both belong to I and I is an interval (hence connected), we must have f (z) ∈ I, no matter what type of interval I might be. This shows that S is a connected subset of [a, b]. Hence S is either a single point, or it is some type of interval contained in [a, b]. Note that S need not be open, even if I is open. Since S is a finite interval, it has one of the following forms: (u, v),

[u, v),

(u, v],

[u, v].

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Since S is contained in the closed interval [a, b], both u and v belong to [a, b], so g is defined at both u and v. Since g is continuous and monotone increasing, it follows that g(S) ⊆ g[u, v] = [g(u), g(v)]. Therefore, |g(S)| ≤ g(v) − g(u) Z v = g′ =

u Z v

(g is absolutely continuous)

f′

(g ′ = f ′ a.e.)

u

≤ f (v−) − f (u+).

(f is monotone).

Now, I is some type of interval, so I has one of the following forms: (c, d),

[c, d),

(c, d],

[c, d].

Given any u < x < v, we have x ∈ S and therefore f (x) ∈ f (S) = f (f −1 (I)) ⊆ I ⊆ [c, d]. Hence f (v−) = lim− f (x) ≤ d, x→v

and f (u+) =

lim f (x) ≥ c.

x→u+

Therefore |g(S)| ≤ f (v−) − f (u+) ≤ d − c = |I|. (d) If we fix ε > 0, then there exists an open set U ⊇ f (A) such that |U | ≤ |f (A)e | + ε. Since U is an open subset of R, we can write U as a union of disjoint open intervals (ak , bk ). For each k, let Ik = (ak , bk ) ∩ [a, b], and let H = ∪Ik . Then and

X k

Now,

f (A) ⊆ H =

S

Ik ,

k

|Ik | = |H| ≤ |U | ≤ |f (A)|e + ε.

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A ⊆ f −1 (f (A)) ⊆ f −1 Therefore g(A) ⊆ and hence |g(A)| ≤ ≤

X k

X k

S

 S

Ik

k



=

S

f −1 (Ik ).

k

g(f −1 (Ik )),

k

|g(f −1 (Ik ))| |Ik |

by part (c)

≤ |f (A)|e + ε. As ε is arbitrary, it follows that |g(A)| ≤ |f (A)|e . (e) Let Z = [a, b]\E. Since f is monotone increasing, it is differentiable a.e., and therefore |Z| = 0. Since g is absolutely continuous, the Banach– Zaretsky Theorem implies that |g(Z)| = 0 as well. Further, g(E) is measurable since E is measurable and g ∈ AC[a, b]. Since g is also monotone increasing, g(E) ∪ g(Z) = g(E ∪ Z) = g[a, b] = [g(a), g(b)]. Hence Z

b

′

f =

a

Z

b

g′

(since f ′ = g ′ a.e.)

a

= g(b) − g(a) = [g(a), g(b)]

(g is absolutely continuous)

= |g(E) ∪ g(Z)| ≤ |g(E)| + |g(Z)| = |f (E)|e + 0 Z f′ ≤ ≤

Z

(by part (d)) (by Lemma 6.2.4)

E

b

a

f′

(as f ′ ≥ 0 a.e.).

Therefore equality must hold on every line of the preceding calculation. In Rb particular, we conclude that a f ′ = |f (E)|e .

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(f) For simplicity of presentation, extend g to R by setting g(x) = g(a) for x < a, and g(x) = g(b) for x > b. Then g is absolutely continuous on every finite interval, and g ′ (x) = 0 for all x ∈ / [a, b]. Note that g maps measurable sets to measurable sets because it is absolutely continuous. Since g is continuous and monotone increasing, the Intermediate Value Theorem implies that if x < y then g[x, y] = [g(x), g(y)], and therefore g[x, y] = [g(x), g(y)] = g(y) − g(x). So

R

Z

′

g =

[x,y]

Z

y

x

′

g ′ = g(y) − g(x) = g[x, y] .

Hence A g = |g(A)| holds when A is a closed finite interval. Now consider an open interval A = (a, b). Note that [a, b] = (a, b) ∪ {a, b}. Since ga, b = {g(a), g(b)} has two elements, it has measure zero. Therefore g[x, y] = g(x, y) ∪ g{x, y} ≤ g(x, y) + 0 ≤ g[x, y] , so g(x, y) = g[x, y] . Hence Z y Z g ′ = g[x, y] = g(x, y) . (A) g′ = x

(x,y)

R

Hence A g ′ = |g(A)| holds when A is a finite open interval. Now let U be any bounded open set, and write U = ∪(ak , bk ), a union of countably many disjoint intervals. If we consider two of these intervals, say (aj , bj ) and (ak , bk ), then one of these intervals must be to the left of the other, say aj < b j < ak < b k . Since g is monotone, g(aj ) ≤ g(bj ) ≤ g(ak ) ≤ g(bk ). Hence [g(aj ), g(bj )] and [g(ak ), g(bk )] are nonoverlapping intervals. Since g(aj , bj ) ⊆ g[aj , bj ] = [g(aj ), g(bj )], it follows that g(aj , bj ) and g(ak , bk ) are nonoverlapping measurable sets. Therefore Z g′ (by Problem 6.2.6) |g(U )| ≤ U

=

XZ k

bk

ak

g′

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=

X g(ak , bk )

(by equation (A))

k

S = g(ak , bk )

(nonoverlapping sets)

k

S  ak , b k = g k

= |g(U )|.

R Hence U g ′ = |g(U )| holds for bounded open sets U. Now let A be any measurable subset of [a, b]. Then there exist bounded nested open sets U1 ⊇ U2 ⊇ · · · ⊇ A such that G = ∩Un satisfies |G\A| = 0. Also, since g(A) is measurable and bounded, there exist bounded nested open sets V1 ⊇ V2 ⊇ · · · ⊇ g(A) such that H = ∩Vn satisfies |V \g(A)| = 0. Note that g −1 (Vn ) is open since g is continuous, and A ⊆ g −1 (g(A)) ⊆ g −1 (Vn ). Therefore Wn = Un ∩ g −1 (Vn ) is an open set with A ⊆ Wn ⊆ Un , so W = ∩Wn is a measurable set that contains A and satisfies |W \A| = 0. By continuity from above, |Wn | → |A| as n → ∞. Hence Z g ′ ≥ |g(A)| (by Problem 6.2.6 since g ∈ AC) A

= lim |Wn | n→∞ Z g′ = lim n→∞

≥ R

Z

A

g′

(continuity from above) (since Wn is open)

Wn

(since g ′ ≥ 0 a.e. and Wn ⊇ A).

Hence A g ′ = |g(A)|. Finally, set Z = [a, b]\E, so Z has measure zero. Then g(Z) has measure zero, so |g(A)| = |g((A ∩ E) ∪ (A ∩ Z))| = |g(A ∩ E)| + |g(A ∩ Z)| = |g(A ∩ E)|.

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′ (g) Since f is differentiable on A ∩ E and R f ≥′ 0 a.e., Growth Lemma II (Lemma 6.2.4) implies that |f (A ∩ E)|e ≤ A∩E f . Therefore Z |f (A ∩ E)|e ≤ f′ (by Growth Lemma II) A∩E

≤ ≤

Z

f′

(f ′ ≥ 0 a.e.)

g′

(f ′ = g ′ + h′ and h′ = 0 a.e.)

A

Z

A

= |g(A ∩ E)|

(by part (f))

≤ |g(A ∩ E)|

(by part (d)).

6.5.2 We fill in one detail in the proof of Theorem 6.5.2. Suppose that t does not belong to the set A defined in the proof. As A = Zg ∪ B, this means that t ∈ / Zg and t ∈ / B. The frac that t does not belong to Zg implies that g is differentiable at t. Since t ∈ / B = g −1 (ZF ), we must have x = g(t) ∈ / ZF . By the definition of ZF , this implies that F ′ (x) must exist and h(x) must equal F ′ (x). Hence F ′ (g(t)) = F ′ (x) = h(x) = h(g(t)). 6.5.9 (a) We are given that f is a strictly increasing bijection of [a, b] onto [c, d]. Its inverse function g : [c, d] → [a, b] is a bijection. Suppose that c ≤ u < v ≤ d. Then u = f (x) and v = f (y) for some x, y ∈ [a, b]. Since u < v and f is strictly increasing, we must have x < y. Therefore g(u) = x < y = g(v). Hence g is a strictly increasing bijection of [c, d] onto [a, b]. The Lemma proved in the solution to Problem 5.1.7 shows that a strictly increasing map of [a, b] onto [c, d] is continuous. Applying that lemma here, it follows that both f and g is continuous. They are also differentiable a.e. since they are monotone increasing. (b) Since f ∈ AC[a, b] (by hypothesis) and g ∈ L∞ [c, d] since it is continuous, we can apply the change of variables formula given in Corollary 6.5.8(b) and conclude that Z d Z f (b) g(t) dt = g(t) dt c

f (a)

= =

Z

Z

b

g(f (t)) f ′ (t) dt

(by Corollary 6.5.8)

tf ′ (t) dt

(since g = f −1 ).

a b a

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Next, recall that by definition we have (f ◦g)(t) = i(t) = t for every t, where i denotes the identity map on [c, d]. Since f is absolutely continuous and g is monotone increasing, we can apply the Chain Rule as given in Corollary 6.5.5(a) to obtain f ′ (g(t)) g ′ (t) = (f ′ ◦ g)(t) g ′ (t) = (f ◦ g)′ (t) = i′ (t) = 1,

a.e. t ∈ [c, d].

(c) This follows by interchanging the roles of f and g in part (b). Alternatively, we can observe that since g ∈ AC[c, d] (by hypothesis) and f ∈ L∞ [a, b] since it is continuous, we can apply Corollary 6.5.8(b) and conclude that Z

b

f (x) dx = a

Z

g(d)

f (x) dx

g(c)

=

Z

d

f (g(t)) g ′ (t) dt

(by Corollary 6.5.8)

tg ′ (t) dt

(since g = f −1 ).

c

=

Z

d

c

Next, recall that by definition we have (g ◦ f )(x) = i(x) = x for every x, where i denotes the identity map on [a, b]. Since g is absolutely continuous and f is monotone increasing, we can apply the Chain Rule as given in Corollary 6.5.5(a) to obtain g ′ (f (x)) f ′ (x) = (g ′ ◦ f )(x) g ′ (x) = (g ◦ f )′ (x) = i′ (x) = 1, a.e. x ∈ [a, b]. Remark: The integration formulas in parts (b) and (c) can also be derived by applying Problem 6.5.12. 6.5.10 The change of variable formulas that we would like to apply is formulated in terms of monotone increasing functions, yet we would like to use it with the monotone decreasing function 1/t. For each monotone increasing result there is an analogous monotone decreasing result, but for preciseness we will will formulate this proof so that it uses the monotone increasing function −1/t instead of 1/t. Fix k ∈ N, and let fk (x) = f (x) x−2k . We know that fk is integrable on [1, ∞) since f is integrable and x−2k is bounded on that interval. Define g(t) = −1/t for t 6= 0, and for t ≤ −1 set

t ≤ −1. hk (t) = fk − 1t t−2 , = f − 1t t2k t−2 ,

Fix 0 < δ < 1, and set d = 1/δ. Then g is monotone increasing on the interval [−1, −δ], and g maps [−1, −δ] onto [1, 1/δ]. The function |fk | is integrable on

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[1, 1/δ]. Corollary 6.5.8 therefore implies that
|hk (t)| = |fk − 1t | t−2 = |fk (g(t))| g ′ (t) is measurable, and Z

−δ

−1

|hk (t)| dt =

Z

−δ

−1

|fk (g(t))| g ′ (t) dt =

Z

1/δ

|fk (x)| dx < ∞.

1

Therefore hk is integrable on [−1, −δ]. Further, the Monotone Convergence Theorem implies that Z

0

−1

|hk (t)| dt = lim

δ→0

= lim

δ→0

Z

−δ

|hk (t)| dt

−1

Z

1/δ

1

|fk (x)| dx =

Z

∞

1

|fk (x)| dx < ∞.

Therefore hk is integrable on [−1, 0]. Applying the same argument without absolute values we obtain Z

−δ

hk (t) dt =

Z

−δ

fk (g(t)) g ′ (t) dt =

1/δ

fk (x) dx.

1

−1

−1

Z

Now that we know that hk is integrable, we can apply the DCT to obtain Z ∞ Z 0 fk (x) dx. hk (t) dt = 1

−1

Now,



hk (t) = fk − 1t t−2 , = f − 1t t2k t−2 , = h0 (t) t2n .

Therefore h0 is an integrable function on [−1, 0] that satisfies Z

0

h0 (t) t

2k

dt =

Z

0

hk (t) dt =

−1

−1

Z

∞

fk (x) dx =

1

Z

∞

f (x) x−2k dx = 0.

1

This is true for every k ∈ N. Consequently, Z

0

t2 h0 (t) t2k dt = 0,

k = 0, 1, 2, . . . .

−1

Therefore t2 h0 (t) = 0 a.e. by Problem 6.4.21(b). (Technically, that problem uses the interval [0, 1] instead of [−1, 0], but an entirely symmetrical argument shows that we can replace [0, 1] by [−1, 0] in Problem 6.4.21.) Therefore h = 0 a.e.

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6.5.11 Let [a, b] = [c, d] = [−1, 1]. Let g = 0, the zero function on [−1, 1], and define fn = nχ{0} and f = 0. Then fn → f pointwise a.e. Given any t ∈ [−1, 1], we have (fn ◦ g)(t) = fn (g(t)) = fn (0) = n. On the other hand, (f ◦ g)(t) = f (g(t)) = f (0) = 0. Hence at no point t does (fn ◦ g)(t) converge to (f ◦ g)(t). 6.5.12 Since f is integrable, there exist bounded functions fn such that fn → f pointwise a.e. and |fn | ≤ |f | for every n. For example, we can set fn (x) = f (x) if |f (x)| ≤ n, and fn (x) = 0 otherwise. Since fn is bounded and g is absolutely continuous, Corollary 6.5.8 implies that if [u, v] ⊆ [a, b], then Z

g(v)

fn =

Z

v

fn (g(t)) g ′ (t) dt.

(A)

u

g(u)

The Dominated Convergence Theorem implies that lim

n→∞

Z

g(v)

g(u)

fn =

Z

g(v)

f.

(B)

g(u)

We would like to apply the DCT to the functions (fn ◦g) g ′ , but we must be careful. Although fn → f pointwise a.e., it need not be true that fn ◦g → f ◦g pointwise a.e. (see Problem 6.5.11). So, let Zf = {x ∈ [c, d] : fn (x) → / f (x)} and Zg = {x ∈ [a, b] : g ′ (x) does not exist}. We know that |Zf | = |Zg | = 0. Let B = g −1 (Zf ) and A = B ∪ Zg . Since g(B) = g(g −1 (Zf )) ⊆ Zf , we have |g(B)| = 0. Corollary 6.2.3 therefore implies that g ′ = 0 a.e. on B. As |Zg | = 0, we also have g ′ = 0 a.e. on A. Therefore, fn (g(t)) g ′ (t) = f (g(t)) g ′ (t)

for a.e. t ∈ A.

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On the other hand, if t ∈ / A then t ∈ / Zg , so g ′ (t) does exist. Further, t ∈ / −1 g (Zf ), so g(t) ∈ / Zf , and therefore fn (t) → f (t). Thus fn (g(t)) g ′ (t) → f (g(t)) g ′ (t)

all t ∈ / A.

Combining the above statements, we see that (fn ◦ g) g ′ → (f ◦ g) g ′ a.e. Furthermore, by hypothesis we have |fn (g(t)) g ′ (t)| ≤ |f (g(t)) g ′ (t)| ∈ L1 [a, b]. Therefore we can apply the DCT to conclude that Z v Z v ′ f (g(t)) g ′ (t) dt. fn (g(t)) g (t) dt = lim n→∞

(C)

u

u

The change of variables formula follows by combining equations (A), (B), and (C). 6.5.13 Note that if g(a) = g(b), then g is constant and g ′ = 0. In this case both sides of equation (6.18) are zero, and so there is nothing to prove. Therefore, we may assume that g(a) < g(b). Also, we claim that it suffices to assume that [c, d] = [g(a), g(b)]. To see why, suppose that the result has been proved for this case, and we then take c < g(a) and d > g(b). Let f be any integrable function on [c, d]. Then f is integrable on [g(a), g(b)]. Therefore, the result holds for the function h = f χ[g(a),g(b)] on the interval [g(a), g(b)]. Now, if t ∈ [a, b], then g(t) ∈ [g(a), g(b)] and consequently f (g(t)) = h(g(t)). Therefore, Z

g(b)

f =

Z

g(b)

h

g(a)

g(a)

=

Z

b

h(g(t)) g ′ (t) dt

(by the result for [g(a), g(b)])

a

=

Z

b

f (g(t)) g ′ (t) dt.

a

Therefore it suffices to assume that c = g(a) and d = g(b). (a) Since g is continuous and monotone increasing, g −1 [u, v] is a closed interval, say [r, s]. Further, even if g is not strictly increasing, since it is continuous it must map [r, s] precisely onto [u, v]. Therefore (f ◦ g)(t) = χ[u,v] (g(t)) = χ[r,s] (t),

(A)

which is measurable. As g ′ is also measurable, it follows that (f ◦ g)(t) g ′ (t) is measurable. Further,

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Z

g(b)

f (x) dx =

g(a)

Z

g(b)

g(a)

χ[u,v] (x) dx = v − u,

while Z

b

f (g(t)) g ′ (t) dt =

a

Z

b

χ[u,v] (g(t)) g ′ (t) dt

a

=

Z

s

g ′ (t) dt

(by equation (A))

r

= g(s) − g(r)

(absolute continuity)

= v − u. Hence equation (6.18) holds for f = χ[u,v] , and therefore f ∈ F.

(c) Assume that f = 0 a.e. on [c, d]. Then we can write f = h χZ , where h is a function on [c, d] and Z ⊆ [c, d] has measure zero, Define B = g −1 (Z). Since g is absolutely continuous, it is differentiable a.e. on B. Further, g(B) = g(g −1 (Z) ⊆ Z,

so |g(B)| = 0. Corollary 6.2.3 therefore implies that g ′ = 0 a.e. on B. Hence f (g(t)) g ′ (t) = 0 a.e. t ∈ B.

On the other hand, if t ∈ / B = g −1 (Z), then g(t) ∈ / Z, and therefore f (g(t)) = h(t) χZ (g(t)) = 0. Hence f (g(t)) g ′ (t) = 0

all t ∈ / B.

Therefore f (g(t)) g ′ (t) = 0 a.e. on [a, b]. Hence (f ◦ g) g is measurable, and we have Z b Z d Z d f (g(t)) g ′ (t) dt. h χZ = 0 = f = c

c

a

Therefore f belongs to F .

(c) We proceed through a series of lemmas. Lemma A. F is closed under finite linear combinations. Proof. If f1 , f2 ∈ F, then f1 + f2 is bounded and measurable, and ((f1 + f2 ) ◦ g) g ′ = (f1 ◦ g) g ′ + (f2 ◦ g) g ′ is measurable. Further, equation (6.18) holds for f1 + f2 by linearity of the integral. Hence F is closed under addition, and similarly it is closed under scalar multiplication. ⊓ ⊔

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Lemma B. If 0 ≤ fk ր f a.e., where each fk ∈ F and f is integrable, then f ∈ F. Proof. Let Z be the set of points where fk (x) does not converge to f (x). Then |Z| = 0 by hypothesis. Define h k = f k χZ C = f k − f k χZ

and

h = f χZ C = f − f χZ .

Then fk = hk a.e., f = h a.e., h is integrable, and 0 ≤ hk (x) ր f (x) for every x. Since fk ∈ F and fk χZ ∈ F, it follows from Lemma A that hk ∈ F. Therefore hk (g(t)) g ′ (t) is measurable. Since hk (g(t)) g ′ (t) → h(g(t)) g ′ (t)

for every t ∈ [a, b],

we know that (h ◦ g) g ′ is measurable. By the Monotone Convergence Theorem and the definition of F , it follows that Z d Z d h(x) dt = lim hk (x) dt (MCT) k→∞

c

= lim

k→∞

=

Z

b

c

Z

b

hk (g(t)) g ′ (t) dt

a

h(g(t)) g ′ (t) dt

(def of F ) (MCT, convergence everywhere),

a

All of these quantities are finite since h is integrable. This shows that h ∈ F. Since f = h + f χZ and f χZ ∈ F, we conclude that f ∈ F. ⊓ ⊔

Lemma C. If 0 ≤ fk → f a.e., where fk ∈ F and f is integrable and there exists some integrable, nonnegative γ ∈ F such that |fk | ≤ γ a.e., then f ∈ F. Proof. The proof is similar to that of Lemma B. Let Z be the set of points where either fk (x) does not converge to x or |fk (x)| > |h(x)|. By hypothesis, |Z| = 0. h k = f k − f k χZ and h = f − f χZ .

Then hk (g(t)) g ′ (t) → h(g(t)) g ′ (t) for every t. Further, since γ and g ′ are nonnegative, for every k we have |hk (g(t)) g ′ (t)| ≤ γ(g(t)) g ′ (t).

But γ(g(t)) g ′ (t) is integrable since it is nonnegative and Z

b a

γ(g(t)) g ′ (t) dt =

Z

c

d

γ(x) dx = kγk1 < ∞.

Therefore we can apply the Dominated Convergence Theorem to obtain

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Z

d

h(x) dt = lim

k→∞

c

= lim

k→∞

=

Z

b

Z

d

hk (x) dt

(DCT)

hk (g(t)) g ′ (t) dt

(def of F )

c

Z

b

a

h(g(t)) g ′ (t) dt

(DCT).

a

Therefore h ∈ F. As f = h + f χZ and f χZ ∈ F by part (b), it follows that f ∈ F. ⊓ ⊔

Lemma D. If U ⊆ [c, d] is open, then χU ∈ F. S Proof. We can write U = [uk , vk ] where the [uk , vk ] are nonoverlapping closed intervals. For each N set fN =

N X

χ[ak ,bk ] .

k=1

We have fN ∈ F by parts (a) and (b). On the other hand, 0 ≤ fN ր χU a.e. and χU is integrable, so it follows from Lemma B that χU ∈ F. ⊓ ⊔

Lemma E. If H ⊆ [c, d] is a Gδ -set, then χH ∈ F. Proof. There exist open sets U1 ⊇ U2 ⊇ · · · ⊃ H such that H = Therefore 0 ≤ χUk ց χH .

T

Uk .

Since χUk ∈ F for every k and χU1 is integrable, it follows from Lemma C that χH ∈ F. ⊓ ⊔ Lemma F. If E ⊆ [c, d] is measurable, χE ∈ F. Proof. We can write E = H \Z where H is a Gδ -set and |Z| = 0. By replacing Z with H ∩ Z, we can assume that H ⊆ Z. Therefore χE = χH − χZ . As both χH and χZ belong to F , it follows that χE ∈ F as well. ⊓ ⊔ (f) Suppose first that f is integrable and nonnegative on [c, d]. Then there exist simple functions φn such that 0 ≤ φn ր f. By combining Lemmas A and F above, each simple function φn belongs to F . Applying Lemma B, it follows that f ∈ F. Finally, if f is a generic integrable function, then we can write f = (f1 − f2 ) + i(f3 − f4 ) where f1 , f2 , f3 , f4 are nonnegative integrable functions. Each fi belongs to F , so we conclude that f ∈ F since F is closed under finite linear combinations. 6.6.2 The base case N = 2 follows from the definition of convexity. Assume that the conclusion holds for some N ≥ 2. Choose points

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x1 , . . . , xN , xN +1 ∈ (a, b) and positive weights t1 , . . . , tN , tN +1 such that t1 + · · · + tN +1 = 1. Without loss of generality, we may assume that x1 < · · · < xN < xN +1 . Define y = xN +1 , t = t1 + · · · + tN , x =

t1 t x1

+ ··· +

tN t

xN .

Since t1 t

+ ··· +

tN t

= 1,

the point x lies between x1 and xN , so it belongs to the interval (a, b). Applying the inductive hypothesis and the definition of convexity, it follows that     N +1 X t1 x1 + · · · + tN xN   = φ tj xj φ t + tN +1 xN +1 t j=1 = φ(tx + (1 − t)y)

≤ t φ(x) + (1 − t) φ(y)   + tN +1 xN +1 = t φ tt1 x1 + · · · + tN t xN ≤ t =



t1 t

N +1 X

φ(x1 ) + · · · +

tN t

 φ(xN ) + tN +1 xN +1

tj φ(xj ).

j=1

6.6.3 We fill in the details of the proof of the final statement of Lemma 6.6.3. Suppose that a < z < x < y < b. Holding z fixed and considering that x < y, it follows as above that φ(x) − φ(z) φ(y) − φ(z) ≤ . x−z y−z

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On the other hand, holding y fixed and considering that z < x, we also have φ(y) − φ(z) φ(y) − φ(x) ≤ . y−z y−x Hence

φ(x) − φ(y) φ(x) − φ(z) ≤ . x−z x−y

6.6.5 We fill in the details of the proof of equation (6.22) in Theorem 6.6.5. Set   a1 a2 M = max , . b1 b2 Then

a1 b1

≤ M and

a2 b2

≤ M, so a1 + a2 b1 M + b2 M ≤ = M. b1 + b2 b1 + bm

The other inequality is similar. 6.6.9 We fill in the details of the argument in the proof of Lemma 6.6.9 for points a < y < x. We are given that L is a line with slope m that passes through (x, φ(x)) and satisfies φ′− (x) ≤ m ≤ φ′+ (x). Equation (6.23) tells us that φ(y) − φ(x) ≤ φ′− (x). y−x Therefore, since y − x is negative and φ′− (x) ≤ m, L(y) = (y − x) m + φ(x) ≤ (y − x) φ′− (x) + φ(x) ≤ (y − x)

φ(y) − φ(x) + φ(x) y−x

= φ(y). Hence the graph of L lies on or below the graph of φ, so L is a supporting line. 6.6.12 The function φ(x) = ex is convex. If a = 0 or b = 0 then the desired inequality is trivial, so it suffices to assume that a, b > 0. Set x = p ln a and y = p′ ln b, so we have a = ex/p Since

1 p

+

1 p′

= 1, it follows that

and

′

b = ey/p .

c Solutions 2019 Christopher Heil x

ab = e p

+ py′

= φ

x p

+

6.6.13 The series

y p′

257 ′




≤

∞ X

ex ey ap bp φ(x) φ(y) + ′ = + ′ = + ′. p p p p p p

2−n an

(A)

n=1

converges to a nonnegative real number since we have 0 < an ≤ 1 for every n. The fact that 0 < an ≤ 1 also impleis that ln an ≤ 0 for every n. Therefore every term of the series ∞ X

2−n ln an

(B)

n=1

is negative. Consequently the series converges in the extended real sense, though the sum could be either −∞ or a nonpositive real number. If the sum is −∞ then there is nothing to prove, so we may assume that the series in equation (B) converges to a finite, nonpositive, real number. Fix any N ∈ N. Since − ln x is convex on (0, ∞), the Discrete Jensen Inequality tells us that ! PN PN −n 2−n ln an 2 a n n=1 ≤ − n=1 . − ln PN P N −n −n n=1 2 n=1 2

Rearranging and evaluating the sum of 2−n , we obtain N X

n=1

2

−n

ln an ≤ (1 − 2

−N

) ln

PN

2−n an 1 − 2−N n=1

!

.

Since the series in equations (A) and (B) both converge and since ln x is a continuous function, it follows that ∞ X

2−n ln an =

n=1

≤

lim

N →∞

N X

n=1

lim

N →∞

= 1 · ln = ln

2−n ln an

(1 − 2

 P∞

∞ X

n=1

−N

) ln

2−n an 1 !

n=1

2−n an



PN

2−n an 1 − 2−N n=1

!!

.

Remark: An alternative approach is use the convexity of ex to prove that

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258 ∞ X

exp

2

−n

ln an

n=1

!

≤

∞ X

2−n an .

n=1

6.6.14 The first inequality follows by applying Jensen’s Inequality using the function φ(x) = ex , which is convex on (−∞, ∞). For the second inequality, recall from Corollary 6.6.6 that φ(x) = − ln x is convex on (0, ∞). We are given that f is integrable and real-valued (although no change is needed if f is complex-valued, since this part of the problem only involves |f |). Therefore |f (x)| ∈ [0, ∞) for every x. Case 1 : Z = {f = 0} has measure zero. In this case ln |f (x)| is defined for a.e. x. Redefining f on a set of measure zero will not change the values of the integrals involved, so we can assume that f is integrable and nonzero at all points. Applying Jensen’s Inequality, we see that   Z Z 1 1 |f | ≤ − ln |f (x)| dx. − ln |E| E |E| E Rearranging then gives the desired inequality.

Case 2 : Z = {f = 0} has positive measure. In this case ln |f | = −∞ on the set Z, which has positive measure. Also, R since f is integrable, ln |f | exists and is a real number. Note that (ln |f |)+ = (ln |f |)χ{|f |≥1} .

Since ln t ≤ t for t ≥ 1 and since f is integrable, we therefore have Z Z Z (ln |f |)+ = ln |f | ≤ |f | < ∞. E

Hence

R

{|f |≥1}

E

ln |f | is well-defined and we have   Z Z 1 1 ln |f | = −∞ ≤ ln |f | . |E| E |E| E

Therefore the desired inequality also holds in this case. 6.6.15 “⇒.” If φ is convex, then we know from Theorem 6.6.7 that φ is continuous. Further, the definition of convexity implies that   φ(x) + φ(y) x+y ≤ (A) φ 2 2 for all points x, y ∈ (a, b). “⇐.” Assume that φ is continuous and equation (A) holds for all points x, y ∈ (a, b).

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Fix [c, d] ⊆ (a, b), and fix any point x ∈ [c, d]. Let (xk )k∈N be the sequence of numbers obtained by starting with [c, d] and dividing dyadically in half so that xk → x. Let L1 be the line joining (c, φ(c)) to (d, φ(d)). To show that φ is convex on (c, d), we must show that φ(x) ≤ L1 (x). By hypothesis,   φ(c) + φ(d) c+d ≤ = L1 (x1 ). φ(x1 ) = φ 2 2 If x = x1 then we are done, so assume that x 6= x1 . If x2 lies to the left of x1 , let L2 be the line joining (c, φ(c)) to (x1 , φ(x1 )), otherwise let L2 be the line joining (x1 , φ(x1 )) to (d, φ(d)). Note that on the half-segment where L2 is defined, we have L2 ≤ L1 , because φ(x1 ) ≤ L1 (x1 ). (A picture would be useful here.) In particular, L2 (x2 ) ≤ L1 (x2 ). For simplicity, assume that x2 is to the right of x1 . Then   φ(x1 ) + φ(d) x1 + d ≤ = L2 (x2 ) ≤ L1 (x2 ). φ(x2 ) = φ 2 2 A similar argument shows that if x2 lies to the left of x1 , then we also have φ(x2 ) ≤ L1 (x2 ). If x = x2 then we are done, otherwise we continue. Continue in this way to construct points xk and lines Lk . If x is equal to any xk then we are done. Otherwise, we continue forever, and for k we have φ(xk ) ≤ Lk (xk ) ≤ Lk−1 (xk ) ≤ · · · ≤ L1 (xk ). Since both φ and L1 are continuous and xk → x, it follows that φ(x) ≤ L1 (x). Therefore φ is convex on (c, d). This is true for all a < c < d < b. It follows from this and the definition of convexity that φ is convex on (a, b). 6.6.16 Since f is integrable, we know that φ is continuous (compare Problem 4.5.17). Fix any points a < x < y < b, and let m = x+y be the midpoint of the 2 interval [x, y]. Let Z y Z m f. f and β = α = m

x

Since f is monotone increasing, we have Z m Z α = f ≤ x

Therefore

Z

m

x

f = α ≤

y

f = β. m

α+β 1 = 2 2

Z

y

x

f.

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Consequently,   Z m x+y φ = φ(m) = f 2 a Z x Z = f + a

f

x

Z Z 1 x 1 y f + f + f 2 a 2 x a Z x Z y  1 f = f + 2 a a 1 ≤ 2

=

Z

m

x

φ(x) + φ(y) . 2

Problem 6.6.15 therefore implies that φ is convex. 6.6.17 Fix [c, d] ⊆ (a, b), and suppose that c ≤ x < y ≤ d. Then Theorem 6.6.7 implies that φ′+ (c) ≤ φ′+ (x) ≤

φ(y) − φ(x) ≤ φ′− (y) ≤ φ′− (d). y−x

Therefore, if we set

then we have

 K = max |φ′+ (c)|, |φ′− (d)| , |φ(y) − φ(x)| ≤ K |y − x|.

As K is independent of the choice of x, y in [c, d], it follows that φ is Lipschitz on [c, d].

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Solutions to Exercises and Problems from Chapter 7 7.1.5 (a) Set f (t) = tθ − θt − (1 − θ). Then f ′ (t) = θtθ−1 − θ. We have f ′ (t) = 0 if and only if t = 1. Also, f is increasing for 0 < t < 1, decreasing for t > 1, and f (1) = 0, so f (t) ≤ 0 for all t > 0, with equality only for t = 1. (b) Note that 1 1 + = 1, p p′

p′ 1 = , p p−1

p , p−1

p′ =

p′ −

p′ = 1. p

′

With t = ap b−p and θ = 1/p, we have by part (a) that ′

a b−p /p =

′

ap b−p


1/p

′

≤ ap b−p

′ ap b−p 1  1 1 = + 1− + ′. p p p p

′

Multiplying through by bp and using the fact that p′ − (p′ /p) = 1, we obtain ′

ab = a b

p′ −p′ /p

bp ap + ′. ≤ p p

′

′

Equality holds if and only if ap b−p = 1. This is equivalent to bp = ap , or ′

b = ap/p = ap−1 . 7.1.8 (a) If x, y ∈ ℓ1 , then kx + yk1 = ≤ =

∞ X

k=1

∞ X

k=1

∞ X

k=1

|xk + yk |
|xk | + |yk | |xk | +

∞ X

k=1

|yk |

= kxk1 + kyk1 . (b) If x, y ∈ ℓ∞ , then for every k we have |xk + yk | ≤ |xk | + |yk | ≤ kxk∞ + kyk∞ . Consequently kx + yk = sup |xk + yk | ≤ kxk∞ + kyk∞ . k∈N

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7.1.17 We are given 0 < p < 1. (a) Let f (t) = (1 + t)p and g(t) = 1 + tp for t > 0. Then f (0) = 1 = g(0). Also, f ′ (t) = p (1 + t)p−1 = p

1 (1 + t)1−p

and

g ′ (t) = ptp−1 = p

1 t1−p

.

Since 0 < 1 − p < 1, we have t1−p < (1 + t)1−p , and therefore f ′ (t) ≤ g ′ (t) for t > 0. Hence g is increasing faster than f, and therefore f (t) ≤ g(t) for all t ≥ 0. Applying this, given a, b ≥ 0 we have p   p   b b p p p = ap + b p . ≤ a 1+ (a + b) = a 1 + a a (b) Using part (a), if x, y ∈ ℓp , then kx + ykpp =

∞ X

k=1

|xk + yk |p ≤

∞ X

k=1

|xk |p + |yk |p




= kxkpp + kykpp .

|xk |p + |yk |p




= kxkpp + kykpp .

This establishes the Triangle Inequality. (c) Using part (b), kx + ykpp =

∞ X

k=1

|xk + yk |p ≤

∞ X

k=1

7.1.19 If p = 1, then ∞ X |xk | k=1

k

≤

∞ X

k=1

|xk | = kxk1 < ∞.

If 1 < p < ∞ then we have 1 < p′ < ∞. Applying H¨ older’s Inequality, it follows that X 1/p′ 1/p′ X 1/p X ∞ ∞ ∞ ∞ X |xk | 1 1 p = kxkp < ∞. ≤ |xk | k k p′ k p′ k=1

k=1

k=1

k=1

If p = ∞ and we set xk = 1 for every k, then ∞ X |xk |

k=1

k

=

∞ X 1 = ∞. k

k=1

7.1.20 If x ∈ ℓ1 , then for every fixed k we have

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|xk | ≤

∞ X j=1

|xj | = kxk1 .

Therefore kxk∞ = sup |xk | ≤ kxk1 . k

Now let xn = (1, 1, . . . , 1, 0, 0, 0, . . . ), where the 1 is repeated n times. Then kxn k∞ = 1 while kxn k1 = n. Since n can be any positive integer, there is no constant B such that the inequality kxn k1 ≤ B kxn k∞ holds for every n.

