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English Pages 348 [349] Year 2014
Loukas Grafakos
Classical Fourier Analysis, Third Edition Solutions of all the exercises – Monograph – January 22, 2014
Springer
v
I would like to express my deep gratitude to the following people who helped in the preparation of the solutions of the books Classical Fourier Analysis, 3rd edition, GTM 249 and Modern Fourier Analysis, 3rd edition, GTM 250. Mukta Bhandari, Jameson Cahill, Santosh Ghimire, Zheng Hao, Danqing He, Nguyen Hoang, Sapto Indratno, Richard Lynch, Diego Maldonado, Hanh Van Nguyen, Peter Nguyen, Jesse Peterson, Sharad Silwal, Brian Tuomanen, Xiaojing Zhang, I remain solely responsible for any error contained in the enclosed solutions.
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1. L p Spaces and Interpolation
Section 1.1. L p and Weak L p Exercise 1.1.1 Suppose f and fn are measurable functions on (X, µ). Prove that (a) d f is right continuous on [0, ∞). (b) If  f  ≤ lim infn→∞  fn  µa.e., then d f ≤ lim infn→∞ d fn . (c) If  fn  ↑  f , then d fn ↑ d f . Hint: Part (a): Let tn be a decreasing sequence of positive numbers that tends to zero. Show that d f (α0 + tn ) ↑ dTf (α0 ) using a convergence theorem. Part (b): Let E = {x ∈ X :  f (x) > α} and En = {x ∈ X :  fn (x) > α}. Use that µ ∞ n=m En ≤ S∞ T∞ lim inf µ(En ) and E j m=1 n=m En µa.e. n→∞
Solution. (a) For all α > 0 let Eα = {x ∈ X :  f (x) > α}. Fix α0 ≥ 0 and let tn be a decreasing sequence of positive numbers that tends to 0. Observe that Eα0 +tn j Eα0 +tn+1 for every n. Therefore χEα0 +tn ↑ χEα0 as n → ∞, so by the Lebesgue monotone convergence theorem it follows that µ(Eα0 +tn ) ↑ µ(Eα0 ) which establishes the desired conclusion. ∞ (b) First note that E ⊆ ∪∞ m=1 ∩n=m En µ−a.e., so we have ∞ d f (α) = µ(E) ≤ µ(∪∞ m=1 ∩n=m En ) ≤ lim inf µ(En ) = lim inf d fn (α). n→∞
n→∞
(c) Let En and E be as above.The clearly we have En ⊆ En+1 and limn→∞ En = E, so the conclusion follows from the Lebesgue monotone convergence theorem as in part (a).
Exercise 1.1.2 (H¨older’s inequality) Let 0 < p, p1 , . . . , pk ≤ ∞, where k ≥ 2, and let f j be in L p j = L p j (X, µ). Assume that 1 1 1 = +···+ . p p1 pk (a) Show that the product f1 · · · fk is in L p and that
f1 · · · fk p ≤ f1 p · · · fk p . L L 1 L k
1
2
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(b) When no p j is infinite, show that if equality holds in part (a), then it must be the case that c1  f1  p1 = · · · = ck  fk  pk µa.e. for some c j ≥ 0.
−1 q . For r < 0 and g > 0 almost everywhere, define kgkLr = g−1 Lr . Show that if g is strictly (c) Let 0 < q < 1 and q0 = q−1 0
positive µa.e. and lies in Lq and f is measurable such that f g belongs to L1 , we have
f g 1 ≥ f q g q0 . L
L
L
Solution. (a) We may assume that all p j < ∞, since otherwise, one can factor from the k f1 · · · fk kL p the L∞ norm of the f j ’s with p j = ∞ and the inequality reduces to the remaining p j ’s which are not infinite. If one k f j kL p1 is 0 or ∞, then the conclusion is obvious. Let fj k f j kL p1
Fj = The convexity of et implies the inequality 1 p
1 p
a1 1 · · · ak k ≤ when 1 < p j < ∞ and p11 + · · · +
1 pk 2
= 1. Using this inequality we obtain 1
1
(F p1 ) p1 · · · (F pk ) pk ≤ Integrating and using p = 1 as follows:
1 p1
a1 ak +···+ p1 pk
 fk  pk  f1  p1 +···+ . p1 k f1 kL p1 pk k fk kL pk
+ · · · + p1 = 1, we obtain k f1 · · · fk kL1 ≤ k f1 kL p1 · · · k fk kL pk . For p 6= 1 reduce the inequality to the case k
1 k f1 · · · fk kL p = k f1 · · · fk  p kLp1 ≤ k f1  p k
p1 L p
· · · k fk  p k
pk L p
1
p
= k f 1 k L p1 · · · k f k k L pk
using the fact that 1/p1 + · · · + 1/pk = 1/p is equivalent to (p1 /p)−1 + · · · + (pk /p)−1 = 1. (b) We note that equality in the inequality 1 p
1 p
a1 1 · · · ak k ≤
a1 ak +···+ p1 pk
holds if and only if a1 = · · · = ak . Indeed, this inequality is a restatement of the fact that t 7→ et is convex. And since et is strictly convex, it follows that equality holds in the preceding inequality if and only if all the a j ’s are equal to each other. Using this a j =  f j  p j /k f j kL p j , we obtain the claim with c j = k f j k−1 p . L j (c) We have k f kLq = k f gg−1 kLq ≤ k f gkL1 kg−1 kLq0  = k f gkL1 kgk−1 , q0 L
1 q
since 1 = +
1 q0
implies
1 q
1 1
= +
1 q0  .
The conclusion follows.
Exercise 1.1.3 Let (X, µ) be a measure space. (a) If f is in L p0 (X, µ) for some p0 < ∞, prove that
lim f L p = f L∞ .
p→∞
(b) (Jensen’s inequality) Suppose that µ(X) = 1. Show that
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f
≥ exp
Lp
Z
log  f (x) dµ(x)
X
for all 0 < p < ∞. (c) If µ(X) = 1 and f is in some L p0 (X, µ) for some p0 > 0, then Z
log  f (x) dµ(x) lim f L p = exp p→0
X
with the interpretation e−∞ = 0. (p−p )/p p /p Hint: Part (a): If 0 < k f kL∞ < ∞, use that k f kL p ≤ k f kL∞ 0 k f kL0p0 to obtain lim sup p→∞ k f kL p ≤ k f kL∞ . Conversely, let (p−p0 )/p p /p Eγ = {x ∈ X :  f (x) > γk f kL∞ } for γ in (0, 1). Then µ(Eγ ) > 0, k f kL p0 (Eγ ) > 0, and k f kL p ≥ γk f kL∞ k f kL0p0 (Eγ ) , 1
hence lim inf p→∞ k f kL p ≥ γk f kL∞ . If k f kL∞ = ∞, set Gn = { f  > n} and use that k f kL p ≥ k f kL p (Gn ) ≥ nµ(Gn ) p to obtain R R lim inf p→∞ k f kL p ≥ n. Part (b) is a direct consequence of Jensen’s inequality X log h dµ ≤ log X h dµ . Part (c): Fix a sequence 0 < pn < p0 such that pn ↓ 0 and define hn (x) =
1 1 ( f (x) p0 − 1) − ( f (x) pn − 1). p0 pn
Use that 1p (t p − 1) ↓ logt as p ↓ 0 for all t > 0. The Lebesgue monotone convergence theorem yields R 1 R pn X pn ( f  − 1) dµ ↓ X log  f  dµ, where the latter could be −∞. Use Z
log  f  dµ
exp
≤
Z
X
to complete the proof.
pn
 f  dµ
1 pn
≤ exp
Z X
X
1 ( f  pn − 1) dµ pn
R
X hn dµ
↑
R X
h dµ, hence
Solution. (a) If k f kL∞ < ∞ use that Z
p
1
Z
p
 f  dµ
p0
f f
=
p−p0
1
p
dµ
X
X
p−p0
p0
≤ k f kL∞p k f kLpp0
from which it follows that lim sup p→∞ k f kL p ≤ k f kL∞ . Conversely, let Eγ = {x ∈ X :  f (x) > γk f kL∞ } for γ ∈ (0, 1). Then µ(Eγ ) > 0 and the integral of  f  p0 over Eγ is positive. For p > p0 we have k f kL p ≥
Z
1
p
p
 f  dµ
≥γ
p−p0 p L∞
p−p0 p
kfk
Eγ
Z
p0
p0 p
 f  dµ
Eγ
and from this it follows that lim inf p→∞ k f kL p ≥ γk f kL∞ . Since γ < 1 was arbitrary, lim inf p→∞ k f kL p ≥ k f kL∞ . Now suppose that k f kL∞ = ∞. Then if Gn = { f  > n}, we must have µ(Gn ) > 0 for all n = 1, 2, . . . . Then 1
k f kL p ≥ k f kL p (Gn ) ≥ nµ(Gn ) p and so lim inf p→∞ k f kL p ≥ n. Thus lim inf p→∞ k f kL p = ∞ = k f kL∞ . (b) By Jensen’s inequality we have that if φ us concave, then Z Z φ gdµ ≥ φ ◦ gdµ, g ≥ 0, a.e. X
Hence
X
Z log X
Let h =  f  p ,then we have
hdµ
≥
Z
log(h)dµ. X
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Z
1
p
p
 f  dµ
log
≥
X
Z
log( f )dµ ⇒ k f k
X
Z
Lp
≥ exp(
log( f ))dµ .
X
(c) By part (b) we have lim sup k f kL p ≥ exp
Z
log  f  dµ .
X
p→0
Fix a sequence 0 < pn < p0 such that pn ↓ 0. By the Lebesgue monotone convergence theorem we know that Z X
 f  p0 − 1  f  pn − 1 − dµ ↑ p0 pn
since f ∈ L p0 . Therefore Z X
ae ≤ ea , ∀a > 0, so
R  f  pn X
pn
R  f  pn −1
≤ exp(
X
Z p→0
X
X
 f  pn − 1 dµ ↓ pn
Z
log  f  dµ.
X
). Therefore
pn
lim inf
 f  p0 − 1 − log( f ) dµ, p0
Z
 f p dµ ≤ exp p
Z
log  f  dµ
X
≤ lim sup k f kL p , p→0
and the conclusion follows.
Exercise 1.1.4 Let a j be a sequence of positive reals. Show that θ (a) ∑∞j=1 a j ≤ ∑∞j=1 aθj , for any 0 ≤ θ ≤ 1. θ (b) ∑∞j=1 aθj ≤ ∑∞j=1 a j , for any 1 ≤ θ < ∞. θ (c) ∑Nj=1 a j ≤ N θ −1 ∑Nj=1 aθj , when 1 ≤ θ < ∞. θ (d) ∑Nj=1 aθj ≤ N 1−θ ∑Nj=1 a j , when 0 ≤ θ ≤ 1. Solution. (a) Consider the numbers
aj ∞ ∑k=1 ak
which are less than 1. Then we have
θ
aj
≥
∑∞ k=1 ak
aj ∞ ∑k=1 ak
since 0 ≤ θ ≤ 1. Summing over j we obtain ∞
∑
j=1
aj ∑∞ k=1 ak
θ ≥ 1.
This is just the claimed inequality (b) This is proved in a way similar to that in (a), with the only difference being that ≥ is replaced with ≤. (c) We have N N N N θ −1 1 1 θ −1 ∑ a j ≤ ∑ 1 θ ∑ aθj θ = N θ ∑ aθj θ j=1
therefore
j=1
j=1
N
∑ aj
j=1
θ
j=1
N
≤ N θ −1 ∑ aθj . j=1
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(d) Here we have N
N
∑ aθj ≤ ∑ 1
N
1−θ
N
θ
= N 1−θ
∑ aj
j=1
j=1
j=1
1
∑ (aθj ) θ )
θ
j=1
and this is the claimed inequality.
Exercise 1.1.5 Let { f j }Nj=1 be a sequence of L p (X, µ) functions. (a) (Minkowski’s inequality) For 1 ≤ p ≤ ∞ show that
N
∑ f j p ≤ L j=1
N
∑ f j L p .
j=1
(b) (Reverse Minkowski inequality) For 0 < p < 1 and f j ≥ 0 prove that N
N
∑ f j L p ≤ ∑ f j L p . j=1
j=1
(c) For 0 < p < 1 show that
N 1−p
∑ f j p ≤ N p L
N
∑ f j L p .
j=1
j=1
1−p p in part (c) is best possible. (d) The constant N Hint: Part (c): Use Exercise 1.1.4 (c). Part (d): Take { f j }Nj=1 to be characteristic functions of disjoint sets with the same measure.
Solution. (a) The proof is as follows: Z
 f  f + g
p−1
dµ ≤
Z
1 Z p
p
 f  dµ
X
X
p−1 p
p
 f + g dµ
,
X
and likewise for g, therefore Z
Z
p
 f + g dµ ≤
p
1
Z
p
 f  dµ
X
X
g dµ
+
1 ! Z p
p
X
p−1 p
p
 f + g dµ
,
X
which implies k f + gkL p ≤ k f kL p + kgkL p . The conclusion for more than two functions follows by induction. (b) Notice that if 0 < p < 1, then in view of Exercise 1.1.2 (c), the first inequality in part (a) is reversed, i.e., we have Z
f ( f + g)
p−1
dµ ≥
Z
X
p
1 Z p
p
p−1 p
( f + g) dµ
f dµ X
,
X
and likewise for g, so adding yields the conclusion. (c) Using Exercise 1.1.4 (c) we obtain Z N X
∑ f j p dµ
!1
p
N
≤
Z
∑
j=1 X
j=1
≤N
1−p p
!1
p
 f j  dµ
N
∑
p
j=1
Z X
p
 f j  dµ
1
p
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=N
N
1−p p
∑ k f j kL p
j=1
(d) Let f j = χA j be characteristic functions of disjoint sets with the same measure. We have N
1
k ∑ f j kL p = (NA1 ) p j=1
and N
1−p p
N
∑ k f j kL p = N
1−p p
1
1
NA1  p = (NA1 ) p .
j=1
Therefore N
1−p p
is the best possible constant in the inequality.
Exercise 1.1.6 (a) (Minkowski’s integral inequality) Let (X, µ) and (T, ν) be two σ finite measure spaces and let 1 ≤ p < ∞. Show that for every nonnegative measurable function F on the product space (X, µ) × (T, ν) we have Z Z T
p 1 1 Z Z p p p F(x,t) dµ(x) dν(t) ≤ F(x,t) dν(t) dµ(x) ,
X
X
T
(b) State and prove an analogous inequality when p = ∞. (c) Prove that when 0 < p < 1, then the preceding inequality is reversed. (d) (Y. Sawano) Consider the example X = T = [0, 1], µ is counting measure, ν is Lebesgue measure, F(x,t) = 1 when x = t and zero otherwise. What is the relevance of this example with the inequalities in (a) and (b)? Hint: Part (a) Split the power p as 1 + (p − 1) and apply H¨older’s inequality with exponents p and p0 . Part (b) Let p → ∞ on subsets of X with finite measure. Solution. (a) Since X and T are σ finite measure spaces, the use of Fubini below is justified. We have
p
Z
F(x, · ) dµ(x)
X
p
Z Z
= T
L (dν)
X
Z Z
= X
≤
Z X
p−1 Z F(x,t) dµ(x) F(x,t) dµ(x) dν(t) X
F(x,t)
p−1 F(x,t) dµ(x) dν(t) dµ(x)
Z X
T
p−1
Z
p0 kF(x, · )kL p (dν) F(x, · ) dµ(x) dµ(x) L (dν) X
Z
p−1
= F(x, · ) dµ(x)
X
Z
L p (dν) X
kF(x, · )kL p (dν) dµ(x),
and the conclusion follows by dividing both sides by the first factor above. (b) We now turn to the case p = ∞. Since X is a σ finite measure space, it is equal to the union of Xn where µ(Xn ) < ∞ for each n = 1, 2, . . . . Then consider the measure spaces Xn with measures µn (A) = µ(A)/µ(Xn ) for A j X. These are probability spaces and the L p (Xn ) norms are increasing. Then for p < ∞ we have
Z
Z
F(x,t)dν(t)
F(x,t) p ≤ dν(t) .
L (Xn ,µn ) T
L p (Xn ,µn )
T
Now let p → ∞ and use Exercise 1.1.3 (a) on the LHS and the Lebesgue Monotone Convergence Theorem on the RHS to obtain
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Z
F(x,t)dν(t)
T
≤
L∞ (Xn )
Z T
F(x,t) ∞ dν(t) . L (Xn )
Estimate the L∞ (Xn ) norm on the RHS by the L∞ (X) norm. Then use the simple fact that on a σ finite measure space we have that kgkL∞ (Xn ) ↑ kgkL∞ (X) , which is valid even if kgkL∞ (X) = ∞, to replace Xn by X on the LHS of the inequality. (c) Using Exercise 1.1.2 (c) we see that the unique inequality in the alignment in (a) can be reversed when 0 < p < 1 and F(x, · ) is not zero νa.e. (d) When X = T = [0, 1], µ =counting measure of X and ν = be Lebesgue measure on T , for F(x,t) = 1 if x = t and zero if x 6= t. We will show that p 1/p F(x,t) dµ(x) dν(t) =1
Z Z T
X
1/p F(x,t) p dν(t) dµ(x) = 0
Z Z X
T
and hence the Minkowski inequality p 1/p Z Z 1/p p F(x,t) dµ(x) dν(t) ≤ F(x,t) dν(t) dµ(x)
Z Z T
X
X
T
fails for this case. To see these assertions, first we observe Z
Z
F(x,t) dµ(x) = [0,1]
X
χ{t} (x) dµ(x)
Z
= [0,1]\{t}
Z
χ{t} (x) dµ(x) +
Z
=
{t}
χ{t} (x) dµ(x)
Z
0 dµ(x) +
1 dµ(x) {t}
[0,1]\{t}
= µ{t} = 1 for all t ∈ [0, 1]. Therefore, Z Z T
p 1/p Z F(x,t) dµ(x) dν(t) =
1/p 1 dt
= 1,
[0,1]
X
and this is also valid when p = ∞. Meanwhile, Z
1/p Z F(x,t) p dν(t) = [0,1]
T
1/p χ{x} (t) dx
= {x}1/p = 0,
even when p = ∞, and hence Z Z X
1/p Z F(x,t) p dν(t) dµ(x) = 0 dµ(x) = 0.
T
X
Exercise 1.1.7 Let f1 , . . . , fN be in L p,∞ (X, µ). (a) Prove that for 1 ≤ p < ∞ we have N
N
∑ f j p,∞ ≤ N ∑ f j p,∞ . L L j=1
(b) Show that for 0 < p < 1 we have
j=1
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N 1
∑ f j p,∞ ≤ N p L
N
∑ f j L p,∞ .
j=1
j=1
Hint: Use that µ({ f1 + · · · + fN  > α}) ≤ ∑Nj=1 µ({ f j  > α/N}) and Exercise 1.1.4(a) and (c). Solution. (a) We have 1
k f kL p,∞ = sup (αd fp (α)), α>0
therefore
! n N o  ∑ f j > α µ ≤ j=1
N
∑µ
j=1
n α o ≤  f j > N
we have αµ
! n N o  ∑ f j > α ≤N
1 p
j=1
N p k f j kLp p,∞ αp j=1 N
∑
!1
p
N
k f j kLp p,∞
∑
N
≤ N ∑ k f j kL p,∞ ,
j=1
j=1
since p ≥ 1which implies the desired result. (b) Similarly, we have αµ
! n N o  ∑ f j > α ≤N
1 p
j=1
N
∑
!1
p
k f j kLp p,∞
1
≤Np
N
∑ k f j kL p,∞
j=1
j=1
since 0 < p < 1 via Exercise 1.1.4 (c), so we obtain the claimed inequality.
Exercise 1.1.8 Let 0 < p < ∞. Prove that L p (X, µ) is a complete quasinormed space. This means that every quasinorm Cauchy sequence is quasinorm convergent. Hint: Let fn be a Cauchy sequence inL p . Pass to a subsequence {ni }i such that k fni+1 − fni kL p ≤ 2−i . Then the series f = p fn1 + ∑∞ i=1 ( f ni+1 − f ni ) converges in L . Solution. Let fn be a Cauchy sequence in L p (p > 1), which means that given ε there is an N such that k fn − fm kL p < ε if n, m > N. Therefore we can choose n1 such that k fn1 − fn kL p < ε2 for all n > n1 . We can choose a subsequence { fn j } of { fn } such that k fn j − fn j+m kL p < 2εj ∀m ≥ 0. Let f = fn1 + ∑∞j=1 ( fn j+1 − fn j ). We want to prove that f ∈ L p , so we need only to prove that ∞
∑  fn j+1 − fn j  ∈ L p .
j=1
By Fatou’s lemma we know that Z
!p
∞
∑  fn j+1 − fn j 
dµ ≤ lim inf k→∞
j=1
so Z
∞
∑  fn j+1 − fn j 
j=1
We can see that
! p ! 1p dµ ≤ lim inf k→∞
Z
k
Z
!p
k
∑  fn j+1 − fn j  ! p ! 1p
∑  fn j+1 − fn j 
k
dµ ≤ lim inf ∑ k fn j+1 − fn j kL p < ε . k→∞ j=1
j=1
∞
k f − fnk kL p = ∑ ( fn j+1 − fn j ) j=k
dµ ,
j=1
Lp
0, x ∈ X : fn (x) → f (x)} j {x ∈ X :  fn (x) − f (x) < ε . m=1 n=m
Solution. Observe that for all ε > 0, {x ∈ X : fn (x) → f (x)} j
∞ \ ∞ [
{x ∈ X :  fn (x) − f (x) < ε}.
m=1 n=m
Since the second set has full measure, its complement must have measure zero which implies that lim µ(∪∞ n=m {x :∈ X :  f n (x) − f (x) > ε}) = 0.
m→∞
The latter implies convergence in measure.
Exercise 1.1.10 Given a measurable function f on (X, µ) and γ > 0, define fγ = f χ f >γ and f γ = f − fγ = f χ f ≤γ . (a) Prove that ( d f (α) when α > γ, d fγ (α) = d f (γ) when α ≤ γ, ( 0 when α ≥ γ, d f γ (α) = d f (α) − d f (γ) when α < γ. (b) If f ∈ L p (X, µ) then Z
p
fγ p = p L
γ
∞
α p−1 d f (α) dα + γ p d f (γ),
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Z
γ p
f p = p L
Z γ
 f  p dµ = p
Z δ
0
γ p and fγ is in Lq (X, µ) for any q < p. Thus L p,∞ j L p0 + L p1 when 0 < p0 < p < p1 ≤ ∞. Solution. (a) For α > γ, we have x ∈ X :  fγ (x) > α = {x ∈ X :  f (x) > α} =⇒ d fγ (α) = d f (α) and for α ≥ γ we have {x ∈ X :  f γ (x) > α} = 0/ =⇒ d fγ (α) = 0 . For α ≤ γ, we have x :  fγ (x) > α = {x :  f (x) > γ} =⇒ d fγ (α) = d f (γ) and for α < γ we have {x :  f γ (x) > α} = {x : γ ≥  f (x) > α} =⇒ d f γ (α) = d f (α) − d f (γ) . Thus we proved the desired results. (b) We have k fγ kLp p = p
Z ∞
k f γ kLp p = p
Z γ
and also
Z
0
 f  p dµ =p
γγ Z ∞
=p0
α p0 −1 d f (α)dα + γ p0 d f (γ)
γ
Z ∞
=p0
α p d f (α)α p0 −p−1 dα + γ p0 d f (γ)
γ
p0 γ p0 −p + k f kLp p,∞ γ p0 −p p − p0 pγ p0 −p =k f kLp p,∞ < ∞. p − p0 ≤k f kLp p,∞
Therefore fγ ∈ L p0 . Similarly f γ ∈ L p1 (p < p1 ≤ ∞). And we can get the desired results.
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Exercise 1.1.11 Let (X, µ) be a measure space and let E be a subset of X with µ(E) < ∞. Assume that f is in L p,∞ (X, µ) for some 0 < p < ∞. (a) Show that for 0 < q < p we have Z q q p  f (x)q dµ(x) ≤ µ(E)1− p f L p,∞ . p−q E (b) Conclude that if µ(X) < ∞ and 0 < q < p, then L p (X, µ) j L p,∞ (X, µ) j Lq (X, µ).
p Hint: Part (a): Use µ E ∩ { f  > α} ≤ min µ(E), α −p f L p,∞ . Solution. 1 (a) Let B = µ(E)− p k f kL p,∞ . We have Z
 f q dµ = q
Z ∞
α q−1 µ(E ∩ { f  > α})dα
(by Proposition 1.1.4)
0
E
≤q
Z ∞ 0
≤q
Z B
α q−1 min(µ(E), α −p k f kLp p,∞ )dα α q−1 µ(E)dα + q
0
Z ∞ B
α q−1 α −p k f kLp p,∞ dα
q k f kLp p,∞ Bq−p (since q < p) p−q q q q = µ(E)1− p k f kqL p,∞ + µ(E)1− p k f kqL p,∞ q− p p q 1− qp = k f kL p,∞ . µ(E) p−q = µ(E)Bq +
(b) The first inclusion is given by proposition 1.1.6. To see the second inclusion let f ∈ L p,∞ (X, µ) and apply part (a) with E = X to see that Z p p k f kqLq =  f (x)q dµ(x) ≤ µ(X)1− q k f kqL p,∞ < ∞ p − q X since µ(X) < ∞ and k f kL p,∞ < ∞. Thus, f ∈ Lq (X, µ).
Exercise 1.1.12 (Normability of weak L p for p > 1) Let (X, µ) be a σ finite measure space and let 0 < p < ∞. Pick 0 < r < p and define f
L p,∞
=
1
1
µ(E)− r + p
sup
Z
 f r dµ
E
0 α} ∩ Xk . Solution. (a) Let E j X such that µ(E) < ∞ and let f ∈ L p,∞ (X, µ). From Exercise 1.1.11 (a) we have Z
 f r dµ
1
r
≤
E
p p−r
1 r
1
1
µ(E) r − p k f kL p,∞ .
Therefore k f k
L p,∞
≤
sup
− 1r + 1p
µ(E)
0 α}, and Eα,N = { f  > α} ∩ XN , R then we clearly have Eα,N  f r dµ ≥ α r µ(Eα,N ) and µ(Eα,N ) < ∞. So f
L p,∞
=
− 1r + 1p
sup
Z
µ(E)
r
1 r
 f  dµ
E
00
If X = {1, 2}, with µ({1}) = 1, µ({2}) = ∞, then clearly X is not σ finite. It is quite easy to check that the function f (x) = 1 which is identically equal to 1 on the whole space does not lie in weak L p with the classical definition, i.e., k1kL p,∞ = ∞, but 1L p,∞ = 1. (c) Choose 0 < ε < p and let r = min(p − ε, 1). For f , g ∈ L p,∞ (X, µ) define d( f , g) = f − grL p,∞ . The fact that d( f , g) = d(g, f ) is obvious from the definition. Now suppose d( f , g) = 0, then by part (b) we know that k f − gkL p,∞ = 0 which means f = g µa.e. by equation (1.1.12) from the text. It remains to show that d satisfies the triangle inequality. For f , g, h ∈ L p,∞ (X, µ) we have d( f , g) =
sup
µ(E)
r p −1
sup
r
µ(E) p −1
sup
r
µ(E) p −1
sup 0 0 let KR = KχB(0,R) , where B(x, R) is the ball of radius
R centered at x. Observe that for x ≤ (1−ε)N, we have B(0, Nε) j B(x, N); R R thus Rn χN (x − y)KNε (y) dy = Rn KNε (y) dy = KNε L1 . Then
KNε ∗ χN p p
K ∗ χN p p
p L (B(0,(1−ε)N) L
p ≥ ≥ KNε L1 (1 − ε)n . p
kχN kL p χN L p
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23
Let N → ∞ first and then ε → 0.
Solution. It is trivial that kT kL p →L p ≤ kKkL1 . To see the converse inequality, fix 0 < ε < 1 and let N be a positive integer. Let χN = χB(0,N) and for any R > 0 let KR = KχB(0,R) , where B(x, R) is the ball of radius R centered at x. Observe that for x ≤ (1 − ε)N, we have B(x, N) ∩ B(0, Nε) = B(0, Nε) thus Z Rn
χN (x − y)KNε (y) dy =
Z
KNε (y) dy = kKNε kL1 .
Rn
kKNε ∗ χN kLp p (B(0,(1−ε)N) kKNε kLp1 B(0, (1 − ε)N) kK ∗ χN kLp p ≥ ≥ , B(0, N) kχN kLp p kχN kLp p Now let N → ∞ first and then ε → 0 to obtain the required conclusion.
Exercise 1.2.10 On the multiplicative group (R+ , dtt ) let T ( f ) = f ∗ K, where K is a positive L1 function and f is in L p , 1 ≤ p ≤ ∞. Prove that the operator norm of T : L p → L p is equal to the L1 norm of K. Deduce that the constants p/(p − 1) and p/b are sharp in Exercises 1.2.7 and 1.2.8. Hint: Adapt the idea of Exercise 1.2.9 to this setting. Solution. It is obvious that kT ( f )kL p ≤ kKkL1 k f kL p . Let B(x, N) = ( Nx , Nx), χN = χB(1,N) , and KR = KχR . Also let R, N > 1, then for x ∈ B(1, N 1−ε ), B(x, N ε ) j B(1, N). We have Z Z x χN KN ε (y)dy = KN ε (y)dy = kKN ε kL1 y R+ R+ for all x ∈ B(1, N 1−ε ). As a result kKN ε ∗ χN kLp p (B(1,N 1−ε )) kK ∗ χN kLp p ≥ ≥ kKN ε kLp1 (1 − ε) kχN kLp p kχN kLp p Let N → ∞ and then ε → 0, we will get the desired result.
Exercise 1.2.11 1 Qk (t) dt = 1 for all k = 1, 2, . . . . Let Qk (t) = ck (1 − t 2 )k for t ∈ [−1, 1] and zero elsewhere, where ck is chosen such that −1 √ (a) Show that ck < k. (b) Use part (a) to show that {Qk }k is an approximate identity on R as k → ∞. (c) Given a continuous function f on R that vanishes outside the interval [−1, 1], show that f ∗ Qk converges to f uniformly on [−1, 1] as k → ∞. function on [−1, 1] can be approximated uniformly by polynomials. (d) (Weierstrass) Prove that every continuous R Hint: Part (a): Estimate the integral t≤k−1/2 Qk (t) dt from below using the inequality (1 − t 2 )k ≥ 1 − kt 2 for t ≤ 1. Part (d): Consider the function g(t) = f (t) − f (−1) − t+1 2 ( f (1) − f (−1)).
R
Solution. (a) We begin with 1 = cj
Z 1 −1
(1 − t 2 ) j dt >
Z t< √1 j
(1 − t 2 ) j dt
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≥2
√1
Z
j
(1 − jt 2 )dt
0
2 2 =√ − √ j 3 j 4 = √ 3 j 1 >√ j (b) We have
R1
−1 Q j (t)dt
= 1 and Q j (t) ≥ 0 for all t ∈ (−1, 1), so it satisfies the condition (i) and (ii). Also Z t>δ
C j (1 − t 2 ) j dt
0 there exists δ > 0 such that  f (x − y) − f (x) < ε Let  f (x) < A ∀x. We have ( f ∗ Q j )(x) − f (x) ≤ ≤
Z 1 −1
j (1 − δ 2 ) j → 0
∀y < δ and j large enough such that
R
t>δ
Q j (t)dt < ε.
 f (x − y) − f (x)Q j (y)dy
Z y 0, T ( f + g) ≤ K(T ( f ) + T (g)) , for all f , g in the domain of T . Prove that in this case, the constant A in (1.3.7) can be taken to be K times the constant in (1.3.8). Solution. Define f0α , f1α as in the proof of Theorem 1.3.2. We can see that dT ( f ) (α) ≤ dT ( f0α ) (α/2K) + dT ( f1α ) (α/2K). Just as in the proof of Theorem 1.3.2 we have the estimate kT ( f )kLp p
≤ p(2KA0 )
p0
Z ∞
α
p−1
α
−p0
Z
p0
 f >δ α
0
 f (x) dµ(x)dα + p(2KA1 )
p1
Z ∞
α
p−1
α
−p1
Z
0
 f (x) p1 dµ(x)dα.
 f ≤δ α
The only difference is the presence of the constant K here, and the constant in the final estimate is have a similar estimate except the final constant is 1 1 p0 p0 p0 p p 1− p0 p p (KA1 )1− p (KA0 ) p = 2K A1 p A0p KA = 2 p − p0 p − p0
instead of A as in the proof of Theorem 1.3.2.
Exercise 1.3.2 Let (X, µ), (Y, ν) be two σ finite measure spaces. Let 1 < p < r ≤ ∞ and suppose that T be a sublinear operator defined on the space L p0 (X) + L p1 (X) and taking values in the space of measurable functions on Y . Assume that T maps L1 (X) to L1,∞ (Y ) with norm A0 and Lr (X) to Lr (Y ) with norm A1 . Let 0 < p0 < p1 ≤ ∞. Prove that T maps L p to L p with norm at most − 1p
8 (p − 1)
1−1 p r 1− 1r
1− 1p 1− 1r
A0
A1
.
p+1 Hint: First interpolate between L1 and Lr using Theorem 1.3.2 and then interpolate between L 2 and Lr using Theorem 1.3.4. Solution. We interpolate between L1 and Lr using Theorem 1.3.2. Since 1 < of T from L
p+1 2
to itself 2
We now interpolate between L
p+1 2
p+1 2 p+1 2 −1
+
p+1 2 r − p+1 2
2 p+1
p+1 2
< p < r we obtain the following bound for the norm
2 1 p+1 − r 1 1− r
A0
2 1− p+1 1 1− r
A1
.
and Lr using Theorem 1.3.4. We obtain the following bound
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27 1
1
2 1 2 1 2 # p−r " 1− p+1 p+1 − p 2 p+1 − r 2 −1 p+1 p+1 p+1 2 − 1r p+1 r 1− 1 1− 1 2 + A0 r A1 r A1p+1 2 p+12 p+1 r− 2 2 −1
which equals 1
1
1−1 p−r " 1− 1p 2 # 2 p r p+1 p+1 p+1 − 1r 1 p+1 1− 1− 1 2 A0 r A1 r . + 2 p+12 p+1 r− 2 2 −1
Since 1 < p < r we have that 1 1 p−r 2 1 p+1 − r
(1 − λ )α}. Part (c): Write f = f0 + f1 , where f0 = f − δ α when f ≥ δ α and zero otherwise. Use that {T ( f ) > α} ≤ {T ( f0 ) > (1 − λ )α} + {T ( f1 ) > λ α} and optimize over δ > 0. Solution. (a) If the conclusion is valid for f ≥ 0, then since kT ( f )kL p ≤ kT ( f )kL p ≤ kT kL p →L p k f kL p the conclusion is valid for all f ∈ L p . We may therefore assume that f ≥ 0. Given α > 0, let f = f0 + f1 , where f0 = f − λ α/A1 if f ≥ λ α/A and 0 otherwise (0 < λ < 1). It’s easy to check that S1 f0 ∈ L p0 and f1 ∈ L p1 . Moreover {T ( f ) > α} j {T ( f0 ) > (1 − λ )α} {T ( f1 ) > λ α} = {T ( f0 ) > (1 − λ )α} since kT ( f1 )kL∞ ≤ A1 k f1 kL∞ ≤ λ α. So we have kT ( f )kLp p =p ≤p
Z ∞
α p−1 dT ( f ) (α)dα
0
Z ∞
α p−1 µ({T ( f0 ) > (1 − λ )α})dα
0
≤
p λ
Z ∞ 0
pA0 = 1−λ = Inserting λ =
p−1 p
Z
α p−2 A0 Z
f >λ α/A1 A1 f λ
Z
f
α
p−2
0
X
( f − λ α/A1 )dµdα λ dαdµ − A1
Z Z
A1 f λ
!
α p−1
dαdµ
X 0
A0 A1p−1 1 k f kLp p . p − 1 (1 − λ )λ p−1
< 1 we obtain that kT ( f )kLp p ≤
p p−1 p k f kLp p p−1 p−1
which is the claimed conclusion. (b) We still need only to consider the nonnegative functions. Let f0 , f1 as in (a). We have kT ( f )kLp p =p ≤p
Z ∞ 0
Z ∞ 0
≤p
Z ∞
α p−1 dT ( f ) (α)dα α p−1 µ({T ( f0 ) > (1 − λ )α})dα p
α p−1
0
A0 0 ((1 − λ )α) p0
p
pA0 0 = (1 − λ ) p0 −1 p
Z
f X
p0
Z 0
A1 f λ
Z f >λ α/A1
( f − λ α/A1 ) p0 dµdα
λ α p0 λ α p−p0 −1 f A1 p−p0 f A1 α 1− d dµ f A1 f A1 λ λ
p−p
pA0 0 A1 0 = B(p − p0 , p0 + 1)k f kLp p . (1 − λ ) p0 λ p−p0 0 Let λ be p−p p can get the desired result. (c) We need only to consider nonnegative functions. Let f = f0 + f1 ,where f0 = f − δ α if f ≥ δ α and 0 otherwise (0 < λ < 1). Therefore
kT ( f )kLp p =p ≤p
Z ∞ 0
Z ∞ 0
α p−1 dT ( f ) (α)dα α p−1 (µ({T ( f0 ) > (1 − λ )α}) + µ({T ( f1 ) > λ α}))dα
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29
≤p
Z ∞
p
α p−1
0
A0 0 ((1 − λ )α) p0
p
=
pA0 0 (1 − λ ) p0 −1
Z
δ p0 −p f p
Z
( f − δ α) p0 dµdα + p
Z ∞
f >δ α
Z
f δ
0
(1 −
0
X
α p−1
A1p1 (λ α) p1
Z
f p1 dµdα f 0. Setting f j = (a j − a j+1 )χS j
i=1 Ai
and aN+1 = 0 we have
N
f=
∑ fj
j=1
and f∗ =
N
N
N
j=1
j=1
j=1
∑ a j χ[µ(A0 )+···+µ(A j−1 ),µ(A0 )+···+µ(A j )) = ∑ (a j − a j+1 )χ[0,µ(A0 )+···+µ(A j )) = ∑ f j∗
where A0 = 0. / We write a finitely simple function g = ∑M k=1 bk χBk analogously. Then we have Z X
f j gk dµ = (a j − a j+1 )(bk − bk+1 )µ
j [
Ai ∩
i=1
Therefore
N
Z
f g dµ = X
k [
Z j k Bl ≤ (a j − a j+1 )(bk − bk+1 ) min ∑ µ(Ai ), ∑ µ(Bl ) = i=1
l=1
M
N
Z
∑∑
j=1 k=1 X
f j gk dµ ≤
M
Z ∞
∑∑
j=1 k=1 0
f j∗ (t)g∗k (t) dt =
Z ∞
l=1
0
∞
f j∗ (t)g∗k (t) dt .
f ∗ (t)g∗ (t) dt .
0
Given general nonnegative measurable functions f , g on X, by σ finiteness, we pick finitely simple functions fn , gn such that fn ↑ f and gn ↑ g. By Proposition 1.4.5 (8) we have that fn∗ ↑ f ∗ and gn ↑ g. Then we obtain the result in general via the Lebesgue monotone convergence theorem.
Exercise 1.4.2 Let (X, µ) be a measure space. Prove that if f ∈ Lq0 ,∞ (X) ∩ Lq1 ,∞ (X) for some 0 < q0 < q1 ≤ ∞, then f ∈ Lq,s (X) for all 0 < s ≤ ∞ and q0 < q < q1 . Solution. We have that k f ksLq,s =
Z ∞
1
(t q f ∗ (t))s
0
Z 1
=
1− 1 q0
(t q
dt t
1
t q0 f ∗ (t))s
0
≤k f ksLq1 ,s
Z 1 0
1− 1 q0
(t q
dt + t
Z ∞
1− 1 q0
(t q
1
t q0 f ∗ (t))s
1
)s−1 dt + k f ksLq0 ,s
Z ∞
dt t
1 − 1 (s−1) q0
tq
dt < ∞ .
1
This implies that f ∈ Lq,s .
Exercise 1.4.3 ([166]) Given 0 < p, q < ∞, fix an r = r(p, q) > 0 such that r ≤ 1, r ≤ q and r < p. For t < µ(X) define
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37
f ∗∗ (t) = sup µ(E)≥t
1 µ(E)
Z
 f r dµ
1/r ,
E
while for t ≥ µ(X) (if µ(X) < ∞) let Z 1/r 1 r f (t) =  f  dµ . t X ∗∗
Also define Z
 f 
L p,q
∞
= 0
q dt t f (t) t 1 p
∗∗
1 q
.
(The function f ∗∗ and the functional f →  f L p,q depend on r.) (a) Prove that the inequality ((( f + g)∗∗ )(t))r ≤ ( f ∗∗ (t))r + (g∗∗ (t))r is valid for all t ≥ 0. Since r ≤ q, conclude that the functional f →  f rL p,q is subadditive and hence it is a norm when r = 1 (this is possible only if p > 1). (b) Show that for all f we have 1/r
p
f p,q .
f p,q ≤ f p,q ≤ L L L p−r (c) Conclude that L p,q (X) is metrizable and normable when 1 < p, q < ∞. Solution. (a) Since r ≤ 1 it follows that 1 µ(E)
Z
1 µ(E)
 f + gr dµ ≤
E
Z
 f r dµ +
E
1 µ(E)
Z
gr dµ ,
E
therefore (( f + g)∗∗ (t))r ≤ ( f ∗∗ (t))r + (g∗∗ (t))r for t ≤ µ(X). Similarly for t > µ(X). So we have r Z ∞ q r ∗∗ r qr dt f + gr p,q = p (t (( f + g) (t)) ) L t 0 r
=kt p (( f + g)∗∗ )r (t)k
q
L r ( dtt )
r
r
≤kt p ( f ∗∗ )r (t) + t p (g∗∗ )r (t)k r
q
L r ( dtt )
r
≤kt p ( f ∗∗ )r (t)k qr dt + kt p (g∗∗ )r (t)k qr dt L (t ) L (t ) r r = f L p,q + g L p,q , where the last inequality is due to the fact that q > r. (b) For t ≤ µ(X) we have s < f ∗ (t) ⇔ d f (s) > t, therefore s=s
1 µ(Es )
1
r
Z
1dµ Es
≤
1 µ(Es )
Z
r
1
 f  dµ
r
≤ f ∗∗ (t) ,
Es
where Es = { f  > s}. So f ∗ (t) ≤ f ∗∗ (t). It’s obviously valid for t > µ(X), therefore k f kL p,q ≤  f L p,q . R 1 We will prove that f ∗∗ (t) ≤ ( 1t 0t ( f ∗ (s))r ds) r . This is obvious in the case µ(X) < t < ∞. If t ≤ µ(E) ≤ µ(X), then 1 µ(E) Let a=
1 t
Z t 0
1  f  dµ ≤ µ(E) E
Z
r
( f ∗ (s)r )ds ,
b=
Z µ(E)
( f ∗ (s))r ds .
0
1 µ(E) − t
Z µ(E) t
( f ∗ (s))r ds .
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Then b < a since f ∗ is decreasing. 1 µ(E)
Z µ(E)
( f ∗ (s))r ds =
0
therefore
ta + (µ(E) − t)b = θ a + (1 − θ )b < a , µ(E)
1 Zt r 1 ∗ r f (s) ds . f (t) ≤ t 0 ∗∗
By Exercise 1.2.8 we have f
L p,q
Zt 1 q 1 r 1 1 dt q ∗ r p t f (s) ds t 0 t 0 Z ∞ Z t q r −1 r q q q r − r −1 ∗ r p = ( f (s)) ds t dt
Z ≤
∞
0
≤ q
0 q r
1 Z r
∞
(f
q p
0 r − p 1 r = k f kL p,q . p−r
∗
q
q
q
q (s))r· r t r + p − r −1 dt
1 q
(c) Set d( f , g) =  f − grL p,q where r ≤ 1, r ≤ q, and r < p. Obviously d is a metric. Moreover, when p > 1, 1 ≤ q < ∞, we can choose r = 1, then N( f ) =  f L p,q is a norm which generates the same topology as k · kL p,q .
Exercise 1.4.4 (a) Show that on a measure space (X, µ) the set of countable linear combinations of simple functions is dense in L p,∞ (X). in L p,∞ (R) for any 0 < p ≤ ∞. (b) Prove that finitely simple functions are not dense −1/p Hint: Part (b): Show that the function h(x) = x χx>0 cannot be approximated by a sequence of simple functions L p,∞ . Given 1 any simple function s which is nonzero on a set A with A > 0, show that ks − hkL p,∞ ≥ sup00
λ >0
This shows that h cannot be approximated by finitely simple functions.
λ >0
Exercise 1.4.5 Let (X, µ) be a nonatomic measure space. (a) If A0 j A1 j X, 0 < µ(A1 ) < ∞, and µ(A0 ) ≤ t ≤ µ(A1 ), show that there exists an Et j A1 with µ(Et ) = t. (b) Given a nonnegative continuous and decreasing function ϕ on [0, ∞) such that ϕ(t) = 0 whenever t ≥ µ(X), prove that there exists a measurable function f on X with f ∗ (t) = ϕ(t) for all t > 0. e of X with µ(A) e = µ(A) (c) Given A j X with 0 < µ(A) < ∞ and g an integrable function on X, show that there exists a subset A such that Z Z µ(A)
g dµ =
(d) If X is σ finite, f ∈ L∞ (X), and g ∈ L1 (X), prove that Z Z sup h g dµ = h: dh =d f
g∗ (s) ds.
0
e A
∞
f ∗ (s)g∗ (s) ds,
0
X
where the supremum is taken over all functions h on X equidistributed with f . Hint: Part (a): Reduce matters to the situation in which A0 = 0. / Consider first the case that for all A j X there exists a 9 1 subset B of X satisfying 10 µ(A) ≤ µ(B) ≤ 10 µ(A). Then we can find subsets of A1 of measure in any arbitrarily small interval, and by continuity the required conclusion follows. Next consider the case in which there is a subset A1 of X such 1 9 that every B j A1 satisfies µ(B) < 10 µ(A1 ) or µ(B) > 10 µ(A1 ). Without loss of generality, normalize µ so that µ(A1 ) = 1. 1 Let µ1 = sup{µ(C) : C j A1 , µ(C) < 10 } and pick B1 j A1 such that 12 µ1 ≤ µ(B1 ) ≤ µ1 . Set A2 = A1 \ B1 and define 1 1 µ2 = sup{µ(C) : C j A2 , µ(C) < 10 }. Continue in this way and define sets A1 k A2 k A3 k · · · and numbers 10 ≥ µ1 ≥ µ2 ≥ 1 1 9 µ3 ≥ · · · . If C j An+1 with µ(C) < 10 , then C ∪ Bn j An with µ(C ∪ Bn ) < 5 < 10 , and hence by assumption we must have T 1 µ(C ∪ Bn ) < 10 . Conclude that µn+1 ≤ 12 µn and that µ(An ) ≥ 45 for all n = 1, 2, . . . . Then the set ∞ n=1 An must be an atom. Part (b): First show that when d is a simple right continuous decreasing function on [0, ∞) there exists a measurable f on X such that f ∗ = d. For general continuous functions, use approximation. Part (c): Let t = µ(A) and define A1 = {x : g(x) > g∗ (t)} e such that A1 j A e j A2 and µ(A) e = t = µ(A) and A2 = {x : g(x) ≥ g∗ (t)}. Then A1 j A2 and µ(A1 ) ≤ t ≤ µ(A2 ). Pick A R µ(A) e
by part (a). Then Ae g dµ = X gχAe dµ = 0∞ (gχAe)∗ ds = 0 g∗ (s) ds. Part (d): Reduce matters to functions f , g ≥ 0. Let f = ∑Nj=1 a j χA j where a1 > a2 > · · · > aN > 0 and the A j are pairwise disjoint. Write f as ∑Nj=1 b j χB j , where b j = (a j −a j+1 ) and B j = A1 ∪· · ·∪A j . Pick Be j as in part (c). Then Be1 j · · · j BeN and the function f1 = ∑Nj=1 b j χBe j has the same distribution function R R as f . It follows from part(c) that X f1 g dµ = 0∞ f ∗ (s)g∗ (s) ds. The case of a general f ∈ L∞ (X) follows by approximation by finitely simple functions. R
R
R
Solution. (a) Assume the assertion is valid for the case A0 = 0/ and 0 < t < µ(A1 ). If µ(A0 ) = t or t = µ(A1 ), it’s valid. If µ(A0 ) < t < µ(A1 ), we can choose a B j A1 \A0 with the measure t = µ(A0 ) such that A0 ∩ B is the required set. So we need only to consider the case that A0 = 0/ and 0 < t < µ(A1 ). Suppose that for all A j X, there exists a B j A 1 9 such that 10 µ(A) ≤ µ(B) ≤ 10 µ(A). Then for any t ≥ 12 µ(A)(the case t < 12 µ(A) can be reduced to this case), we can find a
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set B such that t ≥ µ(B) ≥
t 10 (equivalently,
t − µ(B) ≤
9t 10 ).
If B is a set such that µ(B) > t, then we can choose a subset B1 µ(B1 ) 9 such that < µ(B) ≤ 10 , therefore after n steps like this, we can find a subset Bn of B such that µ(Bn−1 ) > t and µ(Bn ) ≤ t, µ(B ) t 9 n therefore µ(Bn ) ≥ 10n−1 ≥ 10 . Similarly, if we have find a set An such that µ(An ) < t and t − µ(An ) ≤ ( 10 ) t we can always t−µ(An ) 9 n+1 find a set B such that 10 ≤ µ(Bn ) ≤ t − µ(An ), therefore t − µ(An+1 ) ≤ ( 10 ) t. Let An+1 = An ∪ B, {An } is increasing 1 10
set sequence with the property µ(An ) ↑ t, therefore µ(B) = µ(∪An ) = t. Suppose A1 is a set such that every subset of it with 9 1 µ(A1 ) or > 10 µ(A1 ). Without loss of generality, we assume that µ(A1 ) = 1. measure < 10 1 Let µ1 = sup{µ(C) : C j A1 , µ(C) < 10 }, choose B1 j A1 such that 12 µ1 ≤ µ(B1 ) ≤ µ1 , denote A2 = A1 \B1 . Let µ2 = 1 sup{µ(C) : C j A2 , µ(C) < 10 }, choose B2 j A2 such that 12 µ2 ≤ µ(B2 ) ≤ µ2 , denote A3 = A2 \B2 . Continuing in this way, 1 1 1 9 we can get A1 k A2 k A3 k . . . , let C j An+1 with µ(C) ≤ 10 , then µ(C ∪ Bn+1 ) ≤ 10 + 10 < 10 implies that µ(C ∪ Bn ) < 1 µ(C ∪ Bn ) = µ(C) + µ(Bn ) > µn while C ∪ Bn j An . Since C is arbitrary, we have 10 , moreover µ(C) ≤ µ(Bn ), otherwise T 1 µn+1 ≤ 12 µn . Consider the set A = ∞ n=1 An , C is a subset of A with measure less than 10 , therefore µ(C) ≤ µn (since C j An ), 1 9 µn ≤ 2n−1 µ1 ⇒ µ(C) = 0. Meanwhile if C is a subset of A with measure greater than 10 , then µ(A\C) = 0, therefore A is an atom. (b) Suppose that ϕ(0) = 1. Let {t :
j−1 2n
≤ϕ ≤
j 2n }
= [a2n , j , a2n , j−1 ] for n ∈ N, 1 ≤ j ≤ 2n and define 2n
ϕn =
∑
j=1
j−1 χ . 2n [a2n , j ,a2n , j−1 )
These functions are right continuous and ϕn ↑ ϕ. For ϕ choose A2,1 from X such that µ(A2,1 ) = a2,1 − a2,2 = a2,1 , f2 = 12 χA2,1 is the function such that f2∗ = ϕ2 . Suppose we have defined 2n+1 j−1 fn = ∑ χA2n , j−1 n j=1 2 such that fn∗ = ϕn , then divide A2n , j−1 into two parts A2n+1 ,2( j−1) and A2n+1 ,2 j−1 with A2n , j−1 = a2n+1 ,2 j−1 − a2n+1 ,2 j , A2n , j−2 = a2n+1 ,2 j−2 − a2n+1 ,2 j−1 . Define n+1
fn+1 =
∑
j=1
then fn+1 ≥ fn and
∗ fn+1
j−1 χA , 2n+1 2n+1 , j−1
= ϕn+1 , define f = limn fn . Obviously f ∗ = limn fn∗ = limn ϕn = ϕ.
e such that µ(A) e = T := µ(A). Let A1 = {x : g(x) > g∗ (T )}, A2 = {x : g(x) ≥ g∗ (T )}. Then we have µ(A1 ) ≤ (c) Choose A e such that A j A e j A2 and µ(A) e = T . We will prove that g∗ (t) = g∗ (t) for t ≤ T , where T ≤ µ(A2 ). Therefore we can choose A 1 ∗ g1 = gχAe. Recall that g (t) = inf{s : dg (s) ≤ t}. We need only to prove that for s such that dg (s) ≤ t we have dg1 (s) = dg (s). Note that dg (s) ≤ t ⇒ g∗ (t) ≤ s ⇒ s ≥ g∗ (T ). e So dg (s) = dg (s) for dg (s) ≤ t. So g∗ (t) = g∗ (t) for t ≤ T . So Therefore if x is such that g(x) > s0 , then x ∈ A. 1 1 Z µ(A) e
Z
g(x) dµ = 0
e A
g∗1 (t) dt =
Z µ(A)
g∗ (t) dt ,
0
as desired. (d) In the identity Z Z∞ sup h g dµ = f ∗ (s)g∗ (s) ds, X 0
h: dh =d f
if we replace f by  f  nothing changes (since R f  =  f ∗ and d f = d f  ), thus we may assume that f ≥ 0. Moreover, the function R k = hχ{g>0} − hχ{g≤0} satisfies X kg dµ = X hg dµ and k = h, hence it is equidistributed (i.e., has the same distribution function) with h and hence with f . This allows us to replace each given h that appears in the supremum by k and thus also assume that g is replaced by g, that is, g ≥ 0. We assume therefore that f , g ≥ 0.
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e with measure A Assume first that f = χA , where A is a measurable subset of X of finite measure. By part (c), we can find A such that Z Z ∞ g χAe dµ = g∗ (t)(χA )∗ (t) dt , 0
X
thus the claimed identity holds. Consider now the case where f is finitely simple, i.e., f = ∑Ni=1 ai χAi with ∞ > a1 > · · · > aN > 0, then f = ∑Ni=1 fi with fi = (ai − ai+1 )χBi , Bi = ∪ik=1 Ak , f ∗ = ∑Ni=1 fi∗ and µ(Ak ) < ∞ for all k. As in part (c) we can find Be1 j Be2 j · · · j BeN and equimeasurable with the B j ’s such that Z ∞ 0
g∗ (t)(χBi )∗ (t) dt =
therefore Z ∞ 0
∑Ni=1 ai χBei
N
g∗ (t) f ∗ (t) dt = ∑ ai
Z ∞ 0
i=1
Z X
g χBei dµ ,
g∗ (t)(χBi )∗ (t) =
Z X
N
g ∑ ai χBei dµ i=1
∑Ni=1 (ai
and the function = − ai+1 )χBe1 ∪···∪Bei+1 has the same distribution as f . Thus the claim holds for general nonnegative measurable functions g and positive finitely simple functions f . For general nonnegative measurable function f on X, by σ finiteness there exists a sequence { fn }n of nonnegative finitely simple functions such that fn ↑ f . Then fn∗ ↑ f ∗ by Proposition 1.4.5 (8) and by the preceding argument there is a nonnegative function hn equidistributed with fn such that Z ∞
Z X
hn g dµ =
0
g∗ (t) fn∗ (t) dt
Since R 0∞ g∗ (t) fn∗ (t) dt ↑ 0∞ g∗R(t) f ∗ (t) dt as n → ∞ by the Lebesgue monotone convergence theorem, it follows that the expressions X hn g dµ approximate 0∞ g∗ (t) fn∗ (t) dt, thus Z∞ Z f ∗ (s)g∗ (s) ds . sup h g dµ = R
R
h: dh =d f
0
X
Exercise 1.4.6 ([7], [299]) Let K ≥ 1 and let k · k be a nonnegative functional on a vector space X that satisfies kx + yk ≤ K kxk + kyk for all x, y ∈ X. For a fixed α ≤ 1 satisfying (2K)α = 2 show that kx1 + · · · + xn kα ≤ 4 (kx1 kα + · · · + kxn kα ) for to as the AokiRolewicz theorem. all n = 1, 2, . . . and all x1 , x2 , . . . , xn in X. This inequality is referred Hint: Quasilinearity implies that kx1 +· · ·+xn k ≤ max1≤ j≤n [(2K) j kx j k] for all x1 , . . . , xn in X (use that K ≥ 1). Define H : X → R by setting H(0) = 0 and H(x) = 2 j/α if 2 j−1 < kxkα ≤ 2 j . Then kxk ≤ H(x) ≤ 21/α kxk for all x ∈ X. Prove by induction that kx1 + · · · + xn kα ≤ 2(H(x1 )α + · · · + H(xn )α ). Suppose that this statement is true when n = m. To show its validity for n = m + 1, without loss of generality assume that kx1 k ≥ kx2 k ≥ · · · ≥ kxm+1 k. Then H(x1 ) ≥ H(x2 ) ≥ · · · ≥ H(xm+1 ). Assume that all the H(x j )’s are distinct. Then since H(x j )α are distinct powers of 2, they must satisfy H(x j )α ≤ 2− j+1 H(x1 )α . Then kx1 + · · · + xm+1 kα ≤
max (2K) j kx j k
1≤ j≤m+1
α
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≤
≤
α max (2K) j H(x j )
1≤ j≤m+1
α max (2K) j 21/α 2− j/α H(x1 )
1≤ j≤m+1
= 2H(x1 )α ≤ 2(H(x1 )α + · · · + H(xm+1 )α ) . We now consider the case that H(x j ) = H(x j+1 ) for some 1 ≤ j ≤ m. Then for some integer r we must have 2r−1 < kx j+1 kα ≤ kx j kα ≤ 2r and H(x j ) = 2r/α . Next note that kx j + x j+1 kα ≤ K α (kx j k + kx j+1 k)α ≤ K α (2 2r/α )α = 2r+1 . This implies H(x j + x j+1 )α ≤ 2r+1 = 2r + 2r = H(x j )α + H(x j+1 )α . Now apply the inductive hypothesis to x1 , . . . , x j−1 , x j + x j+1 , x j+1 , . . . , xm and use the previous inequality to obtain the required conclusion. Solution. A repeated application of the hypothesis kx + yk ≤ K(kxk + kyk) gives kx1 + · · · + xn k ≤ max [(2K) j kx j k], 1≤ j≤n
for all x1 , . . . , xn in X. (Here we used 2K ≥ 1.) Now define a function H : X → R1 by setting H(0) = 0 and H(x) = 2 j/α if 2 j−1 < kxkα ≤ 2 j . Then we clearly have kxk ≤ H(x) ≤ 21/α kxk for all x ∈ X. The result we want to prove will follow from this inequality and the following claim kx1 + · · · + xn kα ≤ 2(H(x1 )α + · · · + H(xn )α ). The claim is certainly true when n = 1. Suppose it is true when n = m. We will show that it is true when n = m + 1. Without loss of generality assume that kx1 k ≥ kx2 k ≥ · · · ≥ kxm+1 k. Then we must have H(x1 ) ≥ H(x2 ) ≥ · · · ≥ H(xm+1 ). If all the H(x j )’s are distinct, since H(x j )α are distinct powers of 2 and therefore they must satisfy H(x j )α ≤ 2− j+1 H(x1 )α . Then α kx1 + · · · + xm+1 kα ≤ max (2K) j kx j k 1≤ j≤m+1 α ≤ max (2K) j H(x j ) 1≤ j≤m+1 α ≤ max (2K) j 21/α 2− j/α H(x1 ) = ≤
1≤ j≤m+1 2H(x1 )α 2(H(x1 )α + · · · + H(xm+1 )α ),
where the last equality above follows from the choice of α. We now consider the case where H(x j ) = H(x j+1 ) for some 1 ≤ j ≤ m. Then for some integer r we must have 2r−1 < kx j+1 kα ≤ kx j kα ≤ 2r and H(x j ) = 2r/α . Next observe that kx j + x j+1 kα ≤ K α (kx j k + kx j+1 k)α ≤ K α (2 2r )α ≤ 2r+1 , which implies that H(x j +x j+1 )α ≤ 2r+1 = 2r +2r = H(x j )α +H(x j+1 )α . Now apply the inductive hypothesis to x1 , . . . , x j−1 , x j + x j+1 , x j+1 , . . . , xm and use the inequality above to obtain the required conclusion.
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Exercise 1.4.7 (a) ([348]) Let (X, µ) and (Y, ν) be measure spaces. Let Z be a Banach space of complexvalued measurable functions on Y . Assume that Z is closed under absolute values and satisfies k f kZ = k  f  kZ . Suppose that T is a linear operator defined on the space of finitely simple functions on (X, µ) and taking values in Z. Suppose that for some constant A > 0 the following restricted weak type estimate
T (χE ) ≤ Aµ(E)1/p Z holds for some 0 < p < ∞ and for all E measurable subsets of X of finite measure. Show that for all finitely simply functions f on X we have
T ( f ) ≤ p−1 A f p,1 . Z L Consequently T has a bounded extension from L p,1 (X) to Z. (b) ([174]) As an application of part (a) prove that for any U, V measurable subsets of Rn with U, V  < ∞ and any f measurable on U ×V we have Z 1 2
1
f (u, ·) 2 2,1 du ≤ f L2,1 (U×V ) . L (V ) 2 U Hint: Part (a): Let f = ∑Nj=1 a j χE j ≥ 0, where a1 > a2 > · · · > aN > 0, µ(E j ) < ∞ pairwise disjoint. Let Fj = E1 ∪ · · · ∪ E j , B0 = 0, and B j = µ(Fj ) for j ≥ 1. Write f = ∑Nj=1 (a j − a j+1 )χFj , where aN+1 = 0. Then
T ( f ) = T ( f ) Z
Z
N
≤
∑ (a j − a j+1 ) T (χFj ) Z
j=1
N
≤ A ∑ (a j − a j+1 )(µ(Fj ))1/p j=1
N−1
=A
1/p
1/p
∑ a j+1 (B j+1 − B j
)
j=0
= p−1 A f L p,1 , where the penultimate equality follows by a summation by parts; see Appendix F. Solution. (a) Obviously we have kχE kL p,1 =
Z µ(E)
1
tp 0
1 dt = pµ(E) p . t
∑Ni=1 ai χEi , a1
Let f = > a2 > · · · > aN > 0 and Ei are pairwise disjoint measurable subsets of X of finite measure. Then Sj f = ∑ni=1 (ai − ai+1 )χFi , Fi = i=1 Ei and f ∗ = ∑Ni=1 bi χ[0,µ(Bi )) . So
T ( f ) = T ( f ) Z Z N
≤
∑ (a j − a j+1 ) T (χFj ) Z
j=1
N
≤ A ∑ (a j − a j+1 )(µ(Fj ))1/p j=1
N−1
=A
1/p
1/p
∑ a j+1 (B j+1 − B j
j=0
= p−1 A f L p,1 .
)
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Finite simple functions are dense in L p,1 (X) and thus we obtain that T has an extension T on L p,1 (X) that satisfies kT ( f )kZ ≤ A p,1 p k f kL p,1 for all f ∈ L . See Exercise 1.4.17. (b) Let Z be the space U ×V with norm Z 1 2
2
 f  = f (u, ·) L2,1 (V ) du . U
Then Z is a Banach space with the property   f   =  f . Let T be the identity map. To prove the required inequality for all functions in L2,1 (U ×V ), by part (a), it will suffice to prove this inequality for characteristic functions. Let E be a measurable subset of U ×V . One has that χE  = E1/2
2 and that f (u, ·) L2,1 (V ) = Eu , where Eu = {v : (u, v) ∈ E}. Since 1
Z
2
U
Eu  du
1
= E 2
we have that χE  ≤ E1/2 and the the required conclusion follows with A = 1 and p = 2.
Exercise 1.4.8 Let 0 < p, q, α, β < ∞. Also let 0 < q1 < q2 < ∞. (a) Show that the function fα,β (t) = t −α (logt −1 )−β χ[0,e−β /α ) (t) lies in L p,q (R) if and only if either p < 1/α or both p = 1/α and q > 1/β hold. Conclude that the function t 7→ t −1/p (logt −1 )−1/q1 χ[0,e−p/q1 ) (t) lies in L p,q2 (R) but not in L p,q1 (R). (b) Find a necessary and sufficient condition in terms of p, α, β for the function gα,β (t) = (1 + t)−α (log(2 + t))−β χ[0,∞) to lie in L p,q (R). (c) Let ψ(t) be smooth decreasing function on [0, ∞) and let F(x) = ψ(x) for x in Rn , where x is the modulus of x. Show that F ∗ (t) = f ((t/vn )1/n ), where vn is the volume of the unit ball. Use this formula to construct examples showing that L p,q1 (Rn ) $ L p,q2 (Rn ). (d) On a general nonatomic measure space (X, µ) prove that there does not exist a constant C(p, q1 , q2 ) > 0 such that for all f in L p,q2 (X) the following is valid:
f p,q ≤ C(p, q1 , q2 ) f p,q . L 2 L 1 Hint: Parts (a), (b): Use that fα,β and gα,β are equal to their decreasing rearrangements. Part (d): Use Exercise 1.4.5(b) with ϕ(t) = g1/p,1/q1 (t). Solution. (a) On the interval [0, e−β /α ] the function t −α (log 1t )−β is decreasing, and so it is equal to its decreasing rearrangement. We have that ( fα,β )∗ (t) = t −α (log 1t )−β χ[0,e−β /α ) . Consequently k fα,β kqL p,q
Z e−β /α
= 0
1 −β q dt −α t t log t t q p
and this integral is finite if and only if 1/p > α or both 1/p = α and β q > 1 hold. If α = 1/p, by part (a) the function 1
−1
t 7→ t − p (logt −1 ) q1 χ[0,e−p/q1 ) (t) belongs to L p,q (R) if and only if q/q1 > 1. Thus this function lies in L p,q2 (R) but not in L p,q1 (R). (b) The function gα,β is equal to its decreasing rearrangement. Then kgα,β kqL p,q =
Z ∞
q
t p (1 + t)−α (log(2 + t))−β
0
and this integral converges if and only if either 1/p < α or 1/p = α and β q > 1. (c) We have that
q dt t
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F ∗ (t) = inf{s > 0 : d f (s) ≤ t}. But d f (s) = {x : x ≤ f −1 (s)} = vn ( f −1 (s))n . Thus d f (s) ≤ t is equivalent to s ≥ f ((t/vn )1/n ), hence F ∗ (t) = f ((t/vn )1/n ). The radial function G(x) = gn/p,1/q1 (x) therefore satisfies G∗ (t) = (1 + (t/vn )1/n )−n/p (log(2 + (t/vn )1/n ))−1/q1 ≈ (1 + t)−1/p (log(2 + t))−1/q1 . Since α = 1/p, the result in part (b) gives that G lies in L p,q (Rn ) if and only if q/q1 > 1. Thus G lies in L p,q1 (Rn ) but not in L p,q2 (Rn ). (d) In view of Exercise 1.4.5 (b) with ϕ(t) = g1/p,1/q1 (t), there is a function f on X such that f ∗ (t) = g1/p,1/q1 (t). Then f lies in L p,q1 (X) but not in L p,q2 (X), thus the inequality in question cannot be valid.
Exercise 1.4.9 ([347]) Let L p (ω) denote the space of all measurable functions f on Rn such that k f kLp p (ω) = Rn  f (x) p ω(x) dx < ∞, where 0 < ω < ∞ a.e. Let T be a sublinear operator that maps L p0 (ω0 ) to Lq0 ,∞ (ω) and L p1 (ω1 ) to Lq1 ,∞ (ω), where ω0 , ω1 , ω are positive functions and 1 ≤ p0 < p1 < ∞, 0 < q0 , q1 < ∞. Suppose that R
1 1−θ θ = + , pθ p0 p1
1 1−θ θ = + . qθ q0 q1
1−θ p θ θ p p pθ Let Ωθ = ω0 0 ω1 1 . Show that T maps L pθ Ωθ → Lqθ ,pθ (ω) . 1 1 −p Hint: Define L( f ) = (ω1 /ω0 ) p1 −p0 f and observe that for each θ ∈ [0, 1], L maps L pθ Ωθ → L pθ (ω0p1 ω1 0 ) p1 −p0 isomet rically. Then apply Corollary 1.4.24 to the sublinear operator T ◦ L−1 .
Solution. We first note that in view of the definition of pθ we have pθ − p0 θ = pθ p1 − p0 p1 and so Z Rn
ω 1
ω0
1 p1 −p0
p1 − pθ 1−θ = pθ p1 − p0 p0
pθ Z 1 −p f (ω0p1 ω1 0 ) p1 −p0 dx =
Rn
1−θ pθ θ pθ p p  f  pθ ω0 0 ω1 1 dx ,
1−θ p θ p 1 pθ Ω isometrically, where Ω = (ω p1 ω −p0 ) p1 −p0 and Ω = ω p0 θ ω p1 θ . pθ Ω thus for each θ ∈ [0, 1], L maps L → L θ θ 0 1 0 1 Then L−1 maps L p0 Ω → L p0 (ω0 ) and L p1 Ω → L p1 (ω1 ). Consequently T ◦ L−1 maps L p0 Ω to Lq0 ,∞ (ω) and L p1 Ω to Lq1 ,∞ (ω). It is easy to see that T ◦ L−1 is a sublinear operator is defined on L p0 (Ω ) and on L p1 (Ω ) and thus defined on L pθ (Ω ). it also −1 p q ,p p q ,p θ θ θ θ θ θ It follows from Corollary 1.4.24 that T ◦ L maps L Ω to L (ω). Thus T maps L Ωθ to L (ω) with the same norm.
Exercise 1.4.10 ([187], [350]) Let λn be a sequence of positive numbers with ∑n λn ≤ 1 and ∑n λn log( λ1n ) = K < ∞. Suppose all sequences are indexed by a fixed countable set. (a) Let fn be a sequence of complexvalued functions in L1,∞ (X) with k fn kL1,∞ ≤ 1 uniformly in n. Prove that ∑n λn fn lies in L1,∞ (X) with norm at most 2(K + 2). (This property is referred to as the logconvexity of L1,∞ .)
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(b) Let Tn be a sequence of sublinear operators that map L1 (X) to L1,∞ (Y ) with norms kTn kL1 →L1,∞ ≤ B uniformly in n. Use part (a) to prove that ∑n λn Tn maps L1 (X) to L1,∞ (Y ) with norm at most 2B(K + 2). (c) Given δ > 0 pick 0 < ε < δ and use the simple estimate ∞ ∞ µ { ∑ 2−δ n fn > α} ≤ ∑ µ {2−δ n fn > (2ε − 1)2−εn α} n=1
n=1
−δ n to obtain a simple proof of the statement in part (a) when λn = 2 , n = 1, 2, . . . . Hint: Part (a): For fixed α > 0, write fn = un + vn + wn , where un = fn χ fn ≤ α , vn = fn χ fn > 2
u = ∑n λn un , v = ∑n λn vn , and w = ∑n λn wn . Clearly u ≤ Z
w dµ ≤
X
Z
∑ λn
≤
2
Z
α 2λn
α/(2λn )
α/2
n
α 2λn };
Z α/2
n { f n  >
, and wn = fn χ α α/2}) + µ({v 6= 0}) + µ({w > α/2}), deduce the conclusion. Solution. (a) Let fn = un + vn + wn with un = fn χ fn ≤ α , vn = fn χ fn > 2
u = ∑ λn un , α 2
and wn = fn χ α ≤∑ ≤ . 2λn α α n
Moreover, we have Z X
w dµ ≤ ∑ λn n
Z X
wn  dµ
Z
= ∑ λn n
α α ≤ µ({u > }) + µ({v 6= 0}) + µ({w > }) ≤ + (K + 1) = (K + 2), 2 2 α α α n
and consequently,
∑ λn fn n
(b) Consider Tn /B and f with k f kL1 ≤ 1. Then
L1,∞
≤ 2(K + 2) .
2 α.
α 2λn
. Let
Finally,
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47
T
n
B
f
L1,∞
≤
kT kk f kL1 ≤ 1, B
therefore k B1 ∑n λn Tn f kL1,∞ ≤ 2(K + 2) and k ∑n λn Tn kL1,∞ ≤ 2B(K + 2). (c) Obviously, ∞
µ({ ∑ 2−δ n fn > α}) ≤ n=1
∞
∞
n=1
n=1
2−δ n
∑ µ({2−δ n fn > (2ε − 1)2−εn α}) ≤ ∑ (2ε − 1)2−εn α
and this implies that
∞
∑ 2−δ n fn
L1,∞
n=1
1 2ε−δ 2(ε−δ )n = . ε 2ε − 1 1 − 2ε−δ n=1 2 − 1 ∞
≤
∑
Exercise 1.4.11 Let { fn }n be a sequence of measurable functions on a measure space (X, µ). Let 0 < q, s ≤ ∞. (a) Suppose that fn ≥ 0 for all n. Show that
lim inf fn q,s ≤ lim inf fn q,s . L L n→∞
n→∞
(b) Let gn → g in Lq,s as n → ∞. Show that kgn kLq,s → kgkLq,s as n → ∞. Solution. (a) We have
lim inf fn n→∞
s dt 1/s = t (lim inf fn ) (t) n→∞ t 0 Z ∞ s dt 1/s t 1/q lim inf fn∗ (t) (Proposition 1.4.5 (9)) ≤ n→∞ t 0 Z ∞ s dt 1/s ≤ lim inf t 1/q fn∗ (t) (Fatou’s Lemma) n→∞ t 0 Z
Lq,s
∞
∗
1/q
= lim inf k fn kLq,s . n→∞
(b) We have for any ε ∈ (0, 1) 1
1
1
t q fn∗ (t) ≤ t q ( fn − f )∗ (εt) + t q f ∗ ((1 − ε)t) . If s ≥ 1 take Ls ( dtt ) norms and change variables to obtain 1
1
k fn kLq,s ≤ ε − q k fn − f kLq,s + (1 − ε)− q k f kLq,s . If s < 1 raise to the power s and integrate with respect to
dt t
to obtain
s
s
k fn ksLq,s ≤ ε − q k fn − f ksLq,s + (1 − ε)− q k f ksLq,s . Take lim supn→∞ and then let ε → 0 to deduce lim sup k fn kLq,s ≤ k f kLq,s . n→∞
Starting from
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1
1
t q f ∗ (t) ≤ t q ( fn − f )∗ (εt) + t q fn∗ ((1 − ε)t) likewise we obtain
1
1
s
s
k f kLq,s ≤ ε − q k fn − f kLq,s + (1 − ε)− q k fn kLq,s .
when s ≥ 1 and
k f ksLq,s ≤ ε − q k fn − f ksLq,s + (1 − ε)− q k fn ksLq,s .
when s < 1. Take lim infn→∞ and then let ε → 0 to obtain: k f kLq,s ≤ lim inf k fn kLq,s . n→∞
Combining this inequality with the preceding one with the limsup yields the claimed conclusion.
Exercise 1.4.12 (a) Suppose that X is a quasiBanach space and let X ∗ be its dual (which is always a Banach space). Prove that for all T ∈ X ∗ we have
T ∗ = sup T (x) . X x∈X kxkX ≤1
(b) Now suppose that X is a Banach space. Use the Hahn–Banach theorem to prove that for every x ∈ X we have kxkX =
sup T (x) .
T ∈X ∗ kT kX ∗ ≤1
Observe that this result may fail for quasiBanach spaces. For example, if X = L1,∞ , every linear functional on X ∗ vanishes on the set of simple functions. 0 (c) Let 1 < p < ∞, X = L p,1 (Y ), and X ∗ = L p ,∞ (Y ), where (Y, µ) is nonatomic σ finite measure space. Conclude that Z
f p,1 ≈ sup f g dµ , L kgk p0,∞ ≤1 L
f
L p,∞
≈
sup kgk p0,1 ≤1 L
Y
Z f g dµ . Y
Solution. (a) Obviously we have kT kX ∗ ≥ supkxkX ≤1 T (x) by definition. Meanwhile, for any kxkX < ∞ we have T (x) x = T ≤ sup T (x) , kxkX kxkX kxkX ≤1 so kT kX ∗ ≤ sup T (x). kxkX ≤1
Thus we conclude that kT kX ∗ = supkxkX ≤1 T (x). (b) Fix x ∈ X. Obviously, T (x) ≤ kT kX ∗ kxkX and supkxkX ≤1 T (x) ≤ kxkX . Define S such that S(λ x) = λ kxk, it’s a linear continuous functional with norm kSk = 1. By the HahnBanach theorem, there exists an extension S of S on X such that kSkX ∗ = 1, S(x) = kxk; then we have kxkX = S(x) ≤ supkT kX ∗ ≤1 T (x). 0
0
(c) We let X = L p,1R(Y ), X ∗ = L p ,∞ (Y ), where Y is a σ finite nonatomic measure space. For a given g ∈ L p ,∞ define a linear functional Tg ( f ) = Y f (x)g(x)dµ on X. By Theorem 1.4.16 we have kTg kX ∗ ≈ kgkL p0 ,∞ , precisely, for some constants c1 , c2 we have
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c1 ≤
kTg kX ∗ ≤ c2 kgkL p0 ,∞
By part (b) we have k f kL p,1 =
Tg ( f ) ,
sup kTg kX ∗ ≤1
so we have c1 k f kL p,1 = c1
sup kTh kX ∗ ≤1
Th ( f ) ≤
sup kgk p0 ,∞ ≤1 L
Z f (y)g(y)dµ(y) ≤ c2 Y
sup kTh kX ∗ ≤1
Th ( f ) = c2 k f kL p,1 .
Also, by part (a) we have k f kL p,∞ ≈
sup kgk p0 ,1 ≤1 L
Z f (y)g(y)dµ(y) Y
and this proves the other claimed inequality in part (b).
Exercise 1.4.13 Let 0 < p, q < ∞. Prove that any function in L p,q (X, µ) can be written as +∞
f=
∑
cn fn ,
n=−∞
where fn is a function bounded by 2−n/p , supported on a set of measure 2n , and the sequence {ck }k lies in `q and satisfies
1 1 1 1 2− p (log 2) q {ck }k `q ≤ f L p,q ≤ {ck }k `q 2 p (log 2) q . Hint: Let cn = 2n/p f ∗ (2n ), An = {x : f ∗ (2n+1 ) <  f (x) ≤ f ∗ (2n ))}, and fn = c−1 n f χAn . Solution. n For n ∈ Z define An = {x : f ∗ (2n+1 ) <  f (x) ≤ f ∗ (2n ))}, cn = 2 p f ∗ (2n ), fn = c−1 n f χAn . Obviously we have f=
∑ cn fn ,
n∈Z n
 fn (x) ≤ 2− p , and
µ(An ) = d f ( f ∗ (2n=1 )) − d f ( f ∗ (2n )) = 2n .
We have k f kL p,q =
Z
∞
q dt
∗
(t f (t))
=
∞
∑
−∞
Z 2n+1
1 p
q
q dt
∗
(t f (t))
2n
∞ Z 2n+1
1
t
0
≤
1 p
(2
∑
n −∞ 2
n+1 p
1 q
t
f ∗ (2n ))q
dt t
1
1 p
= 2 (log 2) q k{cn }k`q Similarly, 1
1
k f kL p,q ≥ 2− p (log 2) q k{cn }k`q
1 q
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and combining these two inequalities yields the claim.
Exercise 1.4.14 (T. Tao) Let 0 < p < ∞, 0 < γ < 1, A, B > 0, and let f be a measurable function on a measure space (X, µ). (a) Suppose that k f kL p,∞ ≤ A. Then for every measurable set E of finite measure there exists a measurable subset E 0 of E with µ(E 0 ) ≥ γ µ(E) such that f is integrable on E 0 and Z 1 ≤ (1−γ)−1/p A µ(E)1− p . f dµ E0
(b) Suppose that (X, µ) is a σ finite measure space and that f has the property that for any measurable subset E of X with µ(E) < ∞ there is a measurable subset E 0 of E with µ(E 0 ) ≥ γ µ(E) such that f is integrable on E 0 and Z 1 ≤ B µ(E)1− p . f dµ E0
√ Then we have that k f kL p,∞ ≤ B 41/p γ −1 2. (c) Conclude that if (X, µ) is a σ finite measure space then
f
≈
L p,∞
Z
sup
inf 0
−1+ 1p
µ(E)
E jE EjX 0 0, note that the set  f  > α is contained in Re f > √α2 ∪ Im f > √α2 ∪ Re f < − √α2 ∪ Im f < − √α2 . Let En be any of the preceding four sets intersected with Xn , let En0 be a subset of it√with measure at least γ µ(En ) as in the R α hypothesis. Then En0 f dµ ≥ √2 γ µ(En ), from which it follows that α µ(En )1/p ≤ B 2 γ −1 , and let n → ∞. Solution. (a) Let f1 = f χE , we know that f1 and f1∗ have the same distributional function. Consider 1
f1∗ ((1 − γ)µ(E)), ((1 − γ)µ(E)) p f1∗ ((1 − γ)µ(E)) ≤ A 1
1
1
since f ∈ L p,∞ , therefore f1∗ ((1 − γ)µ(E)) ≤ A((1 − γ)µ(E))− p . Let E 0 = {x ∈ E :  f (x) ≤ A(1 − γ)− p µ(E)− p }, therefore µ(E 0 ) ≥ µ(E) − (1 − γ)µ(E) = γ µ(E). Meanwhile Z A(1−γ)− 1p µ(E)− 1p
Z E0
d f χE (β ) dβ
f dµ = 0 1
1
≤ A(1 − γ)− p µ(E)− p µ(E) 1
1
= A(1 − γ)− p µ(E)1− p . (b) Suppose first that µ(X) < ∞. Consider the case that f is positive, and let E = {x ∈ X : f > α}. Then Z 1 1− 1p Bµ(E) ≥ f dµ ≥ αγ µ(E) ⇒ αd f (α) p ≤ B/γ. E0
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Consider the general case that X is σ −finite, then X = ∪n Xn with µ(Xn ) < ∞. Define fn = f χXn , we have fn ↑ f , therefore 1
1
d fn (α) ↑ d f (α). Meanwhile αd fn (α) p ≤ Cγ A/γ by what we have proved, thus αd f (α) p ≤ Cγ A/γ, that is k f kL p,∞ ≤ Cγ A/γ. 1
To remove the condition that f is positive and obtain the result that k f kL p,∞ ≤
Cγ A4 p γ
√ 2
, we note that
n α o n α o n α o α o n { f  > α} j Im f > √ ∪ Im f < − √ ∪ Re f > √ ∪ Re f < − √ . 2 2 2 2 (c) We know from (a) and (b) that k f kL p,∞ < ∞ ⇔
Z
sup
inf
E 0 jE EjX 0 cλ 2α (k0 −k) })+ ∑k≤k0 µ({Tk (χE ) > cλ 2β (k−k0 ) }), where c is a suitable constant and 0 < α 0 < α, 0 < β 0 < β . Apply the restricted weak type (p0 , p0 ) hypothesis on each term of the first sum, the restricted weak type (p1 , p1 ) hypothesis on each term of the second sum, and choose k0 to optimize the resulting expression. Solution. Let T=
∑ Tk . k∈Z 0
Choose α 0 , β 0 such that 0 < α 0 < α, β 0 < β < ∞. Obviously ∑k≥k0 2α (k0 −k) and ∑k≤k0 −1 2β 0 0 ∑k≥k0 c2α (k0 −k) + ∑k≤k0 −1 c2β (k−k0 ) = 1. From the facts that µ({Tk (χE ) > λ }) ≤
A p0 µ(E)2−kα p0 λ p0
µ({Tk (χE ) > λ }) ≤
B p1 µ(E)2−kβ p1 λ p1
and
0 (k−k ) 0
converge. Choose c such that
we obtain µ({T (χE ) > λ }) ≤
0
∑ µ({Tk (χ( E)) > cλ 2α (k0 −k) }) + ∑
µ({Tk (χ( E)) > cλ 2β
})
k≤k0 −1
k≥k0
≤
B p0 µ(E)2−kβ p1 A p0 µ(E)2−kα p0 + ∑ 0 p0 α (k0 −k)p0 p1 β 0 (k−k0 )p1 k≤k0 −1 (cλ ) 2 k≥k0 (cλ ) 2
=
0 0 A p0 2−α k0 p0 µ(E) B p0 2β k0 p1 µ(E) 2−k(α−α )p0 + 2(k−k0 )(β −β )p1 ∑ ∑ p p 0 1 (cλ ) (cλ ) k≥k k≤k −1
∑
0
0
=
0
0 A p0 2−α k0 p0 µ(E)
0 2−k0 (α−α )p0
(cλ ) p0
1 − 2−(α−α )p0
0
+
B p1 2β k0 p1 µ(E) (cλ ) p1
0
=
C1 A p0 µ(E)2−α k0 p0 C2 B p1 µ(E)2β k0 p1 + . λ p0 −p λ p λ p1 −p λ p
Setting
0
A p0 µ(E)2−α k0 p0 B p1 µ(E)2β k0 p1 = λ p0 −p λ p1 −p then we compute that k0 = However, k0 must be an integer, so we let "
p
p −p
p
p −p
log2 A 0 λB p11 0 . β p1 + α p0
log2 A 0 λB p11 0 k0 = β p1 + α p0 Indeed
0 (k−k ) 0
# .
0 2−(β −β )p1
1 − 2−(β −β
0 )p 1
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p −p
p
p −p
log2 A 0 λB p11 0 log2 A 0 λB p11 0 − 1 ≤ k0 ≤ . β p1 + α p0 β p1 + α p0 Therefore
µ({T (χE ) > λ }) ≤ (2
α 0 p0
µ(E) C1 +C2 ) p p λ 0A 0
Let p = (p1 − p0 ) and θ =
α α+β ,
then
1 p
=
1−θ p0
+
θ p1 ,
A p0 λ p1 −p0 B p1
−α p0 β p1 +α p0
−α p2
=C
A
αp p
0 1 p0 − β p +α0p 1 0 B β p1 +α p0 αp
λ
0 +p0 (p1 −p0 ) β p +α p0 1
µ(E) .
α p0 β p1 α p0 + p0 = p1 + p0 , β p1 + α p0 β p1 + α p0 β p1 + α p0
and p0 −
similarly
−α p20 −β p0 p1 β = = p, β p1 + α p0 β p1 + α p0 α +β α α p0 p1 = p, β p1 + α p0 α +β
therefore µ({T (χE ) > λ }) ≤
CA(1−θ )p Bθ p µ(E) , λp
hence T is of restricted weak type (p, p).
Exercise 1.4.17 Let (X, µ), (Y, ν) be measure spaces, 0 < p, r, q, s ≤ ∞ and 0 < B < ∞. Suppose that a sublinear operator T is defined on a dense subspace D of L p,r (X), takes values in the space of measurable functions of another measure space Y , and satisfies T ( f ) ≥ 0 for all f in D. Assume that
T (ϕ) q,s ≤ B ϕ p,r L L for all ϕ in D. Prove that T admits a unique sublinear extension T on L p,r (X) such that
T ( f ) q,s ≤ B f p,r L L p,r for all f ∈ L (X). p,r Hint: Given f ∈ L (X) find a sequence of functions ϕ j in D such that ϕ j → f in L p,r . Use the inequality T (ϕ j ) − T (ϕk ) ≤ T (ϕ j − ϕk ) , to obtain that the sequence {T (ϕ j )} j is Cauchy in Lq,s and thus it has a unique limit T ( f ) which is independent q,s of the choice of sequence ϕ j . Boundedness of T follows by density. To prove that T is sublinear use that convergence in L implies convergence in measure and thus a subsequence of T (ϕ j ) converges νa.e. to T ( f ). Also use Exercise 1.4.11.
Solution. Given f ∈ L p (X) find a sequence of functions ϕ j in D such that ϕ j → f in L p,r . In view of the inequality T (ϕ j ) − T (ϕk ) ≤ T (ϕ j − ϕk ) , it follows that the sequence {T (ϕ j )} j is Cauchy in Lq,s (Y, ν) and thus it has a limit. This limit is independent of the sequence ϕ j , since if ψ j is another sequence converging to f in L p,r , then
T (ϕ j ) − T (ψ j ) q,s ≤ T (ϕ j − ψ j ) q,s ≤ B ϕ j − ψ j p,r , L L L which implies that the sequences {T (ϕ j )} j and {T (ψ j )} j have the same limit, which we call T ( f ). One can easily see that T is a sublinear operator that coincides with T on D. Indeed, if ϕ 1j → f1 and ϕ 2j → f2 in L p,r , then T (ϕ 1j + ϕ 2j ) converges to
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T ( f1 + f2 ) in Lq,s . There is a subsequence jk of j such that T (ϕ 1jk ) converges to T ( f1 ) a.e., then there is a subsequence jk` such that T (ϕ 2jk ) converges to T ( f2 ) a.e. and finally there is a subsequence jk`m such that T (ϕ 1jk + ϕ 2jk ) converges to T ( f1 + f2 ) `m
`
a.e. Since T (ϕ 1jk
`m
`m
+ ϕ 2jk ) ≤ T (ϕ 1jk ) + T (ϕ 2jk ) `m
`m
`m
passing to the limit we obtain the sublinearity of T . Moreover, since {T (ϕ j )} j converge to T ( f ) in Lq,s , it follows from Exercise 1.4.11 (b) that kT (ϕ j )kLq,s converge to kT ( f )kLq,s , thus T is bounded from L p,r to Lq,s with norm at most B.
2. Maximal Functions, Fourier Transform, and Distributions
Section 2.1. Maximal Functions Exercise 2.1.1 A positive Borel measure µ on Rn is called inner regular if for any open subset U of Rn we have µ(U) = sup{µ(K) : K j U, K compact} and µ is called locally finite if µ(B) < ∞ for all balls B. (a) Let µ be a positive inner regular locally finite measure on Rn that satisfies the following doubling condition: There exists a constant D(µ) > 0 such that for all x ∈ Rn and r > 0 we have µ(3B(x, r)) ≤ D(µ) µ(B(x, r)). 1 (Rn , µ) define the uncentered maximal function M ( f ) with respect to µ by For f ∈ Lloc µ
Mµ ( f )(x) = sup
sup
r>0 z: z−x α} is open. Then use the argument of the proof of Theorem 2.1.6 and the inner regularity of µ. Solution. (a) If µ(Rn ) = 0, then Mµ ( f ) = 0 for all f ∈ L1 (Rn , µ) and in this case Mµ maps L1 (Rn , µ) to L1,∞ (Rn , µ) with constant 0. Now suppose that µ(Rn ) > 0, then the assumption µ(3B(x, r)) ≤ D(µ)µ(B(x, r)) implies µ(B(x, r)) > 0 for all x ∈ Rn , r > 0. Consider the operator Z 1  f (y) dµ(y). Mµ ( f )(x) = sup sup r>0 z∈B(x,r) µ(B(z, r)) B(z,r) It is easy to see that Mµ maps L∞ (X, µ) to itself with constant 1. We need to show that Mµ maps L1 (X, µ) to L1,∞ (X, µ) with constant at most D(µ). By applying the Marcinkiewicz interpolation Theorem, we imply that Mµ maps L p (X, µ) to itself with 1/p p norm at most 2 D(µ)1/p . p−1 Indeed, for f ∈ L1 (Rn , µ), denote Eα = x ∈ Rn : Mµ ( f )(x) > α . We will show that Eα is open for α > 0. Take x ∈ Eα , since Mµ ( f )(x) > α, there exist a ball Bx containing x such that 1 µ(Bx )
Z
 f (y) dµ(y) > α.
Bx
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Hence Mµ ( f )(w) > α for all w ∈ Bx . It means that Bx j Eα . Now let K be an arbitrary compact subset of Eα . By the above argument, for each x ∈ K, there exists such a ball Bx such that x ∈ Bx j Eα . The cover of open sets (Bx )x∈K of K has a finite cover, say Bx1 , . . . , Bxk . Then we can choose pairwise disjoint balls Bx j1 , . . . , Bx jl from the later cover such that ∪ki=1 Bxi j ∪li=1 3Bx ji . The measure of K now can be estimated as follows l
l
µ(K) ≤ µ(∪ki=1 Bxi ) ≤ µ(∪li=1 3Bx ji ) ≤ ∑ µ(3Bx ji ) ≤ D(µ) ∑ µ(Bx ji ). i=1
i=1
On the other hand, D(µ) l D(µ) ∑ µ(Bx ji ) < ∑ α i=1 i=1 l
Z Bx j
i
D(µ)  f (y) dµ(y) ≤ α
Z Rn
 f (y) dµ(y).
Taking the supremum over all compact K j Eα and using the inner regularity of µ, we deduce that α µ(Eα ) ≤ D(µ)
Z Rn
 f (x) dµ(x),
α > 0.
This follows that Mµ map L1 (Rn , µ) to L1,∞ (Rn , µ) with constant at most D(µ). (b) The analogue Corollary 2.1.16 is as follows: For any locally integrable function f on Rn (with respect to µ) we have 1 r→0 µ(B(x, r))
Z
f (y) dµ(y) = f (x)
lim
B(x,r)
for almost all x in Rn with respect to µ. Consequently we have  f  ≤ Mµ ( f ) µa.e. Proof: Since Rn is the union of the balls B(0, N) for N = 1, 2, 3 . . . , it suffices to prove the required conclusion for µalmost all x inside a fixed ball B(0, N). Given a locally integrable function f on Rn with respect to µ, consider the function f0 = f χB(0,N+1) . Then f0 lies in L1 (Rn , µ). Let Tε be the operator given by 1 µ(B(x, ε))
Z
f (y) dµ(y) B(x,ε)
for 0 < ε < 1. The corresponding maximal operator T∗ is controlled by the maximal function Mµ ( f ), which maps L1 (Rn , dµ) to L1,∞ (Rn , dµ). It is straightforward to verify that the claim holds for all continuous functions f with compact support. Since this set of functions is dense in L1 (Rn , dµ), and T∗ maps L1 (Rn , dµ) to L1,∞ (Rn , dµ) by part (a), Theorem 2.1.14 implies that required a.e. convergence holds for all functions in L1 (Rn , dµ), in particular for f0 . But for 0 < ε < 1 and x ∈ B(0, N) we have f χB(x,ε) = f0 χB(x,ε) , so it follows that 1 ε→0 B(x, ε)
Z
1 ε→0 B(x, ε)
Z
f (y) dµ(y) = lim
lim
B(x,ε)
B(x,ε)
f0 (y) dµ(y) = f0 (x)
for µalmost all x ∈ Rn , in particular for µalmost all x in B(0, N). But on this set f0 = f , so the required conclusion follows. The last assertion that  f  ≤ Mµ ( f ) a.e. is an easy consequence of limiting identity proved, when the limit is replaced by a supremum.
Exercise 2.1.2 On R consider the maximal function Mµ of Exercise 2.1.1. (a) (W. H. Young ) Prove the following covering lemma.Given a finite set F of open intervals in R, prove that there exist two subfamilies each consisting of pairwise disjoint intervals such that the union of the intervals in the original family is equal to the union of the intervals of both subfamilies. Use this result to show that the maximal function Mµ of Exercise 2.1.1 maps L1 (µ) → L1,∞ (µ) with constant at most 2. 1 (R, µ) we have (b) ([135]) Prove that for any σ finite positive measure µ on R, α > 0, and f ∈ Lloc
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1 α
Z
 f  dµ − µ(A) ≤
A
1 α
Z { f >α}
 f  dµ − µ({ f  > α}) .
Use this result and part (a) to prove that for all α > 0 and all locally integrable f we have µ({ f  > α}) + µ({Mµ ( f ) > α}) ≤
1 α
Z
 f  dµ +
{ f >α}
1 α
Z
 f  dµ
{Mµ ( f )>α}
and note that equality is obtained when α = 1 and f (x) = x−1/p . (c) Conclude that Mµ maps L p (µ) to L p (µ), 1 < p < ∞, with bound at most the unique positive solution A p of the equation (p − 1) x p − p x p−1 − 1 = 0 . (d) ([137]) If µ is the Lebesgue measure show that for 1 < p < ∞ we have
M p p = A p , L →L where A p is the unique positive solution of the equation in part (c). S S Hint: Part (a): Select a subset G of F with minimal cardinality such that J∈G J = I∈F I. Part (d): One direction follows from part (c). Conversely, M(x−1/p )(1) =
0 p γ 1/p +1 p−1 γ+1 ,
where γ is the unique positive solution of the equation
0 p γ 1/p +1 p−1 γ+1
= γ −1/p .
Conclude that M(x−1/p )(1) = A p and that M(x−1/p ) = A p x−1/p . Since this function is not in L p , consider the family 1 +ε fε (x) = x−1/p min(x−ε , xε ), ε > 0, and show that M( fε )(x) ≥ (1 + γ p0 )(1 + γ)−1 ( p10 + ε)−1 fε (x) for 0 < ε < p0 . Solution. (a) Let F be the given finite collection of intervals and let ∪F be the union of all intervals in F . Consider the set S of all subcollections G of F such that ∪G = ∪F . Pick a subcollection F 0 in S with minimal cardinality. Then the union of all intervals in F 0 is equal to ∪F and no interval in F 0 is contained in a union of other intervals in F 0 . Say that F 0 = {(α1 , β1 ), . . . , (αN , βN )}, where α1 < α2 < · · · < αN . Set F1 = {(α j , β j ) : 1 ≤ j ≤ N, j odd} and F2 = {(α j , β j ) : 1 ≤ j ≤ N, j even}. Then (∪F1 ) ∪ (∪F2 ) = ∪F 0 = ∪F and one can easily see that by of the choice of the subfamily F 0 , the intervals in each family F1 , F2 are pairwise disjoint. We now show that the maximal function Mµ of Exercise 2.1.1 maps L1 (µ) → L1,∞ (µ) with constant at most 2. For each R x ∈ Eα = {x : Mµ ( f )(x) > α} find an open interval Ix containing x such that µ(Ix ) < α1 Ix  f (t) dµ(t). Since for every y ∈ Ix we have M( f )(y) > α, it follows that Ix j Eα . Thus the collection {Ix }x∈Eα is an open covering of Eα . Let K be a compact subset of Eα and let F = {Ix1 , . . . , Ixm } be a finite subcover of K. Let F1 and F2 as before. Then µ(K) ≤ µ(∪F ) ≤ µ(∪F1 ) + µ(∪F2 )
α} α { f >α} R
Now fix α > 0 and let Eα = {Mµ ( f ) > α}. If µ({ f > α}) = ∞, then there is nothing to prove. Hence we may assume that µ({ f > α}) < ∞. For every x ∈ Eα there is an open interval Ix containing x such that 1 µ(Ix )
Z
 f (t) dµ(t) > α.
Ix
As in part (a), it is easy to see that Ix j Eα . By Lindel¨of’s theorem there is a countable subcollection I j , j = 1, 2, . . . , such that ∪∞j=1 I j = Eα . Let F = F N = {I j : j = 1, 2, ..., N}. By part (a) we obtain two subcollections F1 and F2 of F , each with pairwise disjoint intervals, a fact that implies
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µ(∪Fi ) =
1 α
µ(I)
α}). α ∪F N α { f >α} We obtain the final conclusion by letting N → ∞ and by using the fact that F N is an increasing sequence of µ−measurable sets whose union is Eα . (c) Let us take f ≥ 0. The previous inequality implies Z R
Mµ ( f ) p dµ +
Z
f p dµ =p
R
Z ∞ 0
≤p
Z ∞
α p−2
Z
and hence
Z R
p p−1
Mµ ( f ) p dµ ≤
Z R
Z
{Mµ ( f )>α}
Z R
α p−1 µ({ f > α}) dα
α p−2
0
Mµ ( f ) p−1 f dµ +
p p−1
0 ∞
f dµ dα + p
0
=
Z ∞
α p−1 µ({Mµ ( f ) > α}) dα + p
p p−1
Mµ ( f ) p−1 f dµ +
Z
Z
f dµ dα { f >α}
f p dµ
R
1 p−1
Z
f p dµ.
R
H¨older’s inequality gives Z R
and hence
Mµ ( f ) p−1 f dµ ≤
Z R
Mµ ( f ) p dµ
(p−1)/p Z
f p dµ
1/p
R
p (p − 1)kMµ ( f )kLp p (µ) ≤ pkMµ ( f )kLp−1 p (µ) k f kL p (µ) + k f kL p (µ)
or equivalently kMµ ( f )k p p−1 kMµ ( f )k p p L (µ) L (µ) −p − 1 ≤ 0. (p − 1) k f kL p (µ) k f kL p (µ) This shows that the ratio kMµ ( f )kL p (µ) /k f kL p (µ) is less than or equal than the unique positive solution A p of the equation (p − 1)x p − px p−1 − 1 = 0. 1
(d) Let f0 (t) = t− p . Homogeneity gives that M( f0 ) is a constant multiple of f0 and thus it suffices to compute M( f0 )(1). An easy calculation gives that 0 p γ 1/p + 1 M( f0 )(1) = , p−1 γ +1 where γ is the unique positive solution of the equation 0
p γ 1/p + 1 = γ −1/p . p−1 γ +1 It is now a matter of simple algebra to show that M( f0 )(1) = γ −1/p is the unique positive root A p of the equation (p − 1)x p − px p−1 − 1 = 0.
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Let fε (x) = x−1/p gε (x), where gε (x) = min(x−ε , xε ). It is not difficult to check that for 0 < ε < 1/p0 we have M( fε )(x) ≥
1 x + γx
0
Z x −γx
t−1/p gε (t) dt ≥
This implies that
1 + γ 1/p +ε x−1/p gε (x). (1/p0 + ε)(γ + 1)
0
kM( fε )kL p 1 + γ 1/p +ε = γ −1/p = A p . lim ≥ lim ε→0 k f ε kL p ε→0 (1/p0 + ε)(γ + 1) The last statement in this exercise is equivalent to the identity 2 + 2A pp =
Z
t−1/p dt +
t≤1
Z p
t−1/p dt
t≤A p
which is equivalent to the fact that A p is the unique positive solution of the equation (p − 1)x p − px p−1 − 1 = 0.
Exercise 2.1.3 Define the centered Hardy–Littlewood maximal function Mcand the uncentered Hardy–Littlewood maximal function Mc using cubes with sides parallel to the axes instead of balls in Rn . Prove that 1≤
M( f ) ≤ 2n , M( f )
1 2n 2n M( f ) ≤ , ≤ n Mc ( f ) vn n 2 vn
2n 1 2n M( f ) ≤ , ≤ n Mc ( f ) vn n 2 vn
where vn is the volume of the unit ball in Rn . Conclude that Mc and Mc are weak type (1, 1) and they map L p (Rn ) to L p (Rn ) for 1 < p ≤ ∞. Solution. 1 R n Let Mc ( f ) = sup Q  f  dx, where Q is any cube with sides parallel to the axes of R . Now suppose B(y, r) be a ball Q Q3x centered at y with radius r. Let Q be a circumscribed cube of the ball, Q = {w ∈ Rn : wi − yi  ≤ 2r, 1 ≤ r ≤ n} . Then we have the estimate Z Z 1 2n 1   f (z) dz. f (z) dz ≤ vn rn B(y,r) vn (2r)n Q 2n Mc ( f ). To obtain the bound below, we take any cube Q centered at y with side length r. Let B be a vn √ circumscribed ball of the cube with radius ρ = r 2 n . The inequality This implies M( f ) ≤
1 rn
√ n Z n 1  f (z) dz ≤ vn  f (z) dz 2 vn ρ n B Q
Z
yields M( f ) ≥
2n n Mc ( f ). vn n 2
Likewise, we get the following inequalities for the centered HardyLittlewood maximal operator M( f ) ≤
2n 2n Mc ( f ) and M( f ) ≥ n Mc ( f ). vn vn n 2
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Exercise 2.1.4 (a) Prove the estimate: {x ∈ Rn : M( f )(x) > 2α} ≤
3n α
Z
 f (y) dy
{ f >α}
and conclude that M maps L p to L p,∞ with norm at most 2 · 3n/p for 1 ≤ p < ∞. Deduce that if f log+ (2 f ) is integrable over a ball B, then M( f ) is integrable over the same ball B. (b) ([375], [338]) Apply Proposition 2.1.20 to  f  and α > 0 and Exercise 2.1.3 to show that with cn = 2n (nn/2 vn )−1 we have {x ∈ Rn : M( f )(x) > cn α} ≥
2−n α
Z
 f (y) dy .
{ f >α}
(c) Suppose that f is integrable and supported in a ball B(0, ρ). Show that for x in B(0, 2ρ) \ B(0, ρ) we have M( f )(x) ≤ M( f )(ρ 2 x−2 x). Conclude that Z Z B(0,2ρ)
M( f ) dx ≤ (4n + 1)
B(0,ρ)
M( f ) dx
and from this deduce a similar inequality for M( f ). (d) Suppose that f is integrable and supported in a ball B and that M( f ) is integrable over B. Let λ0 = 2n B−1 k f kL1 . Use part + −1 (b) to prove that f log (λ0 cn  f ) is integrable over B. Hint: Part (a): Write f = f χ f >α + f χ f ≤α . Part (c): Let x0 = ρ 2 x−2 x for some ρ < x < 2ρ. Show that for R > x − ρ, we have that Z Z  f (z) dz ≤  f (z) dz B(x0 ,R)
B(x,R)
B(x, R) ∩ B(0, ρ) j B(x0 , R).
by showing that M( f )(x) > α} dα.
Part (d): For x ∈ / 2B we have M( f )(x) ≤ λ0 , hence
R∞ 2B M( f )(x) dx ≥ λ0 {x ∈ 2B :
R
Solution. (a) Denote Eα = {x ∈ Rn :  f (x) > α} , f1 = f χEα and f2 = f − f1 . Then M( f ) ≤ M( f1 ) + M( f2 ). It follows that {x ∈ Rn : M( f )(x) > 2α} ≤ {x ∈ Rn : M( f1 )(x) > α} + {x ∈ Rn : M( f2 )(x) > α} . Since  f2 (x) ≤ α for all x ∈ Rn , M( f2 )(x) ≤ α for all x ∈ Rn . Moreover, we have the following inequality 3n α {x∈Rn Z 3n = α {x∈Rn
{x ∈ Rn : M( f1 )(x) > α} ≤
Z
 f1  dx
: M( f1 )(x)>α}
 f  χEα dx ≤
: M( f1 )(x)>α}
Combining together, we get
3n α
Z
3n αp
Z
{x ∈ Rn : M( f )(x) > 2α} ≤
3n α
Z
 f  dx.
Eα
 f  dx.
Eα
From the above inequality, we deduce {x ∈ Rn : M( f )(x) > 2α} ≤ Hence (2α) p {x ∈ Rn : M( f )(x) > 2α} ≤ 3n 2 p
Z Eα
n
 f  p dx.
Eα
 f  p dx ≤ 3n 2 p
Z Rn
 f  p dx.
We deduce that M maps L p to L p,∞ with constant at most 2 · 3 p . (b) Now suppose that f log+ (2  f ) is integrable on a ball B. Denote E = x ∈ Rn :  f (x) ≥ 12 , f1 = f χE and f2 = f − f1 = f χRn \E . First we note that M( f ) ≤ M( f1 ) + M( f2 ). Since  f2  ≤ 21 , M( f2 ) ≤ 21 and hence M( f2 ) is integrable on B. We need to show M( f1 ) is also integrable on B. Indeed, from the above argument, we have
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Z 2 · 3n
{x ∈ B : M( f1 )(x) > α} ≤
α
Eα
 f1  dx,
n αo where Eα = x ∈ B :  f1 (x) > j E. Furthermore, 2 Z ∞
Z B
Note that
R1 0
M( f1 )(x)dx =
0
{x ∈ B : M( f1 )(x) > α} dα.
{x ∈ B : M( f1 )(x) > α} dα ≤ B < ∞. We just only need to show Z ∞ 1
We have the estimate
Z ∞ 1
{x ∈ B : M( f1 )(x) > α} dα < ∞.
{x ∈ B : M( f1 )(x) > α} dα ≤ ≤
Z ∞ Z 2 · 3n
α
1
Z
 f (x)
Z
 f1 (x) dxdα Eα 2 f (x) 2 · 3n α
1
E
≤2 · 3n =2 · 3n
Z ZE
dαdx
 f (x) log(2  f (x))dx  f (x) log+ (2  f (x))dx < ∞.
B
(c) Proposition 2.1.20 shows that there exist disjoint open cubes Q j such that for almost all x ∈ / ∪ j∈N Q j we have  f (x) ≤ α and 1 α < Qj
Z
 f (x) dx < 2n α.
Qj
It follows that {x ∈ Rn :  f (x) > α} ⊂ ∪ j∈N Q j a.e. and Mc ( f )(x) > α for all x ∈ ∪ j∈N Q j . By Exercise 2.1.3 and cn = we can estimate
{x ∈ Rn : M( f )(x) > cn α} ≥ {x ∈ Rn : Mc ( f )(x) > α} Z 1  f (x) dx ≥ ∑ Q j ≥ n ∑ 2 α j∈N Q j j∈N 1 = n 2 α
Hence {x ∈ Rn : M( f )(x) > cn α} ≥
1 2n α
Z
Z
 f (x) dx.
∪ j∈N Q j
 f (x) dx ≥
∪ j∈N Q j
1 2n α
Z {x∈Rn :  f (x)>α}
 f (x) dx.
(d) First, suppose that x ∈ Rn , x > ρ and y ∈ B(0, ρ). We will show that x0 − y < x − y , where x0 = about the ball boundary. Denote λ = the norm of
x0 − y
2n n , vn n 2
x·y , z = λ x, u = x − λ x, δ x2
=
y ρ x . Then λ  ≤ x
ρ2 x x2
is symmetric of x
< δ < 1 and u · x = 0. Now we can compute
and x − y as follows
0 x − y 2 = δ 2 − λ 2 x2 + u2 and x − y2 = (1 − λ )2 x2 + u2 . We need to show that δ 2 − λ < 1 − λ , or δ 2 − λ + λ < 1. This is clear because δ 2 − λ + λ ≤ δ 2 − λ + λ  ≤ δ 2 < 1. Hence B(x, R) ∩ B(0, ρ) j B(x0 , R) for all R > 0. Now if R ≤ x − ρ then B(x, R) ∩ B(0, ρ) = 0. / It follows 1 B(x, R)
1  f (y) dy = 0 ≤ 0 , R) B(x B(x,R)
Z
Consider the case R > x − ρ. Then since B(x, R) ∩ B(0, ρ) j B(x0 , R),
Z B(x0 ,R)
 f (y) dy.
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1 B(x, R)
Z
 f (y) dy =
B(x,R)
1 B(x, R)
Z
 f (y) dy ≤
B(x,R)∩B(0,ρ)
Thus M( f )(x) ≤ M( f )(x0 ) = M( f )( This derives
Z B(0,2ρ)
M( f )(x)dx = ≤
Z B(0,ρ)
Z B(0,ρ)
ρ2
B(x0 ,R)
 f (y) dy.
for all x ∈ Rn , x > ρ.
x),
x2
1 B(x0 , R)
Z
M( f )(x)dx +
Z
M( f )(x)dx +
Z
B(0,2ρ)\B(0,ρ)
B(0,2ρ)\B(0,ρ)
M( f )(x)dx M( f )(
ρ2 x2
x)dx.
Now after two times changing variables, we have Z B(0,2ρ)\B(0,ρ)
M( f )(
ρ2
Z 2ρ
Z
x)dx = x2
Sn−1 ρ
Z ρ
Z
=
Sn−1
≤4n
Z
≤4n
Z
=4n
Z
ρ 2
ρ 2
Sn−1
Z
Z ρ
Sn−1 0
B(0,2ρ)
M( f )(x)dx ≤ (4n + 1)
ρ2 θ )rn−1 drdθ r
M( f )(uθ )
Z ρ
B(0,ρ)
Hence
M( f )(
ρ 2n n−1 u dudθ u2n
M( f )(uθ )un−1 dudθ M( f )(uθ )un−1 dudθ
M( f )(x)dx.
Z B(0,ρ)
M( f )(x)dx.
In general, we have the inequality Z B(0,λ ρ)
M( f )(x)dx ≤ (λ 2n + 1)
Z B(0,ρ)
M( f )(x)dx,
λ > 0.
Moreover, note that ρ2 ρ2 x ≤ 2n M( f ) x , M( f )(x) ≤ 2n M( f )(x) ≤ 2n M 2 x x2
x > ρ.
We get the following inequality for the uncentered maximal HardyLittlewood operator Z
M( f )(x)dx ≤ (8n + 1)
Z
M( f )(x)dx. B(0,ρ)
B(0,2ρ)
(e) Denote E = x ∈ Rn :  f (x) ≥ λ0 c−1 . Since f is supported in the ball B = B(0, ρ), it is easy to check n M( f )(x) ≤
1 B
Z Rn
 f (x) dx,
x∈ / 2B.
This yields M( f )(x) ≤ 2n B−1 k f kL1 = λ0 for x ∈ / 2B. By part (b) and (c), we have
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(1 + 8n )
Z
M( f )(x)dx ≥
B
≥
Z ∞
Z
M( f )(x)dx = Z2B∞
x ∈ 2B : M( f )(x) > α dα
0
x ∈ 2B : M( f )(x) > α dα =
Z ∞
≥
x ∈ Rn : M( f )(x) > cn α dα
λ0 c−1 n Z ∞
cn 2n
x ∈ Rn : M( f )(x) > α dα
λ0
λ0
=cn
Z ∞
λ0 c−1 n
1 α
Z {x∈Rn :  f (x)>α}
 f (x) dxdα ≥
cn 2n
Z ∞ λ0 c−1 n
1 α
Z
 f (x) dxdα
{x∈E :  f (x)>α}
 f (x) 1 cn cn = n  f (x) dαdx = n  f (x) log(λ0−1 cn  f (x))dx 2 E α 2 λ0 c−1 E n Z cn = n  f (x) log+ (λ0−1 cn  f (x))dx. 2 B
Z
Z
Z
Exercise 2.1.5 (A. Kolmogorov) Let S be a sublinear operator that maps L1 (Rn ) to L1,∞ (Rn ) with norm B. Suppose that f ∈ L1 (Rn ). Prove that for any set A of finite Lebesgue measure and for all 0 < q < 1 we have Z A
q S( f )(x)q dx ≤ (1 − q)−1 Bq A1−q f L1 ,
and in particular, for the Hardy–Littlewood maximal operator, Z A
q M( f )(x)q dx ≤ (1 − q)−1 3nq A1−q f L1 .
Hint: Use the identity Z
q
S( f )(x) dx =
Z ∞
q α q−1 {x ∈ A : S( f )(x) > α} dα
0
A
and estimate the last measure by min(A, αB f L1 ). Solution. If A = 0, then we have nothing to prove. Consider 0 < A < ∞. Set λ = A−1 B k f kL1 (Rn ) . Now we split the integral R q A S( f )(x) dx as follows Z
S( f )(x)q dx =q
Z ∞
α q−1 {x ∈ A : S( f )(x) > α} dα
0
A
Z λ
=q
α q−1 {x ∈ A : S( f )(x) > α} dα + q
Z ∞
0
≤q
Z λ
α q−1 A dα + q
0
Z ∞ λ
≤λ q A + =
α q−1 {x ∈ A : S( f )(x) > α} dα
λ
α q−2 B k f kL1 (Rn ) dα
q λ q−1 B k f kL1 (Rn ) 1−q
A1−q Bq k f kqL1 (Rn ) 1−q
.
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Exercise 2.1.6 Let Ms ( f )(x) be the supremum of the averages of  f  over all rectangles with sides parallel to the axes containing x. The operator Ms is called the strong maximal function (a) Prove that Ms maps L p (Rn ) to itself. (b) Show that the operator norm of Ms is Anp , where A p is as in Exercise 2.1.2(c). (c) Prove that Ms is not weak type (1,1). Solution. (a) Let M ( j) denote the onedimensional uncentered HardyLittlewood maximal function acting on the jth variable in Rn . Observe that Ms = M (1) ◦ · · · ◦ M (n) which implies that Ms maps L p (Rn ) into itself with norm at most Anp . (b) To obtain that Anp is indeed the norm of Ms on L p (Rn ) consider the family of functions n
fε (x) = ∏ x j −1/p min(x j ε , x j −ε ). j=1
Repeating the argument in Exercise 2.1.2(d) we obtain that kMs ( fε )kL p /k fε kL p → Anp as ε → ∞. (c) Let f0 = χ[0,1]×[0,1] in R2 . Then for x = (x1 , x2 ) with x1 , x2 > 10 we have Ms ( f0 ) ≥ x1−1 x2−1 . But the function (x1 x2 )−1 is not in L1,∞ (R) where R = (10, ∞) × (10, ∞), since the measure of the set of all (x1 , x2 ) in R with (x1 x2 )−1 > 1 is equal to infinity.
Exercise 2.1.7 Prove that if ϕ(x1 , . . . , xn ) ≤ A(1 + x1 )−1−ε · · · (1 + xn )−1−ε for some A, ε > 0, and ϕt1 ,...,tn (x) = t1−1 · · ·tn−1 ϕ(t1−1 x1 , . . . ,tn−1 xn ), then the maximal operator f 7→
sup  f ∗ ϕt1 ,...,tn 
t1 ,...,tn >0
is pointwise controlled by the strong maximal function. Solution. Let ψ(x) = (1+x)−1−ε . Then ψ ∈ L1 (R). Denote M ( j) is the unentered HardyLittlewood maximal operator with respect to the x j variable. It is clear that the strong maximal operator is the composition of M (n) ◦· · ·◦M (1) . Since ϕ(x) ≤ Aψ(x1 ) . . . ψ(xn ), the maximal operator f 7−→ sup f ∗ ϕt1 ,...,tn t1 ,...,tn >0
is controlled by
A kψknL1 (R) M (n) ◦ · · · ◦ M (1) ;
hence the strong maximal operator controls the above maximal operator.
Exercise 2.1.8 p n → R∞. Prove that for any fixed 1 < p < ∞, the operator norm of M on nL (R ) tends to infinity as n −1 Hint: Let f0 be the characteristic function of the unit ball in R . Consider the averages Bx  Bx f 0 dy, where Bx = B x 1 −1 −1 x ) x , 2 (x + x ) for x > 1.
Solution.
1 2 (x −
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Let f0 = χB(0,1) and Bx = B
1 −1 x 1 −1 2 (x − x ) x , 2 (x + x )
. Then B(0, 1) ∩ Bx has measure at least vn /2, where vn is the
volume of B(0, 1). It follows that M( f0 )(x) ≥ 2n−1 (x + x−1 )−n for x > 1. Using polar coordinates we obtain Z ∞ kM( f0 )kLp p 1 rn−1 (n−1)p ≥ dr , vn + ωn−1 2 np vn k f0 kLp p r=1 (r + 1/r) where ωn−1 = nvn is the surface area of the unit sphere Sn−1 . It suffices to show that the expression Wn = n2(n−1)p
rn−1 dr np r=1 (r + 1/r)
Z ∞
tends to infinity as n → ∞. First change variables r = tan φ to write Wn = n2(n−1)p
Z π/2
(sin φ )np+n−1 (cos φ )np−n−1 dφ
π/4
and then set u = cos2 φ to obtain Wn = 2−p−1 2np n
Z 1/2
np
n
np
n
(1 − u) 2 + 2 −1 u 2 − 2 −1 du.
0
For any 0 < t ≤ 1/2 we estimate Wn from below by 2−p−1 2np n
Z t
np
n
np
np
n
n
(1 − u) 2 + 2 −1 u 2 − 2 −1 du = 2−p (p − 1)−1 2np (1 − t) 2 + 2 −1t
np n 2 −2
0
and select t =
np
n − (np − 1)−1 to maximize the last expression above. One obtains that 2 2 n p 1 p 1 Wn ≥ c p (p + 1) 2 + 2 (p − 1) 2 − 2 p−p
as n → ∞ and since the expression inside the curly brackets above is strictly bigger than 1 when 1 < p < ∞, it follows that the operator norm of M on L p (Rn ) grows exponentially with the dimension.
Exercise 2.1.9 (a) In R2 let M0 ( f )(x) be the maximal function obtained by taking the supremum of the averages of  f  over all rectangles (of arbitrary orientation) containing x. Prove that M0 is not bounded on L p (Rn ) for p ≤ 2 and conclude that M0 is not weak type (1, 1). (b) Let M00 ( f )(x) be the maximal function obtained by taking the supremum of the averages of  f  over all rectangles in R2 of arbitrary orientation but fixed eccentricity containing x. (The eccentricity of a rectangle is the ratio of its longer side to its shorter side.) Using a covering lemma, show that M00 is weak type (1, 1) with a bound proportional to the square of the eccentricity. (c) On Rn define a maximal function by taking the supremum of the averages of  f  over all products of intervals I1 × · · · × In containing a point x with I2  = a2 I1 , . . . , In  = an I1  and a2 , . . . , an > 0 fixed. Show that this maximal function is of weak type (1, 1) with bound independent of the numbers a2 , . . . , an . Hint: Part (b): Let b be the eccentricity. If two rectangles with the same eccentricity intersect, then the smaller one is contained in the bigger one scaled 4b times. Then use an argument similar to that in Lemma 2.1.5. Solution. (a) Let h(x) be the characteristic function of the unit disc B(0, 1) in R2 . Then for x ≥ 10 we have that M0 (h)(x) ≥
R ∩ B(0, 1) R
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where R is the rectangle of dimensions x + 1 and 2 that contains B(0, 1) and has the point x at its shortest side. It follows that for x ≥ 10 we have π M0 (h)(x) ≥ 2(x + 1) and this function does not lie in L p (R2 ) when p ≤ 2. (b) Noting that if I and J are two rectangles with the same eccentricity such √ that I ∩ J 6= 0/ and diag(J) ≤ diag(I). Assume a and b = ra (r ≥ 1) are sides of I, then J is contained in the disc 3B(x0 , a2 1 + r2 ), where x0 is the center of the rectangle I. √ √ Therefore the rectangle 3 2rI contains 3B(x0 , a2 1 + r2 ), hence contains J. Incorporating these elements in the proof for the weak type (1, 1) boundedness of the HardyLittlewood maximal operator, we obtain the estimate kM00 ( f )kL1,∞ ≤ 18r2 k f kL1 . (c) The maximal operator in the problem is bounded above by the strong maximal operator in the Exercise 2.1.6.
Exercise 2.1.10 (a) Let 0 < p, q < ∞ and let X,Y be measure spaces. Suppose that Tε are maps from L p (X) to Lq,∞ (Y ) satisfy Tε ( f + g) ≤ K(Tε ( f ) + Tε (g)) for all ε > 0 and all f , g ∈ L p (X), and also limε→0 Tε ( f ) = 0 a.e. for all f in some dense subspace D of L p (X). Assume furthermore that the maximal operator T∗ ( f ) = supε>0 Tε ( f ) maps L p (X) to Lq,∞ (Y ). Prove that limε→0 Tε ( f ) = 0 a.e. for all f in L p (X). (b) Use the result in part (a) to prove the following version of the Lebesgue differentiation theorem:Let f ∈ L p (Rn ) for some 0 < p < ∞. Then for almost all x ∈ Rn we have 1 B→0 B
Z
lim
g(y) − g(x) p dy = 0 ,
B
B3x
where the limit is taken over all open balls B containing x and shrinking to {x}. 1 (Rn ) and for almost all x ∈ Rn we have (c) Conclude that for any f in Lloc 1 B→0 B
Z
f (y) dy = f (x) ,
lim
B
B3x
where the limit is taken over all open balls B containing x and shrinking to {x}. Hint: (a) Define an oscillation O f (y) = lim supε→0 Tε ( f )(y). For all f in L p (X) and g ∈ D we have that O f (y) ≤ KO f −g (y). Then use the argument in the proof of Theorem 2.1.14. (b) Apply part (a) with Tε ( f )(x) = sup B(z,ε)3x
1 B(z, ε)
Z
p
 f (y) − f (x) dy
1/p ,
B(z,ε)
1−p 1 observing that T∗ ( f ) = supε>0 Tε ( f ) ≤ max(1, 2 p )  f  + M( f  p ) p . (c) Follows from part (b) with p = 1. Note that part (b) can be proved without part (a) but using part (c) as follows: for every rational number a there is a set Ea of Lebesgue measure 1 R p p p zero such that for x ∈ Rn \ Ea we have limB3x,B→0 B B g(y) − a dy = g(x) − a , since the function y 7→  f (y) − a is in
1 (Rn ). By considering an enumeration of the rationals, find a set of measure zero E such for x ∈ Lloc / E the preceding limit exists for all rationals a and by continuity for all real numbers a, in particular for a = g(x).
Solution. (a) Given f ∈ L p (X), we define an oscillation function O f (y) = lim sup Tε ( f )(y) ,
y ∈ Y.
ε→0
It is easy to see that O f is measurable, O f ≤ T∗ ( f ) for f ∈ L p (X). Now we have a claim that for all f ∈ L p (X) and δ > 0
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ν( y ∈ Y : O f (y) > δ ) = 0. Indeed, given η > 0, there exists a function g ∈ D such that k f − gkL p (X) < η. Since lim Tε (g) = 0 for νa.e, it follows that ε→0
Og = 0 νa.e. The definition of the oscillation function implies that O f ≤ K(O f −g + Og ) = KO f −g ,
ν − a.e.
For any δ > 0, we have the estimate ν( y ∈ Y : O f (y) > δ ) ≤ν( y ∈ Y : KO f −g (y) > δ ) ≤ν({y ∈ Y : KT∗ ( f − g)(y) > δ }) ≤(K kT∗ k k f − gkL p (X) )q δ −q ≤(K kT∗ k η)q δ −q . Letting η → 0, the claim is verified. This yields A f = 0 νa.e. for all f ∈ L p (X). Hence lim Tε ( f ) = 0 νa.e. for f ∈ L p (X). ε→0
(b) For ε > 0 and f ∈ L p (Rn ), define the map Tε ( f )(x) = sup B(z,ε)3x
1 B(z, ε)
1 p  f (y) − f (x) dy ,
Z
p
x ∈ Rn .
B(z,ε)
For given functions f , g ∈ L p (Rn ), we have 1 p  f (y) − f (x) p dy
"Z 1 p 1−p [ f (y) + g(y)] − [ f (x) + g(x)] dy ≤ max 1, 2 p
Z
p
B(z,ε)
B(z,ε)
Z
p
g(y) − g(x) dy
+
1 # p
.
B(z,ε)
This yields the inequality for all f , g: Tε ( f + g) ≤ K(Tε ( f ) + Tε (g)),
1−p where K = max 1, 2 p .
Now we need to check T∗ ( f ) = supε>0 Tε ( f ) mapping L p (Rn ) to L p,∞ (Rn ). By applying the same argument as above, we have 1 T∗ ( f )(x) ≤ K  f (x) + (M( f  p )(x)) p ,
1−p K = max 1, 2 p .
For α > 0, the following inequality is obvious n 1 α o n α o p {x ∈ Rn : T∗ ( f )(x) > α} ≤ x ∈ Rn :  f (x) > + x ∈ Rn : M( f  )(x) p > . 2K 2K Since M maps L1 (Rn ) to L1,∞ (Rn ) with constant at most 3n , we imply that n 1 α o αp p p n n p   x ∈ R : M( f )(x) > = x ∈ R : M( f )(x) > p 2K (2K) (2K) p p (2K) p k f kL 1 = 3n k f kLp p ≤3n p α αp and
Z n α o 2K x ∈ Rn :  f (x) > ≤ 2K α {x∈Rn
α :  f (x)> 2K }
(2K) p ≤ α p {x∈Rn
 f (x) dx
Z
α :  f (x)> 2K }
 f (x) p dx ≤
(2K) p k f kLp p . αp
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Therefore we have the estimate 1
1
α {x ∈ Rn : T∗ ( f )(x) > α} p ≤ 2K(1 + 3n ) p k f kL p . 1
Thus T∗ ( f ) maps L p (Rn ) to L p,∞ (Rn ) with constant at most 2K(1 + 3n ) p . For f ∈ C0 (Rn ), the continuity of f implies that for all x ∈ Rn .
lim Tε ( f )(x) = 0,
ε→0
Hence lim sup ε→0 B(z,ε)3x
1 B(z, ε)
Z
p
 f (y) − f (x) dy
1
p
= 0,
ν − a.e x ∈ Rn
B(z,ε)
or equivalently 1 lim B→0 B
Z
 f (y) − f (x) p dy = 0.
B
B3x
(c) This result is reduced from (b).
Exercise 2.1.11 Let f be in L1 (R). Define the right maximal function MR ( f ) and the left maximal function ML ( f ) as follows: 1 x  f (t) dt , r>0 r x−r Z x+r 1 MR ( f )(x) = sup  f (t) dt . r>0 r x Z
ML ( f )(x) = sup
(a) Show that for all α > 0 and f ∈ L1 (R) we have 1  f (t) dt , α {ML ( f )>α} Z 1 {x ∈ R : MR ( f )(x) > α} =  f (t) dt . α {MR ( f )>α} {x ∈ R : ML ( f )(x) > α} =
Z
(b) Extend the definition of ML ( f ) and MR ( f ) for f ∈ L p (R) for 1 ≤ p ≤ ∞. Show that ML and MR map L p to L p with norm at most p/(p − 1) for all p with 1 < p < ∞. (c) Construct examples to show that the operator norms of ML and MR on L p (R) are exactly p/(p − 1) for 1 < p < ∞. (d) Prove that M = max(MR , ML ). (e) Let N = min(MR , ML ). Obtain the following consequence of part (a), Z R
M( f ) p + N( f ) p dx =
p p−1
Z
 f  M( f ) p−1 + N( f ) p−1 dx ,
R
(f) Use part (e) to prove that
p
p−1 p (p − 1) M( f ) L p − p f L p M( f ) L p − f L p ≤ 0 . R Hint: (a) Write the set Eα = {MR ( f ) > α} as a union of open intervals (a j , b j ). For each x in (a j , b j ), let Nx = s ∈ R : xs  f  > R α(s−x) ∩(x, b j ]. Show that Nx is nonempty and that sup Nx = b j for every x ∈ (a j , b j ). Conclude that abjj  f (t) dt ≥ α(b j −a j ), which implies that each a j is finite. For the reverse inequality use that a j ∈ / Eα . Part (d) is due to K. L. Phillips. (e) First obtain a version of the equality with MR in the place of M and ML in the place of N. Then use that M( f )q + N( f )q = ML ( f )q + MR ( f )q for all q. (f) Use that  f  N( f ) p−1 ≤ 1p  f  p + p10 N( f ) p . This alternative proof of the result in Exercise 2.1.2(c) was suggested by J. Duoandikoetxea.
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Solution. (a) We will show that {x : MR ( f )(x) > α} =
1 α
Z
 f (t) dt.
{MR ( f )>α}
The set Eα = {x : MR ( f )(x) > α} can be written as a union of open disjoint intervals (a j , b j ). We claim that for each x in (a j , b j ) the set Z s Nx = s ∈ (x, b j ) :  f (t) dt > α(s − x) x
is nonempty. Since x lies in (a j , b j ) which is a subset of Eα , then there must be a w > x such that w ≥ b j , then Z w bj
 f (t) dt =
Z w
 f (t) dt −
Z bj
x
x
Rw x
 f (t) dt > α(w − x). If
 f (t) dt > α(w − x) − α(b j − x) = α(w − b j ) .
But this implies that b j belongs to Eα which is a contradiction. Thus w ∈ Nx , hence is nonempty. Let now sx = sup Nx . We will prove that sx = b j . If we had sx < b j , then we would also have Z sx
 f (t) dt ≥ α(sx − x) .
x
Indeed, there is a sequence of points ym in Nx with ym ↑ sx such that Z ym
 f (t) dt > α(ym − x) ;
x
and letting m → ∞ we obtain the claimed inequality. Now the set Nsx is nonempty as well and contained in (a j , b j ], so there is a real number y in (sx , b j ] such that Z y
sx
Since
R sx x
 f  ≥ α(sx − x), it follows that
 f (t) dt > α(y − sx ) .
Z y
 f (t) dt > α(y − x)
x
which is a contradiction since y > sx . Hence, for all x ∈ (a j , b j ) we have sx = b j . Continuity (similar to the one previously described) gives Z bj x
 f (t) dt ≥ α(b j − x)
for every x ∈ (a j , b j ). Letting x → a j + we deduce that Z bj aj
 f (t) dt ≥ α(b j − a j ) ,
while the opposite inequality is a consequence of the fact that a j does not belong to Eα . Suppose that a j0 = −∞ for some j0 . Then by what we proved, for each x < b j0 we have that Z bj 0 x
 f (t) dt ≥ α(b j0 − x) .