7.1.21 We certainly have ℓp ⊆ ℓ∞ for every p. Further, the constant sequence x = (1, 1, 1, . . . ) belongs to ℓ∞ but does not belong to ℓp for any finite p, so the inclusion is proper. Suppose 0 < p ≤ q < ∞ and x ∈ ℓp . If kxk∞ = 1, then kxkq =

X ∞

k=1

q

|xk |

1/q

= ≤

X ∞ k=1 X ∞ k=1

p

q−p

|xk | |xk | |xk |p

1/q

1/q

= kxkp/q ≤ kxkp , p

the last inequality following from the fact that p/q ≤ 1 and kxkp ≥ kxk∞ = 1. To extend to a general vector x ∈ ℓp , apply this inequality to the normalized vector x/kxk∞ . To show that the inclusion is strict, set xk = k −1/p . Then since q/p > 1, we have ∞ X 1 < ∞, kxkqq = q/p k k=1 while

kxkpp =

∞ X 1 = ∞. k

k=1

Another example is xk = (k log2 k)−1/q for k ≥ 2. The Integral Test shows that ∞ X 1 kxkqq = < ∞, k log2 k k=2 while

kxkpp = q

∞ X

k=2

1 = ∞. (k log2 k)p/q

7.1.22 Suppose that x ∈ ℓ for some finite q. If x = 0 then kxkp = 0 for every p, so we are done. Therefore, we may assume x 6= 0, which implies kxk∞ 6= 0. By dividing through by kxk∞ , we may further assume that kxk∞ = 1. Then for every p we have 1 = kxk∞ ≤

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kxkp . In particular, |xk | ≤ 1 for every k. Therefore, for every p ≥ q we have |xk |p ≤ |xk |q , so x ∈ ℓp . Further, for p ≥ q, kxk∞ = 1 ≤ kxkp = ≤

X ∞

k=1

X ∞

k=1

p

1/p

q

1/p

|xk |

|xk |

= kxkq/p q → 1 = kxk∞

as p → ∞,

where the limit exists because kxkq is finite and nonzero. Finally, the vector x = (1, 1, 1, . . . ) satisfies kxk∞ = 1, but kxkp = ∞ for every p < ∞. p

p′

7.1.23 Case 1 < p < ∞. By Exercise 7.1.5, equality holds in ab ≤ ap + bp′ if and only if b = ap−1 . For the normalized case kxkp = kykp′ = 1, equality in H¨ older’s Inequality requires that we have equality in equation (7.9), and this will happen if and only if |yk | = |xk |p−1 for each k. This is equivalent to ′

|yk |p = |yk |p/(p−1) = |xk |p . For the nonnormalized case, if x, y 6= 0 then equality holds in H¨ older’s Inequality if and only if it holds when we replace x and y by x/kxkp and y/kykp′ . Therefore, we must have ′

|yk |p

′ kykpp′

=



|yk | kykp′

p′

=



|xk | kxkp

p

=

|xk |p , kxkpp

k ∈ I.

′

Hence α |xk |p = β |yk |p with α = kykpp′ and β = kxkpp . On the other hand, if either x = 0 or y = 0, then we have equality in H¨ older’s Inequality, and we ′ also have α |xk |p = β |yk |p with α, β not both zero. ′ For the converse direction, suppose that α |xk |p = β |yk |p for each k ∈ I, where α, β ∈ C are not both zero. If α = 0, then yk = 0 for every k, and hence we trivially have kxyk1 = 0 = kxkp kykp′ . Likewise, equality holds trivially if β = 0. Therefore, we can assume both α, β 6= 0, and by dividing both sides ′ by β, we may assume that β = 1 and α > 0. Then we have |yk |p = α |xk |p , so X X ′ ′ kykpp′ = |yk |p = α |xk |p = α kxkpp . ′

k∈I

k∈I

If either x = 0 or y = 0 then equality holds trivially in H¨ older’s Inequality, so let us assume both x, y 6= 0. Then we have

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265 ′

|yk |p

′ kykpp′

=

α|xk |p |xk |p . p = αkxkp kxkpp

By the work above, this implies that equality holds in H¨ older’s Inequality. Case p = 1, p′ = ∞. Set M = supk |yk |. Suppose equality holds in H¨ older’s Inequality, i.e.,     X X |xk yk | = |xk | sup |yk | . k∈I

Then

X

|xk yk | =

X

(M − |yk |) |xk | = 0,

k∈I

Hence

k

k∈I

k∈I

X k∈I

M |xk |.

but 0 ≤ M − |yk | for every k, so we must have (M − |yk |) |xk | = 0 for every k. Thus whenever xk 6= 0, we must have |yk | = M. Conversely, if |yk | = M for all k such that xk 6= 0, equality holds in H¨ older’s Inequality. 7.1.24 Case 1: r = 1. This reduces to the ordinary H¨ older Inequality. Case 2: Any one of p, q, r is ∞. If p = ∞ then r = q and the inequality reduces to kxykq ≤ kxk∞ kykq , which is easy to prove. Likewise the case q = ∞ is simple. If r = ∞, then we must have p = q = ∞ as well, again reducing to an easy case. Case 3: The remaining possibilities. In this case we have 1 ≤ p, q, r < ∞. We cannot have p = r, because that forces q = ∞, which is covered by Case 2. Likewise we cannot have q = r. Consequently, if we set p u = , r then 1 < u < ∞. Since 1 1 r r + = + = 1, u q/r p q we have u′ = q/r. Assume x ∈ ℓp and y ∈ ℓq . Let
w = |x|r = |xk |r k∈N

and

z = |y|r =

|yk |r




k∈N

.

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Then kwkuu = and ′

kzkuu′ =

∞ X

|wk |u =

∞ X

|zk |u =

k=1

k=1

′

′

∞ X

|xk |ru =

∞ X

|yk |ru =

k=1

k=1

′

∞ X

|xk |p = kxkpp

∞ X

|yk |q = kykqq .

k=1

k=1

Thus w ∈ ℓu and z ∈ ℓu . By the ordinary H¨ older’s Inequality, we therefore have wz ∈ ℓ1 and ∞ X

kxykrr =

k=1

|xk yk |r =

∞ X

k=1

|wk zk |

= kwzk1 ≤ kwku kzku′ ′

= kxkp/u kykq/u p q = kxkrp kykrq . So the result follows by taking rth roots.

7.1.25 (a) To see that D is closed, suppose that {xn }n∈N is a sequence of points in D and x is a point in ℓp such that xn → x in ℓp -norm. Then, by the Reverse Triangle Inequality, kxkp − kxn kp ≤ kx − xn kp → 0 as n → ∞.

Therefore

kxkp = lim kxn kp ≤ 1, n→∞

so x ∈ D. Thus D is closed. (b) If m 6= n, then

kδm − δn kp = 21/p .

Consequently, if 0 < ε < 21/p then there does not exist an N such that kδm − δn kp < ε for m, n > N. Thus {δn }n∈N is not Cauchy in ℓp . Moreover, if we choose any subsequence {δnk }k∈N , then for every j 6= k we have kδnj − δnk kp = 21/p . (c) Part (a) shows that D is not sequentially compact. Consequently, since ℓp is a normed space, it is not compact. 7.1.29 (a) Fix 1 ≤ p < ∞. Given a sequence x = (x1 , x2 , . . . ) ∈ ℓp , set xN = = (x1 , . . . , xN , 0, 0, . . . ).

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Since p is finite, we therefore have

p kx − xN kp = (0, . . . , 0, xN +1 , xN +2 , . . . ) p ∞ X

=

k=N +1

|xk |p → 0

as n → ∞.

This shows that every vector in ℓp is the limit, in ℓp -norm, of vectors from c00 . Therefore c00 is dense in ℓp (but it is a proper subspace because not every vector in ℓp belongs to c00 ). Now let x = (1, 1, 1, . . . ), which belongs to ℓ∞ . Then kx − yk∞ ≥ 1 for every y ∈ c00 . Consequently c00 is not dense in ℓ∞ .
Next, suppose that y = y(k) k∈N is any vector in c0 . For each n ∈ N, define
yn = y(1), . . . , y(n), 0, 0, . . . .

Then yn ∈ c00 , and ky − yn k∞ → 0 as n → ∞. Therefore every vector in c0 is a limit of vectors from c00 , so c00 is dense in c0 .

(b) Let x = (2−k )k∈N . This vector belongs to ℓp but not to c00 . For each n, set
xn = 2−1 , . . . , 2−n , 0, 0, . . . .

Then {xn }n∈N is a sequence in c00 that is Cauchy in ℓp -norm. However, if xn → y in ℓp -norm then xn converges to y componentwise, which implies that y = x. Since x does not belong to ℓp , it follows that there is no vector y ∈ c00 such that xn → y in ℓp -norm. Hence c00 is incomplete with respect to k · kp . 7.1.26 (a) Suppose that α > 1/p and x = (xk )k∈N satisfies equation (7.17). Then, since αp > 1, we have kxkpp =

∞ X

k=1

|xk |p ≤

∞ X

k=1

C p k −αp < ∞.

Hence x ∈ ℓp . Thus there is a sequence in ℓp that satisfies equation (7.17) if α > 1/p. (b) Set xk = k −1/p . We have kxkpp =

∞ X 1 = ∞, k

k=1

so x does not belong to ℓp . However, xk = k −α , so x trivially satisfies equation (7.17) with C = 1. Now define x1 = 0 and xk = (k ln2 k)−1/p for k > 2. We have

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kxkpp Since

Z

∞

e

=

∞ X

k=2

1 . k ln2 k

∞ 1 1 = 1, dx = − ln x e x ln2 x

the Integral Test implies that kxkpp < ∞. Therefore x ∈ ℓp . Also, |xk | =

1 1 1 1 1 = 1/p ≤ 1/p = Ck −1/p , k k (k ln2 k)1/p (ln2 k)1/p (ln2 2)1/p

where

C = (ln2 2)−1/p .

(c) Set x2j = 2−αj/2 , and let xk = 0 for all other k. Then, since −αp/2 < 0, we have kxkpp =

∞ X

k=1

|xk |p =

∞ X j=1

|x2j |p =

∞ X j=1

2−αpj/2 < ∞.

Hence x ∈ ℓp . On the other hand, for every j we have (2j )α |x2j | = 2αj 2−αj/2 = 2αj/2 → ∞ as j → ∞, so sup k α |xk | = sup (2j )α |x2j | = ∞. k∈N

j∈N

Thus, there cannot be a finite C such that |xk | ≤ C k −α for every k.

(d) Suppose that x = (xk )k∈N is monotonically decreasing. Since x ∈ ℓp , we must have xk → 0, and therefore xk ≥ 0 for every k. Given n ∈ N, we have |xk | ≥ |x2n | for all k ≤ 2n. Consequently, n|x2n |p ≤ Hence |x2n | ≤

2n X

k=n+1

|xk |p ≤

∞ X

k=1

|xk |p = kxkpp .

21/p kxkp kxkp = = A (2n)−1/p , 1/p n (2n)1/p

where A = 21/p kxkp . Also, |x2n+1 | ≤ |x2n | ≤ =

A (2n + 1)1/p 1/p (2n) (2n + 1)1/p A (2n)1/p 1/p (2n + 1) (2n + 1)1/p

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=

A 21/p (2n + 1)1/p

= 21/p A (2n + 1)−1/p . Therefore, for every k, both even and odd, |xk | ≤ 21/p A k −1/p = C k −1/p . Thus, equation (7.17) is satisfied with α = 1/p and C = 21/p A = 22/p kxkp . 7.1.27 The proof is very similar to the proof that ℓp (N) is a Banach space. For example, if {xn }n∈N is a Cauchy sequence in ℓp (I) and we write xn = (xn (i))i∈I , then for each fixed i we have that (xn (i))n∈N is a Cauchy sequence of scalars, and hence converges to some scalar x(i). For a given n, at most countable many components of xn can be nonzero. As a countable union of countable sets is countable, at most countably many components of x can be nonzero. An argument similar to the one used in the proof of Theorem 7.1.15 then shows that xn → x in the norm of ℓp (I), so ℓp (I) is complete. 7.1.30 Suppose that {xn }n∈N is a sequence of vectors in c0 that converges absolutely, i.e., ∞ X kxn k∞ < ∞. n=1

This assumption implies

lim

N →∞

∞ X

n=N +1

kxn k∞ = 0.

(A).

Each xn is a vector in c0 , so let us write the components of xn as

xn = xn (1), xn (2), . . . = xn (k) k∈N .

For each fixed index k ∈ N we have

|xn (k)| ≤ sup |xn (j)| = kxn k∞ . j∈N

Therefore, holding k fixed we compute that ∞ X

n=1

|xn (k)| ≤

∞ X

n=1

kxn k∞ < ∞.

P∞ This tells us that n=1 xn (k) is an absolutely convergent series of scalars. Since the complex plane C is complete, an absolutely convergent series of

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scalars must converge to a scalar. Therefore we can define x(k) =

∞ X

xn (k).

n=1

Each x(k) is a scalar, so we can define a sequence x by x =


x(k) k∈N =




x(1), x(2), . . .

=

X ∞

xn (1),

n=1

∞ X

n=1

 xn (2), . . . .

P∞ We will show that n=1 xn converges to x with respect to the norm of c0 , which is the sup-norm. To do this, we must show that the partial sums sN =

N X

xn =

X N

xn (1),

n=1

n=1

N X

n=1

xn (2), . . .



converge to x in ℓ∞ -norm as N → ∞. We know, just from the definition of x, that sN converges componentwise to x, but our task is to show that kx − sN k∞ → 0 as N → ∞. Noting that x − sN =

 X ∞

xn (1),

n=N +1

∞ X

 xn (2), . . . ,

n=N +1

we compute that kx − sN k∞

X ∞ = sup xn (k) k∈N n=N +1

≤ sup

∞ X

k∈N n=N +1

≤ =

∞ X

|xn (k)|

sup |xn (k)|

(why?)

n=N +1 k∈N ∞ X

n=N +1

kxn k∞

→ 0 as N → ∞, where at the end we have applied equation (A). Thus, the partial P∞ sums sN converge to x in ℓ∞ -norm, and, by definition, this tells us that n=1 xn = x. It only remains to prove that x belongs to c0 . If we fix ε > 0, then we can choose N large enough that

c Solutions 2019 Christopher Heil

271 ∞ X

n=N +1

kxn k∞ < ε.

Since x1 , . . . , xN ∈ c0 , there exists some M such that k > M

|xn (k)|
M, X ∞ N X X ∞ |x(k)| = |xn (k)| + xn (k) ≤ n=1

n=1

≤

n=N +1

N ∞ X X ε + N n=1

n=N +1

|xn (k)|

kxn k∞

< ε + ε = 2ε. Therefore P∞x(k) → 0 as k → ∞, so x ∈ c0 . Thus the absolutely convergent series n=1 xn converges to x, and x is an element of c0 . Theorem 1.2.8 therefore implies that c0 is complete. 7.2.3 The cases p = 1 and p = ∞ are straightforward. Assume 1 < p < ∞. As in the proof of Theorem 7.1.7, for a, b ≥ 0 we have ′

bp ap + ′. ab ≤ p p ′

Consequently, if f ∈ Lp (E) and g ∈ Lp (E) satisfy kf kp = 1 = kgkp′ , then kf gk1 =

Z

E

|f (x) g(x)| dx ≤

Z  E

′

|f (x)|p |g(x)|p + p p′



dx =

1 1 + = 1. p p′

For general nonzero f, g, we apply this result to the normalized vectors f /kf kp and g/kgkp′ to obtain

f kf gk1 g

=

≤ 1. kf kp kgkp′ kf kp kgkp′ 1

7.2.4 The only issue is to prove that the Triangle Inequality holds. This proceeds exactly as in the corresponding proof for ℓp . The result is easy if p = 1 or p = ∞, so assume that 1 < p < ∞. Then

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272

kf + gkpp = ≤ ≤

Z

E

Z

E

Z

|f (x) + g(x)|p−1 |f (x) + g(x)| dx |f (x) + g(x)|p−1 |f (x)| dx +
′ p−1 p

|f (x) + g(x)|

E

Z

+

E

=

Z

E

Z

E

|f (x) + g(x)|p−1 |g(x)| dx

1/p′ Z 1/p dx |f (x)|p dx E


′ p−1 p

|f (x) + g(x)|

1/p′ Z 1/p dx |g(x)|p dx E

(p−1)/p Z 1/p |f (x) + g(x)|p dx |f (x)|p dx Z

+

E

E

(p−1)/p Z 1/p p p |f (x) + g(x)| dx |g(x)| dx E

= kf + gkp−1 kf kp + kf + gkp−1 kgkp , p p where we have applied H¨ older’s Inequality and used the fact p′ = p/(p − 1). Dividing both sides by kf + gkp−1 therefore yields kf + gkp ≤ kf kp + kgkp . p 7.2.7 Suppose that fn → f pointwise a.e. Let Z be the set of points such that fn (x) does not converge to f (x). If gn = fn a.e. then Zn = {fn 6= gn } has measure zero. If g = f a.e. then X = {f 6= g} has measure zero. Hence ∞ S

S = Z ∪ X ∪

Zn

n=1

has measure zero. If x ∈ / S then gn (x) = fn (x) → f (x) = g(x). Hence gn → g pointwise a.e. 7.2.9 (a), (b) This problem parallels the material in Section 7.1.5 in the same way that Exercise 7.3.5 parallels Theorem 7.1.15. (c) For simplicity, consider E = [0, 1]. Let f = 2 χ[0, 12 ) and g = 2 χ[ 21 ,1] . Since p − 1 < 0, we have kf kpp

=

Z

0

1

p

|f (x)| dx = 2

p

Z

1/2

dx = 2p−1 < 1,

0

and a similar calculation shows that kgkpp = 2p−1 < 1. Therefore f, g ∈ B1 (0). The midpoint of the line segment joining f to g is h = Since

f +g = χ[0,1] . 2

c Solutions 2019 Christopher Heil

Z

khkpp =

273 1

0

Z

|h(x)|p dx =

1

dx = 1,

0

we have h ∈ / B1 (0). Therefore B1 (0) is not convex. Given an arbitrary measurable set E with |E| > 0, it follows from Problem 2.3.19 there exist subsets E1 , E2 such that 0 < |E1 | = |E2 | < ∞. Let 

c =

2p−1 |E1 |

1/p

,

and set f = c χE 1 ,

g = cχE2 ,

Then kf kpp = kgkpp = while khkpp =

Z

E1 ∪E2

Z

h =

cp =

E1

Z

E1

c f +g = χE1 ∪E2 . 2 2 2p−1 = 2p−1 , |E1 |

2p−1 1 cp = 2 |E | = 1. 1 2p |E1 | 2p

Therefore B1 (0) is not convex. 7.2.10 We give the details of the proof of Theorem 7.2.10 for the endpoint cases. (a) Case p = 1. By H¨ older’s Inequality, we have Z sup f g ≤ sup kf k1 kgk∞ = kf k1 . kgk∞ =1

(A)

kgk∞ =1

E

To prove that equality holds, fix any nonzero function f ∈ L1 (E). Let |α(x)| = 1 satisfy α(x) f (x) = |f (x)|, and define g(x) = α(x) for all x. Then kgk∞ = 1, and Z Z Z |f (x)| dx = kf k1 . f (x) α(x) dx = fg = E

E

E

This shows that the supremum in equation (A) equals kf k1 . Case p = ∞. By H¨ older’s Inequality, we have Z sup f g ≤ sup kf k∞ kgk1 = kf k∞ . kgk1 =1

E

(B)

kgk1 =1

To prove that equality holds, fix any nonzero function f ∈ L∞ (E) and ε > 0. Then there exists a set A ⊆ E such that 0 < |A| < ∞ and

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274

|f (x)| ≥ kf k∞ − ε, Let |α(x)| = 1 satisfy

α(x) f (x) = |f (x)|,

α(x) χA (x) . |A|

g(x) = Then kgk1 = 1 and

a.e. x ∈ A.

Z

fg =

E

1 |A|

1 = |A| ≥

1 |A|

Z

f (x) α(x) dx

A

Z

A

|f (x)| dx

Z  A

 kf k∞ − ε dx

= kf k∞ − ε.

Since ε is arbitrary, it follows that the supremum in equation (B) equals kf k∞ . 7.2.11 For 1 ≤ p < ∞ and α 6= −1/p, α

kx χ[0,1] kpp =

Z

1

αp

x

dx = lim

t→0+

0

Z

1

xαp dx = lim

t→0+

t

1 − tαp+1 . αp + 1

This is finite if αp + 1 > 0, i.e., α > −1/p. A similar calculation shows that the Lp -norm is infinite if α = −1/p. Hence xα χ[0,1] (x) belongs to Lp (R) if and only if −1/p < α < ∞. Similarly, if 1 ≤ p < ∞ and α 6= −1/p then Z ∞ tαp+1 − 1 αχ p , xαp dx = lim kx [1,∞) kp = t→∞ αp + 1 1 which is finite if αp + 1 < 1, i.e., α < −1/p. The case α = −1/p leads to an infinite Lp -norm. 7.2.12 (a) Case 1 ≤ p < ∞. Assume that fn ∈ Lp (E), fn → f a.e., and C = sup kfn kp < ∞. n

Then by Fatou’s Lemma, Z Z Z p p p |fn |p ≤ C p , lim inf |fn | ≤ lim inf |f | = kf kp = E

so f ∈ Lp (E).

E n→∞

n→∞

E

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Case p = ∞. Assume that fn ∈ Lp (E), fn → f a.e., and C = sup kfn k∞ < ∞. n

Then for almost every x we have |f (x)| = lim |fn (x)| ≤ sup kf k∞ ≤ C, n→∞

n

and therefore kf k∞ ≤ C. (b) Case 1 ≤ p < ∞. Let E = [0, 1], and set fn (x) = x−1/p χ[ n1 ,1] (x). Then fn is bounded and hence belongs to Lp [0, 1]. Further, fn converges pointwise a.e. to f (x) = x−1/p . However, kf kpp = so f ∈ / Lp [0, 1].

Z

0

1

|f |p =

Z

0

1

1 dx = ∞, x

Case p = ∞. Let E = [0, 1], and set fn (x) =

1 χ 1 (x). x [ n ,1]

Then fn is bounded and hence belongs to L∞ [0, 1]. Further, fn converges pointwise a.e. to 1 f (x) = . x However, kf k∞ = ∞, so f ∈ / L∞ [0, 1].

7.2.13 Fix α > 0. Applying Tchebyshev’s Inequality to the function |f |p , we compute that {|f | > α} = {|f |p > αp } Z 1 ≤ p |f |p α |f |p >αp Z 1 = p |f |p α |f |>α Z 1 |f |p . ≤ p α E

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7.2.14 Suppose that |E| < ∞, and let f be a measurable function on E. If kf k∞ = 0 then f = 0 a.e. and we are done. Therefore we may assume that kf k∞ > 0, which implies that |E| > 0. Since Z 1/p Z 1/p p kf kp = |f | ≤ kf k∞ 1 = kf k∞ |E|1/p , E

E

we have lim sup kf kp ≤ kf k∞ lim sup |E|1/p = kf k∞ . p→∞

p→∞

This inequality holds even if kf k∞ = ∞. To prove the opposite inequality, we break into cases. Suppose first that M = kf k∞ < ∞. Then given ε > 0, the set A = {|f | > M − ε} must have positive measure. Therefore kf kp =

Z

E

|f |p

1/p

≥

Z

A

|f |p

1/p

= (M − ε) |A|1/p .

Since |A| > 0, we therefore have lim inf kf kp ≥ (M − ε) lim inf |A|1/p = M − ε. p→∞

p→∞

This is true for every ε > 0, so we conclude that lim inf kf kp ≥ M = kf k∞ . p→∞

This completes the proof under the assumption that M = kf k∞ is finite. Now suppose that kf k∞ = ∞. Then given any R, the set A = {|f | > R} has positive measure. Repeating the same argument as above shows that lim inf p→∞ kf kp ≥ R. Since R is arbitrary, this implies that lim inf kf kp = ∞ = kf k∞ , p→∞

so the proof is complete for this case. To see that the hypothesis |E| < ∞ is necessary, consider the constant function f (x) = 1, which belongs to L∞ (R). We have kf k∞ = 1, but kf kp = ∞ for every finite p. 7.2.15 This problem is very similar to part of Exercise 7.1.5. p′ p By Exercise 7.1.5, equality holds in ab ≤ ap + bp′ if and only if b = ap−1 . Looking at the proof of H¨ older’s Inequality for Lp (E) given in Exercise 7.2.3, for the normalized case kf kp = kgkp′ = 1, we see that equality holds in

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H¨ older’s Inequality if and only if equality holds in the inequality Z

E

|f (x) g(x)| dx ≤

Z  E

′

|f (x)|p |g(x)|p + p p′



dx.

(A)

Since we always have ′

|f (x) g(x)| ≤

|f (x)|p |g(x)|p + , p p′

equality holds in equation (A) holds if and only if ′

|g(x)|p |f (x)|p + a.e. |f (x) g(x)| = p p′ This is equivalent to |g(x)| = |f (x)|p−1 a.e., which is itself equivalent to ′

|g(x)|p = |g(x)|p/(p−1) = |f (x)|p . For the nonnormalized case, if f, g are not the zero function, equality holds in H¨ older’s Inequality if and only if it holds when we replace f and g by f /kf kp and g/kgkp′ . Therefore, we must have ′

|g(x)|p ′ kgkpp′

=



|g(x)| kgkp′

p′

=



|f (x)| kf kp

p

=

|f (x)|p kf kpp

a.e.

′

Hence α |f (x)|p = β |g(x)|p a.e. with α = kgkpp′ and β = kf kpp . On the other hand, if either f = 0 or g = 0, then we have equality in H¨ older’s Inequality, ′ and we also have α |f (x)|p = β |g(x)|p with α, β not both zero. ′ For the converse direction, suppose that α |f (x)|p = β |g(x)|p a.e., where α, β ∈ C are not both zero. If α = 0, then g(x) = 0 a.e., and hence we trivially have kf gk1 = 0 = kf kp kgkp′ in this case. Likewise, equality holds trivially if β = 0. Therefore, we can assume both α, β 6= 0, and by dividing both sides ′ by β, we may assume that β = 1 and α > 0. That is, |g(x)|p = α |f (x)|p a.e., so Z Z p′ p′ kgkp′ = |g(x)| dx = α |f (x)|p dx = α kf kpp . ′

E

E

If either f = 0 or g = 0 then equality holds trivially in H¨ older’s Inequality. On the other hand, if both f, g 6= 0 then ′

|g(x)|p ′ kgkpp′

=

|f (x)|p α|f (x)|p = , p αkf kp kf kpp

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which implies that equality holds in H¨ older’s Inequality. 7.2.16 (a) We are given 0 < |E| < ∞ and 0 < p < q ≤ ∞. If q = ∞ and f ∈ L∞ (E), then Z Z 1 = kf kp∞ |E|. |f |p ≤ kf kp∞ kf kpp = E

E

This gives us the desired norm inequality and proves that L∞ (E) ⊆ Lp (E). Suppose q < ∞ and f ∈ Lq (E). Then 1 ≤ q/p < ∞, so we can apply H¨ older’s Inequality to calculate that kf kpp

=

Z

E

p

|f | · 1 ≤

Hence

Z

E

p q/p

(|f | )

p/q Z

1

E

(q/p)′

= kf kpq |E|

q−p q

.

1

1

kf kp ≤ kf kq |E| p − q .

This gives us the desired norm inequality and proves that Lq (E) ⊆ Lp (E). It remains to prove that Lq (E) is a proper subset of Lp (E). By Problem 2.3.20, there exist disjoint measurable sets Ak ⊆ E such that |Ak | = 2−k |E|, If q = ∞, then set

∞ X

f =

k ∈ N.

2k/(2p) χAk .

k=1

This function is unbounded, but kf kpp

=

∞ Z X k=1

2k/2 = Ak

∞ X

k=1

2k/2 |Ak | =

∞ X

k=1

2−k/2 |E| < ∞.

Therefore f belongs to Lp (E), but does not belong to L∞ (E). On the other hand, if q < ∞ then set ∞ X 1 k/q 2 χA k . 1/q k

f =

k=1

In this case, kf kqq but

=

∞ Z X

k=1

Ak

∞ X 1 k 1 2 = |E| = ∞, k k k=1

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kf kpp = =

279 ∞ Z X

Ak

k=1

1 kp/q 2 k p/q

∞ X 1 kp/q −k 2 2 |E| k p/q

k=1

= |E| ≤ |E|

∞ X 1 k( pq −1) 2 p/q k k=1

∞ X

p

2k( q −1)

k=1

< ∞

(since

p q

− 1 < 0).

Therefore f belongs to Lp (E), but does not belong to Lq (E). (b) Now we are given that |E| = ∞. To show that Lp (E) is not contained in Lq (E), we can adapt the argument from part (a). Let F be a measurable subset of E such that 0 < |F | < ∞. By Problem 2.3.20, there exist disjoint measurable sets Ak ⊆ F such that |Ak | = 2−k |F |,

k ∈ N.

If q = ∞, then we consider f =

∞ X

2k/(2p) χAk ,

k=1

while if q < ∞, then we take f =

∞ X 1 k/q 2 χA k . k 1/q k=1

In either case we obtain a function f that belongs to Lp (E) but does not belong to Lq (E). Now we consider the converse direction. If q = ∞, then the constant function f = 1 belongs to L∞ (E) but does not belong to Lp (E). Suppose that q < ∞. By Problem 2.3.20, there exist disjoint measurable sets Ak ⊆ E such that |Ak | = 1 for every k. Set ∞ X 1 χA k . 1/p k

f =

k=1

Then kf kpp

=

∞ Z X

k=1

Ak

∞ X 1 1 = = ∞. k k k=1

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However, kf kqq because

q p

=

∞ Z X k=1

1

Ak

∞ X 1 < ∞, q/p k

=

k q/p

k=1

q

> 1. Thus f belongs to L (E), but does not belong to Lp (E).

7.2.17 If p = 1 then the result is Tonelli’s Inequality, so suppose 1 < p < ∞. Define Z F (y) = |f (x, y)| dx. E

Then the left-hand side of equation (7.22) can be rewritten as: Z Z F

p 1/p Z 1/p |f (x, y)| dx dy = |F (y)|p dy = kF kp .

E

F

We estimate this as follows: Z F (y)p−1 F (y) dy kF kpp = F

=

Z

p−1

F (y)

F

= ≤ =

E

|f (x, y)| dx dy

F (y)p−1 |f (x, y)| dy dx

F

Z Z

(p−1)p′

F (y)

dy

F

Z Z E

=

E

Z Z E

Z

p

F (y) dy

F

kF kp−1 p

1/p′ Z

1/p′ Z

F

Z Z E

F

p

(Tonelli) p

F

|f (x, y)| dy 1/p

dx

1/p

dx,

|f (x, y)|p dy

|f (x, y)| dy

1/p

1/p

dx

(H¨older)

dx.

Dividing through by kF kp−1 we obtain p kF kp ≤

Z Z E

F

p

|f (x, y)| dy

which is equation (7.22). 7.2.18 (a) Assume that 1 < p < ∞. Given a ≤ x < y ≤ b, we use the Fundamental Theorem of Calculus and H¨ older’s Inequality to compute that

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Z |f (y) − f (x)| =

y

x

≤

Z

′ f (t) dt

y

x

′

p

|f (t)| dt

1/p Z

y

p′

1 dt x

1/p′

′

≤ kf ′ kp |y − x|1/p . This shows that f is H¨ older continuous with H¨ older exponent 1/p′ and H¨ older ′ constant kf kp . The case p = ∞ is similar (and was established earlier in Problem 6.4.10). For this case we have 1/p′ = 1, and H¨ older’s Inequality tells us that Z y Z y ′ ′ |f (y) − f (x)| = f (t) dt ≤ kf k∞ 1 dt = kf ′ k∞ |y − x|. x

x

This shows that f is Lipschitz, which is H¨ older continuity with exponent 1. (b) The function g is given by g(0) = 0 and g(x) = −

1 , ln x

0 0 for every k and ∞ X 1 < ∞. nkp k=1

Since |Enk | > 0, for each k we can choose a set Fk ⊆ Enk that satisfies 0 < |Fk | < ∞. Set F = ∪Fk , and consider the function  nk |Fk |−1/p , x ∈ Fk , f (x) = 0, x∈ / F.

We have

Z

|f |p dx =

but Z

|f φ|p dx =

∞ Z X k=1

Fk

∞ Z X k=1

Fk

∞ X 1 1 = < ∞, nkp |Fk | nkp k=1

∞ Z ∞ X X nkp |φ(x)|p 1 = ∞. ≥ = p nkp |Fk | Fk nk |Fk | k=1

k=1

Thus f ∈ Lp (R) but f φ ∈ / Lp (R).

Case p = ∞. Again we prove the contrapositive statement. If φ ∈ / L∞ (R), ∞ then the constant function f = 1 belongs to L (R) but f φ = φ ∈ / L∞ (R). 7.2.20 We break into endpoint and nonendpoint cases. Endpoint Cases. Endpoint Case 1 : pk = 1 for some k. Without loss of generality, assume that pn = 1. Then 1 ≤

1 1 1 1 1 = + ···+ = + ··· + + 1. r p1 pn p1 pn−1

Consequently we must have r = 1 and p1 = · · · = pn−1 = ∞. Therefore this case reduces to the easy estimate kf1 · · · fr k1 ≤ kf1 k∞ · · · kfpn−1 k∞ kfn k1 . Endpoint Case 2 : pk = ∞ for some k. Without loss of generality, assume that pn = ∞. In this case we have kf1 · · · fn kr ≤ kf1 · · · fn−1 kr kfn k∞ . Since we still have

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1 1 1 + ··· + = , p1 pn−1 r we can replace n with n − 1 and reconsider the problem. If it still contains an endpoint case, we can reduce again, until we reach a problem that contains no endpoint cases. Non-Endpoint Cases. By the above work, we have reduced to the case where 1 < pk < ∞ for every k. This implies that 1 1 1 = + ···+ > 0, r p1 pn so we also have 1 ≤ r < ∞. First proof: Induction. Step 1: Assume r = 1. The base case n = 2 is established in H¨ older’s Inequality. For the inductive step, assume that the result is true for some n ≥ 2. Then we have indices that satisfy 1 < p1 , . . . , pn , pn+1 < ∞ and

1 1 1 + ···+ + = 1. p1 pn pn+1

For simplicity of notation, let p = pn+1 , and let q be the number that satisfies 1 1 1 + ···+ . = q p1 pn Note that 1 < p, q < ∞, and 1 1 + = 1. p q Thus q = p′ . Let gj = |fj |q . Let rj =

pj . q

Then 1 < rj < ∞, and 1 1 q q q + ···+ = + ···+ = = 1. r1 rn p1 pn q Further, kgj krrjj

=

Z

E

gj


rj

=

Z

E

|fj |q


pj /q

(A)

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=

Z

E

|fj |pj = kfj kppjj < ∞,

so gj ∈ ℓrj . Taking roots, this also tells us that kgj krj = kfj kppjj /rj = kfj kqpj .

(B)

Because of equation (A) and the fact that gj ∈ ℓrj , we can apply the inductive hypothesis to obtain g1 · · · gn ∈ ℓ1 and kg1 · · · gn k1 ≤ kg1 kr1 · · · kg1 krn .

(C)

Let f = f1 · · · fn . Our goal is to show that f fn+1 ∈ ℓ1 . We have Z Z q q kf kq = |f | = |f1 · · · fn |q E

E

=

Z

E

|g1 · · · gn |

= kg1 · · · gn k1 < ∞, so f ∈ ℓq . Taking roots and applying equations (B) and (C), we see that 1/q

kf kq = kg1 · · · gn k1

1/q ≤ kg1 k1/q r1 · · · kg1 krn = kf1 kp1 · · · kf1 kpn .

(D)

Since q = p′ and fn+1 ∈ ℓp , by applying the base two-term case of H¨ older’s Inequality and the above equations we see that kf1 · · · fn fn+1 k1 = kf fn+1 k1 ≤ kf kq kfn+1 kp

two term H¨ older

≤ kf1 kp1 · · · kf1 kpn kfn+1 kp

by equation (D)

= kf1 kp1 · · · kf1 kpn kfn+1 kpn+1

by definition of p.

This completes the inductive step. Step 2: Assume 1 < r < ∞. Let fk ∈ Lpk (E) be given, and set gk = |fk |r . Let qk = pk /r. Then 1 1 r r + ···+ = + ··· + = 1. q1 qn p1 pn Also

c Solutions 2019 Christopher Heil

kgk kqk = =

Z

E

|gk |qk

E

|fk |r

Z

Therefore kf1 . . . fn krr =

285

1/qk


pk /r Z

E

r/pk

=

Z

E

|f1 · · · fn |r =

|fk |pk

Z

E

r/pk

= kfk krpk .

|g1 · · · gn |

≤ kg1 kq1 · · · kgn kqn

= kf1 krp1 · · · kfn krpn .