Letting x → −∞ we obtain that f L1 ≥ +∞ which is a contradiction, since we are assuming that f is in L1 . Therefore all a j > −∞ and MR ( f )(a j ) makes sense. The inequality
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Z bj aj
 f (t) dt ≤ α(b j − a j ) ,
is a consequence of the fact that a j does not belong to Eα , so MR ( f )(a j ) ≤ α. (b) Use part (a) and apply Holder inequality, we have Z R
ML ( f )(t) p dt =p
Z ∞ 0
α p−1 {x ∈ R : ML ( f )(x) > α} dα
Z ∞
=p
1 α
p−1
α 0
( f (t) R
p p−1
It follows that R
α p−2 dα)dt
0
p ≤ p−1 Z
 f (t) dtdα
{x∈R : ML ( f )(x)>α}
Z ML ( f )(t)
Z
=p =
Z
ML ( f )(t) p dt
Z R
 f (t) ML ( f )(t) p−1 dt
Z
1 Z p
p
 f (t) dt
R
1
p
≤
R
p p−1
Z
ML ( f )(t)
 f (t) p dt
p
10 p
.
1
p
.
R
(c) Let M be the uncentered HardyLittlewood maximal operator (in one dimension) and ML ( f ) and MR ( f ) the onesided left and right maximal functions. We know that Z {x ∈ Rn : ML ( f )(x) > α} = α −1  f (t) dt {ML ( f )>α}
from which it follows by integration that Z Rb
ML ( f )(x) p dx =
p p−1
Z Rn
 f (x) ML ( f )(x) p−1 dx
and a similar statement for MR ( f ). Since M f (x) = max{ML ( f )(x), MR ( f )(x)} and define N f (x) = min{ML ( f )(x), MR ( f )(x)} we have M f (x) p + N f (x) p = MR ( f )(x) p + ML ( f )(x) p and the same is true when p is replaced by p − 1. Then Z Z Z p p p p p−1 p−1 dx = M( f )(x) dx + N( f )(x) dx =  f  ML ( f )(x) + MR ( f )(x)  f  M( f ) p−1 + N( f ) p−1 dx . p − 1 Rn p − 1 Rn Rn Now use the inequality
0
ab ≤
ap bp + 0 p p
to obtain that Z Rn
M( f )(x) p dx +
Z Rn
N( f )(x) p dx ≤
Finally by H¨older’s inequality we have that
p p−1
Z Rn
 f (x) M( f )(x) p−1 dx +
1 p
Z Rn
 f (x) p dx +
p−1 p
Z Rn
N( f )(x) p dx .
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Z Rn
p−1  f (x) M( f )(x) p−1 dx ≤ f L p M( f ) L p
to deduce that (p − 1)kM( f )k pp − pk f k p kM( f )k p−1 − k f k pp ≤ 0. p This is the sufficient condition. (d) By the continuity of the integral, it is easy to imply that ML ( f ) ≤ M( f ) and MR ( f ) ≤ M( f ), hence max(ML , MR ) ≤ M. Furthermore, for r > 0, Z Z Z 1 1 x 1 x+r 1 x+r  f (t) dt =  f (t) dt +  f (t) dt . 2r x−r 2 r x−r r x This implies M( f )(x) ≤ 12 (ML ( f )(x) + MR ( f )(x)) for all x ∈ R and thus M( f ) ≤ max(ML ( f ), MR ( f )).
Exercise 2.1.12 A cube Q = [a1 2k , (a1 + 1)2k ) × · · · × [an 2k , (an + 1)2k ) on Rn is called dyadic if k, a1 , . . . , an ∈ Z. Observe that either two dyadic cubes are disjoint or one contains the other. Define the dyadic maximal function 1 Q3x Q
Md ( f )(x) = sup
Z
f (y) dy , Q
where the supremum is taken over all dyadic cubes Q containing x. (a) Prove that Md maps L1 to L1,∞ with constant at most one. Precisely, show that for all α > 0 and f ∈ L1 (Rn ) we have {x ∈ Rn : Md ( f )(x) > α} ≤
1 α
Z
f (t) dt . {Md ( f )>α}
(b) Conclude that Md maps L p (Rn ) to itself with constant at most p/(p − 1). Solution. (a) We claim that the set Eα = {x ∈ Rn : Md ( f )(x) > α} is open. Indeed, in the definition of Md we can replace a dyadic cube with its interior and both the measure of a cube as well as the integral over the cube remain unchanged. Given x ∈ Eα , there is an open dyadic cube Qx that contains x such that the average of  f  over Qx is strictly bigger than α. Then Qx is contained in Eα . This proves that Eα is open. Let Q0x be the maximal open dyadic cube that contains x (and Qx ) such that the average of  f  over Q0x is strictly bigger than α. Let {Q0x : x ∈ Eα } = {Q j , j ∈ Z} be the collection of these cubes. These Q j must be disjoint, otherwise if one properly contained another, then for any x in the smaller cube the smaller cube Q0x would not be maximal. Then Eα minus a set of measure zero is equal to the union of cubes Q j . Then we have
∑ Q j  = Eα  j
and also
Z Qj
It follows that
1 Eα  = ∑ Q j  ≤ α j
 f (y) dy > αQ j  . 1  f (y) dy ≤ α Qj
Z
∑ j
This proves that Md maps L1 → L1,∞ with constant at most 1. (b) This is a consequence of part (a) and of the result in Exercise 1.3.3 (a).
Z
 f (y) dy .
Eα
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Exercise 2.1.13 Observe that the proof of Theorem 2.1.6 yields the estimate 1
1
λ {M( f ) > λ } p ≤ 3n {M( f ) > λ }−1+ p
Z
 f (y) dy
{M( f )>λ }
for λ > 0 and f locally integrable. Use the result of Exercise 1.1.12(a) to prove that the Hardy–Littlewood maximal operator M maps the space L p,∞ (Rn ) to itself for 1 < p < ∞. Solution. In the proof of Theorem 2.1.6, we have the estimate {M( f ) > λ } ≤ It follows
3n λ
Z
 f (y) dy.
{M( f )>λ }
1
1
λ {M( f ) > λ } p ≤ 3n {M( f ) > λ }−1+ p hence
Z
 f (y) dy,
{M( f )>λ }
p k f kL p,∞ . p−1
kM( f )kL p,∞ ≤ 3n k f kL p,∞ ≤ 3n Thus M maps the space L p,∞ (Rn ) into itself for 1 < p < ∞.
Exercise 2.1.14 Let K(x) = (1 + x)−n−δ be defined on Rn . Prove that there exists a constant Cn,δ such that for all ε0 > 0 we have the estimate 1 n ε>ε0 ε
sup ( f  ∗ Kε )(x) ≤ Cn,δ sup ε>ε0
Z
 f (y) dy ,
y−x≤ε
n for all f locally integrable on R . Hint: Apply only a minor modification to the proof of Theorem 2.1.10.
Solution. Assume first that f has compact support. For ε0 > 0, set 1 n ε>ε0 ε
Mε0 ( f )(x) = sup
Fix ε > ε0 and denote Br = {x ∈ Rn : x ≤ r} , now we have
Z x−y≤ε
 f (y) dy.
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( f  ∗ Kε )(x) = =
Z ZR
n
 f (x − y) Kε (y)dy  f (x − εy) K(y)dy
Rn ∞ Z
=∑
k=0 Bk+1 \Bk ∞
 f (x − εy) K(y)dy
1 (1 + k)n+δ k=0
≤∑
1
Z Bk+1 \Bk
 f (x − εy) dy ∞
Z
1 − (1 + k)n+δ
1
 f (y) dy + ∑ n+δ N n+δ BN k=1 k Z ∞ kn 1 kn  f (y) dy =∑ − n n+δ n+δ (kε) k (1 + k) y−x≤kε k=1
= lim
N→∞
where Cn,δ = ∑∞ k=1
kn kn+δ
Z
 f (x − εy) dy
Bk
≤Cn,δ Mε0 ( f )(x), kn − < ∞. We used that f has compact support to deduce that the limit is zero. (1 + k)n+δ
Thus we have
1 n ε ε>ε0
sup ( f  ∗ Kε )(x) ≤ Cn,δ sup ε>ε0
Z
 f (y) dy.
x−y≤ε
1 (Rn ). In the inequality To remove the assumption that f has compact support, let fN (x) = f (x)χx≤N for some f ∈ Lloc
1 n ε>ε0 ε
( fN  ∗ Kε )(x) ≤ Cn,δ sup
Z x−y≤ε
1 n ε>ε0 ε
 fN (y) dy ≤ Cn,δ sup
Z
 f (y) dy
x−y≤ε
let N → ∞ and use the Lebesgue monotone convergence theorem to obtain the claimed conclusion. Then take the supremum over ε > ε0 on the left to prove the inequality for general f .
Section 2.2. The Schwartz Class and the Fourier Transform Exercise 2.2.1 (a) Construct a Schwartz function supported in the unit ball of Rn . (b) Construct a C0∞ (Rn ) function equal to 1 on the annulus 1 ≤ x ≤ 2 and vanishing off the annulus 1/2 ≤ x ≤ 4. (c) Construct a nonnegative nonzero Schwartz function f on Rn whose Fourier transform is nonnegative and compactly supported. Hint: Part (a): Try the construction in dimension one first using the C ∞ function η(x) = e−1/x for x > 0 and η(x) = 0 for x < 0. Part (c): Take f = φ ∗ φe2 , where φb is odd, realvalued, and compactly supported; here φe(x) = φ (−x). Solution. (a) The function ϕ(x) = is a Schwartz function with support {x (b) Consider the smooth function
∈ Rn
( exp(−
1 ), 1−x2
0,
x < 1 x ≥ 1
: x ≤ 1} . ( − 1 e x(1−x) η(x) = 0
x ∈ (0, 1) x∈ / (0, 1).
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Set λ = 01 η(x)dx and define the function γ(x) = λ −1 0x η(t)dt, for x ∈ R. It is easy to see that 0 ≤ γ(x) ≤ 1 for all x ∈ R, γ(x) = 0 for x ≤ 0 and γ(x) = 1 for all x ≥ 1. Now fix 41 ≤ c < a ≤ 1 < 4 ≤ b < d ≤ 16 and define the function x−c x − b ψ(x) = γ 1−γ , x ∈ R. a−c d −b R
R
It is clear that 0 ≤ ψ(x) ≤ 1 for all x ∈ R, ψ(x) = 0 for x ∈ (−∞, c] ∪ [d, ∞) and ψ(x) = 1 for x ∈ [a, b]. Finally, the function ϕ(x) = ψ(x2 ),
x ∈ Rn
satisfies the requirement of the problem. (c) Fix u ∈ Rn and define the function φb(x) = (u · x)ϕ(x),
x ∈ Rn ,
where ϕ is the even function constructed in the part (a). Then φb is odd, realvalued and compactly supported. Since φb is odd, then so is φ . Since φb is odd and realvalued, then φ is purely imaginary, that is, it satisfies φ = −φ . Let f = φ ∗ φe2 = (φ ∗ φe)(φ ∗ φe) = (φ ∗ φe)((−φ ) ∗ (−φe)) = (φ ∗ φe)(φ ∗ φe) . Then f is the square of the convolution of two Schwartz functions, hence it is a (nonnegative) Schwartz function itself. Also, recalling that the operations e and b commute we can write Z
fb(ξ ) =
Rn
e e φb(y)φb(y)φ (ξ − y)φb(ξ − y)dy =
Z Rn
φb(y)φb(−y)φb(ξ − y)φb(−ξ + y)dy =
Z Rn
φb(y)2 φb(ξ − y)2 dy ≥ 0
for all ξ ∈ Rn .
Exercise 2.2.2 If fk , f ∈ S (Rn ) and fk → f in S (Rn ), then b fk → fb and fk∨ → f ∨ in S (Rn ). Solution. It is suffice to show that if fk → 0 in S , then b fk → 0 in S . First, fix f ∈ S , we have the estimate (2πiξ )α ∂ β fb(ξ ) =(2πiξ )α ((−2πix)β f (x))b(ξ ) = (∂ α ((−2πix)β f (x)))b(ξ ) Z
= It turns out
Rn
∂ α ((−2πix)β f (x))e−2πiξ ·x dx.
Z sup (2πiξ )α ∂ β fb(ξ ) ≤ sup (1 + xn+1 )∂ α ((−2πix)β f (x))
ξ ∈Rn
x∈Rn
Rn
dx 1 + xn+1
.
Hence ρα,β ( fb) ≤ Cn,α,β ∑γ≤n+1+β ,δ ≤α ργ,δ ( f ). Replacing f by fk and taking the limit as k → ∞, we get b fk → 0 in S . b ∨ Now note that if fk → 0, then e fk → 0. The above argument shows that e fk → 0; hence fk → 0 as k → ∞.
Exercise 2.2.3 Find the spectrum (i.e., the set of all eigenvalues of the Fourier transform), that is, all complex numbers λ for which there exist nonzero functions f such that fb = λ f .
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2 2 Hint: Apply the Fourier transform three times to the preceding identity. Consider the functions xe−πx , (a + bx2 )e−πx , and 2 (cx + dx3 )e−πx for suitable a, b, c, d to show that all fourth roots of unity are indeed eigenvalues of the Fourier transform. Solution. Let λ is an eigenvalue of the Fourier transform, i.e. fb = λ f for some f 6= 0. Apply the Fourier transform three times to the proceeding identity, we get λ 4 f = f . Thus λ 4 = 1, that is λ = ±1 or λ = ±i. ’ Thus the eigenvalues of the Fourier transform are contained in {1, −1, i, −i}. We need to show that all eigenvalues are indeed in this set. 2 2 Since the Fourier transform of e−πx equals e−πξ , we deduce 2
((−2πix)k e−πx )b(ξ ) =
d k −πξ 2 ). (e dξ k
It follows from this that 2 2 e−πx b(ξ ) = 1 · e−πξ 2 2 xe−πx b(ξ ) = −i · e−πξ 1 −πx2 b 1 −πξ 2 x2 − e (ξ ) = −1 · ξ 2 − e 2π 2π 3 −πξ 2 3 −πx2 b x e (ξ ) = i · ξ 3 − ξ e x3 − 2π 2π which shows that the numbers 1, −1, i, −i are indeed eigenvalues of the Fourier transform.
Exercise 2.2.4 Use the idea of the proof of Proposition 2.2.7 to show that if the functions f , g defined on Rn satisfy  f (x) ≤ A(1 + x)−M and g(x) ≤ B(1 + x)−N for some M, N > n, then ( f ∗ g)(x) ≤ ABC(1 + x)−L , where L = min(N, M) and C = C(N, M) > 0. Solution. As in the proof of Proposition 2.2.7 write ( f ∗ g)(x) ≤ AB
Z Rn
(1 + x − y)−M (1 + y)−N dy
and estimate the piece of the integral over the set {y : x/2 ≤ y − x} by Z
AB y−x≥x/2
(1 + x/2)−M (1 + y)−N dy ≤ AB2M (1 + x)−MCN
and the piece of the integral over the set {y : y − x ≤ x/2} by Z
AB y−x≤x/2
(1 + x − y)−M (1 + x/2)−N dy ≤ AB2N (1 + x)−N CM .
The maximum of these estimates gives the claimed decay for f ∗ g.
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Exercise 2.2.5 Show that C0∞ (Rn ) is dense on L p (Rn ) for 0 < p < ∞ but not for p = ∞. Hint: Use a smooth approximate identity when p ≥ 1. Reduce the case p < 1 to p = 1. Solution. In the case 1 ≤ p < ∞ we may use the following argument Let f ∈ L p (Rn ), 1 ≤ p < ∞ with supp( f ) j B(0, R). Then for every η > 0, there exists a function g ∈ C0∞ (Rn ) with supp(g) j B(0, 2R) such that k f − gkL p < η. Indeed, take θ ∈ C0∞ (Rn ) with supp(θ ) j B(0, R) such that θ ≥ 0 and Rn θ (x)dx = 1. Define gε = f ∗ θε , 0 < ε < 1. Then gε converges to f in L p (Rn ) and supp(gε ) j B(0, 2R) for 0 < ε < 1. Now fix f ∈ L p (Rn ). Given η > 0, then there exists R = R(η) > 0 such that k f − fR kL p < η2 , where fR = f χB(0,R) . Since fR is a function in L p (Rn ) with supp( fR ) j B(0, R), the above claim shows that we can find a function g ∈ C0∞ (Rn ) such that supp(g) ∈ B(0, 2R) and kg − fR kL p < η2 . Combining these facts we have k f − gkL p < η. Thus C0∞ (Rn ) is dense in L p (Rn ). R
1
For the case 0 < p < 1, fix f ∈ L p (Rn ). Given η > 0, then there exists R = R(η) > 0 such that k f − fR kL p < 2− p η, where fR = f χAR , with AR = {x ∈ Rn : x < R and  f (x) < R} . Since fR is bounded with supp( fR ) j B(0, R), fR is integrable in Rn . By the above claim, we can find a smooth function g with support in B(0, 2R) such that 1
1− 1p
kg − fR kL1 < 2− p νn
1
(2R)n(1− p ) η,
where νn is the volume of the unit ball in Rn . Apply H¨older inequality to the functions g − fR  and 1 with exponents 1 1−p , we have Z Rn
g − fR  p dx = ≤ ≤
Z B(0,2R)
1 p
and
g − fR  p dx p Z
Z
1−p
g − fR  dx dx B(0,2R) B(0,2R) kg − fR kLp1 νn1−p (2R)n(1−p) −1 p
0. But then for any x ∈ / B(0, R), we have g(x) − f (x) = 1 > 21 . This is a contradiction. L∞ (Rn ).
Exercise 2.2.6 (a) Prove that if f ∈ L1 , then fb is uniformly continuous on Rn . (b) Prove that for f , g ∈ L1 (Rn ) we have Z Z f (x)b g(x) dx = Rn
(c) Take gb(x) = ε −n e−πε Fourier inversion on L1 .
−2 x−t2
Rn
fb(y)g(y) dy .
in (b) and let ε → 0 to prove that if f and fb are both in L1 , then ( fb)∨ = f a.e. This fact is called
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Solution. R R (a) Let M = Rn  f (x) dx < ∞. Given ε > 0, then there exists Rε > 0 such that x>Rε  f (x) < ε4 . Since the function g(t) = e−2πit ε is continuous at t = 0, there exists ρε > 0 such that e−2πit − 1 < 1+2M for all t < ρε . Now set δε = Rρεε , then for ξ , η ∈ Rn such that ξ − η < δε , we have the estimate Z Z Z b  f (x) dx +  f (x) e−2πi(ξ −η)·x − 1 dx  f (x) e−2πiξ ·x − e−2πiη·x dx ≤ 2 f (ξ ) − fb(η) ≤ Rn
x>Rε
Z
ε ε < + 2 1 + 2M
x≤Rε
x≤Rε
 f (x) dx < ε.
Thus fb is uniformly continuous on Rn . (b) Since f , g ∈ L1 it follows that f (x)g(y)e−2πix·y lies in L1 (Rn × Rn ) and hence the interchange of x and y integrals is allowed. The desired equality now follows immediately from Fubini’s theorem. (c) By part (b) we have Z Z −2 −2 2 f (x)ε −n e−πε x−t dx = fb(ξ )e2πiξ ·t e−πεξ  dξ Rn
Rn
as in identity (2.2.15). By Theorem 1.2.19 the left hand side converges to f (t) in L1 as ε → 0. This means we can find a sequence {εk } converging to 0 such that the left hand side converges to f (t) a.e. as k → ∞. Since f ∈ L1 the right hand side converges to ( fb)∨ (t) a.e. as k → ∞ by the Lebesgue dominated convergence theorem.
Exercise 2.2.7 (a) Prove that if f is integrable over Rn and continuous at 0, then Z
lim ε→0
2 fb(x)e−πεx dx = f (0) .
Rn
(b) Prove that if f ∈ L1 (Rn ), fb ≥ 0, and f is continuous at zero, then fb is in L1 and therefore Fourier inversion f (0) = k fbkL1 holds at zero and f = ( fb)∨ a.e. in general. 2 Hint: Part (a): Take g(x) = e−πεx in Exercise 2.2.6 (b). Solution. (a) We have Z Rn
2 fb(x)e−πεx dx =
=
Z ZR
x 2
n
Rn
f (x)ε −n e−π ε  dx 2
f (εy)e−πy dy
(by Exercise 2.2.6(b)) (where y = ε −1 x).
2
Since f ∈ L1 and e−πεx ≤ 1 we have Z
lim ε→0
Rn
2 fb(x)e−πεx dx =
Z
2
Rn
lim f (εy)e−πy dy
(by Lebesgue dominated convergence theorem)
ε→0
Z
= f (0)
Rn
2
e−πy dy
(since f is continuous at 0)
= f (0). (b) Since f ∈ L1 Exercise 2.2.6 (b) and the Lebesgue dominated convergence theorem imply Z
lim
ε→0 Rn
2 fb(x)e−πεx dx =
so since f is continuous at 0, part (a) now implies that
R
Z
Rn
2 lim fb(x)e−πεx dx =
Rn ε→0
Z Rn
fb(x)dx,
fb(x)dx = f (0). Since fb ≥ 0 we have
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Z
k fbkL1 =
fb(x)dx = f (0) < ∞.
Rn
Exercise 2.2.6(c) now implies that ( fb)∨ = f a.e.
Exercise 2.2.8 Given f in L1 (Rn ) ∩ L2 (Rn ), prove that
fb
L2
= f L2 .
1 n b b2 Hint: Let h = f ∗ fe, where fe(x) = f (−x) and the bar indicates complex Z conjugation. Then h ∈ L (R ), h =  f  ≥ 0, and h is continuous at zero. Exercise 2.2.7(b) yields k fbk2 2 = kb hk 1 = h(0) = f (x) fe(−x) dx = k f k2 2 . L
L
Rn
L
Solution. Let fe(x) = f (−x), and let h = f ∗ fe. Then Z
h(x) =
Rn
f (y) fe(x − y)dy.
Observe that limy→0 k f − τ y f kL1 = 0 by the Lebesgue Dominated Convergence Theorem (since f ∈ L1 ), and since the L2 inner product is continuous it follows that h is continuous (since f ∈ L2 ). In particular, h is continuous at 0. Also, since both f and fe are in L1 we know h ∈ L1 . Note that Proposition 2.2.11 (4),(5), and (12) hold for L1 functions so b h = fb bf =  fb2 ; in particular, b h ≥ 0. So we have k fbk2L2 = kb hkL1 = h(0) Z
=
Rn
(by Exercise 2.2.7(b))
f (x) f (x)dx = k f k2L2 .
Exercise 2.2.9 (a) Prove that for all 0 < ε < t < ∞ we have Z t sin(ξ ) dξ ≤ 4. ε ξ (b) If f is an odd L1 function on the line, conclude that for all t > ε > 0 we have Z t
fb(ξ )
ε ξ dξ ≤ 4 f L1 . (c) Let g(ξ ) be a continuous odd function that is equal to 1/ log(ξ ) for ξ ≥ 2. Show that there does not exist an L1 function whose Fourier transform is g. Solution. (a) If ε < t ≤ 1, then Z t sin ξ Z t sin ξ dξ ≤ dξ ≤ t ≤ 1. ξ ξ 0 ε Also for t > 1
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Z t sin ξ
0
ξ
Z t (cos ξ )0
dξ ≤ 1 +
ξ
1
Z t cos ξ
cost cos 1 + dξ ≤ 1 + + t 1
1
ξ2
dξ ≤ 4.
(b) Since f is odd, we have that fb(ξ ) = −i
Z
f (x) sin(2πxξ ) dξ . R
Divide by ξ , and integrate from ξ = 0 to ∞. We obtain Z t Z Z t fb(ξ ) sin(2πxξ ) dξ dx 0 ξ dξ = R f (x) 0 ξ Z 2πtx Z sin(ξ ) ≤  f (x) dξ dx ξ 0 R ≤4k f kL1 , where the first equality above follows from Fubini’s theorem. (c) Assume that such an integrable f exists. Let fe(x) = f (−x). The L1 functions fe and − f have the same Fourier transforms (since fb = g is odd) and thus they coincide. Hence f is odd. But any odd integrable function on the line must satisfy the conclusion of part (b) which clearly fb = g fails to satisfy.
Exercise 2.2.10 Let f be in L1 (R). Prove that Z +∞ −∞
1 dx = f x− x
Z +∞
Hint: For x ∈ (−∞, 0) use the change of variables u = x − 1x or x = √ u = x − 1x or x = 12 u + 4 + u2 .
f (u) du . −∞ 1 2
√ u − 4 + u2 . For x ∈ (0, ∞) use the change of variables
Solution. p Observe that the map x → x − x−1 is a bijection from (0, ∞) onto (−∞, ∞) and its inverse is y → 12 (y + y2 + 4). Similarly p the map x → x − x−1 is a bijection from (−∞, 0) onto (−∞, ∞) and its inverse is y → 21 (y − y2 + 4). Therefore Z ∞ 0
Z 0 −∞
Z p 1 +∞ f (y) dy + d y2 + 4 2 −∞ Z p 1 +∞ f (x − 1/x) dx = f (y) dy − d y2 + 4 . 2 −∞
f (x − 1/x) dx =
Summing the above two identities we obtain the required conclusion.
Exercise 2.2.11 2
(a) Use Exercise 2.2.10 with f (x) = e−tx to obtain the subordination identity 1 e−2t = √ π
Z ∞
e−y−t
2 /y
0
dy √ , y
where t > 0.
(b) Set t = πx and integrate with respect to e−2πiξ ·x dx to prove that (e−2πx )b(ξ ) =
Γ ( n+1 2 ) π
n+1 2
1 (1 + ξ 2 )
n+1 2
.
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This calculation gives the Fourier transform of the Poisson kernel. Solution. (a) Let t > 0 and note
Z ∞
2
e−tx dx =
Z ∞ −∞
−∞
by Exercise 2.2.10. Now apply the change of variables Z ∞
2
e−t(x−1/x) dx
p
y/t = x to the right hand side to get
2
e−t(x−1/x) dx = e2t
Z ∞
−∞
e−tx
2 −t/x2
dx
0
Z ∞ 2 dy 2t 1 1 √ e−y−t /y √ . =e
2
and therefore
t
y
0
∞ 2 1 dy e−2t = √ e−y−t /y √ y π 0 √ √ since the left hand side of the first displayed identity equals π/(2 t). (b) It follows that Z
Z
(e−2πx )b(ξ ) =
e−2πx e−2πiξ ·x dx Z ∞ Z 1 −y−π 2 x2 /y dy √ e = √ e−2πiξ ·x dx y π 0 Rn Z Z ∞ 2 2 1 dy =√ e−y−π x /y−2πiξ ·x √ dx n y π R 0 Z Z ∞ n n 2 2 1 dy =√ e−y−π ∑ j=1 x j /y−2πi ∑ j=1 ξ j x j √ n y π R 0 Z Z ∞ n 2 2 1 dy =√ e−y−∑ j=1 (πx j +iξ y) /y−yξ  √ dx y π Rn 0 Z Z ∞ n 2 1 dy − ∑ j=1 (πx j +iξ j y)2 /y −y(1+ξ  ) =√ e e dx √ n y π 0 R Rn
by part (a) and Fubini’s theorem. Furthermore, −2πx
(e
)b(ξ ) =
1 π n+1/2
y
n+1 − 1 2 2
e
π
n+1 2
(1 + ξ 2 ) n+1
Γ
2
π
n+1 2
−y(1+ξ )2
Z Rn
0
1
= =
Z ∞
Z ∞ n+1 2
z
n+1 − 1 2 2
2
e−w dwdy
e−z dz
0
1
(1 + ξ 2 )
n+1 2
√ by the change of variables w = (πx + iξ y)/ y and z = y(1 + ξ 2 ), and the definition of Γ .
Exercise 2.2.12 Let 1 ≤ p ≤ ∞ and let p0 be its dual index. (a) Prove that Schwartz functions f on the line satisfy the estimate
2
f ∞ ≤ 2 f p f 0 p0 . L L L (b) Prove that all Schwartz functions f on Rn satisfy the estimate
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2
f ∞ ≤ L
α β
∂ f p ∂ f p0 , L L
∑
α+β =n
allRpairs of multiindices α and β whose sum has size n. where the sum is taken over x d 2 Hint: Part (a): Write f (x)2 = −∞ dt f (t) dt. Solution. (a) Since f is a Schwartz function, we have f (x)2 = 2
Z x
f (t) f 0 (t)dt,
x ∈ R.
−∞
By apply the Holder’s inequality, one gets f (x)2 ≤ 2
Z
x
 f (t) p dt
1 Z p
−∞
x
0 p0 f (t) dt
−∞
10 p
≤ 2 k f kL p f 0 L p0 ,
x ∈ R.
This yields
k f k2L∞ ≤ 2 k f kL p f 0 L p0 ,
x ∈ R.
(b) Fix x = (x1 , . . . , xn ) ∈ Rn . Denote I(a) = (−∞, x1 ] × (−∞, x2 ] × · · · × (−∞, xn ]. Since f is a Schwartz function, we have the following identity f (x)2 = f (x1 , . . . , xn )2 = ≤
Z xn
...
∂n ( f (y1 , . . . , yn ))2 dy1 . . . dyn −∞ ∂ yn · · · ∂ y1
Z x1
−∞
Z
∂ α f (y)∂ β f (y)dy
∑
I(x) α+β =n
Z
=
∂ α f (y)∂ β f (y)dy
∑
α+β =n I(x)
≤
Z
∑
α+β =n I(x)
≤
1 Z p ∂ f (y) dy
Z
∑
∑
α
p
I(x)
α+β =n
≤
∂ α f (y) ∂ β f (y) dy
I(x)
p0 p10 β ∂ f (y) dy
k∂ α f kL p ∂ β f L p0 .
α+β =n
This inequality gives the estimate L∞ norm of f k f k2L∞ ≤
∑
k∂ α f kL p ∂ β f L p0 .
α+β =n
Exercise 2.2.13 The uncertainty principle says that the position and the momentum of a particle cannot be simultaneously localized. Prove the following inequality, which presents a quantitative version of this principle:
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2
f 2 L
Z
4π ≤ inf (Rn ) n y∈Rn
2
Rn
1
Z
2
2
x − y  f (x) dx
infn
z∈R
2
Rn
ξ − z  fb(ξ )2 dξ
1 2
,
where f is a Schwartz function on Rn (or an L2 function with sufficient decay at infinity). Hint: Let y be in Rn . Start with Z n
2 ∂
f 2 = 1 f (x) f (x) (x j − y j ) dx , ∑ L n Rn ∂ j=1 x j 2 2 2 b 2 integrate by parts, apply the Cauchy–Schwarz inequality, Plancherel’s identity, and the identity ∑nj=1 ∂d j f (ξ ) = 4π ξ   f (ξ ) for all ξ ∈ Rn . Then replace f (x) by f (x)e2πix·z .
Solution. Fix a y in Rn and f ∈ S (Rn ). Then we have k f k2L2 =
1 n
2 n
2 ≤ n
∂ (x j − y j ) dx j=1 ∂ x j
f (x) f (x) ∑
Rn
=− ≤
n
Z
1 n
n
Z
∑ Rn
f (x)∂ j f (x) + ∂ j f (x) f (x) (x j − y j ) dx
j=1 n
Z
Rn
∑ ∂ j f (x)2
1 2
j=1 n
Z
2
∑ ∂ j f (x) dx Rn
n 1  f (x) ∑ x j − y j 2 2 dx j=1
1 Z 2
2
Rn
j=1
2
1
2
x − y  f (x) dx
.
For some fixed z in Rn now replace f (x) by f (x)e2πix·z in the preceding inequality. Then 2πiz·x b ∂d j f (ξ − z) = −2πi(ξ − z) f (ξ )e
and we have k f k2L2 ≤
2 n
Z
2 n
Z
= =
n 2 j f (ξ − z) dξ ∑ ∂d
1 Z 2
Rn j=1
4π n
Rn
Rn
(2πξ − z)2  fb(ξ )2 dξ
Z Rn
x − y2  f (x)2 dx
x − y2  f (x)2 dx
1 Z 2
Rn
1 Z 2
Rn
1 2
x − y2  f (x)2 dx
ξ − z2  fb(ξ )2 dξ
1 2
1
2
which proves the required estimate after we take the infimum over all y, z ∈ Rn .
Exercise 2.2.14 Let −∞ < α
0. Solution.
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Fix −∞ < α < get Z
n 2
Rn
< β < ∞ and g ∈ L1 (Rn ). By splitting the integral of f on Rn into the unit ball and the complement, one g(x) dx =
Z
g(x) dx +
x≤1
≤
Z
Z
g(x) dx
x>1
x−2α dx
1 2
Z
α
x g(x) 2 n + L (R )
x−2β dx
x>1
x≤1
1
2
β
x g(x)
L2 (Rn )
.
Now let g(λ x), λ > 0, play the role of g(x) in the above inequalities to get Z Rn
g(λ x) dx ≤ C1 xα g(λ x) L2 (Rn ) +C2 xβ g(λ x)
,
L2 (Rn )
where C1 ,C2 are constants depending on α, β and n. It turns out
n n
λ −n kgkL1 (Rn ) ≤ C1 λ −α− 2 xα g(x) L2 (Rn ) +C2 λ −β − 2 xβ g(x)
L2 (Rn )
.
This inequality implies that
n n
kgkL1 (Rn ) ≤ C1 λ −α+ 2 xα g(x) L2 (Rn ) +C2 λ −β + 2 xβ g(x)
L2 (Rn )
and by setting
1
− 1
β −α
λ = xα g(x) L2β(R−αn ) xβ g(x) 2 n , L (R )
we obtain the required inequality.
Section 2.3. The Class of Tempered Distributions Exercise 2.3.1 Show that a positive measure µ that satisfies Z Rn
dµ(x) < +∞ , (1 + x)k
for some k > 0, can be identified with a tempered distribution. Show that if we think of Lebesgue measure as a tempered distribution, then it coincides with the constant function 1 also interpreted as a tempered distribution. Solution. Any such measure defines a linear functional on S via integration. This linear functional is continuous since if fk → f in S , then Z Z dµ(x) k ( fk (x) − f (x)) dµ(x) ≤ sup (1 + x)  fk (x) − f (x) n (1 + x)k n Rn R x∈R and the last expression above is controlled by a constant multiple of finite sum of ρα,β seminorms of fk − f , thus it converges to zero as k → ∞. It follows that µ defines a continuous linear functional on S and it can be identified with an element of S 0 . The actions of Lebesgue measure and of the function 1 (both thought as tempered distributions) coincide.
Exercise 2.3.2 Let ϕ, f ∈ S (Rn ), and for ε > 0 let ϕε (x) = ε −n ϕ(ε −1 x). Prove that ϕε ∗ f → b f in S , where b is the integral of ϕ. Solution.
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Without loss of generality take b = 1. We need to show that for all multiindices α and β we have xα (∂ β (φε ∗ g))(x) converges uniformly to xα (∂ β g)(x) as ε → 0. Setting ∂ β g = h, we see that it suffices to show that xα (φε ∗ h)(x) converges uniformly to xα h(x). This will follow from the fact that ∂ α (φε ∗ h)bconverges to ∂ α b h in L1 . Using Leibniz’s rule we write ∂ α (φbε b h) = φbε ∂ α b h+
h. cγ (∂ γ φbε )∂ α−γ b
∑ 0 0, u ∈ S 0 (Rn ), and f ∈ S (Rn ) we have (δ a f ) ∗ (δ a u) = a−n δ a ( f ∗ u) . Solution. The following steps hold for any x ∈ Rn
af (δ a f ) ∗ (δ a u) (x) = δ a u, τ x δg
= δ a u, τ x (δ a fe)
= δ a u, δ a (τ ax fe)
= u, a−n δ 1/a δ a (τ ax fe)
= a−n u, τ ax fe
= a−n δ a u, τ x fe = a−n δ a ( f ∗ u)(x) ,
hence the claimed identity holds.
Exercise 2.3.4 (a) Prove that the derivative of χ[a,b] is δa − δb . (b) Compute ∂ j χB(0,1) on R2 . (c) Compute the Fourier transforms of the locally integrable functions sin x and cos x. (d) Prove that the derivative of the distribution log x ∈ S 0 (R) is the distribution Z
u(ϕ) = lim ε→0 ε≤x
ϕ(x)
dx . x
Solution. (a) Observe that for f ∈ S we have 0 χ[a,b] (f) = −
(b) We have
Z b a
f 0 (t) dt = − f (b) + f (a) = (δa − δb )( f ) .
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(∂1 χB(0,1) )( f ) = −
Z B(0,1)
Z +1
Z +1 Z +√1−x2 2 ∂1 f dx1 dx2 = √ 2 (∂1 f )(x1 , x2 ) dx1 dx2 −1
Z p f ( 1 − t 2 ,t) dt −
=
+1
−
1−x2
p f (− 1 − t 2 ,t) dt.
−1
−1
The latter can be identified with a measure that lives on the boundary of the circle. Similarly one obtains Z +1
(∂2 χB(0,1) )( f ) =
Z p f (t, − 1 − t 2 ) dt −
−1
+1
f (t,
p
1 − t 2 ) dt.
−1
(c) Since the Fourier transform of x → e2πixy is δy and cos x = 21 (eix + e−ix ), it follows that the Fourier transform of cos x is 1 1 2 (δ1/2π + δ−1/2π ). Similarly the Fourier transform of sin(2πx) is 2i (δ1/2π − δ−1/2π ). (d) Fix f ∈ S (R). We have (log x)0 ( f ) = −
Z
log x f 0 (x) dx = − lim
Z
log x f 0 (x) dx
ε→0 x≥ε
R
= − lim − f (ε) log ε + f (−ε) log ε − ε→0
Z
= lim ε→0 x≥ε
Z x≥ε
1 f (x) dx x
1 f (x) dx x
since for C ∞ functions f the expression ( f (ε) − f (−ε)) log ε is bounded in absolute value by a constant multiple of ε log ε1 and thus tends to zero as ε → 0.
Exercise 2.3.5 Let f ∈ S (Rn ) and let ϕ ∈ C0∞ be identically equal to 1 in a neighborhood of origin. Define ϕk (x) = ϕ(x/k) as in the proof of Proposition 2.3.23. (a) Prove that (τ −he j f − f )/h → ∂ j f in S as h → 0. (b) Prove that ϕk f → f in S as k → ∞. (c) Prove that the sequence ϕk (ϕk f )b converges to fb in S as k → ∞. Solution. (a) We need to show that every ρα,β norm of the expression in part (a) tends to zero as h → 0. Use the mean value theorem twice to write (∂ β f )(x + he j ) − (∂ β f )(x) sup xα − xα (∂ β ∂ j f )(x) = sup xα (∂ j ∂ β f )(x + ξx,h e j ) − (∂ β ∂ j f )(x) h x∈Rn x∈Rn ≤h sup xα (∂ j2 ∂ β f )(x + ξ 0 e j ) x∈Rn
x,h
=A(h), 0  ≤ ξ  ≤ h ≤ 1/2. We have since ξx,h x,h 0 0 (∂ j2 ∂ β f )(x + ξx,h e j ) ≤ CN (1 + x + ξx,h e j )−N ≤ CN0 ( 21 + x)−N ,
and by picking N > α, we estimate the supremum above by a constant, and we conclude that A(h) tends to zero as h → 0. (b) Say that φ is equal to one on B(0, 1) and vanishes off B(0, 2). Then ρα,β (φk f − f ) = sup xα ∂ β (φk f − f )(x) x∈Rn
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= sup xα φk (x)(∂ β f )(x) − (∂ β f )(x) + x∈Rn
∑
cγ (∂ γ φk )(x)(∂ β −γ f )(x)
0N
∑
and the first sum tends to zero as N → ∞, so it follows that uN → u in S 0 . b in S 0 . But Since uN → u in S 0 it follows that uc N →u uc N (ξ ) = and the last claim follows.
∑
ck e−2πiξ ·k = hN (ξ )
k≤N
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Exercise 2.3.9 A distribution in S 0 (Rn ) is called homogeneous of degree γ ∈ C if for all λ > 0 and for all ϕ ∈ S (Rn ) we have
u, δ λ ϕ = λ −n−γ u, ϕ . (a) Prove that this definition agrees with the usual definition for functions. (b) Show that δ0 is homogeneous of degree −n. (c) Prove that if u is homogeneous of degree γ, then ∂ α u is homogeneous of degree γ − α. (d) Show that u is homogeneous of degree γ if and only if ub is homogeneous of degree −n − γ. Solution. (a) If u is a function u(x), then for ϕ ∈ S we have
Z u, δ λ ϕ =
Rn
Z
u(x)ϕ(λ x) dx =
Rn
λ −n u(λ −1 x)ϕ(x) dx =
Z Rn
λ −n−γ u(x)ϕ(x) dx = λ −n−γ u, ϕ .
(b) For ϕ ∈ S and λ > 0 we have
λ −n−γ δ0 , ϕ = δ0 , δ λ ϕ = ϕ(λ 0) = ϕ(0) = δ0 , ϕ and from this it follows that γ + n = 0, thus γ = −n. (c) We have that
α
∂ u, δ λ ϕ = (−1)α u, ∂ α δ λ ϕ = (−1)α λ α u, δ λ ∂ α ϕ = (−1)α λ α λ −n−γ u, ∂ α ϕ = λ −n−γ+α ∂ α u, ϕ , thus ∂ α u is homogeneous of degree γ − α. (d) We have for λ > 0 and ϕ ∈ S −1 λ ϕi = λ −n hu, δ λ ϕi b = λ −n (λ −1 )−n−γ hu, ϕi b = λ −n+(n+γ) hb hb u, δ λ ϕi = hu, δd u, ϕi ,
hence ub is homogeneous of degree −n − γ.
Exercise 2.3.10 (a) Show that the functions einx and e−inx converge to zero in S 0 and D 0 as n → ∞. Conclude that multiplication of distributions is not a continuous operation even when it is defined. √ (b) What is the limit of n(1 + nx2 )−1 in D 0 (R) as n → ∞? Solution. (a) The sequence of functions einx tends to zero in S 0 in view of the RiemannLebesgue Lemma. However the product einx e−inx = 1 does not tend to zero in S 0 . (b) Observe that for f ∈ C0∞ (R) we have √ Z Z √ 1 n f (x) dx = f (y/ n) dy → 0 2 1 + nx 1 + y2 R R √ as n → ∞ in view of the Lebesgue dominated convergence theorem, since f (y/ n)(1 + y2 )−1 → 0 pointwise and is bounded 2 −1 above by k f kL∞ (1 + y ) which is integrable.
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Exercise 2.3.11 (S. Bernstein) Let f be a bounded function on Rn with fb supported in the ball B(0, R). Prove that for all multiindices α there exist constants Cα,n (depending only on α and on the dimension n) such that
α
∂ f ∞ ≤ Cα,n Rα f ∞ . L L Hint: Write f = f ∗ h1/R , where h is a Schwartz function h in Rn whose Fourier transform is equal to one on the ball B(0, 1) and vanishes outside the ball B(0, 2). Solution. Fix a Schwartz function h in Rn whose Fourier transform is equal to one on the ball B(0, 1) and vanishes off the ball B(0, 2). Then fb(ξ ) = fb(ξ )b h(ξ /R). Taking inverse Fourier transforms we obtain f = f ∗ h1/R and thus ∂ α f = Rα f ∗ (∂ α h)1/R . Taking L∞ norms we deduce that
α
∂ f ∞ ≤ Rα f ∞ (∂ α h)1/R 1 = Rα f ∞ ∂ α h 1 = Cα,n Rα f ∞ . L L L L L L
Exercise 2.3.12 b be a C ∞ function that is equal to 1 in a neighborhood of infinity b be a C ∞ function that is equal to 1 in B(0, 1) and let Θ Let Φ 0 and equal to zero in a neighborhood of the origin. Prove the following. (a) For all u in S 0 (Rn ) we have ∨ b ξ /2N ub → u in S 0 (Rn ) as N → ∞. Φ (b) For all u in S 0 (Rn ) we have
∨ b ξ /2N ub → 0 Θ
in S 0 (Rn ) as N → ∞.
Solution. (a) The assertion is equivalent to proving that b ξ /2N ub → ub Φ
in S 0 (Rn ) as N → ∞.
or equivalently b ξ /2N ub → 0 1−Φ
in S 0 (Rn ) as N → ∞.
Pairing with a Schwartz function ϕ the preceding claim is equivalent to showing that b ξ /2N ϕ(ξ b ) → 0 in S (Rn ) as N → ∞. 1−Φ The preceding function is supported in ξ  ≥ 2N . This claim will be a consequence of the fact that h i β b ξ /2N ξ α ϕ(ξ b ) → 0 sup ∂ξ 1 − Φ ξ ∈Rn
b then a term 2−N is produced and the claim easily follows. If no derivative falls on Φ, b then as N → ∞. If a derivative falls on Φ, we must show that h i h i b ξ /2N ∂ β ξ α ϕ(ξ b ξ /2N ∂ β ξ α ϕ(ξ b ) = sup 1 − Φ b ) sup 1 − Φ ξ ξ ξ ∈Rn
ξ ≥2N
β b ) is in S . tends to zero as N → ∞. But this easily follows since the function ∂ξ ξ α ϕ(ξ
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b = 1 − Φ. b (b) This assertion is subsumed in part (a) with Θ
Exercise 2.3.13 Prove that there exists a function in L p for 2 < p < ∞ whose distributional Fourier transform is not a locally integrable function. Hint: Assume the converse. Then for all f ∈ L p (Rn ), fb is locally integrable and hence the map f 7→ fb is a well defined linear operator from L p (Rn ) to L1 (B(0, M)) for all M > 0 (i.e. k fbkL1 (B(0,M)) < ∞ for all f ∈ L p (Rn )). Use the closed graph theorem to deduce that k fbkL1 (B(0,M)) ≤ CM k f kL p (Rn ) for some CM < ∞. To violate this inequality whenever p > 2, take fN (x) = −1 2 2 (1 + iN)−n/2 e−π(1+iN) x and let N → ∞, noting that c fN (ξ ) = e−πξ  (1+iN) . Solution. Suppose that for every f ∈ L p (Rn ) the distributional Fourier transform fb lies in L1 (B(0, M)) for all M > 0. For a given M > 0 define a mapping T (M) : L p (Rn ) → L1 (B(0, M)) such that f 7→ fb B(0,M) . We show that for each M > 0, T (M) has closed graph. Indeed, suppose that fk → f in L p (Rn ) and that b fk B(0,M) → g in L1 (B(0, M)). This is equivalent to ( fk , T (M) ( fk )) → ( f , g) in L p (Rn ) × L1 (B(0, M)). We will show that fb B(0,M) = g a.e., i.e., that T (M) ( f ) = g. To show that fb B(0,M) = g a.e., it will suffice to show that for any smooth function η supported in B(0, M) we have η fb = ηg . If this is established, then we pick a sequence of η’s supported in B(0, M) that converges to 1, to deduce that fb B(0,M) = g a.e. To prove the above identity we fix such an η and we pair with a Schwartz function ϕ. We have
η fb− ηg, ϕ = fb− g, ηϕ = fb− b fk , ηϕ + b fk − g, ηϕ = f − fk , (ηϕ)∨ + b fk − g, ηϕ from which it follows that
η fb− ηg, ϕ ≤ f − fk , (ηϕ)∨ + b fk − g, ηϕ ≤ f − fk L p (ηϕ)∨ L p0 + b fk − g L1 (B(0,M)) ηϕ L∞ (B(0,M)) and this tends to zero as k → ∞. It follows that η fb = ηg a.e. Using the closed graph theorem we deduce that there is a constant CM such that for all f ∈ L p (Rn ) we have k fbkL1 (B(0,M)) ≤ CM k f kL p (Rn ) . −1 2 2 To violate this inequality whenever p > 2, we take fN (x) = (1 + iN)−n/2 e−π(1+iN) x and we note that c fN (ξ ) = e−πξ  (1+iN) . 2 We explain why this is the case. Consider the function e−πzx which is both analytic in z on R+ × R and Schwartz in x on Rn . 1 2 1 2 n n When z = t is positive real, we have that the Fourier transform of this function is t − 2 e−π t x . The function z− 2 e−π z x is also 1 an analytic function on R+ × R, since Re z > 0 implies Re 1z > 0 and since z 2 is a welldefined nonzero analytic function on n
2
1
2
R+ × R. The Fourier transform of e−πzx is also an analytic function on R+ × R and coincides with z− 2 e−π z x when z lies in R+ . It follows by analytic continuation that these two functions coincide for all z with Re z > 0. Now we calculate. We have that
fN
Lp
Z np = (1 + N 2 )− 4
Rn
e
−π p
while k fbkL1 (B(0,M)) = Letting N → ∞ we obtain a contradiction when p > 2.
1 x2 1+N 2
Z ξ ≤M
1
p
dx
n 1
1
= cn,p (1 + N 2 ) − 2 ( 2 − p )
2
e−πξ  dx = cM .
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Section 2.4. More About Distributions and the Fourier Transform Exercise 2.4.1 Suppose that a function f satisfies the estimate  f (x) ≤
C , (1 + x)N
for some C > 0 and N > n + 1. Then fb is C M for all M ∈ Z+ with 1 ≤ M < N − n. Solution. We know that f ∈ L1 (Rn ), so we can write Z
fb(ξ ) =
Rn
f (x)e−2πix·ξ dx
as an absolutely convergent integral. Also, for all α ≤ M we have Z Rn
f (x)(−2πix)α e−2πix·ξ dx
is absolutely convergent and produces a continuous function of ξ , since by hypothesis the function f (x)(−2πix)α is integrable over Rn for all α ≤ M. Let ∂ α fb be the distributional Fourier transform of fb. Then for ϕ ∈ S (Rn ) and α ≤ M we have
α
∂ fb, ϕ = (−1)α fb, ∂ α ϕ Z Z α −2πix·ξ = (−1) f (x)e dx ∂ α ϕ(ξ )dξ Rn Rn Z Z α −2πix·ξ α = (−1) f (x) e ∂ ϕ(ξ ) dξ dx Rn Rn Z Z α −2πix·ξ = f (x)(−2πix) e ϕ(ξ ) dξ dx Rn
Z
=
Rn
Z
=
Rn
Rn
b f (x)(−2πix)α ϕ(x)dx f (x)(−2πix)α b(ξ ) ϕ(ξ ) dξ
and thus the distribution ∂ α fb can be identified with the continuous function f (x)(−2πix)α b. This shows that f ∈ C M since this identification is valid for all α ≤ M.
Exercise 2.4.2 Use Corollary 2.4.3 to prove Liouville’s theorem that every bounded harmonic function on Rn must be a constant. Derive as a consequence the fundamental theorem of algebra stating that every polynomial on C must have a complex root. Solution. Corollary 2.4.3 gives that every harmonic function on Rn is a polynomial. Therefore a bounded harmonic function on Rn has to be a constant. Let us now derive the fundamental theorem of algebra from this. If the nonconstant polynomial P(z) with complex coefficients had no complex roots, then 1/P(z) would be a well defined entire and (thus harmonic function) on C. If
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the degree of P were greater than or equal to 1, then 1/P(z) → 0 as z → ∞ and thus 1/P(z) would have to be bounded on the whole plane. It follows that 1/P(z) would have to be a constant and so would P which contradicts our assumption that P is nonconstant. Therefore P has to have a complex root.
Exercise 2.4.3 x
Prove that ex is not in S 0 (R) but that ex eie is in S 0 (R). Solution. Pick a C ∞ function φ equal to zero in a neighborhood of −∞ and equal to one in a neighborhood of +∞. The function ex is not in S 0 since its integral against the Schwartz function φ (x)e−x does not converge. Now for f ∈ S we have Z +∞
x
f (x)ex eie dx =
Z 0
x
f (x)ex eie dx +
−∞
−∞
Z 1
f (log y)eiy dy.
0
The first integral above converges absolutely and is poitwise controlled by a multiple of k f kL∞ . Integrate by parts twice to write the second integral above as Z ∞ dy − ( f 00 (log y) − f 0 (log y)) eiy 2 y 1 and observe that this integral also converges absolutely and is pointwise controlled by a constant multiple of k f 0 kL∞ + k f 00 kL∞ . x This makes ex eie a continuous linear functional on S .
Exercise 2.4.4 x 7→ sech (πx), x ∈ R, coincides with its Fourier transform. Show that the Schwartz function Hint: Integrate the function eiaz over the rectangular contour with corners (−R, 0), (R, 0), (R, iπ), and (−R, iπ). Solution. Since the function f (x) = sech (πx) is even, we have fb(ξ ) =
Z +∞ cos(2πxξ ) −∞
exπ + e−xπ
dx =
1 π
Z +∞ cos(2ξ x) −∞
ex + e−x
dx.
Consider the meromorphic function G(z) = e2ξ z (ez + e−z )−1 and integrate it over the rectangular contour with corners (−R, 0), (R, 0), (R, iπ), and (−R, iπ). The function G has only one pole inside this contour at the point iπ/2 and the residue at this pole is −ie−πξ /2. The exponential decay of the integrant gives that the parts of integral of G(z) over the two vertical sides of the contour tend to zero as R → ∞. The part of the integral of G over the top horizontal line of the contour tends to e−2πξ π fb(ξ ), while the integral of G over the bottom horizontal line of the contour tends to π fb(ξ ). Cauchy’s residue theorem now gives π fb(ξ ) + e−2πξ π fb(ξ ) + 0 + 0 = 2πi(−ie−πξ /2), which says that fb(ξ ) = sech (πξ ) = f (ξ ).
Exercise 2.4.5 ([176]) Construct an uncountable family of linearly independent Schwartz functions fa such that  fa  =  fb  and  fba  =  fbb  for all fa and fb in the family. Hint: Let w be a smooth nonzero function whose Fourier transform is supported in the interval [−1/2, 1/2] and let φ be a realvalued smooth nonconstant periodic function with period 1. Then take fa (x) = w(x)eiφ (x−a) for a ∈ R.
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Solution. Let w, φ , and fa be as in the hint. Since φ is periodic with period 1 we can expand eiφ (x−a) in Fourier series as eiφ (x−a) =
∑ ck e2πik(x−a) . k∈Z
b − k) and the absolute value of the latter is equal to ∑k∈Z ck w(ξ b − k) since w b is supported Then fba (ξ ) = ∑k∈Z ck e−2πika w(ξ in the interval [−1/2, 1/2]. If φ is chosen so that for any a1 , . . . , am ∈ R there exist x1 , . . . , xm ∈ R such that det(eiφ (x j −ak ) )1≤ j,k≤m 6= 0, then every finite subfamily of the family { fa : a ∈ R} is linearly independent.
Exercise 2.4.6 Let Py be the Poisson kernel defined in (2.1.13). Prove that for f ∈ L p (Rn ), 1 ≤ p < ∞, the function (x, y) 7→ (Py ∗ f )(x) is a harmonic function on Rn+1 + . Use the Fourier transform and Exercise 2.2.11 to prove that (Py1 ∗ Py2 )(x) = Py1 +y2 (x) for all x ∈ Rn . Solution. Let Pt (x) = cn
t n+1
(x2 + t 2 ) 2
be the Poisson kernel. Differentiation gives −(n + 1)t(t 2 − (n + 3)x2j + x2 ) ∂2 t = n+1 ∂ x2j (x2 + t 2 ) n+1 2 (x2 + t 2 )1+ 2 and
t(n + 1)(nt 2 − 3x2 ) ∂2 t = . n+1 ∂t 2 (x2 + t 2 ) n+1 2 (x2 + t 2 )2+ 2
So summing over j = 1, . . . , n and adding the last term we obtain that n
∂2
∂2
∑ ∂ x2 Pt (x) + ∂t 2 Pt (x) = 0 .
j=1
j
To show that the function (x, y) 7→ (Py ∗ f )(x) is harmonic, it suffices to show that one can pass the Laplacian ∆x,y = ∆x + ∂y2 (in x, y) inside the integral Z Rn
Py (x − z) f (z) dz .
The preceding formulas show that any second derivative of Py (x) decays like x−n−1 near infinity, hence it is integrable. Hence we can use the Lebesgue dominated theorem to show that any partial derivative of the preceding integral in (x, y) can be placed inside the integral. We conclude that Z
∆x,y
Rn
Py (x − z) f (z) dz =
Z Rn
∆x,y Py (x − z) f (z) dz = 0 .
Finally, to prove that (Py1 ∗ Py2 )(x) = Py1 +y2 (x), we look at the Fourier transform. Exercise 2.2.11 implies that the Fourier transform of Py is e−2πyξ  . Then for y1 , y2 > 0 we have
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e−2πy1 ξ  e−2πy2 ξ  = e−2π(y1 +y2 )ξ  , hence the claimed identity (Py1 ∗ Py2 )(x) = Py1 +y2 (x) holds.
Exercise 2.4.7 (a) For a fixed x0 ∈ Sn−1 , show that the function v(x; x0 ) =
1 − x2 x − x0 n
is harmonic on Rn \ {x0 }. (b) For fixed x0 ∈ Sn−1 , prove that the family of functions θ 7→ v(rx0 ; θ ), 0 < r < 1, defined on the sphere satisfies Z
lim r↑1
θ ∈Sn−1 θ −x0 >δ
v(rx0 ; θ ) dθ = 0
uniformly in x0 . The function v(rx0 ; θ ) is called the Poisson kernel for the sphere. (c) Show that Z 1 1 (1 − x2 ) dθ = 1 n−1 ωn−1 x − θ n S for all x < 1. (d) Let f be a continuous function on Sn−1 . Prove that the function u(x) =
1 ωn−1
(1 − x2 )
Z Sn−1
f (θ ) dθ x − θ n
solves the Dirichlet problem ∆ (u) = 0 on x < 1 with boundary values u = f on Sn−1 , in the sense limr↑1 u(rx0 ) = f (x0 ) when x  = 1. 0 x −n Hint: Part (c): Apply the mean value property over spheres for the harmonic function y 7→ (1 − x2 y2 ) xy − x . Solution. (a) Direct calculation using the formula 00 f 00 g2 − f gg00 − 2 f 0 g0 g + 2(g0 )2 f f = g g3 gives for (x1 , . . . , xn ) = x 6= x0 = (x01 , . . . , x0n ) n o n o 2n + (1 − x2 ) x − x 2n−4 (n2 + 2n)(x − x )2 − nx − x 2n−2 + 2(x − x )2 + 2x2 − 2x2 nx − x 2n−2 2 2 −2x − x  j j 0 0 0 j 0 0 j 0 j 0 j ∂ 1 − x = . 2 2n 3n x − x0  ∂ x j x − x0  Summing over j from 1 to n yields
∆
1 − x2 x − x0 n
n o n o −2nx − x0 2n + (1 − x2 ) x − x0 2n−4 (n2 + 2n)x − x0 2 − n2 x − x0 2n−2 + 2x − x0 2 + 2x2 − 2x0 2 nx − x0 2n−2 =
x − x0 3n
which is equal to (2n − 2n)x − x0 2n + (n2 + 2n − n2 − 2n)(1 − x2 )x − x0 2n−2 = 0, x − x0 3n since x0  = 1. This shows that the function v(x; x0 ) =
1−x2 x−x0 n
is harmonic on Rn \ {x0 }.
(b) Fix x0 ∈ Sn−1 . When θ − x0  > δ and 1 − r < δ /2 we have
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v(rx0 ; θ ) =
1 − r2 1 − r2 1 − r2 1 − r2 ≤ ≤ ≤ θ − rx0 n θ − x0  − x0 − rx0  δ − (1 − r) δ /2
which tends to zero uniformly in x0 and θ in Sn−1 satisfying θ − x0  > δ . Thus the integral Z θ ∈Sn−1 θ −x0 >δ
v(rx0 ; θ ) dθ → 0
as r ↑ 1 uniformly in x0 . (c) The assertion is valid when x = 0. Now fix x ∈ Rn \ {0} with x < 1 and consider the harmonic function y 7→
1 − y2 x n y − x 
defined on y < 1. It follows that the function 1 − x2 y2 y 7→ xy − x n x is harmonic on the ball B(0, x−1 ) which contains the unit sphere. We apply the meanvalue property of harmonic functions over spheres, in particular over the unit sphere which is centered at zero, and is contained in B(0, x−1 ). We obtain 1=
1 1 − 02 x n = 0 − x  ωn−1
Z Sn−1
1 1 − x2 θ 2 dθ = xθ − x n ωn−1 x
Z Sn−1
1 − x2 dθ θ − x n
2 x 2 since θ − x = 1 + x2 − 2θ · x = xθ − x . (d) Let f be a continuous function on Sn−1 . Clearly the function u(x) =
1 ωn−1
(1 − x2 )
Z Sn−1
1 f (θ ) dθ = x − θ n ωn−1
Z Sn−1
v(x; θ ) f (θ ) dθ
is harmonic on the unit disc by part (a), since the derivatives in x can be passed inside the integral by the Lebesgue dominated convergence theorem. Thus ∆ (u) = 0 on x < 1. Next, we need to show that limr↑1 u(rx0 ) = f (x0 ) when x0  = 1. Fix x0  = 1. Given ε > 0 find δ > 0 such that for all θ ∈ Sn−1 with θ − x0  ≤ δ we have  f (θ ) − f (x0 ) < ε/2 by the continuity of f . Then using the identity in part (c) we write Z 1 Z 1 u(rx0 ) − f (x0 ) = v(x; θ ) f (θ ) dθ − v(x; θ ) f (x0 ) dθ n−1 n−1 ωn−1 S ωn−1 S Z Z 1 1 ≤  f (θ ) − f (x0 )v(rx0 ; θ ) dθ +  f (θ ) − f (x0 )v(rx0 ; θ ) dθ n−1 ∈S ∈Sn−1 ωn−1 θθ−x ωn−1 θθ−x >δ ≤δ 0
≤ 2k f kL∞
1 ωn−1
0
ε 1 v(rx0 ; θ ) dθ + θ ∈Sn−1 2 ωn−1 θ −x >δ
Z
0
Z θ ∈Sn−1 θ −x0 ≤δ
v(rx0 ; θ ) dθ ,
The second term is at most ε/2 while the first term can also be made less less than ε/2 by part (b) provided r is chosen sufficiently close to 1, i.e., r > r0 for some r0 < 1 (depending on δ and k f kL∞ ). Then the preceding displayed sum can be made less than ε for r sufficiently close to 1, i.e., r0 < r < 1 and this shows that claimed convergence.
Exercise 2.4.8 Fix n ∈ Z+ with n ≥ 2 and a real number λ , 0 < λ < n. Also fix η ∈ Sn and y ∈ Rn .
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(a) Prove that
n
Z Sn
ξ − η−λ dξ = 2n−λ
(b) Prove that
π 2 Γ ( n−λ 2 ) Γ (n − λ2 )
.
n
Z
λ
x − y−λ (1 + x2 ) 2 −n dx =
π 2 Γ ( n−λ 2 ) λ 2)
λ
(1 + y2 )− 2 .
Γ (n − Hint: Part (a): See Appendix D.4 Part (b): Use the stereographic projection in Appendix D.6. Rn
Solution. (a) First we note that by applying a rotation we may reduce things to the case ξ = e1 = (1, 0, . . . , 0). To compute the integral Z Sn
dθ , θ − e1 λ
for n ≥ 2 we apply the formula in Appendix D.2 to write Z Sn
dη = η − e1 λ
Z +1 −1
α
(s − 12 + η2 ) 2
√
1−s2 Sn−1
η∈
(1 − s2 )
Z +1
=
=
ωn−1
n−1 2
ds λ √1 − s2 2 (1 − s)2 + 1 − s2
ωn−1
−1
ds √ 1 − s2
dη
Z
n−2 Z +1 (1 − s2 ) 2
ds λ λ 2 2 −1 (1 − s) 2 Z n−2 n−2−λ ωn−1 +1 = λ (1 − s) 2 (1 + s) 2 ds , 2 2 −1 which converges exactly when λ < n. To compute this integral we change variables 1 + s = 2t. This results the integral ωn−1 λ
Z 1
λ
21+n−2− 2
(1 − t)
n−2−λ 2
t
n−2 2
0
22
n−λ n , , dt = ωn−1 2−λ 2n−1 B 2 2
and this is equal to
n
n n−λ 2 2π n/2 −λ n−1 Γ ( n−λ n−λ π Γ ( 2 ) 2 )Γ ( 2 ) 2 2 = 2 . Γ ( 2n ) Γ (n − λ2 ) Γ (n − λ2 )
(b) We use the stereographic projection formula Z Rn
F(Π (x))
where
2 1 + x2
Π (x1 , . . . , xn ) =
n
Z
dx =
Sn
F(ϕ) dϕ ,
2x1 2xn x2 − 1 , . . . , , . 1 + x2 1 + x2 1 + x2
We have the identity Z Rn
λ
x − y−λ (1 + x2 ) 2 −n dx = 2−n
Z
λ
Rn
x − y−λ (1 + x2 ) 2
2 1 + x2
n dx
and the above is equal to −n
2
Z Rn
x2 + y2 − 2x · y 1 + x2
− λ 2
2 1 + x2
n
−n
dx = 2
Z Rn
1 + y2 x2 − 1 1 − y2 2x + − ·y 2 2 1 + x 2 1 + x2
− λ 2
2 1 + x2
n dx .
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Using the the stereographic projection formula above, the above is equal to 2−n
Z Sn
− λ 2 1 − y2 1 + y2 dϕ + ϕn+1 − (ϕ1 , . . . , ϕn ) · y 2 2
where ϕ = (ϕ1 , . . . , ϕn , ϕn+1 ). The above is equal to 2−n
− λ −λ /2 Z 2 2y 1 − y2 dϕ − (ϕ , . . . , ϕ ) · 1 + ϕn+1 n 1 2 2 1 + y 1 + y Sn
1 + y2 2
λ
= 2−n+ 2 1 + y2 = 2−n+λ 1 + y2
−λ /2 Z
−λ /2 Z
= 2−n+λ 1 + y2
− λ 1 − ϕ · Π (y) 2 dϕ
Sn
− λ 2 − 2ϕ · Π (y) 2 dϕ
Sn
−λ /2 Z ϕ − Π (y)−λ dϕ Sn
n
2 −λ /2 n−λ
−n+λ
1 + y
=2
2
π 2 Γ ( n−λ 2 ) Γ (n − λ2 )
n
= 1 + y2
− λ π 2 Γ ( n−λ 2 ) 2 Γ (n − λ2 )
.