The result then follows by taking rth roots. Second proof: Jensen’s Inequality. We are given 1 < p1 , . . . , pn < ∞ and 1 ≤ r < ∞. Set xj = pj ln aj . Then since r r r + ···+ = = 1, p1 pn r we use the Discrete Jensen Inequality and the convexity of ex to compute that (a1 · · · an )r = erx1 /p1 · · · erxn/pn = ex1 /(p1 /r) · · · exn /(pn /r) ≤

ex1 exn + p1 /r pn /r

=

r apnn r ap11 + ··· + . p1 pn

(Jensen)

Choose fk ∈ Lpk (E), and assume first that kfk kpk = 1 for each k. Then  Z Z  r |f1 |p1 r |fn |pn |f1 · · · fn |r ≤ + ··· + p1 pn E E Z Z r r p1 = |f1 | + · · · + |fn |p1 p1 E pn E r r kf1 kpp11 + · · · + kfn kppnn = p1 pn r r + ··· + = p1 pn = 1. Thus, for the normalized case we have

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kf1 · · · fn kr ≤ 1. Consequently, if we let fk be any functions in Lpk (E) then by applying the preceding inequality to the normalized functions fk /kfk kpk , we obtain kf1 · · · fn kr ≤ 1. kf1 kp1 · · · kfn kpn 7.2.21 (a) First note that by applying part (d) of Problem 4.6.21 to the function f p , we have Z ∞ Z p {f > t} dt. |f (x)|p dx = E

0

“⇒.” Assume that f ∈ Lp (E). Then Z |f (x)|p dx ∞ > kf kpp = E ∞

Z

=

0

XZ

=

k∈Z

k∈Z

=

{f > 2k+1 } dt

X


2(k+1)p − 2kp {f > 2k+1 }

k∈Z


2(k+1)p 1 − 2−p ω(2k+1 )

1 − 2−p

=

kf kpp

2(k+1)p

2kp

k∈Z

P

{f p > 2(k+1)p } dt

k∈Z

X

=

{f p > t} dt

2(k+1)p

2kp

XZ

=

2(k+1)p

2kp

XZ

≥

“⇐.” Assume that

{f p > t} dt


X

2kp ω(2k ).

k∈Z

2kp ω(2k ) < ∞. Then Z = |f (x)|p dx

k∈Z

=

E ∞

Z

0

{f p > t} dt

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=

287

XZ

= =

XZ

XZ

{f p > 2kp } dt

2(k+1)p

{f > 2k } dt

k∈Z

2kp

X


2(k+1)p − 2kp {f > 2k }

X k∈Z

=

p {f > t} dt

2(k+1)p

2kp

k∈Z

k∈Z

=

2

2kp

k∈Z

≤

(k+1)p


2kp 2p − 1 ω(2k )

2p − 1

< ∞.


X

2kp ω(2k ).

k∈Z

(b) Applying part (d) of Problem 4.6.21 to the function |f |p and using the change of variables s = t1/p (compare Theorem 6.5.6), we compute that Z Z ∞ p {|f |p > t} dt |f (x)| dx = E

0

=

Z

0

=

Z

0

= p

∞

{|f | > t1/p } dt

∞

Z

{|f | > s} psp−1 ds

∞

ω(s) sp−1 ds.

0

This holds in the extended real sense, i.e., both sides are finite and equal, or both sides are infinite. 7.2.22 Let ω(t) = {|f | > t} be the distribution function of f. Then for each t > 0 we have Z Z ′ ′ f (x) dx ≤ C |{|f | > t}|1/p = C ω(t)1/p . t dx ≤ tω(t) = |f |>t

Since 1 −

1 p′

|f |>t

= p1 , it follows that ′

tω(t)1/p = tω(t) ω(t)−1/p ≤ C,

t > 0,

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and hence ω(t) ≤ C p t−p ,

t > 0.

Similar calculations using q in place of p likewise hold, so we also have ω(t) ≤ C q t−q ,

t > 0.

Fix p < r < q, and note that r − p − 1 > −1 We therefore compute that Z ∞ Z r tr−1 ω(t) dt = r 0

r − q − 1 < −1.

and

1

tr−1 ω(t) dt + r

0

≤ Cp = Cp

Z

Z

1

1

tr−1 t−p dt + C q

0

Z

∞

1

tr−p−1 dt + C q

0

Z

Z

tr−1 ω(t) dt ∞

tr−1 t−q dt

1

∞

tr−q−1 dt

1

< ∞. Applying Problem 7.2.21, we conclude that Z Z ∞ r tr−1 ω(t) dt < ∞, |f (x)| dx = r 0

E

so f ∈ Lr (E).

7.2.23 (a) We are given that f ∈ L1 (E). If kf kp = ∞ then there is nothing to prove, so we may assume that f ∈ Lp (E). Since |E| = 1, we can apply Jensen’s Inequality to the concave function − ln x. Specifically, it follows from Theorem 6.6.10 that 1/p Z p ln kf kp = ln |f | E

Z

1 |f |p ln p E Z 1 ln |f |p ≥ p E Z = ln |f |. =

E

(b) First we establish some facts that will allow us to apply the Dominated Convergence Theorem at the appropriate step.

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p

x −1 = ln x for x > 0. p p→0 Fix x > 0. The limit has the form 0/0, so by applying L’Hopital’s Rule we see that Claim 1 : lim+

lim

p→0+

xp − 1 = lim p p→0+

d p dp (x d dp

− 1)

p

= lim

p→0+

xp ln x = ln x. 1

xp − 1 ≤ x − 1 for x > 0 and 0 < p < 1. p Given a fixed x > 0, set Claim 2 :

g(p) =

xp − 1 , p

0 < p < 1,

We want to show that g(p) ≤ g(1) for 0 < p < 1. In fact, we will show that g(p) is monotone increasing as p increases to 1. We compute that g ′ (p) =

(p xp ln x) − (xp − 1) p2

=

1 − xp + pxp ln x p2

=

1 − xp + xp ln xp . p2

We must show that g ′ (p) ≥ 0. To do this, it suffices to show that xp ln xp ≥ xp − 1. Setting y = xp , this is equivalent to showing that y ln y ≥ y − 1,

(A)

where y = xp > 0. Consider the functions u(y) = y ln y

and

v(y) = y − 1.

We have u(1) = 0 = v(1). Also, u′ (y) =

y + ln y = 1 + ln y y

while v ′ (y) = 1.

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Hence y > 1 =⇒ u′ (y) > v ′ (y) and 0 < y < 1 =⇒ u′ (y) < v ′ (y). Since u(1) = v(1), it follows that u(y) > v(y) for all y > 0, which gives us equation (A). xp − 1 for x > 0 and 0 < p < 1. p To see this, fix 0 < p < 1 and set Claim 3 : ln x ≤

u(x) = ln x

and

v(x) =

xp − 1 . p

We have u(1) = 0 = v(1). Further, u′ (x) = while v ′ (x) =

1 x

pxp−1 1 = xp−1 = 1−p . p x

Note that 1 − p > 0. Therefore, x > 1 =⇒ v ′ (x) =

1 1 < = u′ (x), x1−p x

while 0 < x < 1 =⇒ v ′ (x) =

1 1 > = u′ (x). x1−p x

It follows that u(x) ≤ v(x) for all x > 0. Final Step. Applying Claim 1, for each x ∈ E we have lim+

p→0

|f (x)|p − 1 = ln |f (x)|. p

Further, by Claim 2, for each index 0 < p < 1, |f (x)|p − 1 ≤ |f (x)| − 1 ∈ L1 (E). p Note that the function |f | − 1 is integrable since f is integrable and |E| is finite. In fact, |E| = 1, so by applying the Dominated Convergence Theorem, we see that

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lim+ ln kf kp = lim+

p→0

p→0

= lim

p→0+

= lim+ p→0

=

Z

E

kf kpp R

p

E

Z

E

−1

|f |p − p

(Claim 3) R

E

1

(|E| = 1)

|f |p − 1 p

ln |f |

(DCT).

Hence lim+ kf kp ≤ e

R

E

ln |f |

p→0

.

The opposite inequality follows from part (a). 7.2.24 (a) The fact that ∼ is an equivalence relation is clear. (b) First consider indices p < ∞. If f ∼ g, then f = g a.e., and hence |f |p = |g|p a.e. Therefore Z Z p p |f (x)| dx = |g(x)|p dx = kgkpp . kf kp = E

E

Hence we can define |||fe|||p = kf kp using any representative f of fe. On the other hand, if p = ∞ and f = g a.e., then |f | ≤ M a.e. if and only if |g| ≤ M a.e. Hence, kf k∞ = inf{M : f (x) ≤ M a.e.} = inf{M : g(x) ≤ M a.e.} = kgk∞. Therefore, for this case also we can define |||fe|||∞ = kf k∞ using any representative f of fe. In any case, we see that ||| · |||p is well-defined.

(c) The norm properties of ||| · |||p follow immediately from those of k · kp . In particular, suppose |||fe||| = 0. Then given any representative f of fe we p

have kf kp = |||fe|||p = 0, which implies that f = 0 a.e. But then fe is the fp (E). equivalence class of the zero function, which is the zero element of L

7.3.4 We fill in some details in the proof of Theorem 7.3.4. Assume p < ∞ and fk → f in Lp (E). Choose any ε > 0. Then by Tchebyshev’s Inequality,

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{|f − fk | > ε} = {|f − fk |p > εp } Z 1 ≤ p |f − fk |p ε E =

1 kf − fk kpp → 0 as k → ∞. εp

m

Hence fk → f. The same result also holds for p = ∞. In that case {|f − fk | > ε} = 0 m for all k large enough, so we again conclude that fk → f.

7.3.5 We are given 1 ≤ p < ∞. We will prove that Lp (E) is complete by proving that every Cauchy sequence converges. An alternative proof, which shows that every absolutely convergent series in Lp (E) converges, is given in Problem 7.3.22. Assume that {fn }n∈N is a Cauchy sequence in Lp (E). Choose any ε > 0. Then there exists an N > 0 such that m, n > N

=⇒

kfm − fn kp < ε(p+1)/p .

Consequently, if m, n > N then by applying Tchebyshev’s Inequality we see that {|fm − fn | > ε} = {|fm − fn |p > εp } Z 1 ≤ p |fm − fn |p ε E 1 kfm − fn kpp εp 1 < p εp+1 = ε. ε

=

Hence {fn }n∈N is Cauchy in measure. Problem 3.5.17 therefore implies that there exists a measurable function m f on E such that fn → f. By Lemma 3.5.6, it follows that there exists a subsequence {fnk }k∈N such that fnk (x) → f (x) for a.e. x ∈ E. Choose ε > 0. Then there exists an N such that j, k > N =⇒ kfnj − fnk kpp < ε. Fix j > N. Since fnk (x) → f (x) pointwise a.e., we have by Fatou’s Lemma that Z kf − fnk kpp ≤ lim inf |fnj (x) − fnk (x)|p dx ≤ ε. j→∞

E

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Hence f − fnk belongs to Lp (E). As Lp (E) is closed under addition, it follows that f = (f − fnk ) + fnk ∈ Lp (E).

Furthermore, the argument above shows that fnk → f in Lp (E).

The preceding work shows that {fn }n∈N is a Cauchy sequence that has a subsequence that converges to f. This implies that the full sequence converges, i.e., fn → f in Lp (E), and therefore Lp (E) is complete. To see why convergence follows, fix ε > 0. Then there exists some K > 0 such that kf − fnk kp < ε for all nk > K. Also, there exists an N such that kfm − fn kp < ε for all m, n > N. Suppose that n > N. Then since the nk are strictly increasing, there exists some nk that is greater than both K and N. For this nk we have kf − fn kp ≤ kf − fnk kp + kfnk − fn kp < ε + ε = 2ε. This is true for all n > N, so fn → f.

Case p = ∞. We give the details showing that L∞ (E) is complete. Assume that {fn }n∈N is a Cauchy sequence in L∞ (E). For m, n ∈ N, set  Zmn = x ∈ E : |fm (x) − fn (x)| > kfm − fn k∞ .

Then Z = ∪m,n Zmn has measure zero. For each x ∈ E \Z, we have |fm (x) − fn (x)| ≤ kfm − fn k∞ ,

so {fn (x)}n∈N is a Cauchy sequence of scalars. This sequence therefore converges, so for x ∈ E \Z we can define f (x) = limn→∞ fn (x). Since each function fn is measurable and since |Z| = 0, it follows from Lemma 3.1.7 that f is measurable. Now fix ε > 0. Then there exists an N such that kfm − fn k∞ ≤ ε for all m, n > N. Hence for x ∈ E \Z we have |f (x) − fn (x)| =

lim |fm (x) − fn (x)| ≤

m→∞

lim kfm − fn k∞ ≤ ε.

m→∞

Therefore kf −fn k∞ ≤ ε since |Z| = 0. Consequently f ∈ L∞ (E) and fn → f in L∞ (E), so L∞ (E) is complete. 7.3.10 Suppose that f ∈ Lp (E) where 1 ≤ p < ∞, and fix ε > 0. By Theorem 7.3.9, there exists a compactly supported function g ∈ Lp (E) such that kf − gkp < ε. By Corollary 3.2.15, there exist simple functions φn that converge pointwise to f and satisfy |φn | ≤ |g| a.e. Hence |g − φn |p → 0 a.e., and
p
p |g − φn |p ≤ |g| + |φn | ≤ 2|g| = 2p |g|p ∈ L1 (E). The Dominated Convergence Theorem therefore implies that

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kg − φn kpp =

Z

E

|g − φn |p → 0 as n → ∞.

Hence if we choose n large enough then we will have kf − φn kp ≤ kf − gkp + kg − φn kp ≤ 2ε. Each function φn belong to Sc , so we conclude that Sc is dense in Lp (E). Suppose p = ∞ and f ∈ L∞ (E). Then |f (x)| ≤ kf k∞ except for points x in a set Z that has measure zero. Hence f is bounded on Z C , and therefore Corollary 3.2.15 implies that there exist simple functions that converge uniformly to f on Z C . Consequently these simple functions converge in L∞ -norm to f, so the set of simple functions is dense in L∞ (R). 7.3.11 (a) Step 1. Let f = χE where E is a bounded subset of Rd . If we fix ε > 0, then Theorem 2.1.27 implies that there exists a bounded open set U ⊇ E such that |U \E| < ε. By Problem 2.2.43, there also exists a compact set K ⊆ E such that |E \K| < ε. Applying Urysohn’s Lemma (Theorem 4.5.7), we can find a continuous function θ : Rd → R that satisfies 0 ≤ θ ≤ 1 everywhere, θ = 1 on K, and θ = 0 on Rd \U. This function θ belongs to Cc (Rd ), and we have Z Z p p χ χ |χE − θ|p ≤ |U \K| < 2ε. | E − θ| = k E − θkp = Rd

U \K

Hence χE can be approximated as closely as we like in Lp -norm by an element of Cc (Rd ). Step 2. Let φ be a simple function that has compact PN support. Since φ is zero outside of some compact set, we can write φ = k=1 ak χEk where each set Ek is bounded and each scalar ak is nonzero. Applying Step 1, there exist functions θk ∈ Cc (Rd ) such that

χE − θk < k p

It follows that the function θ = kφ − θkp

ε , |ak | N

PN

k=1

ak θk belongs to Cc (Rd ) and satisfies

N N X

X

ak χEk − ak θk =

k=1

k=1

k = 1, . . . , N.

p

≤

N X

k=1

|ak | kχEk − θk kp < ε.

Hence any compactly supported simple function can be approximated as closely as we like in Lp -norm by elements of Cc (Rd ). Step 3. Now let f be an arbitrary element of Lp (Rd ), and fix ε > 0. By Exercise 7.3.10, there exists a compactly supported simple function φ such that kf − φkp < ε. Applying Step 2, we can find a function θ ∈ Cc (Rd ) such

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that kφ − θkp < ε. Hence kf − θkp < 2ε, so we conclude that Cc (Rd ) is dense in Lp (Rd ). (b) Let θ be any continuous function such that • • •

0 ≤ θ(x) ≤ 1 for every x ∈ Rd , θ(x) = 1 if |x| ≤ 1, and θ(x) = 0 if |x| ≥ 2.

For example, simply take a function η on R that has these properties, and set θ(x) = η(|x|) for x ∈ Rd . Given g ∈ C0 (Rd ), set gn (x) = g(x) θ(x/n). Then gn ∈ Cc (Rd ), and gn → g uniformly as n → ∞. Hence Cc (Rd ) is dense in C0 (Rd ).

7.3.12 Assume 1 ≤ p < ∞. Fix f ∈ Lp (R) and ε > 0. By Exercise 7.3.11, there exists a function g ∈ Cc (R) such that kf − gkp < ε. Let R > 0 be large enough that supp(g) ⊆ [−R, R]. Since g is uniformly continuous, given ε > 0 there exists a 0 < δ < 1 such that |x − y| < δ =⇒ |g(x) − g(y)|
0 such that supp(g) ⊆ [−R, R]d . Since g is uniformly continuous, there exists a δ > 0 such that |a| < δ =⇒ kg − Ta gk∞
k, and set hk = gk · χE∩[−k,k]d . Note that khk kpp =

Z

E

|hk |p =

Z

E∩[−k,k]d

Z

|hk |p ≤

E∩[−k,k]d

k p ≤ (2k)d k p < ∞.

Therefore hk ∈ Lp (E). If q is finite, then a similar argument shows that hk ∈ Lq (E). On the other hand, if q = ∞ then we have hk ∈ L∞ (E) since hk is bounded. In any case, we see that hk ∈ Lp (E) ∩ Lq (E). Additionally, hk → f pointwise, so |f − hk | → 0 pointwise, and we have |f − hk |r ≤ |f |r ∈ L1 (E). Because r is finite, we can apply the Dominated Convergence Theorem to obtain Z lim kf − hk krr = lim |f − hk |r = 0. k→∞

k→∞

r

E

p

That is, hk → f in L -norm. Since hk ∈ L (E) ∩ Lq (E) for every k, it follows that Lp (E) ∩ Lq (E) is dense in Lr (E). 7.3.19 Fix 1 ≤ p < ∞. Since C[a, b] is dense in Lp [a, b], given f ∈ Lp [a, b] and ε > 0, there exists a continuous function g ∈ C[a, b] such that kf − gkp < ε. By the Weierstrass Approximation Theorem, the set of polynomials is dense in C[a, b] with respect to the uniform norm. Therefore, there exists a polynomial q ∈ P such that kg − qk∞ < Consequently,

ε . (b − a)1/p

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kf − qkp ≤ kf − gkp + kg − qkp Z b 1/p p < ε + |g(x) − q(x)| dx a

≤ ε +

Z

a

b

1/p kg − qkp∞ dx

= ε + (b − a)1/p kg − qkp∞ < ε + ε = 2ε. Therefore P is dense in Lp [a, b]. Remark: This result is still valid if 0 < p < 1. Only minor changes are required; specifically, we must use the metric dp (f, g) = kf − gkpp for Lp [a, b]. If p = ∞, then P is dense in C[a, b] with respect to the uniform norm (this is the Weierstrass Approximation Theorem). 7.3.20 The proof is essentially the same as the proof of Theorem 4.5.12. Choose any function f ∈ Lp (R) and any ε > 0. Since Cc (R) is dense in p L (R), there exists a compactly supported, continuous function θ such that kf − θkp < ε. Since θ is compactly supported, if we choose R > 1 large enough then we have θ(x) = 0 for all |x| > R. Since θ is uniformly continuous, there exists some j ∈ N such that |x − y| < 2−j =⇒ |θ(x) − θ(y)|
0 such that Z ′ ′ |F | < δ =⇒ |g|p < εp . F

By Egorov’s Theorem, there is some set A ⊆ B such that |B \ A| < δ and fn → f uniformly on A. Fix n. Then by applying H¨ older’s Inequality, we compute that Z Z Z f g − fn g ≤ |f − fn | |g| =

Z

B\A

Z

|f − fn | |g| +

= kf − fn kp

Z

B\A

A

p′

|g|

+ kf − fn kp

|f − fn | |g| +

1/p′

Z

BC

+ p′

|g|

Z

Z

BC

|f − fn | |g| p

A

|f − fn |

1/p′

1/p

kgkp′

≤ 2M ε + kf − fn k∞ |A|1/p kgkp′ + 2M ε. Since kf − fn k∞ → 0, it follows that Z Z lim sup f g − fn g ≤ 4M ε. n→∞

Since ε > 0 is arbitrary, we therefore have Z Z lim sup f g − fn g = 0. n→∞

Case p = 1. The result does not hold when p = 1. Consider fnR= χ[n,n+1] ∈ L (R). Then fnR → 0 a.e., but if we take g = 1 ∈ L∞ (R), then fn g = 1 for every n, while f g = 0. 1

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7.3.22 Suppose that {fn }n∈N is an absolutely convergent series in Lp (R), i.e., ∞ X B = kfn kp < ∞. n=1

Fix a representative of fn . Each representative fn is defined almost everywhere on E. By changing fn on a set of measure zero, we may assume that fn is defined everywhere on E. For example, we can define fn (x) to be zero at any point where it is undefined. This gives us a new representative of fn that is defined at every point. (We do this just for convenience; we could instead simply keep track of the sets Zn where fn is undefined.) Set N ∞ X X gN (x) = |fn (x)| and g(x) = |fn (x)|. n=1

n=1

These are series of nonnegative scalars, so they converge pointwise a.e. in the extended real sense. By the Triangle Inequality, for each N we have kgN kp ≤

N X

n=1

kfn kp ≤ B.

Since |gN |p ր |g|p , the Monotone Convergence Theorem implies that Z Z |g|p = lim kgkpp = |gN |p = lim kgN kpp ≤ B. E

N →∞

N →∞

E

Therefore g ∈ Lp (R). Hence g is finite a.e., and consequently the series f (x) =

∞ X

fn (x)

n=1

converges pointwise a.e. Since |f | ≤ g, we have f ∈ Lp (R). Also, if we set hN (x) =

N X

fn (x),

n=1

then hN → f pointwise a.e. and
p |f − hN |p ≤ |f | + |hN | ≤

g + gN


p

≤

2g


p

∈ L1 (R).

Therefore hN →Pf in Lp (R) by the Dominated Convergence Theorem. Hence p the series f = ∞ n=1 fn converges in L (R). The function f is independent of the choice of representatives fn in the sense that if we choose different representatives of fn , then we will obtain a new function f that differs from the previous one only on a set of measure zero.

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(b) Part (a) shows that every absolutely convergent series in Lp (E) converges. Theorem 1.2.8 therefore implies that Lp (E) is complete. (c) We use the same notation as in part (a). By definition of infinite series we have that hN (x) → f (x) a.e. Further, |hN | ≤ g ∈ L1 (E). Consequently, the Dominated Convergence Theorem implies that Z Z f = lim hN (DCT) N →∞

E

=

lim

N →∞

=

lim

N →∞

E

Z X N

fn

E n=1

N Z X

n=1

fn .

E

By definition of infinite series, this says that Z

E

f =

∞ Z X

n=1

P∞ R

n=1 E

fn converges, and

fn .

E

7.3.23 “⇒.” Assume that fn → f in Lp (Rd ). We must show that each of conditions (a), (b), and (c) hold. (a) Tchebyshev’s Inequality implies that every Lp -convergent sequence must converge in measure (see Theorem 7.3.4). (b) Fix ε > 0. Since |f |p is integrable, Exercise 4.5.5 implies that there exists a δ0 > 0 such that Z ε |E| < δ0 =⇒ kf χE kpp = |f |p < p+1 . 2 E Further, there is some N > 0 such that n > N =⇒ kf − fn kpp
N and |E| < δ0 then we have Z |fn |p = k(fn − f + f ) χE kpp

.

E

≤ 2p k(fn − f ) χE kpp + 2p kf χE kpp ≤ 2p kfn − f kpp + 2p
0 such that Z p χ |fn |p < ε. |E| < δn =⇒ kfn E kp = E

Therefore if we take δ = min{δ0 , δ1 , . . . , δN }, then we have shown that statement (b) holds. (c) Fix ε > 0. Since |f |p is integrable, by setting E = Br (0) with r large enough we will have Z ε p χ C |f |p < p+1 . kf E kp = 2 C E There is some N > 0 such that n > N =⇒ kf − fn kpp
N we have kf χE C kpp ≤ k(fn − f + f ) χE C kpp ≤ 2p k(fn − f ) χE C kpp + 2p kf χE C kpp ≤ 2p kfn − f kpp + 2p
0. By statement (c), there is a set E such that for every n we have Z εp |fn |p < p . 2 EC m

Since fn → f, there exists a subsequence such that fnk → f a.e. Consequently, by Fatou’s Lemma, Z Z εp |f |p ≤ lim inf |fnk |p ≤ p . k→∞ 2 EC EC Therefore for every n we have

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Z

p

EC

|f − fn |

1/p

≤

Z

p

EC

Hence

|f |

Z

EC

1/p

+

Z

EC

p

|fn |

1/p

≤

ε ε + = ε. 2 2

|f − fn |p ≤ εp .

R By statement (b), there is some δ > 0 such that if |A| < δ then A |fn |p < εp . As above, by applying Fatou’s Lemma and the Triangle Inequality, it follows from this that Z |A| < δ =⇒ |f − fn |p ≤ εp . A

Let An m

 = |f − fn | >

 ε . |E|1/p

Since fn → f, there is some N > 0 such that |An | < δ for all n > N. Consequently, for n > N we have Z |f − fn |p ≤ εp . An

Putting this all together, for n > N we have Z Z Z p p p kf − fn kp = |f − fn | + |f − fn | + E∩An

≤

Z

An

E \ An

|f − fn |p +

≤ εp + |E \An |

Z

E \ An

εp + |E|

Z

EC

EC

|f − fn |p

|f − fn |p

p

ε + εp |E|

≤ εp + εp + εp = 3εp .

Thus kf − fn kp ≤ 31/p ε for all n > N, so fn → f in Lp -norm. 7.3.24 (a) Suppose that f = g a.e. Then Y = {f 6= g} has measure zero. Suppose that F is a closed set and f = 0 a.e. on F C . Then there exists a set Z with measure zero such that f (x) = 0 for every x ∈ F C \ Z. Consequently g(x) = 0 for every x ∈ F C \ (Z ∪ Y ). Since Z ∪ Y has measure zero, it follows that g = 0 a.e. on F C . Conversely, if g = 0 a.e. off of a closed set F, then f = 0 a.e. on F C . Hence the set of F whose intersection forms supp(f ) is precisely the same as the set of F whose intersection forms supp(g). Therefore supp(f ) = supp(g). (b) “⇒.” Suppose that f is compactly supported in the sense of Definition 7.3.8, i.e., there is a compact set K such that f = 0 a.e. outside of K. Then

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K is one of the sets F that are intersected to form supp(f ), so supp(f ) ⊆ K. By definition supp(f ) is also closed, so it is a closed subset of the compact set K. Therefore supp(f ) is a compact set. “⇐.” We give a contrapositive argument. Suppose that f is not compactly supported. By definition, supp(f ) is a closed set, and therefore it is compact if any only if it is bounded. Given N ∈ N, we will show that supp(f ) is not contained in [−N, N ], and therefore is an unbounded set. By part (a), we know that supp(f ) is independent of the choice of representative, so let f denote any fixed representative of f. Since f is not compactly supported and [−N, N ] is a compact set, it is not true that f = 0 a.e. on [−N, N ]C . Therefore, there must exist a set EN ⊆ [−N, N ]C such that |EN | > 0 and f (x) 6= 0 for x ∈ EN . Suppose that F is any closed set such that f (x) = 0 for a.e. x ∈ / F. Let Z = {x ∈ F C : f (x) 6= 0}. Then f (x) = 0 for all x ∈ F C \Z. Therefore, if x ∈ EN , then x 6∈ F C \Z = F C ∩ Z C = (F ∪ Z)C , which implies that x ∈ F ∪ Z. Hence EN ⊆ F ∪ Z, and therefore EN \Z ⊆ F. Taking the intersection over all such sets F, we obtain EN \Z ⊆ supp(f ). Since EN has positive measure and Z has zero measure, EN \Z is nonempty. Therefore, there exists some point xN ∈ EN that belongs to supp(f ). By definition of EN , we must have |xN | > N. Hence supp(f ) is unbounded and therefore not compact. (c) Suppose that f is continuous. Let E be the usual definition of the support of f, i.e., the closure of the set of points where f is nonzero, and let K be the set supp(f ) that is defined in the problem statement. Since E is closed and f (x) = 0 for all x ∈ / E, the set E is one of the sets that is intersected to form K. Therefore K ⊆ E. Suppose that F is a closed set and f (x) = 0 for a.e. x ∈ / F. Choose any point x ∈ / F such that f (x) 6= 0. Since F C is open, there is an open set U that contains x and is contained in F C . There is also an open set V containing x on which f is nonzero. Therefore U ∩ V is a nonempty open set contained in F C , and f is nonzero on U ∩ V. This implies that it is not true that f = 0 on F C , which is a contradiction. Therefore f (x) = 0 for every point x ∈ / F. Hence F contains the set of points where f is nonzero, and therefore F is one of the sets that is intersected to form E. This shows that E ⊆ F. Since K is the intersection of all such sets F, it follows that E ⊆ K. 7.3.25 (a) The fact that k · k is a norm on Lp (E) ∩ Lq (E) is clear.

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To show completeness, suppose that {fn }n∈N is a Cauchy sequence in Lp (E) ∩ Lq (E). Since kfm − fn kp ≤ kfm − fn k, it follows that {fn }n∈N is a Cauchy sequence in Lp (E). Hence there exists a function f ∈ Lp (E) such that kf − fn kp → 0. Similarly, there exists a function g ∈ Lp (E) such that kg − fn kq → 0. Moreover, in each case we can find a subsequence that converges pointwise a.e., so we have f = g a.e. Therefore, fn → f in both the Lp and Lq norms, and consequently kf − fn k = kf − fn kp + kf − fn kq → 0

as n → ∞.

Therefore Lp (E) ∩ Lq (E) is complete. (b) First we show containment for the case 1 ≤ p < r < q < ∞. In this case there is some 0 < t < 1 such that r = tp + (1 − t)q. Let u = 1t . Since 1 < u < ∞, its dual index is 1 u = . u−1 1−t

u′ =

Choose any f ∈ Lp (E) ∩ Lq (E). Applying H¨ older’s Inequality, we compute that Z Z r r |f |tp |f |(1−t)q |f | = kf kr = E

E

≤ =

Z

E

Z

|f |tpu p

E

|f |

1/u Z

′

E

t Z

E

q

|f |

|f |(1−t)qu 1−t

1/u′

(1−t)q = kf ktp . p kf kq

This is finite, so we have f ∈ Lr (E). Further, taking rth roots, we obtain kf kr ≤ kf ktp/r kf k(1−t)q/r . p q If we set θ = then

(A)

tp r

(1 − t)q r tp − = , r r r so we can rewrite equation (A) as 1−θ =

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. kf kr ≤ kf kθp kf k1−θ q Finally, we observe that θ 1−θ t 1−t 1 + = + = . p q r r r This completes the proof for the case 1 ≤ p < r < q < ∞. Next we show containment for the case 1 ≤ p < r < q = ∞. Suppose that f ∈ Lp (E) ∩ L∞ (E). Then kf kr =

Z

r

E

|f |

1/r

=

Z

E

≤

p

r−p

|f | |f |

kf k(r−p)/r ∞

Z

1/r

E

p

|f |

1/r

(r−p)/r = kf kp/r < ∞. p kf k∞

Hence f ∈ Lr (E), so Lp (E) ∩ L∞ (E) ⊆ Lr (E). Set θ = p/r. Then 1−θ = 1−

r−p p = , r r

so (r−p)/r kf kr ≤ kf kp/r = kf kθp kf k1−θ ∞ . p kf k∞

Further, θ 1 1 1−θ + = + 0 = . p ∞ r r 7.3.26 (a) First we show that d is a metric on M(E). We have 0 ≤ d(f, g) ≤ 1 for each f, g ∈ M(E). If d(f − g) = 0, then |f − g| = 0 a.e. The metric d is trivially symmetric. To prove the Triangle Inequality, we need the following lemma. Lemma. If a, b, c ≥ 0 and a ≤ b + c, then a b c ≤ + . 1+a 1+b 1+c Proof. Since a ≤ b + c, a ≤ b + bc + c + bc + abc.

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Therefore a + ab + ac + abc ≤ b + ab + bc + abc + c + ac + bc + abc. Factoring, a (1 + b + c + bc) ≤ b (1 + a + c + ac) + c (1 + a + b + ab). Factoring again, a (1 + b) (1 + c) ≤ b (1 + a) (1 + c) + c (1 + a) (1 + b). Dividing both sides by (1 + a) (1 + b) (1 + c) then gives the result. ⊓ ⊔ Now we prove the Triangle Inequality. Given f, g, h ∈ M(E), we have |f − h| ≤ |f − g| + |g − h|. Applying the lemma above, we see that Z |f − h| d(f, h) = 1 + |f − h| E  Z  |f − g| |g − h| ≤ + 1 + |f − g| 1 + |g − h| E = d(f, g) + d(g, h). (b) Now we show that the metric d induces convergence in measure. m

“⇒.” Suppose that fk → f, and fix ε > 0. Then Z |f − fk | d(f, fk ) = 1 + |f − fk | E Z Z |f − fk | |f − fk | + = 1 + |f − f | 1 + |f − fk | k |f −fk |≤ε |f −fk |>ε Z Z ε ≤ 1 + |f −fk |>ε |f −fk |≤ε 1 ≤ {|f − fk | > ε} + ε |E|.

Consequently, Z   |f − fk | lim sup ≤ lim sup {|f − fk | > ε} + ε |E| = ε|E|. k→∞ k→∞ E 1 + |f − fk |

Since |E| < ∞ and ε is arbitrary, we conclude that d(f, fk ) → 0 as k → ∞.

“⇐.” Assume that d(f, fk ) → 0, and fix ε > 0. Since function of x, we have

x x+1

is an increasing

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ε≤x

309

x ε ≤ . 1+ε 1+x

=⇒

Therefore {|f − fk | > ε} = 1 + ε ε ≤

1+ε ε

Z

|f −fk |>ε

Z

|f −fk |>ε

ε 1+ε |f − fk | 1 + |f − fk |

≤ d(f, fk ) → 0 as k → ∞. m

Hence fk → f. (c) Finally, we will show that M(E) is complete with respect to the metric defined in the problem statement. Suppose that {fk }k∈N is Cauchy with respect to the metric d. Fix any ε > 0. Then there is some N such that d(fj , fk ) < ε for all j, k > N. Using precisely the same calculations that appear in the argument for the second half of part (b), we see that for j, k > N we have {|fj − fk | > ε} ≤ d(fj , fk ) < ε.

It therefore follows from Problem 3.5.17 that there exists a measurable funcm tion f such that fk → f. Since convergence in measure corresponds to convergence with respect to the metric d, it follows that the sequence {fk }k∈N converges with respect to d. Hence M(E) is complete. 7.4.5 Case 1 < p ≤ ∞. R We are given f ∈ Lp (R) such that f φ = 0 for every φ ∈ Cc (R). Suppose ′ that g is any function in Lp (R). Since 1 ≤ p′ < ∞, we know that Cc (R) ′ is dense in Lp (R). Therefore there exist functions φk ∈ Cc (R) such that kg − φk kp′ → 0 as k → ∞. Applying the hypotheses and H¨ older’s Inequality, we compute that Z ∞ Z ∞ Z ∞ 0 ≤ f g ≤ f φk + f (g − φk ) −∞

−∞

−∞

≤ 0 + kf kp kg − φk kp′

→ 0

as k → ∞.

R ′ Therefore f g = 0 for every g ∈ Lp (R). Applying the Converse to H¨ older’s Inequality, we conclude that

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kf kp = Therefore f = 0 a.e.

Z sup

kgkp′ =1

∞ −∞

f g = 0.