Exercise 2.4.9 Prove the following beta integral identity: n−α1 n Γ dt 2 Γ =π2 α α α 1 2 x − t y − t Γ 21 Γ
Z Rn
n−α2 Γ α1 +α2 2 −n 2 α2 α1 +α2 2 Γ n− 2
x − yn−α1 −α2 ,
where 0 < α1 , α2 < n, α1 + α2 > n. Hint: Reduce to the case y = 0, interpret the integral as a convolution, and use Theorem 2.4.6. Solution. Replacing x by x + y matters reduce to proving −α1
(t
−α2
∗ t
Z
)(x) =
Rn
n−α2 Γ α1 +α2 2 −n 2 α2 α1 +α2 2 Γ n− 2
n−α1 Γ n Γ dt 2 2 =π α1 x − tα1 tα2 Γ 2 Γ
xn−α1 −α2 .
Taking the Fourier transform in x and using Theorem 2.4.6 the left hand side becomes n−α1 2
Γ π
n−α1 2
π
−n+α1
ξ 
Γ
n+(−n+α1 ) 2
n+(−n+α1 ) 2
n−α2 2
Γ π
n−α2 2
π
−n+α2
ξ 
Γ
n+(−n+α2 ) 2
n+(−n+α2 ) 2
n−α n−α π α1 +α2 Γ 2 1 Γ 2 2 ξ −n+(α1 +α2 −n) = πn Γ α21 Γ α22
By assumption we have 0 < α1 + α2 − n < n and so we can use Theorem 2.4.6 again to compute the inverse Fourier transform of the preceding expression. This is equal to
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n−α2 Γ α1 +α2 2 −n 2 α2 α1 +α2 2 Γ n− 2
xn−α1 −α2 .
Exercise 2.4.10 (a) Prove that if a continuous integrable function f on Rn (n ≥ 2) is constant on the spheres r Sn−1 for all r > 0, then so is its Fourier transform. (b) If a continuous integrable function on Rn (n ≥ 3) is constant on all (n − 2)–dimensional spheres orthogonal to e1 = (1, 0, . . . , 0), then its Fourier transform has the same property. Solution. (a) The hypothesis implies that f (rx) = f (rx0 ) for all x, x0 ∈ Sn−1 and all r > 0. Given x, x0 ∈ Sn−1 , there is an orthogonal matrix A such that Ax = x0 . So given any orthogonal matrix A and any x ∈ Sn−1 we have f (rx) = f (rAx) . Taking the Fourier transform in x and using Proposition 2.2.11 (8) and (13) we obtain r−n fb(r−1 ξ ) = r−n fb(r−1 Aξ ) and this implies that fb is constant on spheres r−1 Sn−1 for all r > 0. (b) The hypothesis implies that f (x1 , sx0 ) = f (x1 , sx00 ) for all x1 ∈ R, all x0 , x00 ∈ Sn−2 and all s > 0. This implies that for any orthogonal (n − 1) × (n − 1) matrix B and all x0 ∈ Sn−2 we have f (x1 , sBx0 ) = f (x1 , sx0 ) . Consider the n × n matrix
1 0 Q= , 0 sB
where the zeros in the matrix are vectors of length n − 1. The inverse of Q is 1 0 −1 Q = , 0 s−1 Bt the transpose of Q−1 is (Q−1 )t =
1 0 , 0 s−1 B
and the determinant of Q is sn−1 . Let ξ = (ξ1 , ξ 0 ) ∈ Rn . For any x ∈ Rn and s > 0 we have that f (x) = f (Qx), hence fb(ξ1 , sξ 0 ) = =
Z ZR
0
n
Rn
f (x)e−2πix·(ξ1 ,sξ ) dx 0
f (Qx)e−2πix·(ξ1 ,sξ ) dx
−1 0 1 = f (x)e−2πiQ x·(ξ1 ,sξ ) dx  det Q Rn Z −1 t 0 1 = n−1 f (x)e−2πix·(Q ) (ξ1 ,sξ ) dx s Rn Z 0 1 = n−1 f (x)e−2πix·(ξ1 ,Bξ ) dx s Rn
Z
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101 0 0 1 f (x1 , x0 )e−2πi(x1 ξ1 +x ·Bξ ) dx0 dx1 sn−1 R Rn−1 Z Z −1 0 0 1 = n−1 f (x1 , x0 )e−2πi(x1 ξ1 +s x ·sBξ ) dx0 dx1 s R Rn−1
Z Z
=
Z Z
=
Rn−1
R
f (x1 , x00 )e−2πi(x1 ξ1 +x
00 ·sBξ 0 )
dx00 dx1
= fb(ξ1 , sBξ 0 ) . It follows that fb(ξ1 , sξ 0 ) = fb(ξ1 , sBξ 0 ) ,
hence fb is also constant on all spheres orthogonal to e1 .
Exercise 2.4.11 ([138]) Suppose that 0 < d1 , d2 , d3 < n satisfy d1 + d2 + d3 = 2n. Prove that for any distinct x, y, z ∈ Rn we have the identity Z Rn
x − t−d2 y − t−d3 z − t−d1 dt 3 d Γ n − 2j n =π2 ∏ x − yd1 −n y − zd2 −n z − xd3 −n . dj Γ j=1 2
Hint: Reduce matters to the case that z = 0 and y = e1 . Then take the Fourier transform in x and use that the function h(t) = t − e1 −d3 t−d1 satisfies b h(ξ ) = b h(A−2 ξ ) for all ξ 6= 0, where Aξ is an orthogonal matrix with Aξ e1 = ξ /ξ . ξ Solution. Clearly both sides of the identity we wish to prove are invariant under translations, dilations, and rotations of x, y, z. Therefore, by a translation we can assume that z = 0, by a dilation that y = 1, and by a rotation that y = e1 = (1, 0, . . . , 0). Let us set Γ d2 c(d) = π n/2−d Γ n−d 2 In view of Theorem 2.4.6 we have d−n ∧
(x
) (ξ ) = π
n/2−d
Γ Γ
d 2 n−d 2
ξ −d = c(d)ξ −d
and in the case 0 < d < n both functions are locally integrable. After these reductions, we prove the claimed identity by showing that the Fourier transform of both sides coincide. The right hand side of the claimed identity is the function π
n 2
3
∏
Γ n−
dj 2
Γ
j=1
dj 2
x − e1 d1 −n xd3 −n
which has Fourier transform
π
n 2
3
∏
j=1
Γ n− Γ
dj 2
dj 2
d3 −n
x
d1 −n ∧
x − e1 
n 2
3
Γ n−
dj 2
c(d3 )c(d1 )ξ −d3 dj j=1 Γ 2 3 d Z Γ n − 2j n 2 =π c(d3 )c(d1 ) ξ dj Rn j=1 Γ 2
(ξ ) =π
∏ ∏
∗ ξ −d1 e−2πiξ ·e1
− η−d3 η−d1 e−2πiη·e1 dη
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Contents n
3
=π 2
∏
Γ n−
j=1
dj 2
c(d3 )c(d1 )ξ n−d1 −d3
dj 2
Γ
Z Rn
ξ 0 − t−d3 t−d1 e−2πiξ t·e1 dt,
where ξ 0 = ξ /ξ . Now, for given ξ find a rotation Aξ so that Aξ e1 = ξ 0 . Clearly ξ 0 − t = e1 − A−1 t, t = A−1 t, and ξ ξ t · e1 = t · A−1 ξ 0 = A−1 t · A−2 ξ 0. ξ ξ ξ Hence, changing variables s = A−1 t the last expression in the alignment is equal to ξ n−d1 −d3
c(d3 )c(d1 )ξ 
Z
e1 − s−d3 s−d1 e
Rn
−2πis·A−2 ξ ξ
h(A−2 ξ ), ds = c(d3 )c(d1 )ξ n−d1 −d3 b ξ
where h(t) = t − e1 −d3 t−d1 . Next we claim that b h(ξ ) = b h(A−2 ξ) ξ
for all ξ ∈ Rn \ {0}
a fact that we will prove late. Thus the Fourier transform of the right hand side of the identity we wish to prove is π
n 2
3
∏
j=1
Γ n−
dj 2
dj 2
Γ
d1 −n ∧
d3 −n
x
x − e1 
(ξ ) = π
n 2
3
∏
j=1
Γ n− Γ
dj 2
dj 2
c(d3 )c(d1 )ξ n−d1 −d3 b h(ξ )
On the other hand, we denote by Z
g(x) =
Rn
x − t−d2 t − e1 −d3 t−d1 dt = (h ∗  · −d2 )(x)
the left hand side of identity we wish to prove when z = 0 and y = (1, 0, . . . , 0). We have that gb(ξ ) h ∗ t−d2
∧
(ξ ) = c(n − d2 )b h(ξ )ξ d2 −n .
Then we verify the claimed identity by checking that π
n 2
3
∏
j=1
Γ n− Γ
dj 2
c(d3 )c(d1 )ξ n−d1 −d3 b h(ξ ) = c(n − d2 )b h(ξ )ξ d2 −n
dj 2
Since d1 + d2 + d3 = 2n the powers of ξ  match. Moreover, notice that c(n − d2 )−1 = c(d2 ) and thus we need to verify that π
n 2
3
∏
Γ n−
j=1
Γ
dj 2
dj 2
c(d3 )c(d1 )c(d2 ) = 1 .
Γ d2 and d1 + d2 + d3 = 2n. But this is certainly true since c(d) = π n/2−d n−d Γ 2 We now return to proving that b h(ξ ) = b h(A−2 ξ ) for all ξ ∈ Rn \ {0}. We use ξ
the fact that if a function is reflectioninvariant with respect to a given hyperplane then so is its Fourier transform. Modulo rotations and translations it is enough to check this for hyperplanes of the form x j = 0. But the function h is constant along circles orthogonal to e1 ; in particular h is reflection invariant with respect to the hyperplanes x j = 0, for j = 2, 3, . . . , n, and hence so is b h. But A−2 ξ can be obtained from ξ by ξ −2 b b finitely many reflections with respect to the above hyperplanes, and this proves that h(ξ ) = h(A ξ ) for all ξ ∈ Rn \ {0}. ξ
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Exercise 2.4.12 2
(a) Integrate the function eiz over the contour consisting of the three pieces P1 = {x + i0 : 0 ≤ x ≤ R}, P2 = {Reiθ : 0 ≤ θ ≤ π4 }, π and P3 = {r ei 4 : 0 ≤ r ≤ R} (with the proper orientation) to obtain the Fresnel integral identity: Z R
lim
R→∞ 0
√ 2π 4 (1 + i) .
2
eix dx =
2
πn
2
(b) Use the result in part (a) to show that the Fourier transform of the function eiπx in Rn is equal to ei 4 e−iπξ  . 4 2 2 Hint: Part (a): On P2 we have e−R sin(2θ ) ≤ e− π R θ , and the integral over P2 tends to 0. Part (b): Try first n = 1. Solution. (a) The integral over P1 is equal to
Z R
2
eix dx .
0
The integral over P2 is equal to
π 4
Z
eiR
2 ei2θ
Rieiθ dθ
0
which is bounded in absolute value by Z
π 4
−R2 sin(2θ )
e
R dθ ≤
Z
0
π 4
e
−R2 π2 2θ
R dθ ≤
Z ∞
0
2
e− π 2θ
0
1 dθ R
which tends to zero as R → ∞. Here we uses that sin(t) ≥ π2 t when 0 < t < π2 . The integral over the path −P3 (i.e., P3 oriented in the opposite direction) is equal to Z 0
eir
π 2 ei 2
π
ei 4 dr = −
Z R
2
π
e−r dr = −ei 4
Z R
0
R
2
e−r dr
0
2
Since the function z 7→ eiz is analytic on the entire plane, its integral is equal to zero over the union P1 ∪ P2 ∪ P3 . Thus we have Z Z R
0=
2
Z R
ix2
Z R
i π4
e dx = −E(R) + e
−r2
e
0
0
as R → ∞. So we obtain
2
e−r dr .
0
0
It follows that
R
π
eix dx + E(R) − ei 4
Z R
lim
R→∞ 0
2
π
eix dx = ei 4
√ π = 2
√ π dr → 0 + e 2 i π4
√ 2π 4 (1 + i) .
2
(b) We compute the distributional Fourier transform of eiπx . For ϕ ∈ S we have
d2 iπx2 eiπx , ϕ = e , ϕb Z
=
eiπx
2
R
Z
= lim
Z
ϕ(ξ )e−2πixξ dξ dx
R +R
R→∞ −R
eiπx
Z Z +R
2
Z
ϕ(ξ )e−2πixξ dξ dx
R 2
= lim
eiπx ϕ(ξ )e−2πixξ dx dξ
= lim
eiπ(x−ξ ) dx e−iπξ ϕ(ξ ) dξ
=
eiπ(x−ξ ) dx e−iπξ ϕ(ξ ) dξ
R→∞ R −R Z Z +R R→∞ R −R Z Z +R
lim
R R→∞ −R
2
2
2
2
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Z +R0
Z
=
lim 0
−R0
R R →∞
2 dx 2 ei(x−ξ ) √ e−iπξ ϕ(ξ ) dξ π
Z +R0 +ξ
2 dx 2 eix √ e−iπξ ϕ(ξ ) dξ π R R →∞ Z √π √π 1 π 2 2 √ e−iπ ξ ϕ(ξ ) dξ = ei 4 + 2 2 π R Z i π4 −iπ 2 ξ 2 = e e ϕ(ξ ) dξ R
π 2 2 = ei 4 e−iπ ξ , ϕ
Z
=
lim 0
−R0 +ξ
and thus the claim holds for n = 1. The passage of the limit inside the integral is justified by the Lebesgue dominated convergence theorem. Extending this result in higher dimensions is trivial in view of the multiplicative nature of the function 2
2
2
eiπx = eiπx1 · · · eiπxn .
Section 2.5. Convolution Operators on L p Spaces and Multipliers Exercise 2.5.1 Prove that if f ∈ Lq (Rn ) and 0 < q < ∞, then
h
τ ( f ) + f q → 21/q f q L L
as h → ∞.
Solution. If φ ∈ C0∞ , then for h large τ h φ has disjoint support from φ and therefore kτ h φ + φ kqLq = kφ kqLq + kτ h φ kqLq = 2kφ kqLq . Now given f ∈ Lq and ε > 0 find a φ ∈ C0∞ such that k f − φ kLq < ε. Now find a h0 > 0 such that for h > h0 we have + φ kLq = 21/q kφ kLq . Then for h > h0 we have
kτ h φ
kτ h f + f kLq ≤ 2ε + kτ h φ + φ kLq ≤ (2 + 21/q )ε + 21/q k f kLq and similarly kτ h f + f kLq ≥ 21/q k f kLq − (2 + 21/q )ε.
The assertion follows.
Exercise 2.5.2 h
Prove Proposition 2.5.14. Also prove that if δ j j is a dilation operator in the jth variable (for instance δ1h1 f (x) = f (h1 x1 , x2 , . . . , xn )), then
h
δ 1 · · · δnhn m = m . 1 Mp Mp Solution.
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105
(a) We have Z
Tτ x m ( f )(t) =
fb(ξ )m(ξ − x)e2πit·ξ dξ
Rn
Z
=
fb(y + x)m(y)e2πit·(y+x) dy
Rn
Z
= e2πit·x
(where y = ξ − x)
τ −x fb(y)m(y)e2πit·y dy
Rn
= e2πit·x Tm (e−2πit·x f )(t), so kTτ x m ( f )kL p = kTm (e2πi(·)·x f )kL p , and since the mapping f 7→ e2πi(·)·x f is an L p isometry it follows that kTτ x m kL p →L p = kTm kL p →L p . (b) We have that Z
Tδ h m ( f )(t) =
fb(ξ )m(hξ )e2πit·ξ dξ
Rn
= h−n Z
= =δ
Z
−1
Rn
δh
−1 fb(y)m(y)e2πi(h t)·ξ dξ
−1 h f (y)m(y)e2πi(h t)·y dy δd
Rn h−1
Tm (δ h f )(t).
Observe that kδ h f kL p = h−n/p k f kL p for every f ∈ L p so we have kTδ h m kL p →L p = hn/p kTm ◦ δ h kL p →L p . We also know kTm ◦ δ h kL p →L p =
sup
kTm (δ h f )kL p
f ∈L p ,k f k=1
but k f kL p = 1 if and only if kδ h f kL p = h−n/p . So it follows that kTm ◦ δ h kL p →L p = h−n/p kTm kL p →L p , and therefore kTδ h m kL p →L p = kTm kL p →L p . (c) We verify that Z
Tme ( f )(t) =
fb(ξ )m(−ξ )e2πit·ξ dξ
Rn
Z
=
fb(−ξ )m(ξ )e2πi(−t)·ξ dξ
Rn
^ = Tm ( fe)(t) Therefore kTme f kL p = kTm f kL p since f 7→ fe is an L p isometry. (d) We check that Z
Te2πi(·)·x m ( f )(t) =
Rn
fb(ξ )e2πiξ ·x m(ξ )e2πiξ ·t dξ
y = hξ
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Z
=
Rn
−x f (ξ )m(ξ )e2πiξ ·t τ[
= Tm (τ −x f )(t) = τ −x Tm ( f )(t). Since f 7→ τ −x f is an L p isometry it follows that kTeeπi(·)·x m kL p →L p = kTm kL p →L p . (e) We have Z
Tm◦A ( f )(t) =
Rn
Z
=
Rn
fb(ξ )m(Aξ )e2πit·ξ dξ t fb(At ξ )m(ξ )e2πi(A t)·ξ dξ
= Tm ( f ◦ At )(At t). Since A is orthogonal f 7→ f ◦ At is an L p isometry, so kTm◦A kL p →L p = kTm kL p →L p . (f) First observe that δ1h1 · · · δnhn fb(ξ1 , ..., ξn ) =
Z Rn
f (x1 , ..., xn )e−2πi(x1 h1 ξ1 +···+xn hn ξn ) dx1 · · · dxn
= (h1 · · · hn )−1 =
Z
f ( hy11 , ..., hynn )e2πi(y1 ξ1 +···+yn ξn ) dy1 · · · dyn Rn h−1 h−1 (h1 · · · hn )−1 (δ1 1 · · · δn n f )∧ (ξ1 , ..., ξn ).
Also observe that kδ1h1 · · · δnhn f kL p = (h1 · · · hn )−1/p k f kL p for every f ∈ L p . An argument identical to part (b) now establishes the desired result.
Exercise 2.5.3 Let m ∈ M p (Rn ) where 1 ≤ p < ∞. (a) If ψ is a function on Rn whose inverse Fourier transform is an integrable function, then prove that
ψm ≤ ψ ∨ 1 m . Mp L Mp (b) If ψ is in L1 (Rn ), then prove that
ψ ∗ m ≤ ψ 1 m . Mp L Mp Solution. (a) Suppose ϕb = ψ and ub = m. The ϕd ∗ u = ϕbub = ψm. Therefore Tψm ( f ) = f ∗ (ϕ ∗ u) = ( f ∗ u) ∗ ϕ = Tm ( f ) ∗ ϕ. Since ϕ ∈ L1 , Minkowski’s inequality yields kTψm ( f )kL p = kTm ( f ∗ ϕ)kL p ≤ kϕkL1 kTm ( f )kL p ≤ kϕkL1 kTm kL p →L p k f kL p . (b) First note that Z
Tψ∗m ( f ) =
Rn
ψ(y)(m(ξ − y) fb(ξ ))∨ dy.
Therefore kTψ∗m ( f )kL p ≤
Z Rn
ψ(y) k(τ y m(ξ ) fb(ξ ))∨ kL p dy
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107
≤
Z Rn
ψ(y) kmkM p k f kL p dy
(by Exercise 2.5.2(a))
= kψkL1 kmkM p k f kL p .
This proves the claimed inequality.
Exercise 2.5.4 Fix a multiindex γ. (a) Prove that the map T ( f ) = f ∗ ∂ γ δ0 maps S continuously into S . (b) Prove that when 1/p − 1/q 6= γ/n, T does not extend to an element of the space M p,q . Solution. (a) Since ρα,β (∂ γ f ) = ρα,β +γ ( f ), for all multiindices α and β , it follows that every finite sum of seminorms of ∂ γ f is controlled by a finite sum of seminorms of f . The conclusion now follows immediately. 2 (b) Assume that k∂ γ f kLq ≤ Ck f kL p for some 0 < C < ∞ and all f ∈ S (Rn ). Let f0 (x) = e−πx and apply this inequality to the functions (δ λ f0 )(x) = f0 (λ x) with λ > 0. Since ∂ γ δ λ f0 = λ γ δ λ ∂ γ f0 one obtains λ γ λ −n/q k∂ γ f0 kLq ≤ Cλ −n/p k f0 kL p . If 1/q − 1/p 6= γ/n, then by sending λ to zero or infinity we obtain ∞ ≤ Ck f0 kL p < ∞ or 0 < k∂ γ f0 kLq ≤ 0 which are both impossible.
Exercise 2.5.5 Let Kγ (x) = x−n+γ , where 0 < γ < n. Use Theorem 1.4.25 to show that the operator Tγ ( f ) = f ∗ Kγ ,
f ∈S ,
extends to a bounded operator in M p,q (Rn ), where 1/p − 1/q = γ/n, 1 < p < q < ∞. This provides an example of a nontrivial operator in M p,q (Rn ) when p < q. Solution. Observe that the function Kγ (x) = x−n+γ is in Lr,∞ with r = (n − γ)/γ. Theorem 1.4.25 gives that the operator f → f ∗ Kγ maps L p → Lq when 1 < p < q < ∞ and 1/p + 1/r = 1/q + 1. This is the same as 1/p − 1/q = γ/n.
Exercise 2.5.6 (a) Use the ideas of the proof of Proposition 2.5.13 to show that if m j ∈ M p , 1 ≤ p < ∞, km j kM p ≤ C for all j = 1, 2, . . . , and m j → m a.e., then m ∈ M p and
m
n ≤ lim inf m j n ≤C. M p (R )
j→∞
M p (R )
(b) Prove that if m ∈ M p , 1 ≤ p < ∞, and the limit m0 (ξ ) = lim m(ξ /R) exists for all ξ ∈ Rn , then m0 is a radial function in R→∞
M p (Rn ) and satisfies km0 kM p ≤ kmkM p . (c) If m ∈ M p (R) has left and right limits at the origin, then prove that
m ≥ max(m(0+), m(0−)). M p (R) (d) Suppose that for some 1 ≤ p < ∞, mt ∈ M p (Rn ) for all 0 < t < ∞. Prove that
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dt
mt < ∞ =⇒ m(ξ ) = M p (Rn ) t
Z ∞ 0
Z ∞
mt (ξ )
0
dt ∈ Mp . t
Solution. (a) Fix f ∈ S . Since m j → m a.e. and the m j are uniformly bounded, it follows that L∞ norm of m is bounded and Z
Tm j ( f )(x) =
Rn
Z
fb(ξ )m j (ξ )e2πix·ξ dξ →
Rn
fb(ξ )m(ξ )e2πix·ξ dξ = Tm ( f )(x)
for almost x ∈ Rn , by the Lebesgue dominated convergence theorem. Fatou’s lemma also implies that Z Rn
1
Z
p
p
Tm ( f )(x) dx
=
Rn
Z =
Rn
p 1 p lim inf Tm (x) dx j j→∞
p lim inf Tm j (x) dx
1
p
j→∞
1 p p ≤ lim inf Tm j (x) dx j→∞ Rn
≤ lim inf Tm j L p →L p k f kL p . Z
j→∞
Hence kmkM p ≤ lim inf j→∞ km j kM p and m ∈ M p . (b) For each R > 0, define mR (ξ ) = m(R−1 ξ ), ξ ∈ Rn . Changing variables, we obtain Z
TmR ( f )(x) =
Rn
fb(ξ )m(R−1 ξ )e2πix·ξ dξ =
Z Rn
−1 fb(Rξ )m(ξ )e2πiRx·ξ Rn dξ = Tm (δ R f )(Rx).
Taking the L p norm, we have Z Rn
TmR ( f )(x) p dx
1
Z
p
=
Rn
Tm (δ R f )(Rx) p dx
1
p
p 1p −1 Tm (δ R f )(x) dx Rn
−1 n
≤R− p kmkM p δ R f n
=R− p
Z
Lp
n
≤R− p kmkM p
Z Rn
f (R−1 x) p dx
= kmkM p k f kL p . This says kmR kM p = kmkM p , for all R > 0. Applying part (a), we deduce km0 kM p ≤ lim inf kmR kM p = kmkM p . R→∞
(c) Given m ∈ M p (R) we have ( m(0+) m0 (ξ ) = lim m(ξ /R) = R→+∞ m(0−)
when ξ > 0 when ξ < 0
since we are assuming that these limits exist. Then
m ≥ m L∞ (R) = max(m(0+), m(0−)) M p (R) which proves the desired inequality.
1
p
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(d) First of all notice that the integral Z ∞
m(ξ ) = 0
mt (ξ )
dt t
converges in M p and hence in L∞ . Thus m(ξ ) is a well defined L∞ function and it remains to prove that it lies in M p . Take f ∈ S . We have Z Tm ( f )(x) = fb(ξ )m(ξ )e2πix·ξ dξ Rn
∞ dt fb(ξ )e2πix·ξ mt (ξ ) dξ t Rn 0 Z ∞ Z dt 2πix·ξ b = f (ξ )mt (ξ )e dξ n t 0 R Z ∞ dt = Tmt ( f )(x) . t 0 Fubini’s theorem can be used to interchange the above integrals because the following integral converges absolutely Z ∞ Z Z ∞Z dt b fb(ξ )e2πix·ξ mt (ξ ) dξ dt = f (ξ ) mt (ξ ) dξ dt t t 0 Rn 0 Rn Z ∞Z dt b ≤ f (ξ ) kmt kL∞ dξ dt t 0 Rn Z Z∞ dt b kmt kM p ≤ < ∞. f (ξ ) dξ t Rn 0
Z
Z
=
Applying Minkowski’s inequality, one gets Z Rn
p
Tm ( f )(x) dx
1
p
Z ∞ 1p dt p Tmt ( f )(x) dx = t Rn 0 1p Z Z ∞ dt p Tmt ( f )(x) dx ≤ t Rn 0 1 Z ∞ Z p dt p Tm ( f )(x) dx ≤ t 0 Rn Z ∞ dt kmt kM p . ≤ k f kL p t 0 Z
This implies m ∈ M p with kmkM p ≤
Z ∞ 0
kmt kM p
dt < ∞. t
Exercise 2.5.7 Let 1 ≤ p < ∞ and suppose that m ∈ M p (Rn ) satisfies m(ξ ) ≥ c (1 + ξ )−N for some c, N > 0. Prove that the operator T ( f ) = ( fbm−1 )∨ satisfies kT ( f )kL p ≥ c p k f kL p for all f ∈ S (Rn ), where c p = kmk−1 Mp . Solution. Simply observe that k f kL p = k( fbm m1 )∨ kL p ≤ kmkM p k( fb m1 )∨ kL p = kmkM p kT ( f )kL p . Notice that the condition on m(ξ )−1  ≤ c−1 (1 + ξ )N makes the product fb m1 an integrable function and hence its inverse Fourier transform is a well defined function.
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Exercise 2.5.8 (a) Prove that if m ∈ L∞ (Rn ) satisfies m∨ ≥ 0, then for all 1 ≤ p < ∞ we have
m = m∨ 1 . L Mp (b) (L. Colzani and E. Laeng ) On the real line let ( −1 for ξ > 0 m1 (ξ ) = 1 for ξ < 0,
( min(ξ − 1, 0) m2 (ξ ) = max(ξ + 1, 0)
for ξ > 0 for ξ < 0.
Prove that
m1 = m2 Mp Mp for all 1 < p < ∞. Hint: Part (a): Use Exercise 1.2.9. Part (b): Use part (a) to show that km2 m−1 1 kM p = 1. Deduce that km2 kM p ≤ km1 kM p . For the converse use Exercise 2.5.6(c). Solution. (a) Follows directly from Exercise 1.2.9. (b) Observe that Exercise 2.5.6 gives that km2 kM p ≥ km1 kM p . Now m1 m2 = χ[−1/2,1/2] ∗ χ[−1/2,1/2] and since χ[−1/2,1/2] is even, it follows that (χ[−1/2,1/2] )∨ is real and that (m1 m2 )∨ = (χ[−1/2,1/2] )∨
2
2 = (χ[−1/2,1/2] )∨ ≥ 0.
By part (a), km1 m2 kM p =
Z R
(χ[−1/2,1/2] )∨ 2 (t) dt = 1,
in view of Plancherel’s Theorem. Thus km2 kM p = km2 m1 m1 kM p ≤ km2 m1 kM p km1 kM p = km1 kM p .
Exercise 2.5.9 ([95]) Let 1 < p < ∞, {am }m∈`1 (Zn ) , and 0 < A < ∞. Prove that the following are equivalent: (a) The operator f 7→ ∑m∈Zn am f (x − m) is bounded on L p (Rn ) with norm A. (b) The M p norm of the function ∑m∈Zn am e−2πim·x is exactly A. (c) The operator given by convolution with the sequence {am } is bounded on ` p (Zn ) with norm A. Solution. Assertions (a) and (b) are clearly equivalent, since the multiplier of the operator f 7→ ∑m∈Zn am f (x − m) is
∑ n am e−2πim·x .
m∈Z
Let A be the norm of the operator in (a) and C the norm of the operator in (c). We can obtain easily that C ≤ A by applying the operator in (a) to step functions. For the converse write
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p p Z Z a f (x + k − m) dx ∑ am f (x − m) dx = ∑ ∑ m [0,1]n k∈Zn m∈Zn Rn m∈Zn ≤C
Z
p
∑
[0,1]n k∈Zn
 f (x − k) p dx
= C p k f kLp p (Rn ) , which implies that A ≤ C.
Exercise 2.5.10 ([179]) Let m(ξ ) in M p (Rn ) be supported in [0, 1]n . Then the periodic extension of m in Rn , M(ξ ) =
∑
m(ξ − k) ,
k∈Zn
is also in M p (Rn ). Solution. Since m is in L∞ and is compactly supported, it lies in L2 and hence it has a Fourier expansion. Write m(ξ ) =
∑
a j e2πi j·ξ
∑
a j e2πi j·ξ
j∈Zn
in Fourier series for ξ ∈ [0, 1]n . Then M(ξ ) =
j∈Zn
for ξ ∈ Rn . Let h ∈ C0∞ to be determined later and let Q = [−1/2, 1/2]n be the unit cube centered at the origin. Define operators E({c j } j ) =
ck χQ (x + k)
∑
k∈Zn
Z
R( f )( j) =
f (x) dx − j+Q
Let Tmh be the operator with multiplier mh. An easy calculation shows that (RTmh E)({ck }k )( j) =
∑
ck q j−k ,
k∈Zn
where
Z
qj =
Q
∨ (χc Q mh) (x − j)dx =
Z
∑n
k∈Z
ak
Q
2 2πi(k− j)·ξ (χc dξ = a j , Q ) (ξ )h(ξ )e
if we pick a function h equal to = ∏nj=1 π 2 ξ 2 sin−2 (πξ ) on Q. Therefore the operator RTmh E is given by convolution with the fixed sequence {a j } j . Since both R and E are contractions, we have C0∞ (Rn )
−2 (χc Q ) (ξ )
kRTmh Ekl p →l p ≤ kTmh kL p →L p ≤ kh∨ kL1 kmkM p by Exercise 2.5.3. The conclusion follows from the Exercise 2.5.9.
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Exercise 2.5.11 Suppose that u is a C ∞ function on Rn \ {0} that is homogeneous of degree −n + iτ, τ ∈ R. Prove that the operator given by convolution with u maps L2 (Rn ) to L2 (Rn ). Solution. In view of Exercise 2.3.9 the distributional Fourier transform ub of u is a homogeneous distribution of degree −iτ. But Proposition 2.4.8 gives that ub is a C ∞ function on Rn \ {0}, and in particular bounded on the sphere Sn−1 . Since ub(ξ ) = ukL∞ < ∞ and the operator given by convolution with u ξ −iτ ub( ξξ  ) it follows that ub is a bounded function on Rn \ {0}. Thus kb is L2 bounded.
Exercise 2.5.12 0
([144]) Let m1 ∈ Lr (Rn ) and m2 ∈ Lr (Rn ) for some 2 ≤ r ≤ ∞. Prove that m1 ∗ m2 ∈ M p (Rn ) when 1p − 21 = 1r and 1 ≤ p ≤ 2. Hint: Prove that the trilinear operator (m1 , m2 , f )7→ (m1 ∗ m2 ) fb ∨ is bounded from L2 × L2 × L1 → L1 and L∞ × L1 × L2 → L2 . Apply trilinear complex interpolation (Corollary 8.2.11 in [132]) to deduce the required conclusion for 1 ≤ p ≤ 2. Solution. We have that the trilinear operator S(m1 , m2 , f ) = (m1 ∗ m2 ) fb
∨
= m1 ∗ m2
∨
∗ f = m1∨ m2∨ ∗ f
maps L2 × L2 × L1 → L1 with norm at most 1 and also L∞ × L1 × L2 → L2 with norm at most 1. It follows from Corollary 8.2.11 in [132] that it maps L p1 × L p2 × L p → L p with norm at most 1, where 1 ≤ p ≤ 2 and 1−θ θ 1 θ 1 = + = − , p1 2 ∞ 2 2 It follows from these that r = p1 satisfies
1 p
− 12 =
1 1−θ θ 1 θ = + = + , p2 2 1 2 2 1 r
1 1−θ θ θ = + = 1− . p 1 2 2
and p2 = r0 thus the claimed inequality holds.
Exercise 2.5.13 2
(M. Peloso) Show that the function eiξ  is an L p Fourier multiplier on Rn if and only if p = 2. 2 2 ∨ Hint: By Exercise 2.4.12 the inverse Fourier transform of eiξ  is in L∞ , thus the operator f 7→ fb(ξ )eiπξ  maps L1 to L∞ . 0 2 2 p p Since this operator also maps L to L , it should map L to L for all 1 ≤ p ≤ 2. Solution. 2 2 By dilation we may work with the multiplier eiπξ  whose Fourier transform is cn e−iπx by Exercise 2.4.12. Since the kernel 2 cn e−iπx is in L∞ , it follows that the operator 2 ∨ T ( f ) = fb(ξ )eiπξ  0
maps L1 to L∞ . Since T maps L2 to L2 , then T also maps L p to L p for all 1 ≤ p ≤ 2. The same conclusion is also valid for the operator 2 ∨ T 0 ( f ) = fb(ξ )e−iπξ  .
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0
Suppose that T mapped L p to L p for some 1 < p < 2. Since T 0 maps L p to L p , the composition T 0 T would map L p to L p . 0 But the composition T 0 T is the identity operator and this would map L p to L p . This is impossible since we can find an L p 0 p function which is not in L if p 6= 2.
Section 2.6. Oscillatory integrals Exercise 2.6.1 Suppose that u is a realvalued C k function defined on the line that satisfies u(k) (t) ≥ c0 > 0 for some k ≥ 2 and all t ∈ (a, b). Prove that for λ ∈ R \ {0} we have Z b −1/k iλ u(t) e dt ≤ 12 k (λ c0 ) a and that the same conclusion is valid when k = 1, provided u0 is monotonic. Solution. Denote λ0 = λ c0 and u0 = u0 . We have
1 c0 u.
(k)
Then u0  ≥ 1 for k ≥ 2 and for all t ∈ (a, b). Apply Proposition 2.6.7 to λ0 and the function Z b 1 eiλ0 u0 (t) dt ≤ 12k λ0 − k . a
Now, replacing λ0 , u0 back to λ , u, we have
Z b 1 eiλ u(t) dt ≤ 12k λ c0 − k . a
The case k = 1 is treated by the same change of variables.
Exercise 2.6.2 Show that if u0 is not monotonic in part (c) of Proposition 2.6.7, then the conclusion may fail. Hint: Let ϕ(t) be a realvalued smooth function that is equal to 2t on intervals [2πk + εk , 2π(k + 21 ) − εk ] and equal to t on intervals [2π(k + 21 ) + εk , 2π(k + 1) − εk ], where 0 ≤ k ≤ N, for some N ∈ Z+ . Show that the absolute value of the integral of eiϕ(t) over the interval [ε0 , 2π(N + 1) − εN ] tends to infinity as N → ∞. Solution. Let ϕ be a smooth function such that ( t, t ∈ [2πk + εk , 2π(k + 12 ) − εk ] ϕ(t) = 2t, t ∈ [2π(k + 12 ) + εk , 2π(k + 1) − εk ] where εk ∈ (0, π4 ] will be chosen later and 0 ≤ k ≤ N, for some N ∈ Z+ . Then we have Z 2π(N+1)−ε Z 2π(N+1)−ε N N iϕ(t) e dt ≥ sin(ϕ(t))dt ε ε 0 0 N Z 2π(l+ 21 )−εl N Z 2π(l+1)−εl N Z 2π(l+ 12 )+εl = ∑ sin(t)dt + ∑ sin(2t)dt + sin(ϕ(t))dt ∑ 2π(l+ 1 )−ε 1 )+ε l=0 2πl+εl 2π(l+ l l l=0 l=0 2 2 N Z π−εl N Z π−εl N Z 2π(l+ 21 )+εl = ∑ sin(t)dt + ∑ sin(2t)dt + ∑ sin(ϕ(t))dt 1 l=0 εl l=0 π+εl l=0 2π(l+ 2 )−εl
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N N Z 2π(l+ 12 )+εl = ∑ 2 cos(εl ) + ∑ sin(ϕ(t))dt 1 )−ε l=0 2π(l+ l l=0 2 N
N
≥ ∑ 2 cos(εl ) − ∑ 2εl . l=0
l=0
Now assume that part (c) of Proposition 2.6.7 is still true for this function ϕ. Then choosing the sequence (εN )N∈Z+ such that ∞
∑ εk ≤ k=0
π 2
yields Z 2π(N+1)−ε N iϕ(t) e dt ≥ C≥ ε 0
N
N
∑ 2 cos εl − ∑ 2εl ≥ l=0
√ 2N −π
l=0
for all N ≥ 1. This contradicts the statement of Proposition 2.6.7 (c), in particular the fact that the constant C is independent of the interval (a, b).
Exercise 2.6.3 Prove that the dependence on k of the constant inpart (b) of Proposition 2.6.7 is indeed linear. Hint: Take u(t) = t k /k! over the interval (0, k!). Solution. Consider Z k!
e 1
iλt k /k!
k Z eiλt /k! 1−k k! (k − 1)!(1 − k) k! iλt k /k! −k dt = t + e t dt iλ /(k − 1)! iλ 1 1
Taking the absolute value, we have Z k! iλt k /k! e dt ≈ 1+ 0
1 (k − 1)! (k − 1)! (k − 1)! 1 )≈ + + (1 − , λ λ λ (k!)k−1 λ /(k − 1)! (k!)k−1
provided we choose (k − 1)!/λ = kλ −1/k or λ= Then
(k − 1)! k
k k−1
.
Z k! iλt k /k! dt ≈ kλ −1/k . 1 e
Exercise 2.6.4 Follow the steps below to give an alternative proof of part (b) of Proposition 2.6.7. Assume that the statement is known for some k ≥ 2 and some constant C(k) for all intervals [a, b] and all C k functions satisfying u(k) ≥ 1 on [a, b]. Fix a C k+1 function u such that u(k+1) ≥ 1 on an interval [a, b]. Let c be the unique point at which the function u(k) attains its minimum in [a, b]. (a) If u(k) (c) = 0, then for all δ > 0 we have u(k) (t) ≥ δ in the complement of the interval (c − δ , c + δ ) and derive the bound Z b iλ u(t) e dt ≤ 2C(k)(λ δ )−1/k + 2δ . a
(b) If u(k) (c) 6= 0, then we must have c ∈ {a, b}. Obtain the bound
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Z b iλ u(t) dt ≤ C(k)(λ δ )−1/k + δ . a e (c) Choose a suitable δ to optimize and deduce the validity of the statement for k + 1 with C(k + 1) = 2C(k) + 2, hence C(k) = 3 · 2k−1 + 2k − 2, since C(1) = 3. Solution. Let u ∈ C k+1 be a function such that u(k+1) ≥ 1 on [a, b]. We will show that Z b eiλ u(t) dt ≤ C(k + 1)λ −1/k . a
(a) Suppose u(k) (c) = 0 for some c ∈ (a, b). For any δ > 0, we have two possibilities u(k) (x) = u(k) (x) − u(k) (c) = u(k+1) (ξ )(x − c) > δ ,
x > c+δ
or −u(k) (x) = u(k) (c) − u(k) (x) = u(k+1) (ξ )(c − x) > δ , By apply the induction hypothesis, we have Z b Z eiλ u(t) dt ≤
a
a
c−δ
e
iλ u(t)
x < c−δ.
Z b Z c+δ iλ u(t) iλ u(t) e dt e dt + dt + c+δ c−δ
≤C(k)(λ δ )−1/k + 2δ +C(k)(λ δ )−1/k =2C(k)(λ δ )−1/k + 2δ . (b) If u(k) (a) ≥ 0, then for δ > 0, we have u(k) (x) = u(k) (x) − u(k) (a) + u(k) (a) = u(k) (a) + u(k+1) (ξ )(x − a) > δ , Therefore
x > a+δ.
Z b Z a+δ Z b iλ u(t) iλ u(t) eiλ u(t) dt ≤ e dt ≤ δ +C(k)(λ δ )−1/k . e dt + a a a+δ
(c) If u(k) (b) ≤ 0, then for δ > 0, we have −u(k) (x) = u(k) (b) − u(k) (x) − u(k) (b) = −u(k) (b) + u(k+1) (ξ )(b − x) > δ , Therefore
x < b−δ.
Z b Z b−δ Z b iλ u(t) iλ u(t) eiλ u(t) dt ≤ e dt + e dt ≤ C(k)(λ δ )−1/k + δ . a a b−δ
Thus, in all cases, we always have Z b eiλ u(t) dt ≤ 2C(k)(λ δ )−1/k + 2δ . a 1
Now choose δ = λ − k+1 , we have Z b 1 1 eiλ u(t) dt ≤ (2C(k) + 2)λ − k+1 = C(k + 1)λ − k+1 . a Where C(k + 1) = 2C(k) + 2. Notice that since C(1) = 3 we have C(k) = 3 · 2k−1 + 2k − 2.
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Exercise 2.6.5 (a) Prove that for some constant C and all λ ∈ R and ε ∈ (0, 1) we have Z iλt dt e ≤C. ε≤t≤1 t (b) Prove that for some C0 < ∞ , all λ ∈ R, k > 0, and ε ∈ (0, 1) we have Z iλt±t k dt e ≤ C0 . ε≤t≤1 t (c) Show that there is a constant C00 such that for any 0 < ε < N < ∞, for all ξ1 , ξ2 in R, and for all integers k ≥ 2, we have Z i(ξ1 s+ξ2 sk ) ds e ≤ C00 . ε≤s≤N s Hint: Part (a): For λ  small use the inequality eiλt − 1 ≤ λt. If λ  is large, split the domains of integration into the regions t ≤ λ −1 and t ≥ λ −1 and use integration by parts in the second case. Part (b): Write k
k
ei(λt±t ) − 1 e±it − 1 eiλt = eiλt + t t t and use part (a). Part (c): When ξ1 = ξ2 = 0 it is trivial. If ξ2 = 0, ξ1 6= 0, change variables t = ξ1 s and then split the domain of integration into the sets t ≤ 1 and t ≥ 1. In the interval over the set t ≤ 1 apply part (b) and over the set t ≥ 1 use integration by parts. In the case ξ2 6= 0, change variables t = ξ2 1/k s and split the domain of integration into the sets t ≥ 1 and k −1/k t±t k ) = k! ≥ 1. t ≤ 1. When t ≤ 1 use part (b) and in the case t ≥ 1 use Corollary 2.6.8, noting that d (ξ1 ξ2dt Solution. (a) Notice that if λ  ≤ 1, then Z Z dt Z iλt dt iλt ε≤t≤1 e t ≤ ε≤t≤1 e − 1 t ≤ ε≤t≤1 λ  dt ≤ λ  (1 − ε) ≤ 1. Now, consider the case λ  > 1. Fix ε ≤ a ≤ 1. Then Z Z dt Z iλt iλt dt iλt dt ε≤t≤1 e t = ε≤t≤a e − 1 t + a≤t≤1 e t Z dt eiλt −a eiλt 1 Z eiλt iλt ≤ dt e − 1 + + + 2 iλt −1 iλt a t ε≤t≤a a≤t≤1 iλt ≤
Z
λ  dt +
ε≤t≤a
2 2 + + λ  λ a
dt 2 a≤t≤1 λ t
Z
4 ≤ λ  (a − ε) + . λ  a Now, there are some possibilities of choosing a depending on λ and ε. If λ −1 ≤ ε then choose a = ε. In this case we have λ  (a − ε) +
4 4 ≤ ≤ 4. λ  a λ  ε
Otherwise, ε < λ −1 < 1. Now choosing a = λ −1 yields λ  (a − ε) +
4 4 ≤ λ  a + = 5. λ  a λ  a
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Thus, in all cases, we still have
Z iλt dt ε≤t≤1 e t ≤ 5.
k
(b) We can rewrite eiλt±t as k k eiλt±t = eiλt e±t − 1 + eiλt . Therefore, we can estimate
Z 1 Z 1 ±t k − 1 Z 1 k dt e iλt dt iλt iλt±t e ≤ e e + ε ε t t t ε Z 1 Z 1 iλt dt dt e ≤ + ε 1−k t ε t 1 ≤ + 5. k Notice that we used the estimate in part (a) for the last term. (c) The trivial case ξ1 = ξ2 = 0 is omitted. Now assume ξ1  + ξ2  6= 0. If ξ2 = 0, then by the part (a), Z 1 Z N i(ξ1 s+ξ2 sk ) ds eiNξ1 s ds ≤ C. = e ε ε s s N Now if ξ2 6= 0, then we have variable and use the part (b) to have Z N Z Nξ 1/k 2 −1/k k ) ds k ) ds ξ  i(ξ s+ξ s i(ξ s±s 1 2 1 2 = e e . ε s εξ2 1/k s We may suppose ε1 = ε ξ2 1/k < 1 because otherwise, we just consider the integral for s > ε1 ≥ 1. Set N1 = N ξ2 1/k . Then we split the above integral Z N Z Nξ2 1/k Z 1 1 i(ξ ξ −1/k s±sk ) ds i(ξ1 ξ2 −1/k s±sk ) ds 1 2 ei(ξ1 ξ2 −1/k s±sk ) ds + . e ≤ e εξ2 1/k s ε1 s 1 s From the part (b) we have Z 1 ei(ξ1 ξ2 −1/k s±sk ) ds ≤ C0 . ε s 1
−1/k
Denote ϕ(s) = ±ξ1 ξ2 
1 s + sk and ψ(s) = that satisfy all conditions of Proposition 2.6.7. Therefore s Z N 1 i±(±ξ ξ −1/k s+sk ) ds 1 2 ≤ C00 . 1 e s
Exercise 2.6.6 (a) Show that for all a > 0 and λ > 0 the following is valid: Z aλ iλ logt e dt ≤ a . 0
(b) Prove that there is a constant c > 0 such that for all b > λ > 10 we have
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Z b c iλt logt dt ≤ . 0 e λ log λ Hint: Part (b): Consider the intervals (0, δ ) and [δ , b) for some δ . Apply Proposition 2.6.7 with k = 1 on one of these intervals and with k = 2 on the other. Then choose a suitable δ . Solution. (a) By changing variables, we have Z Z iλ log a Z iλ logt iλ logt iλ logt e dt . e dt = a dt = a e 0 10. Note that δ < λ < b. Hence the splitting makes sense. Both quantities (λ /δ )−2 and (λ log δ )−1 are comparable to (λ log λ )−1 , since log λ − log log λ is comparable to log λ . Thus the claimed conclusion follows.
Exercise 2.6.7 Show that there is a constant C < ∞ such that for all nonintegers γ > 1 and all λ , b > 1 we have Z b C iλt γ 0 e dt ≤ λ γ .
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Hint: On the interval (0, δ ) apply Proposition 2.6.7 with k = [γ] + 1 and on the interval (δ , b) with k = [γ]. Then optimize by choosing δ = λ −1/γ . Solution. For δ > 0 we split the integral into two parts Z Z b eiλt γ dt ≤ 0
δ
e
iλt γ
0
Z b iλt γ dt + e dt . δ
On (0, δ ), the function u(t) = t γ satisfies u(k) (t) ≥ δ γ−k ,
k = [γ] + 1.
On the other hand, for t ∈ (δ , b) u(l) (t) ≥ δ γ−l ,
l = [γ].
Applying Proposition 2.6.7 we obtain Z b 1 1 eiλt γ dt ≤C(γ) λ δ γ−[γ]−1 − [γ]+1 + λ δ γ−[γ] − [γ] 0 − 1 − 1 =C(γ)δ λ δ γ [γ]+1 + λ δ γ [γ] . 1
Now choosing δ = λ − γ yields Z b 1 eiλt γ dt ≤ C(γ)λ − γ . 0
Chapter 3. Fourier Analysis on the Torus
Section 3.1. Fourier Coefficients Exercise 3.1.1 Identify T1 with [−1/2, 1/2] and let h(t) be an integrable function on T1 . (a) If h(t) ≥ 0 is even, show that b h(m) is real and b h(m) ≤ b h(0) for all m ∈ Z. (b) If h(t) is odd and h(t) ≥ 0 on [0, 1/2), then i b h(m) is real and b h(m) ≤ i m b h(1) for all m ∈ Z. Solution. (a) We have Z 1/2
b h(m) = since h is even. Then b h(m) ≤
Z 1/2
−1/2
h(x) dx =
−1/2
h(x)e−2πimx dx =
Z 1/2
h(x) cos(2πmx) dx −1/2
Z 1/2 −1/2
h(x) dx = b h(0), since h ≥ 0.
(b) If h is odd and h(t) ≥ 0 on [0, 1/2), then Z 1/2
b h(m) =
Z 1/2
h(x)e−2πimx dx = −i
−1/2
h(x) sin(2πmx) dx = −2i
Z 1/2
−1/2
h(x) sin(2πmx) dx 0
thus i b h(m) is real. Then b h(m) ≤ 2
Z 1/2
h(x)  sin(2πmx)dx ≤ 2m
0
Z 1/2
h(x)  sin(2πx)dx = 2m
0
Z 1/2 0
h(1) . h(x) sin(2πx)dx = imb
Exercise 3.1.2 Suppose that h is a periodic integrable function on [−1/2, 1/2) with integral zero. Define another periodic function H on T1 by setting Z x
H(x) =
h(t) dt −1/2
b Compute H(m) in terms of b h(m) for m ∈ Z. Solution. We have
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Z 1/2 Z x
b H(m) =
h(t) dte−2πimx dx =
−1/2 −1/2
b if m 6= 0, i.e., H(m) =
Z 1/2
Z 1/2
h(t) −1/2
1 b 2πim h(m),
e−2πimx dx dt =
Z 1/2
h(t) −1/2
t
b while the same calculation gives that H(0) =−
e−πim − e−2πimt dt = −2πim
Z 1/2
h(t) −1/2
e2πimt dt −2πim
Z 1/2
th(t) dt. −1/2
Exercise 3.1.3 Suppose that {gε }ε>0 is an approximate identity on Rn as ε → 0 and let Gε (x) =
∑ gε (x + `) .
`∈Zn
Show that the family {Gε }ε>0 is an approximate identity on Tn . Solution. R R We have that kGε kL1 (Tn ) = kgε kL1 (Rn ) ≤ C and Tn Gε dx = Rn gε dx = 1 for all ε > 0. Thus properties (i) and (ii) of the definition hold. Let δ > 0. Then Z
G (x) dx ≤ x∈Tn ε
x≥δ
Z
∑
x∈Tn x≥δ `∈Zn
gε (x + `) dx ≤
Z
gε (y) dy y∈Rn y≥δ `∈Zn
∑
→0
as ε → 0. Thus property (iii) holds.
Exercise 3.1.4 On T1 define the de la Vall´ee Poussin kernel VN (x) = 2F2N+1 (x) − FN (x) . (a) Show that the sequence VN is an approximate identity. c c (b) Prove that V N (m) = 1 when m ≤ N + 1, and V N (m) = 0 when m ≥ 2N + 2. Solution. (a) Properties (i) and (iii) of approximate identities follow from the corresponding properties for the Fej´er kernel FN . Since Z T1
FN (x) dx = 1 for all N, it follows that this is also true for VN , hence property (ii) of approximate identities follows.
 j c (b) We know that F N ( j) = 1 − N+1 when  j ≤ N + 1 and zero otherwise. Therefore
 j  j 2 1 − 2N+2 − 1 − N+1 = 1  j c V N ( j) = 2 1 − 2N+2 0
when  j ≤ N + 1, when 2N + 1 ≥  j > N + 1, when  j > 2N + 1.
Exercise 3.1.5 (a) Show that for all t ≤
π 2
we have
1 1 2 − ≤ 1− . sin(t) t π
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(b) Let DN be the Dirichlet kernel on T1 . Prove that for N ∈ Z+ we have N
4 N 1
DN 1 ≤ 3 − 2 + 4 ∑ 1 . ≤ ∑ L π 2 k=1 k π π 2 k=1 k
Conclude that the numbers kDN kL1 grow logarithmically as N → ∞ and therefore the family {DN }∞ N=1 is not an approximate 1 . The numbers kD k , N = 1, 2, . . . are called the Lebesgue constants identity on T 1 N L 1 − 1t is nonnegative on (0, π2 ], equivalently prove that tan(t) sin(t) ≥ t 2 on Hint: Part (a): Show that the derivative of sin(t) p 1 1 (0, π2 ]; this is a consequence of the inequality sin(t) tan(t) ≥ 2( sin(t) + tan(t) )−1 = 2 tan( 2t ) ≥ t. Part (b): Replace DN (t) by sin((2N+1)πt) and estimate the difference using part (a). πt Solution. (a) The derivative of
1 sin(t)
− 1t is equal to p
(b) Now consider kDN kL1 = 2
R
1 2
0
1 t2
cos(t) π 2 − sin 2 (t) . This is nonnegative if tan(t) sin(t) ≥ t on (0, 2 ] which is true since
t 2 2 sin(t) = = 2 tan ≥t. 1 1 1 + cos(t) 2 sin(t) + tan(t)
sin(t) tan(t) ≥
dx. We will use throughout that for 0 ≤ k ≤ N − 1  sin((2N+1)πx) sin πx k+1 2N+1
Z
 sin((2N + 1)πx) dx =
2 1 π 2N + 1
 sin((2N + 1)πx) dx =
1 1 π 2N + 1
k 2N+1
while Z
1 2 N 2N+1
which follow by a change of variables. Then we have k+1 Z 1 Z 1 N−1 Z 2N+1 sin((2N + 1)πx) 2 sin(2N + 1)πx 2 1 sin(2N + 1)πx kDN kL1 = dx = ∑ dx + N dx. k 2 sin(πx) sin(πx) sin(πx) 0 k=0 2N+1 2N+1 On [0, 1/2] we have
1 sin(πx)
≥
1 πx ,
thus
N−1 1 2N + 1 kDN kL1 ≥ ∑ 2 k=0 π(k + 1)
Z
k+1 2N+1 k 2N+1
 sin(2N + 1)πx dx +
2 π
Z
1 2 N 2N+1
 sin(2N + 1)πx dx
N−1
2 2 1 1 + π (2N + 1)π k=0 π(k + 1) π N 2 1 1 = 2 ∑ + . π k=1 k 2N + 1 =
∑
Therefore kDN kL1
4 ≥ 2 π
N
1 1 ∑ + 2N + 1 . k=1 k
Also 1 2
sin(2N + 1)πx sin(2N + 1)πx sin(2N + 1)πx dx − + sin(πx) πx πx 0 1 k+1 Z Z Z N−1 2N+1 sin(2N + 1)πx 1 2 dx + ∑ 2N+1 sin(2N + 1)πx dx + ≤ 1− + k 2 π πx πx 0
1 kDN kL1 ≤ 2
Z
k=1
2N+1
1 2
sin(2N + 1)πx dx N πx 2N+1
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≤
2 2N + 1 N−1 2 1 1 1 1 1 1 1− + + +∑ . N k 2 π 2N + 1 k=1 π 2N + 1 π 2N+1 π 2N + 1 π 2N+1
So kDN kL1
4 2 ≤ 3− + 2 π π
N
1 1 ∑ − 2N . k=1 k
Exercise 3.1.6 Let DN be the Dirichlet kernel on T1 . Prove that for all 1 < p < ∞ there exist two constants C p , c p > 0 such that
0 0 c p (2N + 1)1/p ≤ DN L p ≤ C p (2N + 1)1/p . Hint: Consider the two closest zeros of DN near the origin and split the integral into the intervals thus obtained. Solution. One can easily see that DN (x) has N roots in the interval (0, 1/2), the smallest one being x = (2N + 1)−1 . The maximum of DN attained at zero is 2N + 1 and we therefore have DN (x) ≤ 2N + 1 while the estimate DN (x) ≤ π/(2x) is always true. We have Z 1 Z 1 2N+1 2 πp kDN kLp p ≤ 2 dx ≤ C p (2N + 1) p−1 . (2N + 1) p dx + 2 1 p (2x) 0 2N+1 The estimate from below follows by observing that when x ≤ 1/2(2N + 1) we have 2(2N + 1)x πx
DN (x) ≥ by Appendix E and thus kDN kLp p ≥ 2
Z
1 2(2N+1)
0
(2(2N + 1)/π) p dx ≥ c p (2N + 1) p−1 .
Exercise 3.1.7 The Poisson kernel on Tn is the function Pr1 ,...,rn (x) =
m 
m 
∑ n r1 1 · · · rn n e2πim·x
m∈Z
and is defined for 0 < r1 , . . . , rn < 1. Prove that Pr1 ,...,rn can be written as 1 + r j e2πix j Pr1 ,...,rn (x1 , . . . , xn ) = ∏ Re 1 − r j e2πix j j=1 n
n
=∏
j=1
1 − r2j 1 − 2r j cos(2πx j ) + r2j
,
and conclude that Pr,...,r (x) is an approximate identity as r ↑ 1. Solution. For r1 , . . . , rn < 1 the geometric series that defines Pr1 ,...,rn converges absolutely and uniformly for x ∈ Tn . Because of the multiplicative structure of the kernel Pr1 ,...,rn , it suffices to show the required identity for n = 1. First observe that
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1 + re2πix (1 + re2πix )(1 − re−2πix ) 1 − r2 + 2i sin(2πx) = = , 1 − re2πix 1 − re2πix 2 1 − 2r cos(2πx) + r2 which implies the second identity by taking real parts. Now ∞ ∞ 1 + re2πix 2πix j 2πi jx = (1 + re ) r e = 1 + 2 r j e2πi jx ∑ ∑ 1 − re2πix j=0 j=1
and the first identity also follows by taking real parts. Let us now prove that the Poisson kernel Pr gives an approximate identity as r ↑ 1. The last identity above gives that Pr is integrable and that its integral is equal to 1. Since Pr is positive, this proves properties (i) and (ii) of the definition of approximate identities. To prove property (iii) of approximate identities, let us fix a δ > 0. If x ≤ 1/2, then 1 − e2πix  ≥ 2 · 2πx/π by Appendix E. Hence 1 − e2πix  ≥ 4δ if 1/2 ≥ x ≥ δ . It follows that 1 − re2πix  ≥ r − re2πix  − (1 − r) ≥ 4δ r − (1 − r) and thus Z 1/2≥x≥δ
Pr (x) dx ≤ (1 − 2δ )
1 − r2 . (4δ r − (1 − r))2
Now this last expression tends to zero as r ↑ 1 (take r > (4δ + 1)−1 ). We now extend this result to Tn which we identify with [−1/2, 1/2)n . Since Pr,...,r (x1 , . . . , xn ) = Pr (x1 ) . . . Pr (xn ), properties (i) and (ii) of approximate identities follow by multiplying each coordinate separately. To prove property (iii), let Vδ = {x ∈ Tn : x j  < δ , 1 ≤ j ≤ n} be the cubic neighborhood of 0. For x = (x1 , . . . , xn ) ∈ / Vδ we have that x j  ≥ δ for some 1 ≤ j ≤ n. Therefore Z Z Z n Pr (x j ) dx j Pr,...r (x1 , . . . , xn ) dx ≤ ∑ ∏ Pr (xk ) dxk Vδc
j=1
1 k6= j T
≤ n (1 − 2δ )
1/2≥x j ≥δ
1 − r2 (4δ r − (1 − r))2
by the previous argument, and the last expression above tends to 0 as r ↑ 1.
Section 3.2. Reproduction of Functions from Their Fourier Coefficients Exercise 3.2.1 On T1 let P be a trigonometric polynomial of degree N > 0. Show that P has at most 2N zeros. Construct a trigonometric polynomial with exactly 2N zeros. Solution. Given a trigonometric polynomial P(x) = ∑Nj=−N a j e2πi jx consider the meromorphic function Q(z) = ∑Nj=−N a j z j on C. Then j Q(z) = z−N R(z) where R(z) = ∑2N j=0 a j−N z is a polynomial with at most 2N zeros on the plane. Therefore Q(z) has at most 2N zeros on the plane and so does P(x) on the circle. The trigonometric polynomial e2πiNx + e−2πiNx = 2 cos(2πx) has exactly 2N zeros on the circle, namely xk = (2k + 1)/4N, where k ∈ {0, 1, . . . , 2N − 1}.
Exercise 3.2.2 0
(Hausdorff–Young inequality) Prove that when f ∈ L p (Tn ), 1 ≤ p ≤ 2, the sequence of Fourier coefficients of f is in ` p (Zn ) and 0 0 1/p ≤ f p n . ∑  fb(m) p m∈Zn
Also observe that 1 is the best constant in the preceding inequality.
L (T )
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Solution. This is a consequence of the RieszThorin interpolation theorem. Interpolate between the estimates k{ fb(k)}k k`∞ ≤ k f kL1 and k{ fb(k)}k k`2 ≤ k f kL2 to obtain for 1 < p < 2 the estimate k{ fb(k)}k k` p0 ≤ k f kL p for all f simple functions. Use density to extend this to all functions. To see that the norm of the operator f → { fb(k)}k∈Zn is 1, take f (x) = 1, in which case b 1(0) = 1 and b 1(m) = 0 if m 6= 0.
Exercise 3.2.3 Use without proof that there exists a constant C > 0 such that for all t ∈ R we have N √ ∑ eik log k eikt ≤ C N,
N = 2, 3, 4, . . . ,
k=2
to construct an example of a continuous function g on T1 with
∑ bg(m)q = ∞
m∈Z
for all q < 2. Thus the Hausdorff–Young inequality of Exercise 3.2.2 fails for p > 2. eik log k 2πikx . For a proof of the previous estimate, see Zygmund [388, Theorem (4.7) p. 199]. Hint: Consider g(x) = ∑∞ k=2 k1/2 (log k)2 e Solution. Let
N
gN (x) =
eik log k ∑ k1/2 (log k)2 e2πikx k=2
for x ∈ [0, 1]. Set bk = k−1/2 (log k)−2 for k ≥ 2 and b1 = 0. Also set ak = eik(log k+2πx) for k ≥ 1. Summation by parts gives N
N−1
∑ ak bk = AN bN − ∑ Ak (bk+1 − bk ), k=1
k=1
where Ak = ∑kj=1 a j . Since bk − bk+1  =  f 0 (x0 ) ≤ ck−3/2 (log k)−2 (where f (x) = x−1/2 (log x)−2 and k ≤ x0 ≤ k + 1) and Ak  ≤ Ck1/2 , the Weirstrass Mtest gives that the sequence of functions gN converges uniformly to a continuous function g on [0, 1]. Because of the uniform convergence, one can interchange the integration with the limit below Z
lim
N→∞ T1
gN (x)e−2πix·ξ dx =
to obtain that the Fourier coefficients gb(k) of g are have ∑k∈Z b g(k)q = ∞ for all q < 2.
Z
lim gN (x)e−2πix·ξ dx,
T1 N→∞
eik log k when k ≥ 2 and zero otherwise. For this sequence we certainly k1/2 (log k)2
Exercise 3.2.4 (S. a trigonometric polynomial of degree N on T1 . Prove that kP0 kL∞ ≤ 4πNkPkL∞ . Bernstein) Let P(x) be Hint: Prove first that P0 (x)/2πiN is equal to (e−2πiN(·) P)∗ FN−1 (x) e2πiNx − (e2πiN(·) P)∗ FN−1 (x) e−2πiNx and then take L∞ norms.
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Solution. 2πixk if P(x) = N 2πikx . b b Note that P0 (x) = ∑Nk=−N P(k)2πike ∑k=−N P(k)e We will check that P0 (x) = ((e−2πiN(·) P) ∗ FN−1 )(x)e2πiNx − ((e2πiN(·) P) ∗ FN−1 )(x)e−2πiNx . 2πiN We have that N−1
k 2πik(x−y) −2πiNy 1 − e e P(y)dy N x≤ 21 k=−N+1 Z N−1 k 2πikx =e2πiNx ∑ 1− e e−2πi(N+k)y P(y)dy 1 N x≤ k=−N+1 2 N−1 k b = ∑ 1− P(N + k)e2πi(k+N)x N k=−N+1 2N−1 k − N b = ∑ 1− P(k)e2πikx N k=1 N 2N−1 N −kb k−N b = ∑ 1− P(k)e2πikx + ∑ 1 − P(k)e2πikx N N k=1 k=N+1
((e−2πiN(·) P) ∗ FN−1 )(x)e2πiNx =e2πiNx
Z
∑
N
kb P(k)e2πikx + 0 , k=1 N
=∑
b since P has degree N and P(m) = 0 if m ≥ N + 1. k b 2πiN(·) 2πikx . Similarly we have ((e P) ∗ FN−1 )(x)e−2πiNx = − ∑−1 k=−N N P(k)e Therefore our claim is true. Consequently, kP0 (x)kL∞ ≤2πNk((e−2πiN(·) P) ∗ FN−1 )(x)e2πiNx kL∞ k((e2πiN(·) P) ∗ FN−1 )(x)e−2πiNx kL∞ ≤2πN(kPkL∞ kFN−1 kL1 + kPkL∞ kFN−1 kL1 ) =4πNkPkL∞
Exercise 3.2.5 (Fej´er and F. Riesz) Let P(ξ ) = ∑Nk=−N ak e2πikξ be a trigonometric polynomial on T1 of degree N such that P(ξ ) > 0 for all ξ . Prove that there exists a trigonometric polynomial Q(ξ ) of the form ∑Nk=0 bk e2πikξ such that P(ξ ) = Q(ξ )2 . Hint: Since P ≥ 0 the complexvariable polynomial R(z) = ∑Nk=−N ak zk+N must satisfy R(z) = z2N R(1/z), and thus it must have N zeros inside the unit circle and the other N outside. Therefore we may write R(z) = aN ∏sk=1 (z − zk )rk (z − 1/zk )rk for some 0 < zk  < 1 and rk ≥ 1 with ∑sk=1 rk = N. Then take z = e2πiξ . Solution. Define R(z) = ∑Nk=−N ak zk+N where z ∈ C. Since P is positive we have ak = a−k and hence R(z) = z2N R(1/z). It is easy to see that the zeros of R are away from the origin and away from the unit circle. Therefore there exist 0 < zk  < 1 and rk ≥ 1 with ∑sk=1 rk = N such that s
R(z) = aN ∏ (z − zk )rk (z − 1/zk )rk , k=1
Since z − 1/w = −z(1/z − w)/w, for z = e2πiξ on the unit circle we have P(ξ ) = e−2πiNξ R(e2πiξ )
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= aN e−2πiξ
s
∏ (e2πiξ − zk )rk (e2πiξ − 1/zk )rk
k=1 s −2πiNξ
∏ (e2πiξ − zk )rk (e−2πiξ − zk )rk (−1)rk e2πirk ξ
= aN e
k=1 s
1 zk rk
s
= aN (−1)N B ∏ ∏ e2πiξ − zk 2rk k=1 k=1
2 s 2πiξ rk − zk ) , = B0 ∏ (e k=1
where B = ∏sk=1 zk −rk and B0 is a complex square root of aN (−1)N B. Finally set Q(ξ ) to be the polynomial (of degree N) inside the last absolute value above.
Exercise 3.2.6 Let g be a function on Rn that satisfies g(x) + b g(x) ≤ C(1 + x)−n−δ for some C, δ > 0 and all x ∈ Rn . Prove that λn
α
∑
m∈Zn
gb(λ m + α)e2πix·(m+ λ ) =
x+k k·α g ∑n λ e−2πi λ k∈Z
for any x, α ∈ Rn and λ > 0. Solution. We define a function f on Rn by setting
x x·α e−2πi λ f (x) = g λ
that is, g(x) = f (λ x)e2πix·α . We observe that in view of the conditions on g, the function f satisfies the hypotheses of Theorem 3.2.8 (Poisson summation formula), that is,  f (x) ≤ max(1, λ n+δ )(1 + x)−n−δ and ∑m∈Zn  fb(m) < ∞. Moreover, a simple change of variables gives Z Z x x·α fb(ξ ) = e−2πi λ e−2πix·ξ dx = λ n g g(x)e−2πix·(λ ξ +α) dx = λ n gb(λ ξ + α) . λ Rn Rn Applying Theorem 3.2.8 we obtain
∑
fb(m)e2πim·x =
m∈Zn
∑
which translates into λn
∑ n gb(λ m + α)e2πix·m = ∑n g
m∈Z 2πi x·α λ
Multiplying by e
f (x + k)
k∈Zn
k∈Z
x+k (x+k)·α e−2πi λ . λ
we obtain the claimed identity.
Exercise 3.2.7 Verify the following identity when 0 < r < 1 and x ∈ Rn Γ ( n+1 2 ) π
n+1 2
∑n
k∈Z
1 2π 1 ( 2π
log
log 1r
1 2 2 r ) + x − k
n+1 = 2
∑ n rm e2πim·x .
m∈Z
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In the special case n = 1 and x ∈ R we have 1 1 1 − r2 1 2π log r = ∑ 1 π k∈Z ( 2π 1 − 2r cos(2πx) + r2 log 1r )2 + x − k2
Hint: Use identity (3.2.4) and Exercise 3.1.7 when n = 1. Solution. Start with identity (3.2.4) Γ ( n+1 2 ) π
n+1 2
ε
∑n
n+1
2 2 k∈Z (ε + k + x ) 2
= e−2πε
=
∑ n e−2πεm e−2πim·x
m∈Z 1 2π
in which we set r < 1 and we replace x by −x. Then ε = log 1r and the claimed identity follows immediately. In the special case when the dimension n = 1, we use the identity 1 − r2
∑ rm e2πim·x = 1 − 2r cos(2πx) + r2
m∈Z
proved in Exercise 3.1.7 to obtain the claimed result.
Exercise 3.2.8 Let γ ∈ R and λ > 0. Show that cos(2πkγ) π cosh(2πλ (γ − [γ] − 12 )) = . 2 2 λ sinh(πλ ) k∈Z λ + k
∑
Hint: Use Exercise 3.2.6 when n = 1 with x = 0, α = −γλ , g(x) =
1 1 π 1+x2
and sum in m.
Solution. We start with the identity λ π
e2πikγ
∑ λ 2 + k2 = ∑ e−2πλ m−γ
k∈Z
m∈Z
that can be obtained from Exercise 3.2.6 when n = 1 with x = 0, α = −γλ , g(x) = in k, the sine term drops out. Thus we have λ π
1 1 π 1+x2 ,
gb(ξ ) = e−2πξ  . Since λ 2 + k2 is even
cos(2πkγ) = ∑ e−2πλ m−γ = ∑ e−2πλ (m−γ) + ∑ e−2πλ (γ−m) . 2 2 m>γ m≤γ m∈Z k∈Z λ + k
∑
But m > γ is equivalent to m ≥ [γ] + 1 and m ≤ γ is equivalent to m ≤ [γ] since m is an integer. We now compute 1
∑
e−2πλ (m−γ) +
m≥[γ]+1
∑ m≤[γ]
e−2πλ (γ−m) =
1
(e−2πλ )[γ]+1 2πλ γ (e−2πλ )−[γ] −2πλ γ e2πλ (γ−[γ]− 2 ) + e−2πλ (γ−[γ]− 2 ) (e )+ (e )= −2πλ −2πλ 1−e 1−e eπλ − e−πλ
where the last identity follows by multiplying numerator and denominator by eπλ . But this is exactly equal to cosh(2πλ (γ − [γ] − 12 )) sinh(πλ ) and multiplying by
π λ
yields the claimed assertion.
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Section 3.3. Decay of Fourier Coefficients Exercise 3.3.1 Given a sequence {an }∞ n=0 of positive numbers such that an → 0 as n → ∞, find a nonnegative integrable function h on [0, 1] such that Z 1 h(t)t m dt ≥ am . 0
Use this result to deduce a different proof of Lemma 3.3.2. ∞ Hint: Try h = e ∑ (sup a j − sup a j )(k + 2)χ[ k+1 ,1] . k=0 j≥k
Solution.
k+2
j≥k+1
∞
Let h(t) = e ∑ (Mk − Mk+1 )(k + 2)χ[ k+1 ,1] (t), where Mk = sup j≥k a j cm =
R1
k+2
k=0
0
h(t)t m dt. We claim that {cm } satisfies the three
properties: (1) cm ≥ am ; (2) cm ↓ 0; (3) cm+2 + cm ≥ 2cm . Let’s prove (2) first. We have Z 1 0
∞
h(t)dt = e ∑
Z 1
∞
(Mk − Mk+1 )(k + 2)dt = e ∑ (Mk − Mk+1 ) = eM0 , k+1
k=0 k+2
k=0
so h is integrable. Also, Z 1
h(t)t m dt ≤
Z 1
h(t)dt < ∞.
0
0
Since t m ≥ t m+1 for t ∈ [0, 1] we have cm ≥ cm+1 . For any ε > 0, we can find δ > 0 s.t.
R1
1−δ
h(t) dt < ε.Then we have
Z 1−δ Z 1 Z 1 Z 1−δ Z 1 m m m m ≤ h(t)t dt + h(t)t dt ≤ (1 − δ ) h(t)dt + h(t)dt ≤ eM0 (1 − δ )m + ε = 2ε, h(t)t dt 0 0 1−δ 0 1−δ provided m is larger than some m0 . (3) follows trivially from the inequality t m + t m+2 ≥ 2t m+1 . Finally we prove (1). By the definition of e and the Bernoulli’s inequality we know that k+1 m+1 1 − ( k+2 ) 1 (k + 2) ≥ m+1 e
for k ≥ m. So ∞
cm =e ∑ (Mk − Mk+1 )
Z 1 k+1 k+2
k=0 ∞
=e ∑ (Mk − Mk+1 ) k=0
m+1 1 − ( k+1 k+2 ) (k + 2) m+1
∞
≥e
∑ (Mk − Mk+1 ) k=m ∞
≥
∑ (Mk − Mk+1 ) k=m
=Mm
(k + 2)t m dt
k+1 m+1 1 − ( k+2 ) (k + 2) m+1
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≥am .
Exercise 3.3.2 Prove that given a positive sequence {dm }m∈Zn with dm → 0 as m → ∞, there exists a positive sequence {a j } j∈Z with am1 · · · amn ≥ d(m1 ,...,mn ) and a j → 0 as  j → ∞. Solution. 1/n For j ∈ Z define a j = supk≥ j d(k1 ,...,kn ) for j ≥ 0 and a j = a− j for j < 0. Then a j is decreasing for j > 0 and tends to zero as  j → ∞. Moreover, n
a j1 · · · a jn ≥ ∏ sup d(k1 ,...,kn )
1/n
r=1 k≥ jr
≥
sup k≥max( j1 ,..., jn )
d(k1 ,...,kn ) ≥ d( j1 ,..., jn ) .
Exercise 3.3.3 (a) Use the idea of the proof of Lemma 3.3.3 to prove that if a twice continuously differentiable function f ≥ 0 is defined on (0, ∞) and satisfies f 0 (x) ≤ 0 and f 00 (x) ≥ 0 for all x > 0, then limx→∞ x f 0 (x) = 0. R (b) Suppose that a continuously differentiable function g is defined on (0, ∞) and satisfies g ≥ 0, g0 ≤ 0, and 1∞ g(x) dx < +∞. Prove that lim xg(x) = 0. x→∞
Solution. (a) The function f is decreasing and bounded below by zero, thus it has a limit as x → ∞. Then ( f (x) − f (x/2)) → 0 as x → ∞. We have Z x −( f (x) − f (x/2)) = − f 0 (t) dt ≥ − f 0 (x)x/2 ≥ 0 , x/2
since
− f 0 (t)
is decreasing and nonnegative. Now let x → ∞ and use the squeeze law to deduce that lim f 0 (x)x = 0. x→∞
Z x
(b) Set f (x) =
g(t) dt. Then f ≥ 0, f 0 ≥ 0, and f 00 = g0 ≤ 0. Also f has limit at infinity the number
1
R∞ 1
g(t) dt. Thus part (a)
gives the required conclusion. Z ∞
Here is an alternative solution. Given ε > 0 find x1 such that Z ∞
integrating from x1 to ∞ we would obtain g(x0 ) ≤
ε . Now for x > x0 we have 2x0
x1
g(t) dt < x1
ε ε . If for all t > x1 we had g(t) ≥ , then 2 2t
g(x) dx = +∞, which is a contradiction. Therefore for some x0 > x1 we have
Z x
x f (x) = x0 g(x0 ) +
x0
ε ε g(t) + tg0 (t) dt ≤ + = ε 2 2
since the second term in the integral above is negative. This proves part (b).
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Exercise 3.3.4 .
.
Prove that for 0 < γ < δ < 1 we have k f kΛ. ≤ Cn,γ,δ k f kΛ. for all functions f and thus Λδ is a subspace of Λγ . γ
δ
Solution.
 f (x + h) − f (x) In the definition k f kΛ. γ = sup we may identify Tn with [−1/2, 1/2]n . Given h outside the ball of radius n hγ x,h∈T 1√ 1√ 0 n 0 n 0 2 n, there is an h in [−1/2, 1/2] such that h − h ∈ Z . Then h  ≤ 2 n ≤ h and we have sup sup
x∈Tn h≥ 1 √n 2
 f (x + h) − f (x)  f (x + h0 ) − f (x)  f (x + h0 ) − f (x) = sup sup ≤ sup sup . γ γ h h h0 γ x∈Tn h≥ 1 √n x∈Tn 0 1 ≥ 2 k ≤ N and k ≥ N + 1 in the sum. Solution. Let N be the integer such that 2N h > 1 ≥ 2N−1 h, for a given h. So we have  f (x + h) − f (x) ∞
k
k
= ∑ 2−αk e2πi2 x (e2πi2 h − 1) k=0 N
∞
k
≤ ∑ 2−αk e2πi2 h − 1 + k=0 ∞
∑
k
∑
2−αk e−2πi2 h − 1 .
k=N+1 k
2−αk e−2πi2 h − 1 ≤ 2
2−αN
∞
∑ 2−αk = 2 1 − 2−α
= Cα 2−αN < Cα hα .
N+1
k=N+1
Since α < 1, we have 1 − α > 0 thus, N
N
k
∑ 2−αk e2πi2 h − 1 = ∑ 2−αk 2π2k h k=0
k=0
N
=2π =2π
∑ 2(1−α)k h k=0 2(1−α)(N+1) − 1
21−α − 1
h
2π (2N+1 )1−α h −1 2π 4 1−α ≤ 1−α h 2 − 1 h =Cα0 hα . ≤
.
So f ∈ Λα .
21−α
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.
Finally, we conclude that there does not exist positive β > α such that for all f in Λα we have supm∈Z mβ  fb(m) < ∞, since for the given f we have that  fb(m) = m−α for infinitely many m (m = 2k ).
Exercise 3.3.8 Use without proof that there exists a constant C > 0 such that N √ ik log k ikt sup ∑ e e ≤ C N,
N = 2, 3, 4, . . . ,
t∈R k=2
to prove that the function eik log k 2πikx ∑ k e k=2 ∞
g(x) =
.
is in Λ1/2 (T1 ) but not in A(T1 ). Conclude that the restriction s > 1/2 in Theorem 3.3.16 is sharp. Hint: Estimate the difference g(x + h) − g(x) using the summation by parts identity in Appendix F, taking the sums of the 2πikh sequence eik log k e2πikx and the differences of the sequence e k −1 . Solution. Obviously ∞
∞
1
∑ bg(k) = ∑ k = ∞,
k=2
so g
∈ / A(T1 ).
k=2
We have the following summation by parts identity M
M−1
∑ ak bk = AM bM − AN−1 bN + ∑ Ak (bk − bk+1 ) . k=N
We will consider
k=N
∞ ik log k e g(x + h) − g(x) = ∑ (e2πik(x+h) − e2πikx ) . k k=2
Given h 6= 0, pick M ∈ Z+ such that M ≈ h−1 . We let ak = eik log k e2πikx , then
bk =
e2πikh − 1 k
M M−1 M−1 1 1 1 h 2 ∑ ak bk = ∑ Ak (bk − bk+1 ) + AM bM  ≤ ∑ CC1 √ + √ ≤ C2 M 2 h + 2M − 2 = C3 h 2 . M k k=2 k=2 k=2
Since e2πikh − 1 e2πi(k+1)h − 1 (k + 1)e2πikh − (k + 1) − ke2πi(k+1)h + k kh h bk − bk+1  = − = ≤ C1 2 = C1 . k k k(k + 1) k k Now let ak = eik log k (e2πik(x+h) − e2πikx ), bk = 1k , then
∞
∞
∞
∑ ak bk = ∑ Ak (bk − bk+1 ) ≤ ∑ 2CC1
k=M
1 since bk − bk+1  =  1k − k+1 =
1 k(k+1)
k=M
k=M
.
≤ C1 k12 . Therefore g ∈ Λ 1 (T1 ). 2
1 k
3 2
1
1
= C2 M − 2 ≤ C20 h 2 ,
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Exercise 3.3.9 Show that there exist sequences {am }m∈Zn that tend to zero as m → ∞ for which there do not exist functions f in L1 (Tn ) with fb(m) = am for all m. Hint: Suppose the contrary. Then the open mapping theorem would imply the inequality k f kL1 (Tn ) ≤ Ak fbk`∞ (Zn ) for some A > 0. To contradict it, fix a smooth nonzero function h equal to 1 on B(0, 14 ) and supported in B(0, 21 ). For b > 0 deR 2 fine gb (x) = h(x)e−π(1+ib)x and extend gb to a 1periodic function in each variable on Rn . Use that gbb (m) = Rn b h(y)(1 + π m−y2 − 1+ib −n/2 dy, and let b → ∞ in the inequality kgb kL1 (Tn ) ≤ Akgbb k`∞ (Zn ) to obtain a contradiction. ib) e Solution. 2 Consider the function e−πzx which is both analytic in z on R+ × R and Schwartz in x on Rn . When z = t is positive real, we 1 2 1 2 n n have that the Fourier transform of this function is t − 2 e−π t x . The function z− 2 e−π z x is also an analytic function on R+ × R, 1 since Re z > 0 implies Re 1z > 0 and since z 2 is a welldefined nonzero analytic function on R+ × R. The Fourier transform 1
n
2
2
of e−πzx is also an analytic function on R+ × R and coincides with z− 2 e−π z x when z lies in R+ . It follows by analytic continuation that these two functions coincide for all z with Re z > 0. Let c0 be the space of all sequences on Zn that tend to zero at infinity equipped with the L∞ norm. The Fourier transform defines an injective continuous map from L1 (Tn ) into c0 . If this map were onto, then by the open mapping theorem it would be open, and since it is invertible, its inverse would be continuous. Then there would exist a positive constant A such that for all {am }m in c0 we would have k f{am } kL1 (Tn ) ≤ Ak{am }m k`∞ (Zn ) where f{am } is the unique L1 (Tn ) whose Fourier coefficients are am . In particular, the inequality k f kL1 (Tn ) ≤ Ak fbk`∞ (Zn ) would be true for all f ∈ L1 (Tn ). We will show that this inequality is false. Fix a smooth nonzero function h equal to 1 on the 2 ball B(0, 41 ) and supported in B(0, 21 ). Define gb (x) = h(x)e−π(1+ib)x , b > 0, and extend gb to a periodic function on Rn with 2
π
2
period one in each variable. The Fourier transform of e−π(1+ib)x is (1 + ib)−n/2 e− 1+ib ξ  . The Fourier coefficients of gb are Z
gbb (m) =
Rn
π 2 b h(y)(1 + ib)−n/2 e− 1+ib m−y dy ,
obtained by restricting the Fourier transform of gb on Zn . It follows that kgbb k`∞ ≤ kb h kL1 (1 + b2 )−n/4 which tends to zero as 1 )2 1 −π( b → ∞. Since kgb kL1 ≥ B(0, 4 )e 4 > 0, this contradicts the inequality k f kL1 ≤ Ak fbk`∞ .
Section 3.4. Pointwise Convergence of Fourier Series Exercise 3.4.1 Identify T1 with [−1/2, 1/2) and fix 0 < b < 1/2. Prove the following: m
(a) The mth Fourier coefficient of the function x is i (−1) 2πm when m 6= 0 and zero when m = 0. when m 6= 0 and 2b when m = 0. (b) The mth Fourier coefficient of the function χ[−b,b] is sin(2πbm) mπ x sin2 (πbm) (c) The mth Fourier coefficient of the function 1 − b + is bm2 π 2 when m 6= 0 and b when m = 0. m
(−1) (d) The mth Fourier coefficient of the function x is − 2m12 π 2 + 2m 2 π 2 when m 6= 0 and
(e) The mth Fourier coefficient of the function (f) The mth Fourier coefficient of the function (g) The mth Fourier coefficient of the function
x2
is
(−1)m 2m2 π 2
when m 6= 0 and
m sinh π cosh(2πx) is (−1) π . 1+m2 im(−1)m sinh π sinh(2πx) is 1+m2 π .
1 12
1 4
when m = 0.
when m = 0.
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Solution. R1 (a) Let f (x) = x, then fb(0) = 2 1 xdx =
1
x2 2 2 − 1
−2
Z
fb(m) =
= 0.
2
1 2
xe−2πixm dx = 1
−2
Z
1 2
x 1
−2
de−2πimx xe−2πimx 21 =  + −2πim −2πim − 12
Z
1 2
− 12
e−2πixm (−1)m dx = i . 2πim 2πm
(b) We have Z b
e−2πixm dx =
−b
e−2πibm − e2πibm sin(2πmb) e−2πixm b −b = = . −2πim −2πim πm
(c) We have 1 2
Z b x −2πixm x −2πixm 1 − e dx = 1 − e dx b + b −b − 12 Z 0 Z b x x −2πixm = 1 + e−2πixm dx + 1− dx e b b + −b 0 Z b x = 1 − 2 cos(2πxm)dx b 0 Z b x sin(2πxm) b sin(2πxm) = 1− dx + b πm bπm 0 0 sin2 (πmb) = . bπ 2 m2
Z
(d) Z 0
−xe−2πimx dx + 1
−2
Z
1 2
xe−2πimx dx =2
1 2
Z
0
x cos(2πmx)dx
0 1 2
d sin(2πmx) πm 0 cos(2πmx) 12 = 2π 2 m2 0 (−1)m − 1 = 2π 2 m2 Z
=
x
(e) If m 6= 0, then Z
1 2
x2 e−2πimx dx = 1
−2
Z 1 −2πixm 2 e e−2πixm 2 12 x 1+ 1 xdx −2πim − 2 − 2 πim
1 2π 2 m2 cos πm = 2 2 2π m (−1)m = 2 2 2π m =
For m = 0, the integral is equal to (f) Z
1 2
− 12
Z
1 2
− 12
xde−2πimx
1 12 .
e2πx + e−2πx −2πimx 1 e dx = 2 2 =
Z
1 2
e−2π(im−1)x dx + 1
−2 (−1)m sinh π
−2π(im − 1)
+
1 2
Z
1 2
− 12
(−1)m sinh π 2π(im + 1)
e−2π(im+1)x dx
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137
=
(−1)m sinh π . π(1 + m2 )
(g) The calculation is similar to (f). Indeed, Z
1 2
− 12
1 e2πx − e−2πx −2πimx e dx = 2 2 =
Z
1 2
−2π(im−1)x
e
− 12 (−1)m sinh π
−
1 dx − 2
Z
1 2
− 12
e−2π(im+1)x dx
(−1)m sinh π 2π(im + 1)
−2π(im − 1) (−1)m im sinh π = . π(1 + m2 )
Exercise 3.4.2 Use Exercise 3.4.1 and Proposition 3.4.2 to prove that 1
∑ (2k + 1)2
k∈Z
=
π2 , 4
π2 (−1)k+1 = , k2 6 k∈Z\{0}
∑
Solution. (i) Since x =
1 π2 = , 2 k 3 k∈Z\{0}
∑
(−1)k
2π
∑ k2 + 1 = eπ − e−π .
k∈Z
1 −1 + (−1)m 2πimx 1 −2 + ∑ e = +∑ e2πimx . Let x = 0, then 4 m∈Z\{0} 2π 2 m2 4 k∈Z 2(2k + 1)2 π 2 1 1 1 π2 ⇒∑ = =∑ . 2 2 2 4 k∈Z (2k + 1) π 4 k∈Z (2k + 1)
(ii) x2 =
1 (−1)m 2πimx 1 (−1)m + ∑ e , so + = 0. Then we can have ∑ 12 m∈Z\{0} 2π 2 m2 12 m∈Z\{0} 2π 2 m2 1 −1 1 +∑ +∑ = 0, 2 2 2 2 12 k 2(2k + 1) π k6=0 2(2k) π
therefore (iii)
1 π2 = . k2 3 k∈Z\{0}
∑
1 (−1)m + ∑ = 02 , so 12 m∈Z\{0} 2π 2 m2 (−1)k+1 π2 = . k2 6 k∈Z\{0}
∑
(iv) We have (−1)k sinh π (−1)k 2π = cosh(0) = 1 ⇒ ∑ k2 + 1 π ∑ 1 + k2 = eπ − e−π . k∈Z k∈Z
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Exercise 3.4.3 Let M > N be given positive integers. (a) For f ∈ L1 (T1 ), prove the following identity: (DN ∗ f )(x) =
N +1 M+1 (FM ∗ f )(x) − (FN ∗ f )(x) M−N M−N M+1  j − fb( j)e2πi jx . 1− M − N N 0 there exists an a > 1 and a k0 > 0 such that for all k ≥ k0 we have ∑  fb(m) < ε. k 0 there exists δ > 0 such that τ < δ implies
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f (· − τ) − f
L1
0 such that for all M, N, and x 6= 0 we have C .  e2πiN( · ) (g ∗ DN ) ∗ DM (x) ≤ x (c) Show that there exists a constant C1 > 0 such that 1 2πirx sup sup (g ∗ DN )(x) = sup sup ∑ e ≤ C1 < ∞ . r 1 1 N>0 x∈T N>0 x∈T 1≤r≤N k
(d) Let λk = 1 + ee . Define
∞
f (x) =
1
∑ k2 e2πiλk x (g ∗ Dλk )(x)
k=1
T1
on and that its Fourier series converges at every x 6= 0, but lim and prove that f is econtinuous supM→∞ ( f ∗ DM )(0) = ∞. m Hint: Take M = e with m → ∞. The inequality in part (b) follows by a summation by parts. Solution. R (a) If m = 0, then gb(0) = 01 −2πi(x − 21 )dx = 0. For m 6= 0, Z 1
gb(m) =
0
Z 1
=
1 −2πi(x − )e−2πixm dx 2 −2πixe−2πixm dx
0
=
Z 1 x 0
=
m
de−2πimx
1 m
(b) g1 (y) = e2πiNy (g ∗ DN )(y) =
1 1 2πiy(m+N) e = ∑ e2πiym m m − N 0≤m≤2N 1≤m≤N
∑
m6=N
Therefore, if M ≥ 2N, then g1 ∗ DM = g1 = e2πiyN If M < 2N, then
1 2πiyr e . r 1≤r≤N
∑
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g1 ∗ DM = e2πiyN
1 2πiyr e . −N≤r≤M−N r
∑
r6=0
(c)
1 2πirx r e t is convergent uniformly for x and 0 ≤ t < δ for any 0 ≤ δ < 1. r 1≤r 1 and log 2 > 0.
Exercise 3.6.3 Let ak ≥ 0 for all k ∈ Z+ and 1 ≤ p < ∞. Show that there exist constants C p , c p such that for all N ∈ Z+ we have cp
N 2
∑ ak 
1
2
k=1
Z 1 N p 1p N 1 2 2πi2k x ≤ ≤ C p ∑ ak 2 , ∑ ak e dx 0
while
k=1
k=1
N k sup ∑ ak e2πi2 x =
x∈[0,1] k=1
Solution. Let
N
N
∑ ak  . k=1
k
∑ ak e2πi2 x .
g(x) =
k=1
Then obviously
∑Nk=1 ak  = ∑Nk=1 ak
= g(0) ≤ g(x) ≤ ∑Nk=1 ak , kgkL2 =
thus the last conclusion follows. We clearly have that
N
∑ ak 2
1
2
k=1
and in view of Theorem 3.6.4 all equivalence of norms follows.
Lp
norms of g are comparable with constants depending only on A = 2 and on p. Thus the
Exercise 3.6.4 Suppose that 0 < λ1 < λ2 < · · · is a lacunary sequence and let f be a bounded function on the circle that satisfies fb(m) = 0 whenever m ∈ Z \ {λ1 , λ2 , . . . }. Suppose also that sup t6=0
 f (t) − f (0) =B n and f = ∑m∈Zn fb(m)e2πimx . Thus
fb(m)e2πix·m kL p ≤
m: ∃ j m j >N
∑
m: ∃ j m j >N
c ≤ (1 + m)s
n
∑
j=1
c0 ∑ (1 + m j )s/n m j >N
1 →0 ∏∑ s/n k6= j mk ∈Z (1 + mk )
as N → ∞. ◦ The proof is the same for DnN except that in this case the summation is over the set m ≥ N and the sum c →0 (1 + m)s m>N
∑
as N → ∞.
Exercise 4.1.2 Prove that kP+ kL2 (T1 )→L2 (T1 ) = kP− kL2 (T1 )→L2 (T1 ) = kW kL2 (T1 )→L2 (T1 ) = 1 , where W ( f ) = fe is the conjugate function on the circle. Moreover, show that the mappings f 7→ W ( f ) + fb(0) and f 7→ W ( f ) − fb(0) are isometries on L2 (T1 ). Solution. 2πim·x , so b We have P+ ( f ) = ∑∞ m=1 f (m)e kP+ ( f )k2L2 =
∑  fb(m)2 ≤ k f k2L2 ⇒ kP+ kL2 →L2 ≤ 1.
m>0
Similarly, we can also show that kP− kL2 →L2 ≤ 1. Moreover, the function e2πix shows that the norm of P+ is equal to 1, while the function e−2πix shows that the norm of P− is equal to 1.
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Finally, fe(x) = −i
∑ sgn(m) fb(m)e2πim·x , thus k fekL2 ≤ k f kL2 for any f ∈ L2 . And k fekL2 = k f kL2 for f = e2πix , so W has
m∈Z
norm equal to 1. Also
fe± fb(0) 2 2 = ∑ −i sgn(k) fb(k)e2πikx ± fb(0) 2 2 = L L k6=0
∑  fb(k)2 = k f k2L2 . k∈Z
Exercise 4.1.3 Let −∞ ≤ a j < b j ≤ +∞ for 1 ≤ j ≤ n. Consider the rectangular projection operator defined on C ∞ (Tn ) by P( f )(x) =
fb(m)e2πi(m1 x1 +···+mn xn ) .
∑
a j ≤m j ≤b j
Prove that when 1 < p < ∞, P extends to a bounded operator from L p (Tn ) to itself with bounds independent of the a j , b j . Hint: Express P in terms of the Riesz projection P+ . Solution. Let N(a j ) be [a j ] if a j ≥ 0 and [a j + 1] if a j < 0. Let Fa j ( f )(x) = e2πiN(a j )x j f (x) be an operator defined on Tn ; this operator has norm 1. Then n
P( f ) =
∏ Fb j ◦ P− ◦ F−b j ◦ Fa j ◦ P+ ◦ F−a j
(f)
j=1
So kP( f )kL p ≤ K 2n k f kL p , where K = kP+ kL p →L p = kP− kL p →L p .
Exercise 4.1.4 Let Pr (t) be the Poisson kernel on T1 as defined in Exercise 3.1.7. For 0 < r < 1, define the conjugate Poisson kernel Qr (t) on the circle by +∞
Qr (t) = −i
∑
sgn (m) rm e2πimt .
m=−∞
(a) For 0 < r < 1, prove the identity Qr (t) =
2r sin(2πt) . 1 − 2r cos(2πt) + r2
(b) Prove that fe(t) = limr→1 (Qr ∗ f )(t) whenever f is smooth. Conclude that if f is realvalued, then so is fe. (c) Let f ∈ L1 (T1 ). Prove that the function z 7→ (Pr ∗ f )(t) + i(Qr ∗ f )(t) is analytic in z = re2πit on the open unit disc {z ∈ C : z < 1}. (d) Let f ∈ L1 (T1 ). Conclude that the functions z 7→ (Pr ∗ f )(t) and z 7→ (Qr ∗ f )(t) are conjugate harmonic functions of z = re2πit in the region z < 1. The term conjugate Poisson kernel stems from this property. Solution.
∞
(a) It’s easy to see that 21 Qr (t) is equal to the imaginary part of
∑ rm e2πint , which is
m=1
re2πit (1 − re−2πit ) r sin(2πt) re2πit Im = Im = . 1 − re2πit (1 − re2πit )(1 − re−2πit ) 1 − 2r cos(2πt) + r2
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(b) For f smooth we have ((rm e2πim·(·) ) ∗ f )(t) = rm fb(m)e2πimt , and (Qr ∗ f )(t) = −i
∑
sgn(m)rm fb(m)e2πimt .
m=−∞ ∞
lim (Qr ∗ f )(t) = −i
r→1
∑
sgn(m) fb(m)e2πimt = fe(t).
m=−∞
Therefore if f is realvalued, then so is fe, since each Qr is realvalued. (c) With z = re2πit we have Pr (t) + iQr (t) = 1 + 2
1+z
∑ zm = 1 − z ,
m≥1
and Pr and Qr are real and imaginary parts of this function respectively. This function is analytic for z < 1. Convolution with f produces the function Z Z 1 + e−2πis z z 7→ f (s)(Pr (t − s) + iQr (t − s)) ds = f (s) ds 1 − e−2πis z T1 T1 which is also analytic (holomorphic) in z < 1 for every f ∈ L1 (T1 ). To check analyticity, one simply passes the differentiation inside the integral via the Lebesgue dominated convergence theorem. (d) Since the function z 7→ (Pr ∗ f )(t) + i(Qr ∗ f )(t) is analytic on the open unit disc, its real and imaginary parts are conjugate harmonic functions.
Exercise 4.1.5 .
Let f be in Λα (T1 ) for some 0 < α < 1. Prove that the conjugate function fe is well defined and can be written as Z
f (x − t) cot(πt) dt
fe(x) = lim
ε→0 ε≤t≤1/2
Z
f (x − t) − f (x) cot(πt) dt.
= t≤1/2
Hint: Use part (b) of Exercise 4.1.4 and the fact that Qr has integral zero over the circle to write ( f ∗ Qr )(x) = ( f − f (x)) ∗ Qr (x), allowing use of the Lebesgue dominated convergence theorem. Solution. Since Qr has integral zero over the circle, we can write ( f ∗ Qr )(x) =
Z
f (x − t) − f (x) Qr (t) dt.
t≤1/2
Now observe that for 1/2 ≤ r < 1 we have 1 + r2 − 2r cos(2πt) = (1 − r)2 + 2r(1 − cos(2πt)) ≥ 1 − cos(2πt) ≥
22πt2 = 8t 2 . π2
Therefore Qr (t) ≤ π2 t−1 and clearly Qr (t) → cot(πt) as r ↑ 1. Since f is smooth, we have that  f (x − t) − f (x) ≤ Ct and the Lebesgue dominated convergence theorem implies that lim ( f ∗ Qr )(x) =
r→1
Z
f (x − t) − f (x) cot(πt) dt.
t≤1/2
Finally, it quite simple to see that the latter is equal to Z
lim ε→0 ε≤t≤1/2
f (x − t) cot(πt) dt.
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Exercise 4.1.6 Suppose that f is a realvalued function on T1 with  f  ≤ 1 and 0 ≤ λ < π/2. (a) Prove that Z 1 e eλ f (t) dt ≤ . cos(λ ) T1 (b) Conclude that for 0 ≤ λ < π/2 we have Z
eλ  f (t) dt ≤ e
T1
2 . cos(λ )
Hint: Part (a): Consider the analytic function F(z) on the disk z < 1 defined by F(z) = −i(Pr ∗ f )(θ ) + (Qr ∗ f )(θ ), where z = re2πiθ . Then Re eλ F(z) is harmonic and its average over the circle z = r is equal to its value at the origin, which is e cos(λ f (0)) ≤ 1. Let r ↑ 1 and use that for z = e2πit on the circle we have Re eλ F(z) ≥ eλ f (t) cos(λ ). Solution. (a) Let F(z) = −i((Pr ∗ f )(θ ) + (Qr ∗ f )(θ )), where z = re2πiθ , so F(z) is analytic for z < 1, then consider the function eλ F(z) , which is harmonic on the open unit disc. Then Z Z R Re eλ F(z) dt = Re eλ F(z) = Re(e−iλ T1 f (t)dt ) = cos λ f (t)dt ≤ 1. T1
T1
Meanwhile Re eλ F(z) = eλ (Qr ∗ f )(θ ) cos(λ (Pr ∗ f )(θ )), so
λ (Qr ∗ f )(θ ) cos(λ (P ∗ r T1 e
R
f )(θ ))dθ ≤ 1. We also have
lim cos(λ (Pr ∗ f )(θ ) = cos(λ f (θ )) ≥ cos(λ ),
r→1
since  f  ≤ 1 for 0 ≤ λ ≤ π2 . Letting r → 1 we obtain Z
1 . cos λ
eλ f (t) dt ≤ e
T1
(b) Decompose T1 as the union of A = {t ∈ T1 : fe(t) ≥ 0}, B = T1 \A, then Z
λ fe(t)
e
dt ≤
Z
and
Z
e
−λ fe(t)
dt ≤
eλ f (t) dt ≤ e
T1
A
Z
1 cos λ
e−λ f (t) dt ≤ e
T1
B
1 . cos λ
For the second inequality use the same method as in part (a) to obtain it with −F(z) in place of G(z). So Z
eλ  f (t) dt ≤
Z
eλ f (t) dt +
e
T1
Z
e
A
e−λ f (t) dt ≤ e
B
2 . cos λ
Exercise 4.1.7 Prove that for 0 < α < 1 there is a constant Cα such that
fe . 1 ≤ Cα f . Λ (T ) Λ α
α (T
1)
.
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Hint: Using Exercise 4.1.5, for h ≤ 1/10 write fe(x + h) − fe(x) as Z
f (x − t) − f (x + h) cot(π(t + h)) dt
t≤5h
−
Z
f (x − t) − f (x) cot(πt) dt
t≤5h
Z
f (x − t) − f (x) cot(π(t + h)) − cot(πt) dt
+ 5h≤t≤1/2
Z + f (x) − f (x + h)
cot(π(t + h)) dt .
5h≤t≤1/2
You may use the fact that cot(πt) =
1 πt
+ b(t), where b(t) is a bounded function when t ≤ 1/2. The case h ≥ 1/10 is easy.
Solution. The function Z
fe(x) =
.
t≤ 21
( f (x − t) − f (x))
cos(πt) dt sin πt
is bounded for f ∈ Λα (T1 ), so we need only to consider the case where h is small enough. Let h ≤ fe(x + h) − fe(x) =
Z t≤ 12
Z
( f (x + h − t) − f (x + h)) cot(πt)dt −
Z t≤ 21
( f (x − t) − f (x + h)) cot(π(t + h))dt − 1
= Z
t≤ 21
( f (x − t) − f (x)) cot(πt)dt
( f (x − t) − f (x + h)) cot(π(t + h))dt
= t≤5h
−
Z
( f (x − t) − f (x)) cot(πt)dt
t≤5h
Z
+
5h≤t≤ 21
( f (x − t) − f (x))(cot(π(t + h)) − cot(πt))dt
+ ( f (x) − f (x + h))
Z 5h≤t≤ 12
cot(π(t + h))dt
=A1 + A2 + A3 + A4 , where we denote these four parts by A1 , A2 , A3 , A4 respectively. A1  ≤ C
t + hα dt = C1 hα t≤5h t + h
Z
A2  ≤ C
tα dt = C2 hα t≤5h t
Z
1 dt sin (t + sh) Z 1 ≤Ch dt tα 1 (t − h)2 5h≤t≤ 2
A3  ≤
Z
5h≤t≤ 21
≤C1 h
tα h
Z 5h≤t≤ 12
tα−2 dt
hα−1 +C2 1−α ≤C3 hα . =C1 h
2
then
( f (x − t) − f (x)) cot(πt)dt
Z
t≤ 2
1 10 ,
where(0 ≤ s ≤ 1)
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Assume h > 0, and by that for t ≤ 21 , cot(πt) =
1 πt
+ b(t), b(t) is bounded, we can have
Z
Z
cot(π(t + h))dt = 1
5h≤t≤ 2
1 2 +h
Z −4h
cot(πt)dt +
6h
− 12 +h
cot(πt)dt
Z −4t
=
cot(πt)dt −6t Z −4t
1 + b(t) dt −6t πt Z −4t 1 2 = log + b(t)dt π 3 −6t =
Similarly for the case h < 0, so Z 5h≤t≤ 1 cot(π(t + h)) dt ≤ C , 2
so A4  ≤ C1
hα .
Exercise 4.1.8 The beta function is defined in Appendix A.2. Derive the identity tα =
1 B(α −β , β +1)
Z t
(t − s)α−β −1 sβ ds
0
2 α and show that the function KRα (x) = ∑m≤R 1 − m e2πim·x satisfies (4.1.17). R2 2 2 2 Hint: Take t = 1 − m and change variables s = r −m in the displayed identity. R2 R2
Solution. The identity tα =
1 B(α −β , β +1)
Z t
(t − s)α−β −1 sβ ds
0
is easily derived from 1=
1 B(α −β , β +1)
Z 1
(1 − u)α−β −1 uβ du
0
by setting s = tu. We now prove (4.1.17). It suffices to prove that 1 Z m2 α 2Γ (α + 1) 1 R r2 α−β −1 r2 β + 2 m2 β ∑ 1 − R2 = Γ (α − β )Γ (β + 1) R 0 1 − R2 ∑ 1 − r2 dr R2 m≤r m≤R 2Γ (α + 1) 1 = ∑ Γ (α − β )Γ (β + 1) R m≤R
r2 1− 2 R
Z R m
α−β −1
r2 R2
β + 1 2 m2 β 1− 2 dr . r
Given m ∈ Zn with m ≤ R we need only the following computation: r2 α−β −1 r2 β + 12 m2 β r 1 − dr let u = 2 2 2 R R r R m Z 1 2 β m = m (1 − u2 )α−β −1 u2β 1 − 2 2 du2 let y = u2 R u R 2 R
Z R
1−
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m2 β m2 R2 (1 − y)α−β −1 y − 2 dy let t = y − 2 R R R2 − m2 Z 1 R2 − m2 m2 α−β −1 R2 − m2  β R2 − m2 t + t dt = 1− R2 R2 R2 R2 0 Z 1 R2 − m2 α = (1 − t)α−β −1t β dt R2 0 m2 α =B(α − β , β + 1) 1 − 2 . R Z 1
=
m2 R2
Noting that Γ (α + 1) 1 = , B(α − β , β + 1) Γ (α − β )Γ (β + 1)
the proof is complete.
Section 4.2. A. E. Divergence of Fourier Series and BochnerRiesz means Exercise 4.2.1 Using Theorem 4.2.1 construct a function F on Tn such that lim sup (DnN ∗ F)(x1 , . . . , xn ) = ∞ N→∞
for almost all (x1 , . . . , xn ) ∈ Tn . Solution. Given a function f on T1 such that lim supN→∞ (DnN ∗ f )(x1 ) = ∞ for almost all x1 ∈ L1 , define F(x1 , . . . , xn ) = f (x1 ). Then b 1 , . . . , mn ) = fb(m1 )δm =0 · · · δmn =0 and so F(m 2 (DnN ∗ F)(x1 , . . . , xn ) = (DN ∗ f )(x1 ) Then for all x2 , . . . , xn and for almost all x1 ∈ T1 , the limsup of these expressions as N → ∞ is equal to infinity.
Exercise 4.2.2 For any 0 ≤ α < ∞ and R > 0 consider the BochnerRiesz kernel KRα (x) =
2
∑
m≤R
1 − m R2
α
e2πim·x .
Use Exercise 4.1.8 to obtain that if for some x0 ∈ Tn we have lim sup KRα (x0 ) < ∞ , R→∞
then for all β > α we have β
sup KR (x0 ) < ∞ .
R>0 n−1 2 ,
Conclude that whenever 0 ≤ α ≤ the Bochner–Riesz means of order α of the function f constructed in the proof of Theorem 4.2.5, in particular the circular (spherical) Dirichlet means of this function, diverge a.e.
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Solution. We recall identity (4.1.17) BαR ( f ) =
1 2Γ (α + 1) Γ (α − β )Γ (β + 1) R
Z R
1−
0
r2 R2
α−β −1
r2 R2
α−β −1
r2 R2
β + 1
r2 R2
β + 1
2
Bβr ( f ) dr ,
which is valid for α > β . This can be written equivalently as KRα (x) =
2Γ (α + 1) 1 Γ (α − β )Γ (β + 1) R
Z R
1−
0
2
Krβ (x) dr ,
for any x ∈ Tn . In particular, for the point x0 if β > α we have r2 β −α−1 r2 α+ 12 α K (x )dr 1− 2 0 r R 0 R R2 Z 1 R r2 β −α−1 r2 α+ 12 ≤ dr 1− 2 R 0 R R2
1 β KR (x0 ) = C
Z R
Z 1
=C2
(1 − t)β −α−1t α dt
0
=C2 B(β − α, α + 1) < ∞ .
Exercise 4.2.3 (a) Show that for M, N positive integers we have FM (x) for M ≤ N, (FM ∗ DN )(x) = F (x) + M−N 2πikx for M > N. k e ∑ N (M+1)(N+1) k≤N
(b) Prove that for some constant c > 0 we have T1
Z
∑
k≤N
k e2πikx dx ≥ c N log N
as N → ∞. Hint: Part (b): Show that for x ∈ [− 12 , 12 ] we have
∑
k e2πikx = (N + 1)(DN (x) − FN (x))
k≤N
and use the result of Exercise 3.1.5. Solution. (a) By the definition of DN , we know that FM ∗ DN = FM if M ≤ N. If M > N, then FM ∗DN =
 j
 j
 j
 j
M−N
∑ (1− M + 1 )e2πi jx = ∑ (1− N + 1 )e2πi jx + ∑ ( N + 1 − M + 1 )e2πi jx = FN (x)+ (N + 1)(M + 1) ∑
 j≤N
(b) We have
 j≤N
 j≤N
k≤N
ke2πikx .
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DN (x) − FN (x) =
∑
e2πikx −
k
1
∑ (1 − N + 1 )e2πikx = N + 1 ∑
k≤N
k≤N
ke2πikx ,
k≤N
so Z

2πikx
∑
T1 k≤N
ke
dx ≥ (N + 1)
Z T1
DN (x)dx − (N + 1)
Z T1
FN (x)dx
N 1 ≥ (N + 1) ∑ dx − (N + 1) k=1 k N
≥ (N + 1) ∑
Z k+1 1
x
k=2 k
dx
= (N + 1)(log(N + 1) − log 2) ≥ CN log N .
Exercise 4.2.4 Given the integrable functions ∞
f1 (x) =
∞
∑ 2− j F222 j (x) ,
f2 (x) =
j=0
∑
j=1
1 F j (x) , j2 22
x ∈ T1 ,
show that k f1 ∗ DN kL1 → ∞ and k f2 ∗ DN kL1 → ∞ as N → ∞. 2j j Hint: Let M j = 22 or M j = 22 depending on the situation. For fixed N let jN be the least integer j such that M j > N. Then Conclude that k f1 ∗ DN kL1 and k f2 ∗ DN kL1
M j −N M j +1
≥ 12 . Split the summation indices into the sets j ≥ jN and j < jN . tend to infinity as N → ∞ using Exercise 4.2.3.
for j ≥ jN + 1 we have M j ≥ M 2jN > N 2 ≥ 2N + 1, hence
Solution. 2j We start with f1 . Let M j = 22 . For a given N, we can find a jN s.t. M jN > N ≥ M jN −1 . Now for j > jN , M j ≥ M 2jN > N 2 ≥ 2N + 1, then
M j −N M j +1
≥ 12 . Consider the L1 norm of f ∗ DN .
Z T1
 f1 ∗ DN dx = ≥
Z T1
Z T1
∞ ∑ 2− j (FM j ∗ DN )(x) dx j=0
∑
j≥ jN
Z 2− j (FM j ∗ DN ) dx −
∑
T1 j< jN
2− j FM j (x)dx
Mj − N ke2πikx ) dx − 1 ∑ (M jN + 1)(N + 1) k≤N T1 j≥ jN Z Mj − N 2πikx −j ≥ ke − 2 F (x) ∑ 2− j dx dx − 1 N ∑ ∑ (M jN + 1)(N + 1) k≤N T1 j≥ jN j≥ jN ≥
Z
≥C
log N 2 jN
→ ∞, so lim
Z
N→∞ T1
2− j (FN +
1 2− j −C1 2 j> jN
≥C2 However
∑
∑
log N −C1 2 jN
 f1 ∗ DN dx = ∞. j
The proof of f2 is similar with the only difference being that we chose M j = 22 .
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Z T1
∞ 1 ∑ 2 (FM j ∗ DN )(x) dx 1 j T j=0 Z Z 1 1 ≥ ∑ 2 (FM j ∗ DN ) dx − ∑ 2 FM j (x)dx T1 j≥ jN j T1 j< jN j Z Mj − N 1 ≥ ke2πikx dx − 1 ∑ 2 FN + ∑ 1 (M jN + 1)(N + 1) k≤N T j≥ jN j Z Mj − N 1 1 ≥ ke2πikx dx − ∑ 2 FN (x) dx − 1 ∑ 2 ∑ 1 j (M j + 1)(N + 1) T j≥ jN jN j≥ jN k≤N
 f2 ∗ DN dx =
Z
≥C
∑
j> jN
≥C2 However
log N jN
→ ∞ as N → ∞, so lim
1 1 −C1 j2 2
log N −C1 jN Z
N→∞ T1
 f2 ∗ DN dx = ∞.
Section 4.3. Multipliers, Transference, and Almost Everywhere Convergence Exercise 4.3.1 2 α n α n Let α ≥ 0. Prove that the function (1 − ξ  )+ is in M p (R ) if and only if the function (1 − ξ )+ is in M p (R ). n Hint: Use that smooth functions with compact support lie in M p (R ).
Solution. Smooth functions with compact support lie in M p (Rn ). Choose ϕ smooth with suppϕ = {x : with suppψ = {x : x ≤ 34 } and ϕ + ψ = 1 for x ≤ 1.
1 2
≤ x ≤ 32 }, and ψ smooth
(1 − ξ 2 )α+ = (1 − ξ 2 )α+ ψ(ξ ) + (1 − ξ )α+ ϕ(ξ )(1 + ξ )α Both functions (1 − ξ 2 )α+ ψ(ξ ) and ϕ(ξ )(1 + ξ )α lie in M p (Rn ) since they are both smooth functions with compact support. So if (1 − ξ )α+ ∈ M p (Rn ), then (1 − ξ 2 )α+ ∈ M p (Rn ). Conversely, if (1 − ξ 2 )α+ ∈ M p (Rn ) we write (1 − ξ )α+ = (1 − ξ )α+ ψ(ξ ) + (1 − ξ 2 )α+
ϕ(ξ ) (1 + ξ )α
ϕ(ξ ) and use that both (1 − ξ )α+ ψ(ξ ) and (1+ξ are smooth functions with compact support we deduce the other direction; here )α we used that a product of functions in M p (Rn ) is also in M p (Rn ).
Exercise 4.3.2 The purpose of this exercise is to introduce distributions on the torus. The set of test functions on the torus is C ∞ (Tn ) equipped with the following topology. Given f j , f in C ∞ (Tn ), we say that f j → f in C ∞ (Tn ) if
α
∂ f j − ∂ α f ∞ n → 0 as j → ∞, ∀ α. L (T )
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Under this notion of convergence, C ∞ (Tn ) is a topological vector space with topology induced by the family of seminorms ρα (ϕ) = supx∈Tn (∂ α f )(x), where α ranges over all multiindices. The dual space of C ∞ (Tn ) under this topology is the set of all distributions on Tn and is denoted by D 0 (Tn ). The definition implies that for u j and u in D 0 (Tn ) we have u j → u in D 0 (Tn ) if and only if
u j , f → u, f as j → ∞ for all f ∈ C ∞ (Tn ). The following operations can be defined on elements of D 0 (Tn ): differentiation (as in Definition 2.3.6), translation and reflection (as in Definition 2.3.11), convolution with a C ∞ function (as in Definition 2.3.13), multiplication by a C ∞ function (as in Definition 2.3.15), the support of a distribution (as in Definition 2.3.16). Use the same ideas as in Rn to prove the following: (a) Prove that if u ∈ D 0 (Tn ) and f ∈ C ∞ (Tn ), then ( f ∗ u)(x) = hu, τ x ( fe)i is a C ∞ function. (b) In contrast to Rn , the convolution of two distributions on Tn can be defined. For u, v ∈ D 0 (Tn ) and f ∈ C ∞ (Tn ) define
u ∗ v, f = u, f ∗ ve . Check that convolution of distributions on D 0 (Tn ) is associative, commutative, and distributive. (c) Prove the analogue of Proposition 2.3.23, i.e., that C ∞ (Tn ) is dense in D 0 (Tn ). Solution. (a) It is straightforward that ( f ∗ u)(x) = u(τ x ( fe)). Then
α Z 1
∂ f (x + he j ) − ∂ α f (x)
α
∂ α ∂ j f (x + θ he j ) − ∂ α ∂ j f (x) ∞ n dθ → 0 as h → 0
≤ − ∂ ∂ f (x) j
∞ n
L (T ,dx) h 0 L (T ,dx) and so by the definition of the convergence in D 0 (Tn ), we obtain τ (·)+he j ( fe) − τ (·) ( fe) u(τ (·)+he j ( fe)) − u(τ (·) ( fe)) =u → u(∂ j f ) . h h The same case is also valid for any other highorder derivatives, so f ∗ u ∈ C ∞ . (b) We have ^ e ∗ ve) >=< g ∗ v, f ∗ w e >=< v, ge∗ ( f ∗ w) e >=< v, (e e >=< v ∗ w, ge∗ f >=< g ∗ (v ∗ w), f >=< g, f ∗ (v < g, ( f ∗ w g∗ f)∗w ∗ w) > for all g ∈ C ∞ . This implies that ^ e ∗ ve = f ∗ (v ( f ∗ w) ∗ w). So ^ e >=< u, ( f ∗ w) e ∗ ve >=< u, f ∗ (v < (u ∗ v) ∗ w, f >=< u ∗ v, f ∗ w ∗ w) >=< u ∗ (v ∗ w), f > and this operation is associative. It’s easy to check that δ ∗ u = u, so < u ∗ v, f >=< u, f ∗ ve >=< δ , ue ∗ ( f ∗ ve) >=< δ , f ∗ ue ∗ ve >=< v ∗ u, f > and the operation is also commutative. Finally, < (u1 + u2 ) ∗ v, f >=< u1 + u2 , f ∗ ve >=< u1 , f ∗ ve > + < u2 , f ∗ ve >=< u1 ∗ v + u2 ∗ v, f > hence it is also distributive. (c) Given u ∈ D 0 (Tn ), choose an approximate identity { fk } ∈ C ∞ (Tn ). Then for any g in C ∞ (Tn ) we have g ∗ fk → g in C ∞ (Tn ) ⇒< e fk ∗ u, g >=< u, g ∗ fk >→< u, g > . This implies that e fk ∗ u ∈ C ∞ (Tn ) and e fk ∗ u → u in D 0 (Tn ).
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Exercise 4.3.3 For u ∈ D 0 (Tn ) and m ∈ Zn define the Fourier coefficient ub(m) by
ub(m) = u(e−2πim·( · ) ) = u, e−2πim·( · ) . Prove properties (1), (2), (4), (5), (6), (8), (9), (11), and (12) of Proposition 2.3.22 regarding the Fourier coefficients of distributions on the circle. Moreover, prove that for any u, v in D 0 (Tn ) we have (u ∗ v)b(m) = ub(m) vb(m). In particular, this is valid for finite Borel measures. Solution. Property (1). We have \ (u + v)(m) = (u + v)(e−2πim·(·) ) = u(e−2πim·(·) ) + v(e−2πim·(·) ) = ub(m) + vb(m). Properties (2),(4),(5),(6),(8) are trivial. Property (11). We have < f ∗ u, e−2πim·(·) >=< u, ( fe∗ e−2πim·(·) )(·) >=< u, fb(m)e−2πim·(·) >= ub(m) fb(m) . Property (12). We have < f u, e−2πim·(·) >=< u, f e−2πim·(·) >= ∑ < u, fb(k)e−2πim·(·) >= ∑ fb(k)b u(m − k). k
k
For ub ∗ vb, notice that (e−2πim·(·) ∗ ve)(x) = ve(e2πim·((·)−x) ) = b ve(−m)e−2πim·x = vb(m)e−2πim·x , so it follows that (u ∗ v)b(m) =< u ∗ v, e−2πim·(·) >=< u, e−2πim·(·) ∗ ve >=< u, vb(m)e−2πim·(·) >= vb(m) < u, e−2πim·(·) >= ub(m) vb(m) for all m ∈ Zn .
Exercise 4.3.4 Let µ be a finite Borel measure on Rn and let ν be the periodization of µ, that is, ν is a measure on Tn defined by ν(A) =
∑
µ(A + m)
m∈Zn
for all measurable subsets A of Tn . Prove that the restriction of the Fourier transform of µ on Zn coincides with the sequence of the Fourier coefficients of the measure ν. Solution. For any k ∈ Zn we have Z
b (k) = µ
Rn
e−2πik·x dµ =
Z
∑n
m∈Z
Tn
e−2πik·x dµ(x + m) =
Z Tn
e−2πik·x dν = νb(k).
Exercise 4.3.5 Let vn be the volume of the unit ball in Rn and e1 = (1, 0, . . . , 0). Prove that
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163
lim
1
Z
ε→0 vn ε n
Conclude that the function
x−e1 ≤ε
χx≤1 dx =
1 χeB(0,1) (x) = 1/2 0
1 . 2
when x < 1, when x = 1, when x > 1
is regulated. Solution. We use coordinates θ = (θ1 , θ2 , 0, . . . , 0). Obviously (by geometric considerations) we have that lim sup ε→0
1 vn ε n
Z x−e1 ≤ε
1 . 2
χx≤1 dx ≤
2
To obtain the reverse inequality we note that the spheres θ − e1 2 = ε 2 and θ 2 = 1 intersect at the plane θ1 = 1 − ε2 and 2
over the line 1 − ε ≤ θ1 ≤ 1 − ε2 the ball θ − e1 2 = ε 2 is interior to the ball θ 2 = 1. The volume of every slice of the first 2
ball over the line segment 1 − ε ≤ θ1 ≤ 1 − ε2 is vn−1 (ε 2 − θ1 − 12 ) Z 1− ε 2 2
Z
lim inf ε→0
x−e1 ≤ε
χx≤1 dx ≥ lim inf
vn ε n
1−ε
n−1 2
Thus we can write for ε < 1
vn−1 (ε 2 − θ1 − 12 )
Z ε ε2 2
= lim inf
vn−1 (ε 2 − t 2 )
Z 1 ε 2
= lim inf
vn−1 (1 − t 2 )
dt
n−1 2
dt
vn
ε→0
=
n−1 2
vn ε n
ε→0
0
dθ1
vn ε n
ε→0
Z 1
n−1 2
vn−1 (1 − t 2 )
n−1 2
dt =
vn
1 , 2
noticing that Z 1
vn = We deduce that
−1
p vn−1 ( 1 − t 2 )n−1 dt .
R
lim ε→0
x−e1 ≤ε
vn
χx≤1 dx εn
1 = , 2
hence the function χeB(0,1) is regulated at the point e1 . By a rotation, it is regulated at every other point on the unit sphere.
Exercise 4.3.6 2
Let Lε (x) = e−πεx be defined for ε > 0 and x ∈ Rn and let 1 < q < ∞. Prove that there is a constant C(n, q) < ∞ such that for any 1periodic continuous function g on Rn we have
n sup ε 2q g Lε/q Lq,1 (Rn ) ≤ C(n, q)kgkLq,1 (Tn ) .
0 0. But for each such lattice point, there is at most one set of measure E in the intersection, so the total measure of the intersection is at most r n r √ q 1 n q 1 2 0 0 dSε (χE ) (λ ) ≤ Cn n+ E . log E ≤ Cn E +Cn log πε λ πε λ \ [ ε 2 dSε (g) (λ ) = {x ∈ Rn : e−π q x > λ } (k + E) = B 0,
Inserting this estimate in the preceding integral yields
1
Sε (χE ) q,1 n ≤ C(n, q)E q L (R ) uniformly in 0 < ε < 1.
Exercise 4.3.7 Let 0 < C0 < ∞. Suppose that { ft }t∈R+ is a family of measurable functions on a measure space X that satisfies
sup  ft  p ≤ C0 L t∈F
for every finite subset F of R+ . (a) Suppose that for each x ∈ X, the function t 7→ ft (x) is continuous. Show that
sup  ft  p ≤ C0 . L t>0
(b) Prove that for any t > 0 there is a measurable function fet on X that is a.e. equal to ft such that
sup  fet  p ≤ C0 . L t∈R+
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Hint: Part (a): Notice that in view of the Lebesgue monotone convergence theorem, we have supt∈Q  ft  L p ≤ C0 . Also, for each x ∈ X we have supt∈Q  ft (x) = supt∈R+  ft (x) by continuity. Part (b): Let a = supF k supt∈F  ft kL p ≤ C0 , where the supremum is taken over all finite subsets F of R. Pick an increasing sequence of finite sets Fn such that k supt∈Fn  ft kL p → a as n → ∞. Let g = supn supt∈Fn  ft  and note that kgkL p = a. Then for any s ∈ R we have
max( fs , sup  ft ) p ≤ a . L
t∈Fk
This implies k max( fs , g)kL p ≤ a = kgkL p , so that  fs  ≤ g a.e. for all s ∈ R.
Solution.
S (a) Using the Lebesgue monotone convergence theorem, we show supt∈Q  ft  L p ≤ C0 , since we can write Q = ∞ k=1 Fk of finite sets of size k. Also, for each x ∈ X we have supt∈Q  ft (x) = supt∈R+  ft (x) by continuity; given any t ∈ R pick a sequence of positive rationals rk → t, then  frk (x) →  ft (x) and thus supt∈R+  ft (x) ≤ supr∈Q  fr (x). Thus the assertion follows. (b) Let a = sup k sup  ft kL p F
where the supremum is taken over all finite subsets F of
R+ .
t∈F
Choose Fn such that lim k sup  ft kL p = a and define n
t∈Fn
g = sup sup  ft  . n t∈Fn
Then kgkL p ≥ lim sup k sup  ft kL p = a. n
t∈Fn
Also kgkL p ≤ lim k sup sup  ft kL p ≤ a. So we have kgkL p = a. Notice that n
k≤n t∈Fn
k max( fs , sup sup  ft )kL p ≤ a 1≤k≤n t∈Fn
since {s}
Sn
k=1 Fk
is a finite set, and k max( fs , g)kL p = lim k max( fs , sup sup  ft )kL p ≤ a = kgkL p . n→∞
1≤k≤n t∈Fn
Since max( fs , g) ≥ g a.e. it follows that k max( fs , g)kL p ≥ kgkL p , hence k max( fs , g)kL p = kgkL p and consequently max( fs , g) = g a.e. It follows from this that  fs  ≤ g a.e. for any s ∈ R+ . Define ft if  ft  ≤ g e ft = g if  ft  > g Then fet = ft a.e.,  fet  ≤ g everywhere, and k sup  fet kL p = kgkL p = a. t
Exercise 4.3.8 Show that for f ∈ L2 (T2 ) we have that
∑
m1 ≤N m2 ≤N 2
fb(m1 , m2 )e2πi(m1 x1 +m2 x2 ) → f (x1 , x2 )
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for almost all (x1 , x2 ) in T2 . Hint: Use the splitting fb(m1 , m2 ) = fb(m1 , m2 )χm2 ≤m1 2 + fb(m1 , m2 )χm2 >m1 2 and apply the idea of the proof of Theorem 4.3.16. Solution. The claimed assertion is clearly valid for functions f ∈ C ∞ (T2 ). So, it will suffice to obtain the estimate sup +
Z
T2 N∈Z
2 2 2πi(m1 x1 +m2 x2 ) b f (m1 , m2 )e dx1 dx2 ≤ Ck f kL2 (T2 )
∑
m1 ≤N m2 ≤N 2
and from this standard arguments imply the a.e. convergence for general functions in L2 (T2 ). Every bounded sequence is an element of M2 (Z2 ). The bounded sequence {am1 ,m2 }m1 ,m2 defined by am1 ,m2 = 1 when m2  < m1 2 ≤ L2 and zero otherwise, lies in M2 (Z2 ) with norm independent of L in Z+ . Likewise the bounded sequence {bm1 ,m2 }m1 ,m2 defined by bm1 ,m2 = 1 when m1 2 ≤ m2  ≤ L2 and zero otherwise, lies in M2 (Z2 ) with norm independent of L in Z+ . This means that for some constant B we have the following inequality for all f in L p (T2 ): 2 T
Z
∑
∑
m2 ∈Z m1 ∈Z m2 ≤L2 m1 2 ≤m2 
2
2πi(m1 x1 +m2 x2 ) 2 2 b f (m1 , m2 )e dx2 dx1 ≤ B f L2 (T2 ) ,
where B is independent of L ∈ Z+ . Likewise, there is a constant A > 0 such that for all f in L p (T2 ): 2 T
Z
∑
∑
m1 ∈Z m2 ∈Z m1 ≤L m2  0 as follows: 0
Z ∞ Z ∞ ∞ Z ∞ sin(x/a) −x sin x −ax e dx = e dx = ∑ (−1)k 0
x
x
0
(−1)k 2k+1 k=0 a ∞
0 k=0
Z ∞
=∑
0
x 2k+1 a (2k + 1)!x
e−x dx
∞ x2k (−1)k 1 2k+1 1 π e−x dx = ∑ = arctan = − arctan(a). (2k + 1)! 2k + 1 a a 2 k=0
(c) Integrating by parts, we write ∞ Z ∞ Z ∞ Z ∞ (1 − cos x)e−ax sin x −ax (1 − cos x)e−ax 1 − cos x −ax e dx = − + e dx − a dx 2 0
x
x
x
0
0
x
0
=I(a) + aJ(a), where J(a) = −
Z ∞ (1 − cos x)e−ax
x
0
dx.
It suffices to prove that J(a) =
1 a2 log , a > 0. 2 1 + a2
Changing variable y = ax, we have J(a) = −
Z ∞ (1 − cos ay )e−y
y
0
dy = −
Z ∞ y (2 sin2 2a )e−y
y
0
Taking the absolute value, we get J(a) ≤
Z ∞ y (2 sin2 2a )e−y 0
Thus lima→∞ J(a) = 0.
y
dy ≤
1 2a2
Z ∞ 0
ye−y dy.
dy.
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Apply the same argument as above, we can show that J(a) is differentiable for a > 0 and J 0 (a) =
Z ∞
(1 − cos x)e−ax dx =
0
Hence J(a) = Since lima→∞ J(a) = 0, we imply that c = 0, i.e J(a) =
1 . a(1 + a2 )
1 a2 + c. log 2 1 + a2
1 a2 . log 2 1 + a2
Exercise 5.1.2 (a) Let ϕ be a compactly supported C m+1 function on R for some m in Z+ {0}. Prove that if ϕ (m) is the mth derivative of ϕ, then H(ϕ (m) )(x) ≤ Cm,ϕ (1 + x)−m−1 S
for some Cm,ϕ > 0. (b) Let ϕ be a compactly supported C m+1 function on Rn for some m ∈ Z+ . Show that R j (∂ α ϕ)(x) ≤ Cn,m,ϕ (1 + x)−n−m for some Cn,m,ϕ > 0 and all multiindices α with α = m. (c) Let I be an interval on the line and assume that a function h is equal to 1 on the left half of I, is equal to −1 on the right half of I, and vanishes outside I. Prove that for x ∈ / 2I we have H(h)(x) ≤ 4I2 x − center(I)−2 . Hint: Use that when t ≤
1 2
we have log(1 + t) = t + R1 (t), where R1 (t) ≤ 2t2 .