Case p = 1, first proof. R We are given a function f ∈ L1 (R) that satisfies fRφ = 0 for every φ ∈ Cc (R). Let g ∈ Cc (R) be any function that satisfies g = 1. Set gh = (1/h) g(x/h). If we fix t ∈ R, then gh (t − x) ∈ Cc (R), so by hypothesis we have Z ∞ (f ∗ gh )(t) = f (x) gh (t − x) dx = 0. −∞

But Problem 5.5.14 tells us that f ∗ gh → f in L1 -norm, so this implies that f = 0 a.e Case p = 1, second (more difficult) proof. R We are given a function f ∈ L1 (R) that satisfies f φ = 0 for every φ ∈ Cc (R). If f is complex-valued, then for every real-valued φ ∈ Cc (R) we have Z Z Z 0 = fφ = fr φ + i fi φ,

R R so fr φ = 0 = fi φ for every real-valued φ ∈ Cc (R). Therefore it suffices to consider the real case, i.e., f is extended real-valued (but finite a.e.) and φ is real-valued. (If we like, we can choose a representative of f that is finite at every point, but it is not necessary for this proof.) Let feh be defined by equation (5.26), i.e., feh (x) = (f ∗ χh )(x) =

Z

∞

−∞

f (x − t) χh (t) dt =

1 2h

Z

x+h

f (t) dt,

x−h

where

1 χ[−h,h] . 2h Then feh is continuous and integrable, and feh → f in L1 -norm as h → 0. Given any function φ ∈ Cc (R), if we assume that the interchange of integrals can be justified then we compute that  Z ∞ Z ∞ Z ∞ e fh φ = f (x − t) χh (t) dt φ(x) dx χh =

−∞

−∞

=

Z

−∞

∞

χh (t)

−∞

=

Z

∞

−∞

∞

−∞

Z

χh (t)

Z

∞

−∞

 f (x − t) φ(x) dx dt

 f (x) φ(x + t) dx dt

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=

Z

311

∞

−∞

χh (t) · 0 dt

= 0, where we have used the fact that a translate of φ is just another function in Cc (R). To justify the interchange, we note that Z

∞ −∞

Z

∞

−∞

|χh (t) f (x − t) φ(x)| dx dt =

Z

h

−h

Z

≤ kφk∞ = kφk∞

∞

−∞

Z

|f (x − t) φ(x)| dx dt

h

−h

Z

Z

∞

−∞

|f (x − t)| dx dt

h

−h

kf k1 dt

= 2h kφk∞ kf k1 < ∞. Therefore Fubini’s Theorem justifies the interchange. Suppose now there is some point x such that feh (x) > 0. Then there is a δ > 0 such that feh (y) > 0 for all y ∈ (x − δ, x + δ). If we choose any nonnegative function φ that is supported within (x − δ, x + δ) but is not identically zero, then we have 0 =

Z

∞ −∞

feh (y) φ(y) dy ≥

Z

x+δ

x−δ

feh (y) φ(y) dy > 0,

which is a contradiction. Similarly we cannot have feh (x) < 0 at any point, so feh (x) = 0 for every x. But feh → f in L1 -norm, so this implies that f = 0 a.e. 7.4.6 Suppose that {fn }n∈N is a complete sequence in a normed space X, and let X  N S = rn fn : N > 0, Re(rn ), Im(rn ) ∈ Q . n=1

Then S is countable, and we claim it is dense in X. Without loss of generality, we may assume that each vector fn is nonzero. Choose any f ∈ X and fix ε > 0. Since span{fn } is dense in X, there exists a vector N X g = cn fn ∈ span{fn } n=1

such that kf − gk < ε. For each n ∈ N, choose a scalar rn with real and imaginary parts such that

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|cn − rn | < and set h =

N X

ε , N kfn k rn fn .

n=1

Then h ∈ S and kg − hk ≤

N X

n=1

|cn − rn | kfn k
0. Since P is dense in Lp [a, b], there exists some polynomial

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q(x) =

N X

ck xk

k=0

that satisfies

kf − qkp < ε. Let R = max{|a|, |b|}. Then k

kx kp = Let

Z

a

b

1/p Z |x | dx ≤ k p

b

R

a

kp

1/p dx = Rk (b − a)1/p .

 C = max Rk (b − a)1/p : k = 0, . . . , N .

For each k = 0, . . . , N, let rk be a rational scalar such that |ck − rk |
0 is fixed. Then there exists an M such that |f (x)| < ε for all |x| ≥ M. By the Weierstrass Approximation PN Theorem, there is some polynomial p = k=0 ck ek such that k(f − p) χ[−M−1,M+1] ku < ε.

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Then kf − p θM ku < 2ε. Choose rational scalars rk such that |ck − rk | < Set q =

PN

k=0 rk ek .

ε , N (M + 1)k

k = 0, . . . , N.

Then

kq θM − p θM ku ≤

N X

k=0

|ck − rk | (M + 1)k < ε.

Hence kf − q θM ku < 3ε, and q θM ∈ S, so we conclude that S is dense in C0 (R). An alternative approach could proceed as follows. Define the space of all continuous functions that are nonzero outside of [−N, N ]:  C[−N,N ](R) = f ∈ C(R) : supp(f ) ⊆ [−N, N ] ,

and show that C[−N,N ](R) is separable. A minor point to observe is that we cannot identify C[−N,N ] (R) with the space of continuous functions whose domain is [−N, N ]. That is the space  C[−N, N ] = f : [−N, N ] → C : f is continuous .

A function f ∈ C[−N,N ] (R) must satisfy f (−N ) = f (N ) = 0, so C[−N,N ](R) can be identified with the subspace of C[−N, N ] that consists of functions that vanish at the endpoints −N and N. The space of continuous, compactly supported functions is a countable union of the separable spaces C[−N,N ] (R): Cc (R) =

∞ S

C[−N,N ] (R).

N =1

Show that Cc (R) is separable. As Cc (R) is dense in C0 (R), the final step is to show that if S is a dense, separable subset of a normed space X then X is separable. 7.4.10 (a) First show that Lp (Rd ) is separable, then it is easy to restrict to measurable sets E ⊆ Rd . (b) Since |E| > 0, Problem 2.3.20 implies that there exist infinitely many disjoint measurable sets E1 , E2 , . . . contained in E, each with positive and finite measure. Let S be the set of all functions that only take the values 0 or 1 and are constant on each set Ek . That is, a function f belongs to S if and only if ∞ X f = c k χE k k=1

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where every ck is either 0 or 1. Then S is an uncountable subset of L∞ (E), and if f 6= g ∈ S then kf − gk∞ = 1. Theorem 7.4.4 therefore implies that L∞ (E) is not separable. PN 7.4.11 (a) Given x = (xk )k∈N in ℓp , set sN = k=1 xk δk . Then lim kx − sN kℓp =

N →∞

lim

N →∞

∞ X

k=N +1

|xk |p = 0,

P so we have xk δk with convergence of this series in ℓp -norm. Further, P x= if x = ck δk for some scalars ck , then we must have ck = xk , because sN converges in norm to x and therefore must converge componentwise to x. (b) Let x = (xk )k∈N be any vector in c0 . If we set sN =

N X

xk δk = (x1 , . . . , xN , 0, 0, . . . ),

k=1

then lim kx − sN k∞ =

N →∞

lim sup |xk | = lim sup |xk | = 0.

N →∞ k>N

k→∞

That is, the partial sums sN converge to x. By the definition of an infinite series, we therefore have ∞ X x = xk δk , k=1

where this series converges with respect to the norm of c0 . Furthermore, because convergence in ℓ∞ -norm implies componentwise convergence, the scalars xn in this representation are unique. P∞ Thus every vector x in c0 can be uniquely expressed in the form x = k=1 xk δk , with convergence of the series in norm. This shows that E is a Schauder basis for c0 . (c) We are given vectors yn = (1, . . . , 1, 0, 0, . . . ), where the 1 is repeated n times. Choose any sequence x = (xn )n∈N ∈ c0 . Set cn = xn − xn+1 ,

n ∈ N.

Note that cn → 0 since x ∈ c0 , and therefore N X

n=1

Hence

P

cn =

N X

n=1

(xn − xn+1 ) = x1 − xN +1 → x1 .

cn converges, and for 1 ≤ k ≤ N we have

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xk − Set SN x =

PN

N X

n=k

cn = xk − (xk − xN +1 ) = xN +1 → 0.

n=1 cn yn .

SN x =

N X

Then

cn y n =

n=1

X N

cn ,

n=1

N X

n=2

 cn , . . . , cN , 0, 0, . . . ,

and therefore x − SN x =

  N N X X x1 − cn , x2 − cn , . . . , xN − cN , xN +1 , xN +1 , . . . n=1

n=2

= (xN +1 , xN +1 , . . . ). Since each component of x − SN x is xN +1 and x ∈ c0 , it follows that kx − SN xkℓ∞ = |xN +1 | → 0. That is, the partial sums SN x converge to x in ℓ∞ -norm. This says that x =

∞ X

cn y n ,

n=1

with convergence of the infinite series in ℓ∞ -norm. P∞ Next we show that the representation x = n=1 cn yn is unique. SupP pose that we also had x = dn yn , with convergence of the series in ℓ∞ norm. Since ℓ∞ -norm convergence implies componentwise convergence, we have both X  ∞ ∞ ∞ X X x = cn y n = cn , cn , . . . n=1

and

x =

c1 =

∞ X

n=1

cn −

n=2

X ∞

∞ X

∞ X

dn yn =

∞ X

cn = x1 − x2 =

n=1

Therefore

n=1

n=2

dn ,

n=1

n=2

∞ X

n=1

 dn , . . . .

dn −

∞ X

dn = d1 .

n=2

Iterating this argument, by induction we obtain ck = dk for every k. Hence the representation of x in terms of the sequence {yn }n∈N is unique, and therefore {yn }n∈N is a Schauder basis for c0 .

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Although not asked for, we will prove that this Schauder basis is conditional, i.e., if we reorder the elements then it no longer is a Schauder basis. It remains to show that this basis is conditional. Consider the sequence x =

(−1)k+1 /k




1 1 1 = (1, − , , − , . . . ). 2 3 4

This sequence belongs to c0 , and its representation in the basis {yn } is x = However, the series



1+

1 1 1 y1 − y2 + · · · . + 2 2 3

 1 1 1 1+ y1 + y2 + · · · + 2 2 3

does not converge componentwise (consider the first component), and therefore it cannot converge in the norm of c0 . Hence the basis representation of this x does not converge unconditionally, so {yn } is not an unconditional basis for c0 . (d) Equation (7.25) means that the partial sums of the series converge to x in the norm of X, i.e.,

N X

lim x − cn (x) xn

= 0.

N →∞

n=1

PN

Each partial sum sN = n=1 cn (x) xn belongs to span{xn }n∈N , so this shows that span{xn }n∈N is dense in X. To prove linear independence, suppose that two finite linear combinations of the xn were equal, say x =

N X

an xn =

n=1

N X

bn xn .

n=1

Setting an = bn = 0 for n > N, it follows that x =

∞ X

an xn =

n=1

∞ X

bn xn .

n=1

However, by the definition of a Schauder basis, there are unique coefficients cn (x) such that N X x = cn (x) xn . n=1

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Therefore an = cn (x) = bn for every n. In particular, an = bn for n = 1, . . . , N. Hence {xn }n∈N is finitely linearly independent. (e) The finite span of M is P, so our goal is to show that P is dense in C[a, b]. Choose any function f ∈ C[a, b]. Then for each n ∈ N, the Weierstrass Approximation Theorem implies that there exists a polynomial pn ∈ P such that kf − pn ku < n1 . Hence lim kf − pn ku = 0.

n→∞

Therefore P is dense in C[a, b]. To show that PN M is independent, suppose that some finite linear combination p(x) = k=0 ck xk of elements of M is zero (i.e., the zero function), but not every ck is zero. Without loss of generality, we can assume that cN 6= 0, so p has degree N. The Fundamental Theorem of Algebra tells us that p can have at most N roots, i.e., there are at most N distinct values of x such that p(x) = 0. This contradicts the fact that p(x) = 0 for every x. Consequently every ck must in fact be zero. Therefore M is finitely linearly independent. Suppose that M was a Schauder basis for C[0, 1], and fix any f ∈ C[0, 1]. Then, by the definition of a Schauder basis, we would be able to write f as a unique “infinite linear combination” of elements of M. That is, there would exist scalars ck such that f (x) =

∞ X

ck xk ,

(7.36)

k=0

where this series converges uniformly on [0, 1]. Since uniform convergence implies pointwise convergence, this implies that the series in equation (7.36) converges when x = 1. The properties of power series imply that the series in equation (7.36) converges for every x ∈ (−1, 1), and furthermore f is infinitely differentiable on (−1, 1), and hence is infinitely differentiable on the smaller interval (0, 1). This implies that every function in C[0, 1] is infinitely differentiable on (0, 1). But this simply is not true—there are continuous functions on [0, 1] that are not infinitely differentiable on the interior of the interval (give an example). Therefore we have reached a contradiction.

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Solutions to Exercises and Problems from Chapter 8 8.1.2 (a) This follows from linearity in the first variable and the fact that hx, yi = hy, xi. (b) Given x, y ∈ H, kx + yk2 = hx + y, x + yi = hx, xi + hx, yi + hy, xi + hy, yi = kxk2 + 2 Rehx, yi + kyk2 . (c) This follows immediately from part (b). (d) Given x, y ∈ H, kx + yk2 + kx − yk2 = kxk2 + hx, yi + hy, xi + kyk2 + kxk2 − hx, yi − hy, xi + kyk2 = 2 kxk2 + 2 kyk2. 8.1.5 (a) Suppose xn → x and yn → y. Then M = sup kxn k < ∞, so |hx, yi − hxn , yn i| ≤ |hx − xn , yi| + |hxn , y − yn i| ≤ kx − xn k kyk + kxn k ky − yn k ≤ kx − xn k kyk + M ky − yn k → 0. (b) Suppose that the series x =

P∞

sN =

n=1

xn converges in H, and let

N X

xn

n=1

denote the partial sums of this series. Then, by definition, sN → x in H. Hence, given y ∈ H, we have ∞ X

n=1

hxn , yi = =

lim

X N

N →∞

lim

n=1

X N

N →∞

n=1

 hxn , yi xn , y



=

lim hsN , yi = hx, yi,

N →∞

where at the last step we have used the continuity of the inner product.

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8.1.9 “⇒.” The proof of the Cauchy–Bunyakovski–Schwarz (CBS) Inequality shows that for every real t we have 0 ≤ kx − αtyk2 = kxk2 − 2t |hx, yi| + t2 kyk2 ,

(A)

where |α| = 1 satisfies hx, yi = α |hx, yi|. This is a quadratic polynomial in the real variable t, and since it is everywhere nonnegative, it can have at most one real root. This requires that the discriminant be at most zero, i.e., (−2 |hx, yi|)2 − 4 kxk2 kyk2 ≤ 0, which yields the CBS Inequality. Suppose now that equality holds in CBS, i.e., |hx, yi| = kxk kyk for some x and y. Then we have (−2 |hx, yi|)2 − 4 kxk2 kyk2 = 0, which means that the discriminant above is zero. This implies that the polynomial in equation (A) does have a real root, i.e., there is a t such that kx − αtyk2 = kxk2 − 2t |hx, yi| + t2 kyk2 = 0. “⇐.” Suppose that kax + byk = 0 for some scalars a, b ∈ C, not both zero. Suppose that a = 0. Then b 6= 0 but |b| kyk = kax + byk = 0, so kyk = 0. Hence 0 ≤ |hx, yi| ≤ kxk kyk = 0, so equality holds in CBS. Similarly, equality holds if b = 0, so we can assume a, b 6= 0. By the Polar Identity, 0 ≤ kax + tbyk2 = kaxk2 + 2t Rehax, byi + t2 kbyk2 .

(B)

If |b| kyk = kbyk = 0, then we have for t = 1 that 0 = kax + byk2 = kaxk2 + 2 Rehax, byi. Hence 0 ≤ kaxk2 = −2 Rehax, byi ≤ 2 |hax, byi| ≤ 2 kaxk kbyk = 0. Thus in this case we have that |ab| |hx, yi| = 0 = |ab| kxk kyk. Since ab 6= 0, it follows that equality holds in CBS. On the other hand, if kbyk = 6 0 then equation (B) is a true quadratic polynomial with at most one real root. Further, t = 1 is a root, so the discriminant is zero. Therefore

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2 2 Rehax, byi − 4 kaxk2 kbyk2 = 0,

or

|Rehax, byi| = kaxk kbyk. Hence |ab| kxk kyk = kaxk kbyk ≤ |hax, byi| = |ab| |hx, yi| ≤ |ab| kxk kyk. Since ab 6= 0, equality in CBS holds. 8.1.10 “⇒.” If xn → x, then kxn k2 → kxk2 by the continuity of the norm, and hxn , yi → hx, yi by the continuity of the inner product. “⇐.” Assume that kxn k2 → kxk2 and hxn , yi → hx, yi for every vector y. Applying the Polar Identity, we see that kx − xn k22 = hx − xn , x − xn i = kxk22 − 2 Re hxn , xi + kxn k22 → kxk22 − 2 Re hx, xi + kxk22 = kxk22 − 2kxk22 + kxk22 = 0. Therefore xn → x. 8.1.11 Since |E| > 0, there exists some measurable subset F such that 0 < |F | < ∞. Applying Problem 2.3.19, there exists some measurable set A ⊆ F such that |A| = 12 |F |. Consequently the set B = F \A has measure |B| = 21 |F |. Assume first that p < ∞. Let f = χA and g = χB . Then kf + gk2p + kf − gk2p = =

Z

E

Z

|χA + χB |p χF

E

2/p

+

2/p Z

+

Z

E

χF

E

|χA − χB |p

2/p

= 2|F |2/p = 2(2|A|)2/p = 21+2/p |A|2/p . On the other hand, 2

kf k2p

+

kgk2p




= 2

Z

E

|χA |p

2/p

= 2|A|

2/p

+ 2 2/p

+ 2|B|

Z

E

|χB |p

2/p

= 4|A|2/p .

Since p 6= 2, we have 21+2/p 6= 4, so the Parallelogram Law fails.

2/p

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For the case p = ∞, we simply observe that kf + gk2∞ + kf − gk2∞ = 12 + 12 = 2, while 2 kf k2∞ + kgk2∞




= 2 12 + 12




= 4.

Hence the Parallelogram Law fails for this case as well. Consequently, in either case, the norm on Lp (E) cannot be induced from an inner product, because if it was then the Parallelogram Law would have to be satisfied. 8.1.12 Since f is monotone increasing, its derivative f ′ exists a.e., and f ′ ≥ 0. If f ′ = 0 on any set of positive measure, then 1/f ′ = ∞ on a set of positive measure and we are done. Therefore, we may assume that f ′ > 0 a.e. Fix any a < b. Applying the Cauchy–Bunyakovski–Schwarz Inequality, we see that  Z b Z b ′ 1/2 2  Z b f (x) 1 2 ′ dx dx ≤ (b − a) = f (x) dx ′ ′ 1/2 a a f (x) a f (x) Z b 
1 = f (b) − f (a) dx ′ a f (x) Z b 1 2 ≤ Cb dx. ′ a f (x) Therefore 1 (b − a)2 = lim ≤ lim b→∞ b→∞ C Cb2

Z

a

b

1 dx = f ′ (x)

Z

a

∞

1 dx. f ′ (x)

(A)

Now, if 1/f ′ was integrable, then it would follow from the Dominated Convergence Theorem that Z ∞ 1 lim dx = 0. ′ a→∞ a f (x) However, this would contradict equation (A), so 1/f ′ cannot be integrable. 8.1.13 It is clear that h·, ·i is an inner product on H. To show that H is complete, suppose that {fn }n∈N is a Cauchy sequence in H. Then {fn }n∈N is a Cauchy sequence in L2 [a, b], so there is some function f ∈ L2 [a, b] such that fn → f in L2 -norm. Also, {fn′ }n∈N is a Cauchy sequence in L2 [a, b], so there is some function g ∈ L2 [a, b] such that fn′ → g in L2 -norm. For future reference, note that since [a, b] has finite measure, we can use the CBS Inequality to compute that for each function F ∈ L2 [a, b] we have

c Solutions 2019 Christopher Heil

kF k1 =

Z

a

Z

b

|F | ≤

323 b

a

12

1/2 Z

a

b

|F |2

1/2

= (b − a)1/2 kF k2 .

This problem would be significantly easier if we knew that {fn (a)}n∈N was a Cauchy sequence. However, we have not yet established that (and I do not see an easy way to infer it). So, recall that fn is absolutely continuous, and define Z x hn (x) = fn (x) − fn (a) =

a

fn′ ,

x ∈ [a, b].

For each x we therefore have

Z x ′ ′ |hm (x) − hn (x)| = (fm − fn ) a Z x ′ |fm − fn′ | ≤ a

′ ≤ kfm − fn′ k1

′ ≤ (b − a)1/2 kfm − fn′ k2 .

Consequently ′ khm − hn ku = sup |fm (x) − fn (x)| ≤ (b − a)1/2 kfm − fn′ k2 . x

Since {fn′ }n∈N is Cauchy in L2 -norm, we conclude that {hn }n∈N is Cauchy with respect to the uniform norm. Since each fn is continuous and C[a, b] is a Banach space with respect to k · ku , this implies that there exists some continuous function h such that hn → h uniformly. Since fn → f in L2 -norm, there is a subsequence such that fnk → f a.e. If x is such that fnk (x) → f (x), then we also have hnk (x) → h(x). Therefore fnk (a) converges, because   C = lim fnk (a) = lim fnk (x) − fnk (x) + fnk (a) k→∞ k→∞     = lim fnk (x) − lim fnk (x) − fnk (a) k→∞

k→∞

= f (x) − h(x).

Hence f = h + C a.e. Thus f is equal a.e. to the continuous function h + C. Since f is only defined up to a set of measure zero, we can redefine f on a set of measure zero and take f = h + C. That is, we choose a representative of f such that f = h + C. Since [a, b] has finite measure, g is integrable and we can define Z x G(x) = g(t) dt + C, x ∈ [a, b]. a

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This function G is absolutely continuous and G′ = g a.e. Also, given x ∈ [a, b] we have |f (x) − G(x)| = lim |fnk (x) − G(x)| k→∞ Z x Z ′ = lim fnk + fnk (a) − k→∞

a

a

x

f − C

Z x ≤ lim (fn′ − g) + lim fnk (a) − C k→∞ k→∞ a

≤ lim kfn′ − gk1 + 0 k→∞

≤ lim (b − a)1/2 kfn′ − gk2 k→∞

= 0. Hence G = f. Therefore f is absolutely continuous, f ′ = G′ = g a.e., and kf − fn k = kf − fn k2 + kf ′ − fn′ k2 = kf − fn k2 + kg − fn′ k2

→ 0

as n → 0.

Therefore fn → f in the norm of H, and hence H is complete. 8.1.14 (a) Because f is square-integrable, it is integrable on any compact Rx set (see Problem 7.2.16). Hence 0 f (t) dt exists for each x. Also, since t1/2 Rx is bounded on any finite interval, the integral 0 t1/2 |f (t)|2 dt also exists for every x. Applying CBS (H¨older’s Inequality for p = 2), we see that Z

0

x

2 Z f (t) dt =

x

0

≤

Z

2 t−1/4 t1/4 f (t) dt

x

0

= 2x1/2

t−1/2 dt Z

0

x

2 Z

x

0

t1/2 |f (t)|2 dt



t1/2 |f (t)|2 dt.

(b) The function F is well-defined because f is locally integrable. Applying part (a) and using Tonelli’s Theorem as in Problem 4.6.10, we compute that

c Solutions 2019 Christopher Heil

kF k22 = =

Z

∞

0

Z

∞

0

325

|F (x)|2 dx 1 x2

Z

x

f (t) dt

0

2

dx

Z x 1 1/2 2x t1/2 |f (x)|2 dt dx ≤ x2 0 0 Z ∞Z ∞ = 2 x−3/2 t1/2 |f (x)|2 dx dt Z

∞

0

= 2

Z

t

∞

t

0

= 2

Z

0

∞

1/2

2

|f (x)|

Z

∞

−3/2

x t

 dx dt

t1/2 |f (x)|2 2t−1/2 dt

= 4 kf k22. 8.1.15 This is the p = 2 case of Exercise 9.1.13, but we give the proof for the p = 2 here. An easier proof can also be constructed by applying Minkowski’s Integral Inequality, but the point of having this problem here is to exhibit an application of the CBS Inequality. All integrals below are over Rd . We need to first restrict to nonnegative functions in order to show that f ∗ g is measurable (at least, this is the easiest way to prove this that I see). We can use Tonelli since the functions are nonnegative. Because we do not have that f is integrable, there seems to be no good way to try to go directly to the general case by using Fubini. In any case, assuming f, g ≥ 0, we have that f (y) g(x − y) is measurable and nonnegative of R2d . Tonelli’s Theorem therefore implies that Z (f ∗ g)(x) = |f (y) g(x − y)| dy is a measurable function of x. Now we use CBS to estimate (f ∗ g)(x). We include absolute values below because exactly the same computation will be repeated later using functions that need not be nonnegative. By CBS, Z |f (y) g(x − y)| dy |(f ∗ g)(x)| ≤ = ≤

Z   f (y) g(x − y) 1/2 g(x − y) 1/2 dy Z

2

|f (y)| |g(x − y)| dy

1/2 Z

|g(x − y)| dy

1/2

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326

=

Z

2

|f (y)| |g(x − y)| dy 1/2

= kgk1

Z

1/2 Z

|f (y)|2 |g(x − y)| dy

|g(y)| dy

1/2

1/2

.

Using the inequality from above and interchanging integrals by Tonelli’s Theorem, we compute that Z 2 kf ∗ gk2 = |(f ∗ g)(x)|2 dx ≤ kgk1 = kgk1 = kgk1 = kgk1 = kgk1

Z Z

Z Z Z

Z

Z

|f (y)|2 |g(x − y)| dy dx |f (y)|2 |g(x − y)| dx dy 2

|f (y)|

|f (y)|2

Z

Z

 |g(x − y)| dx dy  |g(x)| dx dy

|f (y)|2 kgk1 dy

= kgk21 kf k22 < ∞. For the general case, we write f = (f1 − f2 ) + i(f3 − f4 ) and g = (g1 − g2 ) + i(g3 − g4 ) with fi and gi nonnegative. Then f ∗ g is a finite linear combination of fi ∗ gj , so is measurable and belongs to L2 (Rd ). Repeating then exactly the same calculations as above we see that kf ∗ gk2 ≤ kf k2 kgk1 . 8.1.16 (a) Applying the CBS Inequality, we compute that Z x |f − fn | |F (x) − Fn (x)| ≤ a

≤ (a − x)1/2 1/2

≤ (b − a)

Z

a

x

|f − fn |2

1/2

kf − fn k2 .

Therefore kF − Fn ku ≤ (b − a)1/2 kf − fn k2 → 0, so Fn converges uniformly to F. (b) Choose a ≤ x < y ≤ b. Applying the CBS Inequality, we see that

c Solutions 2019 Christopher Heil

Z |F (y) − F (x)| =

a

327

y

f −

Z

x a

Z f =

y

x

f ≤ kf k2 |y − x|1/2 .

Therefore F is H¨ older continuous with exponent 1/2 and constant kf k2 . Similarly, for each n ∈ N we have |Fn (y) − Fn (x)| ≤ kfn k2 |y − x|1/2 . These facts also follow directly from Problem 7.2.18. (c) Fix x ∈ [a, b]. Since fn converges weakly to f we have by hypothesis that hfn , gi → hf, gi for all g ∈ L2 [a, b]. In particular, g = χ[0,x] ∈ L2 [a, b], so Fn (x) =

Z

a

x

fn

= hfn , χ[0,x] i → hfn , χ[0,x] i =

Z

x

f = F (x).

a

That is, Fn → F pointwise on [a, b]. Let M = sup kfn k2 . n

Then by part (a), for x, y ∈ [a, b] we have |Fn (x) − Fn (y)| ≤ kfn k2 |x − y|1/2 ≤ M |x − y|1/2 . Since Fn converges pointwise to F, we see that we also have |F (x) − F (y)| = lim |Fn (x) − Fn (y)| ≤ M |x − y|1/2 . n→∞

Thus the functions Fn and F are all H¨ older continuous with exponent 1/2 and the same constant M. Then given any x, y ∈ [a, b], we have |F (y) − Fn (y)| ≤ |F (y) − F (x)| + |F (x) − Fn (x)| + |Fn (x) − Fn (y)| ≤ M |x − y|1/2 + |F (x) − Fn (x)| + M |x − y|1/2 = 2M |x − y|1/2 + |F (x) − Fn (x)|. Fix ε > 0, and choose δ > 0 small enough that 2M δ 1/2 < ε. Explicitly, this means that we take δ < ε2 /(4M ). Now partition [a, b] with a mesh size less than δ, say a = x0 < x1 < · · · < xk = b where xj − xj−1 < δ/2. Then

c Solutions 2019 Christopher Heil

328 k S

j=0

(xj − δ, xj + δ) ⊇ [a, b].

Let N be large enough that n>N

=⇒

|F (xj ) − Fn (xj )| < ε for j = 0, . . . , k.

Fix any n > N. Given any y ∈ [a, b], there is some j such that |y − xj | < δ. Consequently, |F (y) − Fn (y)| ≤ 2M |xj − y|1/2 + |F (xj ) − Fn (xj )| ≤ 2M δ 1/2 + ε < ε + ε = 2ε. This is true for all y ∈ [a, b], so kF − Fn ku ≤ 2ε. And this is true for all n > N, so we have shown that Fn converges uniformly to F. 8.2.2 We write out the details of the final claim made in the proof of Lemma 8.2.2. We have established that
(A) 2 Re λhx, yi ≤ |λ|2 kyk2

for every scalar λ. Consider λ = it where t > 0. Since Re(−iz) = Im(z), equation (A) becomes

t Imhx, yi = Im thx, yi = Re −ithx, yi ≤ |it|2 kyk2 .

Dividing both sides by the positive number t therefore gives Im hx, yi ≤ t kyk2 . Letting t → 0+ , we obtain Imhx, yi ≤ lim t kyk2 = 0. t→0+

Now consider λ = −it where t > 0. Since Re(iz) = −Im(z), equation (A) becomes

−t Imhx, yi = −Im thx, yi = Re ithx, yi ≤ | − it|2 kyk2 .

Dividing both sides by the positive number t therefore gives −Imhx, yi ≤ t kyk2 , and hence Im hx, yi ≥ t kyk2 .

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Letting t → 0+ , we obtain Imhx, yi ≥ lim+ t kyk2 = 0. t→0

8.2.5 E and O are clearly subspaces. If fn ∈ E and fn → f in L2 -norm, then there exists a subsequence such that fnk → f a.e. Therefore for almost every x we have f (−x) = lim fnk (−x) = lim fnk (x) = f (x), k→∞

k→∞

so f ∈ E. Therefore E is closed, and similarly O is closed. If f ∈ E and g ∈ O, then f g is odd, and therefore hf, gi = 0. This shows that E ⊆ O⊥ and O ⊆ E ⊥ . Suppose that f ∈ O⊥ . Given t ∈ R and h > 0, the function χt = χ[t,t+h] − χ[−t−h,−t] is odd, so 0 = hf, χt i = Therefore 1 h

Z

t+h

t

Z

t+h

f −

t

1 f = h

Z

Z

−t

f.

−t−h

−t

f.

−t−h

Applying the Lebesgue Differentiation Theorem, it follows that for almost every t we have 1 h→0 h

f (t) = lim

Z

t+h

1 h→0 h

f = lim

t

Z

−t

f = f (−t).

−t−h

Therefore f is even. This shows that O⊥ ⊆ E, and a similar argument shows that E ⊥ ⊆ O. 8.2.10 (b) We give the details of a direct proof that M is closed. Suppose that {xk }k∈N is a Cauchy sequence in H. By part (a), for each k ∈ N we can write d X xk = hxk , en i. n=1

The Pythagorean Theorem therefore implies that kxk k2 =

d X

n=1

|hxk , en i|2 .

Fix any index 1 ≤ n ≤ d. Then for all j, k ∈ N we have

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330

|hxj , en i − hxk , en i| = |hxj − xk , en i| X 1/2 d ≤ |hxj − xk , en i|2 = kxj − xk k. n=1



This implies that hxk , en i k∈N is a Cauchy sequences of scalars. Hence this sequence must converge, say to the scalar cn . Let y =

d X

cn e n .

n=1

Then y ∈ M, since M = span{e1 , . . . , ed }. Also, y − xk =

d X


cn − hxk , en i en ,

n=1

so by applying the Pythagorean Theorem again we see that lim ky − xk k2 = lim

k→∞

k→∞

=

d X

n=1

d  X

n=1

|cn − hxk , en i|2

lim |cn − hxk , en i|2

k→∞



= 0.

Thus xk → y as k → ∞. This shows that every Cauchy sequence in M converges to an element of M. Hence M is complete. But H is a Hilbert space, so it is complete, and therefore a subspace of H is complete if and only if it is closed. So, we conclude that M is closed. 8.2.11 We give the details of the proofs of the remaining implications in Theorem 8.2.11. (b) ⇔ (c). This equivalence follows from the fact that e ∈ M ⊥ if and only if e ⊥ M. (c) ⇒ (a). Suppose that x = p + e where p ∈ M and e ⊥ M. Choose any vector y ∈ M. Then p − y ∈ M, so x − p = e ⊥ p − y. Therefore the Pythagorean Theorem implies that kx − yk2 = k(x − p) + (p − y)k2 = kx − pk2 + kp − yk2 ≥ kx − pk2 . Hence p is the point in M that is closest to x. (c) ⇒ (d). Suppose that x = p + e where p ∈ M and e ∈ M ⊥ . If y ∈ (M ⊥ ) then hp, yi = 0, so p ∈ (M ⊥ )⊥ . Therefore x = e + p where e ∈ M ⊥ and

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p ∈ (M ⊥ )⊥ . The equivalence of statements (a) and (b) therefore implies that e is the orthogonal projection of x onto M ⊥ . (d) ⇒ (b). Suppose that x = p + e where e is the orthogonal projection of x onto M ⊥ . Then by the equivalence of statements (a)–(c) we have that p ∈ (M ⊥ )⊥ . We must show that p ∈ M. Write p = q + z where q ∈ M and z ∈ M ⊥ . Since p ∈ (M ⊥ )⊥ we have hp, zi = 0. Since we also have hq, zi = 0, it follows that 0 = hp, zi = hq, zi + hz, zi = 0 + kzk2 . Hence z = 0, so p = q ∈ M. 8.2.13 Suppose that x, y ∈ span(A) are given. Then there exist vectors xn , yn ∈ span(A) such that xn → x and yn → y in norm. Therefore xn + yn → x+y in norm. As xn +yn ∈ span(A) for every n, it follows that x+y belongs to the closure of span(A), which is span(A). Therefore span(A) is closed under vector addition, and a similar argument shows that it is closed under scalar multiplication. Therefore span(A) is a subspace of X. By definition span(A) is a closed set, so it is a closed subspace. (b) Suppose that M is a closed subspace of X and A ⊆ M. Since M is closed under vector addition and scalar multiplication, it follows that span(A) ⊆ M. Since M is closed under limits, it follows that M contains every limit of elements of span(A). The set of all such limits is the closure of the span, so we have shown that span(A) ⊆ M. 8.2.15 Suppose that x ⊥ A, i.e., PNx is orthogonal to every vector in A. If y ∈ span(A), then y = k=1 ck ak for some integer N ∈ N, vectors ak ∈ A, and scalars ck . Hence hy, xi =

N X

k=1

ck hak , xi = 0.

This shows that x ⊥ span(A). Now choose any vector z ∈ span(A). Then there exist vectors yn ∈ span(A) such that yn → z as n → ∞. Since x ⊥ yn for every n, the continuity of the norm therefore implies that hz, xi = lim hyn , xi = 0. n→∞

This shows that x ⊥ span(A). Let M = span(A). Then A ⊆ M, so the inclusion M ⊥ ⊆ A⊥ follows directly. On the other hand, if x ⊥ A, then x ⊥ M by the work above. This shows that A⊥ ⊆ M ⊥ . Therefore A⊥ = M ⊥ . (A⊥ )⊥ = (M ⊥ )⊥ = M. 8.2.19 Choose finitely vectors x1 , . . . , xn ∈ S. Suppose that c1 , . . . , cn Pmany n are scalars such that k=1 ck xk = 0. Then, by the Pythagorean Theorem,

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332

0 = k

n X

k=1

ck xk k

2

=

n X

k=1

kck xk k

2

=

n X

k=1

|ck |2 kxk k2 .

Hence |ck | kxk k = 0 for every k. By hypothesis, kxk k = 6 0, so this implies that ck = 0 for every k. Hence S is linearly independent. 8.2.21 Suppose that gn ∈ M and gn → g ∈ L2 (Rd ). Then there exists a subsequence gnk → g pointwise a.e. Each gnk is zero a.e. outside of M, so it follows that g = 0 a.e. outside of M as well. Therefore g ∈ M, and hence M is closed. If f ∈ L2 (Rd ), then p = f χE ∈ M. Let e = f − p, and choose any function g ∈ M. Note that e(x) = 0 for a.e. x ∈ E. On the other hand, if g ∈ M then g(x) = 0 for a.e. x ∈ / E. Therefore e(x) g(x) = 0 for a.e. x. Hence Z he, gi = e(x) g(x) dx = 0. Rd

This shows that e ∈ M ⊥ . Since we have written f = p + e where p ∈ M and e ∈ M ⊥ , one of the characterizations of orthogonal projections tells us that p is the orthogonal projection of f onto M. 8.2.24 (a) ⇒ (b). For each m, let Em = span{xn }n6=m . We can write xm = ⊥ pm + qm for a unique choice of vectors pm ∈ Em and qm ∈ Em . In particular, qm ⊥ xn for every n 6= m. If hxm , qm i = 0, then 0 = hxm , qm i = hpm , qm i + hqm , qm i = 0 + kqm k2 . But this implies that xm = pm ∈ Em , which is a contradiction. Therefore we must have hxm , qm i = 6 0. On the other hand, if n 6= m then xn ∈ Em and ⊥ qm ∈ Em , so hxn , qm i = 0 when n 6= m. Define qm . ym = hqm , xm i Since ym is a multiple of qm we have ym ⊥ xn for every n 6= m. And since hxm , ym i = 1, the sequence {ym }m∈N has the required properties.

(b) ⇒ (a). Suppose that a biorthogonal sequence {yn }n∈N exists. With Em as above, we have by linearity and continuity of the inner product that ym ⊥ Em . Since hxm , ym i = 1 6= 0, the vector xm cannot belong to Em . (a) + completeness ⇒ (b) + uniqueness. Suppose that (a) holds and that {xn }n∈N is complete. Suppose that {yn }n∈N and {zn }n∈N are each biorthogonal to {xn }n∈N . Fix any n. Then for any m we have hxm , yn − zn i = hxm , yn i − hxm , zn i = δmn − δmn = 0.

c Solutions 2019 Christopher Heil

333

Since {xm }m∈N is complete, this implies that yn −zn = 0. Hence the biorthogonal sequence is unique. (b) + uniqueness ⇒ (a) + completeness. Suppose that a biorthogonal sequence {yn }n∈N exists and is unique. Suppose that z ∈ H is orthogonal to every xn . Then hxn , ym + zi = hxn , ym i + hxn , zi = δmn + 0 = δmn . Thus {ym + z}m∈N is also biorthogonal to {xn }n∈N . By uniqueness, we therefore have z = 0. Hence {xn }n∈N is complete. 8.2.25 Let c(x) = cos 2πx and s(x) = sin 2πx. A direct computation shows that hc, si = 0. Suppose there is an f ∈ L2 [0, 1] such that kc − f k22
1, Z Z hf, χEn i = (χA − χB ) χE1 = 0 = 0. E

E

Thus f ⊥ en for every n > 1. Thus f is orthogonal to every vector en . This implies that f is a nonzero vector orthogonal to the closed span of the en , so this closed span is not all of H. Therefore E is not complete, and hence is not an orthonormal basis for L2 (E).

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337

8.3.25 (a) We are P given an orthonormal basis {en }n∈N and a sequence {yn }n∈N such that ken − yn k < 1. Suppose that hx, yn i = 0 for every n. If x 6= 0 then we have by the Parseval Equality that kxk2 =

∞ X

n=1

∞ X

|hx, en i|2 =

n=1 ∞ X

=

n=1 ∞ X

≤

n=1

|hx, en − yn i + hx, yn i|2 |hx, en − yn i|2 kxk2 ken − yn k2

< kxk2 , which is a contradiction. Therefore x = 0, so Corollary 8.2.18 implies that {yn }n∈N is complete. Remark: {yn }n∈N need not be a Schauder basis for H. P (b) If we take y1 = x2 and yn = xn for n > 1, then kxn − yn k2 = 1 but {yn } is not complete. 8.3.26 Orthonormality follows “from inspection,” or by a direct calculation. Inspection also shows that the function w(t) = R1 (t) R2 (t) is orthogonal to Rn for every n ≥ 0. Therefore the Rademacher system is not complete.

8.3.27 (a) Let M = {xk }k≥0 . Then span(M) = P, the set of all polynomials. Problem 7.3.19 proved that P is dense in L2 [a, b], so M is complete in L2 [a, b] by definition. (b) Suppose that f ∈ L2 [a, b], and f ⊥ xk for every integer k ≥ N. Let g(x) = xN f (x), and note that g ∈ L2 [a, b] since xN is bounded on [a, b]. Given any k ≥ 0 we have k

hg, x i =

Z

a

b

k

g(x) x dx =

Z

a

b

f (x) xk+N dx = hf, xk+N i = 0.

Thus g ⊥ xk for every k ≥ 0. But M is complete, so Corollary 8.2.18 implies that g = 0 a.e. Since xN 6= 0 a.e., this implies that f = 0 a.e. Applying Corollary 8.2.18 again, we conclude that {xk }k≥N is complete in L2 [a, b]. (c) The Legendre polynomials are complete because they are an orthogonal basis for L2 [−1, 1]. If J is a proper subset of {0, 1, 2, . . . } and k ∈ / J, then Pk ⊥ Pj for every j ∈ J, yet Pk is not the zero function. Therefore {Pj }j∈J is not complete by Corollary 8.2.18. (d) Suppose that f ∈ L2 [0, 1] and f ⊥ x2k for every integer k ≥ 0. Define g on [−1, 1] by

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338

g(x) =

(

f (x), x ≥ 0, f (−x), x < 0.

That is, g is obtained by extending f evenly to the interval [−1, 1]. Given any integer k ≥ 0, the inner product of g and x2k as elements of L2 [−1, 1] is 2k

hg, x iL2 [−1,1] = =

Z

1

g(x) x2k dx

−1

Z

1

g(x) x2k dx +

0

=

Z

Z

0

g(x) x2k dx

−1

1

f (x) x2k dx +

0

=

Z

Z

0

g(−y) (−y)2k (−dy)

1

1

f (x) x2k dx +

0

Z

1

(y = −x)

f (y) y 2k dy

0

= hf, x2k iL2 [0,1] + hf, x2k iL2 [0,1] = 0. Thus g ⊥ x2k in L2 [−1, 1]. On the other hand, g is even and x2k+1 is odd, so g ⊥ x2k+1 in L2 [−1, 1]. Therefore, as elements of L2 [−1, 1] we have g ⊥ xk for every k ≥ 0. However, {xk }k≥0 is complete in L2 [−1, 1] by part (a). Corollary 8.2.18 therefore tells us that the only function orthogonal to every xk is the zero function. Hence g = 0 a.e. But this implies that f = 0 a.e. In summary, the only function f ∈ L2 [0, 1] that satisfies f ⊥ x2k for all k ≥ 0 (using the inner product in L2 [0, 1]) is f = 0 a.e. Corollary 8.2.18 therefore implies that {x2k }k≥0 is complete in L2 [0, 1]. (e) The same argument used in part (b) works here. Specifically, Suppose that f ∈ L2 [0, 1], and f ⊥ xk for every integer k ≥ N. Let g(x) = x2N f (x), and note that g ∈ L2 [0, 1] since x2N is bounded on [0, 1]. Given any k ≥ 0 we have Z b Z b f (x) x2k+2N dx = hf, x2k+2N i = 0. g(x) x2k dx = hg, x2k i = a

a

Thus g ⊥ x2k for every k ≥ 0. But part (c) tells us that {x2k }k≥0 is complete, so Corollary 8.2.18 implies that g = 0 a.e. Since x2N 6= 0 a.e., this implies that f = 0 a.e. Applying Corollary 8.2.18 again, we conclude that {x2k }k≥N is complete in L2 [0, 1]. 8.3.28 “⇒.” If E = {en }n∈N is complete, then we have by Plancherel’s Equality that

c Solutions 2019 Christopher Heil ∞ Z x X

n=1



a

339

2 ∞ X en (t) dt = |hχ[a,x] , en i|2 n=1

= kχ[a,x] k22 Z b = |χ[a,x] (t)|2 dt = x − a. a

“⇐.” Suppose that ∞ Z X

a

n=1

Then,

∞ X

n=1

x

2 en (t) dt = x − a,

x ∈ [a, b].

|hχ[a,x] , en i|2 = x − a = kχ[a,x] k22 .

Thus, the Plancherel Equality holds for χ[a,x] , so Theorem 8.3.6 implies that χ[a,x] ∈ span(E). This is true for every x ∈ [a, b], so χ[x,y] = χ[a,y] − χ[a,x] ∈ span(E) for every x < y. But  span χ[x,y] : a ≤ x < y ≤ b

is the set of “really simple functions,” which is dense in L2 [a, b] by Exercise 7.3.12. Hence span(E) = L2 [a, b], as desired. 8.3.29 “⇒.” If {fn }n∈N is complete, then Problem 8.3.28 implies that ∞ Z X

n=1

b

a

Z

a

“⇐.” Suppose that ∞ Z X

n=1

a

x

2 Z fn (t) dt =

b

|hχ[a,x] , fn i|2 dx =

b

(x − a) dx =

a

∞ Z X

n=1

a

b

Z

x a

(b − a)2 . 2

2 (b − a)2 . fn (t) dt = 2

Since {fn }n∈N is orthonormal, Bessel’s Inequality implies that ∞ X

n=1

Therefore

|hχ[a,x] , fn i|2 ≤ kχ[a,x] k22 = x − a.

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340

0 ≤ =

=

Z

Z

Z

  ∞ X 2 2 χ χ k [a,x] k2 − |h [a,x] , fn i| dx

b a

n=1

b

Z

kχ[a,x] k22 −

a b a

b

a

∞ X

n=1

|hχ[a,x , fn i|2 dx

(b − a)2 = 0. 2

(x − a) dx −

Consequently, x − a = kχ[a,x] k22 =

∞ X

n=1

|hχ[a,x] , fn i|2 a.e. x.

Now, kχ[a,x] k22 = x − a is a continuous function of x. By Bessel’s Inequality and the Triangle Inequality on ℓ2 , |||f ||| =

X ∞

n=1

|hf, fn i|2

1/2

is a seminorm on L2 [a, b] and therefore is continuous with respect to L2 -norm. If y → x, then χ[a,y] → χ[a,x] in L2 -norm, so F (x) =

∞ X

n=1

|hχ[a,x] , fn i|2

is a continuous function of x. Since we have shown that F (x) = x − a a.e., it follows that F (x) = x − a for every x. Problem 8.3.28 therefore implies that {fn }n∈N is complete.

8.3.30 We are given an orthonormal basis {fn }n∈N for L2 [a, b]. The function f1 is not the zero element of L2 [a, b], and it is finite a.e., so S 1 { n < |f1 | < n} = {0 < |f1 | < ∞}. n∈N

Therefore there is some positive n such that E = { n1 < |f1 | < n} has positive (and finite) measure. Write E = ∪Fk disjointly where each Fk has positive measure. Define ( |Fk |1/2 |f1 (t)|, t ∈ Fk , m(t) = 1, otherwise. Then for t ∈ Fk we have |m(t)| ≤ |Fk |1/2 n ≤ n (b − a)1/2 ,

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341

so m ∈ L∞ [a, b]. Note that f1 /m is defined a.e., and Z

a

b

X Z f1 (t) 2 XZ X f1 (t) 2 1 dt ≥ dt = dt = 1 = ∞, m(t) m(t) Fk Fk |Fk | k∈N

k∈N

k∈N

so f1 /m ∈ / L2 [a, b]. Since m ∈ L∞ [a, b], we have mfn ∈ L2 [a, b] for every n. Now suppose that g ∈ L2 [a, b] satisfies hg, mfn i = 0 for every n ≥ 2. Then (since m is real-valued) 0 = hg, mfn i =

Z

b

a

g(t) m(t) fn (t) dt = hgm, fn i,

n ≥ 2.

Since gm ∈ L2 [a, b] and {fn }n∈N is an orthonormal basis for L2 [a, b], we conclude that ∞ X gm = hgm, fn i fn = hgm, f1 i f1 . n=1

Thus, gm = cf1 where c = hgm, f1 i. Now, if c 6= 0 then since m(t) 6= 0 a.e. we have g = cf1 /m ∈ / L2 [a, b], which is a contradiction. Hence we must have c = 0. But then hgm, fn i = 0 for every n, so gm = 0 a.e. Since m(t) 6= 0 a.e., we conclude that g = 0 a.e., and therefore {mfn }n≥2 is complete.

8.3.32 (a) Since L2 (E) is a separable infinite-dimensional Hilbert space, Theorem 8.3.17 implies that there exists a unitary operator U : H → ℓ2 . Likewise there is a unitary operator V : K → ℓ2 . The inverse mapping V −1 : ℓ2 → K is also unitary, and hence V −1 U : H → K is unitary. 8.4.3 A direct calculation shows that  {Ekn }k,n∈Z = e2πinx χ[k,k+1] k,n∈Z

is orthonormal in L2 (R). Suppose hf, Ekn i = 0 for every k and n, and fix k. Restricted to the domain [k, k + 1], the sequence {Ekn }n∈Z is an orthonormal basis for L2 [k, k + 1]. The function f (restricted to [k, k + 1]) is orthogonal to every element of this orthonormal basis, so we must have f = 0 a.e. on [k, k + 1]. This is true for each integer k ∈ Z, so f = 0 a.e. on R. Hence {Ekn }k,n∈Z is a complete orthonormal system in L2 (R) and therefore is an orthonormal basis for L2 (R). √ √ 8.4.4 Set T = {1} ∪ { 2 sin 2πnt}n∈N ∪ { 2 cos 2πnt}n∈N . We have eiθ = cos θ + i sin θ. Therefore e−iθ = cos θ − i sin θ. Write 2 cos 2πnt = e2πint + e−2πint = en (t) + e−n (t). Then for m, n ≥ 0 we have

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342



 4 cos 2πmt, cos 2πnt = em + e−m , en + e−n

= hem , en i + hem , e−n i + he−m , en i + he−m , e−n i

= δmn + δm,−n + δ−m,n + δ−m,−n = 2δmn + 2δm,−n = 2δmn . A similar argument applies to the other inner products, so we conclude that T is orthogonal. Since each element is normalized, E is orthonormal. Completeness: First Proof. Since span(T ) = span{Tn }n∈Z , it follows that E is an orthonormal basis for complex L2 [0, 1]. Hence every vector f in real L2 [0, 1] can be represented in the orthonormal basis E. Since each element of E is real-valued, the scalars in this representation, which are inner products of f with the elements of E, will be real. Hence f has a real-valued series representation, and therefore E is an orthonormal basis for real L2 [0, 1].

Completeness: First Proof. Let f be a function in real-L2 [0, 1] that is orthogonal to every element of E. Since we have sin 2π(−n)x = sin 2πnx and cos 2π(−nx) = − cos 2πnx, it follows that f is orthogonal to sin 2πnx and cos 2πnx for every integer n ∈ Z, both positive and negative. The function f belongs to L2 [0, 1] (the space of complex-valued squareintegrable functions on [0, 1]). The inner product on real-L2 [0, 1] is simply the restriction of the inner product on L2 [0, 1]. Therefore, given n ∈ Z, we have hf, e2πinx i = hf, cos 2πnx + i sin 2πnxi = hf, cos 2πnxi − ihf, sin 2πnxi = 0. Since the trigonometric system is complete in L2 [0, 1], it follows that f = 0 a.e. Therefore E is complete in real-L2 [0, 1]. √ 8.4.5 Problem 8.4.4 showed that { 2 cos 2πnx}n∈N is an orthonormal sequence in real-L2[0, 1]. As this space is contained in L2 [0, 1] and has the same √ inner product, { 2 cos 2πnx}n∈N is also an orthonormal sequence in L2 [0, 1]. Therefore, Bessel’s Inequality implies that ∞ X

n=1

|hf,

√

2 cos 2πnxi|2 ≤ kf k22 .

Since the norm of f is finite, it follows that the terms of series converge to zero. Therefore Z 1 lim f (x) cos 2πnx dx = lim hf, cos 2πnxi = 0. n→∞

0

n→∞

8.4.6 (a) Let f = χ[0,1/2) − χ[1/2,1) . For n = 0 we have

c Solutions 2019 Christopher Heil

For n 6= 0, fb(n) =

fb(0) = Z

1

Z

1

343

f (x) e−2πi0x dx =

0

Z

1

f (x) dx = 0.

0

f (x) e−2πinx dx =

0

Z

1/2

0

e−2πinx dx −

Z

1

e−2πinx dx

1/2

1/2 1 e−2πinx e−2πinx = − −2πin 0 −2πin 1/2

e−πin − 1 1 − e−πin − −2πin −2πin ( 2(−1)n − 1 0, n even, = = 2i −2πin , n odd. − πn =

Therefore, 1 = kf k22 =

X

n∈Z

|fb(n)|2 =

X

n∈Z, n odd

Rearranging, we see that

2 2i = 2 πn

X

n∈N, n odd

4 π 2 n2

.

∞ X 1 π2 = . 8 (2k − 1)2 k=1

(b) Using part (a), we have ∞ ∞ ∞ ∞ X X X 1 1 1 1X 1 π2 = + = + . n2 (2n)2 (2n − 1)2 4 n=1 n2 8 n=1 n=1 n=1

Consequently

8.4.7 For n = 0 we have

∞ 3X 1 π2 = . 4 n=1 n2 8

fb(0) =

Z

1

x dx = 0

1 . 2

For n 6= 0, we use integration by parts with u = x and dv = e−2πinx dx to compute that

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fb(n) =

Z

1

xe

−2πinx

0

Now,

1 Z 1 −2πinx e xe−2πinx − dx dx = −2πin 0 0 −2πin 1 e−2πin − 0 1 e−2πin = = − − 0. −2πin (−2πin)2 0 −2πin

kf k22 =

Z

1

x2 dx =

0

1 x3 1 = . 3 0 3

Applying the Plancherel Equality, we therefore have X 1 = kf k22 = |fb(n)|2 3 n∈Z

= |fb(0)|2 +

X

n6=0

|fb(n)|2

X 1 1 + 4 4π 2 n2

=

n6=0

∞

X 1 1 + 2 . 4 4π 2 n2 n=1

= Therefore

∞ 1 1 2 X 1 1 , = − = 12 3 4 4π 2 n=1 n2

which after rearranging yields

∞ X 1 4π 2 π2 = = . 2 n 2 · 12 6 n=1

8.4.8 (a) ⇔ (b). We will apply Vitali’s criterion with [a, b] = [0, 1]. Note that |e2πinx − 1|2 = (e2πinx − 1)(e−2πinx − 1) = 1 − e2πinx − e−2πinx + 1 = 2 − 2 cos 2πnx. Therefore X Z

n∈Z

0

x

Z 2 e2πint dt =

0

x

2 X e2πinx − 1 2 1 dt + 2πin n6=0

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X 2 − 2 cos 2πnx = x2 + 4π 2 n2 n6=0

= x2 +

∞ X 1 − cos 2πnx . π 2 n2 n=1

(A)

Vitali’s criterion tells us that {e2πinx }n∈Z is complete in L2 [0, 1] if and only if 2 X Z x 2πint x ∈ [0, 1]. (B) e dt = x, n∈Z

0

Substituting equation (A), equation (B) reduces to x2 +

∞ X 1 − cos 2πnx = x. π 2 n2 n=1

(b) ⇔ (c). This follows by using Euler’s formula

P

1 n2

=

π2 6 .

8.4.9 Assume that f ∈ L2 [0, 1] and fb ∈ ℓ1 (Z). Since en (x) = e2πinx ∈ C[0, 1] is continuous and the series X g = fb(n) en n∈Z

converges absolutely with respect to the uniform norm, it follows that g is continuous. However, we also know that X f = fb(n) en , n∈Z

where this series coverges in L2 -norm. Letting sN = lows that we have both sN → g in L∞ -norm

and

Pn

n=−N

fb(n) en , it fol-

sN → f in L2 -norm.

Consequently f = g a.e. (see Problem 7.3.14). This is exactly what we mean when we say that an element of L2 [0, 1] is continuous. 8.4.10 For simplicity of notation, let ebn (x) = e2πbnx . (a) This follows from the fact that {e2πinx }n∈Z is an orthonormal basis for L [0, 1] combined with a change of variables. 2

(b) We will show that if b > 1 then {ebn }n∈Z is incomplete in L2 [0, 1]. Each exponential ebn is 1b -periodic. Let ε be small enough that both [0, ε] and [ 1b , ε] are contained within [0, 1]. Define f = χ[0,0+ε] − χ[ 1b , 1b +ε] .

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Then hf, ebn i =

Z

0

ε

e−2πibnx dx −

Z

1 b +ε

e−2πibnx dx = 0.

1 b

Thus f is orthogonal to every ebn , and therefore {ebn }n∈Z is incomplete. Alternatively, since each function ebn is 1b -periodic, any finite linear combination will be 1b -periodic. As limits of 1b -periodic functions are still 1b -periodic,
it follows that every function in span {ebn }n∈Z is 1b -periodic. However, not every function in L1 [0, 1] is 1b -periodic, e.g., consider f (x) = x. Therefore {ebn }n∈Z is incomplete. (c) In the language of frame theory, we want to show that if 0 < b < 1 then {ebn }n∈Z is a tight frame for L2 [0, 1]. By part (a), {b1/2 ebn }n∈Z is an orthonormal basis for the space L2 [0, 1b ]. Therefore, X |hf, ebn i|2 = kf k22 . (A) ∀ f ∈ L2 [0, 1b ], b n∈Z

Note that these are inner products in L2 [0, 1b ], i.e., kf k22 =

Z

0

1/b

|f (x)|2 dx

and

hf, ebn i =

Z

1/b

f (x) e−2πibnx dx.

0

If we take f ∈ L2 [0, 1], we can regard it as an element of L2 [0, 1b ] by setting f (x) = 0 for 1 < x ≤ 1b . Then we can apply equation (A), but because we have extended by zero, the norm and inner product are from L2 [0, 1]. In other words, equation (A) holds for f ∈ L2 [0, 1], so {ebn }n∈Z is a tight frame for L2P [0, 1] with frame bounds A = B = 1/b. For the same reason, we have f = b n∈Z hf, ebn i ebn with convergence of the series in L2 [0, 1]. An alternative approach is to note that P : L2 [0, 1b ] → L2 [0, 1] given by P f = f χ[0,1] is the orthogonal projection of L2 [0, 1b ] onto L2 [0, 1] if we regard this latter space as a subspace of L2 [0, 1b ]. Hence if f ∈ L2 [0, 1] then P f = f and therefore if we write out the orthonormal basis expansion of f and rearrange we obtain X  f = P (P f ) = P hP f, b1/2 ebn iL2 [0, 1b ] b1/2 ebn n∈Z

= b

X

hf, P ebn iL2 [0, 1b ] P ebn

n∈Z

= b

X

hf, ebn iL2 [0,1] P ebn ,

n∈Z

with convergence in L2 [0, 1b ]. However, since both f and P ebn are zero outside of [0, 1], we conclude that

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f = b

347

X

hf, ebn iL2 [0,1] ebn

(B)

n∈Z

with convergence in L2 [0, 1]. In either approach, the convergence of the series in equation (B) is unconditional because the orthonormal basis representation of a vector converges unconditionally. Note that {ebn }n∈Z is not an orthogonal sequence, because he2πimbx , ebn i =

Z

1

e2πi(m−n)bx dx

0

1 e2πi(m−n)bx = 2πi(m − n)b 0 =

e2πi(m−n)b 1 − 6= 0. 2πi(m − n)b 2πi(m − n)b

More precisely, this quantity is not zero for every choice of m and n, e.g., consider m = 1 and n = 0. Now we show directly that the constant function f = 1 does not have a unique representation in terms of the exponentials {ebn }. Since f (x) = e2πi0bx = e0 (x), one expansion is X f (x) = δmn ebn . n∈Z

Another expansion is provided by the tight frame property: X f = b hf, ebn i ebn . n∈Z

We must check whether this is the same expansion we found before. So we check the coefficients:  Z 1 n = 0, b, −2πibnx −2πibn e dx = b hf, ebn i = b 1 − e 0 + , n 6= 0. 2πin 2πin

Since these are not the same values as δmn we have indeed found two different ways to write f = 1 in terms of the frame elements ebn . Therefore this system is not a Schauder basis for L2 [0, 1].

8.4.11 (a) Since g ∈ L2 (R) is supported within [0, b−1 ], the translated function g(x − ak) belongs to L2 (Ik ), where Ik = [ak, ak + b−1 ]. Futher, since Ik has length b−1 , it follows from Problem 8.4.10(a) that  1/2 2πibnx b e (A) n∈Z

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is an orthonormal basis for L2 (Ik ). To show that the tight frame equality holds, let us first consider a function f ∈ Cc (R). Since f is bounded, the product f (x) g(x − ak) belongs to L2 (Ik ). Applying the Plancherel Equality using the orthonormal basis given in equation (A) to this function, we therefore have that Z ∞ Z 2 f (x) g(x − ak) 2 dx |f (x) g(x − ak)| dx = −∞

Ik

= f · g(x − ak) L2 (I

k)

X

2  f · g(x − ak), b1/2 e2πibnx 2 = L (I ) k

n∈Z

X Z = b

Ik

n∈Z

= b

X Z

∞

−∞

n∈Z

= b

f (x) g(x − ak) e

−2πibnx

2 dx

2 f (x) e2πibnx g(x − ak) dx

X

 f, gkn 2 .

n∈Z

Hence, using Tonelli’s Theorem to interchange the sum and integral, it follows that  X

X X

  f, b1/2 gkn 2 = f, gkn 2 b k,n∈Z

n∈Z

k∈Z

=

XZ k∈Z

=

Z

=

−∞

|f (x) g(x − ak)|2 dx

∞

−∞

Z

∞

∞

−∞

|f (x)|2

X

k∈Z

|g(x − ak)|2 dx

|f (x)|2 dx = kf k22 .

Thus, the desired Parseval Equality holds for all functions f that lie in the dense subspace Cc (R). Since Cc (R) is dense in L2 (R), this equality can then be extended to all functions f ∈ L2 (R). To see this, let f be an arbitrary function in L2 (R). Then there exist functions fj ∈ Cc (R) such that fj → f in L2 -norm as j → ∞. Our work above shows that for each fixed j, we have X |hfj , b1/2 gkn i|2 = kfj k22 . k,n∈N

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By applying the continuity of the inner product and Fatou’s Lemma for series (Problem 4.2.18), we can therefore compute that X X |hf, b1/2 gkn i|2 = lim |hfj , b1/2 gkn i|2 k,n∈N

k,n∈N

j→∞

≤ lim inf j→∞

X

k,n∈N

|hfj , b1/2 gkn i|2

= lim inf kfj k22 j→∞

= kf k22 . That is, for every function f ∈ L2 (R) we have X |hf, b1/2 gkn i|2 ≤ kf k22 .

(A)

k,n∈N

Proving the opposite inequality takes more work. Fix f ∈ L2 (R) and ε > 0. Then there exists some function g ∈ Cc (R) such that kf − gk2 < ε. This function g satisfies X |hg, b1/2 gkn i|2 = kgk22 . k,n∈N

Applying equation (A) to the function g − f, we also have X |hg − f, b1/2 gkn i|2 ≤ kg − f k22 < ε2 . k,n∈N

Combining these facts with the Triangle Inequality for the ℓ2 -norm, we see that kf k2 ≤ kf − gk2 + kgk2 X 1/2 1/2 2 < ε + |hg, b gkn i| k,n∈N

≤ ε +

X

k,n∈N

< ε + ε +

|hg − f, b

X

k,n∈N

1/2

2

gkn i|

1/2

|hf, b1/2 gkn i|2

1/2

As this is true for every ε > 0, it follows that

+

X

k,n∈N

.

|hf, b

1/2

2

gkn i|

1/2

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350

kf k2 ≤

X

|hf, b

k,n∈N

1/2

2

gkn i|

1/2

.

This finishes the proof that G is a tight frame. (b) As long as ab < 1, it is possible to construct an example where g is continuous. For example, take a = 1 and b = 1/2, and let g 2 be the hat function supported on [0, 2]. That is,   0 ≤ x ≤ 1, x, 2 g(x) = 2 − x, 1 ≤ x ≤ 2,   0, otherwise.

Then g is continuous, supp(g) = [0, 2], and X X g(x − ak)2 = g(x − k)2 = 1. k∈Z

k∈Z

Hence, it follows from part (a) that the Gabor system G = {b1/2 gkn }k,n∈Z = {2−1/2 eπinx g(x − k)}k,n∈Z is a tight frame. However, since g is nonnegative and the supports of g and g(x − 1) overlap, these two elements of G are not orthogonal. 8.4.12 First note that if ξ 6= η then, since |eξ | ≤ 1, 1 heξ , eη i = lim T →∞ 2T

Z

T

e2πiξt e−2πiηt dt

−T

T 1 e2πi(ξ−η)t = lim T →∞ 2T 2πi(ξ − η) −T

e2πi(ξ−η)T − e−2πi(ξ−η)T = 0, T →∞ 4T πi(ξ − η)

= lim and

1 heξ , eη i = lim T →∞ 2T

Z

T

e2πiξt e−2πiξt dt = 1.

−T

Hence, once we show that h·, ·i is an inner product, we can conclude that {eξ }ξ∈R is an orthonormal system. Suppose that f, g, h ∈ H and the limits defining hf, hi and hg, hi exist. Then Z T 1 (f (t) + g(t)) h(t) dt = hf, hi + hg, hi hf + g, hi = lim T →∞ 2T −T

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351

exists as well, and similarly hcf, hi = chf, hi and hh, f i = hf, hi. PM In particular, given f, g ∈ H we can write f = m=1 cm eξm and g = PN n=1 dn eξn , so we have that hf, gi =

M X N X

m=1 n=1

M X N X

cm dn heξ , eη i =

cm dn δξη

m=1 n=1

exists. This shows that h·, ·i is well-defined on H. Note we have that hf, f i ≥ 0 for every f ∈ H, as 1 hf, f i = lim T →∞ 2T

Z

T

−T

|f (t)|2 dt.

Thus h·, ·i is a semi-inner product on H. The Pythagorean Theorem holds for semi-inner products. If f ∈ H, then we can write f =

N X

ck eξk

k=1

for some N > 0, ck ∈ C, and ξk ∈ R. Applying orthogonality, hf, f i = kf k2 =

N X

k=1

kck eξk k2 =

N X

k=1

|ck |2 .

Therefore hf, f i = 0 if and only if f = 0 (note that this calculation works because f is a finite linear combination of the functions eξ ). Therefore we conclude that h·, ·i is an inner product on H, and {eξ }ξ∈R is an uncountable orthonormal system in H. 8.4.13 (a) The norm of fn satisfies kfn k22

=

Z

1

|xe

0

2πinx 2

| dx =

Z

1

0

x2 dx =

1 . 3

For gn , note first that en (x) + e−n (x) = e2πinx + e−2πinx = 2 cos 2πnx and recall that

1 − cos 2θ = 2 sin2 θ.

Therefore, kgn k22 =

Z

1 0

2πinx 2 e − 1 dx x

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352

Z

=

Z

=

Z

=

Z

=

Z

1

e2πinx − 1 e2πinx − 1 dx x x

1

1 − e2πinx − e−2πinx + 1 dx x2

1

2 − 2 cos 2πnx dx x2

1

4 sin2 πnx dx x2

0

0

0

0

πn

4 sin2 u du (u/(πn))2 πn 0 Z πn sin2 u du. = 4πn u2 0 =

This is finite, so g ∈ L2 [0, 1]. However, lim

n→∞

Z

πn

0

sin2 u du = u2

Z

∞

0

(u = πnx)

sin2 u π du = . u2 2

In fact, we do not even need to know that the value of the integral is π/2; the only important point is that 

sin u u

2

is integrable on [0, ∞), and therefore its integral is finite (and nonzero). Since 4πn → ∞ as n → ∞, it therefore follows that Z πn sin2 u lim kgn k2 = lim 4πn du = ∞. n→∞ n→∞ u2 0 (b) Assume that m and n are both nonzero. Then hfm , gn i =

Z

1 0

xe2πinx

e−2πimx − 1 dx x

( Z 1  0, m 6= n, 2πi(m−n)x 2πinx dx = e −e = 1, m = n. 0

Hence F and G are biorthogonal.

(c) Let m 6= 0 be fixed. By part (b), we have gm ⊥ fn for all n 6= m. Consequently gm is orthogonal to every vector in the closed span of these fn , i.e.,
gm ⊥ span {fn }n6=m,n6=0 .

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If fm belonged to this closed span then we would have gm ⊥ fm . But we know from part (b) that hfm , gm i = 1, so this would be a contradiction. Therefore fm does not belong to the closed span of the remaining fn . As a consequence, fm does not belong to the finite linear span of the remaining fn . Thus fm is not equal to any finite linear combination of the remaining fn . This shows that F is finitely linearly independent. (d) This is essentially a special case of the Boas and Pollard result (Problem 8.3.30), but we will give a direct proof. Suppose that f ∈ L2 [0, 1] is such that f ⊥ fn for every n 6= 0. Let g(x) = xf (x), and note that g ∈ L2 [0, 1]. For every n 6= 0 we have hg, en i =

Z

1

xf (x) e

−2πinx

dx =

0

Z

0

1

f (x) fn (x) dx = hf, fn i = 0.

But since g belongs to L2 [0, 1] and {en }n∈Z is an orthonormal basis for L2 [0, 1], this implies that X g = hg, en i en = hg, e0 i e0 . n∈Z

As e = 1 is constant, this tells us that g is a constant. If we set c = hg, e0 i = R1 0 0 g, then we have xf (x) = g(x) = c. If c 6= 0 then this implies that f (x) =

c 6∈ L2 [0, 1], x

which is a contradiction. Therefore we must have c = 0, so g = 0 a.e. Consequently F is complete in L2 [0, 1] by Corollary 8.2.18. P (e) Suppose that the series f = n6=0 cn fn converges. Implicitly taking this sum to be ordered as stated in the problem statement, we then have X  X X hf, gm i = cn hfn , gm i = cn f n , g m = cn δmn = cm . n6=0

n6=0

n6=0

P Let sn denote the nth partial sum of the series f = n6=0 cn fn , with respect to the ordering specified in the problem statement. Then s2n = c1 f1 + c−1 f−1 + · · · + cn fn + c−n f−n and s2n−1 = c1 f1 + c−1 f−1 + · · · + cn fn .

Since the partial sums converge to f in L2 -norm, we have

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ks2n − s2n−1 k2 ≤ ks2n − f k2 + kf − s2n−1 k2 → 0 as n → ∞. But s2n − s2n−1 = c−n f−n = hf, g−n i f−n , so ks2n − s2n−1 k2 = |hf, g−n i kf−n k2 = 3−1/2 |hf, g−n i = 3−1/2 |c−n |. Therefore c−n → 0 as n → ∞. A similar argument shows that cn → 0 as n → ∞. (f) The constant function 1 belongs to the closed span because span(F ) = L2 [0, 1]. P If we could write 1 = n6=0 cn fn , then the partial sums of this series would converge to 1 in L2 -norm. Let sn denote the nth partial sum of the series. Then we have s2n = c1 f1 + c−1 f−1 + · · · + cn fn + c−n f−n and s2n−1 = c1 f1 + c−1 f−1 + · · · + cn fn .

Since the partial sums converge to 1 in L2 -norm, we have ks2n − s2n−1 k2 ≤ ks2n − 1k2 + k1 − s2n−1 k2 → 0 as n → ∞. But s2n − s2n−1 = c−n f−n = h1, g−n i f−n , so ks2n − s2n−1 k2 = |h1, g−n i kf−n k2 = 3−1/2 |h1, g−n i|. If we prove that |h1, g−n i| does not converge to zero, then we will have a contradiction. Using the substitution u = 2πnx, we have du/u = dx/x, so h1, g−n i =

Z

1

0

e2πinx − 1 dx x

Z 1 sin 2πnx cos 2πnx − 1 dx + i dx x x 0 0 Z 2πn Z 2πn cos u − 1 sin u = du + i du. u u 0 0 =

Z

1

By Problem 4.6.19, the second integral integral converges as n → ∞. Specifically, Z 2πn π sin u du = . lim n→∞ 0 u 2

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This limit exists even though sinu u is not integrable, because this function decreases in absolute value and alternates sign as u → ∞. Thus the behavior is analogous to that of an alternating series. Since the modulus of a complex number dominates the absolute value of its real or imaginary parts, it follows that Z 2πn π sin u du = . lim inf |h1, g−n i| ≥ lim inf n→∞ n→∞ u 2 0

Therefore |h1, g−n i| does not converge to zero, so we have obtained a contradiction. We obtain an even more striking fact by considering the real part of h1, g−n i. Since cos u − 1 ≤ 0, all x > 0, u there is no alternating behavior in the integral of this function. The same types of calculations that prove that Z ∞ sin u u du = ∞ 0

also show that

cos u−1 u

is not integrable on [0, ∞). That is, lim

n→∞

Z

0

2πn

cos u − 1 du = −∞. u

This also shows that |h1, g−n i| does not converge as n → ∞.

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Solutions to Exercises and Problems from Chapter 9 9.1.2 Let χ = χ[− 12 , 12 ] , and set W = χ ∗ χ. If x ∈ [0, 1], then W (x) = (χ ∗ χ)(x) = = =

Z

Z

Z

∞ −∞

χ(y) χ(x − y) dy

∞ −∞

χ[− 1 , 1 ] (y) χ[− 1 , 1 ] (x − y) dy 2 2

2 2

∞ −∞

χ[x− 1 , 1 ] (y) dy 2 2

  1 1 = = 1 − x. − x− 2 2 Similar calculations show that if x ∈ [−1, 0] then W (x) = x− 1, and if |x| > 1 then W (x) = 0. 9.1.4 (a) Given any fixed x ∈ Rd , H¨ older’s Inequality implies that the function f (·) g(x − ·) is integrable on Rd . Hence f ∗ g is well-defined at every point. Further, by making the change of variables z = x − y we see that Z Z f (y) g(x − y) dy = f (x − z) g(z) dz = (g ∗ f )(x). (f ∗ g)(x) = Rd

Rd

(b) Part (a) shows that f ∗ g exists at every point. H¨ older’s Inequality implies that f ∗ g is bounded, since Z |f (y) g(x − y)| dy |(f ∗ g)(x)| ≤ Rd

≤

Z

Rd

|f (y)|p dy

1/p Z

Rd

′

|g(x − y)|p dy

1/p′

= kf kp kgkp′ . Thus, kf ∗ gk∞ ≤ kf kp kgkp′ < ∞. (c), (d) Assume that p′ is finite, i.e., 1 < p ≤ ∞. To prove that f ∗ g is continuous, fix x ∈ Rd . Then given any h ∈ Rd , using the translation operator Th g(x) = g(x − h) we can write |(f ∗ g)(x) − (f ∗ g)(x − h)| Z   = f (y) g(x − y) dy − f (y) g(x − h − y) dy Rd

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Z

Rd

357

|f (y)| |g(x − y) − g(x − h − y)| dy

Z

Rd

= kf kp

|f (y)|p dy Z

1/p Z

Rd

′

|g(x − y) − g(x − h − y)|p dy p′

Rd

|g(t) − g(t − h)| dt

1/p′

1/p′

= kf kp kg − Th gkp′ → 0

as h → 0.