Solution. (a) Let ϕ be a C m+1 function on R for some m ≥ 0 with support containing in [−R, R] for some R > 1. To estimate the L∞ norm of its Hilbert transform, we consider two cases. First, suppose x > 2R and compute the integral H(ϕ (m) )(x) =
1 lim π ε→0
Z x−y≥ε
ϕ (m) (y) 1 dy = x−y π
ϕ (m) (y) (−1)m m! dy = π y 2R and compute the Riesz transform of ∂ α ϕ as follows Z (x j − y j )(∂ α ϕ)(y) R j (∂ α ϕ)(x) =cn lim dy ε→0 x−y≥ε x − yn+1 Z (x j − y j )(∂ α ϕ)(y) =cn dy y 0. e − e−πα S Hint: First take E = Nj=1 (a j , b j ), where b j < a j+1 . Show that the equation H(χE )(x) = α has exactly one root ρ j in each open interval (a j , b j ) for 1 ≤ j ≤ N and exactly one root r j in each interval (b j , a j+1 ) for 1 ≤ j ≤ N, (aN+1 = ∞). Then {x ∈ R : H(χE )(x) > α} = ∑Nj=1 r j − ∑Nj=1 ρ j , and this can be expressed in terms of ∑Nj=1 a j and ∑Nj=1 b j . Argue similarly for the set {x ∈ R : H(χE )(x) < −α}. For a general measurable set E, find sets En such that each En is a finite union of intervals and that χEn → χE in L2 . Then H(χEn ) → H(χE ) in measure; thus H(χEnk ) → H(χE ) a.e. for some subsequence nk . The Lebesgue dominated convergence theorem gives dH(χEn ) → dH(χE ) . See Figure 5.1. k
Solution. First consider E = ∪Nj=1 (a j , b j ), where b j < a j+1 . Then we have 1 H(χE )(x) = π
N
x−aj ∑ log x − b j .
j=1
It is easy to see that limx→a j H(χE )(x) = −∞, limx→b j H(χE )(x) = ∞ and
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H(χE )0 (x) =
1 π
N
∑
j=1
1 1 . − x−aj x−bj
It follows that H(χE )0 (x) < 0 for all x ∈ (b j , a j+1 ), aN+1 = ∞ and H(χE )0 (x) > 0 for all x ∈ (a j , b j ), since π H(χE )0 (x) =
N−1 1 1 1 1 +∑ − − . x − a1 j=1 x − b j x − a j+1 x − bN
Thus fix α > 0, the equation H(χE )(x) = α has exactly one root ρ j in each open interval (a j , b j ) for 1 ≤ j ≤ N and exactly one root r j in each interval (b j , a j+1 ) for 1 ≤ j ≤ N. Then we have N
{x ∈ R : H(χE )(x) > α} =
N
∑ r j − ∑ ρ j.
j=1
j=1
Noting that r j , 1 ≤ j ≤ N, are roots of a polynomial N
N
∏ (x − a j ) = eπα ∏ (x − b j ) j=1
j=1
and ρ j are roots of N
N
∏ (x − a j ) = −eπα ∏ (x − b j ). j=1
Hence
N
1
N
∑ r j = eπα − 1 ∑ (b j − a j )
j=1
j=1 N
and
j=1
N
1
∑ ρ j = eπα + 1 ∑ (b j − a j ).
j=1
j=1
This yields N
{x ∈ R : H(χE )(x) > α} =
N
2
∑ r j − ∑ ρ j = eπα − e−πα E .
j=1
j=1
Likewise, we have the identity {x ∈ R : H(χE )(x) < −α} =
eπα
2 E . − e−πα
Combining together, we have 4 E , α > 0. − e−πα Now suppose E is an arbitrary finite measure set of R. For each n ∈ N, there exists an open set En such that E ⊂ En and En \ E < 1/n. Since En is an open set, it can be expressed as a union of finite disjoint intervals En = ∪Nj=1 (anj , bnj ). By the above argument, we have 4 En  . dH(χEn ) (α) = πα e − e−πα It is easy to see that Z 1 χEn − χE 2 dx ≤ ; n R dH(χE ) (α) =
eπα
hence χEn → χE in L2 (R). Since the Hilbert transform maps L2 to L2 , H(χEn ) → H(χE ) in L2 . It turns out H(χEn ) → H(χE ) in measure; thus H(χEnk ) → H(χE ) a.e. for some subsequence nk . Finally, the Lebesgues convergence Theorem gives that dH(χEn ) → dH(χE ) as k → ∞. Hence k
4 Enk 4 E dH(χE ) (α) = lim πα = πα . −πα k→∞ e −e e − e−πα
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Exercise 5.1.5 Let 1 ≤ p < ∞ and let T be a linear operator defined on the space of Schwartz functions that commutes with dilations, i.e., T (δ λ f ) = δ λ T ( f ) for all f ∈ S (Rn ) and all λ > 0. (Here δ λ ( f )(x) = f (λ x).) Suppose that there exists a constant C > 0 such that for all f ∈ S (Rn ) with L p norm one we have {x : T ( f )(x) > 1} ≤ C. Prove that T admits a bounded extension from L p (Rn ) to L p,∞ (Rn ) with norm at most C1/p . Hint: Try functions of the form λ −n/p f (λ −1 x)/k f kL p with λ > 0. Solution. We know that T (δ α f )(x) = T ( f )(αx) for all α > 0 and ∀ f ∈ L p (R)
{x ∈ R : T ( f )(x) > 1} ≤ C, for some C > 0. Fix f ∈ S (R) and λ > 0. Denote g(x) = λ −1 f (β x), where β =
k f k p Lp
λ
. Then, by the hypothesis,
T (g)(x) = λ −1 T ( f )(β x). It follows that {x ∈ R : T (g)(x) > 1} = {x ∈ R : T ( f )(β x) > λ } = β −1 {x ∈ R : T ( f )(x) > λ } . This implies 1
1
1
1
1
1
λ {x ∈ R : T ( f )(x) > λ } p = λ β p {x ∈ R : T (g)(x) > 1} p ≤ λ β p C p = C p k f kL p (R) for all λ > 0. It follows that T maps L p to weak L p,∞ with norm at most C1/p , when restricted to Schwartz functions. Since S is dense on L p for 1 ≤ p < ∞, it follows that T admits an extension from L p to weak L p,∞ with the same norm.
Exercise 5.1.6 Let ϕ be in S (R). Prove that e2πiNx ϕ(x) dx = ϕ(0)πi, N→∞ x R Z 2πiNx e lim p.v. ϕ(x) dx = − ϕ(0)πi. N→−∞ x R Z
lim p.v.
Solution.
e2πiNx ϕ(x)dx = W0 , e2πiNx ϕ x R ∨
2πiNx ϕ) = W0 , (e\
= W0∨ , τ N ϕb = τ −N W0∨ , ϕb
Z
p.v.
Z
=π R
b )dξ isgn(ξ + N)ϕ(ξ
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Z = πi
∞
−N
b )dξ − ϕ(ξ
Z −N −∞
b )dξ ϕ(ξ
which converges to Z
πi R
as N → ∞ and −πi
Z R
b )dξ = πiϕ(0) ϕ(ξ b )dξ = −πiϕ(0) ϕ(ξ
as N → −∞.
Exercise 5.1.7 Let Tα , α ∈ R, be the operator given by convolution with the distribution whose Fourier transform is the function uα (ξ ) = e−πiα sgn ξ . (a) Show that the Tα ’s are isometries on L2 (R) that satisfy (Tα )−1 = T2−α . (b) Express Tα in terms of the identity operator and the Hilbert transform. Solution. (a) Fix f ∈ L2 (R). Then
kTα ( f )kL2 = \ Tα ( f ) L2 = fbuα L2 = fb L2 = k f kL2 . Thus Tα is an isometric map. (b) For f ∈ L2 (R), we have
∨ Tα ( f )(x) = fb(ξ )e−πiα sgn(ξ ) (x).
It is easy to check that T2−α ◦ Tα ( f ) = Tα ◦ T2−α ( f ) = f for all f ∈ L2 (R). (c) We have
∨ Tα ( f )(x) = fb(ξ )e−πiα sgn(ξ ) (x) ∨ = fb(ξ ) cos(πα sgn(ξ )) − i sin(πα sgn(ξ )) (x) = cos(πα) f (x) + sin(πα)H( f )(x).
Exercise 5.1.8 ( j)
Let Qy be the jth conjugate Poisson kernel of Py defined by ( j)
Qy (x) =
Γ ( n+1 2 ) π
Prove that
n+1 2
xj (x2 + y2 )
n+1 2
.
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185 ( j)
(Qy )∧ (ξ ) = −i
ξ j −2πyξ  e . ξ 
( j)
( j)
Conclude that R j (Py ) = Qy and that for f in L2 (Rn ) we have R j ( f )∗Py = f ∗Qy . These results are analogous to the statements cy (ξ ) = −i sgn(ξ )Pby (ξ ), H(Py ) = Qy , and H( f ) ∗ Py = f ∗ Qy . Q Solution. Denote ϕ(ξ ) = e−2πξ  . Using Exercise 2.2.11, we have Γ ( n+1 2 )
ϕ ∨ (x) =
π Since
−i
1
n+1 2
2
(x + 1)
n+1 2
.
Γ ( n+1 ξ j −2πξ  ∨ xj 2 ) e = x j ϕ ∨ (x) = n+1 n+1 , 2 ξ  π 2 (x + 1) 2
we deduce that b ) = −i ψ(ξ where ψ(x) =
Γ ( n+1 2 ) π
Now from the representation
ξ j −2πξ  e , ξ 
n+1 2
xj 2
(x + 1)
n+1 2
.
x ( j) Qy (x) = y−n ψ( ), y
we have ξj d ( j) b Qy (ξ ) = ψ(yξ ) = −i e−2πyξ  . ξ  ( j)
( j)
This identity implies that R j (Py ) = Qy and R j ( f ) ∗ Py = f ∗ Qy for all f ∈ L2 (Rn ).
Exercise 5.1.9 Fix n ≥ 2. Let f0 , f1 , . . . , fn be in L2 (Rn ) and, for 0 ≤ j ≤ n, let u j (x, x0 ) = (Px0 ∗ f j )(x) be the Poisson integrals of f j where x = (x1 , . . . , xn ) ∈ Rn and x0 > 0. Show that a necessary and sufficient condition for f j = R j ( f0 ),
j = 1, . . . , n,
is that the following system of generalized CauchyRiemann equations holds: n
∂uj
∑ ∂ x j (x, x0 ) =
0,
j=0
∂uj ∂ uk (x, x0 ) = (x, x0 ) , ∂ xk ∂xj
0 ≤ j 6= k ≤ n .
Solution. In progress of proving the necessary condition, we need to note that ∂d u0 ∂d u0 ∂ ∂ b b (ξ ) = (ξ ) = ub0 (ξ ) = Py f0 (ξ ) = −2π ξ  e−2πyξ  fb0 (ξ ). ∂ x0 ∂y ∂y ∂y Fix 1 ≤ j ≤ n, we have
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ξ j2 ∂d uj ξj b f0 (ξ ) = 2π e−2πyξ  fb0 (ξ ). (ξ ) = (2πiξ j )b u j (ξ ) = (2πiξ j )e−2πyξ  − i ξ  ξ  ∂xj Sum the above identities, we get n
ξ j2
n
∂d uj
∑ ∂ x j (ξ ) = ∑ 2π ξ  e−2πyξ  fb0 (ξ ) − 2π ξ  e−2πyξ  fb0 (ξ ) = 0
j=0
j=1
which implies the first identity in the necessary condition. Furthermore, we have ∂d uj ξk ξ j −2πyξ  b f0 (ξ ). (ξ ) = (2πiξk )b u j (ξ ) = 2π e ξ  ∂ xk Interchanging k and j, implies the second equality of necessary condition. We now turn to the converse implication. We have ∂d uj (ξ ) = (2πiξ j )e−2πyξ  fbj (ξ ), 1 ≤ j ≤ n ∂xj and ∂d u0 (ξ ) = −2π ξ  e−2πyξ  fb0 (ξ ). ∂ x0 Hence the identity ∑nj=0
∂uj = 0 implies that ∂xj n
ξj
∑ i ξ  fbj (ξ ) = fb0 (ξ ).
j=1
Fix 1 ≤ k ≤ n. Then we have
n
ξ j ξk b ξk b f (ξ ) = − i f0 (ξ ). ∑ 2 j ξ  j=1 ξ  In addition, the equation
∂uj ∂ uk = yields ∂ xk ∂xj ξk fbj (ξ ) = ξ j fbk (ξ ).
Thus we have ξk b fbk (ξ ) = − i f0 (ξ ) ξ  which turns out that fk = Rk ( f0 ), 1 ≤ k ≤ n.
Exercise 5.1.10 Prove the distributional identity ∂ j x−n+1 = (1 − n)p.v.
xj . xn+1
Then take Fourier transforms of both sides and use Theorem 2.4.6 to obtain another proof of Proposition 5.1.14. Solution. Let ϕ be a Schwartz function which is split into odd and even parts ϕ = ϕ− + ϕ+ . We have
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D E D E ∂ j x−n+1 , ϕ = − x−n+1 , ∂ j ϕ =− =−
Z Rn
Z Rn
x−n+1 ∂ j ϕ− (x)dx −
Z Rn
x−n+1 ∂ j ϕ+ (x)dx
x−n+1 ∂ j ϕ− (x)dx Z
x−n+1 ∂ j ϕ− (x)dx = − lim ε→0 x≥ε Z Z = lim (1 − n)x j x−n−1 ϕ− (x)dx − ε→0
x≥ε
Thus
x
x=ε
Z
xj
ε→0 x≥ε
xn+1
=(1 − n) lim
−n+1
ϕ− (x)dσ (x)
ϕ(x)dx.
∂ j x−n+1 = (1 − n) p.v.
xj xn+1
.
Now take the Fourier’s transform of the above identity, the Riesz’s Fourier transform is written as Γ ( n+1 2 ) π
n+1 2
Γ ( n+1 xj \ 2 ) 1 x−n+1 (ξ ) p.v. n+1 (ξ ) = ∂ j\ n+1 x π 2 1−n Γ ( n+1 2 ) 2πiξ j \ x−n+1 (ξ ) = n+1 1 − n 2 π n−1
=
2 Γ ( n+1 2 ) 2πiξ j π ξ −1 n+1 n−1 π 2 1−n Γ( 2 )
= −i
ξj , ξ 
since yΓ (y) = Γ (y + 1). Therefore, we have another proof for Riesz’s Fourier transform Γ ( n+1 2 ) π
n+1 2
ξj xj \ p.v. n+1 (ξ ) = −i . ξ  x
Exercise 5.1.11 (a) Prove that if T is a bounded linear operator on L2 (R) that commutes with translations and dilations and anticommutes with the reflection f (x) 7→ fe(x) = f (−x), then T is a constant multiple of the Hilbert transform. (b) Prove that if T is a bounded operator on L2 (R) that commutes with translations and dilations and vanishes when applied to ∨ functions whose Fourier transform is supported in [0, ∞), then T is a constant multiple of the operator f 7→ fbχ(−∞,0] . Solution. (a) Since T commutes with translations, we imply that T ( f ) = f ∗ W for some distribution W in L2 (R). Taking the Fourier’s transform, one get b T[ ( f ) = fbW for all f ∈ L2 (R). From the hypothesis that T commutes with dilations deduces that b (ξ ), b ξ =W W λ b (ξ ) = W b (1) for ξ > 0 and W b (ξ ) = W b (−1) for ξ < 0. Thus W
ξ ∈ R, λ > 0.
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e b = −W b ; hence W b (ξ ) = −W b (ξ ) for all ξ ∈ R. Especially, The anticommuting with the reflections of T implies that W b b b b W (1) = −W (−1) = a and W (0) = 0. It therefore forces W (ξ ) = a sgn(ξ ) for all ξ ∈ R. More specifically, T is a constant multiple of the Hilbert transform. (b) Apply the same argument as in the previous part, since T vanishes when applied to functions whose Fourier transform is b (ξ ) = 0 for all ξ > 0 and thus W b = cχ(−∞,0] for some constant c ∈ R. Hence supported in [0, ∞), we have W T ( f ) = c fbχ(−∞,0]
∨
for all f ∈ L2 (R).
Exercise 5.1.12 ([284]) Fix 1 < p ≤ 2. (a) Show that the function G(x, y) = Re (x + iy) p is subharmonic on R2 . (b) Let u(x, y), v(x, y) be realvalued functions on R2 such that u + iv is a holomorphic function of x + iy. Prove that G(u, v) is a subharmonic function on R2 . (c) Prove that there is a constant B p such that for all a and b reals we have π p p a − B p Re (a + ib) p . b p ≤ tan 2p (d) Prove that for f in C0∞ (R) we have
Z
Re ( f (x) + iH( f )(x)) p dx ≥ 0.
R
(e) Combine the results in parts (d) and (c) with a = f (x), b = H( f )(x) to obtain that
H p p ≤ tan π . L →L 2p (f) To deduce that this constant is sharp, take π/2p0 < γ < π/2p and let fγ (x) = (x + 1)−1 x + 12γ/π x − 1−2γ/π cos γ. Then 1 x+1 2γ/π sin γ when x > 1, x+1 x−1 H( fγ )(x) = 2γ/π x+1 − 1 sin γ when x < 1. x+1 x−1 Hint: Part (d): Let CR be the circle of radius R centered at (0, R) in R2 . Use that the integral of the subharmonic function G((Py ∗ f )(x), Qy ∗ f )(x)) over CR is at least 2πR Re ((PR ∗ f )(0) + i(QR ∗ f )(0)) p and let R → ∞. Part (f): The formula for H( fγ ) is best derived by considering the restriction of the analytic function −1
F(z) = (z + 1)
iz + i z−1
2γ/π
on the real line. Solution. (a) For any z = x + iy in the complex plane, let G(z) = Re [(x + iy) p ]. Obviously, G is continuous on R2 . It is enough to verify that for all z0 = x0 + iy0 ∈ C, there exists an r0 > 0 such that for all r ∈ (0, r0 ), Z 1 π G(z0 ) ≤ G(z0 + reiθ ) dθ (0.0.1) 2π −π We prove (0.0.1) by considering the following four cases.
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Case 1: x0 > 0. In this case, we choose a positive r0 small enough such that the ball B(z0 , r0 ) is in the right half plane of C. Thus, for any z = x + iy ∈ B(z0 , r0 ), we have x > 0 and G(z) = Re [(x + iy) p ] = Re z p . Since z p is holomorphic in B(z0 , r0 ), it follows that Re z p is harmonic in B(z0 , r0 ). Consequently, for all r ∈ (0, r0 ), G(z0 ) =
1 2π
Z π −π
G(z0 + reiθ ) dθ .
Thus, (0.0.1) holds. Case 2: x0 < 0. Choose a positive r0 small enough such that B(z0 , r0 ) is in the left half plane of C. Thus, for any z = x + iy ∈ B(z0 , r0 ), we have x < 0 and G(z) = Re [(−x + iy) p ] = Re [(−x − iy) p ] = (−1) p Re z p . As in Case 1, (0.0.1) holds. Case 3: z0 = 0. Write 1 2π
Z π
1 2π
G(reiθ ) dθ =
−π
Z
− π2
π 2
Z
+
− π2
−π
Z π
+
π 2
G(reiθ ) dθ .
Observing that by a changing of variable t = π − θ , Z π π 2
G(reiθ ) dθ =
Z π π 2
Z
=
π 2
Re [(r cos θ  + ir sin θ ) p ] dθ Re [(r cost + ir sint) p ] dt
0
Z
=
π 2
G(reit ) dt.
0
Likewise, by a changing of variable t = −π − θ gives that Z −π 2
Z 0
iθ
G(re ) dθ =
−π
− π2
G(reit ) dt.
Therefore, 1 2π
1 G(re ) dθ = π −π
Z π
iθ
Z
π 2
− π2
G(reiθ ) dθ
π
1 2 p r cos(pθ ) dθ π − π2 2r p pπ = sin( ) pπ 2 ≥0 Z
=
= G(0). Case 4: x0 = 0, y0 6= 0. Set
H(z) = Re z p .
Notice that H(z) = G(z) if z is in the closed righthalf complex plane. Next, we will show that G(z) ≥ H(z) when z belongs to the open lefthalf complex plane. If z = reiθ with r > 0 and π2 < θ < π, then 0 < π −θ
0 and −π < θ < − π2 , then G(z) = Re (r cos θ  + ir sin θ ) p = r p Re (− cos θ + i sin θ ) p = r p Re (cos(−π − θ ) + i sin(−π − θ )) p = r p Re (ei(−π−θ )p ) = r p cos[(π + θ )p], and hence
G(z) − H(z) = r p (cos[(π + θ )p] − cos[pθ ]) = −2r p sin(p(θ + π2 )) sin pπ 2 ≥ 0.
(0.0.3)
Combining (0.0.2), (0.0.3) and the fact that H is harmonic on R2 , we see that for 0 < r < y0 , G(iy0 ) = H(iy0 ) Z 1 π H(iy0 + reiθ ) dθ = 2π −π Z 1 π ≤ G(iy0 + reiθ ) dθ . 2π −π Summarizing the arguments in Cases 1,2,3,4 yields (0.0.1). This concludes the proof of (a). (b) Suppose first that F is a C 2 subharmonic function on R2 and u, v are realvalued functions on R2 such that u + iv is holomorphic. Then we prove that F(u, v) is subharmonic. To prove this we note first that by the chain rule we have: ∆ (F(u, v)) = (∂12 F)(u, v)(∂ u/∂ x)2 + (∂1 ∂2 F)(u, v)(∂ u/∂ x)(∂ v/∂ x)+ (∂1 ∂2 F)(u, v)(∂ u/∂ x)(∂ v/∂ x) + (∂22 F)(u, v)(∂ v/∂ x)2 + (∂12 F)(u, v)(∂ u/∂ y)2 + (∂1 ∂2 F)(u, v)(∂ u/∂ y)(∂ v/∂ y)+ (∂1 ∂2 F)(u, v)(∂ u/∂ y)(∂ v/∂ y) + (∂22 F)(u, v)(∂ v/∂ y)2 . Now in view of the CauchyRiemann equations we have ∂ u/∂ x = ∂ v/∂ y
∂ u/∂ y = −∂ v/∂ x
and replacing ∂ v/∂ y by ∂ u/∂ x and ∂ v/∂ x by −∂ u/∂ y, we notice that the cross terms cancel while the remaining terms give ∆ (F(u, v)) = (∆ F)(u, v) ∇u2 ,
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which is nonnegative when F is subharmonic and C 2 . We will now show that if G is subharmonic (and merely continuous but not necessarily C 2 ) and u + iv is holomorphic (u, v realvalued), then G(u, v) is subharmonic. We consider the function G ∗ φε , where φε is an nonnegative smooth approximate identity. Given point z0 = (x0 , y0 ) and a radius r we have (G ∗ φε )(z0 ) =
Z Z R R
G(z0 − (t, s))φε (t, s) dtds
1 π G(z0 − (t, s) + reiθ ) dθ φε (t, s) dtds ≤ R R 2π −π Z Z Z 1 π G(z0 − (t, s) + reiθ ) φε (t, s) dtds dθ ≤ 2π −π R R Z 1 π = (G ∗ φε )(z0 + reiθ ) dθ 2π −π Z
Z Z
thus G ∗ φε is subharmonic for any ε > 0. Now by the previous result we have that (G ∗ φε )(u, v) is subharmonic. But G ∗ φε converges to G uniformly on compact sets (by Theorem about approximate identities), in particular, it converges pointwise to G on R2 . Replacing z0 by (u(z0 ), v(z0 )) we obtain (G ∗ φε )(u(z0 ), v(z0 )) ≤
1 2π
Z π −π
(G ∗ φε )((u(z0 ), v(z0 )) + reiθ ) dθ
and letting ε → 0 and using the uniform convergence of G ∗ φε to G (on compact sets), we deduce that G(u, v) is subharmonic. (c) Fix a and b real numbers and 1 < p < 2. We show that the following inequality is true: b p ≤ tan p (π/2p)a p − B p Re [(a + ib) p ],
(0.0.4)
where B p = sin p−1 (π/2p) sec(π/2p) is a positive constant. To prove this, by scaling, we may assume that a2 + b2 = 1, and by symmetry we may take a, b > 0, so we may set a = cos(x),
b = sin(x)
for some 0 < x < π/2. So (0.0.4) reduces to sin p (x) ≤ tan p (π/2p) cos p (x) − B p cos(px) To prove this, consider the function F(x) = whose derivative is F 0 (x) = p
sin p (x) + B p cos(px) cos p (x)
sin p−1 (x) sin((p − 1)x) 1 − B p cos p+1 (x) sin p−1 (x)
Since the function inside the square brackets has a derivative of a constant sign, it must be strictly monotonic. Then the function inside the square brackets vanishes only when x = π/(2p) and consequently x = π/(2p) is the only zero of F 0 (x). Since sin((p − 1)x)/ sin p−1 (x) tends to zero as x → 0, F 0 (x) is positive for small values of x. It follows that F has a maximum attained for the unique value of x = π/(2p). This maximum is F(π/(2p)) and the conclusion follows. (d) Consider the holomorphic extension of f (x) + iH( f )(x) on the upper half space given by u(z) + iv(z) =
i π
Z +∞ f (t) −∞
We make a few observations about u + iv. First we note that
z−t
dt ,
u, v realvalued.
(0.0.5)
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u(x + iy) + v(x + iy) ≤
Cf . 1 + x + y
(0.0.6)
This is obvious for x ≥ 2C0 , where [−C0 ,C0 ] contains the support of f . Also for x ≤ 2C0 we have Z +∞ f (x − t)
t + iy
−∞
=
Z C0 +x −(C0 +x)
dt =
Z C0 +x −(C0 +x)
f (x − t) − f (x) dt + t + iy
Z C0 +x
and this is at most C0f
Z t≤3C0
t 3C0 + 2 tan−1 t + y y
f (x − t) dt t + iy
≤
−(C0 +x)
C00f 1 + y
f (x) dt t + iy
≤
C000 f 1 + x + y
.
We conclude from (0.0.6) that C pf
p
G(u(x + iy), v(x + iy)) ≤ (u(x + iy) + v(x + iy)) ≤
(1 + x + y) p
(0.0.7)
Since u(z) + iv(z) is holomorphic and g(x, y) is subharmonic, it follows that G(u(z), v(z)) is subharmonic on the upper half space. The boundary values of G(u(z), v(z)) are G(u(x + i0), v(x + i0)) = Re [( f (x) + iH( f )(x)) p ].
(0.0.8)
For R > 100, consider the circle with center (0, R) and radius R − R−1 and denote by CRU = (0, R) + {(R − 1/R)eiφ : −π/4 ≤ φ ≤ 5π/4} and CRL = (0, R) + {(R − 1/R)eiφ : 5π/4 ≤ φ ≤ 7π/4}. It follows from the subharmonicity of G(u(z), v(z)) that Z
Z
CRU
G(u(z), v(z)) ds +
CRL
G(u(z), v(z)) ds ≥ 2π(R − R−1 ) G(u(iR), v(iR)).
(0.0.9)
C →0 (1 + R) p
(0.0.10)
Clearly (0.0.7) implies that (R − R−1 ) G(u(iR), v(iR)) ≤ (R − R−1 ) and that
Z C −1 CU G(u(z), v(z)) ds ≤ (R − R ) (1 + R) p → 0
as R → ∞,
as R → ∞,
(0.0.11)
R
since R(1 + sin φ ) ≥ R/2 when −π/4 ≤ φ ≤ 5π/4. Letting R → ∞ in (0.0.9), and using (0.0.8), (0.0.10), and (0.0.11), we obtain the assertion in part (d), provided Z CRL
G(u(z), v(z)) ds →
Z +∞
Re [( f (x) + iH( f )(x)) p ]dx
−∞
as R → ∞. To show the last assertion, using parametric equations, we write the integral on the left as Z 7π/4
G(u(R0 cos φ , R + R0 sin φ ), v(R0 cos φ , R + R0 sin φ ))R0 dφ
5π/4
where R0 = R − R−1 . The change of variables x = R0 cos φ makes the preceding expression equal to
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Z
√ R0 2/2
√ −R0 2/2
G u x + iR − iR0
r 1−
x2 , v x + iR − iR0 R02
r 1−
x2 R02
!
dx q
(0.0.12)
2
1 − Rx02
and an application of the Lebesgue dominated convergence theorem gives that (0.0.12) converges to Z +∞
Z ∞
G(u(x + i0), v(x + i0)) dx = −∞
Re [( f (x) + i(H f )(x)) p ] dx .
−∞
The Lebesgue dominated convergence theorem is used in the fact that the integrand in (0.0.12) is bounded by the integrable q p x2 −p function C f (1 + x) since 1 − R02 is bounded from below by 1/2 in the range of integration. (e) Fix a C ∞ function f on the real line and apply (0.0.4) with a = f (x) and b = H( f )(x). Then integrate (0.0.4) with respect to x to obtain Z +∞ Z +∞ π Z +∞  f (x) p dx − B p Re [( f (x) + iH( f )(x)) p ]dx. H( f )(x) p dx ≤ tan p 2p −∞ −∞ −∞ Since B p ≥ 0, the assertion follows since Z +∞
Re [( f (x) + iH( f )(x)) p ] dx ≥ 0 ,
−∞
as shown in part (d). (f) Consider the analytic function −1
F(z) = (z + 1)
iz + i z−1
2γ/π
on the upper half space. Set z = x + iy. The restriction of F(z) on the real line is 2γ/π 1 x + 1 2γ/π 1 x + 1 2γ/π 1 x + 1 2γ/π πi x + 1 2γ/π 2 Q , i = = i sgn e x x+1 x−1 x+1 x−1 x−1 x+1 x−1 πi
πi
πi
πi
where Qx = 1 when x > 1 and Qx = −1 if x < 1. If x > 1 we have e 2 Qx = e 2 and if x < 1 we have e 2 Qx = e− 2 . It follows that x + 1 2γ/π 1 (cos(γ) + i sin(γ)) x+1 x−1 when x > 1 and 1 x + 1 2γ/π (cos(γ) − i sin(γ)) x+1 x−1 when x < 1. Thus F(x + i0) =
1 x+1 2γ/π sin γ x+1 x−1 fγ (x) + i − 1 x+1 2γ/π sin γ x+1 x−1
when x > 1, when x < 1
and since this is equal to the boundary values of a holomorphic function on the upper half plane, it follows that 1 x+1 2γ/π sin γ when x > 1, x+1 x−1 H( fγ )(x) = − 1 x+1 2γ/π sin γ when x < 1, x+1 x−1
2γ
2γ
as claimed. Notice that the function (x − 1− π x + 1 π −1 ) p is integrable over the entire line, since π/2p0 < γ < π/2p. It follows that kH( fγ )kL p sin(γ) = = tan(γ) k fγ kL p cos(γ)
194
and letting γ ↑
Contents π 2p
we obtain the sharpness of the constant in part (e).
Section 5.2. Homogeneous Singular Integrals and the Method of Rotations Exercise 5.2.1 Show that the directional Hilbert transform Hθ is given by convolution with the distribution wθ in S 0 (Rn ) defined by Z +∞
1 ϕ(tθ ) wθ , ϕ = p.v. dt. π t −∞
Compute the Fourier transform of wθ and prove that Hθ maps L1 (Rn ) to L1,∞ (Rn ). Solution. Fix f ∈ S and x ∈ Rn . We have Z +∞ D E 1 f (x − tθ ) = Hθ ( f )(x). f ∗ wθ (x) = wθ , τ x fe = p.v. π t −∞
We now compute the Fourier transform of wθ . Take ϕ ∈ S , then b bθ , ϕi = hwθ , ϕi hw Z b ) 1 ϕ(tθ = lim dt π ε→0 t≥ε t Z Z 1 = lim e−2πitθ ·ξ ϕ(ξ )dξ dt π ε→0 t≥ε Rn Z Z 1 = lim 1 e−2πitθ ·ξ ϕ(ξ )dξ dt π ε→0 ε ≥t≥ε Rn =
1 lim π ε→0
=−
Rn
Z
ϕ(ξ )
Rn
2i lim π ε→0
Z
=
Z
Z Rn
1 ε ≥t≥ε Z 1 ε
ϕ(ξ )
ε
cos(tθ · ξ ) − i sin(tθ · ξ ) dtdξ t
sin(tθ · ξ ) dtdξ t
(−i sgn(θ · ξ ))ϕ(ξ )dξ .
bθ (ξ ) = − sgn(θ · ξ ). It follows that Hence w Hθ ( f )(x) = (−i sgn(θ · ξ ) fb(ξ ))∨ (x). It is easy to check the identity HA(e1 ) ( f )(x) = He2 ( f ◦ A)(A−1 x), for all A ∈ O(n). Fix θ ∈ Sn−1 then there exists A ∈ O(n) such that θ = Ae1 . Take f ∈ L1 (Rn ) and set f1 = f ◦ A. We have {x ∈ Rn : Hθ ( f )(x) > λ } = A({x ∈ Rn : He1 ( f ◦ A)(y) > λ }). It follows {x ∈ Rn : Hθ ( f )(x) > λ } = {x ∈ Rn : He1 ( f1 )(x) > λ } . Since He1 maps L1 (R) to L1,∞ (R), we have {x1 ∈ R :  f1 (x1 , x2 , . . . , xn ) > λ } ≤
C λ
Z R
 f1 (x1 , x2 , . . . , xn ) dx1 ,
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for all x2 , . . . , xn ∈ R. This inequality and Fubini’s theorem yield {x ∈ Rn :  f1 (x) > λ } =
Z
χ{ f1 >λ } dx Z = χ{x1 ∈R :  f1 (x1 ,x2 ,...,xn )>λ } dx1 dx2 · · · dxn n−1 ZR R {x1 ∈ R :  f1 (x1 , x2 , . . . , xn ) > λ } dx2 · · · dxn = Rn−1 Z C Z  f (x1 , x2 , . . . , xn ) dx1 dx2 · · · dxn . ≤ Rn−1 λ R ZR
n
It turns out that λ {x ∈ Rn :  f1 (x) > λ } ≤ C k f1 kL1 (Rn ) = C k f kL1 (Rn ) . Hence Hθ maps L1 (Rn ) to L1,∞ (Rn ).
Exercise 5.2.2 Extend the definitions of WΩ and TΩ to Ω = dµ a finite signed Borel measure on Sn−1 with mean value zero. Compute the Fourier transform of Wdµ and find a necessary and sufficient condition on measures dµ so that Tdµ is L2 bounded. Notice that the directional Hilbert transform Hθ is a special case of such an operator Tdµ . Solution. Let dµ be a finite signed Borel measure on Sn−1 with mean value zero. Define the distribution Wdµ by hWΩ , ϕi = lim
Z ∞Z Sn−1
ε→0 ε
ϕ(rθ )dµ(θ )
dr , r
ϕ ∈S
and define the corresponding singular integral operator Tdµ ( f ) = f ∗Wdµ ,
f ∈S.
Next, we will compute the Fourier transform of Wdµ . Fix ϕ ∈ S , then
d b W dµ , ϕ = Wdµ , ϕ Z ∞Z
= lim
Sn−1
ε→0 ε 1 ε
b )dµ(θ ) ϕ(rθ
dr r
dr r Z Z Z 1 −2πirξ θ ·ξ 0 ε e − cos(2πr ξ ) = lim ϕ(ξ ) drdµ(θ )dξ ε→0 Rn r Sn−1 ε Z Z Z 1 −irθ ·ξ 0 − cos(r) 2πεξ  e = lim ϕ(ξ ) drdµ(θ )dξ ε ε→0 Rn r Sn−1 2πξ  Z Z 1 πi 0 = ϕ(ξ ) log − sgn(θ · ξ ) dµ(θ )dξ θ · ξ 0  2 Rn Sn−1 Z Z 1 πi = ϕ(ξ ) log − sgn(θ · ξ ) dµ(θ )dξ . θ · ξ  2 Rn Sn−1 Z
= lim
ε→0 ε
Z
Z
Sn−1 Rn
It follows that
e−2πirθ ·ξ ϕ(ξ )dξ dµ(θ )
Z
d W dµ (ξ ) =
Sn−1
log
1 πi − sgn(θ · ξ ) dµ(θ ). θ · ξ  2
(ξ = ξ  .ξ 0 )
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Observe, however, that
Z πi Sn−1 2 sgn(θ · ξ )dµ(θ ) < ∞.
Hence, if
Z 0 ess.sup log(θ · ξ )dµ(θ ) < ∞, 0 n−1 Sn−1 ξ ∈S
∞ n 2 2 d then W dµ ∈ L (R ) and thus Tdµ maps L to L . Finally, we notice that the directional Hilbert transform
Hθ ( f )(x) =
1 p.v. π
Z +∞
f (x − tθ )
−∞
dt t
in the direction θ is a special case of such a measure with dµ = π1 δθ − π1 δ−θ .
Exercise 5.2.3 Use the inequality AB ≤ A log A + eB for A ≥ 1 and B > 0 to prove that if Ω satisfies (5.2.24) then it must satisfy (5.2.16). Conclude that ifRΩ  log+ Ω  is in L1 (Sn−1 ), then TΩ is L2 bounded. Hint: Use that Sn−1 ξ · θ −α dθ converges when α < 1. See Appendix D.3. Solution. For real numbers x, y > 0, applying the inequality AB ≤ A log(A) + eB , A ≥ 1, B ≥ 0, we have √ √ x log(y) ≤ 2 max (x, 1) log(max ( y, 1/ y)) √ √ ≤ 2 max(x, 1) log( y + 1/ y) h √ √ i ≤ 2 max(x, 1) log(max(x, 1)) + y + 1/ y √ √ = 2 x log+ (x) + y + 1/ y . In particular, Ω (θ ) log As a sequence, Z Ω (θ ) log Sn−1
1/2 −1/2 1 ≤ 2 Ω (θ ) log+ (Ω (θ )) + θ · ξ 0 + θ · ξ 0 . 0 θ · ξ 
Z Z 1 1 + θ · ξ 0 1/2 + θ · ξ 0 −1/2 dθ < ∞. Ω Ω dθ (θ ) (θ ) log (Ω (θ )) + ≤ log ≤ 2 θ · ξ 0  θ · ξ 0  Sn−1 Sn−1
Exercise 5.2.4 Let Ω be a nonzero integrable function on Sn−1 with mean value zero. Let f ≥ 0 be nonzero and integrable over Rn . Prove that TΩ ( f ) is not in L1 (Rn ). Hint: Show that T\ Ω ( f ) cannot be continuous at zero. Solution. If TΩ ( f ) were in L1 (Rn ), then T\ Ω ( f ) would be continuous at zero. But
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b T\ Ω ( f )(ξ ) = f (ξ )
Z
Sn−1
1 πi − sgn(θ · ξ ) dθ log θ · ξ  2
and while fb(ξ ) is continuous at zero (since f ∈ L1 ) we have that Z
b (ξ ) = Ω
Sn−1
log
πi 1 − sgn(θ · ξ ) dθ θ · ξ  2
is not continuous at zero, since it is a non constant homogeneous function of degree zero, a contradiction. b (ξ ) to be continuous would be for Ω b (ξ ) to be constant. But if Ω b (ξ ) were a constant, then Ω would The only possibility for Ω be equal to a multiple of Dirac mass at zero, and thus it would zero on the sphere, which cannot happen by the assumption of the exercise.
Exercise 5.2.5 Let θ ∈ Sn−1 . Use an identity similar to (5.2.18) to show that the maximal operators 1 a>0 a
Z a
sup
1 a>0 2a
 f (x − rθ ) dr ,
Z +a
sup
0
 f (x − rθ ) dr
−a
are L p (Rn ) bounded for 1 < p < ∞ with norm at most 3 p (p − 1)−1 . Solution. First, we need to show that the maximal operator 1 a>0 2a
Mθ ( f )(x) = sup
Z a
 f (x − rθ ) dr
−a
is L p (Rn ) bounded for 1 < p < ∞. In fact, we observe that MA(e1 ) ( f )(x) = Me1 ( f ◦ A)(A−1 x), for all matrices A ∈ O(n). It follows that L p boundedness of Mθ can be reduced to that of Me1 and we imply that Mθ is L p bounded for 1 < p < ∞ with norm at most that of Me1 . Let M1 be the centered HardyLittlewood maximal operator with respect to first variable x1 . Then the following inequality is valid Me1 ( f )(x1 , x2 , . . . , xn ) ≤ M1 ( f )(x1 , x2 , . . . , xn ). It turns out Z Z Z p p p Me1 ( f )(x1 , x2 , . . . , xn ) dx1 ≤ f (x1 , x2 , . . . , xn ) dx1 . M1 ( f )(x1 , x2 , . . . , xn ) dx1 ≤ (3p) p (p − 1)−p R
R
Hence
Z
... R
R
Z Z p p f (x1 , x2 , . . . , xn ) dx1 . . . dxn . Me1 ( f )(x1 , x2 , . . . , xn ) dx1 . . . dxn ≤ (3p) p (p − 1)−p . . .
Z R
Thus Mθ is bounded in Next, denote
R
L p (Rn )
with constant at most
R
3p(p − 1)−1 . 1 a>0 a
Nθ ( f )(x) = sup
Z a
 f (x − rθ ) dr,
0
then it is clear that Nθ ( f )(x) ≤ 2Mθ ( f )(x). Hence Nθ is also bounded on L p (Rn ) for 1 < p < ∞.
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Exercise 5.2.6 For Ω ∈ L1 (Sn−1 ) and f locally integrable on Rn , define 1 n R>0 vn R
Z
MΩ ( f )(x) = sup
Ω (y/y)  f (x − y) dy .
y≤R
Apply the method of rotations to prove that MΩ maps L p (Rn ) to itself for 1 < p < ∞. Solution. Let Nθ be as in the previous exercise. For R > 0, we have 1 νn Rn
Z
R 1 Ω (θ )  f (x − rθ ) rn−1 drdθ n νn R 0 Sn−1 Z Z 1 R 1 Ω (θ )  f (x − rθ ) drdθ ≤ νn Sn−1 R 0 Z 1 Ω (θ ) Nθ ( f )(x)dθ . ≤ νn Sn−1
Z
Ω (y/ y)  f (x − y) dy =
y≤R
Z
This implies MΩ ( f )(x) ≤
1 νn
Z Sn−1
Ω (θ ) Nθ ( f )(x)dθ
for all x ∈ Rn . Applying Minkowski’s inequaltity, we have kMΩ ( f )kL p (Rn ) ≤
1 νn
Z
Ω (θ ) kNθ ( f )kL p (Rn ) dθ ≤
Sn−1
Thus MΩ maps L p (Rn ) to itself with constant at most
Cp νn
Z Sn−1
Cp νn
Z Sn−1
Ω (θ ) dθ .
Exercise 5.2.7 Let Ω (x, θ ) be a function on Rn × Sn−1 satisfying (a) Ω (x, −θ ) = −Ω (x, θ ) for all x and θ . (b) supx Ω (x, θ ) is in L1 (Sn−1 ). Use the method of rotations to prove that Z
TΩ ( f )(x) = p.v.
Rn
Ω (x, y/y) f (x − y) dy yn
is bounded on L p (Rn ) for 1 < p < ∞. Solution. For 0 < ε < N < ∞, denote (ε,N)
Hθ and
(∗∗)
Hθ (∗∗)
Z
f (x − rθ )
( f )(x) = ε≤r≤N
( f )(x) =
Then Hθ maps L p (Rn ) to itself. Furthermore, for a given function in S (Rn ) we have
sup 0 0 such that for all f of class C 2m with compact support and all differential monomials ∂xα of order α = 2m we have
α
∂x f p ≤ C p ∆ m f p . L L ∆m
Solution. (a) Denote g = ∆ f . Then we have ∂ j ∂k f = −R j Rk (g), where R j , Rk are Riesz transform. From the bound of the Riesz transforms, we obtain that
∂ j ∂k f p n ≤ A p kgk p n = A p k∆ f k p n . L (R ) L (R ) L (R ) (b) By induction on m, via the preceding argument, we can show that k∂xα f kL p (Rn ) ≤ C p k∆ m f kRn ,
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for all functions f of class C 2m with compact support and all differential monomials ∂xα of order α = 2m.
Exercise 5.2.11 Use the same idea as in Lemma 5.2.5 to show that if f is continuous on [0, ∞), differentiable in (0, ∞), and satisfies lim
Z Na f (u)
du = 0
u
N→∞ N
for all a > 0, then lim
Z N f (at) − f (t)
ε→0 ε N→∞
t
1 dt = f (0) log . a
Solution. For 0 < ε < N, we have Z N f (at) − f (t)
t
ε
Z NZ a
dt = ε
Z
1 aZ N
= 1
=
f 0 (tr)drdt f 0 (tr)dtdr
ε
Z a f (Nr) − f (εr)
r
1
=
Z aN f (u)
u
N
du −
dr
Z a f (εr) 1
r
dr.
Taking the limit, we obtain that lim
Z N f (at) − f (t)
ε→0 ε N→∞
t
1 dt = f (0) log . a
Exercise 5.2.12 Let Ωo be an odd integrable function on Sn−1 and Ωe an even function on Sn−1 that satisfies (5.2.24). Let f be a function supported in a ball B in Rn . Prove that (∗∗) (a) If  f  log+  f  is integrable over a ball B, then TΩo ( f ) and TΩo ( f ) are integrable over B. (∗∗)
(b) If  f (log+  f )2 is integrable over a ball B, then TΩe ( f ) and TΩe ( f ) are integrable over B. Hint: Use Exercise 1.3.7. Solution. (a) Let f be a function supported in a ball B ⊂ Rn . Then the operator f 7−→ TΩo ( f ) B maps functions supported in B to L p (B). Since Ωo is odd and Remark 4.2.9, the norm of this restriction operator is bounded by A(p − 1)−1 , (1 < p ≤ 2) for some constant A. Apply the Yano’s extrapolation, we have Z hZ i 1 1 2 TΩo ( f )(x) dx ≤ 6A(1 + B) 2  f (x) log+  B f (x) dx +C + < ∞. 1 2 B
B
(b) Apply again the Yano’s extrapolation for the restriction operator f 7−→ TΩe ( f ) B with constant at most A(p − 1)−2 , (1 < p ≤ 2), we have Z hZ i 1 1 2 2 TΩe ( f )(x) dx ≤ 6A(1 + B) 2  f (x) (log+  B f (x)) dx +C + < ∞. 2 2 B
B
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Exercise 5.2.13 ([326]) Let Ω be integrable on Sn−1 with mean value zero. Use Jensen’s inequality to show that for some C > 0 and every radial function f ∈ L2 (Rn ) we have
TΩ ( f ) 2 ≤ C f 2 . L L This inequality subsumes that TΩ is well defined on radial L2 (Rn ) functions. Solution. Since Ω is an integrable function with mean value zero, the Fourier transform of WΩ is given by Z
cΩ (ξ ) = W
Sn−1
Ω (θ ) log
1 iπ 0 sgn(ξ .θ ) dθ . − ξ 0 .θ  2
Now fix a radical function f ∈ L2 (Rn ). We have 2 1 iπ 0 dξ + sgn(ξ .θ ) dθ ξ 0 .θ  2 L2 (Rn ) Rn Sn−1 Z ∞ Z Z 2 2 1 iπ b n−1 0 Ω (θ ) dθ dξ 0 dr. log + sgn(ξ .θ ) ≤ f (re ) r 1 0 ξ .θ  2 0 Sn−1 Sn−1 Z b 2 f (ξ )
Z
\
TΩ ( f )
=
Ω (θ ) log
We may now assume that kΩ kL1 (Sn−1 ) = 1. Then the Jensen’s inequality implies that 2 iπ 1 0 dξ + sgn(ξ .θ ) dθ ξ 0 .θ  2 L2 (Rn ) Rn Sn−1 2 Z ∞ Z Z 2 1 iπ b n−1 0 log 0 + sgn(ξ .θ ) Ω (θ ) dθ dξ 0 dr ≤ f (re1 ) r n−1 n−1 ξ .θ  2 0 S S 2 Z ∞ Z Z 2 1 iπ b n−1 0 dξ 0 dθ dr Ω ≤ f (re ) r log (θ ) + sgn(ξ .θ ) 1 0 ξ .θ  2 0 Sn−1 Sn−1
\
TΩ ( f )
Z
=
Z b 2 f (ξ )
Ω (θ ) log
≤C k f k2L2 (Rn ) , since Z Sn−1
log
2 Z 1 iπ 0 dξ 0 ≤ 2 + sgn(ξ .θ ) ξ 0 .θ  2 Sn−1
log
! 1 2 π 2 + dξ 0 < ∞. ξ 0 .θ  4
Section 5.3. The Calder´on–Zygmund Decomposition and Singular Integrals Exercise 5.3.1 Let A1 be defined in (5.3.4). Prove that 1 1 A1 ≤ sup 2 R>0 R thus the expressions in (5.3.6) and (5.3.4) are equivalent.
Z x≤R
K(x) x dx ≤ 2A1 ;
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Solution. Suppose Z
sup R>0 R≤x≤2R
K(x) dx = A1 < ∞.
Fix R > 0, then we have 1 2R
Z
K(x) x dx ≥
x≤2R
It follows that
1 R>0 R
Z
1 2R
Z
K(x) x dx ≥
R≤x≤2R
K(x) x dx ≥
sup
x≤R
1 sup 2 R>0
Z
1 2
Z
K(x) dx.
R≤x≤2R
K(x) dx =
R≤x≤2R
A1 . 2
Furthermore, we have the following estimate 1 R
Z
∞
K(x) x dx ≤
x≤R
1 ∑R j=0
∞
Z 2− j−1 R0 R≤x≤2R
and
Z
sup y6=0 x≥2y
K(x) dx = A1 < ∞
K(x − y) − K(x) dx = A2 < ∞.
Then for y 6= 0, we have the estimate Z x≥2y
Z K(x − y)χx−y≥ε − K(x)χx≥ε dx ≤
x≥2y
K(x − y)χx−y≥ε − K(x)χx−y≥ε dx
Z
+ x≥2y
≤
Z
K(x)χx−y≥ε − K(x)χx≥ε dx
K(x − y) − K(x) dx
x≥2y
Z
+ x≥2y
K(x)χx−y≥ε − K(x)χx≥ε dx
Z
≤A2 +
x≥2y
K(x)χx−y≥ε − K(x)χx≥ε dx.
Noting that for A, B and C arbitrary subsets of Rn , if (A ∪ B) \ (A ∩ B) j C, then χA − χB  ≤ χC . Hence Z x≥2y
Z K(x − y)χx−y≥ε − K(x)χx≥ε dx ≤A2 +
x≥2y
≤A2 +
K(x) χε+y≥x≥ε−y dx
Z ε+y≥x≥ 12 (2y+ε−y)
K(x) dx ≤ A1 + A2 .
(b) Apply the same argument as above, we have Z x≥2y
Z K(x − y)χx−y≤N − K(x)χx≤N dx ≤A2 +
x≥2y
≤A2 + ≤A2 +
Z x≥2y
K(x)χx−y≤N − K(x)χx≤N dx K(x) χN−y≤x≤N+y dx
Z 1 (2y+N−y)≤x≤N+y 2
K(x) dx ≤ A1 + A2 .
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(c) Notice that χε≤x α ≤ x ∈ Rn : T (∗∗) (g)(x) > + x ∈ Rn : T (∗∗) (b)(x) > 2 2 n o r 2 α ≤ r kT (g)krLr + ∪ j Q∗j + x ∈ / ∪ j Q∗j : T (∗∗) (b)(x) > α 2 o n 2r α ≤ r Br kgkrLr + ∑ Q∗j + x ∈ / ∪ j Q∗j : T (∗∗) (b)(x) > . α 2 j The first two terms can be estimated as √ ∗ (2n+1 Bγ)r 2r r (5 n)n r kgk k k k f kL1 . B + Q f + ≤ ∑ j Lr L1 αr 2n γα γα j For estimating of the last term, we need to bound T (ε,N) (b) uniformly in ε and N. To get this we use the inequality T (ε,N) (b) ≤ T (ε) (b) + T (N) (b). We are now dealing with T (ε) and note that T (N) can be treated similarly. Fix x ∈ / ∪ j Q∗j and ε > 0. Define J1 (x, ε) = j : x − y < ε, ∀y ∈ Q j , J2 (x, ε) = j : x − y > ε, ∀y ∈ Q j , J3 (x, ε) = j : x − y = ε, for some y ∈ Q j . Since T (ε) (b j )(x) =
Z Qj
K(x − y)b j (y)χ{x−y≥ε} (y)dy,
/ ∪ j Q∗j and j ∈ J1 (x, ε). Also T (ε) (b j )(x) = T (b j )(x) whenever x ∈ / ∪ j Q∗j and it is easy to see that T (ε) (b j )(x) = 0 whenever x ∈ (ε) ∗ j ∈ J2 (x, ε) and T (b j )(x) = T (b j χ{x−·≥ε} )(x) whenever x ∈ / ∪ j Q j and j ∈ J3 (x, ε). Together with noting that b = ∑ j b j , we have (ε) T (b)(x) ≤ ∑ T (ε) (b j )(x) ≤ ∑ T (b j )(x) + ∑ T (b j χ{x−·≥ε} )(x) , j
j∈J2 (x,ε)
but it is clear that
∑ j∈J2 (x,ε)
where
T (b j )(x) =
∑ j∈J2 (x,ε)
j∈J3 (x,ε)
Z Q K(x − y)b j (y)dy ≤ E1 (x), j
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Z
E1 (x) = ∑ j
K(x − y) − K(x − y j ) b j (y) dy
Qj
and y j is the center of the cube Q j . It remains to estimate the term ∑ T (b j χ{x−·≥ε} )(x) . j∈J3 (x,ε)
Noting that for x ∈ / ∪ j Q∗j and j ∈ J3 (x, ε) there exists y0 ∈ Q j such that x − y0  = ε. It follows √ 1 1 √ ε = x − y0  ≥ (`(Q∗j ) − `(Q j )) = (5 n − 1)`(Q j ) ≥ 2 n`(Q j ) 2 2 and for any y ∈ Q j we have
√ ε x − y ≥ x − y0  − y0 − y ≥ ε − n`(Q j ) ≥ , 2 √ 3ε x − y ≤ x − y0  + y − y0  ≤ ε + n`(Q j ) ≤ . 2
It follows that [ j∈J3 (x,ε)
3ε ε \ B x, . Q j j B x, 2 2
Denote
1 c j (x, ε) = b j (y)χ{x−y≥ε} (y)dy Qj Qj and note from Calder´onZygmund decomposition that c j (x, ε) ≤ 2n+1 γα. Then we have the estimate Z Z ∑ K(x − y)b j (y)χ{x−y≥ε} (y)dy ≤ ∑ K(x − y) − K(x − y j ) b j (y)χ{x−y≥ε} (y) − c(x, ε) dy j∈J3 (x,ε)
Z
Qj
j∈J3 (x,ε)
Qj
+
c(x, ε)
∑ Z
∑
Qj
j
K(x − y)dy
Qj
j∈J3 (x,ε)
≤
Z
K(x − y) − K(x − y j ) b j (y) + 2n+1 γα dy
+ 2n+1 γα
Z ε ≤x−y≤ 3ε 2 2
K(x − y) dy
≤ E1 (x) + 2n+1 γαE2 (x) + 2n+2 A1 γα, where
Z
E2 (x) = ∑ j
Qj
K(x − y) − K(x − y j ) dy.
In turn, we have that T (ε) (b)(x) is bounded uniformly in ε as following (ε) T (b)(x) ≤ 2E1 (x) + 2n+1 γαE2 (x) + 2n+2 A1 γα. This implies the estimate T (∗∗) (b)(x) ≤ 4E1 (x) + 2n+2 γαE2 (x) + 2n+3 A1 γα for all x ∈ (∪ j Q∗j )c . Now we go back to estimate n α o / ∪ j Q∗j : T (∗∗) (b)(x) > x∈ . 2 α Set γ = 2−n−5 (A1 + A2 + B)−1 . Then we have 2n+3 A1 γα < ; hence 3
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α o n n α o n α o / ∪ j Q∗j : 4E1 (x) > / ∪ j Q∗j : 2n+2 γαE2 (x) > / ∪ j Q∗j : T (∗∗) (b)(x) > ≤ x ∈ + x ∈ x∈ 2 12 Z 12 Z 48 ≤ E1 (x)dx + 2n+6 γ E2 (x)dx, α (∪ j Q∗j )c (∪ j Q∗j )c α α since α2 = α3 + 12 + 12 . Furthermore, the two last integrals above can be estimated as
Z (∪ j Q∗j )c
E1 (x)dx ≤ ∑ j
Z Qj
Z b j (y)
(Q∗j )c
K(x − y) − K(x − y j ) dxdy
Z b j (y) K(x − y) − K(x − y j ) dxdy x−y j ≥2y−y j  j Qj Z b j (y) dy ≤ A2 ∑
≤∑
Z
j
Qj
= A2 ∑ b j L1 j
n+1
≤2
A2 k f kL1 ,
where we used the fact that x − y j  ≥ 2y − y j  for all x ∈ / Q∗j and y ∈ Q j . We also obtain the same estimate for E2 (x) as A2 k f kL1 . E2 (x)dx ≤ A2 ∑ Q j ≤ γα (∪ j Q∗j )c j
Z
Combining together, we then get the inequality n k f kL 1 k f kL1 α o / ∪ j Q∗j : T (∗∗) (b)(x) > + 2n+6 A2 , x∈ ≤ 3 · 2n+5 A2 2 α α and therefore n o r √ k f kL1 x ∈ Rn : T (∗∗) ( f )(x) > α ≤ 2n+1 B2−n−5 (A1 + A2 + B)−1 2−n 2n+5 + (5 n)n 2n+5 + 3 · 2n+5 + 2n+6 (A1 + A2 + B) α √ n n+5 k f kL1 n+5 n+6 5 +3·2 + 2 )(A1 + A2 + B) . ≤(2 + (5 n) 2 α Hence T (∗∗) maps L1 (Rn ) to L1,∞ (Rn ).
Exercise 5.3.5 Assume that T is a linear operator acting on measurable functions on Rn such that whenever a function f is supported in a cube Q, then T ( f ) is supported in a fixed multiple of Q. (a) Suppose that T maps L p to itself for some 1 < p < ∞ with norm B. Prove that T extends to a bounded operator from L1 to L1,∞ with norm a constant multiple of B. (b) Suppose that T maps L p to Lq for some 1 < q < p < ∞ with norm B. Prove that T extends to a bounded operator from L1 to Ls,∞ with norm a multiple of B, where 1 1 1 + = . p0 q s Solution. Assume that T ( f ) is supported in Q∗ = ρQ (ρ > 0 fixed) for any measurable function f supported in the cube Q.
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(a) Fix α > 0 and γ > 0 to be chosen later. Use the Calder´onZygmund decomposition of an arbitrary integrable function f at height γα, we then have f = g + b, where g ∈ L1 ∩ L∞ and b = ∑ j b j with b j supported in some cube Q j . Denote Q∗j = ρQ j . Then we have the following inequalities n α o α o n {x ∈ Rn : T ( f )(x) > α} ≤ x ∈ Rn : T (g)(x) > + x ∈ Rn : T (b)(x) > 2 2 [ n α o 2p / ∪ j Q∗j : T (b)(x) > ≤ p B p kgkLp p + Q∗j + x ∈ j α 2 ∗ 2 Z 2p p p kgk B + ∑ Q j + α (∪ j Q∗ )c T (b)(x) dx Lp αp j j Z (2n+1 Bγ) p ρ n k f k α L1 T (b j )(x) dx ≤ + + ∑ n 2 γ γ α 2 j (Q∗j )c (2n+1 Bγ) p ρ n k f k L1 = + , 2n γ γ α ≤
since T (b j ) is supported in Q∗j = ρQ j . Choosing γ = 2−n−1 B−1 yields the estimate {x ∈ Rn : T ( f )(x) > α} ≤ (2 + 2n+1 ρ)B
k f kL1 . α
Thus T maps L1 to L1,∞ with constant multiple of B. (b) The preceding argument can be modified as follows if T is assumed to map L p to Lq instead of L p to L p : n α o α o n {x ∈ Rn : T ( f )(x) > α} ≤ x ∈ Rn : T (g)(x) > + x ∈ Rn : T (b)(x) > 2 2 [ n 2q α o / ∪ j Q∗j : T (b)(x) > ≤ q kT (g)kqLq + Q∗j + x ∈ j α 2 ∗ 2 Z 2q q q T (b)(x) dx kgk B p +∑ Qj + L αq α (∪ j Q∗j )c j Z 1 q 1 2q ρ n k f kL1 α T (b j )(x) dx ≤ q Bq (2αγ) p0 k f kLp1 + + ∑ α γ α 2 j (Q∗j )c 1 1 q 2q ρ n k f kL1 , = q Bq (2αγ) p0 k f kLp1 + α γ α ≤
since T (b j ) is supported in Q∗j = ρQ j . Choosing γ = α s−1 k f k1−s B−s L1 and inserting this value of γ, the preceding expression becomes CB since
q− sq p0
q
α p0
−q+(s−1) pq0
q q p −(s−1) p0
k f kL1
+ ρ n Bs
s k f ksL1 0 k f kL1 s = C B αs αs
1 1 1 + = . p0 q s
This proves that T maps L1 to Ls,∞ with norm at most a multiple of B.
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Exercise 5.3.6 (a) Let 1 < q < ∞. Show that the good function g in Theorem 5.3.1 lies in the Lorentz space Lq,1 and that kgkLq,1 ≤ 0 1/q Cn,q α 1/q k f kL1 for some constant Cn,q . (b) Let 1 < r < ∞. Obtain a generalization of Theorem 5.3.3 in which the assumption that T maps Lr to Lr is replaced by that T maps Lr,1 to Lr,∞ with norm B. (c) Let 1 < r < ∞. Obtain a further generalization of Theorem 5.3.3 in which the assumption that T maps Lr to Lr is replaced by that it is of restricted weak type (r, r), i.e., it satisfies E αr
{x : T (χE )(x) > α} ≤ Br for all subsets E of Rn with finite measure. Solution. (a) Let g∗ be the decreasing rearrangement of g. Then
dt t 0 Z γ Z ∞ 1 1 dt dt = t q g∗ (t) + t q g∗ (t) t t 0 γ Z γ Z ∞ 1 1 dt dt − ≤ kgkL∞ t q + γ q0 tg∗ (t) t t γ 0 1 q1 − q10 = kgkL∞ γ + γ kgkL1 q
kgkLq,1 =
Z ∞
1
t q g∗ (t)
Choosing γ = qkgkL1 /kgkL∞ so that these terms become equal, we obtain that − q10
kgkLq,1 ≤ 2 q
1 0
1
kgkLq1 kgkLq∞ .
Recalling that kgkL∞ ≤ 2n α and kgkL1 ≤ k f kL1 we deduce that kgkLq,1 ≤ 2 q
− q10
n
1
1
2 q0 k f kLq1 α q0 .
(b) In Theorem 5.3.3, we replace the assumption that T maps Lr to Lr by that that it maps Lr,1 to Lr,∞ . Recall that we have applied the Calder´onZygmund decomposition to a fixed function f ∈ L1 (Rn ) at height αγ. The proof proceeds as in the book with the only exception being the treatment of the good function g. The modification of this argument is as follows: x ∈ Rn : T ( f )(x) > α n α o n α o ≤ x ∈ Rn : T (g)(x) > + x ∈ Rn : T (b)(x) > 2 2 n [
r α o 2r [ ≤ r T (g) Lr,∞ + Q∗i + x ∈ / Q∗i : T ∑ b j (x) > α 2 j i i [ n [
r 2r α o = r T (g) Lr,∞ + Q∗i + x ∈ / Q∗i :  ∑ T (b j )(x) > α 2 j i i Z 2r r 2 ≤ r Br g Lr,1 + ∑ Q∗i  + T (b j )(x) dx α α (Si Q∗i )c ∑ i j
√ n f L1 r 2r r r 2 0
r + 2n+1 A2 f L1 ≤ r Cn,r B (γα) f L1 + (2 n) α γα α
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≤
√
f 1 (2Cn,r Bγ)r (2 n)n n+2 L + + 2 A2 . γ γ α
0 B. Choosing γ = (2Cn,r B)−1 , we deduce that T is of weak type (1, 1) with constant at most a multiple of Cn,r r,∞ (c) In view of the result of Exercise 1.1.12 (c), the space L is normable since r > 1. Under this equivalent norm Lr,∞ becomes a Banach space, let us call it Z with norm k · kZ which satisfies
khkLr,∞ ≤ khkZ ≤
r khkLr,∞ . r−1
We now use the result of Exercise 1.4.7 which says that if a linear map T satisfies {x : T (χE )(x) > α} ≤ Br or equivalently kT (χE )kZ ≤ then T maps Lr,1 to Z with constant obtain the claimed conclusion.