The convergence on the final line follows from Problem 7.3.16, which states ′ that translation is a strongly continuous family of operators on Lp (Rd ) when ′ ′ p is finite. This argument does not apply if p = ∞, but in this case we can simply interchange the roles of f and g, since f ∗ g = g ∗ f by part (a). Hence in any case we have shown that f ∗ g is continuous and bounded, so f ∗ g ∈ Cb (Rd ).

9.1.6 If f ∈ L1 (R) and g ∈ C0 (R) then g is bounded, so it follows from Exercise 9.1.4 that f ∗g is continuous and bounded. Therefore it only remains to prove that f ∗ g decays at infinity. First Proof. Assume first that g ∈ Cc (R). Then supp(g) ⊆ [−R, R] for some R > 0. Since f is integrable, it follows that |(f ∗ g)(x)| ≤

Z

x+R

x−R

≤ kgk∞

|f (y)| |g(x − y)| dy Z

x+R

x−R

|f (y)| dy → 0

as |x| → ∞.

This implies that f ∗ g ∈ C0 (R). Now we extend to arbitrary g by density. Choose any function g ∈ C0 (R). Since Cc (R) is dense in C0 (R), we can find functions gn ∈ Cc (R) such that gn → g uniformly. By our previous case we know that f ∗ gn ∈ C0 (R) for every n. Also, kf ∗ g − f ∗ gn k∞ = kf ∗ (g − gn )k∞ ≤ kf k1 kg − gn k∞ → 0

as n → ∞,

so f ∗ gn → f ∗ g uniformly. Since f ∗ gn ∈ C0 (R) for every n and since C0 (R) is a Banach space with respect to the uniform norm, it follows that f ∗ g ∈ C0 (R).

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Second Proof. Fix ε > 0. Since g ∈ C0 (R), there exists some R > 0 such that |g(x)| ≤ ε for all |x| > R. Since f is integrable, there is some M such that Z |f (x)| dx < ε. |x|>M

If y > x + R then x − y < −R and hence |g(x − y)| < ε. Similarly, if y < x − R then x − y > R and therefore |g(x − y)| < ε. Hence if |x| > M + R then we compute that |(f ∗ g)(x)| ≤

Z

x+R

x−R

≤ kgk∞

|f (y)| |g(x − y)| dy + Z

Z

[x−R,x+R]C

x+R

x−R

|f (y)| |g(x − y)| dy + ε

Z

|f (y)| |g(x − y)| dy

[x−R,x+R]C

|f (y)| dy

≤ ε kgk∞ + ε kf k1. This shows that (f ∗ g)(x) → 0 as |x| → ∞.

9.1.7 (a) Case 1: Real-Valued Functions. Suppose that f ∈ L1 (R) and g ∈ Cb1 (R) are real-valued (we can allow f to be extended real-valued). Then g ∈ L∞ (R), so f ∗ g is continuous by Exercise 9.1.4. Let x be fixed. Given h ∈ R, we have Z ∞ (f ∗ g)(x + h) − (f ∗ g)(x) g(x + h − y) − g(x − y) f (y) = dy. h h −∞ The integrand converges pointwise a.e. to f (y) g ′ (x − y) as h → 0. Since g ′ is continuous, the Mean Value Theorem implies that there exists a point c (depending on h and y) such that g(x + h − y) − g(x − y) = g ′ (c). h Recalling that x is fixed, our functions depend on y, and h is an index that will be converging to 0, we have f (y) g(x + h − y) − g(x − y) = |f (y) g ′ (c)| ≤ |f (y)| kg ′ k∞ ∈ L1 (R). h

Since the dominating function |f (y)| kg ′ k∞ is integrable and does not depend on h, we can apply the Lebesgue Dominated Convergence Theorem. We find that

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(f ∗ g)(x + h) − (f ∗ g)(x) h→0 h Z ∞ g(x + h − y) − g(x − y) = lim dy f (y) h→0 −∞ h Z ∞ = f (y) g ′ (x − y) dy

(f ∗ g)′ (x) = lim

−∞

= (f ∗ g ′ )(x).

Thus f ∗ g is differentiable. Furthermore, (f ∗ g)′ = f ∗ g ′ is continuous since f ∈ L1 (R) and g ′ ∈ L∞ (R).

(b) Case 2: Complex-Valued Functions. Suppose that f ∈ L1 (R) and g ∈ are complex-valued. Then g ∈ L∞ (R), so f ∗g is continuous by Exercise 9.1.4. Let x be fixed. Given h ∈ R, we have Z ∞ g(x + h − y) − g(x − y) (f ∗ g)(x + h) − (f ∗ g)(x) f (y) = dy. h h −∞ Cb1 (R)

The integrand converges pointwise a.e. to f (y) g ′ (x − y) as h → 0. We must show that the integrand is bounded by an integrable function. Write g = gr + igi , where gr and gi are real-valued. Since g ′ is continuous and bounded, so are gr′ and gi′ . Set C = kgr′ k∞ + kgi′ k∞ . Since gr and gi are differentiable and real-valued, the Mean Value Theorem implies that there exist points ξ and η (which depend on h and y) such that gr (x + h − y) − gr (x − y) = gr′ (ξ) h and

gi (x + h − y) − gi (x − y) = gi′ (η). h Recall that x is fixed. Therefore the independent variable in our functions is y, while h is an index that will be converging to 0. As a function of y, we see that f (y) g(x + h − y) − g(x − y) h gr (x + h − y) − gr (x − y) gi (x + h − y) − gi (x − y) = f (y) + f (y) h h = |f (y) gr′ (ξ) + f (y) gi′ (η)|

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  ≤ |f (y)| kgr′ k∞ + kgi′ k∞

= C |f (y| ∈ L1 (R).

Since the dominating function C |f (y)| is integrable and does not depend on h, we can apply the Dominated Convergence Theorem. We find that (f ∗ g)(x + h) − (f ∗ g)(x) h→0 h Z ∞ g(x + h − y) − g(x − y) = lim f (y) dy h→0 −∞ h Z ∞ = f (y) g ′ (x − y) dy

(f ∗ g)′ (x) = lim

−∞

= (f ∗ g ′ )(x).

Thus f ∗ g is differentiable. Furthermore, (f ∗ g)′ = f ∗ g ′ is continuous since f ∈ L1 (R) and g ′ ∈ L∞ (R). (b) We proceed by induction. The base step n = 1 is established in part (a). Inductive step. Assume that the result holds for some n ≥ 1. Suppose that f ∈ L1 (R) and g ∈ Cbn+1 (R). By the inductive hypothesis, (f ∗g)(n) = f ∗g (n) . Since f is integrable and g (n) is bounded, f ∗ g (n) is continuous. Also, g (n) ∈ Cb1 (R). Therefore, by the case n = 1, f ∗ g (n) is differentiable. Further, by the case n = 1 and by the inductive hypothesis,
′ f ∗ g (n+1) = f ∗ (g (n) )′ = (f ∗ g (n) )′ = (f ∗ g)(n) = (f ∗ g)n+1 .

Hence (f ∗ g)(n+1) = f ∗ g (n+1) . Further, this function is continuous since f ∈ L1 (R) and g (n+1) ∈ L∞ (R). (c) This follows from part (c).

9.1.10 The Fej´er function w is integrable because it is bounded and decays like 1/x2 . To compute its integral we make a change of variables, use the half-angle formula sin2 x = (1 − cos 2x)/2, and integrate by parts using u = 1 − cos 2x,

dv =

1 dx, 2x2

v =−

du = 2 sin 2x dx, This gives us Z

πR

−πR



sin πx πx

2

1 dx = π

Z

R

−R

sin2 x dx x2

1 . 2x

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= = Since

cos 2x−1 x

1 π

Z

R

−R

1 dx 2x2 Z R sin 2x + dx. −R πx

(1 − cos 2x)

cos 2x − 1 R πx −R

belongs to C0 (R), we therefore have Z

∞

−∞



sin πx πx

2

dx = lim

R→∞

= lim

R→∞

= lim

R→∞

Z

R

−R

Z



R

−R

Z

sin πx πx

2

dx

sin 2x dx πx

2R

−2R

sin x dx πx

= 1. The computation of the improper Riemann integral at the final step is given in Problem 4.6.19. 9.1.13 If f and g are nonnegative, then f (x − y) g(y) is a measurable, nonnegative function on R2 , so Z ∞ Z ∞ f (x − y) g(y) dy dx −∞

−∞

exists. Tonelli’s Theorem implies that Z ∞ f (x − y) g(y) dy (f ∗ g)(x) = −∞

is defined for a.e. x and is a measurable function of x. Using H¨ older’s Inequality, Z ∞ |f (y) g(x − y)| dy |(f ∗ g)(x)| ≤ −∞

=

Z

∞

−∞

=

Z

  ′ f (y) g(x − y) 1/p g(x − y) 1/p dy

∞

−∞

p

|f (y)| |g(x − y)| dy

1/p Z

∞

−∞

|g(x − y)| dy

1/p′

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=

Z

∞

−∞

|f (y)| |g(x − y)| dy

1/p′

= kgk1

p

Z

∞

−∞

1/p Z

∞

−∞

|f (y)|p |g(x − y)| dy

|g(y)| dy

1/p

1/p′

.

Using the inequality from above and interchanging integrals by Tonelli’s Theorem, we compute that Z ∞ |(f ∗ g)(x)|p dx kf ∗ gkpp = −∞

≤

p/p′ kgk1

=

p/p′ kgk1 p/p′

= kgk1

∞

−∞

Z

∞

−∞

Z

∞

−∞

′

p/p

= kgk1

Z

p/p

= kgk1

1+ pp′

Z

Z

∞

−∞

Z

|f (y)|p |g(x − y)| dy dx

∞

−∞

|f (y)|p |g(x − y)| dx dy

|f (y)|p

∞

−∞

′

= kgk1

Z

∞

−∞

|f (y)|p

Z

∞

−∞

Z

∞

−∞

 |g(x − y)| dx dy  |g(x)| dx dy

|f (y)|p kgk1 dy

kf kpp

= kgkp1 kf kpp < ∞. The result therefore follows upon taking pth roots. Note that we used the fact that p p(p − 1) 1+ ′ = 1+ = 1 + p − 1 = p. p p For the general case, we write f = (f1 − f2 ) + i(f3 − f4 ) and g = (g1 − g2 ) + i(g3 − g4 ) with fi and gi nonnegative. Then f ∗ g is a finite linear combination of fi ∗ gj , so is measurable and belongs to Lp (Rd ). Repeating then exactly the same calculations as above we see that kf ∗ gkp ≤ kf kp kgk1 . 9.1.16 (a) The proof is similar to that of Exercise 9.1.7. For simplicity of presentation, we assume that f and g are real-valued. The modifications for complex-valued functions are similar to those laid out in the solution to Exercise 9.1.7. Base step m = 1. Suppose that f ∈ Lp (R) with p finite and g ∈ Cc1 (R). Then supp(g) ⊆ [−R, R] for some R > 0, and f ∗ g is continuous by Exercise 9.1.4. We have that

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(f ∗ g)(x + h) − (f ∗ g)(x) = h

Z

f (y)

g(x + h − y) − g(x − y) dy. h

The integrand converges pointwise a.e. to f (y) g ′ (x − y) as h → 0. Further, g ′ is bounded since it is continuous and compactly supported. Therefore, by the Mean Value Theorem, given x, y, and h there exists a point c such that g(x + h − y) − g(x − y) = |g ′ (c)| ≤ kg ′ k∞ . h If |h| < 1, x − y ∈ [−R, R], and x + h − y ∈ [−R, R] then x−R ≤ y ≤ x+R

and

x + h − R ≤ y ≤ x + h + R.

Hence, if y does not belong to the interval Ix = [x − R − 1, x + R + 1], then for |h| < 1 we have g(x + h − y) − g(x − y) = 0. Therefore, if x ∈ R and |h| < 1 then (as a function of y), f (y) g(x + h − y) − g(x − y) ≤ |f (y)| χIx (y) kg ′ k∞ ∈ L1 (R). h

In more detail, since Ix is compact and f ∈ Lp (R) we have f χIx ∈ Lp (Ix ) ⊆ L1 (Ix ),

which implies that f χIx is integrable on all of R since it is zero outside of Ix . The Lebesgue Dominated Convergence Theorem therefore applies, and we find that (f ∗ g)(x + h) − (f ∗ g)(x) h Z g(x + h − y) − g(x − y) = lim f (y) dy h→0 h Z = f (y) g ′ (x − y) dy = (f ∗ g ′ )(x).

(f ∗ g)′ (x) = lim

h→0

Thus f ∗ g is differentiable. It remains to show that f ∗ g ∈ C01 (R). There are two cases. First, if 1 < p < ∞, then f ∗ g ∈ C0 (R) because f ∈ Lp (R)

and

′

g ∈ Cc (R) ⊆ Lp (R),

so we can apply Theorem 9.1.5. Similarly, (f ∗ g)′ = f ∗ g ′ ∈ C0 (R) because f ∈ Lp (R)

and

′

g ′ ∈ Cc (R) ⊆ Lp (R).

Since both f ∗ g and (f ∗ g)′ belong to C0 (R), we conclude that f ∗ g ∈ C01 (R).

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Second, if p = 1, then f ∗ g ∈ C0 (R) because f ∈ L1 (R)

and

g ∈ Cc (R) ⊆ C0 (R),

so we can apply Exercise 9.1.6. Similarly, (f ∗ g)′ = f ∗ g ′ ∈ C0 (R) because f ∈ Lp (R)

and

g ′ ∈ Cc (R) ⊆ C0 (R).

The combination of these two facts tells us that f ∗ g ∈ C01 (R). Inductive step. Assume that the result holds for some m ≥ 1. Suppose that f ∈ Lp (R) and g ∈ Ccm+1 (R). By the inductive hypothesis, we have f ∗ g ∈ C0m (R). Also, g (m) ∈ Ccm (R), so by the case m = 1 we have that f ∗ g (m) is differentiable, and that f ∗ g (m+1) = f ∗ (g (m) )′ = (f ∗ g (m) )′ ∈ C0 (R). However, we also have by the inductive hypothesis that
′ f ∗ g (m+1) = (f ∗ g (m) )′ = (f ∗ g)(m) = (f ∗ g)m+1 .

Hence (f ∗ g)m+1 ∈ C0 (R), so in the end we have f ∗ g ∈ C0m+1 (R). Modifications for p = ∞. In this case, we cannot conclude that f ∗g belongs to C0m (R), but instead only obtain f ∗ g ∈ Cbm (R). For example, if f (x) = 1 and g ∈ Ccm (R) then Z Z (f ∗ g)(x) = f (x − y) g(y) dy = g(y) dy, so f ∗ g is constant. (b) The proof is similar to the R proof of Theorem 9.1.12. Let k be any function in k ∈ Cc∞ (R) such that k = 1. If we define kN (x) = N k(N x), then {kN }N ∈N is an approximate identity, and furthermore we have kkN k1 = kkk1 for every N. Given f ∈ Lp (R), define fN = (f χ[−N,N ] ) ∗ kN ,

N ∈ N.

Since f χ[−N,N ] and kN each have compact support, so does fN . Also note that f χ[−N,N ] ∈ L1 (R), so Exercise 9.1.7 implies that fN is infinitely differentiable. Thus fN ∈ Cc∞ (R). The Dominated Convergence Theorem implies that f χ[−N,N ] → f in Lp norm. Also, Theorem 9.1.15 tells us that f ∗ kN → g in Lp -norm. Using the Triangle Inequality and Young’s Inequality, we compute that

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kf − fN kp ≤ kf − f ∗ kN kp + kf ∗ kN − (f χ[−N,N ]) ∗ kN kp = kf − f ∗ kN kp + k(f − f χ[−N,N ] ) ∗ kN kp ≤ kf − f ∗ kN kp + kf − f χ[−N,N ] kp kkN k1 = kf − f ∗ kN kp + kf − f χ[−N,N ] kp kkk1 → 0

as N → ∞.

Therefore Cc∞ (R) is dense in Lp (R). 9.1.17 (a) Fix ε > 0. Since f is uniformly continuous, there exists a δ > 0 such that |t| < δ =⇒ kf − Tt f k∞ < ε. By definition of approximate identity, there exists an N0 such that Z |kN (t)| dt < ε. N > N0 =⇒ |t|≥δ

Therefore, if N > N0 then for every x ∈ R we have |f (x) − (f ∗ kN )(x)| Z ∞ Z = f (x) kN (t) dt − ≤

≤

Z

≤ ε

−∞

−∞ ∞

−∞

Z

∞

|f (x) − f (x − t)| |kN (t)| dt

|t| N0 we have kf − f ∗ kN k∞ ≤ εK + 2kf k∞ ε. This says that f ∗ kN → f uniformly.

(b) If we assume that f ∈ C0 (R) then we have f ∈ L∞ (R). The proof of Exercise 9.1.16 can be followed without change except that we only obtain f ∗ g ∈ Cbm (R). However, we do have (f ∗ g)(k) = f ∗ g (k) for each k = 0, 1, . . . , m. On the other hand, we can apply Exercise 9.1.6. That exercise tells us that since f ∈ C0 (R) g ∈ Cc (R) ⊆ L1 (R), we have f ∗ g ∈ C0 (R). But we also

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have f ∈ C0 (R) and g ′ ∈ Cc (R) ⊆ L1 (R), so (f ∗ g)′ = f ∗ g ′ ∈ C0 (R), and so forth. This tells us that f ∗ g ∈ C0m (R). (c) The proof is similar to that of Exercise 9.1.16(b). In this part all of our functions will be continuous, so the uniform norm and the L∞ -norm will coincide. R Let k be any function in k ∈ Cc∞ (R) such that k = 1. If we define kN (x) = N k(N x), then {kN }N ∈N is an approximate identity, and furthermore we have kkN k1 = kkk1 for every N. Given f ∈ C0 (R), define fN = (f χ[−N,N ] ) ∗ kN ,

N ∈ N.

Since f χ[−N,N ] and kN each have compact support, so does fN . Also note that f χ[−N,N ] ∈ L1 (R), so Exercise 9.1.7 implies that fN is infinitely differentiable. Thus fN ∈ Cc∞ (R). The fact that f vanishes at infinity implies that f χ[−N,N ] → f uniformly. Also, Theorem 9.1.15 tells us that f ∗ kN → g uniformly. Using the Triangle Inequality and Young’s Inequality, we compute that kf − fN k∞ ≤ kf − f ∗ kN k∞ + kf ∗ kN − (f χ[−N,N ]) ∗ kN k∞ = kf − f ∗ kN k∞ + k(f − f χ[−N,N ] ) ∗ kN k∞ ≤ kf − f ∗ kN k∞ + kf − f χ[−N,N ] k∞ kkN k1 = kf − f ∗ kN k∞ + kf − f χ[−N,N ] k∞ kkk1 → 0

as N → ∞.

Therefore Cc∞ (R) is dense in C0 (R). 9.1.18 (a) This is similar to the proof of Young’s Inequality given in the solution to Exercise 9.1.13. Alternatively, we can adapt Minkowski’s Inequality. For each k ∈ Z let cj be the sequence whose components are cjk = |aj bk−j |,

k ∈ Z.

Since the components of cj are nonnegative, we can form (in the extended real sense) the series X cj , j∈Z

and we have

X j

c

j∈Z

p

=

N

X j

c lim

N →∞ j=−N

p

≤

lim

N →∞

N X

j=−N

kcj kp =

X j∈Z

kcj kp .

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Therefore, considering the case 1 < p < ∞, we compute that ka ∗ bkp =

X k∈Z

|(a ∗ b)k |p

1/p

p 1/p X X = aj bk−j k∈Z j∈Z

X X p 1/p cjk ≤ k∈Z j∈Z



X j

= c

p

j∈Z

≤ =

X

kcj kp

j∈Z

X

|aj |

j∈Z

X k∈Z

p

|bk−j |

1/p′

= kak1 kbkp . The cases p = 1 or p = ∞ are straightforward.

(b) If c = (ck )k∈Z ∈ ℓp (Z), then X (c ∗ δ)(n) = c(n − k) δ(k) = c(n), k∈Z

so c ∗ δ = c. 9.1.19 If f, g ≥ 0, then we can apply Tonelli’s theorem to compute that Z ∞ |(f ∗ g)(x)| dx kf ∗ gk1 = −∞

Z ∞ dx f (x − y) g(y) dy −∞ −∞ Z ∞ Z ∞ = f (x − y) g(y) dy dx

=

Z

=

Z

∞

−∞ ∞

−∞

=

Z

∞

−∞

−∞

Z

∞

−∞

Z

f (x − y) g(y) dx dy

∞ −∞

 f (x − y) dx g(y) dy

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Z

=

∞

−∞

kf k1 g(y) dy

= kf k1 kgk1 . 2

To prove that strict inequality can hold, consider h(x) = xe−x . This function is odd and is nonnegative on [0, ∞), so khk1 = 2

Z

∞

2

xe−x dx = 2

0

2 ∞ −e−x 0+1 = 2 = 1. 2 0 2 2

By Problem 4.6.25, we have (h ∗ h)(x) = (π/2)1/2 (x2 − 1) e−x /2 /4. This function is even and is negative on (0, 1) and positive on (1, ∞). Hence (integration via Mathematica), kh ∗ hk1

 1/2 Z 1  Z ∞ 1 π 2 −x2 /2 2 −x2 /2 = 2 (x − 1) e dx (1 − x ) e dx + 4 2 1 0  1/2 π = ≈ 0.76+. 2e

Therefore kh ∗ hk1 < khk21 . 9.1.20 We use Minkowski’s Integral Inequality to compute that kf ∗ gkp =

Z

Z

=

∞

−∞ ∞

−∞

≤

=

Z

∞

−∞

Z

Z

∞

−∞

p 1/p f (y) g(x − y) dy dx

f (y) g(· − y) dy

p

|f (y)| kTy gkp dy

∞

−∞

|f (y)| kgkp dy

= kf k1 kgkp . 9.1.21 (a), (b) Assume first that 1 < p, q, r < ∞. Note that p1 − 1r = 1 − 1q and q1 − 1r = 1 − p1 , so if we define p1 and p2 by 1 1 1 = − , p1 p r then 1 < p1 , p2 < ∞. Also

1 1 1 = − , p2 q r

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p p q q + = 1 = + r p1 r p2

1 1 1 + + = 1. r p1 p2

and

Therefore, by H¨ older’s Inequality, |(f ∗ g)(x)| Z ≤ |f (y) g(x − y)| dy = ≤

Z 

     |f (y)|p/r |g(x − y)|q/r · |f (y)|p/p1 · |g(x − y)|q/p2 dy

Z 

|f (y)|p/r |g(x − y)|q/r ×

=

Z

=

|g(x − y)|q/p2

|f (y)|p |g(x − y)|q dy ×

Z

Z  Z

q

|f (y)| |g(x − y)| dy

p2

dy

1/r Z 

dy

1/p2

1/r Z

|g(x − y)|q dy

p

r

1/p2

1/r

|f (x)|p/p1

|f (y)|p dy

p1

dy

1/p1

1/p1

2 1 . kgkq/q kf kp/p q p

Hence (interchanging integrals by Tonelli’s Theorem), Z kf ∗ gkrr = |(f ∗ g)(x)|r dx ≤ = =

Z Z

Z

Z

 2 1 kgkrq/q |f (y)| |g(x − y)| dy dx kf krp/p q p p

p

|f (y)|

Z

q



2 1 kgkrq/q |g(x − y)| dx dy kf krp/p q p

q

 1 2 |f (y)|p kgkqq dy kf krp/p kgkrq/q p q

1 2 = kf kpp kgkqq kf krp/p kgkrq/q = kf krp kgkrq . p q

Taking rth roots therefore finishes the proof. (c) If one of p, q, or r is 1 or ∞, then one or more of p, q, r, p1 , p2 may be ∞. The proof is the same in principle, just with changes at appropriate points. 9.1.22 We argue very similar to the proof of Theorem 9.1.12.

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p q ∞ R Fix f ∈ L (R) ∩ L (R). Let k ∈ Cc (R) be any function that satisfies k = 1, and let gn = (f · χ[−n,n] ) ∗ kn .

Because f · χ[−n,n] is integrable and kn is infinitely differentiable, gn is also infinitely differentiable. As f · χ[−n,n] is zero a.e. outside of [−n, n] and kn is identically zero outside of some interval [a, b], a direct calculation shows that their convolution, which is gn , is identically zero outside of the interval [−n + a, n + b]. Therefore gn is compactly supported, so it belongs to Cc∞ (R). Now, Theorem 9.1.15 tells us that f ∗ kn → f in Lp -norm and in Lq -norm. Further, the Dominated Convergence Theorem implies that f · χ[−n,n] → f in Lp -norm and Lq -norm. Letting r denote either p or q, it follows that kf − gn kr ≤ kf − f ∗ kn kr + kf ∗ kn − (f · χ[−n,n] ) ∗ kn kr = kf − f ∗ kn kr + k(f − f · χ[−n,n] ) ∗ kn kr ≤ kf − f ∗ kn kr + kf − f · χ[−n,n] kr kkn k1 = kf − f ∗ kn k1 + kf − f · χ[−n,n] kr kkk1 → 0

as n → ∞.

9.1.23 Suppose f is continuous at x, and fix ε > 0. Then there exists some δ > 0 such that |t| < δ =⇒ |f (x) − f (x − t)| < ε. By definition of approximate identity, there exists some N0 > 0 such that Z |kN (t)| dt < ε. N > N0 =⇒ |t|>δ

Then for N > N0 we have |f (x) − (f ∗ kN )(x)| Z Z = f (x) kN (t) dt − f (x − t) kN (t) dt dx ≤

Z

≤ ε

|t| 0, then γ(x) = e−1/x , which is strictly positive. (b) For x > 0, γ ′ (x) = x−2 e−1/x =

p1 (x) −1/x e , x2

where p1 (t) = 1 is a polynomial of degree 0. Suppose that pn (x) γ (n) (x) = γ(x), x2 n where pn is a polynomial of degree n − 1. Then for x > 0,   x2n p′n (x) − 2nx2n−1 pn (x) −1/x pn (x) −1/x 1 e + e γ (n+1) (x) = x2n x2 x4n   pn (x) p′n (x) 2npn (x) = e−1/x + 2n − 2n+1 x2n+2 x x =

pn (x) + x2 p′n (x) − 2nxpn (x) −1/x e x2n+2

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x2 p′n (x) + (1 − 2nx) pn (x) −1/x e x2n+2 pn+1 (x) −1/x e , = x2n+2

=

where pn+1 (x) = x2 p′n (x) + (1 − 2nx) pn (x).

Since p′n has degree n − 2 and pn has degree n − 1, it follows that pn+1 has degree n. (c) Since e−t converges to zero faster than t2n grows, pn (x) −1/x e x2n    e−1/x = lim pn (x) lim x→0+ x→0+ x2n

lim γ (n) (x) = lim

x→0+

x→0+

= pn (0) lim t2n e−t t→∞

= 0. Therefore γ (n) is differentiable at x = 0, and γ (n) (0) = 0. (b), (c) Alternative Solution. We will show that for each n ∈ N there is a polynomial pn of degree 2n such that −1

γ (n) (x) = pn (x−1 ) e−x ,

x > 0.

For x > 0, γ ′ (x) = x−2 e−x

−1

−1

= p1 (x−1 ) e−x ,

where p1 (t) = t2 is a polynomial of degree 2. −1 Suppose that γ (n) (x) = pn (x−1 ) e−x for x > 0, where pn is a polynomial of degree 2n. Then for x > 0, γ (n+1) (x) = pn (x−1 ) e−x

−1

x−2 + p′n (x−1 ) x−2 e−x

−1

−1

= pn+1 (x−1 ) e−x ,

where pn+1 (t) = t2 pn (t) + t2 p′n (t). Since t2 pn (t) has degree 2n + 2 while t2 p′n (t) has degree 2n + 1, we conclude that their sum has degree 2n + 2. Finally, since pn is a polynomial, lim+ γ (n) (x) = lim+ pn (x−1 ) e−x

x→0

x→0

−1

= lim pn (t) e−t = 0. t→∞

Therefore γ (n) is differentiable at x = 0, and γ (n) (0) = 0.

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(d) The function 1 − x2 is infinitely differentiable. Since γ is infinitely differentiable, the composition β(x) = γ(1 − x2 ) is infinitely differentiable as well. Note that γ(y) > 0 for all y > 0. If |x| < 1, then 1 − x2 > 0 and therefore β(x) = γ(1 − x2 ) > 0. On the other hand, if |x| ≥ 1 then 1 − x2 ≤ 0, so β(x) = γ(1 − x2 ) = 0.

9.1.27 Let χ = χ[−2,2] . Since χ and k are both integrable and compactly supported, it follows that θ = χ ∗ k is integrable and compactly supported. Also, Exercise 9.1.7 implies that θ is infinitely differentiable. Since χ and k are both nonnegative, it follows that θ = χ ∗k is nonnegative. By hypothesis, k is supported within [−1, 1]. Also, χ is only nonzero on [−2, 2], so it follows that θ(x) = (χ ∗ k)(x) =

Z

∞

−∞

χ(y) k(x − y) dy =

Z

2

−2

k(x − y) dy

In order for this to be nonzero, there must be some y ∈ [−2, 2] such that x − y ∈ [−1, 1]. That is, −2 ≤ y ≤ 2 and −1 ≤ x − y ≤ 1, which implies that −3 ≤ −1 + y ≤ x ≤ 1 + y ≤ 3. Thus θ(x) can only be nonzero when |x| ≤ 3, which tells us that θ(x) = 0 for all |x| > 3. On the other hand, suppose that x ∈ [−1, 1]. Then k(x − y) = 0 for all |y| > 2. Consequently, Z ∞ χ(y) k(x − y) dy χ θ(x) = ( ∗ k)(x) = −∞

=

Z

2

−2

=

Z

k(x − y) dy

∞

−∞

=

Z

k(x − y) dy

∞

k(y) dy = 1.

−∞

Thus, θ = 1 on the interval [−1, 1]. 9.1.28 (a) The Chain Rule implies that θn′ (x) =

d θ dx

Therefore for every n ∈ N we have

x n




=

1 ′ θ n

x n


.

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1 ′ kθ k∞ ≤ kθ′ k∞ . n

kθn′ k∞ ≤

(b) Recall that 0 ≤ θ ≤ 1 everywhere, θ = 1 on [−1, 1], and θ = 0 outside [−3, 3]. Therefore θn = 1 on [−n, n], and θn = 0 outside of [−3n, 3n]. Consequently, f ′ θn → f ′ pointwise, and |f ′ θn | ≤ |f ′ | ∈ L1 (R).

Therefore, the Dominated Convergence Theorem implies that f ′ θn → f ′ in L1 -norm. It follows from part (a) that 1 ′ kθ k∞ → 0. n

kθn′ k∞ ≤ Therefore f θn′ → 0 pointwise. Also,

|f θn′ | ≤ |f | kθn′ k∞ ≤ |f | kθ′ k∞ = C|f | ∈ L1 (R), so we can apply the Dominated Convergence Theorem and conclude that f θn′ → 0 in L1 -norm.

(c) Since f is differentiable everywhere on [−3n, 3n] and f ′ ∈ L1 [−3n, 3n], it follows from Corollary 6.3.3 that f ∈ AC[−3n, 3n]. Also, θn ∈ AC[−3n, 3n] since θn ∈ C 1 [−3n, 3n]. Therefore we can apply integration by parts (Problem 6.4.9) to obtain Z

∞

−∞

f θn′ =

Z

3n

−3n

f θn′

= f (3n) θ(3n) − f (−3n) θ(−3n) − = 0 − 0 −

Z

∞

n→∞

−∞

3n

f ′ θn −3n

f ′ θn .

−∞

Combining the preceding parts, we see that Z ∞ Z ∞ Z f ′ = lim f ′ θn = − lim −∞

Z

n→∞

∞

−∞

f θn′ = 0.

9.1.29 Since K is compact and R\U is closed, the distance d between these sets is strictly positive. Set θ = k ∗ χV . Since k and χV are both compactly supported, their convolution is also compactly supported, and hence it follows from Exercise 9.1.7 that θ ∈ Cc∞ (R). Since χV and k are both nonnegative, it follows that θ = χ ∗ k is nonnegative. Further,

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θ(x) =

Z

V

Z

k(x − y) dy ≤

∞

k = 1,

−∞

so 0 ≤ θ ≤ 1 everywhere. Also, if x ∈ K and y ∈ / V then |x − y| ≥ k(x − y) = 0. Therefore for x ∈ K we have Z Z ∞ θ(x) = k(x − y) dy = k(x − y) dy = 1. V

d 3

and so

−∞

Similarly if x ∈ / U then it follows that θ(x) = 0. ′

9.1.30 Since f ∈ Lp (R) and g ∈ Lp (R), Exercise 9.1.4 implies that f ∗ g ∈ Cb (R). Likewise, since ∂p f ∈ Lp (R) we have ∂p f ∗ g ∈ Cb (R). Given any x ∈ R, (f ∗ g)(x + a) − (f ∗ g)(x) − (∂p f ∗ g)(x) a T−a (f ∗ g)(x) − (f ∗ g)(x) − (∂p f ∗ g)(x) = a (T−a f ∗ g)(x) − (f ∗ g)(x) = − (∂p f ∗ g)(x) a   f − T−a f − ∂p f ∗ g(x) = −a

f − T−a f

≤ − ∂p f (by Young’s Inequality)

kgkp′ −a p → 0 as a → 0.

Hence f ∗ g is differentiable at x, and (f ∗ g)′ (x) = (∂p f ∗ g)(x).

9.1.31 Suppose m = 0. Since f ∈ L1 (R) and g ∈ L∞ (R), we know from Exercise 9.1.4 that f ∗ g ∈ Cb (R). We will prove the result for m > 0 by induction. Base Step m = 1. Assume that f ∈ Cc1 (R) and g ∈ L∞ (R) are given. Since f is compactly supported, we can choose R > 0 large enough that supp(f ) ⊆ [−R, R]. Let h = g χ[−2R,2R] . Then h is integrable and f belongs to Cc1 (R), so Exercise 9.1.7 implies that f ∗ h ∈ Cb1 (R) and (f ∗ h)′ = f ′ ∗ h. Fix any x with |x| < R. Note that |y| > 2R

=⇒

|x − y| > R

=⇒

f ′ (x − y) = 0.

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Therefore (f ∗ h)(x) = =

Z

∞

−∞

Z

2R

−2R

=

Z

f (x − y) h(y) dy f (x − y) g(y) dy

∞

−∞

f (x − y) g(y) dy = (f ∗ g)(x).

Thus f ∗ g = f ∗ h on (−R, R), and a similar calculation shows that f ′ ∗ g = f ′ ∗ h on (−R, R). Therefore (f ∗ g)′ (x) = (f ∗ h)′ (x) = (f ′ ∗ h)(x) = (f ′ ∗ g)(x),

|x| < R.

Thus f ∗ g is differentiable on (−R, R) and its derivative on this interval is f ′ ∗ g. Since R is arbitrary, we conclude that f ∗ g is differentiable at every point and (f ∗ g)′ = f ′ ∗ g everywhere. Finally, since f ′ ∈ Cc (R) ⊆ L1 (R) and g ∈ L∞ (R), Exercise 9.1.4 implies that (f ∗ g)′ = f ′ ∗ g ∈ Cb (R). Inductive Step. Assume that the result holds for some m ∈ N. Suppose that f ∈ Ccm+1 (R) and g ∈ L∞ (R) are given. Then f ∈ Ccm (R), so the inductive hypothesis implies that f ∗ g ∈ Cbm (R) and (f ∗ g)(k) = f (k) ∗ g for k = 1, . . . , m. Further, since f m ∈ Cc1 (R), the base step implies that f (m) ∗ g ∈ Cb1 (R) and (f (m) ∗ g)′ = f (m+1) ∗ g. Therefore (f ∗ g)(m+1) =

(f ∗ g)(m)


′

= (f (m) ∗ g)′ = f (m+1) ∗ g.

Finally, since f (m) ∗ g ∈ Cb1 (R), we have (f (m) ∗ g)′ ∈ Cb (R), so we end with (f ∗ g)(m+1) ∈ Cb (R),

and therefore f ∗ g ∈ Cbm+1 (R). This completes the induction.

9.1.32 First proof, covering 1 ≤ p < ∞. The convolution-based solution to Problem 7.4.5 for the case p = 1 can be extended to finite p as follows. R We are given f ∈ Lp (R) such that f φ = 0 for every φ ∈ Cc∞ (R). Let R k ∈ Cc∞ (R) be such that k = 1, and set kN (x) = N k(N x). If we fix t ∈ R,

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then kN (t − x) ∈ Cc∞ (R), so by hypothesis we have Z ∞ (f ∗ kN )(t) = f (x) kN (t − x) dx = 0. −∞

But f ∗ kN → f in Lp -norm, so this implies that f = 0 a.e.

Second proof, covering 1 < p ≤ ∞. The solution to Problem 7.4.5 can essentially be repeated for this range of p. For completeness, we give the details below. R We are given f ∈ Lp (R) such that f φ = 0 for every φ ∈ Cc∞ (R). Sup′ pose that g is any function in Lp (R). Since 1 ≤ p′ < ∞ (this is why we ′ assumed p > 1), we know that Cc∞ (R) is dense in Lp (R). Therefore there exist functions φk ∈ Cc∞ (R) such that kg − φk kp′ → 0 as k → ∞. Applying the hypotheses and H¨ older’s Inequality, we compute that Z ∞ Z ∞ Z ∞ 0 ≤ f g ≤ f φk + f (g − φk ) −∞

−∞

−∞

≤ 0 + kf kp kg − φk kp′

→ 0

as k → ∞.

R ′ Therefore f g = 0 for every g ∈ Lp (R). Applying the Converse to H¨ older’s Inequality, we conclude that Z ∞ kf kp = sup f g = 0. kgkp′ =1

−∞

Therefore f = 0 a.e.

9.1.33 We are given a function f ∈ L∞ (R) such that kTa f − f k∞ → 0

as a → 0.

Let kn = 2nχ[− n1 , n1 ] . Then {kn }n∈N is an approximate identity. In particular, kkn k1 = 1. Set gn = f ∗ kn . Since kn belongs to L1 (R) and f ∈ L∞ (R), we know that gn is uniformly continuous by Exercise 9.1.4. We can also see this directly, because the uniform norm of Ta gn − gn satisfies kTa gn − gn ku = kTa (f ∗ kn ) − (f ∗ kn )ku = kTa f ∗ kn − f ∗ kn ku

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≤ k(Ta f − f ) ∗ kn k∞ ≤ kTa f − f k∞ kkn k1 → 0 as a → 0. We claim that the family {gn }n∈N is pointwise bounded and equicontinuous. Pointwise boundedness follows from the fact that for each x ∈ R we have sup |gn (x)| = sup |(f ∗ kn )(x)| n

n

≤ sup kf ∗ kn k∞ n

≤ sup kf k∞ kkn k1 n

= kf k∞ < ∞. For equicontinuity, fix any ε > 0. Then there exists some δ > 0 such that |a| < δ =⇒ kTa f − f k∞ < ε. Choose any n, and suppose that |x − y| < δ. Then y = x − a with |a| < δ. Hence, for every n, |gn (y) − gn (x)| = |(f ∗ kn )(x − a) − (f ∗ kn )(x)| = |(Ta f ∗ kn )(x) − (f ∗ kn )(x)| ≤ k(Ta f ∗ kn )(−(f ∗ kn )k∞ = k(Ta f − f ) ∗ kn k∞ ≤ kTa f − f k∞ kkn k1 < ε. Thus |x − y| < δ =⇒ sup |gn (x) − gn (y)| < ε, n

which says that {gn }n∈N is equicontinuous. Set hn = gn |[−1,1] , i.e., hn is gn restricted to the interval [−1, 1]. Since [−1, 1] is a compact set and {hn }n∈N is both pointwise bounded and equicontinuous on [−1, 1], the Arzel´ a–Ascoli Theorem implies that there exists some subsequence {hnk }n∈N that converges uniformly on [−1, 1]. The function F1 = lim hnk (x) k→∞

is therefore uniformly continuous on [−1, 1]. Now, f is essentially bounded, and therefore is integrable on [−1, 1]. The Lebesgue Differentiation Theorem therefore implies that

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gn (x) = (f ∗ kn )(x) → f (x)

for a.e. x.

Consequently hnk → f a.e. on [−1, 1] as k → ∞. As hnk → F1 uniformly on [−1, 1], it follows that f = F1 a.e. on [−1, 1]. A similar argument shows that there exists a function F2 that is uniformly continuous on [−2, 2] and satisfies f = F2 a.e. on [−2, 2]. Hence F1 = F2 a.e. on [−1, 1]. As both of these functions are continuous, this implies that F1 = F2 everywhere on [−2, 2]. Continuing in this way, for each N ∈ N we can construct a function FN that is uniformly continuous on [−N, N ], satisfies FN = f a.e. on [−N, N ], and also satisfies FN +1 = FN on [−N, N ]. We can therefore uniquely define a function F on R by setting F |[−N,N ] = FN . This function is uniformly continuous and equals f a.e. 9.1.34 (a) If x ∈ R and n ∈ N, then f (nx) = f (x) + · · · + f (x) = nf (x). Therefore

nf (x/n) f (nx/n) 1 = = f (x). n n n In particular, for every m, n ∈ N we have f (x/n) =

f (mx/n) =

m 1 f (mx) = f (x). n n

Also, since f (0) = f (0 + 0) = f (0) + f (0), we have f (0) = 0. Therefore f (x) + f (−x) = f (x − x) = f (0) = 0. Hence for every m, n ∈ N we have f (−mx/n) = −f (mx/n) = −

m f (x). n

Combining the above results, we conclude that f (rx) = rf (x),

x ∈ R, r ∈ Q.

(b) “⇐.” The function f (x) = cx is additive and continuous. “⇒.” Suppose that f is additive and continuous. If we set c = f (1), then part (a) implies that f (r) = rf (1) = cr,

all r ∈ Q.

Since f is continuous, it follows that f (x) = cx for every x ∈ R.

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(c) Let {xi }i∈I be a basis for R over the field Q. Choose any countable subsequence J = {j1 , j2 , . . . } of I. Define f (xjn ) = n for n ∈ N and f (xi ) = 0 for i ∈ I\J0 . Given a nonzero x ∈ R, by definition of Hamel basis we can PN write x = k=1 ck xik for some unique i1 , . . . , iN ∈ I and unique nonzero PN rational scalars c1 , . . . , cN . Define f (x) = k=1 ck f (xik ). This f is a Qlinear function on R, i.e., f (x + ry) = f (x) + rf (y) for all x, y ∈ R and r ∈ Q. If f was R-linear, then we would have f (x) = ax for some a ∈ R. Hence m = f (xjm ) = axjm and n = f (xjn ) = axjn , so a 6= 0 and a a xj − xj = 0. m m n n Therefore

so {xi }i∈I R-linear.

1 1 xj − xj = 0, m m n n is not Q-independent, which is a contradiction. Hence f cannot be

(d) Problem 2.1.42 tells us that C + C = [0, 2]. Therefore, if we fix any z ∈ [0, 2], then there exist points x, y ∈ C such that z = x + y. Consequently, f (z) = f (x + y) = f (x) + f (y) = 0 + 0 = 0. Hence f is identically zero on [0, 2]. Applying part (a), we conclude that f is identically zero on R. 9.1.35 (a) Since g is bounded and φ ∈ Cc1 (R), we can apply Exercise 9.1.31 and conclude that φ ∗ g ∈ Cb1 (R) and (φ ∗ g)′ = φ′ ∗ g. Using the additivity of f, we compute that g(x + y) = e2πif (x+y) = e2πif (x)+2πif (y) = e2πif (x) e2πif (y) = g(x) g(y). Given φ ∈ Cc (R), we compute that Z ∞ φ(y) g(x − y) dy (φ ∗ g)(x) = −∞

=

Z

∞

−∞

= g(x)

φ(y) g(x) g(−y) dy Z

∞

φ(y) e−2πif (y) dy −∞

= Cφ g(x), where

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Cφ =

Z

∞

φ(y) e−2πif (y) dy =

−∞

Z

∞

φ(y) g(−y) dy.

−∞

Since φ is integrable and g is bounded, Cφ is a well-defined scalar. Suppose that Cφ = 0 for every φ ∈ Cc1 (R). Since g ∈ L∞ (R), Problem 9.1.32 implies that g = 0 a.e. This contradicts the fact that |g(x)| = 1 for every x. Therefore we must have Cφ 6= 0 for some φ ∈ Cc1 (R). If we choose φ ∈ Cc1 (R) so that Cφ 6= 0, then g =

1 (φ ∗ g) ∈ Cb1 (R). Cφ

Therefore g and g ′ are both continuous. Further, since φ′ is simply another function in Cc (R), we have g′ =

1 1 ′ Cφ′ (φ ∗ g)′ = (φ ∗ g) = g = βg. Cφ Cφ Cφ

Alternatively, we can show that differentiability of g is a consequence of continuity of g and the functional equation that g satisfies. In other words, we can derive differentiability of g even if we only consider functions φ ∈ Cc (R). To see why, observe that any function φ ∈ Cc (R) is integrable. Since g ∈ L∞ (R), it follows from Exercise 9.1.4 that φ ∗ g ∈ Cb (R). If we choose φ ∈ Cc (R) such R a that Cφ 6= 0, then this shows that g is continuous. We cannot have 0 g = 0 for every a, so there must exist some a > 0 such that Ra C = 0 g 6= 0. We then compute that Cg(x) = g(x) =

Z

Z

a

g(t) dt

0

a

g(x + t) dt

0

=

Z

x+a

g(t) dt

x

=

Z

0

x+a

g(t) dt −

Z

x

g(t) dt.

0

Since g is integrable on compact sets, we can apply the Fundamental Theorem of Calculus and conclude that g is differentiable and Z x+a Z x d d ′ Cg (x) = g(t) dt − g(t) dt dx 0 dx 0 = g(x + a) − g(x)

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Setting

383

= g(x) g(a) − g(x)
= g(x) g(a) − 1 . β =

we see that


g(a) − 1 C,

g ′ (x) = βg(x).

In any case, we have shown that g is a differentiable function that satisfies g ′ = βg. Standard undergraduate calculus tells us that the only solutions to this equation are g(x) = Ceβx where C ∈ C. However, since g(0) = e2πif (0) = e0 = 1, we have C = 1. Also, |g(x)| = 1 for every x, so β must be purely imaginary. Therefore β = 2πiα for some real α. Thus g(x) = e2πiαx for x ∈ R. (b) Let h(x) = ⌊x⌋, where ⌊x⌋ is the integer n that satisfies n ≤ x < n + 1. Then e2πih(x) = 1 for every x. This is continuous, even though h is not continuous. (c) We have e2πif (x) = g(x) = e2πiαx , so e2πi(f (x)−αx) = 1,

x ∈ R.

Therefore n(x) = f (x) − αx ∈ Z,

x ∈ R.

Given any x, y ∈ R, we compute that n(x + y) = f (x + y) − α(x + y) = f (x) + f (y) − αx − αy = n(x) + n(y). Part (a) of Exercise 9.1.34 therefore implies that x ∈ R, r ∈ Q.

n(rx) = rn(x),

Consequently, if n(x) 6= 0 for some x, then n(rx) is not an integer for some r ∈ Q. This contradicts the fact that n is integer-valued, so we must have n(x) = 0 for every x. Therefore f (x) = αx for every x. 9.2.6 Note that ZZ Z |f (y) g(x − y) e−2πiξx | dx dy = =

∞ −∞

Z

|f (y)|

∞

−∞

Z

∞

−∞

|f (y)| dy

|g(x − y)| dx dy

 Z

∞

−∞

 |g(x)| dx

< ∞. Therefore Fubini’s Theorem allows us to interchange integrals in the following calculation:

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384 ∧

(f ∗ g) (ξ) =

Z

∞

−∞

ZZ

(f ∗ g)(x) e−2πiξx dx

f (y) g(x − y) e−2πiξx dy dx Z ∞  Z ∞ −2πiξy −2πiξ(x−y) = f (y) e g(x − y) e dx dy −∞ −∞ Z ∞  Z ∞ = f (y) e−2πiξy g(x) e−2πiξx dx dy −∞ −∞ Z ∞ = f (y) e−2πiξy gb(ξ) dy =

−∞

= fb(ξ) gb(ξ).

9.2.11 We justify the use of Fubini’s Theorem in the proof of Lemma 9.2.11. We compute that Z ∞Z ∞ |f (y) e−2πiξy W (ξ/N ) e2πiξx | dξ dy −∞

−∞

=

Z

∞

−∞

≤

Z

Z

∞

−∞

|f (y)| W (ξ/N )| dξ dy

∞

−∞

|f (y)|

Z

N

1 dξ dy

−N

= 2N kf k1 < ∞. 9.3.6 (a) The proof is similar to the proof of Young’s Inequality given in the solution to Exercise 9.1.13. We write out the details. First we show that f ∗ g exists and is measurable if f, g ∈ L1 (T). Since g is 1-periodic, for any y we have Z 1 Z 1 |g(x)| dx = kgk1 , |g(x − y)| dx = 0

0

and therefore Z 1Z 1 0

0

|f (y) g(x − y)| dy dx = =

Z 1 Z 0

Z

0

1 0

 |g(x − y)| dx |f (y)| dy

1

kgk1 |f (y)| dy

= kgk1 kf k1 < ∞. Hence, it follows from Fubini’s Theorem that

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Z

(f ∗ g)(x) =

1

f (y) g(x − y) dy

0

exists for almost every x and is an integrable function of x. Since g is 1periodic, f ∗ g is 1-periodic as well. The cases p = 1 and p = ∞ are straightforward, so we concentrate on the case 1 < p < ∞. Choose f ∈ Lp (T) and g ∈ L1 (T). Since Lp (T) ⊆ L1 (T), the above work tells us that f ∗ g exists. Applying H¨ older’s Inequality with exponents p and p′ and using the periodicity together with a change of variables, we compute that Z 1 |(f ∗ g)(x)| ≤ |f (y) g(x − y)| dy 0

Z 1 ′  f (y) g(x − y) 1/p g(x − y) 1/p dy = 0

≤

=

Z

1

0

Z

1

0

p

dy

|f (y)|p |g(x − y)| dy

1/p Z

1/p′

1+

1/p Z

|f (y)| |g(x − y)|

= kgk1 Note that

p/p

Z

|f (y)|p |g(x − y)| dy

0

|g(y)| dy

1/p

p′ /p′

|g(x − y)|

0

1

0

1

1

1/p′

.

p(p − 1) p = 1+ = 1 + p − 1 = p. p′ p

Therefore, interchanging integrals by Tonelli’s Theorem, kf ∗ gkpp =

Z

0

1

|(f ∗ g)(x)|p dx p/p′

≤ kgk1 = =

p/p′ kgk1 p/p′ kgk1 p/p′

= kgk1

Z

1

0

Z

0

1

1

0

|f (y)|p |g(x − y)| dy dx p

p

|f (y)|

0

Z

1

|f (y)|

0

Z

Z

Z

1

0

Z

0

1

 |g(x − y)| dx dy

 |g(x)| dx dy

1

|f (y)|p kgk1 dy

dy

1/p′

c Solutions 2019 Christopher Heil

386 1+ pp′

= kgk1

kf kpp

= kgkp1 kf kpp , so the result follows upon taking pth roots. 9.3.8 (a), (b) The proof is identical to the solution of Theorem 9.1.15, except that we integrate over [0, 1] instead of R, and note that translation is strongly continuous on Lp (T), just as it is on Lp (R). 9.3.9 (a) We have σN =

N N n X X X 1 1 sn = ak N + 1 n=0 N + 1 n=0 k=−n

=

N X 1 N +1

N X

ak

k=−N n=|k|

=

N X N − |k| + 1 ak = N +1

k=−N

N  X

k=−N

1−

|k|  ak . N +1

(b) Let ω = e2πix , and let s =

N X

m=−N

e2πimx =

N X

ωm.

m=−N

Then we have sω = s − ω −N + ω N +1 , so s(ω − 1) = ω N +1 − ω −N .

Multiplying both sides by ω −1/2 = e−πix , we obtain s(ω 1/2 − ω −1/2 ) = ω N +1/2 − ω −N −1/2 . Now, ω 1/2 − ω −1/2 = eπix − e−πix = 2i sin πx, and likewise ω N +1/2 − ω N −1/2 = e2πi(N +1/2)x − e−2πi(N +1/2)x = 2i sin(2N + 1)πx, so s =

sin(2N + 1)πx . sin πx

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(c) We use parts (a) and (b) to compute that N  X

n=−N

=

|n|  2πinx e N +1

1−

N n 1 X X 2πikx e N + 1 n=0 k=−n

=

=

=

= = = =

N X

sin(2N + 1)πx 1 N + 1 n=0 sin πx N 1 X e(2n+1)πix − e−(2n+1)πix N + 1 n=0 2i sin πx

  N N X X 1 1 eπix e2nπix − e−πix e−2nπix N + 1 2i sin πx n=0 n=0   2πi(N +1)x −2πi(N +1)x 1 e −1 −1 1 −πix e eπix − e N + 1 2i sin πx e2πix − 1 e−2πix − 1  2πi(N +1)x  1 1 e −1 e−2πi(N +1)x − 1 − N + 1 2i sin πx eπix − e−πix e−πix − eπix  2πi(N +1)x  1 e − 2 + e−2πi(N +1)x 1 N + 1 2i sin πx eπix − e−πix  πi(N +1)x 2  1 1 (e − e−πi(N +1)x) N + 1 2i sin πx 2i sin πx

1 (2i sin π(N + 1)x)2 N +1 (2i sin πx)2  2 sin π(N + 1) 1 . = N +1 sin πx =

(d) The first requirement to be an approximate identity follows from the fact that Z

1

wN (x) dx =

0

1−

|n|  N +1

N  X

1−

|n|  δ0n = 1. N +1

n=−N

=

Z

N  X

n=−N

1

e2πinx dx

0

Since wN ≥ 0, the second requirement follows trivially.

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Finally, choose 0 < δ < 1/2. Note that sin πx is increasing on [0, 1/2]. Therefore, Z

2 N +1

wN (x) dx =

δ≤|x| 0, let g ∈ L1 (R) be any function such that b g(ξ) = 1. Then given any scalars a1 , . . . , aN and c1 , . . . , cN , we have N N



X X



cj M−aj fb cj e−2πiaj ξ fb(ξ) ≤ b 1 = b g − g (ξ) − j=1

j=1

∞

N

X

cj Taj f . ≤ g − 1

j=1


Hence g ∈ / span {Ta f }a∈R .

9.2.28 Since f and fb are integrable we have f ∈ C0 (R). Since g is integrable, ∧ it follows that f g ∈ L1 (R), and therefore (f g) ∈ C0 (R). Further, since we 1 1 have both fb ∈ L (R) and g ∈ L (R), we can apply the Inversion Formula and Fubini’s Theorem to compute that Z ∧ (f g) (ξ) = f (x) g(x) e−2πiξx dx = = = =

Z


∨ fb (x) g(x) e−2πiξx dx

Z Z Z

Z

(Inversion)

 2πiηx b f (η) e dη g(x) e−2πiξx dx

fb(η)

Z

g(x) e−2πi(ξ−η)x dx dη

fb(η) gb(ξ − η) dη

= (fb ∗ b g )(ξ).

(Fubini)

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9.2.29 Since fb is continuous, the hypotheses imply that fb ∈ L1 (R). Therefore the Inversion Formula applies, so |f (x + h) − f (x)|

= |T−h f (x) − f (x)|
∨ ∨ = |(Mh fb) (x) − fb (x)| Z   Mh fb(ξ) e2πiξx − fb(ξ) e2πiξx dξ = Z Z |fb(ξ)| |e2πihξ − 1| dξ + ≤ |ξ|≤1/|h|

≤

Z

|ξ|≤1/|h|

≤ 2πC|h| = 4πC|h|

Z

|fb(ξ)| |2πihξ| dξ + −α

|ξ|≤1/|h|

Z

1/|h|

|ξ|

|ξ|>1/|h|

Z

|ξ|>1/|h|

dξ + 2C

ξ −α dξ + 4C

Z

Z

|fb(ξ)| 2 dξ

|ξ|>1/|h|

∞

|fb(ξ)| |e2πihξ − 1| dξ

|ξ|−1−α dξ

ξ −1−α dξ

1/|h|

0

1/|h| ∞ ξ 1−α ξ −α = 4πC|h| + 4C 1 − α 0 −α 1/|h| = 4πC|h|

= 4πC

|h|α |h|α−1 + 4C 1−α α

|h|α |h|α + 4C 1−α α

= C ′ |h|α , where

4πC 4C + . 1−α α Hence f is H¨ older continuous with exponent α. C′ =

9.2.30 Let χ = χ[−1/2,1/2] be the box function, so that

Let

χ b (ξ) = sinc(ξ) =

sin πξ . πξ

f (x) = e−2π|x|

c Solutions 2019 Christopher Heil

395

be the dilated two-sided exponential. By making a change of variables in Problem 9.2.19, we have that fb(ξ) =

1 . π(ξ 2 + 1)

Since both f and fb are integrable, the Inversion Formula applies to f. Let g = fb. Since f and g are even, we have gb = f.

Since χ ∗ g ∈ L1 (R) and (χ ∗ g) = (sinc) · b g ∈ L1 (R), inversion also applies to χ ∗ g. Therefore, Z ∞ Z ∞ sin πξ −2π|ξ|+πiξ e dξ = π sinc(ξ) f (ξ) eπiξ dξ ξ −∞ −∞ Z ∞ χ b (ξ) gb(ξ) e2πiξ(1/2) dξ = π ∧

−∞

b gb)∨ = π (χ

1 2

= π (χ ∗ g) Z ∞ χ = π −∞

Z

1




1 2




1 2


− y g(y) dy

1 dy 2 + 1) π(y 0 1 = tan−1 y = π

0

π = . 4

9.2.31 (a) Although sinx x is not integrable, because it oscillates sign, the sum of the (signed) areas of the regions under its graph behaves like an alternating series. Consequently, Z b Z π/2 π sin x sin x K = sup dx = dx < 1 · < ∞. x x 2 01 −1 Z ∞ 1 dx ≤ 2 kxm f (n) (x)k∞ + kxm+2 f (n) (x)k∞ 2 2 x 1 = 2 kxm f (n) (x)k∞ + 2 kxm+2 f (n) (x)k∞ .

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Also, Cc∞ (R) ⊆ S(R). Since Cc (R) is dense in L1 (R), it follows that the Schwartz space is also dense in L1 (R). One example of an element of S(R) is 2 the Gaussian function φ(x) = e−x . (c) Choose any f ∈ S(R). Then, by the product rule, we have for any m, n ≥ 0 that n   X
n n m Dj (−2πix)m f (n−j) (x) ∈ L1 (R). D (−2πix) f (x) = j j=0 Hence, 

(2πiξ)n Dm fb(ξ) =


∧ Dn (−2πix)m f (x) (ξ) ∈ L∞ (R).

Since this is true for every m and n, we conclude that fb ∈ S(R).

(d) Part (c) shows that F maps S(R) into S(R). If f ∈ S(R), then part (c) ∨

implies that g = f ∈ S(R). Therefore b g ∈ S(R) by part (c). However, f = ∨

∧

( f ) = gb by the Inversion Formula. This shows that F is surjective. Finally, the Uniqueness Theorem implies that the Fourier transform is injective on L1 (R), so its restriction to S(R) is also injective.

9.3.2 A technique similar to that used in the proof of Theorem 9.2.5 can be employed. Functions in L1 (T) are 1-periodic on R. Since e−πi = −1, we have for n 6= 0 that Z 1 f (x) e−2πinx dx fb(n) = 0

= −

= − = −

Z

1

1

f (x) e−2πinx e−2πin( 2n ) dx

0

Z

1

1

f (x) e−2πin(x+ 2n ) dx

0

Z

0

1

 1  −2πinx e dx. f x− 2n

Averaging the first and last lines in the equalities above, we obtain Z  1 1 1  −2πinx b f (n) = e dx. f (x) − f x − 2 0 2n

Problem 9.3.20 shows that translation is strongly continuous on L1 (T). Using this we compute that

c Solutions 2019 Christopher Heil

|fb(n)| ≤

1 2

Z

1 0

as |n| → ∞.

399

 1 1  1 f k1 → 0 dx = kf − T 2n f (x) − f x − 2n 2

9.3.4 Since f is differentiable, we can apply integration by parts to conclude that Z 1 fb′ (n) = f ′ (x) e−2πinx dx 0

Z 1 = f (x) e−2πinx − 0

1

f ′ (x) e−2πinx (−2πin) dx

0

= 0 + 2πinfb(n).

Therefore, for n 6= 0 we have

|fb(n)| ≤

|fb′ (n)| kf ′ k1 ≤ . 2π |n| 2π |n|

The result extends to higher derivatives by induction. 9.3.6 The cases p = 1 and p = ∞ are easier, so we will concentrate on the case 1 < p < ∞. First we show that f ∗ g exists and is measurable if f, g ∈ L1 (T). Since g R1 R1 is 1-periodic, for any y we have 0 |g(x − y)| dx = 0 |g(x)| dx = kgkL1 , and therefore  Z 1Z 1 Z 1 Z 1 |g(x − y)| dx |f (y)| dy |f (y) g(x − y)| dy dx = 0

0

0

=

Z

0

0

1

kgkL1 |f (y)| dy

= kgkL1 kf kL1 < ∞.

R1 Hence, it follows from Fubini’s Theorem that (f ∗ g)(x) = 0 f (y) g(x − y) dy exists for almost every x and is an integrable function of x. Since g is 1periodic, f ∗ g is 1-periodic as well. Now suppose that 1 < p < ∞, and choose f ∈ Lp (T) and g ∈ L1 (T). Since p L (T) ⊆ L1 (T), the above work tells us that f ∗ g exists. Applying H¨ older’s Inequality with exponents p and p′ and making a change of variables, we have Z 1 |(f ∗ g)(x)| ≤ |f (y) g(x − y)| dy 0

Z 1 ′  f (y) g(x − y) 1/p g(x − y) 1/p dy = 0

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400

≤ =

Z

1

0

Z

1

0

=

p

1+

1/p Z

|f (y)| |g(x − y)|

dy

|f (y)|p |g(x − y)| dy

1/p Z

1/p′ kgkL1

Note that

p/p

Z

1

|f (y)| |g(x − y)| dy

0

p′ /p′

|g(x − y)|

0

1

0

p

1

|g(y)| dy

1/p

dy

1/p′

1/p′

.

p(p − 1) p = 1+ = 1 + p − 1 = p. p′ p

Therefore, interchanging integrals by Tonelli’s Theorem, Z 1 |(f ∗ g)(x)|p dx kf ∗ gkpLp = 0

p/p′

≤ kgkL1

1

0

p/p′

= kgkL1

Z

1

p/p′

Z

1

0

p/p′

= kgkL1

Z

0

p ′

Z

1

0

|f (y)|p |g(x − y)| dy dx

|f (y)|p

0

= kgkL1

1+

Z

1

|f (y)|p

Z

1

0

Z

1

0

 |g(x − y)| dx dy  |g(x)| dx dy

|f (y)|p kgkL1 dy

= kgkL1 p kf kpLp = kgkpL1 kf kpLp , so the result follows upon taking pth roots. Finally, since [0, 1] has finite measure, both f and g belong to L1 (T). Using Fubini’s Theorem to interchange of the order of integration, we compute that Z 1 ∧ (f ∗ g) (n) = (f ∗ g)(x) e−2πinx dx 0

Z

1

Z

1

f (y) g(x − y) dy e−2πinx dx Z 1  Z 1 −2πiny −2πin(x−y) = f (y) e g(x − y) e dx dy

=

0

0

0

=

Z

0

1

f (y) e

−2πiny

Z

0

1

g(x) e 0

−2πinx

 dx dy

c Solutions 2019 Christopher Heil

=

Z

0

1

401

f (y) e−2πiny gb(n) dy

= fb(n) gb(n).

9.3.9 (a) We have

σN =

N N n X X X 1 1 sn = ak N + 1 n=0 N + 1 n=0 k=−n

=

N X 1 N +1

N X

ak

k=−N n=|k|

=

N X N − |k| + 1 ak N +1

k=−N

=

N  X 1−

k=−N

|k|  ak . N +1

(b) Let ω = e2πix , and let N X

s =

m=−N

e2πimx =

N X

ωm.

m=−N

Then we have sω = s − ω −N + ω N +1 , so s(ω − 1) = ω N +1 − ω −N .

Multiplying both sides by ω −1/2 = e−πix , we obtain s(ω 1/2 − ω −1/2 ) = ω N +1/2 − ω −N −1/2 . Now, ω 1/2 − ω −1/2 = eπix − e−πix = 2i sin πx, and likewise ω N +1/2 − ω N −1/2 = e2πi(N +1/2)x − e−2πi(N +1/2)x = 2i sin(2N + 1)πx, so s = (c) Given N ∈ N, define

sin π(2N + 1)x . sin πx

c Solutions 2019 Christopher Heil

402 ∨

χN (x) =

N X

e2πimx .

m=−N

Using part (a), we compute that wN (x) =

N  X

n=−N

1−

|n|  2πinx e N +1

∨

= =

=

∨

χ0 (x) + · · · + χN (x) N +1

N 1 X sin(2N + 1)πx N + 1 n=0 sin πx N 1 X e(2n+1)πix − e−(2n+1)πix N + 1 n=0 2i sin πx

  N N X X 1 1 πix 2nπix −πix −2nπix e e − e e = N + 1 2i sin πx n=0 n=0   2πi(N +1)x −2πi(N +1)x 1 −1 −1 1 πix e −πix e e − e = N + 1 2i sin πx e2πix − 1 e−2πix − 1  2πi(N +1)x  1 1 e −1 e−2πi(N +1)x − 1 = − N + 1 2i sin πx eπix − e−πix e−πix − eπix  2πi(N +1)x  1 e − 2 + e−2πi(N +1)x 1 = N + 1 2i sin πx eπix − e−πix  πi(N +1)x 2  1 1 (e − e−πi(N +1)x) = N + 1 2i sin πx 2i sin πx 1 (2i sin π(N + 1)x)2 N +1 (2i sin πx)2  2 1 sin π(N + 1) = . N +1 sin πx

=

9.3.19 (a) Suppose that a ∈ ℓ1 (Z). Set

n WN (k) = max 1 −

Then for each k we have

o |k| ,0 . N +1

lim WN (k) ak = ak

N →∞

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403

and WN (k) |ak | ≤ |ak |. Since a is summable, the Dominated Convergence Theorem for series implies that lim σN =

N →∞

=

=

N  X

lim

N →∞

k=−N ∞ X

lim

N →∞ ∞ X

∞ X

|k|  ak N +1

WN (k) ak

k=−∞

lim WN (k) ak

k=−∞

=

1−

N →∞

ak .

k=−∞

More generally, suppose that we know that the partial sums sN converge to L. Choose any ε > 0. Then there exists an M such that N > M =⇒ |L − sN | < ε. Let C = sup |sN |. Then we have |L| ≤ C, so (N + 1)L − (s0 + · · · + sN ) lim sup |L − σN | = lim sup N +1 N →∞ N →∞  X  N M |L − sk | X |L − sk | ≤ lim sup + N +1 N +1 N →∞ k=M+1

≤ lim sup N →∞

 X N

k=M+1

k=0

(M + 1) 2C ε + N +1 N +1

≤ ε + 0 = ε. This is true for every ε > 0, so σN → L. (b) We have s2N = 1 and s2N +1 = 0, so σ2N = σ2N +1 = and therefore σN →

1 2

N s0 + · · · + s2N = , 2N + 1 2N + 1 s0 + · · · + s2N +1 N = , 2N + 2 2N + 2

as N → ∞.



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9.3.20 (a) Fix f ∈ C(T) and ε > 0. Since f is uniformly continuous, there exists 0 < δ < 1 such that |x − y| < δ =⇒ |f (x) − f (y)| < ε. Fix |a| < δ. Then |f (x) − Ta f (x)| = |f (x) − f (x − a)| < ε. Thus kf − Ta f k∞ ≤ ε whenever |a| < δ, so kTa f − f k∞ → 0. (b) We will reduce the problem to the point where we can apply facts about the denseness of Cc (R) in Lp (T). Fix 1 ≤ p < ∞, and choose f ∈ Lp (T) and ε > 0. Applying the Dominated Convergence Theorem, there must exist some 0 < δ < 12 such that g = f · χ[2δ,1−2δ] satisfies kf − gkp =

Z

1

0

p

|f − g|

1/p

< ε.

Although f is 1-periodic, the function g is identically zero outside of the interval [2δ, 1 − 2δ], so g belongs to Lp (R). Since Cc (R) is dense in Lp (R), there exists a function θ ∈ Cc (R) such that kg − θkp =

Z

∞

−∞

|g − θ|p

1/p

< ε.

Since g is identically zero outside of [2δ, 1 − 2δ], we can modify θ so that • • •

θ is unchanged on [2δ, 1 − 2δ], θ = 0 outside of [δ, 1 − δ], kg − θkp < ε.

Since θ(0) = θ(1), we can take θ on the interval [0, 1) and extend it 1periodically to R to obtain a continuous, 1-periodic function on R. This function θ belongs to C(T), and, computing the integrals on the domain [0, 1), kf − θkp ≤ kf − gkp + kg − θkp < 2ε. Therefore C(T) is dense in Lp (T). Now we will show that translation is strongly continuous on Lp (T). Fix 1 ≤ p < ∞, and choose f ∈ Lp (T). Given ε > 0, we can find g ∈ C(T) such that kf − gkp < ε. Since g is uniformly continuous, there exists a δ > 0 such that |a| < δ =⇒ kg − Ta gk∞ < ε. Therefore, for such a we have

c Solutions 2019 Christopher Heil

kg − Ta gkpp =

Z

1

0

405

|g(x) − Ta g(x)|p dx ≤

Z

1

εp dx = εp .

0

p

Since translation is isometric on L (T), we therefore have for |a| < δ that kf − Ta f kp ≤ kf − gkp + kg − Ta gkp + kTa g − Ta f kp ≤ ε + ε + ε = 3ε. Hence Ta f → f in Lp (T) as a → 0. 9.3.21 The Fej´er kernel {wN }N ∈N is a sequence of infinitely differentiable, 1-periodic functions. Lemma 9.3.10 tells us that if f ∈ Lp (T) with 1 ≤ p < ∞, then f ∗ wN → f in Lp -norm as N → ∞. By equation (9.41), N X

f ∗ wN =

n=−N

WN (n) fb(n) en .

Thus f ∗ wN is a finite linear combination of complex exponentials, and hence is infinitely differentiable. Thus, f ∗wN ∈ C ∞ (T) and f ∗wN → f in Lp -norm, so it follows that C ∞ (T) is dense in Lp (T). Lemma 9.3.10 also tells us that f ∗ wN → f uniformly if f ∈ C(T), so we likewise conclude that C ∞ (T) is dense in C(T) with respect to the uniform norm. 9.3.22 Suppose that there was a function δ ∈ L1 (T) that is an identity for convolution. Then by part (a), for every f ∈ L1 (T) and n ∈ Z we would have ∧ b fb(n) δ(n) = (f ∗ δ) (n) = fb(n).

In particular, if we fix n and take f (x) = en (x) = e2πinx then fb = δn and b therefore fb(n) = 1, so we must have δ(n) = 1 for every n. As δ ∈ L1 (T), contradicts the Riemann–Lebesgue Lemma. 9.3.23 We compute that Z (f ∗ en )(x) =

∞

f (y) e2πin(x−y) dy

−∞

= e2πinx

Z

∞

−∞

f (y) e−2πiny dy = en (x) fb(n).

9.3.24 (a) Suppose that f ∈ L1 (T) and fb ∈ ℓ2 (Z). Then fb ∈ ℓ1 (Z), so the Inversion Formula applies. In particular, f is continuous and bounded on T, so f ∈ L2 (T). (b) If f ∈ L2 (T), then we know that the Plancherel Equality holds.

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So, suppose that f ∈ L1 (T) \ L2 (T). Then we must have fb ∈ / ℓ2 (Z) by part (a). Hence we have both kf k2 = ∞ and kfbk2 = ∞, so again the Plancherel Equality holds. 9.3.25 First note that integration by parts shows that  Z 1  i , n 6= 0, 2πn x e−2πinx dx =  0 1/2, n = 0, and

Z

1

0

  πin + 1 , 2π 2 n2 x2 e−2πinx dx =  1/3,

n 6= 0, n = 0.

(a) The Fourier coefficients of

 1 g(x) = 2π 2 x2 − x + 6

(extended 1-periodically) are

and, for n 6= 0, gb(n) = 2π 2

gb(0) = 2π 2  πin + 1 2π 2 n2

−

1

3

−

1 1 = 0 + 2 6

 πin + 1 1 i  πin  = 2π 2 − 2 2 = 2. 2 2 2πn 2π n 2π n n

Hence b g ∈ ℓ1 (Z), and therefore the Fourier series for f converges uniformly on [0, 1] to the continuous function g. Combining positive and negative terms, we find that ∞ X 2 cos 2πnx = n2 n=1

=

X

n∈Z,n6=0

X

n∈Z

1 2πinx e n2

g (n) e2πinx b

= g(x)  1 , = 2π 2 x2 − x + 6

where the series converges uniformly on [0, 1]. Since f and each term in the series is 1-periodic, we also have uniform convergence on any compact subset of R, except that we must remember that g is periodic, and hence is given by the formula g(x) = 2π 2 (x2 − x + 16 ) only for x ∈ [0, 1]. Rearranging the above equality, we find that

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∞ X

1 cos 2πnx = x2 − x + , 2 n2 π 6 n=1

x ∈ [0, 1].