1 r−1
E αr
r B E , r−1
B. Then T maps Lr,1 to Lr,∞ with norm at most
1 r−1
B. Applying the result in part (b) we
Exercise 5.3.7 Let K satisfy (5.3.12) for some A2 > 0, let W ∈ S 0 (Rn ) be an extension of K on Rn as in (5.3.7), and let T be the operator given by convolution with W . Obtain the case r = ∞ in Theorem 5.3.3. Precisely, prove that if T maps L∞ (Rn ) to itself with constant B, then T has an extension on L1 + L∞ that satisfies
T 1 1,∞ ≤ Cn0 (A2 + B), L →L and for 1 < p < ∞ it satisfies
T p p ≤ Cn L →L
1 (A2 + B), (p − 1)1/p
where Cn ,Cn0 are constants that depend only on the dimension. Hint: Apply the Calder´on–Zygmund decomposition f = g + b at height αγ, where γ = (2n+1 B)−1 . Since g ≤ 2n αγ, observe that {x : T ( f )(x) > α} ≤ {x : T (b)(x) > α/2} . For the interpolation use the result of Exercise 1.3.2. Solution. Apply the Calder´onZygmund decomposition to f at height γα, where γ = 2−n−1 B−1 , then we have f = g + b with kgkL∞ ≤ n 2 α and b = ∑ j b j , where b j is L1 function with supported in some cube Q j . The following estimate satisfies n α o {x ∈ Rn : T ( f )(x) > α} ≤ x ∈ Rn : T (b)(x) > . 2 √ Let Q∗j = 2 nQ j . We can estimate the last term in the inequality above as n n α o α o / ∪ j Q∗j : T (b)(x) > x ∈ Rn : T (b)(x) > ≤ ∪ j Q∗j + x ∈ 2 2 √ n Z (2 n) k f kL1 2 T (b)(x) dx ≤ + γ α α (∪ j Q∗j )c √ Z (2 n)n k f kL1 2 T (b j )(x) dx, ≤ + ∑ γ α α j (Q∗j )c
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R since b = ∑ j b j . It suffices to estimate the last sum ∑ j (Q∗ )c T (b j )(x) dx. Denote y j is the center of the cube Q j , we have j
Z dx T (b j )(x) dx = ∑ b (y) K(x − y) − K(x − y ) dy j j ∗ c Q (Q∗j )c j j (Q j ) Z Z K(x − y) − K(x − y j dx dy b j (y) ≤∑ (Q∗j )c j Qj Z Z K(x − y) − K(x − y j dx dy b j (y) ≤∑ x−y j ≥2y−y j  j Qj
≤A2 ∑ b j 1 ≤ 2n+1 A2 k f k 1 ,
Z
∑ j
Z
j
L
L
where we used the fact that x − y j ≥ 2 y − y j for all x ∈ / Q∗j and y ∈ Q j . Combining together we get k f kL1 √ k f kL1 {x ∈ Rn : T ( f )(x) > α} ≤ (2 n)n 2n+1 B + 2n+1 A2 ≤ Cn (A2 + B) , α α √ where Cn = (2 n)n 2n+1 + 2n+1 . Thus T maps L1,∞ with constant Cn (A2 + B); hence it has an extension on L1 + L∞ . The last part of the problem is deduced from the interpolation between the estimates L1 → L1,∞ and L∞ → L∞ . The interpolation (given in Exercise 1.3.2) gives the bound 1 8 (p − 1)− p Cn B for the norm of T from L p to L p .
Exercise 5.3.8 (Calder´on–Zygmund decomposition on Lq ) Fix a function f ∈ Lq (Rn ) for some 1 ≤ q < ∞ and let α > 0. Then there exist functions g and b on Rn such that (1) f = g + b. n
(2) kgkLq ≤ k f kLq and kgkL∞ ≤ 2 q α. (3) b = ∑ j b j , where each b j is supported in a cube Q j . Furthermore, the cubes Qk and Q j have disjoint interiors when j 6= k. (4) kb j kqLq ≤ 2n+q α q Q j . (5)
R Qj
b j (x) dx = 0.
(6) ∑ j Q j  ≤ α −q k f kqLq . (7) kbkLq ≤ 2
n+q q
k f kLq and kbkL1 ≤ 2α 1−q k f kqLq .
Hint: Imitate the basic idea of the proof of Theorem 5.3.1, but select a cube Q if in the proof of Theorem 5.3.1.
1/q 1 R q Q Q  f (x) dx
> α. Define g and b as
Solution. Apply the same argument as in Calder´onZygmund decomposition of a L1 function, we can find a countable family of cubes Q j such that 1 Z q 1 q  f (x) dx > α, ∀ j Q j Q j
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and that for each cube Q j , there exists unique a parent cube Q0j of Q j with double size of Q j such that
1 Q0j 
Z
1
q
q
Q0j
 f (x) dx
≤ α.
Then it is easy to see that Z −q ∑ Qj ≤ α ∑ j
q
 f (x) dx = α
−q
Z ∪ jQ j
Qj
j
 f (x)q dx ≤ α −q k f kqLq (Rn )
thus (6) holds; also we have 1 1 Z 1 2n Z 1 1 Z n q q q q  f (x)q dx ≤  f (x)q dx ≤  ≤ 2 q α. f (x) dx 0 Q j  Q j Q j  Q0j Q j  Q0j Now define
Z 1 b j = f − f (x)dx χQ j , Qj Qj
b = ∑bj j
and g = f − b. We need to check that these functions satisfy the requirement of the problem. Conditions (1), (3) and (5) hold by construction. We have these estimate 1 Z
b j q n ≤ f χQ q n + f (x)dx χQ j Lq (Rn ) j L (R ) L (R ) Q j Q j Z 1 n 1 1  f (x) dx ≤2 q α Q j q + Q j q Qj Qj Z 1 1 1 n 1 1 q  f (x)q dx Q j q0 ≤2 q α Q j q + Q j q Qj Qj 1
n
≤2 q +1 α Q q , which implies (4), i.e.,
q
b j q n ≤ 2n+q α q Q j . L (R )
Furthermore, take arbitrarily x ∈ Rn . If x ∈ Q j then 1 g(x) ≤ Qj
1  f (x) dx ≤ Qj Qj
Z
Z
 f (x)q dx
Qj
1 1 n q Q j q0 ≤ 2 q α.
(k) Otherwise, if x ∈ / ∪ j Q j then for each k = 0, 1, . . . there exists unique cube Q˜ j such that
1 Z 1 q q  f (x) dx ≤α (k) Q˜ (k) ˜ Q  j j
and intersection of the closures of these cubes is the singleton {x} . This implies 1 (k) Q˜  j
Z
 f (y) dy ≤ (k)
Q˜ j
1 (k) Q˜  j
Z
1 q (k) 1 q  f (y) dy Q˜ j  q0 ≤ α. (k)
Q˜ j
The Lebesgue differentiation theorem tields that  f (x) ≤ α for a.e. x ∈ Rn \ ∪ j Q j ; hence g(x) =  f (x) ≤ α for a.e. x ∈ Rn \ ∪ j Q j . Thus we have n kgkL∞ ≤ 2 q α. This is contained in (2). Now we estimate the Lq norm of g as follows
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q Z Z 1 Z q q g(x) dx =  f (x) dx + ∑ f (x)dx Q j Q Q Rn Rn \∪ j Q j j j j Z Z q 1  f (x)q dx + ∑ q Q j  f (x)q dx Q j q0 ≤ Rn \∪ j Q j Qj j Qj ≤
Z Rn \∪ j Q j
Z
=
Rn
 f (x)q dx + ∑
Z
j
 f (x)q dx
Qj
q
 f (x) dx.
Then we get the inequality kgkL p (Rn ) ≤ k f kL p (Rn ) . Thus both assertions in (2) hold. Finally, we prove (7). First we show that b belongs to Lq (Rn ). Indeed, we have Z Rn
b(x)q dx = ∑
Z
j
Qj
b j (x) q dx ≤ 2n+q α q ∑ Q j ≤ 2n+q k f kq q n , L (R ) j
which is equivalent to
n
kbkLq (Rn ) ≤ 2 q +1 k f kLq (Rn ) . Moreover, the L1 norm of b can be estimated by Z Rn
Z
b(x) dx = ∑
Qj
j
b j (x) dx
Z
≤ 2∑
 f (x) dx
Qj
j
Z
 f (x) dx
=2 ∪ jQ j
Z ≤2
 f (x)q dx
1 q
∪ jQ j
≤ 2 k f kLq (Rn ) α =
q10 ∑ Q j j
− qq0
q q0 Lq (Rn )
kfk
2α 1−q k f kqLq (Rn ) .
Exercise 5.3.9 Let f ∈ L1 (Rn ). Then for any α > 0, prove that there exist disjoint cubes Q j in Rn such that the set Eα = {x ∈ Rn : Mc ( f )(x) > S 1 R α α} is contained in j 3Q j and 4n < Q j  Q j  f (t) dt ≤ 2αn . Hint: For given α > 0, select all maximal dyadic cubes Q j (α) such that the average of f over them is bigger than α. Given x ∈ Eα , pick a cube R that contains x such that the average of  f  over R is bigger than α and find a dyadic cube Q such that R 2−n Q < R ≤ Q and that R∩Q  f  dx > 2−n αR. Conclude that Q is contained in some Qk (4−n α) and thus R is contained in 3Qk (4−n α). The collection of all Q j = Q j (4−n α) is the required one. Solution. For given α > 0, select all maximal dyadic cubes Q j (α) such that the average of f over them is bigger than α. As the hint suggests we consider the collection {Q j } j of all cubes of the formSQ j (4−n α) over all indices j. To show that Eα = {x ∈ Rn : Mc ( f )(x) > α} is contained in j 3Q j we fix x ∈ Eα . We pick a cube R that contains x such that the average of  f  over R is bigger than α. Let k be an integer such that
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2−(k+1)n < R ≤ 2−kn . Consider the dyadic grid of cubes of side length 2−k . There is either none of exactly one point in 2−k Zn in the interior of R. Consequently, there is at least one and at most 2n dyadic cubes Q of side length 2−k that intersect the interior of R. Since 1 R n R R  f  dx > α, for at least one of the aforementioned at most 2 cubes Q that intersect the interior of R we have Z
 f  dx > 2−n αR .
R∩Q
Since R ≤ Q < 2n R, it follows that 1 Q
Z
 f  dx >
Q
1 Q
Z
 f  dx > 4−n α .
R∩Q
Then Q is contained in some Qk (4−n α) and thus R is contained in 3Qk (4−n α). It follows that x∈Rj
[
3Q j (4−n α) =
j
and thus {Mc ( f ) > α} is contained in
S
j 3Q j .
[
3Q j
j
Now each Q j = Q j (4−n α) satisfies 1 α < 4n Q j 
Z
 f (t) dt
Qj
by selection and 1 Q j 
Z
 f (t) dt ≤ 2n
Qj
α α = n n 4 2
by the maximality of the Q j ’s.
Exercise 5.3.10 Let K(x) be a function on Rn \ {0} that satisfies K(x) ≤ Ax−n . Let η(x) be a smooth function equal to 1 when x ≥ 2 and vanishing when x ≤ 1. For f ∈ L p , 1 ≤ p < ∞, define truncated singular integral operators T (ε) ( f )(x) =
Z
K(y) f (x − y) dy,
y≥ε
Z
(ε)
Tη ( f )(x) =
Rn
η(y/ε)K(y) f (x − y) dy .
Show that the truncated maximal singular integral T (∗) ( f ) = supε>0 T (ε) ( f ) is L p bounded for 1 < p < ∞ if and only if the (∗) (ε) smoothly truncated maximal singular integral Tη ( f ) = supε>0 Tη ( f ) is L p bounded. Formulate an analogous statement for p = 1. Solution. Fix ε > 0. Notice that (ε)
T (ε) ( f )(x) − Tη ( f )(x) =
Z
K(y) f (x − y)dy −
y≥ε
Z Rn
η(y/ε)K(y) f (x − y)dy =
Z Rn
(1 − η(y/ε))χε≤y≤2ε K(y) f (x − y)dy.
Since K(y) ≤ A y−n , y 6= 0, it follows that (ε)
T (ε) ( f )(x) − Tη ( f )(x) ≤ Aε −n
Z Rn
(1 − η(y/ε))χε≤y≤2ε y/ε−n  f (x − y) dy = A(F (ε) ∗  f )(x) ≤ A kFkL1 (Rn ) M( f )(x),
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where M( f ) is the maximal HardyLittlewood operator and F(x) = (1 + kηkL∞ )χ1≤x≤2 x−n . Thus we have (ε)
T (ε) ( f )(x) − Tη ( f )(x) ≤ (1 + kηkL∞ )( A log 2) M( f )(x) = C M( f )(x) for some constant C ∈ (0, ∞). This inequality ensures that T (∗) maps L p (Rn ) to L p (Rn ) and maps L1 (Rn ) to L1,∞ (Rn ). Furthermore, the following inequalities (∗) T (∗) ( f ) ≤Tη ( f ) +C M( f ), (∗)
Tη ( f ) ≤T (∗) ( f ) +C M( f ) (∗)
imply that T (∗) maps L p (Rn ) to L p (Rn ) if and only if Tη
is L p bounded. (∗)
Likewise, T (∗) maps L1 (Rn ) to L1,∞ (Rn ) if and only if Tη does so.
Exercise 5.3.11 (M. Mastyło) Let 1 ≤ p < ∞. Suppose that Tε are linear operators defined on L p (Rn ) such that for all f ∈ L p (Rn ) we have Tε ( f ) ≤ A ε −a k f kL p for some 0 < a, A < ∞. Also suppose that there is a constant C < ∞ such that the maximal operator T∗ ( f ) = supε>0 Tε ( f ) satisfies T∗ (h)kL p ≤ C khkL p for all h ∈ S (Rn ). Prove that the same inequality is valid for all f ∈ p n L (R ). Hint: For a fixed δ > 0 define Sδ ( f ) = supε>δ Tε ( f ), which is a subadditive functional on L p (Rn ). For a fixed f0 ∈ L p (Rn ) define a linear space X0 = {λ f0 : λ ∈ C} and a linear functional T0 on X0 by setting T0 (λ f0 ) = λ Sδ ( f0 ). By the Hahn–Banach theorem there is an extension Te0 of T0 that satisfies Te0 ( f ) ≤ Sδ ( f ) for all f ∈ L p (Rn ). Since Sδ is L p is bounded on Schwartz functions with norm at most C, then so is Te0 . But Te0 is linear and by density it is bounded on L p (Rn ) with norm at most C; consequently, kSδ ( f0 )kL p = kT0 ( f0 )kL p = kTe0 ( f0 )kL p ≤ C k f0 kL p . The required conclusion for T∗ follows by Fatou’s lemma. Solution. For a fixed δ > 0 define a subadditive operator on L p (Rn ) by setting Sδ ( f ) = sup Tε ( f ) . ε>δ
By assumption we have Tε ( f ) ≤ A ε −a k f kL p , so Sδ is well defined and satisfies Sδ ( f ) ≤ A δ −a k f kL p for any f ∈ L p (Rn ), i.e., Sδ is a subadditive functional on L p (Rn ). For a fixed f0 ∈ L p (Rn ) define a linear space X0 = {λ f0 : λ ∈ C} and a linear functional T0 on X0 by setting T0 (λ f0 ) = λ Sδ ( f0 ). By the Hahn–Banach theorem there is an extension Te0 of T0 that satisfies Te0 ( f ) ≤ Sδ ( f ) for all f ∈ L p (Rn ). Since Sδ is L p is bounded on Schwartz functions with norm at most C, then so is Te0 . But Te0 is linear and by density it is bounded on L p (Rn ) with norm at most C; consequently, kSδ ( f0 )kL p = kT0 ( f0 )kL p = kTe0 ( f0 )kL p ≤ C k f0 kL p . The required conclusion for T∗ follows by Fatou’s lemma since kT∗ ( f0 )kL p = k lim inf Sδ ( f0 )kL p ≤ lim inf kSδ ( f0 )kL p ≤ C k f0 kL p . δ →0
δ →0
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Section 5.4. Sufficient Conditions for L p Boundedness Exercise 5.4.1 Let T be a convolution operator that is L2 bounded. Suppose that a given function f0 ∈ L1 (Rn ) ∩ L2 (Rn ) has vanishing integral and that T ( f0 ) is integrable. Prove that T ( f0 ) also has vanishing integral. Solution. Since T be a convolution operator that is L2 bounded, there is an L∞ function m such that T[ ( f )(ξ ) = fb(ξ )m(ξ ) for all suitable functions f . Since f0 and T ( f0 ) are integrable functions, it follows that fb0 and T[ ( f0 ) are continuous functions and, moreover fb0 (0) = 0. To show that T ( f0 ) has vanishing integral, it will suffice to show that T[ ( f0 )(0) = 0. By continuity we have T[ ( f0 )(0) = lim T[ ( f0 )(ξ ) = lim fb0 (ξ )m(ξ ) . ξ →0
ξ →0
But the expression on the right is a product of a bounded function times a function that tends to zero as ξ → 0. It follows that lim fb0 (ξ )m(ξ ) = 0 ξ →0
and the conclusion follows.
Exercise 5.4.2 Let K satisfy (5.4.1), (5.4.2), and (5.4.3) and let W ∈ S 0 be an extension of K on Rn . Let f be a compactly supported C 1 function on Rn with mean value zero. Prove that the function f ∗W is in L1 (Rn ). Solution. Conditions (5.4.1), (5.4.2), and (5.4.3) are the following: Z
sup R>0 R≤x≤2R
Z
sup y6=0 x≥2y
sup 0 0 for all l ∈ I and ak,l = 0 for l ∈ Z \ I, and I is a finite subset of Z. The space S0 (X) × `r (C) is dense L p (`r ) in view of Proposition 5.5.6 (a) and in view of the fact that finitely supported sequences are dense in `r (C) for all r < ∞. We need to control Z
~T (~f ) q s = ~ ~ sup T ( f )(y) ·~g(y) dν(y) , L (Y,` ) 0 Y ~g∈S0 (Y )×`s (C) k~g k q0 s0 ≤1 L (` )
where the · denotes the inner product between sequences. The supremum is taken over functions of the form n
~g =
∑ χB j
b j,l eiβ j,l l∈Z ,
j=1
where b j,l > 0, β j,l are real, and B j are pairwise disjoint subsets of Y of finite measure with k~g kLq0 (`s0 ) ≤ 1. Let P(z) =
p p (1 − z) + z, p0 p1
Q(z) =
q0 q0 (1 − z) + 0 z , 0 q0 q1
R(z) =
r r (1 − z) + z, r0 r1
S(z) =
s0 s0 (1 − z) + 0 z . 0 s0 s1
For z in the closed strip S = {z ∈ C : 0 ≤ Re z ≤ 1}, define m
~fz =
∑ χAk k=1
P(z)
ak,l
eiαk,l R(z)−P(z) k{ak,l }l k`r
n
,
~gz =
l∈Z
j=1
Then we have T (χ )(y) ∑ Ak k=1
We also define
Z
F(z) = Y
Q(z)
b j,l
eiβ j,l S(z)−Q(z) k{bk,l }l k s0
P(z)
m
~T (~fz )(y) =
∑ χB j
ak,l
eiαk,l R(z)−P(z) k{ak,l }l k`r
`
l∈Z
. l∈Z
~T (~fz )(y) ·~gz (y) dν(y) .
The expression Z Y
T (χAk )(y) χB j (y) dν(y)
is a finite constant, being is an absolutely convergent integral via H¨older’s inequality with exponents q0 and q00 . By linearity we have Q(z) P(z) Z m n b j,l ak,l iαk,l iβ j,l F(z) = ∑ ∑ ∑ e e T (χAk )(y) χB j (y) dν(y) . R(z)−P(z) S(z)−Q(z) Y k{bk,l }l k s0 k=1 j=1 l∈Z k{ak,l }l k`r `
Obviously, F(z) is an analytic function. By the disjointness of the sets Ak we observe that
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~fit p r = L 0 (` 0 )
1 m p0 p 1p p p p −p0 (Re (R(it)−P(it))) P(it) r0 r0 r r k{ak,l }l k`r = ∑ Ak  ∑ ak,l  = ~f L p0 (`r ) , ∑ Ak  ∑ ak,l  m
k=1 P(it)
l
k=1
l
r r
since ak,l  = ak,l0 . Using the same idea and the disjointness of the B j ’s we have q0
~git
0
0
q s L 0 (` 0 )
q0 = ~g 00
0
Lq (`s )
,
q0
since
q0 Q(it) b j,k  = b j,k0 .
Thus H¨older’s inequality and the hypothesis give
F(it) ≤ ~T (~fit ) Lq0 (`s0 ) ~git q00 s00 L (` )
≤ M0 ~fit p r ~git q0 s0 L 0 (` 0 )
p p0 L p (`r )
= M0 ~f By similar calculations we have
L 0 (` 0 )
q0 q00 0 0 Lq (`s )
~g
(0.0.14)
.
p
~f1+it p r = ~f pp1 r L (` ) L 1 (` 1 )
and
q0
~g1+it
0
0
q s L 1 (` 1 )
q0 = ~g q10
0
L (`s )
.
Also, in a way analogous to that we obtained (0.0.14) we deduce that q0
pp q0 F(1 + it) ≤ M1 ~f L p1 (`r ) ~g q10
0
L (`s )
.
(0.0.15)
We observe that F is analytic in the open strip S and continuous on its closure. Also, F is bounded on the closed unit strip (by some constant that depends on f and g). Therefore, (0.0.14), (0.0.15), and Lemma 1.3.5 give 0
p qq0 p F(z) ≤ M0 ~f L p0 (`r ) ~g q00 L
0
0 (`s )
1−θ p qq0 p M0 ~f L p0 (`r ) ~g q00 L
θ 0 (`s )
= M01−θ M1θ ~f L p (`r ) ~g Lq0 (`s0 ) ,
when Re z = θ . Observe that P(θ ) = Q(θ ) = 1 and hence Z
F(θ ) =
~T (~f )(y) ·~g(y) dν .
Y 0
Taking the supremum over functions ~g in S0 (Y ) × Lq (Y ) of norm less than or equal to one, we conclude the proof.
Exercise 5.5.3 Prove the following version of the Marcinkiewicz interpolation theorem. Let (X, µ) and (Y, ν) be σ finite measure spaces and let 0 < p0 < p < p1 ≤ ∞ and 0 < θ < 1 satisfy 1−θ θ 1 + = . p0 p1 p Let B1 , B2 be Banach spaces and ~T be defined on L p0 (X, B1 ) + L p1 (X, B1 ) such that for every y ∈ Y and for all F, G ∈ L p0 (X, B1 ) + L p1 (X, B1 ) we have
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~T (F + G)(y) ≤ ~T (F)(y) + ~T (G)(y) . B B B 2
2
2
(a) Suppose that ~T maps L p0 (X, B1 ) to L p0 ,∞ (Y, B2 ) with norm A0 and L p1 (X, B1 ) to L p1 ,∞ (Y, B2 ) with norm A1 . Show that ~T 1 p θ + p1p−p p A1−θ maps L p (X, B1 ) to L p (Y, B2 ) with norm at most 2 p−p 0 A1 . 0 (b) Let p0 = 1. If ~T is linear and maps L1 (X, B1 ) to L1,∞ (Y, B2 ) with norm A0 and L p1 (X, B1 ) to L p1 (Y, B2 ) with norm A1 , −1/p 1−θ θ show that ~T maps L p (X, B1 ) to L p (Y, B2 ) with norm at most 72 p − 1 A0 A1 . Hint: Part (a): Copy the proof of Theorem 1.3.2. Part (b): Use Exercise 5.5.1. Solution. (a) Assume first that p1 < ∞. Fix a function F in L p (X, B1 ) and α > 0. We split F = F0α + F1α , where F0α is in L p0 (X, B1 ) and F1α is in L p1 (X, B1 ). The splitting is obtained by cutting F at height δ α for some δ > 0 to be determined later. Set ( F(x) for kF(x)kB1 > δ α, α F0 (x) = 0 for kF(x)kB1 ≤ δ α, ( F(x) for kF(x)kB1 ≤ δ α, F1α (x) = 0 for kF(x)kB1 > δ α. It is easy to verify that F0α lies in L p0 (X, B1 ) and that F1α lies in L p1 (X, B1 ). Indeed, since p0 < p, we have Z
α p0
F p 0 L 0 (B ) = 1
kFkB1 >δ α
and similarly, since p < p1 ,
p p −p p kF(x)kB kF(x)kB01 dµ(x) ≤ (δ α) p0 −p F L p (B 1
1)
α p1
p1 −p p
F p F L p (B ) . 1 L 1 (B ) ≤ (δ α) 1
1
In view of the subadditivity of ~T we obtain for every y ∈ Y k~T (F)(y)kB2 ≤ k~T (F0α )(y)kB2 + k~T (F1α )(y)kB2 , and this implies that {y ∈ Y : k~T (F)(y)kB2 > α} j {y ∈ Y : k~T (F0α )(y)kB2 > α/2} ∪ {y ∈ Y : k~T (F1α )(y)kB2 > α/2}, and therefore d~T (F) (α) ≤ d~T (F α ) (α/2) + d~T (F α ) (α/2) . 0
1
(0.0.16)
Using the boundedness of ~T from L p0 (B1 ) → L p0 ,∞ (B2 ) with norm A0 and from L p1 (B1 ) → L p1 ,∞ (B2 ) with norm A1 we obtain Z Z p A0 0 A1p1 p0 p1 kF(x)k dµ(x) + kF(x)kB dµ(x). d~T (F) (α) ≤ B1 1 (α/2) p0 kFkB1 >δ α (α/2) p1 kFkB1 ≤δ α In view of the last estimate and Proposition 1.1.4, we obtain that
~T (F) p p L (B
2)
≤ p(2A0 ) p0
Z ∞
=
Z X
Z ∞
p
kFkB1 >δ α
0
+ p(2A1 ) p1 = p(2A0 ) p0
Z
α p−1 α −p0
α p−1 α −p1
0 p
kF(x)kB01
+ p(2A1 ) p1
Z
p(2A0 ) p0 1 p − p0 δ p−p0
Z
Z
kF(x)kB01 dµ(x) dα
Z kFkB1 ≤δ α
1 kF(x)k B1 δ
p1 dµ(x) dα kF(x)kB 1
α p−1−p0 dα dµ(x)
0
Z ∞
X
p1 kF(x)kB 1
X
kF(x)kB01 kF(x)kB1 0 dµ(x)
p
1 kF(x)k B1 δ
p−p
α p−1−p1 dα dµ(x)
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p(2A1 ) p1 1 p−p1 p1 dµ(x) kF(x)kB kF(x)kB 1 1 p1 − p δ p−p1 X
(2A0 ) p0 1 (2A1 ) p1 p1 −p
F p p =p , + δ L (B1 ) p−p 0 p − p0 δ p1 − p Z
+
and the convergence of the integrals in α is justified from p0 < p < p1 , while the interchange of the integrals (Fubini’s theorem) uses the hypothesis that (X, µ) is a σ finite measure space. We pick δ > 0 such that (2A0 ) p0
1 = (2A1 ) p1 δ p1 −p , δ p−p0
and observe that the last displayed constant is equal to the pth power of the constant in (1.3.8). We have therefore proved the theorem when p1 < ∞. We now consider the case p1 = ∞. Write F = F0α + F1α , where ( F(x) for kF(x)kB1 > γα, α F0 (x) = 0 for kF(x)kB1 ≤ γα, ( F(x) for kF(x)kB1 ≤ γα, F1α (x) = 0 for kF(x)kB1 > γα. We have
~T (F α ) ∞ ≤ A1 F α ∞ ≤ A1 γα = α/2 , 1 L 1 L provided we choose γ = (2A1 )−1 . It follows that the set {y ∈ Y : k~T (F1α )(y)kB2 > α/2} has measure zero. Therefore, d~T (F) (α) ≤ d~T (F α ) (α/2). 0
Since ~T maps L p0 (cb1 ) to L p0 ,∞ (B2 ) with norm at most A0 , it follows that
p Z (2A0 ) p0 F0α L0p0 (2A0 ) p0 p d~T (F α ) (α/2) ≤ = k~F(x)kB01 dµ(x). 0 α p0 α p0 kFkB1 >γα Using this inequality and Proposition 1.1.4, we obtain Z
~T (F) p p = p L
∞
0
≤p
Z ∞ 0
≤p
Z ∞ 0
α p−1 d~T (F) (α) dα α p−1 d~T (F α ) (α/2) dα 0
(2A0 ) p0 α p−1 α p0
= p(2A0 ) p0 =
p(2A1
Z
Z
p
kFkB1 >α/(2A1 ) p
kF(x)kB01
X ) p−p0 (2A
p 0) 0
Z 2A1 kF(x)kB 1
Z
p − p0
kF(x)kB01 dµ(x) dα
X
α p−p0 −1 dα dµ(x)
0 p dµ(x) . kF(x)kB 1
This proves the claim with constant A=2
p p − p0
1
p
p 1− p0
A1
p0
A0p
which coincides with the previous constant if p1 = ∞. (b) We apply the idea of Exercise 1.3.2. We interpolate first between L1 and Lr using the result in part (a). Since 1 < p+1 p+1 we obtain the following bound for the norm of ~T from L 2 (B1 ) to L 2 (B2 ) by
p+1 2
< p 1 we obtain that 9·2
p+1 2p
p+11 1 1 1 p p 2 ≤ 9 · 2·2·2 = 72 (p − 1)− p , p−1 p−1
as claimed.
Exercise 5.5.4 ~ For all x ∈ Rn let K(x) be a bounded linear operator from B1 to B2 and let Q ⊗ B1 the space of all finite linear combinations of n + elements of the form F = ∑m i=1 χEi ui , where Ei are disjoint measurable subsets of R of finite measure, ui in B1 , and m ∈ Z . ~ (a) Suppose that K satisfies Z
~ K(x) dx = C1 < ∞ . B1 →B2
Rn
Prove that the operator ~T (F)(x) =
Z
~ − y)(F(y)) dy , K(x
Rn p L (Rn , B1 )
initially defined on Q ⊗ B1 , has an extension that maps to L p (Rn , B2 ) with norm at most C1 for 1 ≤ p < ∞. ~ satisfies (b) (Young’s inequality) Let 1 ≤ p, q, s ≤ ∞ be such that p < ∞, s > 1, 1/q + 1 = 1/s + 1/p. Suppose that K
~
K( · ) B →B s n = C2 < ∞ . 1
2
L (R )
Prove that the ~T defined in part (a) has an extension that maps L p (Rn , B1 ) to Lq (Rn , B2 ) with norm at most C2 . ~ satisfies (c) (Young’s inequality for weak type spaces) Suppose that 1 < p, s < ∞, 1/q + 1 = 1/s + 1/p, and that K
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229
1 →B2
~
K( · ) B
Ls,∞ (Rn )
= C3 < ∞ .
Prove that the ~T defined in part (a) has an extension that maps L p (Rn , B1 ) to Lq (Rn , B2 ). Solution. Let Q ⊗ B1 be the space of all finite linear combinations of elements of the form m
F = ∑ χEi ui i=1
where Ei are disjoint measurable subsets of Rn of finite measure, ui ∈ B1 , and m ∈ Z+ . Then for any y ∈ Rn we have m
kF(y)kB1 = ∑ χEi (y)kui kB1 ≤ sup kui kB1 < ∞ 1≤i≤m
i=1
(a) Let F ∈ Q ⊗ B1 . For any x ∈ Rn we have Z Z
~
K(x − y)(F(y)) dy ≤ B2
Rn
hence
~
R
Rn K(x − y)(F(y)) dy
Rn
~ K(x − y) B
kF(y)kB1 dy ≤ C1 sup kui kB1 < ∞ ,
1 →B2
1≤i≤m
is B2 Bochner integrable and satisfies Z
Z
~
~ K(x − y)(F(y)) dy ≤
K(x − y)(F(y)) B2
Rn
B2
Rn
dy .
Thus T (~F)(x) is well defined for F in Q ⊗ B1 . Thus for all x ∈ Rn we have Z
~T (F)(x) ≤ B 2
~ K(x − y) B
1 →B2
Rn
kF(y)kB1 dy .
Using the scalar version of the Minkowski’s inequality (Theorem 1.2.10) which says that convolution with an integrable kernel maps L p (Rn ) to L p (Rn ) we obtain that
Z
~T (F)
~
≤ K(x) dx
kFkB1 p n = C1 kFkB1 p n ,
B2 B1 →B2 Rn
L p (Rn )
L (R )
L (R )
for all F in Q ⊗ B1 . Since Q ⊗ B1 is dense in L p (Rn , B1 ), it follows that ~T has an extension that maps ~T : L p (Rn , B1 ) → L p (Rn , B2 ) with norm at most C1 . (b) Let F ∈ Q ⊗ B1 . For any x ∈ Rn we have Z Z
~
K(x − y)(F(y)) dy ≤ B2
Rn
~ K(x − y) B
1 →B2
Rn
≤
Z Rn
s
~ K(x − y) B
1 →B2
m
= C2
kF(y)kB1 dy
0
∑ Ei  kui ksB1
1 Z s dy
Rn
10 s
< ∞.
i=1
Hence
R
~
Rn K(x − y)(F(y)) dy
is B2 Bochner integrable and satisfies Z
Z
~ ~ − y)(F(y)) dy K(x K(x − y) B →B kF(y)kB1 dy .
≤ 1 2 Rn
B2
Rn
0
kF(y)ksB1 dy
10 s
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Thus T (~F)(x) is well defined for F in Q ⊗ B1 . Thus for all x ∈ Rn we have Z
~T (F)(x) ≤ B 2
Rn
~ K(x − y) B
1 →B2
kF(y)kB1 dy .
Using the scalar version of Young’s inequality (Theorem 1.2.12) which says that convolution maps L p (Rn ) to Lq (Rn ) when 1/q + 1 = 1/ + 1/s, we obtain that
2
~T (F)
B
≤
Lq (Rn )
Z Rn
s
~ K(x)
B1 →B2
dx
1
s
kFkB1
L p (Rn )
= C2 kFkB1
L p (Rn )
,
for all F in Q ⊗ B1 . Since Q ⊗ B1 is dense in L p (Rn , B1 ), it follows that ~T has an extension that maps ~T : L p (Rn , B1 ) → Lq (Rn , B2 ) with norm at most C2 . ~ To show this we argue as follows: Let F ∈ Q ⊗ B1 . (c) We show that ~T is well defined on Q ⊗ B1 under the assumption on K. For any x ∈ Rn , using Theorem 1.4.16 (vi) we write Z Z
~
~ K(x − y) B →B kF(y)kB1 dy
K(x − y)(F(y)) dy ≤ 1 2 n n B R R 2
~
kF(y)kB ≤ cs K(x − y) 1
B1 →B2 Ls,∞ Ls,1
= cs C3
kF(y)kB1 < ∞ , Ls,1
since kF(y)kB1 = ∑m of characteristic functions of sets of finite measure lies i=1 χEi (y)kui kB1 and every finite linear combination R s,1 ~ in the Lorentz space L (see formula in Example 1.4.8). Hence Rn K(x − y)(F(y)) dy is B2 Bochner integrable and satisfies Z
Z
~ ~ K(x − y)(F(y)) dy ≤ K(x − y) B →B kF(y)kB1 dy .
1 2 B2
Rn
Rn
Then the proof proceeds as that in case (b) with the only difference being that Theorem 1.2.12 is replaced by Theorem 1.2.13, i.e. the scalar version of Young’s inequality for weak type spaces. (d) For F ∈ Q ⊗ B1 and x ∈ Rn , using the hypothesis, we can write Z Z
~
~
K(x − y)(F(y)) dy =
K(y)(F(x − y)) dy ≤ C F(x − ·) B ≤ C sup kui kB1 < ∞ , Rn
hence
R
~
Rn K(x − y)(F(y)) dy
B2
B2
Rn
1
is B2 Bochner integrable and satisfies Z
Z
~
~ K(x − y)(F(y)) dy ≤
K(x − y)(F(y)) B2
Rn
B2
Rn
1≤i≤m
dy .
Thus T (~F)(x) is well defined for F in Q ⊗ B1 . Thus for all x ∈ Rn we have Z
~T (F)(x) ≤ B 2
Rn
~ K(x − y) B
1 →B2
kF(y)kB1 dy .
and using the scalar version of the Minkowski’s inequality (Theorem 1.2.10) we obtain the desired inequality, as in part (a).
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231
Exercise 5.5.5 Use the inequality for the Rademacher functions in Appendix C.2 instead of Lemma 5.5.2 to prove part (a) of Theorem 5.5.1 in the special case p = q. Notice that this approach does not yield a sharp constant. Solution. Fix a positive integer N. We reindex the Rademacher functions so that they are indexed by the set of integers, not the set of nonnegative integers. Then, by the inequality in Appendix C.2 we have !1
∑
p ! 1 p ∑ T ( f j )r j (t) dt  j≤N
2
T ( f j ) 2
≤B p
Z 1
=B p
Z 1
0
 j≤N
0
T
∑
 j≤N
!1 p p f j r j (t) dt .
Now raising to the power p and integrating the above inequality, we obtain
p ! 1 !1
Z 1 p
2 2
∑ T ( f j )
≤B p
dt T ( f )r (t) j j
0 ∑
 j≤N
p
p  j≤N L L
p ! 1
Z 1 p
≤B p
T ∑ f j r j (t) dt
0  j≤N Lp
p ! 1
Z 1 p
≤B p A
∑ f j r j (t) dt
0  j≤N Lp
p ! 1p Z Z 1 =B p A ∑ f j (x)r j (t) dtdx X 0  j≤N !p
2
Z
≤B pC p A
=B pC p A
X
∑
∑
f j (x) 2
1p dx
 j≤N
2 f j
 j≤N
!1 2
.
p L
Letting N → ∞ we obtain the assertion in part (a) of Theorem 5.5.1, via the use of the Lebesgue monotone convergence theorem. Thus we proved part (a) of Theorem 5.5.1 in the special case p = q. The constant we obtained is B pC p A is not sharp, in fact it is bigger than A.
Exercise 5.5.6 Let 0 < p 6= 2 ≤ ∞ and suppose that T j are uniformly bounded linear operators from L p (R) to L p (R). Show that the inequality
∑ Tj ( f j )2
j∈Z
1 2
Lp
≤ Cp
∑  f j 2
j∈Z
1 2
Lp
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may fail. Hint: Let T j (g)(x) = g(x − j). When p > 2 take f j (x) = χ[− j,1− j] for j = 1, 2, . . . , N. When p < 2 take f j = χ[0,1] for j = 1, 2, . . . , N. Solution. Let T j (g)(x) = g(x − j). Obviously, T j maps from L p to L p with norm 1 for all 0 < p ≤ ∞. 1 2 When p > 2 take f j (x) = χ[− j,1− j] , for j = 1, 2, . . . , N. Then we have ∑Nj=1  f j 2 = χ[−N,0] and
N 1
2 2
∑  f j
j=1
1 = χ[−N,0] L p = N p .
Lp
On the other hand, T j ( f j ) = χ[0,1] for all j = 1, 2, . . . , N. Hence
N 1
1 1
2 2
∑ T j ( f j )
= N 2 χ[0,1] L p = N 2 .
j=1
p L
This implies that the inequality
N 1
2 2
∑ T j ( f j )
j=1
Lp
fails for p > 2. When p < 2 take f j = χ[0,1] for j = 1, 2, . . . , N. Then
N 1
2 2
∑ T j ( f j )
j=1
N 1
2 2 ≤ Cp ∑  f j 
j=1
Lp
1 = χ[1,N+1] L p = N p
Lp
but
N 1
2 2
∑  f j
j=1
1
= N 2 χ[0,1]
Lp
Lp
1
= N2.
So the inequality cannot hold for p < 2.
Exercise 5.5.7 Suppose that T is a linear operator that takes realvalued functions to realvalued functions. Use Theorem 5.5.1 (a) with p = q to prove that
T ( f ) p
T ( f ) p L
L . sup sup
f p = f complexvalued
f p f realvalued f 6=0
L
f 6=0
L
Solution. Let f1 , f2 be realvalued functions and consider the function f1 + i f2 . Then T ( f1 + i f2 ) = T ( f1 ) + iT ( f2 ), where each of T ( f1 ), T ( f2 ) are realvalued. Then q T ( f1 + i f2 ) = T ( f1 )2 + T ( f2 )2 . Let kT kL p →L p be the norm of T when restricted to realvalued L p functions. It follows from Theorem 5.5.1 (a) with p = q (in which case C p,q = 1 that
q
q
2 + T ( f )2 2 +  f 2
T ( f1 + i f2 ) p = p →L p T ( f ) ≤ kT k  f 
p = kT kL p →L p f1 + i f2 L p . L 1 2 1 2 L p L
L
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233
This proves the ≥ inequality, while the other one is trivial.
Section 5.6. VectorValued Singular Integrals Exercise 5.6.1 (a) For all j ∈ Z, let I j be an interval in R and let T j be the operator given on the Fourier transform by multiplication by the characteristic function of I j . Prove that there exists a constant C > 0 such that for all 1 < p, r < ∞ and for all square integrable functions f j on R we have
1 r
∑ T j ( f j )r
L p (R)
j
1 r
∑ T j ( f j )r
L1,∞ (R)
j
1 1 1 r
≤ C max r, max p,
∑  f j r p , r−1 p−1 L (R) j 1 1 r
≤ C max r,
∑  f j r 1 . r−1 L (R) j
(b) Let R j be arbitrary rectangles on Rn with sides parallel to the axes and let S j be the operators given on the Fourier transform by multiplication by the characteristic functions of R j . Prove that there exists a dimensional constant Cn < ∞ such that for all indices 1 < p, r < ∞ and for all square integrable functions f j in L p (Rn ) we have
1 r
∑ S j ( f j )r
1 1 n 1 n
r r max p,  f  ≤ C max r,
n ∑ j L p (Rn ) . r−1 p−1 L p (Rn ) j
j
Hint: Part (a). Use Theorem 5.5.1 and the identity T j = 2i M a HM −a − M b HM −b , if I j is χ(a,b) , where M a ( f )(x) = f (x)e2πiax and H is the Hilbert transform. Part (b). Apply the result in part (a) in each variable. Solution. Case 1: n = 1. If I j = (a j , b j ) for all j ∈ Z, then (b f j χI j )∨ (x) = i
e2πixa j H(e−2πi(·)a j f j ) − e2πixb j H(e−2πi(·)b j f j ) (x) 2
where H is the Hilbert transform. In view of this identity, to prove the required inequality, it will suffice to show that
1 2
∑ H(e−2πi(·)a j f j )2
L p (R)
j
1 2
2
≤ c(p) ∑  f j 
,
L p (R)
j
for families of square integrable functions f j on Rn . But this inequality is equivalent to
1 2
∑ H( f j )2
j
L p (R)
1 2
2
≤ c(p) ∑  f j 
L p (R)
j
The above identity is trivial when p = 2 since H is an isometry on L2 (R). We have now showed that the operator { f j } j∈Z 7→ H( f j ) j∈Z maps L2 (`2 (Z), `2 (Z)) to itself. This operator has the nice kernel ~ K(x)({c j } j∈Z ) =
nc o j
x
j∈Z
,
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which obviously satisfies the conditions of Theorem 5.6.1. It follows that for every p ∈ (1, ∞) the operator { f j } j∈Z 7→ H( f j ) j∈Z maps L p (`2 (Z), `2 (Z)) to itself with norm at most C max(p, (p − 1)−1 ). Case 2: n ≥ 2. Write the rectangle R(t) as R j = (a j , b j ) × R0j . We apply the inequality in the first variable freezing the remaining variables. We obtain Z R
∑ ( bf j χR j )∨ (x1 , x2 , . . . , xn )2 j
p 2
p h ip Z Z 2 u −1 t 2 ( f j χR0j ) (x1 , x2 , . . . , xn ) dx1 ≤ C max(p, (p−1) ) dx1 , R
T
where the the cup and inverse cup denote the Fourier and inverse Fourier transforms on the right in the variables x2 , . . . , xn . Next we freeze the x3 , . . . , xn variables and we apply the inequality in the x2 variable. We write R0j = (d j , e j ) × R00j and applying the inequality in part (a) in the second variable we obtain Z Z R R
∨
∑ ( bf j χR j )
2
p 2
(x1 , x2 , . . . , xn )
p h i2p Z Z Z 2 u −1 t 2 dx1 dx2 ≤ C max(p, (p−1) ) dx1 dx2 , ( f j χR00j ) (x1 , x2 , . . . , xn ) R R
j
T
where the cup and inverse cup now denote the Fourier and inverse Fourier transforms in the variables x3 , . . . , xn . Continue in this way to obtain Z Rn
∑ ( bf j χR j )∨ (x1 , x2 , . . . , xn )2
p 2
h inp Z −1 dx1 · · · dxn ≤ C max(p, (p−1) )
Rn
j
2
∑  f j (x1 , x2 , . . . , xn )
p 2
dx1 · · · dxn .
j
This procedure yields the inequality in all dimensions.
Exercise 5.6.2 Let (T, dµ) be a σ finite measure space. For every t ∈ T , let R(t) be a rectangle in Rn with sides parallel to the axes such that the map t 7→ R(t) is measurable. Then there is a constant Cn > 0 such that for all 1 < p < ∞ and for all families of square integrable functions { ft }t∈T on Rn such that t 7→ ft (x) is measurable for all x ∈ Rn we have
Z 1 2
∨ 2
b
T ( ft χR(t) )  dµ(t)
Lp
Z 1 2
2 ≤ Cn max(p, (p−1) )  ft  dµ(t)
p, T −1 n
L
Hint: When n = 1 reduce matters to an L p (L2 (T, dµ), L2 (T, dµ)) inequality for the Hilbert transform, via the hint in the preceding exercise. Verify the inequality p = 2 and then use Theorem 5.6.1 for the other p’s. Obtain the ndimensional inequality by iterating the onedimensional. Solution. Case 1: n = 1. If R(t) = (at , bt ) for all t ∈ T , where t 7→ at , t 7→ at are measurable, then ( bft χR(t) )∨ (x) = i
e2πixat H(e−2πi(·)at ft ) − e2πixbt H(e−2πi(·)bt ft ) (x) 2
where H is the Hilbert transform. In view of this identity, to prove the required inequality, it will suffice to show that
Z 1 2
−2πi(·)at 2
H(e f ) dµ(t) t
T
L p (R)
Z 1 2
2
≤ c(p)  ft  dµ(t)
T
L p (Rn )
,
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235
for families of square integrable functions ft on Rn such that t 7→ ft (x) is measurable in t for all x ∈ Rn . But this inequality is equivalent to
Z
Z 1 1 2 2
2 2
T H( ft ) dµ(t) p ≤ c(p) T  ft  dµ(t) p n , L (R)
L (R )
We first show the above identity when p = 2. This is trivial and can be easily obtained by Fubini’s theorem (which is applicable since T is σ finite) and the fact that H is an isometry on L2 (R). We have now showed that the operator { ft }t∈T 7→ H( ft ) t∈T maps L2 (L2 (T, dµ), L2 (T, dµ)) to itself. This operator has a nice kernel which obviously satisfies the conditions of Theorem 5.6.1. It follows that for every p ∈ (1, ∞) the operator { ft }t∈T 7→ H( ft ) t∈T maps L p (L2 (T, dµ), L2 (T, dµ)) to itself with norm at most a constant multiple of max(p, (p − 1)−1 ). Case 2: n ≥ 2. Write the rectangle R(t) as R(t) = (at , bt ) × R0 (t) . We apply the inequality in the first variable freezing the remaining variables. We obtain Z Z R
T
p p h ip Z Z 2 2 u −1 t 2 ∨ 2 b ( ft χR0 (t) ) (x1 , x2 , . . . , xn ) dµ(t) dx1 , ( ft χR(t) ) (x1 , x2 , . . . , xn ) dµ(t) dx1 ≤ C max(p, (p−1) ) R
T
where the Fourier and inverse Fourier transforms on the right are in the variables x2 , . . . , xn . Next we freeze the x3 , . . . , xn variables and we apply the inequality in the x2 variable. Continue in this way to obtain Z Rn
Z T
p h inp Z 2 ∨ 2 b ( ft χR(t) ) (x1 , x2 , . . . , xn ) dµ(t) dx1 · · · dxn ≤ C max(p, (p−1)−1 )
Rn
Z T
p 2  ft (x1 , x2 , . . . , xn ) dµ(t) dx1 · · · dxn . 2
This procedure yields the inequality in all dimensions.
Exercise 5.6.3 Let Φ be a function on Rn that satisfies supx∈Rn xn Φ(x) ≤ A and Z Rn
Φ(x − y) − Φ(x) dx ≤ η(y),
Z x≥R
Φ(x) dx ≤ η(R−1 ) ,
for all R ≥ 1, where η is a continuous increasing function on [0, 2] that satisfies η(0) = 0 and 02 η(t) t dt < ∞ . (a) Prove that (5.6.27) holds. (b) Show that if Φ lies in L1 (Rn ), then the maximal function f 7→ sup j∈Z  f ∗ Φ2 j  maps L p (Rn ) to itself for 1 < p ≤ ∞. Hint: Part (a). Modify the calculation in the proof of Theorem 5.6.6. Part (b). Use Theorem 5.6.1 with r = ∞. R
Solution. (a) We show that (5.6.27) holds, i.e. we show that Z
sup
sup Φ2 j (x − y) − Φ2 j (x) dx = Cn < ∞ ,
y∈Rn \{0} x≥2y j∈Z
We have Z
sup Φ2 j (x − y) − Φ2 j (x) dx
x≥2y j∈Z
≤
Z
∑
j∈Z x≥2y
Φ2 j (x − y) − Φ2 j (x) dx
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≤
Z
∑
2 j >y x≥2y
≤ ≤
Φ2 j (x − y) − Φ2 j (x) dx +
Z
∑
2 j ≤y x≥2y
Φ(x0 − 2− j y) − Φ(x0 ) + 2
∑ j
0 j+1 2 >y x ≥2 y
∑
η(2− j y) + 2
∑
η(2− j y) + 2
2 j >y
≤
Z
Z
∑
−j 2 j ≤y x≥2 y
∑
(Φ2 j (x − y) + Φ2 j (x)) dx
Z
∑ j
2 ≤y x≥y
Φ2 j (x) dx
Φ(x) dx
η(2 j y−1 )
2 j ≤y
2 j >y
Now we pick j0 ∈ Z such that 2 j0 −1 ≤ y < 2 j0 . We estimate the preceding expression by Z 2 j+1
2
∑
j≥ j0
2j
η(y/t)
dt +4 ∑ t j< j0
Z 2 j+1 2j
η(t/y)
dt =2 t
Z ∞ 2 j0 −1
η(y/t)
Z 2y/2 j0
=2
η(t) 0
≤6
Z 2
η(t) 0
dt +4 t
dt +4 t
Z 2 j0
η(t/y) 0
Z 2 j0 /y
η(t) 0
dt t
dt t
dt < ∞. t
(b) We consider the Banachvalued operator f 7→ sup  f ∗ Φ2 j  j∈Z
which maps L∞ (Rn ) to L∞ (Rn , `∞ (Z)), since Φ is integrable. The kernel of this operator is ~ Φ (x)(c) = {c Φ2 j (x)}i∈Z . K Clearly we have
~ KΦ (x) C→`∞ (Z) = sup Φ2 j (x) . j∈Z
~ Φ . Also condition (5.6.1) holds since by assumption Now the conclusion in part (a) implies condition (5.6.2) for the kernel K we have Φ(x) ≤ Ax−n , hence sup Φ2 j (x) ≤ A 2− jn 2− j x−n = A x−n j∈Z
and also condition (5.6.3) holds since for every j ∈ Z we have Z
lim ε→0 ε≤y≤1
Z
Φ2 j (y) dy =
y≤1
Φ2 j (y) dy = L j .
Then Theorem 5.6.6 applies and yields the L p (Rn ) to L p (Rn , `∞ (Z)) boundedness of the operator f 7→ sup j∈Z  f ∗ Φ2 j .
Exercise 5.6.4 (a) On R, take f j = χ[2 j−1 ,2 j ] to prove that inequality (5.6.25) fails when p = ∞ and 1 < r < ∞. (b) Again on R, take N > 2 and f j = χ[ j−1 , j ] for j = 1, 2, . . . , N to prove that (5.6.25) fails when 1 < p < ∞ and r = 1. N
N
Solution. Note that (5.6.25) says that
1 r
∑ M( f j )r j
Lp
1 r
≤ Cn c(p, r) ∑  f j r p .
We need to disprove this inequality in certain endpoint cases.
j
L
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237
(a) Suppose p = ∞ and 1 < r < ∞. 1 On R, take f j = χ[2 j−1 ,2 j ] . We have (∑Nj=1 f jr ) r = χ[1,2N ] which has L∞ norm 1. However, for j = 1, 2, . . . , N we have Z x+t
1 2t
M(χ[2 j−1 ,2 j ] )(x) = sup t>0
x−t
1 χ[2 j−1 ,2 j ] (y)dy ≥ χ[0,2 j ] (x), 4
(t = 2 j ).
This estimate implies that !1 r
N r
N
≥
∑ M(χ[2 j−1 ,2 j ] )
j=1
1 r j] ∑ 4 χ[0,2 j=1
!1
1
r
Nr χ . 4 [0,2]
≥
1 r Therefore, L∞ norm of ∑Nj=1 M(χ[2 j−1 ,2 j ] )r tends to infinity when N approaches infinity. (b) Suppose r = 1 and 1 < p < ∞. Take f j = χ[ j−1 , j ] for j = 1, 2, . . . , N. Then ∑Nj=1 f j = χ[0,1] has L p norm equal to 1. In contrast, N
N
1 M(χ[ j−1 , j ] )(x) = sup N N 2t t>0 Therefore
Z x+t x−t
χ[ j−1 , j ] (y)dy ≥ N
N
N
1 χ (x), 2[N(x − 1) + 1] [1,∞) N
, j )≥ ∑ M(χ[ j−1 N ,N ] 2[N(x − 1) + 1]
(t = x −
j−1 N ).
x ≥ 1.
j=1
Taking L p norm we have
N
∑ M(χ[ j−1 , j ] )
j=1 N N
≥
Z
∞
1
Lp
dx [(x − 1) + N −1 ] p
1
Z
p
= 0
∞
dx [x + N −1 ] p
1
p
=N
p−1 p
which tends to infinity as N → ∞.
Exercise 5.6.5 Let K be an integrable function on the real line and assume that the operator f 7→ f ∗K is bounded on L p (R) for some 1 < p < ∞. Prove that the vectorvalued inequality
1 q
∑ K ∗ f j q
Lp
j
1 q
≤ C p,q ∑  f j q
Lp
j
may fail in general when q < 1. Hint: Take K = χ[−1,1] and f j = χ[ j−1 , j ] for 1 ≤ j ≤ N. N
N
Solution. q Take K = χ[−1,1] and f j = χ[ j−1 , j ] for j = 1, 2, . . . , N. Then we have ∑Nj=1 f j = χ[0,1] and thus N
N
N 1 q
∑  f j q j=1
Lp
=1
for all N = 1, 2, . . . . On the other hand, K ∗ f j (x) =
Z
K(x − y) f j (y)dy =
Z
j N j−1 N
χ[−1,1] (x − y)dy =
Z
j N j−1 N
χ[x−1,x+1] (y)dy ≥
1 χ j−1 j (x). N [ N ,N ]
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Therefore
!1 q
N
q ∑ K ∗ f j
N
1 ∑ N q χ[qj−1 j N ,N ] j=1
≥
j=1
!1
q
≥N
1−q q
χ[0,1] ,
and taking L p norms over R we obtain that
1 q
∑ K ∗ f j q
Lp
j
≥N
1−q q
→∞
as N approaches infinity, since q < 1. This provides a counterexample.
Exercise 5.6.6 Let {Q j } j be a countable collection of cubes in Rn with disjoint interiors. Let c j be the center of the cube Q j and d j its diameter. For ε > 0, define the Marcinkiewicz function with the family {Q j } j as follows: Mε (x) = ∑ j
d n+ε j x − c j n+ε + d n+ε j
.
Prove that for some constants Cn,ε,p and Cn,ε one has 1
Mε p ≤ Cn,ε,p ∑ Q j  p , L
p>
j
Mε
n L n+ε ,∞
≤ Cn,ε
n+ε
∑ Q j 
n
n , n+ε
,
j
and consequently Z Rn
Mε (x) dx ≤ Cn,ε ∑ Q j  . j
Hint: Verify that d n+ε j x − c j and use Theorem 5.6.6.
n+ε
+ d n+ε j
≤ CM(χQ j )(x)
n+ε n
Solution.
1 n+ε n+ε Fix x ∈ Rn . Let R = 2 x − c j + d n+ε . The definition of the HardyLittlewood maximal function gives us j
Z B(x, R) ∩ Q j 1 M(χQ j )(x) ≥ n χQ (y)dy ≥ . R ωn B(x,R) j Rn ωn Notice that R ≥ 2 max x − c j , d j ≥ x − c j + d j . Therefore Q j ⊂ B(x, R). This implies that Q j d nj M(χQ j )(x) ≥ n = n . n+ε R n+ε Cn x − c j + d n+ε j Raising to the power of
n+ε n
on both sides of the above inequality, we have
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239
d n+ε n+ε j ≤ Cn,ε M(χQ j (x) n . n+ε n+ε x − c j +dj Thus, we have just proved that n+ε Mε (x) ≤ Cn,ε ∑ M(χQ j (x) n . j
We now apply Theorem 5.6.6 to deduce that
n n
n+ε kMε kLn+ε
p(n+ε) p = Mε L n
! n
n+ε
n+ε
n M(χ ) ≤ Qj
∑
j
p(n+ε) L n
n !
n+ε
n+ε
n ≤Cn,ε,p
∑(χQ j )
j
p(n+ε) L n
n !
n+ε
=Cn,ε,p
∑ χQ j
j
p(n+ε) L n
n n+ε
=Cn,ε,p ∑ χQ j
j
p L
for p >
n n+ε .
Hence kMε kL p
≤ Cn,ε,p ∑ χQ j
j
!1 p ∑ Qj .
= Cn,ε,p
j
Lp
In particular, Z
Mε (x)dx ≤ Cn,ε ∑ Q j .
Rn
j
To prove the inequality
Mε
n L n+ε ,∞
≤ Cn,ε
n+ε
∑ Q j 
n
j
we use (5.6.24) combined with the fact that 1
kgkLr,∞ = kgr kLr 1,∞ with r =
n n+ε .
Chapter 6. LittlewoodPaley Theory and Multipliers
Section 6.1. LittlewoodPaley Theory Exercise 6.1.1 b (2− j ξ )2 = 1 for all ξ ∈ Rn \ {0} and whose Fourier transform is Construct a Schwartz function Ψ that satisfies ∑ j∈Z Ψ 6 . supported in the annulus 7 ≤ ξ  ≤ 2 and is equal to 1 on the annulus 1 ≤ ξ  ≤ 14 −1/2 7 ∞ n −k 2 b Hint: Set Ψ (ξ ) = η(ξ ) ∑k∈Z η(2 ξ ) for a suitable η ∈ C (R ) . 0
Solution. Consider the smooth function
( − 1 e x(1−x) η(x) = 0
x ∈ (0, 1) x∈ / (0, 1).
Set λ = 01 η(x)dx and define the function γ(x) = λ −1 0x η(t)dt, for x ∈ R. It is easy to see that 0 ≤ γ(x) ≤ 1 for all x ∈ R, γ(x) = 0 for x ≤ 0 and γ(x) = 1 for all x ≥ 1. Now fix 0 < c < a < b < d < ∞ and define the function x − b x−c 1−γ , x ∈ R. ψ(x) = γ a−c d −b R
R
It is clear that 0 ≤ ψ(x) ≤ 1 for all x ∈ R, ψ(x) = 0 for x ∈ (−∞, c] ∪ [d, ∞) and ψ(x) = 1 for x ∈ [a, b]. Finally, the C ∞ function ϕ(x) = ψ(x2 ),
x ∈ Rn √ √ √ √ is equal to 1 on the annulus a ≤ x ≤ b and vanishes off the annulus c ≤ x ≤ d. Now fix a smooth function η(ξ ) supported in the annulus 76 ≤ ξ  ≤ 2 and is equal to 1 on the annulus 1 ≤ ξ  ≤ b (ξ ) = η(ξ ) Ψ
∑ η(2−k ξ )2
14 7 .
Set
−1/2
k∈Z
and notice that ∑
η(2−k ξ )2
k∈Z =1 ∑ Ψb (2− j ξ )2 = ∑k∈Z η(2−k ξ )2
j∈Z
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Exercise 6.1.2 0
b (ξ ) ≤ B min(ξ ε , ξ −ε ) for some ε 0 , ε > 0. Show that for Suppose that Ψ is an integrable function on Rn that satisfies Ψ some constant Cε,ε 0 < ∞ we have Z sup ξ ∈Rn
0
∞
b (tξ )2 dt Ψ t
1 2
+ sup
∑ Ψb (2− j ξ )2
ξ ∈Rn
1 2
≤ Cε,ε 0 B .
j∈Z
Solution. b (0) = 0, so both the sum and the integral vanish. So we may assume Fix a x ∈ Rn . If ξ = 0, the hypothesis implies that Ψ that ξ 6= 0. We split the integral in the region where t < ξ −1 and where t ≥ ξ −1 . We obtain Z ∞ 0
b (tξ )2 Ψ
dt ≤ B2 t
Z ty−1 x≥2y
Ψ2− j (x − y) −Ψ2− j (x) dx =
Z
2 jn Ψ (2 j x − 2 j y) −Ψ (2 j x)dx
∑
2 j >y−1 x≥2y
Z
=
∑
x ≥2y 2j
2 j >y−1
≤
Ψ (x − 2 j y) −Ψ (x) dx
Z
j
∑
j j 2 j >y−1 x−2 ≥2 y
≤2
Z
∑
j 2 j >y−1 x≥2 y
2B
∑
2 j >y−1
x≥2 j+1 y
Ψ (x) dx
Ψ (x) dx
Z
≤
Ψ (x − 2 y) dx +
Z
x≥2 j y
x−n−ε dx
j −ε 2 y ≤ 2B ∑ ε 2 j >y−1 ≤
2B ε
∞
1
∑ 2 jε
j=0
Then (6.1.4) and (6.1.5) follow from Theorem 6.1.2, since we already have the p = 2 case. b (ξ ) as in part (a). In case where ξ  ≥ 1 the estimate is the same, that is Ψ b (ξ ) ≤ Bξ −ε 0 . (c) We return to the estimate for Ψ Suppose now that ξ  < 1. Then Z Z ε −2πix·ξ Bx−n−ε 2πx ξ  dx ≤ B(2π)ξ  ≤ B 2π ξ  2 , Ψ (x)(e − 1) dx ≤ x≤1
x≤1
where the integral converges since −n − ε + 1 > −n. Also, the last inequality uses that ξ  < 1 and 1 > ε2 . Also Z Z Z ε ε ε ε ε −2πix·ξ 1− ε2 ≤ 2 dx ≤ B21− 2 (2πξ ) 2 Ψ Ψ (x)(e − 1) dx (x) 2 x−n−ε x 2 dx = C0 B ξ  2 . (2πxξ ) x≥1
x≥1
x≥1
This proves that ε
b (ξ ) ≤ c0 0 Bξ min(1, 2 ) Ψ n,ε,ε b (ξ ) ≤ Bξ −ε 0 which is valid for ξ  ≥ 1, we obtain when ξ  < 1. Combining this with Ψ 0
b (ξ ) ≤ cn,ε,ε 0 B min(ξ ε/2 , ξ −ε ) Ψ for some constant cn,ε,ε 0 . Thus Exercise 6.1.2 applies. This gives that the inequality in the case p = 2 holds, while for the remaining p’s is the same.
Exercise 6.1.4 Let Ψ be an integrable function on Rn with mean value zero that satisfies Ψ (x) ≤ B(1 + x)−n−ε ,
Z Rn
for some B, ε 0 , ε > 0 and for all y 6= 0. Let Ψt (x) = t −nΨ (x/t). (a) Prove that there are constants cn , c0n such that
0
Ψ (x − y) −Ψ (x) dx ≤ Byε ,
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Z
∞
0
dt dx Ψt (x) t 2
Z
Z
∞
sup y∈Rn \{0}
x≥2y
0
1 2
≤ cn B x−n ,
Ψt (x − y) −Ψt (x)2
dt t
1 2
dx ≤ c0n B .
(b) Show that there exist constants Cn ,Cn0 such that for all 1 < p < ∞ and for all f ∈ L p (Rn ) we have ∞
Z
0
 f ∗Ψt 2
dt 12
p n ≤ Cn B max(p, (p − 1)−1 ) f L p (Rn ) t L (R )
and also for all f ∈ L1 (Rn ) we have
Z
∞
0
(c) Under the additional hypothesis that 0
y. In the second region use that Ψ is bounded to replace the square by the first power. Part (b): Use Exercise 6.1.2 and part (a) of Exercise 6.1.3 and todeduce the inequality when p = 2. Then apply Theorem 5.6.1. Part (c): Prove the inequality first for f ∈ S (Rn ) using duality. Solution. (a) Obviously we have Z
∞
0
Ψt (x)2
dt dx t
1 2
Z ≤B
∞
0
1 Z ∞ 1 1 x −2n−2ε dt 2 1 1 −2n−2ε dt 2 −n 1 + = B 1 + x = cn B x−n , 2n t 2n t t t t 0 t
where the convergence of the integral is obtained by splitting it in t ≤ 1 and t ≥ 1. Next, we must show that Z ∞ 1 Z 2 2 dt Ψt (x − y) −Ψt (x) dx ≤ Cn0 B . t x≥2y 0 To do this, estimate the left hand side by the CauchySchwarz inequality as follows: Z x≥2y
Z 0
∞
Ψt (x − y) −Ψt (x)2
dt t
1/2 dx ≤
Z
x−n−ε dx
x≥2y ε
1 Z 2
x≥2y
xn+ε
Z ∞ 0
Ψt (x − y) −Ψt (x)2
The first term of the product is equal to a multiple of y− 2 . To prove the inequality it will suffice to show that Z ∞ Z dt n+ε 2 x Ψt (x − y) −Ψt (x) dx ≤ C0 B2 yε . t 0 x≥2y Changing variables z = x/t, the above can be written as
dt dx t
1 2
.
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Z ∞ Z 0
dt y n+ε 2 ≤ C0 B2 yε . z Ψ (z − ) −Ψ (z) dz tε t t z≥2 yt 
Since z ≥ 2 yt  we have Ψ (z − yt ) −Ψ (z) ≤ Ψ (z − yt ) + Ψ (z) ≤ Bz − yt −n−ε + Bz−n−ε ≤ B( 12 z)−n−ε + Bz−n−ε = c B z−n−ε . We use this estimate in the part of the integral where t ≤ y. We obtain Z y Z 0
z≥2 yt 
zn+ε Ψ (z − yt ) −Ψ (z)2 dzt ε
dt . B2 t
Z y Z z≥2 yt 
0
dt ≈ B2 t
zn+ε z−2n−2ε dzt ε
Z y y dt  −(2ε−ε)t ε ≈ B2 yε . 0
t
t
In the part of the integral where t ≥ y we use that Ψ is bounded by B to write Z ∞Z y
z≥2 yt 
Ψ (z − yt ) −Ψ (z)2 dzt ε
dt ≤ 2B t
Z ∞Z z≥2 yt 
y
Ψ (z − yt ) −Ψ (z) dzt ε
dt ≤ 2B t
Z ∞ ε 0 dt y B t ε ≈ B2 yε , y
t
t
where we used the second assumption on Ψ . (b) As in the Exercise 6.1.3 we obtain 0
ε
b (ξ ) ≤ cn,ε,ε 0 B min(ξ min(1, 2 ) , ξ −ε ) . Ψ Then Exercise 6.1.2 yields that Z sup 0
ξ ∈Rn
∞
b (tξ )2 dt Ψ t
1 2
≤ Cε,ε 0 B .