(b) Taking x = 0 in part (a), which is allowed since the series converges uniformly, we see that ∞ X

1 1 = 0−0+ . 2 n2 π 6 n=1 Rearranging gives Euler’s Formula. (c) Taking 1/04 = 0 for this calculation, the Plancherel Equality implies that 2

∞

 1  2 X 1

= kb g k2ℓ2 (Z) = kgk22 . =

2 4 n n n∈Z ℓ2 (Z) n=1

We compute that

Z

 1 2 dx (2π 2 )2 x2 − x + 6 0 Z 1 4x2 x 1 x4 − 2x3 + = 4π 4 dx − + 3 3 36 0   1 1 1 4 1 4 − + − + = 4π 5 2 9 6 36

kgk22 =

1

1 π4 = . 180 45

= 4π 4 Therefore

∞ X 1 1 π4 = kgk22 = . 4 n 2 90 n=1

9.3.26 The given function f is continuous and 2π-periodic. Given n ∈ Z, we compute that fb(n) = =

Z

1

f (x) e−2πinx dx

0

π eπiα sin πα

π eπiα = sin πα π eπiα = sin πα

Z

1

e−2πiαx e−2πinx dx

0

Z

1

e−2πi(α+n)x dx

0



e−2πi(α+n)x −2πi(n + α)

1 0

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= = = = = Since

π eπiα e−2πi(α+n) − 1 sin πα −2πi(n + α)

π eπiα e−2πiα e−2πin − eπiα sin πα −2πi(n + α) e−πiα − eπiα π sin πα −2πi(n + α)

(e−2πin = 1 for n ∈ Z)

−2i sin πα π sin πα −2πi(n + α)

(¯ z − z = −2i Re(z))

1 . n+α

πiα πe π |f (x)| = e−2πiαx = , sin πα | sin πα|

which is a constant, we have Z Z 1 |f (x)|2 dx = kf k22 =

0

0

1

π2 π2 dx = . 2 sin πα sin2 πα

The Plancherel Equality therefore implies that ∞ X

1 = (n + α)2 n=−∞

∞ X

n=−∞

|fb(n)|2 = kf k22 =

π2 . sin2 πα

9.3.27 (a) Assume that f ∈ L1 (T) and g ∈ C(T). Then g is uniformly continuous, so Z 1
|(f ∗ g)(x) − (f ∗ g)(x − a)| = f (y) g(x − y) − g(x − a − y) dy ≤

Z

0

0

1

|f (y)| kg − Ta gk∞ dy

= kf kL1 kg − Ta gk∞ → 0

as a → 0.

(b) Assume that f ∈ L1 (T) and g ∈ C 1 (T). We have Z (f ∗ g)(x + h) − (f ∗ g)(x) g(x + h − y) − g(x − y) = f (y) dy. h h The integrand converges pointwise a.e. to f (y) g ′ (x − y) as h → 0. Further, g ′ is bounded since it is continuous and periodic. Therefore, by the Mean Value Theorem, given x, y, and h there exists a point c such that

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g(x + h − y) − g(x − y) = |g ′ (c)| ≤ kg ′ k∞ . h

Therefore (as a function of y), f (y) g(x + h − y) − g(x − y) ≤ |f (y)| kg ′ k∞ ∈ L1 (T). h

The Lebesgue Dominated Convergence Theorem therefore applies, and we find that (f ∗ g)(x + h) − (f ∗ g)(x) h→0 h Z g(x + h − y) − g(x − y) = lim f (y) dy h→0 h Z = f (y) g ′ (x − y) dy = (f ∗ g ′ )(x).

(f ∗ g)′ (x) = lim

Thus f ∗ g is differentiable, and furthermore (f ∗ g)′ = f ∗ g ′ ∈ C(T) by part (a). Hence f ∗ g ∈ C 1 (T). 9.3.28 (a) Since f is absolutely continuous, we can apply integration by parts to compute that fb′ (n) =

Z

1

f ′ (x) e−2πinx dx

0

= f (x) e

Z 1 −

−2πinx

0

1

f ′ (x) e−2πinx (−2πin) dx

0

= 0 + 2πinfb(n).

Because f is absolutely continuous, we have f ′ ∈ L1 (T). Applying the Riemann–Lebesgue Lemma to f ′ , it follows that fb′ (n) = 0. lim |nfb(n)| = lim 2π |n|→∞ |n|→∞

R1 (b) Since 0 f = 0, we have fb(0) = 0. Applying the Plancherel Equality and part (a) to f ′ , we therefore have Z

0

1

|f ′ (x)|2 dx = =

X

n∈Z

X

n∈Z

|fb′ (n)|2

|2πinfb(n)|2

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≥ 4π 2 = 4π 2

X

n∈Z

Z

1

0

|fb(n)|2

|f (x)|2 dx.

(since fb(0) = 0)

If equality holds, then X

(n2 − 1) |fb(n)|2 = 0.

n6=0

Since each term is nonnegative, this implies that (n2 − 1) |fb(n)|2 = 0 for every n, and hence fb(n) = 0 except possibly for n = ±1. Since we are given that fb(0) = 0, it follows that f (x) = ae2πix + be−2πix

for some scalars a, b ∈ C. We can rewrite this as f (x) = ae2πix + be−2πix =

a + b 2πix a − b 2πix (e + e−2πix ) + (e − e−2πix ) 2 2

= (a + b) cos(2πx) + i (a − b) sin(2πx). 9.3.29 Suppose f is H¨ older continuous on T. Using the inequality derived in the solution to Exercise 9.3.2, we see that  α Z Z  1 1 1 1 1 α 1  1 1 b . |f (n)| ≤ dx ≤ f (x) − f x − dx = 2 0 2n 2 0 2n 2 2|n|

9.3.30 First consider any integers Nk . Since kwN k1 = 1 for every N, we have ∞ ∞ X X

−k

2 wN = 2−k = 1 < ∞. k 1 k=1

Therefore the series f =

P∞

k=1

k=1

2

−k

wNk converges absolutely in L1 (T).

Next, if N ∈ N then we have w d N (n) ≥ 0 for all n and X

n∈Z

w d N (n) =

 N X 1−

n=−N

|n| N +1

= (2N + 1) −



N 2 X n N + 1 n=1

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=

2 N (N − 1) (2N + 1)(N + 1) − N +1 N +1 2

=

2N 2 + 3N + 1 − N 2 + N N +1

=

N 2 + 4N + 1 N2 N ≥ = . N +1 2N 2

Therefore, if we take Nk = 2k then the function f defined above satisfies X

n∈Z

∞ XX

|fb(n)| =

n∈Z k=1

=

−k

∞ X

2

9.3.31 We are given that

X

n∈Z

k=1

Hence fb ∈ / ℓ1 (Z).

2−k w d 2k (n)

P

n∈Z

w2k (n) ≥

∞ X

2−k

k=1

2k = ∞. 2

|ncn | < ∞. Note that

X e−2πin(ξ+h) − e−2πinξ c (ξ + h) − b b c (ξ) = . cn h h n∈Z

Since e−n (ξ) = e−2πinξ is differentiable, the summand converges pointwise as h → 0: lim cn

h→∞

e−2πin(ξ+h) − e−2πinξ d = cn e−2πinξ = −2πincn e−2πinξ . h dξ

Further, since |1 − eiθ | ≤ |θ|, we have −2πinh e−2πin(ξ+h) − e−2πinξ − 1 cn = |cn | |e−2πinh | e h h ≤ |cn | |e−2πinh |

2π|nh| |h|

= 2π|ncn | ∈ ℓ1 (Z). Alternatively, we can obtain the same estimate by applying the Mean-Value Theorem to e−2πinξ . The Dominated Convergence Theorem for series therefore allows us to interchange the sum and integral in the following calculation:

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X e−2πin(ξ+h) − e−2πinξ b c (ξ + h) − b c (ξ) = lim cn h→0 h→0 h h lim

n∈Z

=

X

e−2πin(ξ+h) − e−2πinξ h→0 h

cn lim

n∈Z

= −2πi

X

ncn e−2πinξ .

n∈Z

Since (ncn )n∈Z ∈ ℓ1 (Z), we conclude that cb ′ (ξ) exists and where d = (−2πincn )n∈Z .

b b c ′ (ξ) = d(ξ)

9.3.32 Suppose that f = g ∗ h where g, h ∈ L2 (T). Exercise 9.1.4 showed ′ us that the convolution of a function in Lp (R) with a function in Lp (R) is continuous. The same result holds on the torus, so we conclude that f = g ∗ h ∈ C(T) ⊆ L1 (T). Also, b g, b h ∈ ℓ2 (Z), so we have fb = b gb h ∈ ℓ1 (Z). Therefore f ∈ A(T). Conversely, suppose that f ∈ A(T). Then fb ∈ ℓ1 (Z). For each n ∈ Z, let gn be any complex number such that gn2 = fb(n). Then X X |gn |2 = |fb(n)| < ∞. n∈Z

n∈Z

Since {e2πinx }n∈Z is an orthonormal basis for L2 (T), it follows that the function X g(x) = gn e2πinx n∈Z

2

belongs to L (T) and satisfies gb(n) = gn . Further, g ∗ g ∈ C(T) ⊆ L1 (T), and ∧ (g ∗ g) (n) = b g(n) gb(n) = fb(n),

n ∈ Z.

By uniqueness, we must have f = g ∗ g ∈ L2 (T) ∗ L2 (T).

9.3.33 Step 1. Consider en (x) = e2πinx and g ∈ L∞ (T). Making the change of variables y = mx and using the 1-periodicity of en and g, we compute that Z m Z 1 dx en (x/m) g(x) en (x) g(mx) dx = m 0 0 m−1 Z
1 X 1 en (x + k)/m g(x + k) dx = m 0 k=0

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m−1 Z 1 X 1 2πinx/m 2πink/m e e g(x) dx m 0

=

k=0

Z

=

   m−1 1 X 2πink/m . e e2πinx/m g(x) dx m

1

0

k=0

By the Lebesgue Dominated Convergence Theorem, we have Z 1 Z 1 lim e2πinx/m g(x) dx = 1 · g(x) dx = gb(0). m→∞

0

0

To evaluate the limit of the other factor, set z = e2πin/m . If n is an integer multiple of m then z = 1 and m−1 m−1 1 X 2πink/m 1 X k = e 1 = 1. m m k=0

k=0

Otherwise

m−1 m−1 1 X 2πink/m 1 X k 1 zm − 1 . e = z = m m m z−1 k=0

k=0

However, if n 6= 0 is fixed then n is not a multiple of m when m is large, so m−1 1 X 2πink/m e = m→∞ m

1 zm − 1 = 0, m→∞ m z − 1 lim

lim

k=0

If n = 0 then

n 6= 0.

m−1 1 X 2πi0·k/m lim e = 1. m→∞ m k=0

Combining these facts, we see that Combining these facts, we see that lim

m→∞

=

Z

1

en (x) g(mx) dx

0



lim

m→∞

Z

1

e

2πinx/m

0

  m−1 1 X 2πink/m e g(x) dx lim m→∞ m k=0

= b g(0) δn0 = gb(0) c en (0).

Step 2. By forming finite linear combinations, it follows that if p is any trigonometric polynomial and g ∈ L∞ (T), then lim

m→∞

Z

0

1

p(x) g(mx) dx = gb(0) pb(0).

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Step 3. Now fix f ∈ L1 (T) and g ∈ L∞ (T). Since {e2πinx }n∈Z is complete in L1 (T), there exists some trigonometric polynomial p such that kf −pk1 < ε. By part (a) there exists some m0 such that Z 1 m > m0 =⇒ pb(0) b g(0) − p(x) g(mx) dx < ε. 0

Therefore for m > m0 we have Z 1 fb(0) gb(0) − f (x) g(mx) dx 0

Z 1 b g(0) − p(x) g(mx) dx g(0) − pb(0) gb(0) + pb(0) b ≤ f (0) b 0 Z 1 Z 1 + p(x) g(mx) dx − f (x) g(mx) dx 0

0

Z Z
≤ |b g(0)| f (x) − p(x) dx + ε +

1

0

|p(x) − f (x)| |g(mx)| dx

≤ kb g k∞ kf − pk1 + ε + kb g k∞ kp − f k1 < Hence

R1 0


2 kb g k + 1 ε.

f (x) g(mx) dx → fb(0) gb(0) as m → ∞.

9.3.34 Suppose that there are infinitely many integers 0 < n1 < n2 < · · · such that √ sin 2πnk x ≥ 0 for every x ∈ E. Since { 2 sin 2πnx}n∈Z is an orthonormal sequence in L2 [0, 1] and since P 1 k2 < ∞, the series ∞ X sin 2πnk x f (x) = k k=1

2

converges in L -norm to a function in L2 [0, 1]. Consequently there is a subsequence of partial sums that converge pointwise a.e. on [0, 1] (alternatively, we could appeal to the Carleson–Hunt Theorem). That is, there exists an increasing sequence of integers Nj such that sNj (x) =

Nj X sin 2πnk x k=1

k2

→ f (x)

for a.e. x.

Let Z be the set of measure zero on which pointwise convergence does not occur. If x ∈ E \Z, then

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f (x) = lim

j→∞

415

Nj X sin 2πnk x

≥ lim inf

k

k=1

j→∞

Nj X δ = ∞. k

k=1

Hence |f | = ∞ on E \Z, which is a set of positive measure. This contradicts the fact that f ∈ L2 [0, 1].

9.3.35 We have dN ∈ L1 (T) since it is continuous on T, and also Z

Z

1

dN =

0

N X

1

e2πinx dx = 1.

0 n=−N

However, we will show that sup kdN kL1 = ∞. Using the fact that Z

1 0

1 2 − cos πx = , sin πx dx = π π 0

together with the estimate | sin x| ≤ |x|, we have 1 kdN kL1 = 2 ≥ = ≥ =

Z

Z

Z

1/2 0 1/2

sin(2N + 1)πx dx sin πx

| sin(2N + 1)πx| dx |πx|

0

N + 21 0

| sin πx| dx π|x|

N −1 Z k+1 X k=0

k

| sin πx| dx π|x|

N −1 Z 1 X k+1 | sin πx| dx π k+1 k k=0

Z N −1 1 X 1 k+1 = sin πx dx π k+1 k k=0

=

N −1 2 X 1 π2 k+1 k=0

=

N 2 X1 dx → ∞ as N → ∞. π2 k k=1

In fact, we can estimate using the Integral Test as follows:

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kdN kL1 ≥

Z N N 4 4 4 X1 1 ≥ dx = 2 ln N. 2 2 π k π 1 x π k=1

To obtain an upper bound, the fact that f (x) =

1 1 − sin πx πx

is odd and increasing on [−1/2, 1/2] means that |f (x)| reaches its maximum value at x = 1/2. Hence we have

1 2 1 − = |f (x)| ≤ f (1/2) = 1 − , sin πx πx π

|x| ≤

1 . 2

Consequently,

 1 2 1 , ≤ + 1− | sin πx| π|x| π

|x| ≤

1 . 2

0.4

0.2

-0.4

0.2

-0.2

0.4

-0.2

-0.4 Fig. 9.10 Graph of f (x).

Arguing similarly to before, we have Z 1/2 sin(2N + 1)πx 1 dx kdN k1 = 2 sin πx 0 Z Z 1/2  2  1/2 | sin(2N + 1)πx| dx + 1 − | sin(2N + 1)πx| dx ≤ π|x| π 0 0 Z N + 21  2 1 | sin πx| ≤ dx + 1 − π|x| π 2 0

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≤

Z

1

0

N Z 1 1 X k+1 | sin πx| 1 sin πx dx + dx + − πx π k 2 π k

≤ α + = α+

417

k=1

2 π2

N X

k=1

1 1 1 + − k 2 π

N 1 2 X1 1 − + 2 , 2 π π k k=1

where α =

Z

1

0

Hence

kdN k1 ≤ 2α + 1 −

sin πx dx ≈ 0.58949 < 1. πx

N N 2 4 X1 4 X1 + 2 ≈ 2 + 1.54236 . . . . π π k π k k=1

k=1

Now, for every n we have N X

k=1

≤ 1 + ln N,

and in fact Euler’s constant is lim

X N

N →∞

k=1

− ln N



= γ ≈ 0.577 . . . .

Hence kdN k1 ≤ 2α + 1 −

N 4 X1 2 + 2 π π k k=1

2 4 + (1 + ln N ) π π2 2 4 4 = 2α + 1 − + 2 + 2 ln N, π π π ≤ 2α + 1 −

where γ ≈ 0.577 . . . is Euler’s constant. Numerically, kdN k1 ≤ 2α + 1 −

2 4 4 4 + 2 + 2 ln N ≈ 2 ln N + 1.94764 . . . . π π π π

9.4.9 The fact that A is linear immediately implies that range(A) is a subspace of Y.

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Suppose that vectors yn ∈ range(A) converge to a vector y ∈ Y. By the definition of the range, for each n there is some vector xn ∈ X such that Axn = yn . As A is linear and isometric, we therefore have kxm − xn k = kA(xm − xn )k = kAxm − Axn k = kym − yn k.

(A)

But {yn }n∈N is Cauchy in Y (because it converges), so equation (A) implies that {xn }n∈N is Cauchy in X. Since X is complete, there is some x ∈ X such that xn → x. As A is bounded and therefore continuous, this implies that Axn → Ax. By assumption we also have Axn = yn → y, so the uniqueness of limits implies that y = Ax. Thus y ∈ range(A). This shows that range(A) contains every limit of points from range(A), so it is a closed set. Remark: We did not need to assume that Y is a Banach space, we only needed X to be complete. Hence the result is still true if X is a Banach space and Y is a normed space. 9.4.10 First part. Suppose that f ∈ L2 (R) and fb ∈ L1 (R). Note that
∨ we also have fb ∈ L2 (R). The inverse L2 -Fourier transform of fb is fb =
∨ f a.e. The inverse L1 -Fourier transform of fb satisfies fb ∈ C0 (R). Since these two Fourier transforms agree almost everywhere, we conclude that f =
∨ fb ∈ C0 (R) in the sense of identifying functions that are equal almost
∨ everywhere. Since f = fb a.e. and fb ∈ L1 (R), we therefore have that
∨ kf k∞ = k fb k∞ ≤ kfbk1 .

b . Then s ∈ L2 (R) \ L1 (R), and Second part. Let χ = χ[−1,1] and s = χ 1 sb = χ ∈ L (R). 9.4.11 (a) Proof 1. Suppose that f ∈ L1 (R) and fb ∈ L2 (R). Then g = f ∗ fe ∈ L1 (R), and b g = |fb|2 ∈ L1 (R), so the Inversion Formula applies to g. In particular, we have g, gb ∈ C0 (R), so g(0) = kf k22 is finite.

Proof 2. Let {vλ }λ>0 be the de la Vall´ee Poussin kernel. Then vbλ is a continuous function with supp(vbλ) ⊆ [−2λ, 2λ], 0 ≤ vbλ ≤ 1 everywhere, and and vbλ = 1 on [−λ, λ]. Suppose f ∈ L1 (R) is such that fb ∈ L2 (R). Since f ∗ vλ → f in L1 -norm, ∧ we know that fb vbλ = (f ∗ vλ ) → fb in L∞ -norm, where fb is the Fourier 1 transform of f as an L function. But we also know that fb and fb vbλ belong to L2 (R), and we have Z ∞ kfb − fb vbλ k22 = |fb(ξ)|2 |1 − vbλ(ξ)|2 dξ −∞

≤

Z

|ξ|>λ

|fb(ξ)|2 dξ → 0

as λ → ∞,

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so fbvbλ → fb in L2 -norm. Let g be the inverse Fourier Transform of fb as a ∨ function in L2 (R). Then, since the L2 Fourier transform is unitary, (fb vbλ) → g in L2 -norm. However, fb vbλ belongs to L2 (R) and has compact support, so it also belongs to L1 (R). Therefore the L1 Inversion Formula applies, i.e., ∨ (fb vbλ) = f ∗ vλ . Thus, we have both f ∗ vλ → f in L1 -norm and f ∗ vλ → g in L2 -norm. Since Lp -convergence implies the existence of a subsequence that converges pointwise a.e., we conclude that f = g a.e., so f ∈ L2 (R).

(b) Suppose that f ∈ L1 (R). If kf k2 < ∞ then f ∈ L2 (R), and hence fb ∈ L2 (R) and kf k2 = kfbk2 by the Plancherel Equality. If kfbk2 < ∞, then part (a) shows that f ∈ L2 (R), and hence again we have the equality kf k2 = kfbk2 . The remaining possibility is f, fb ∈ / L2 (R), in which case both kf k2 and kfbk2 are ∞.

(c) Let f (x) = x−1/2 χ(0,1) (x). Then we have f ∈ L1 (R) \ L2 (R). If we had fb ∈ L1 (R) then the Inversion Formula would apply and we would obtain f ∈ C0 (R), which is a contradiction. Consequently fb cannot be integrable.

(d) Let F be any function in L2 (R) such that F (ξ) does not converge to zero as ξ → ±∞. Since the Fourier transform maps L2 (R) onto itself, there is a function f ∈ L2 (R) such that fb = F. The Riemann–Lebesgue Lemma does not hold for f.

9.4.12 (a) Proof 1. If f, g ∈ L2 (R), then f g ∈ L1 (R). Hence (f g) ∈ A(R) ⊆ C0 (R). On the other hand, since fb, gb ∈ L2 (R), we have fb ∗ gb ∈ C0 (R) by Exercise 9.1.4. Let fn , gn ∈ Cc∞ (R) be such that fn → f and gn → g in L2 -norm. Then fn gn → f g in L1 -norm. To see why, let C = sup kfn k2 < ∞. Then the Cauchy-Bunyakovski-Schwarz Inequality implies that ∧

kf g − fn gn k1 ≤ kf g − fn gk1 + kfn g − fn gn k1 ≤ kf − fn k2 kgk2 + kfn k2 kg − gn k2 ≤ kf − fn k2 kgk2 + C kg − gn k2 → 0. ∧

∧

Therefore (fn gn ) → (f g) uniformly. On the other hand, recall from Exercise 9.1.4 that kF ∗ Gk∞ ≤ kF k2 kGk2 for any functions F, G ∈ L2 (R). Combining this with the isometric nature of the Fourier transform on L2 (R), we see that b b − fbn ∗ gb k∞ + kfbn ∗ gb − fbn ∗ gc kfb ∗ b g − fbn ∗ gc n k∞ n k∞ ≤ kf ∗ g g − gc g k2 + kfbn k2 kb ≤ kfb − fbn k2 kb n k2

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≤ kf − fn k2 kgk2 + C kg − gn k2 → 0. b g uniformly. But (fn gn )∧ = fbn ∗ gc Hence fbn ∗ gc n , so this implies that n → f ∗b ∧ (f g) = fb ∗ b g. ∨

Similarly, we have (f g) = f ∗ g . Since this is true for all f, g ∈ L2 (R),
∨
∨ ∨ we therefore also have that (fbgb ) = fb ∗ b g = f ∗ g. ∨

∨

Proof 2. Suppose f, g ∈ L2 (R), and recall that (e g ) (ξ) = gb(ξ). Using the unitarity of the Fourier transform, we therefore compute that

∧ (f ∗ g)(x) = hf, Tx e g i = fb, (Tx ge )

 = fb, M−xb ge Z ∨ fb(ξ) e2πixξ gb(ξ) dξ = (fbgb ) (x). = ∧

∨ Thus f ∗ g = (fbgb ) .

∨

Likewise, f ∗g = ( f g ) . Since this is true for all f, g ∈ L2 (R), we therefore also have
∨
∨ ∧ ∧ fb ∗ b g = ( fb b g ) = (f g) . ∨

∧

(b) Since fb, gb ∈ L2 (R), we can apply part (a) to these functions to conclude that
∨
∨ ∨ (fb gb ) = fb ∗ gb = f ∗ g ∈ C0 (R).

Consequently, if it is the case that f ∗ g ∈ L2 (R), then (fb gb ) ∈ L2 (R) as well. Since fb gb ∈ L1 (R), this implies by Problem 9.4.11 that fb b g ∈ L2 (R). Hence ∧ ∨∧ (f ∗ g) = (fb b g) = fb b g. ∨

In particular, if f ∈ L1 (R) and g ∈ L2 (R), then f ∗ g ∈ L2 (R), so this applies.

(c) Suppose that f, g ∈ L2 (R). Then fb, b g ∈ L2 (R), so fbgb ∈ L1 (R). ∨ Therefore, by part (a), we have f ∗ g = (fb b g ) ∈ A(R). Conversely, suppose that F ∈ A(R). Then F = fb for some f ∈ L1 (R). Define a square root function Sz = z 1/2 on C by (reiθ )1/2 = r1/2 eiθ/2 ,

r > 0, 0 ≤ θ < 2π.

Suppose that B is any open ball in C. Under this square root mapping, S −1 (B) need no longer be an open set, but it will either be open or a “halfopen” set. In any case, S −1 (B) is a Borel set, so S is a Borel measurable mapping. Therefore, the function g = f 1/2 is Lebesgue measurable and belongs to L2 (R). Moreover, by part (a) we have

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∧ F = fb = (g g) = gb ∗ gb ∈ L2 (R) ∗ L2 (R).

9.4.13 (a) Suppose that f, g ∈ L2 (R). Then we know that f ∗ g exists ∨ and belongs to C0 (R). By Problem 9.4.12, we have that f ∗ g = (fb b g) . By Problem 9.4.11, we know that the Plancherel Equality holds even for L1 ∨ functions. In particular, fb b g ∈ L1 (R), so k(fb gb ) k2 = kfb b g k2 . Therefore, kf ∗ gk22 = k(fb gb ) k22 ∨

= kfb gb k22

by Plancherel

≤ kfb 2 k2 kb g 2 k2

by Cauchy–Bunyakovski–Schwarz

= k(fb 2 ) (b g 2 )k1

= kfb fbk2 kb gb g k2

= k(fb fb) k22 k(b g gb ) k22 ∨

∨

= kf ∗ f k2 kg ∗ gk2

by Plancherel by Problem 9.4.12.

(b) Suppose that it was true that kf ∗ gk21 ≤ kf ∗ fRk1 kg ∗ gk1 held for all f, g ∈ L1 (R). Let kλ (x) = λk(λx) where k ≥ 0 and k = 1. Since kλ is nonnegative, Z Z kkλ ∗ kλ k1 = kλ (y) kλ (x − y) dy dx = kkλ k21 = 1. Hence, given any f ∈ L2 (R), we have kf k21 = lim kf ∗ kλ k21 ≤ lim sup kf ∗ f k1 kkλ ∗ kλ k1 = kf ∗ f k1 ≤ kf k21 . λ→∞

λ→∞

Thus, kf ∗ f k1 = kf k21 must hold for all f ∈ L1 (R). However, Problem 9.1.19 gives an example of an f ∈ L1 (R) for which this equality fails.

9.4.14 Let E be a measurable set with |E| < ∞. Then χE ∈ L1 (R) ∩ L2 (R). ∧ Set f = χE , so f 2 = f. Then Problem 8.1.15 implies that fb∗ fb = (f f ) = fb, and fb is a nontrivial element of L2 (R).

9.4.15 (a) χT = χ[−T,T ] belongs to L1 (R) ∩ L2 (R), so its L1 and L2 inverse ∨ ∨
∧ Fourier transforms agree and we have χT = d2πT . Therefore χT = χT = ∧ (d2πT ) .

(b) Let χT = χ[−T,T ] . Since fb ∈ L2 (R) and χT ∈ L2 (R), Problem 9.4.12 implies that (fbχT )

∨

=

fb


∨

∨

∗ χT = f ∗ d2πT ∈ C0 (R),

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the final conclusion following since f, d2πT ∈ L2 (R). But we also have that fb ∈ L2 (R) and χT ∈ L∞ (R), so fb · χT ∈ L2 (R), and therefore f ∗ d2πT = ∨ ∧ (fb · χT ) ∈ L2 (R). Hence (f ∗ d2πT ) = fb · χT , and this converges to fb in 2 L -norm by the Lebesgue Dominated Convergence Theorem.

∧ (c) By part (b), (f ∗ d2πT ) → fb in L2 -norm. Since the Fourier transform is unitary, this implies that f ∗ d2πT → f in L2 -norm as T → ∞.

9.4.16 (a) Let f ∈ L1 (R) be any function such that supp(fb) ∩ [−T, T ] = ∅. ∧ Then by Problem 8.1.15, we have (f ∗ d2πT ) = fb χ[−T,T ] = 0. Therefore f ∗ d2πT = 0 since the Fourier transform is unitary on L2 (R). (b) By the Plancherel Equality, Z

∞

−∞

sin2 t dt = t2

Z



∞

−∞

sin πx πx

2

π dx

= π kdπ k22 = π kdbπ k22

= π kχ[−1/2,1/2] k22 = π. (c) We are given j ∈ N and r ≥ j. Let χ = χ[−1/2,1/2] , and let χr = χ[−r/2,r/2] . Then χ ∗· · ·∗ χ (j times) is supported in [−j/2, j/2] ⊆ [−r/2, r/2]. Therefore, by the change of variable t = πx and the Parseval Equality, we have Z  Z  sin t j sin rt sin πx j sin πrx dt = π dx t t πx πx Z = π dπ (x)j dπr (x) dx 

= π djπ , dπr

c  dπr = π djπ , d

 = π χ ∗ · · · ∗ χ, χr = π

Z

r/2

−r/2

= π

Z

(χ ∗ · · · ∗ χ)(x) dx

(χ ∗ · · · ∗ χ)(x) dx

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= π (χ ∗ · · · ∗ χ) (0) ∧

= π dπ (0)j = π. 9.4.17 Set p(x) = Z

1 π(x2 +1) .

Then pb(ξ) = e−2π|ξ| by Problem 9.2.19, so

1 dx = 2 2 2 2 2 2 (x + a )(x + b ) a b

Z

(x/a)2

dx

+ 1 (x/b)2 + 1

Z π2 p(x/a) p(x/b) dx a2 b 2 Z π2 p1/a (x) p1/b (x) dx = ab

=

=

 π2

p1/a , p1/b ab

π2

∧ ∧ (p1/a ) , (p1/b ) ab Z π2 = e−2πa|ξ| e−2πb|ξ| dx ab Z π 2 ∞ −2π(a+b)ξ = 2 e dx ab 0 ∞ π 2 e−2π(a+b)ξ · = 2 ab −2π(a + b) 0 =

π2 0−1 · ab −2π(a + b) π . = (a2 b + b2 a) = 2

9.4.18 “⇒.” Suppose that there is a set E of positive measure on which fb is zero. Since the sets Ek = E ∩ [k, k + 1) are disjoint, at least one set Ek must have positive measure. Then χEk ∈ L2 (R) is not the zero function, so ∨ g = (χEk ) ∈ L2 (R) is nonzero as well. By the Parseval Equality we have for every a that Z b χ hTa f, gi = hM−a f , Ek i = e−2πiaξ fb(ξ) χEk (ξ) dξ = 0. Hence {Ta f }a∈R is not complete in L2 (R).

“⇐.” Suppose that fb(ξ) 6= 0 a.e., and g ∈ L2 (R) is such that hTa f, gi = 0 for every a ∈ R. Then fbb g ∈ L1 (R), and we have for every a that

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0 = hTa f, gi = hM−a fb, gb i =

Z

∧ e−2πiaξ fb(ξ) gb(ξ) dx = (fb gb ) (a).

By the Uniqueness Theorem, we therefore have fb gb = 0 a.e. Since fb(ξ) 6= 0 a.e., this implies that b g = 0 a.e., and hence g = 0 a.e. Therefore {Ta f }a∈R is complete in L2 (R).

9.4.19 “⇒.” Suppose that {Tk g}k∈Z is an orthonormal in L2 (R). P sequence 2 1 2 Because |b g| ∈ L (R), the periodization G(ξ) = |b g (ξ − k)| converges absolutely in L1 (T). Because of this, we can compute the Fourier coefficients of G as follows: Z 1 b G(n) = G(ξ) e−2πinξ dξ 0

=

Z

1

X

0 k∈Z

= = =

Z

Z

|b g(ξ − k)|2 e−2πin(ξ−k) dξ

|b g (ξ)|2 e−2πinξ dξ gb(ξ) e−2πinξ gb(ξ) dξ



 M−n gb, b g = Tn g, g = δ(n).

b = δ, so we must have G = 1 a.e. by the uniqueness of Fourier That is, G coefficients. P “⇐.” Suppose that G(ξ) = |b g(ξ − k)|2 = 1 a.e. Then given n ∈ Z we have Z



 Tn g, g = M−n gb, gb = g(ξ) e−2πinξ gb(ξ) dξ b =

=

Z

|b g (ξ)|2 e−2πinξ dξ

XZ k∈Z

=

Z

1

1

0

X

0 k∈Z

=

Z

0

1

|b g (ξ − k)|2 |b g (ξ − k)|2 e−2πin(ξ−k) dξ

b G(ξ) e−2πinξ dξ = G(n) = δ(n),

the interchange of summation and integration allowed by Fubini’s Theorem. Therefore {Tk g}k∈Z is orthonormal.

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9.4.20 (a) The image of W(ψ, a, b) under the Fourier transform is  −n/2 2πibka−n ξ b −n ξ) a e ψ(a k,n∈Z  n/2 2πibkan ξ n b ξ) = a e ψ(a . k,n∈Z

c a, b) = W(ψ,

Since the Fourier transform is unitary, our goal is to show that this collection is a tight frame for L2 [0, ∞). Note that, with n fixed,  n/2 1/2 2πibkan ξ a b e k∈Z

is an orthonormal basis for L2 (I) where I is any interval of length a−n b−1 . Let f be any function in Cc (R). For simplicity let us take c = 1, but the same argument applies to any positive c. Let d = 1 + b−1 , so ψb ∈ L2 (R) is supported within [1, d], which is an interval of length b−1 . The dilated b n ξ) belongs to L2 (In ) where In = [a−n , a−n c], an interval of function ψ(a −n b n ξ) length a (d − 1) = a−n b−1 . Since f is bounded, the product f (ξ) · ψ(a also belongs to L2 (In ). Applying the Plancherel Equality, we therefore have Z ∞

b n ξ)|2 dξ = b n ξ) |f (ξ) ψ(a

f (ξ) ψ(a

2 L (In )

0

E X D b n ξ), an/2 b1/2 e2πibkan ξ = f (ξ) ψ(a

L2 (In )

k∈Z

X Z = a b n

k∈Z

In

X Z n = a b k∈Z

2 −2πibkan ξ n b f (ξ) ψ(a ξ) e

∞

0

2 −2πibkan ξ n b f (ξ) ψ(a ξ e .

E X D b n ξ) e2πibkan ξ f (ξ), an/2 ψ(a

= b

k∈Z

L2 [0,∞)

Hence, using Tonelli’s Theorem to interchange the sum and integral, E X D b n ξ) e2πiban ξ f (ξ), an/2 ψ(a

k,n∈Z

= b

−1

XZ

n∈Z

= b−1

Z

0

∞

0

∞

b n ξ)|2 dξ |f (ξ) ψ(a

|f (ξ)|2

X

n∈Z

L2 [0,∞)

 b n ξ)|2 dξ ψ(a

2

2

2 .

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= Ab−1

Z

0

∞

|f (ξ)|2 dξ

= b−1 kf k2L2[0,∞) . for all f ∈ Cc (R). Since Cc (R) is dense in L2 (R), it follows (exercise) that W(ψ, a, b) is a tight frame with frame bound Ab−1 .

Remark: We can choose a, b, and ψ so that ψb is as smooth as we like, even infinitely differentiable. We just need to make sure that the dilations of the interval [c, c + b−1 ] have nontrivial overlaps, i.e., we need to choose a, b so that ac < c + b−1 .

(b) Parseval frame means a tight frame where the constant in the tight frame equality is 1. Once we have a tight frame, we can obtain a Parseval frame simply by rescaling the elements of a frame. We take a = 2 and b = 1. Define ψ1 ∈ L2 (R) by   0, x < 1/2,       linear, 1/2 ≤ x ≤ 3/4,   ψb1 (ξ)2 = 1, 3/4 ≤ x ≤ 1,     linear, 1 ≤ x ≤ 3/2,     0, x > 3/2.

P If we set c = 1/2 then we have c + b−1 = 3/2. Also, n∈Z |ψb1 (ξ)|2 = 1 on 2 (R) by part (a). (0, ∞), so W(ψ1 ) is a Parseval frame for H+ 2 b Now let ψ2 ∈ L (R) be the function such that ψb2 (ξ) = ψ(−ξ). Then 2 W(ψ1 ) ∪ W(ψ2 ) is a Parseval frame for L (R). Explicitly, we just have to set ψ2 (x) = ψ(−x), for then Z Z −2πiξx b b ψ2 (ξ) = ψ(−x) e dx = ψ(x) e2πiξx dx = ψ(−ξ).

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