Thus the inequality holds when p = 2. To obtain boundedness for the remaining p’s we consider this operator: ~T ( f )(x) = { f ∗Ψt (x)}t>0 . We must prove that ~T maps L p (Rn , C) to L p (Rn , L2 ((0, ∞), dtt )) and L1 (Rn , C) to L1,∞ (Rn , L2 ((0, ∞), dtt )) with the appropriate constants. We will achieve this via Theorem 5.6.1. We have that ~T maps L2 (Rn , C) to L2 (Rn , L2 ((0, ∞), dtt )). We also have Z Z ~T ( f )(x) = ~ − y) f (y)dy , Ψt (x − y) f (y)dy = K(x Rn
Rn
t>0
~ is given by K(x)(a) ~ where the kernel K = {Ψt (x)a}t>0 . We see that ~ kK(x)k C→L2 ((0,∞), dt ) =
Z
t
2 dt
∞
0
Ψt (x)
1/2
t
.
In part (a) we proved Z
sup y6=0 x≥2y
kK(x − y) − K(x)kC7→L2 ((0,∞), dt ) dx ≤ c0n B t
also −n ~ kK(x)k C→L2 ((0,∞), dt ) ≤ cn B x , t
and also Z
~ K(y) dy =
lim
Z 0
ε→0 ε≤y≤1
~ 0 acts on a c ∈ C via and K
1
Ψt (y) dy t>0
Z1 ~ 0 (c) = c Ψt (y) dy K 0
t>0
.
~0 =K
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Thus the hypotheses of Theorem 5.6.1 are satisfied. We are now apply Theorem 5.6.1 and the conclusion follows. (c) We are assuming that Z ∞ b (tξ )2 dt = c0 Ψ t 0 for all ξ ∈ Rn . Let f , g be Schwartz functions. Then all the interchanges of the integrals in the displayment below are justified. We have Z Z b g(ξ ) dξ Rn f (x)g(x) dx = Rn f (ξ )b Z Z ∞ 1 2 dt b b = Ψ (tξ ) f (ξ )b g(ξ ) dξ n c0 R 0 t Z Z 1 ∞ dt 2b b = Ψ (tξ ) f (ξ )b g(ξ ) dξ c0 0 Rn t Z ∞ Z 1 dt b b b = Ψ (tξ ) f (ξ )Ψ (tξ ) gb(ξ ) dξ c0 0 Rn t Z ∞ Z 1 dt = (Ψt ∗ f )(x)(Ψt ∗ g)(x) dx c0 0 Rn t Z Z ∞ 1 dt = Qt ( f )(x) Qt (g)(x) dx c0 Rn 0 t Z Z ∞ 1 Z ∞ 1 dt dt 12 2 ≤ Qt ( f )(x)2 Qt (g)(x)2 dx c0 Rn 0 t t 0
Z ∞ Z ∞ 1 dt 12 dt 12
≤ Qt ( f )2 Qt (g)2
p
p0 c0 t t L L 0 0
Z ∞ 1
dt 2
≤ Qt ( f )2
p Cn B max p0 , (p0 − 1)−1 g L p0 , t L 0 having used the definition of the adjoint, the Cauchy–Schwarz inequality, H¨older’s inequality, and the upper estimate obtained 0 in part (b). Taking the supremum over all g ∈ S (Rn ) with L p norm at most one, we obtain that
f
Lp
Z
≤ Cn B max p, (p − 1)−1
0
∞
Qt ( f )2
dt 12
p. t L
for all f ∈ S (Rn ). We now extend this inequality to all functions f ∈ S (Rn ). Let S( f ) = find a sequence of functions fk ∈ S (Rn ) such that k f − fk kL p → 0 as k → ∞. Then
R∞ 0
Qt ( f )2 dtt
1
2
. Given f ∈ L p (Rn )
k f kL p ≤ k f − fk kL p + k fk kL p ≤ k f − fk kL p +Cn B max p, (p − 1)−1 kS( fk )kL p ≤ k f − fk kL p +Cn B max p, (p − 1)−1 kS( fk − f )kL p +Cn B max p, (p − 1)−1 kS( f )kL p h i2 ≤ k f − fk kL p + Cn B max p, (p − 1)−1 k fk − f kL p +Cn B max p, (p − 1)−1 kS( f )kL p and letting k → ∞ we obtain the claimed inequality.
Exercise 6.1.5 Prove the following generalization of Theorem 6.1.2. Let A > 0. Suppose that {K j } j∈Z is a sequence of locally integrable functions on Rn \ {0} that satisfies 1 2 sup xn ∑ K j (x)2 ≤ A , x6=0
j∈Z
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Z
sup y∈Rn \{0} x≥2y
∑ K j (x − y) − K j (x)2
1 2
dx ≤ A < ∞ ,
j∈Z
and for each j ∈ Z there is a number L j such that Z
lim
εk ↓0 εk ≤y≤1
K j (y) dy = L j .
If the K j coincide with tempered distributions W j that satisfy cj (ξ )2 ≤ B2 , ∑ W
j∈Z
then the operator f→
∑ K j ∗ f 2
1 2
j∈Z
maps
L p (Rn )
to itself and is weak type (1, 1) norms at most multiples of A + B.
Solution. ~ ∈ L(C, `2 (Z)) defined as follows: We define a vectorvalued operator ~T defined on L2 (Rn , C) with kernel K ~ K(x)(a) = aK j (x) j . Then we have ~ `2 (Z)→`2 (Z) = kKk
∑ Kcj (ξ )2
1
2
j∈Z
and by hypothesis we have cj (ξ )2 ≤ B2 ∑ W
j∈Z
which implies by Plancherel’s theorem that
!1/2
2
∑ K j ∗ f 
j∈Z
. k f kL 2
L2
i.e., ~T is bounded from L2 (Rn , C) to L2 (Rn , `2 (Z)). The other conditions of Theorem 5.6.1 are also satisfied, hence the conclusion of this theorem implies the claim of the exercise.
Exercise 6.1.6 Suppose that H is a Hilbert space with inner product h · , · iH . Let A > 0 and 1 < p < ∞. Suppose that an operator T from L2 (Rn ) → L2 (Rn , H ) is a multiple of an isometry, that is,
T (g) 2 n = A g 2 n L (R ,H )
L (R )
for all g ∈ L2 (Rn , H ). Then the inequality kT ( f )kL p (Rn ,H ) ≤ C p k f kL p (Rn ) for all f ∈ S (Rn ) implies
f
0
L p (Rn )
≤ C p0 A−2 T ( f ) L p0 (Rn ,H )
n for all in f ∈ S (R ). Hint: Use the inner product structure and polarization to obtain
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Z Z D E T ( f )(x), T (g)(x) dx f (x)g(x) dx = A2 H Rn Rn and then argue as in the proof of inequality (6.1.8). Solution. We first show that
Z Z hT ( f )(x), T (g)(x)iH dx . f (x)g(x) dx = A Rn Rn 2
We see that A2 which implies Z A2
Rn
 f 2 dx +
Z Rn
Z Rn
Z
( f + g)( f + g) dx =
Rn
hT ( f + g), T ( f + g)iH dx ,
Z hT ( f ), T (g)iH dx . g2 dx + 2Re h f , gi = kT ( f )k2L2 (Rn ,H ) + kT (g)k2L2 (Rn ,H ) + 2Re Rn
This implies A2 Re h f , gi = Re
Z Rn
hT ( f ), T (g)iH dx.
The imaginary case follows similarly by considering f + ig in place of f + g. We can now use a duality argument to solve the problem. For f , g ∈ S (Rn ) we have Z Z = 1 hT ( f )(x), T (g)(x)iH dx f (x)g(x) dx Rn A 2 Rn Z 1 ≤ 2 kT f (x)kH kT g(x)kH dx A Rn 1 ≤ 2 kT ( f )kL p0 (Rn ,H ) kT (g)kL p (Rn ,H ) A 1 ≤ 2 kT ( f )kL p0 (Rn ,H ) kT (g)kL p (Rn ,H ) A Taking the supremum over all g ∈ S (Rn ) with kgkL p ≤ 1 gives us the claimed inequality.
Exercise 6.1.7 Suppose that {m j } j∈Z is a sequence of bounded functions supported in the intervals [2 j , 2 j+1 ]. Let T j ( f ) = ( fbm j )∨ be the corresponding multiplier operators. Assume that for all sequences of functions { f j } j the vectorvalued inequality
1 2
∑ T j ( f j )2
Lp
j
1 2
≤ A p ∑  f j 2 j
Lp
is valid for some 1 < p < ∞. Prove there is a C p > 0 such that for all finite subsets S of Z we have
∑ m j ≤ Cp A p . j∈S
Hint: Use that ∑ j∈S T j ( f ), g = ∑ j∈S ∆ #j T j ( f ), ∆ #j (g) . Solution. By definition, we have
Mp
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∑ m j j∈S
Mp
!∨
b = sup f ∑ m j
p
k f kL p ≤1 S L Z ∨ b = sup sup ( f ∑ m j ) g dx k f kL p ≤1 kgk p0 ≤1 Rn S L Z ( fbm j )∨ g dx = sup sup ∑ k f kL p ≤1 kgk p0 ≤1 S Rn L Z # # ∆ j T j ( f ) ∆ j (g)dx = sup sup ∑ k f kL p ≤1 kgk p0 ≤1 S Rn L Z # 2 1/2 # 2 1/2 ≤ sup sup (∑ ∆ j T j ( f ) ) (∑ ∆ j (g) ) dx k f kL p ≤1 kgk p0 ≤1 Rn S S L
≤ sup sup (∑ ∆ #j T j ( f )2 )1/2 p (∑ ∆ #j (g)2 )1/2 p0 k f kL p ≤1 kgk p0 ≤1 L
≤
L
S
S
L
sup (∑ T j ( f )2 )1/2 p C p g L p0
sup
k f kL p ≤1 kgk p0 ≤1 L
S
L
≤ A pC p , where we used Theorem 6.1.5 and Exercise 5.6.1 in the penultimate inequality.
Exercise 6.1.8 ∨ Let m be a bounded function on Rn that is supported in the annulus 1 ≤ ξ  ≤ 2 and define T j ( f ) = fb(ξ )m(2− j ξ ) . Suppose 1/2 that the square function f 7→ ∑ j∈Z T j ( f )2 is bounded on L p (Rn ) for some 1 < p < ∞. Show that for every finite subset S of the integers we have
≤ C p,n f p n
∑ T j ( f ) j∈S
L p (Rn )
L (R )
for some constant C p,n independent of S. Solution. The proof is just like the one in the preceding exercise with the only difference being that the role of ∆ #j is played by another LittlewoodPaley operator ∆ j associated with a fixed bump Ψ , whose Fourier transform is equal to one on the annulus 1 ≤ ξ  ≤ 2 and vanishes off a slightly bigger annulus 67 ≤ ξ  ≤ 15 7 . # We have that T j ( f ) = ∆ j T j ( f ), hence
∑ m j j∈S
Mp
!∨
b
= f ∑mj
p n S L (R ) Z ∨ b = sup ( f ∑ m j ) g dx kgk p0 ≤1 Rn S L Z = sup ∑ ( fbm j )∨ g dx kgk p0 ≤1 S Rn L Z = sup ∑ ∆ j T j ( f ) ∆ j (g)dx kgk 0 ≤1 S Rn Lp
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Z 2 1/2 2 1/2 ≤ sup (∑ ∆ j T j ( f ) ) (∑ ∆ j (g) ) dx kgk p0 ≤1 Rn S S L
≤ sup (∑ ∆ j T j ( f )2 )1/2 p n (∑ ∆ j (g)2 )1/2 kgk p0 ≤1 L
≤
L (R )
S
S
0
L p (Rn )
sup (∑ T j ( f )2 )1/2 p C p,n g L p0 (Rn )
kgk p0 ≤1 L
L
S
≤ A pC p,n k f kL p (Rn ) ,
where in the penultimate inequality we used Theorem 6.1.5 and (6.1.22).
Exercise 6.1.9 Fix a nonzero Schwartz function h on the line whose Fourier transform is supported in the interval − 81 , 18 . For {a j } a sequence of numbers, set ∞
f (x) =
j
∑ a j e2πi2 x h(x) .
j=1
Prove that for all 1 < p < ∞ there exists a constant C p such that
f
L p (R)
≤ Cp
∑ a j 2 j
1 2 h Lp .
j Hint: Write f = ∑∞j=1 ∆ j (a j e2πi2 (·) h), where ∆ j is given by convolution with ϕ2− j for some ϕ whose Fourier transform is 6 10 supported in the interval 8 , 8 and is equal to 1 on 78 , 89 . Then use (6.1.21). Solution. We first consider a Schwartz function ϕ whose Fourier transform is supported in the interval 68 , 10 8 , and is equal to 1 7 9 10 6 on 8 , 8 . We consider the even extension of ϕ on the interval − 8 , − 8 , so we have a function ϕ that is supported in 10 6 6 10 − 8 , − 8 ∪ 8 , 8 . We consider the LittlewoodPaley operator ∆ j , where ∆ j ( f ) = f ∗ ϕ2− j and ϕ2− j (x) = 2 j ϕ(2 j x). We first show that f = ∑∞j=1 ∆ j ( f j ) for all x. We can do this by showing that b f j = ∆\ j ( f j ) for all j, since these are both Schwarz and the Fourier transform is bijective on the Schwartz class. Now, by the properties of the Fourier transform, we have: jx \ b f j (ξ ) = a j e2π2 h(ξ ) = a j b h(ξ − 2 j )
as well as: j b −j b b d ∆\ ξ) j ( f j )(ξ ) = f \ j ∗ ϕ2− j (ξ ) = f (ξ ) ϕ 2− j (ξ ) = a j h(ξ − 2 )ϕ(2
Now, we need to show ϕ(2− j ξ ) = 1 on the support of b h(ξ − 2 j ). j 7 j 9 j − j We know that ϕ(2 ξ ) = 1 on 2 8 , 2 8 = Ξ , and supp(τ 2 b h) ⊂ − 18 + 2 j , 18 + 2 j = Y . We only need to show that Y ⊂ Ξ . First, it’s easy to see that − 81 + 2 j > 2 j 78 , since this is tantamount to − 18 > −2 j 18 for all j ∈ Z+ . Likewise, it’s clear that 1 j j9 8 + 2 < 2 8 . So we have that Y ⊂ Ξ . Thus, we have b f j = ∆\ j ( f j ), so f j = ∆ j ( f j ). Since we have that ϕ is radial, has a realvalued Fourier transform, and satisfies ϕ(x) ≤ B(1 + x)−n−1 for some B > 0 by the virtue of the fast decay of Schwartz functions. Applying the result in Remark 6.1.3 we obtain the relation:
!
∞ ∞
−1 2 k f kL p = k ∑ ∆ j ( f j )kL p ≤ Cn B max p, (p − 1)
∑  f j
p j=1 j=1 L
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Setting C p = Cn B max p, (p − 1)−1 , we get the following relation:
!2
∞
j 2πi2 x
h(x) k f kL p ≤ C p ∑ a j e
j=1 p
!2
∞
= Cp
∑ a j h(x)
j=1 p !1/2 ∞
∑ a j 2
= C p kh(x)
kp
j=1
!1/2
∞
2
khkL p .
∑ a j 
= Cp
j=1
Exercise 6.1.10 Let Ψ be a Schwartz function whose Fourier transform is supported in the annulus 21 ≤ ξ  ≤ 2 and that satisfies (6.1.7). Define a Schwartz function Φ by setting ( b (2− j ξ ) when ξ 6= 0, ∑ j≤0 Ψ b Φ(ξ ) = 1 when ξ = 0. Let S0 be the operator given by convolution with Φ. Let 1 < p < ∞ and f ∈ L p (Rn ). Show that
f
Lp
∞ 1
2
≈ S0 ( f ) L p + ∑ ∆ j ( f )2 p . L
j=1
Hint: Use Theorem 6.1.2 together with the identity S0 + ∑∞j=1 ∆ j = I. Solution. The inequality ∞
kS0 ( f )kL p + k(∑ ∆ j ( f )2 )1/2 kL p ≤ C p k f kL p 1
follows from Theorem 6.1.2 and the simple fact that S0 has an integrable kernel. To show the other inequality we use the inner product and a duality argument: ∞
∞
h f , gi = hS0 ( f ) + ∑ ∆ j ( f ), S0 (g) + ∑ ∆k (g)i j=1
k=1 ∞
= hS0 ( f ), S0 (g)i + hS0 ( f ), ∆0 (g)i + h∆0 ( f ), S0 (g)i + ∑ h∆ j ( f ), ∆ j−1 (g) + ∆ j (g) + ∆ j+1 (g)i j=1
1
= hS0 ( f ), S0 (g)i + hS0 ( f ), ∆0 (g)i + hS0 ( f ), ∆0 (g)i + h∆1 ( f ), ∆0 (g)i + ∑
Z
r=0
∞
Z
∑ ∆ j ( f )∆ j+r (g) dx + Rn j=1
∞
∑ ∆ j+1 ( f )∆ j (g) dx Rn j=2
The we apply the CaucgySchwarz inequality to estimate the sum by a square function, then H¨older’s inequality, and we obtain:
∞
∞
h f , gi . kS0 ( f )kL p kgkL p0 + kS0 ( f )kL p kgkL p0 + k∆1 ( f )kL p kgkL p0 + ( ∑ ∆ j ( f )2 )1/2 L p ( ∑ ∆ j (g)2 )1/2 L p0 j=1
j=1
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Using the fact that
∞
( ∑ ∆ j (g)2 )1/2 p0 ≤ C p0 kgk p0 . L L j=1
and taking the supremum over kgkL p0 ≤ 1 yields the inequality ∞
2 1/2 p C−1 kL . p k f kL p ≤ kS0 ( f )kL p + k( ∑ ∆ j ( f ) ) j=1
Section 6.2. Two Multiplier Theorems Exercise 6.2.1 Let ψ(ξ ) be a smooth function supported in [3/4, 2] ∪ [−2, −3/4] and equal to 1 on [1, 3/2] ∪ [−3/2, −1] that satisfies ∑ j∈Z ψ(2− j ξ ) = 1 for all ξ 6= 0. Let 1 ≤ k ≤ n. Prove that m ∈ M p (Rn ) if and only if (6.2.1) is satisfied with mj (ξ ) replaced by the function m(ξ )ψ(2− j1 ξ1 ) · · · ψ(2− jk ξk ). Hint: To prove one direction, partition Zk in 2k sets such that for every j = ( j1 , . . . , jk ) in each of these sets, ji has a fixed remainder modulo 2. For the other direction, use Theorem 6.1.6 in the variables x1 , . . . , xk . Also use the inequality k f kL p (Rn ) ≤ C p k(∑j∈Zk ( fbχRj )∨ 2 )1/2 kL p (Rn ) , Rj = ([−2, − 21 ] ∪ [ 12 , 2])k × Rn−k , which can be derived by duality from the iden tity ∑j∈Zk χRj = 2k . Solution. Here we define j = ( j1 , . . . , jk ) and we define mj (ξ ) = m(ξ )ψ(2− j1 ξ1 ) . . . ψ(2− jk ξk ). We need to show that m ∈ M p (Rn ) if and only if
∑ ( bfj mj )∨ 2 j∈Zk
1 2
Lp
≤ Cp
∑  fj 2 j∈Zk
1 2
Lp
(0.0.17)
for all sequences of functions fj in L p (Rn ) indexed by j = ( j1 , . . . , jk ) ∈ Zk . I. Suppose that m ∈ M p (Rn ). We partition Z into even and odd integers, which we call A1 and A2 . Then for any k1 , k2 ∈ A` (where ` ∈ {1, 2}) we have that ψ(2−k1 x) and ψ(2−k2 x) have disjoint support, where x ∈ R. So it’s clear that if we take (s1 , . . . , sk ) ∈ A` × Zk−1 and (t1 , . . . ,tk ) ∈ A` × Zk−1 (where ` ∈ {1, 2}), we have that ψ(2−s1 ξ1 ) · · · ψ(2−sk ξk ) and ψ(2−t1 ξ1 ) · · · ψ(2−tk ξk ) will have disjoint supports. We introduce functions m(t1 ,...,tk ) (ξ ) = m(ξ ) ∑ · · · ∑ ψ(2− j1 ξ1 ) . . . ψ(2− jk ξk ) j1 ∈At1
jk ∈Atk
where t1 , . . .tk ∈ {1, 2} and we notice that
∑
m(t1 ,...,tk ) = m
t1 ,...tk ∈{1,2}
We also observe that each m(t1 ,...,tk ) is a sum of disjointly supported functions (in different rectangles) and that each m(t1 ,...,tk ) lies in M p (Rn ). To verify the last assertion we use Corollary 6.2.5, which applies since ∂1 · · · ∂k ∑ · · · ∑ ψ(2− j1 ξ1 ) · · · ψ(2− jk ξk ) ≤ C ξ1 −1 · · · ξk −1 . j1 ∈At1
jk ∈Atk
Thus ∑ j1 ∈At · · · ∑ jk ∈At ψ(2− j1 ξ1 ) · · · ψ(2− jk ξk ) lies in M p (Rn ) and consequently, 1
k
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(t ,...,t )
m 1 k ≤ C0 m M p (Rn ) . M p (Rn ) If we prove (0.0.17) for each m(t1 ,...,tk ) in place of m, then summing over all sequences of (t1 , . . . ,tk ) we obtain (0.0.17) for m. For j ∈ Zk define (t ,...,t ) mj 1 k (ξ ) = m(ξ )ψ(2− j1 ξ1 ) · · · ψ(2− jk ξk ) . (t ,...,tk )
Fix (t1 , . . . ,tk ) in {1, 2}k . The important observation is for each j ∈ Zk there is a rectangle Rj 1 (t ,...,tk )
Rj 1
in Rn of the form
= I1 × · · · × Ik × R × · · · × R
where I1 , . . . , Ik are intervals, such that (t ,...,tk )
mj 1
=χ
(t ,...,tk )
Rj 1
m(t1 ,...,tk )
Then Exercise 5.6.1 gives the first inequality below
∑ (χRj(t1 ,...,tk ) m(t1 ,...,tk ) bfj )∨ 2
j∈Zk
1 2
Lp
≤ Cp
∑ (m(t1 ,...,tk ) bfj )∨ 2 j∈Zk
1 2
Lp
≤ C p m(t1 ,...,tk ) M p (Rn )
∑  fj 2 j∈Zk
1 2
p, L
while the second inequality follows from Theorem 5.5.1. (Observe that when p = q in Theorem 5.5.1, then C p,q = 1.) This proves (0.0.17) with m(t1 ,...,tk ) in place of m. II. Conversely, suppose that (0.0.17) holds for all sequences of functions fj . Fix a function f and apply (0.0.17) to the sequence ( fbχRj )∨ , where Rj = ([−2, −1] ∪ [−1, − 12 ] ∪ [ 21 , 1] ∪ [1, 2])k which contains the support of mj . Notice that Rj is equal to the union of 4k disjoint dyadic rectangles, which we call Rsj . Applying (0.0.17) we obtain
∑ j∈Zk
( fbmχRj )∨ 2
1 2
L
≤ C p p
∑
( fbχRj )∨ 2
j∈Zk
1 2
L
k ≤ C p 4 p
4k
∑
∑ ( fbχRsj )∨ 2
j∈Zk s=1
1 2
L
4k
k/2 ≤ C 4 p ∑ p s=1
∑ j∈Zk
( fbχRsj )∨ 2
1 2
p. L
Using Theorem 6.1.6 in the variables x1 , . . . , xk (and simply integrating over the remaining variables), the last expression is bounded by k f kL p (Rn ) . So we proved:
∑ ( fbmχRj )∨ 2 j∈Zk
1 2
Lp
≤ C(p, k)k f kL p .
(0.0.18)
We now show that the expression on the left is bigger than a constant times ( fbm)∨ L p .
( fbm)∨ p ≤ c p f p , L L To prove (0.0.19) we need the following fact:
∑ χRj = 2k . j∈Zk
Then we write for f , g ∈ S (Rn ). h f , gi = h fb, gbi = 2−2k
∑ ∑ h fbχRj , gbχRj0 i
j∈Zk j0 ∈Zk
= 2−2k
4k
∑ ∑ h fbχRj , gbχRsj i
j∈Zk s=1 4k
= 2−2k ∑
Z
∑ ( fbχRj )∨ (bg χRsj )∨ dx .
n s=1 R j∈Zk
(0.0.19)
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Using the CauchySchwarz and then H¨older’s inequality, the preceding expression is bounded by 4k
2−2k ∑
Z
n s=1 R
∑
( fbχRj )∨ 2
1/2
∑ j∈Zk
j∈Zk
(b gχRsj )∨ 2
1/2
dx ≤ 2−2k
∑
( fbχRj )∨ 2
j∈Zk
1/2
4k
∑
L p s=1
∑ j∈Zk
(b gχRsj )∨ 2
1/2
Lp
which is bounded by
C p0
∑ ( fbχRj )∨ 2 j∈Zk
1/2
p kgkL p0 . L
0
Taking the supremum over all g ∈ L p with norm at most 1, for f ∈ S (rn) we obtain that
1/2
k f kL p (Rn ) ≤ C p0 ∑ ( fbχRj )∨ 2
p n . L (R )
j∈Zk
This inequality extends to all f ∈ L p (Rn ). Using this inequality with ( fbm)∨ in place of f we obtain that
( fbm)∨ p ≤ c p
L
∑ ( fbmχRj )∨ 2 j∈Zk
1 2
p, L
which combined with (0.0.18) yields the converse inequality in (0.0.17), in other words m ∈ M p (Rn ).
Exercise 6.2.2 Let ϕ be a smooth function on the real line supported in the interval [−1, 1]. Let ψ(t) be a smooth function on the real line 2 that is equal to 1 when t ≥ 10 and vanishes when t ≤ 9. Show that for the function m(ξ1 , ξ2 ) = eiξ2 /ξ1 ϕ(ξ2 )ψ(ξ1 ) lies in 2 M p (R ), 1 < p < ∞, using Theorem 6.2.4. Also show that Theorem 6.2.7 does not apply. Solution. We make the observation that on the support of m(ξ1 , ξ2 ) we have ξ1  ≥ 9 and ξ2  ≤ 1. This implies ξ1  ≈ ξ1  + ξ2  . We have the following inequalities: ∂1 m(ξ1 , ξ2 ) . ξ1 −2 . ξ1 −1 ∂2 m(ξ1 , ξ2 ) . ξ1 −1 ≈ (ξ1  + ξ2 )−1 . ξ2 −1 ∂1 ∂2 m(ξ1 , ξ2 ) . ξ1 −2 ≈ (ξ1  + ξ2 )−2 . ξ1 −1 ξ2 −1 . It follows from Corollary 6.2.5 that m lies in M p (R2 ). Theorem 6.2.7 does not apply since ∂22 m(ξ1 , ξ2 ) ≈ ξ1 −1 ≈ (ξ1  + ξ2 )−1 but for the theorem to apply we would need to have that ∂22 m(ξ1 , ξ2 ) ≈ (ξ1  + ξ2 )−2 .
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Exercise 6.2.3 Consider the differential operators L1 = ∂1 − ∂22 + ∂34 , L2 = ∂1 + ∂22 + ∂32 . Prove that for every 1 < p < ∞ there exists a constant C p < ∞ such that for all Schwartz functions f on R3 we have
∂2 ∂32 f p ≤ C p L1 ( f ) p , L
L
∂1 f p ≤ C p L2 ( f ) p . L
L
Hint: Use Corollary 6.2.5 and the idea of Example 6.2.6. Solution. Notice that for f ∈ S (R3 ) we have ∂2 ∂32 f
∧
(ξ1 , ξ2 , ξ3 ) = (2πiξ2 )(2πiξ3 )2 fb(ξ1 , ξ2 , ξ3 ) ∧
(ξ1 , ξ2 , ξ3 ) = (2πiξ1 ) fb(ξ1 , ξ2 , ξ3 ) (L1 f )∧ (ξ1 , ξ2 , ξ3 ) = (2πiξ1 ) − (2πiξ2 )2 + (2πiξ3 )4 fb(ξ1 , ξ2 , ξ3 ) = P1 (ξ1 , ξ2 , ξ3 ) fb(ξ1 , ξ2 , ξ3 ) (L2 f )∧ (ξ1 , ξ2 , ξ3 ) = (2πiξ1 ) + (2πiξ2 )2 + (2πiξ3 )2 fb(ξ1 , ξ2 , ξ3 ) = P2 (ξ1 , ξ2 , ξ3 ) fb(ξ1 , ξ2 , ξ3 ) . ∂1 f
Notice that the polynomials P1 and P2 vanish only at the origin in R3 . The inequalities are equivalent to showing that the functions ξ2 ξ32 , P1 (ξ1 , ξ2 , ξ3 )
ξ1 P2 (ξ1 , ξ2 , ξ3 )
lie in M p (R3 ). Dropping the constants via a change of variables, we are looking at the function m1 (ξ1 , ξ2 , ξ3 ) =
ξ2 ξ32 iξ1 + ξ22 + ξ34
which is C0∞ in R3 \ {0} and is homogeneous of degree zero with respect to the group of dilations (ξ1 , ξ2 , ξ3 ) 7→ (λ 4 ξ1 , λ 2 ξ2 , λ ξ3 ) . It follows that for any λ > 0 we have k
k
k
∂1k1 ∂2k2 ∂3 3 m1 (ξ1 , ξ2 , ξ3 ) = ∂1k1 ∂2k2 ∂3 3 m1 (λ 4 ξ1 , λ 2 ξ2 , λ ξ3 ) = λ 4k1 +2k2 +k3 (∂1k1 ∂2k2 ∂3 3 m1 )(λ 4 ξ1 , λ 2 ξ2 , λ ξ3 ) For each (ξ1 , ξ2 , ξ3 ) pick λ such that
(λ 4 ξ1 , λ 2 ξ2 , λ ξ3 ) = 1 . k
Then λ 4 ξ1  ≤ 1 which implies that λ 4k1 ≤ ξ1 −k1 . Likewise we obtain λ 2k2 ≤ ξ1 −k2 and λ k3 ≤ ξ3 −k3 . Since ∂1k1 ∂2k2 ∂3 3 m1 is bounded on the the unit sphere in R3 by Ck1 ,k2 ,k3 , we conclude that k
∂1k1 ∂2k2 ∂3 3 m1 (ξ1 , ξ2 , ξ3 ) ≤ λ 4k1 +2k2 +k3 Ck1 ,k2 ,k3 ≤ Ck1 ,k2 ,k3 ξ1 −k1 ξ2 −k2 ξ3 −k3 . For the function
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m2 (ξ1 , ξ2 , ξ3 ) =
ξ1 iξ1 − ξ22 − ξ32
argue similarly using the group of dilations (ξ1 , ξ2 , ξ3 ) 7→ (λ 2 ξ1 , λ ξ2 , λ ξ3 ) . to obtain that
k
∂1k1 ∂2k2 ∂3 3 m2 (ξ1 , ξ2 , ξ3 ) ≤ λ 2k1 +k2 +k3 Ck0 1 ,k2 ,k3 ≤ Ck1 ,k2 ,k3 ξ1 −k1 ξ2 −k2 ξ3 −k3 .
In both cases Corollary 6.2.5 applies and yields m1 , m2 lie in M p (R3 ).
Exercise 6.2.4 Suppose that m(ξ ) is a realvalued function that satisfies either (6.2.9) or ∂ α m(ξ ) ≤ Cα ξ −α for all multiindices α with n n im(ξ ) lies in M (Rn ) for any 1 < p < ∞. p α ≤ [ 2 ] + 1 and all ξ ∈ R \ {0}. Show that e Hint: Prove by induction and use that ∂ α eim(ξ ) = eim(ξ )
1
∑j
k
cβ 1 ,...,β k (∂ β m(ξ ))l1 · · · (∂ β m(ξ ))lk ,
l j ≥0,β ≤α l1 β 1 +···+lk β k =α
where the sum is taken over all partitions of the multiindex α as a linear combination of multiindices β j with coefficients l j ∈ Z+ ∪ {0}. Solution. (a) Suppose that ∂ α m(ξ ) ≤ Cα ξ −α for all multiindices α with α ≤ [ 2n ] + 1 and all ξ ∈ Rn \ {0}. Then there are constants cβ 1 ,...,β k such that ∂ α eim(ξ ) = eim(ξ )
1
∑j
k
cβ 1 ,...,β k (∂ β m(ξ ))l1 · · · (∂ β m(ξ ))lk ,
l j ≥0,β ≤α l1 β 1 +···+lk β k =α
where the sum is taken over all partitions of the multiindex α as a linear combination of multiindices β j with coefficients l j ∈ Z+ ∪ {0}. This can be proved by induction. i.e., if it is true for α, then it is true for α + e j0 . One obtains the bound α im(ξ ) 1 k ∂ e ≤ C0 (ξ −β  )l1 · · · (ξ −β  )lk = Cα0 ξ −α ∑ l j ≥0,β j ≤α l1 β 1 +···+lk β k =α
which shows that eim(ξ ) satisfies Mihlin’s condition and hence eim(ξ ) lies in M p for 1 < p < ∞. (b) Suppose that m(ξ ) satisfies (6.2.9). Let q ∈ {1, . . . , n}. We consider the special case in the preceding formula when α is sequence of zeros and ones and the ones are exactly at the entries: j1 , . . . , jq , for some fixed numbers in {1, 2, . . . , n}. Then we write 1 k ∂ j1 · · · ∂ jq eim(ξ ) = eim(ξ ) ∑ cβ 1 ,...,β k ∂ β m(ξ ) · · · ∂ β m(ξ ) . 0≤β j ≤α β 1 +···+β k =α
Using condition (6.2.9) (∂ j · · · ∂ j m)(ξ1 , . . . , ξn ) ≤ A ξ j −1 · · · ξ j −1 q q 1 1 and writing β j = (β1j , . . . , βnj ), j = 1, . . . , k, we obtain
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∂ j · · · ∂ j eim(ξ ) ≤ q 1
k k 1 1 cβ 1 ,...,β k  ξ1 −β1 · · · ξn −βn · · · ξ1 −β1 · · · ξn −βn
∑
0≤β j ≤α β 1 +···+β k =α
=
cβ 1 ,...,β k  ξ1 −α1 · · · ξn −αn
∑j
0≤β ≤α β 1 +···+β k =α
=
∑j
cβ 1 ,...,β k  ξ j1 −1 · · · ξ jq −1
0≤β ≤α β 1 +···+β k =α
= C00 ξ j1 −1 · · · ξ jq −1 . We used the notation that α = (α1 , . . . , αn ) and that αs = 1 if and only if s ∈ { j1 , . . . , jq }, while αk = 0 otherwise. This shows that eim(ξ ) satisfies (6.2.9) and thus it lies in M p (Rn ).
Exercise 6.2.5 Suppose that ϕ(ξ ) is a smooth function on Rn that vanishes in a neighborhood of the origin and is equal to 1 in a neighborhood −1 of infinity. Prove that the function eiξ j ξ  ϕ(ξ ) is in M p (Rn ) for 1 < p < ∞. Solution. We use the formula of the preceding exercise: ∂ α eim(ξ ) = eim(ξ )
1
k
cβ 1 ,...,β k (∂ β m(ξ ))l1 · · · (∂ β m(ξ ))lk .
∑j
l j ≥0,β ≤α l1 β 1 +···+lk β k =α
In this case m j (ξ ) =
ξj ξ 
1
which is a homogeneous function of degree zero. Then ∂ β m j (ξ ) is a homogeneous function of degree
−β 1  and since it is smooth away from the origin, it satisfies 1
1
∂ β m j (ξ ) ≤ Cβ 1 ξ −β  . Likewise for the remaining derivatives. Inserting this estimate in the preceding identity, we obtain α im (ξ ) ∂ e j ≤ C0
1
k
(ξ −β  )l1 · · · (ξ −β  )lk ≈ ξ −α .
∑j
l j ≥0,β ≤α l1 β 1 +···+lk β k =α −1
We now obtain a similar estimate for eiξ j ξ  ϕ(ξ ). We have ∂ α eim j (ξ ) ϕ(ξ ) =
∑ cγ,α ∂ γ
eim j (ξ ) ∂ α−γ ϕ(ξ )
γ≤α
where γ ≤ α means γ j ≤ α j . We obtain α im (ξ ) ∂ e j ϕ(ξ ) ≤
∑ c0γ,α ξ −γ
∂ α−γ ϕ(ξ )
γ≤α
and in all terms of the sum we have that ξ  ≥ c > 0 since ϕ vanishes in a neighborhood of the origin. Now if γ = α, then we have the bound ξ −α . If γ < α, then some derivative falls on ϕ, and since ϕ is equal to 1 near infinity, we have that ξ  ≈ 1. Then for all the terms with γ < α, we have ξ −γ ≈ 1 ≈ ξ −α . Thus all terms of the sum are bounded by a constant multiple of ξ −α and this implies that α im (ξ ) ∂ e j ϕ(ξ ) ≤ Cα ξ −α .
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Thus Theorem 6.2.7 applies, hence eim j (ξ ) ϕ(ξ ) lies in M p (Rn ) for 1 < p < ∞.
Exercise 6.2.6 Let τ, τ1 , . . . , τn be real numbers and ρ1 , . . . , ρn be even natural numbers. Prove that the following functions are L p multipliers on Rn for 1 < p < ∞: ξ1 iτ1 · · · ξn iτn , (ξ1 ρ1 + · · · + ξn ρn )iτ , (ξ1 −ρ1 + ξ2 −ρ2 )iτ . Solution. (a) For λ > 0 we have with s = τ1 + · · · + τn λ ξ1 iτ1 · · · λ ξn iτn = λ i(τ1 +···+τn ) ξ1 iτ1 · · · ξn iτn = λ is ξ1 iτ1 · · · ξn iτn thus the identity in Example 6.2.6 holds and it follows that α ∂ ξ1 iτ1 · · · ξn iτn ≤ Cα ξ1 −α1 · · · ξn −αn for all multiindices. Thus Corollary 6.2.5 applies. (b) For the second example, we set k j = 1/ρ j for 1 ≤ k ≤ n, and s = τ. This gives us: iτ λ 1/ρ1 ξ1 ρ1 + · · · + λ 1/ρn ξn ρn = λ iτ (ξ1 ρ1 + · · · + ξn ρn )iτ thus the identity in Example 6.2.6 holds and it follows that α ∂ (ξ1 ρ1 + · · · + ξn ρn )iτ ≤ Cα ξ1 −α1 · · · ξn −αn for all multiindices. Notice that the function in question is smooth everywhere except at the origin since ρ j are even natural numbers. In particular, it is smooth on the axes. (c) For the third example, we set k j = ρ j and again s = τ, and get: iτ λ ρ1 ξ1 1/ρ1 + · · · λ ρn ξn 1/ρn  = λ iτ ξ1 1/ρ1 + · · · + ξn 1/ρn Using the same idea we obtain α iτ ∂ ≤ Cα ξ1 −α1 ξ2 −α2 ξ1 −ρ1 + ξ2 −ρ2 for all multiindices. This function is not smooth on the axes but this is allowed in Corollary 6.2.5.
Exercise 6.2.7 Let ζb(ξ ) be a smooth function on Rn is supported in a compact set that does not contain the origin and let a j be a bounded sequence of complex numbers. Prove that the function m(ξ ) =
∑ a j ζb(2− j ξ )
j∈Z
is in M p (Rn ) for all 1 < p < ∞.
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Solution. We pick numbers 0 < R1 < R2 such that the support of ζb is contained in the annulus R1 < ξ  < R2 and has a small distance from the boundary of the annulus. Suppose we are given a ξ0 in the real line which is nonzero. To differentiate m(ξ ) near ξ0 we consider a very small neighborhood of ξ and we note that only finitely many terms in the sum
∑ a j ζb(2− j ξ )
j∈Z
are nonzero. Precisely, the terms indexed by j such that ξR02 < 2 j < ξR01 are the terms that appear in the sum. Differentiation gives ∂ α ∑ a j ζb(2− j ξ ) = ∑ a j 2− jα ∂ α (ζb)(2− j ξ ) j∈Z
j∈Z
for ξ near ξ0 . Then we evaluate at ξ = ξ0 to obtain ∂
α
− jα α b −j −j b ∑ a j ζ (2 ξ ) (ξ0 ) = ∑ a j 2 ∂ (ζ )(2 ξ0 ) j∈Z
j∈Z
and recalling that the sum is finite and over the indices j with
ξ0  R2
< 2j
0
is
Lp
bounded for all 1 < p < ∞.
Solution. This follows by applying Corollary 2.1.12. We see that we have: ∨ Mm ( f )(x) = sup fb(ξ )m(tξ ) (x) t>0 = sup ( f ∗ m∨ )t (x)(x) t>0
≤ Cn,δ M( f )(x) since the function m∨ (x) is pointwise bounded by the function (1 + x)−n−δ which is integrable over Rn . Since M maps L p to itself, we have that Mm : L p 7→ L p .
Exercise 6.5.2 Suppose that the function m is supported in the annulus R ≤ ξ  ≤ 2R and is bounded by A. Show that the gfunction Z
∞
G( f )(x) = 0
(m(tξ ) fb(ξ ))∨ (x)2
dt t
1 2
√ maps L2 (Rn ) to L2 (Rn ) with bound at most A log 2. Solution. We apply Fubini’s theorem, Plancherel’ theorem, and then Fubini’s theorem again to write kG( f )k2L2 (Rn ) =
Z
Z ∞
Rn 0
m(tξ ) fb(ξ )2
dt dξ t
Now for a given ξ ∈ Rn \ {0} we have Z ∞ 0
dt m(tξ ) fb(ξ )2 ≤ t
Z 2R/ξ  R/ξ 
dt A2  fb(ξ )2 ≤  fb(ξ )2 A2 log(2) t
and inserting this estimate (which holds for almost all ξ ∈ Rn ) in the preceding expression we obtain
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Z
Z ∞
Rn 0
m(tξ ) fb(ξ )2
dt dξ ≤ A2 (log 2) t
Z Rn
 fb(ξ )2 dξ = A2 (log 2)k f k2L2 .
Hence the result immediately follows.
Exercise 6.5.3 ([304]) Let A, a, b > 0 with a + b > 1. Use the idea of Lemma 6.5.2 to show that if m(ξ ) satisfies m(ξ ) ≤ A (1 + ξ )−a and ∇m(ξ ) ≤ A (1 + ξ )−b for all ξ ∈ Rn , then the maximal operator ∨ Mm ( f )(x) = sup fb(ξ ) m(tξ ) (x) t>0 2 n is bounded from L (R ) to itself. Hint: Use that ∞
Mm ≤
∑ Mm, j ,
j=0
where Mm, j corresponds to the multiplier ϕ j m; here ϕ j is as in (6.5.8). Show that
1
1 1−(a+b)
Mm, j ( f ) 2 ≤ C ϕ j m 2∞ ϕ j m e L2∞ f L2 ≤ C 2 j 2 f L2 , L L e ) = ξ · ∇m(ξ ). where m(ξ Solution. We can construct a sequence of functions (ϕ j )∞j=0 as in the proof of Theorem 6.5.1, such that ϕ0 is supported in ξ  ≤ 2 and identical to 1 on the unit ball, ϕ j is supported in 2 j−1 ≤ ξ  ≤ 2 j+1 , j ≥ 1 and that ∑∞j=0 ϕ j = 1. Set m j = ϕ j m, j ≥ 0. Then we have m = ∑∞j=0 m j and hence Mm ≤
∞
∑ Mm, j ,
j=0
where
∨ b Mm, j f (x) = sup f m j (t·) (x) . t>0
At this point, we have ∞
kMm ( f )kL2 ≤ kMm,0 ( f )kL2 + ∑ Mm, j ( f ) L2 . j=1
Since the function m0 has compact support, we have that Mm,0 maps L p to itself for all 1 < p ≤ ∞. It’s enough to show that
Mm, j ( f ) ≤ C j k f k , (0.0.20) 2 2 for which ∑∞j=1 C j < ∞. Define Gm ( f )(x) =
!1 ∨ 2 dt 2 fbm j (t·) (x) . t
Z ∞ 0
e j (ξ ) = ξ · ∇m j (ξ ). Then since m j is supported away from the origin, we have and denote m
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∨ 2 Z t d ∨ 2 fbm j (t·) (x) = b f m j (s·) (x) ds 0 ds Z t ∨ ∨ ds e j (s·) (x) =2 fbm j (s·) (x) fbm s 0 ≤2 Gm j ( f )(x)Gme j ( f )(x). This implies that Mm, j f (x)2 ≤ 2Gm j f (x)Gme j f (x); hence, by the CauchySchwarz inequality,
Mm, j ( f ) 2 2 ≤ 2 Gm ( f ) 2 Gme ( f ) 2 . j j L L L e j with noting that they are both supported in annuli of width about 2 j and Now apply Exercise 6.5.2 to the functions m j and m satisfy
m j ≤ CA2− ja ,
m e j ≤ CA2− j(b−1) ; ∞
∞
we obtain the estimates
Mm, j ( f ) 2 2 ≤ CA2− ja k f k 2 CA2− j(b−1) k f k 2 = C0 A2 2− j(a+b−1) k f k2 2 . L L L L Thus, inequality (0.0.21) has been just established with C j = a + b > 1. Therefore, Mm maps L2 to itself.
√ j C0 A2− 2 (a+b−1) . Obviously, the series ∑∞j=1 C j is convergent for
Exercise 6.5.4 Let A, c > 0, a > 1/2, 0 < b < n. Follow the idea of the proof of Theorem 6.5.1 to obtain the following more general result: c ) ≤ A (1 + ξ )−a for all ξ ∈ Rn and If dµ is a finite Borel measure supported in the closed unit ball that satisfies dµ(ξ dµ(B(y, R)) ≤ c Rb for all R > 0, then the maximal operator Z f 7→ sup f (x − ty) dµ(y) t>0
Rn
maps L p (Rn ) to itself when p > 2n−2b+2a−1 n−b+2a−1 . 1 Hint: Using the notation of the preceding exercise, show that kMm, j ( f )kL2 ≤ C 2 j( 2 −a) k f kL2 and that kMm, j ( f )kL1,∞ ≤ C 2 j(n−b) k f kL1 for all j ∈ Z+ , where C is a constant depending on the given parameters. Solution. c ). Let ϕ be as in Exercise 6.5.3. Using all notation as in Exercise 6.5.3, we have that the maximal operator Let m(ξ ) = dµ(ξ in question is Mm which satisfies Mm ≤
∞
∑ Mm, j ;
j=0
hence, ∞
kMm ( f )kL p ≤ kMm,0 ( f )kL p + ∑ Mm, j ( f ) L p . j=1
Since the function m0 has compact support, we have that Mm,0 maps L p to itself for all 1 < p ≤ ∞. It’s enough to show that
Mm, j ( f ) p ≤ C j k f k p , (0.0.21) L L where C j satisfy ∑∞j=1 C j < ∞. Notice that since m is the Fourier transform of a compactly supported measure, m(ξ ) is a C ∞ function Also we have m j (ξ ) = m(ξ )ϕ(2− j ξ ) and e j (ξ ) = ∇(m j (ξ )) · ξ = ∇(ϕ(2− j ξ )m(ξ )) · ξ = 2− j ∇ϕ(2− j ξ ) · ξ m(ξ ) + ϕ(2− j ξ )∇m(ξ ) · ξ m
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e j kL∞ . 2 j 2− ja . It follows that km j kL∞ . 2− ja and km Therefore, from the solution of the Exercise 6.5.3 we have the estimate
Mm, j ( f ) 2 ≤ C 2 j( 12 −a) k f k 2 , L L
j ≥ 1.
(0.0.22)
j ≥ 1.
(0.0.23)
Now we claim that
Mm, j ( f ) 1,∞ ≤ C 2 j(n−b) k f k 1 , L L
If (0.0.23) is established, interpolating between (0.0.22) and (0.0.23), we deduce
1 2 2
Mm, j ( f ) p ≤ C 2 j( 2 −a)(2− p )+ j(n−b)( p −1) k f k p L L 2
2
1
1 < p < 2.
p n The series ∑ j≥1 2 j( 2 −a)(2− p )+ j(n−b)( p −1) converges exactly when p > 2n−2b+2a−1 n−b+2a−1 ; hence M maps L (R ) to itself whenever p ∞ lies in this range. The remaining p’s follow by interpolation with the trivial L estimate. We now prove (0.0.23). Let K ( j) = (ϕ j )∨ ∗ dµ = Φ2− j ∗ dµ, where Φ is a Schwartz function. Setting
(K ( j) )t (x) = t −n K ( j) (t −1 x) we have that M j ( f ) = sup (K ( j) )t ∗ f  . t>0
We will prove that M j ( f ) ≤ C 2 j(n−b) M( f ) , where M is the centered HardyLittlewood maximal operator with respect to balls. It suffices to show that for any M > n there is a constant CM < ∞ such that K j (x) = (Φ2− j ∗ dµ)(x) ≤
CM 2 j(n−b) . (1 + x)M
(0.0.24)
Then Theorem 2.1.10 yields the required conclusion. Using the fact that Φ is a Schwartz function, we have for every N > 0, (Φ2− j ∗ dµ)(x) ≤ CN
2n j dµ(y) . j N y≤1 (1 + 2 x − y)
Z
We pick an N to depend on M (6.5.12); in fact, any N > M suffices for our purposes. We split the last integral into the regions S−1 (x) = B(0, 1) ∩ {y ∈ Rn : 2 j x − y ≤ 1} and for r ≥ 0, Sr (x) = B(0, 1) ∩ {y ∈ Rn : 2r < 2 j x − y ≤ 2r+1 } . We use that dµ(B(0, R)) ≤ c Rb . This implies that the dµ measure of each Sr (x) is at most c 2(r+1− j)b , an estimate that is useful only when r ≤ j. Using this observation, together with the fact that for y ∈ Sr (x) we have x ≤ 2r+1− j + 1, we obtain the following estimate for the expression (Φ2− j ∗ dµ)(x): j
∑
r=−1
∞ CN 2n j dµ(y) CN 2n j dµ(y) + ∑ j N j N Sr (x) (1 + 2 x − y) r= j+1 Sr (x) (1 + 2 x − y) j ∞ dµ(S (x)) χ µ(Sr (x))χB(0,3) (x) r B(0,2r+1− j +1) (x) ≤ CN0 2n j ∑ + ∑ 2rN 2rN r=−1 r= j+1 j ∞ µ(Rn ) χ cn 2(r+1− j)b χB(0,3) (x) B(0,2r+2− j ) (x) 0 nj ≤ CN 2 + ∑ ∑ 2rN 2rN r=−1 r= j+1
Z
Z
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∞ 1 (1 + 2r+2− j )M ∑ rN (1 + x)M r= j+1 2 ∞ 1 0 j(n−b) jn − jM r(M−N) ≤ CM,n 2 +2 2 ∑ 2 (1 + x)M r= j+1
≤ CN,n 2 j(n−b) χB(0,3) (x) + 2n j
≤
00 2 j(n−b) CM,n , (1 + x)M
where we used that N > M > n. This establishes (0.0.24).
Exercise 6.5.5 Show that Theorem 6.5.1 is false when n = 1, that is, show that the maximal operator M1 ( f )(x) = sup t>0
 f (x + t) + f (x − t) 2
is unbounded on L p (R) for all p < ∞. Solution. Let H ∈ C0∞ such that h(x) ≥ χ[−1,1] . Then for x > 1 we have M1 (h)(x) = sup t>0
h(x + t) + h(x − t) h(x + x) + h(x − x) h(0) 1 ≥ ≥ = . 2 2 2 2
This shows that M1 (h) cannot lie in any L p (R) for p < ∞.
Exercise 6.5.6 n Show that when n ≥ 2 and p ≤ n−1 there exists an L p (Rn ) function f such that M ( f )(x) = ∞ for all x ∈ Rn . Hence Theorem 6.5.1 is false is this case. Hint: Choose a compactly supported and radial function equal to y1−n (− log y)−1 when y ≤ 1/2.
Solution. We define the following radial function with compact support as suggested by the hint: ( 1−n y y ≤ 1/2 f (y) = − log y 0 y > 1/2 We will show that k f kL p < ∞ converges and kM ( f )kL p = ∞ for n ≥ 2, p ≤ n Note that if p < n−1 , then p(1 − n) > −n and consequently Z
n n−1 .
 f (y) p dy < ∞ .
y≤1/2
Moreover, if p =
n n−1 ,
then p(1 − n) = −n and y(1−n)p dy = ωn−1 1 y≤1/2  log y  p
Z
Z 1/2 (1−n)p r 0
log(1/r) p
rn−1 dr
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r−1
Z 1/2
= ωn−1
n dr log(1/r) n−1 Z ∞ 1 dr = ωn−1 < ∞. n 2 log(r) n−1 r
0
Next we examine M ( f )(x). For a fixed x ∈ Rn with x 6= 0 we have M ( f )(x) ≥ sup t>0
1
Z {θ ∈Sn−1 : x−tθ ≤ 12 }
x − tθ n−1 log
1 x−tθ 
dθ
and the expression inside the supremum is nonzero only if x − 12 ≤ t ≤ x + 12 . We pick t = x, we set x0 = x/x, and we note that Z 1 1 M ( f )(x) ≥ dθ =∞ 0 n−1 1 n−1 0 log 2 {θ ∈S : x −θ ≤ 2x } x − θ  in view of the result in Appendix D.4. Now if x = 0, then M ( f )(x) ≥ sup t>0
cnt n−1 1 1 dθ = sup = ∞. 1 n−1 log 1t {θ ∈Sn−1 : tθ ≤ 12 } tθ n−1 log tθ  t>0 t
Z
Section 6.6. Wavelets and Sampling Exercise 6.6.1 (a) Let A = [−1, − 12 ) [ 21 , 1). Show that the family {e2πimx }m∈Z is an orthonormal basis of L2 (A). (b) {e2πim·x }m∈Zn in L2 (An ). Obtain the same conclusion for the family 2 Hint: To show completeness, given f ∈ L (A), define h on [0, 1] by setting h(x) = f (x − 1) for x ∈ [0, 21 ) and h(x) = f (x) for x ∈ [ 21 , 1). Observe that b h(m) = fb(m) for all m ∈ Z and expand h in Fourier series. S
Solution. (a) First we show {e2πimx }m∈Z are orthonormal for L2 (A). Since e2πimx is 1periodic, we have he2πimx , e2πinx i =
Z −1/2
e2πi(m−n)x dx +
Z 1
−1
Z 1/2
=
e2πi(m−n)x dx +
0
Z 1
=
e2πi(m−n)x dx
1/2
Z 1
e2πi(m−n)x dx
1/2
e2πi(m−n)x dx
0
=0 for m 6= n. Clearly, when m = n, we get he
2πimx
2πinx
,e
i=
Z −1/2
Z 1
dx = 1,
dx + −1
1/2
and we conclude {e2πimx }m∈Z is an orthonormal set. Next we show completeness. Given f ∈ L2 (A), let ( f (x − 1) for x ∈ [0, 1/2) h(x) = f (x) for x ∈ [1/2, 1).
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Then clearly h ∈ L2 [0, 1), and by periodicity Z −1/2
fb(m) =
f (x)e
−2πimx
Z 1 1/2
Z 1/2
f (x − 1)e−2πimx dx +
=
Z 1
0
Z 1
=
f (x)e−2πimx dx
dx +
−1
f (x)e−2πimx dx
1/2
h(x)e−2πimx dx
0
=b h(m). Thus if h f , e2πimx i = fb(m) = 0 for all m ∈ Z, then b h(m) = 0 giving h(x) = 0 a.e. by uniqueness (Proposition 3.1.13). It follows that f (x) = 0 a.e., and we conclude {e2πimx }m∈Z is complete. (b) The arguments of part (a) directly generalize to higher dimensions without any issues since e2πim·x is 1periodic in each axis. We now have in L2 (An ), he2πip·x , e2πiq·x i =
Z An
Z
=
e2πi(p−q)·x dx
···
A n
Z A
Z
=∏
e2πi(p j −q j )x j dx j
j=1 A n Z 1
=∏
j=1 0
e2πi(p1 −q1 )x1 · · · e2πi(pn −qn )xn dx1 . . . dxn
e2πi(p j −q j )x j dx j
which equals 0 for p 6= q and 1 for p = q. This generalization works for completeness as well; define h as before but apply the translated halfinterval to every axis. We can do this step by step if we define n functions where we shift in the first j variables: ( f (x1 − 1, . . . , x j − 1, x j+1 . . . , xn ) for xr ∈ [0, 1/2), r ≤ j h j (x) = f (x) else . The argument is now similar: Z
fb(m) =
An
Z
f (x)e−2πim·x dx Z
f (x)e−2πim·x dx1 . . . dxn Z Z Z Z 1/2 = ··· f (x1 − 1, x2 , . . . , xn )e−2πim·x dx1 +
=
···
A
A
A
Z
A
0
1
1/2
f (x)e−2πim·x dx1 dx2 . . . dxn
Z Z 1
h1 (x)e−2πim·x dx Z Z Z 1 Z 1/2 Z = ··· h1 (x1 , x2 − 1, . . . , xn )e−2πim·x dx2 +
=
···
A 0
A
A 0
A
Z
=
···
0
Z Z 1Z 1 A 0
A
0
1/2
h2 (x)e−2πim·x dx
.. . Z
=
[0,1]n
hn (x)e−2πim·x dx
= hbn (x).
1
h1 (x)e−2πim·x dx2 dx1 dx3 . . . dxn
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As before, if h f , e2πim·x i = fb(m) = 0 for all m ∈ Z, then 0 = hn (x) = f (x) a.e., and we conclude {e2πimx }m∈Z is complete.
Exercise 6.6.2 Let g be an integrable function on Rn . (a) Suppose that g is supported in [−b, b]n for some b > 0 and that the sequence {b g(k/2b)}k∈Zn lies in `2 (Zn ). Show that g(x) = (2b)−n
k
∑n gb( 2bk )e2πi 2b ·x χ[−b,b]n ,
k∈Z
where the series converges in L2 (Rn ) and deduce that g is in L2 (Rn ). (b) Suppose that g is supported in [0, b]n for some b > 0 and that the sequence {b g(k/b)}k∈Zn lies in `2 (Zn ). Show that g(x) = b−n
k
∑n gb( bk )e2πi b ·x χ[0,b]n ,
k∈Z
where the series converges in L2 (Rn ) and deduce that g is in L2 (Rn ). b S b (c) When n = 1, obtain the same as the conclusion in part (b) for x ∈ [−b, − 2 ) [ 2 , b), provided g is supported in this set. Hint: Part (c): Use the result in Exercise 5.6.1. Solution. (a) We dilate g, and reconstruct its dilation via its Fourier series. Using the notation δ a f (x) = f (ax), we see δ 2b g has support in [−1/2, 1/2]n . We also have (δ 2b g) b(k) = (2b)−n gb(k/(2b)) so that δ 2b g(x) =
∑n (δ 2b g) b(k)e2πix·k = (2b)−n ∑n gb(k/(2b))e2πix·k χ[−1/2,1/2]n .
k∈Z
k∈Z
Since the series has squareintegrable coefficients, it follows that δ 2b g is square integrable over [−1/2, 1/2]n . Finally, dilating the above equation by 1/(2b), we arrive at 1
g(x) = δ 2b δ 2b g(x) = (2b)−n
k
∑n gb(k/(2b))e2πi 2b ·x χ[−b,b]n .
k∈Z
(b) We proceed just as in part (a) except this time we dilate by b. Now δ b g is supported in [0, 1]n , and we have δ b g(x) =
∑n (δ b g)b(k)e2πix·k = b−n ∑n gb(k/b)e2πix·k χ[0,1]n .
k∈Z
k∈Z
Dilating by 1/b, we arrive at 1
g(x) = δ b δ b g(x) = b−n
k
∑n gb(k/b)e2πi b ·x χ[0,b]n .
k∈Z
The squareintegrability of g follows in the same way. (c) Consider δ b g which is now supported in A = [−1, −1/2) ∪ [1/2, 1). By Exercise 5.6.1, {e2πimx }m∈Z is an orthonormal basis for L2 (A), and thus the problem is no different than part (b). We dilate, expand in the orthonormal basis, and dilate the result precisely as in part (b): k 1 g(x) = δ b δ b g(x) = b−1 ∑ gb(k/b)e2πi b ·x . k∈Z
Exercise 6.6.3 Show that the sequence of functions
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n sin π(2Bx − k ) n j j 2 Hk (x1 , . . . , xn ) = (2B) ∏ , π(2Bx j − k j ) j=1
k ∈ Zn ,
2 n is orthonormal in L (R ). Hint: Interpret the functions Hk as the Fourier transforms of known functions.
Solution. Since the Fourier transform in an isometry on L2 (Rn ), it suffices to show orthonormality of the inverse Fourier transforms Hk ∨ (x1 . . . xn ). Let f (x1 , . . . , xn ) = χ[−1/2,1/2]n . Notice in one dimension, we have Z χ[−1/2,1/2] b(ξ ) =
1/2
e−2πixξ dx =
−1/2
eπiξ − e−πiξ sin(πξ ) −1 −2πixξ 1/2 = e = . 2πiξ 2πiξ πξ −1/2
Clearly then fb(ξ1 , . . . , ξn ) = ∏nj=1 sin(πξ j )/(πξ j ). The functions Hk are simlar to translated and dilated versions of fb. Specifically, we see h i i h k 1 k e2πix· 2B δ 2B f (x) b(ξ ) = e2πix· 2B χ[−B,B]n b(ξ ) kj n sin 2Bπ ξ j − 2B = (2B)n ∏ k j=1 2Bπ ξ j − 2Bj n
sin (π (2Bξ j − k j )) . π (2Bξ j − k j ) j=1
= (2B)n ∏ k
So now define hk (x) = e2πix· 2B χ[−B,B]n . We now check the orthogonality of these functions: Z Rn
k
k
e2πix· 2B χ[−B,B]n e2πix· 2B χ[−B,B]n dx =
Z
k−`
e2πix· 2B dx
[−B,B]n
= (2B)n
Z [−1/2,1/2]n
e2πix·(k−`) dx
=0 for k 6= `, and these functions are orthogonal. From the above, we also see when k = `, khk k2 = (2B)n . We normalize the hk to make them orthonormal, and this gives the result. Letting Hk (x) = hbk (x)/ khk k, we have Hk (x) =
n sin (π (2Bξ − k )) hbk (x) j j n/2 = (2B) ∏ (2B)n π (2Bξ − k ) j j j=1
as an orthonormal sequence.
Exercise 6.6.4 Prove the following spherical multidimensional version of Theorem 5.6.9. Suppose that fb is supported in the ball ξ  ≤ R. Show that 1 J n2 (2πRx + 2k ) k f (x) = ∑ f − 2R , n 2n Rx + 2k  2 k∈Zn where Ja is the Bessel function of order a. Solution. Since fb is supported in the ball B(0, R) ⊂ [−R, R]n , we can use exercise 5.6.2 to reconstruct fb. We have
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k ·ξ k k 1 k 1 b 2πi 2R b f e e2πi 2R ·ξ . fb(ξ ) = f − = ∑n 2R ∑n (2R)n k∈Z (2R)n k∈Z 2R Then taking the inverse Fourier transform, we have Z
f (x) =
Rn
fb(ξ )e2πix·ξ dξ
k 1 k = e χB(0,R) e2πi 2R ·ξ dξ f − ∑ n n (2R) k∈Zn 2R R Z k k 1 = ∑ f − χB(0,R) e2πi 2R ·ξ e2πix·ξ dξ n n 2R (2R) R n k∈Z Z 1 k k 1 (δ R χB(0,1) )e2πi(x+ 2R )·ξ dξ = ∑ f − n n 2R (2R) R k∈Zn Jn/2 (2π R(−x − k/(2R))) k 1 Rn = ∑ f − n 2R (2R) R(−x − k/(2R))n/2 k∈Zn 1 Jn/2 (2π Rx + k/2) k = ∑ f − , 2R 2n Rx + k/2n/2 k∈Zn Z
2πix·ξ
where the second to last equality follows from the derivation of the Fourier transform of radial functions in Appendix B.5.
Exercise 6.6.5 Let {ak }k∈Zn be in ` p for some 1 < p < ∞. Show that the partial sums n
sin(2πBx j − πk j ) 2πBx j − πk j j=1
∑n ak ∏
k∈Z k≤N
converge in S 0 (Rn ) as N → ∞ to an L p function on Rn whose Fourier transform is supported in [−B, B]n . Here k = (k1 , . . . , kn ). Moreover, the L p norm of A is controlled by a constant multiple of the ` p norm of {ak }k . Solution. We have already calculated Z 1/2
[χ(−1/2,1/2) ]b(ξ ) =
e
−2πixξ
−1/2
eπiξ − e−πiξ sin(πξ ) −1 −2πixξ 1/2 = dx = e = , 2πiξ 2πiξ πξ −1/2
so then [sin(πx)/πx] b = [χ(−1/2,1/2) ]bb(x) = χ(−1/2,1/2) (−x) = χ(−1/2,1/2) (x). Define
n
AN (x) =
∑
k≤N
and
sin(2πBx j − πk j ) 2πBx j − πk j j=1
ak ∏ n
A(x) =
sin(2πBx j − πk j ) . 2πBx j − πk j j=1
∑n ak ∏
k∈Z
The sum defining A(x) converges absolutely for any fixed x ∈ Rn since it is bounded by ∑k∈Zn ak  ∏nj=1 (1 + 2Bx j − k j )−1 which is convergent by H¨older’s inequality, since {ak }k lies in ` p (Zn ). Notice that for any Schwartz function φ on Rn ,
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cN , φbi hAN , φ i = hA Z
=
∑
Rn k≤N
ak −2πiξ · k b 2B χ e [−B,B]n (ξ )φ (ξ )dξ (2B)n
=
ak b [φ (ξ )χ[−B,B]n (ξ )] ∨ (−k/2B) n (2B) k≤N
=
ak ψ(−k/2B) n (2B) k≤N
∑
∑
where ψ(x) = [φb(ξ )χ[−B,B]n (ξ )] ∨ (x). To show that AN converges in S 0 (Rn ) to A, we show that the above sum is absolutely convergent and bounded by a multiple of kφ kL p0 . Letting N → ∞ and using Holder’s inequality, we have ak 1 ∑n (2B)n ψ(−k/2B) ≤ (2B)n k{ak }k k` p k{ψ(k/2B)}k k` p0 . k∈Z 0 Then by Theorem 5.6.1.2 and the L p boundedness of φ 7→ [φb(ξ )χ[−B,B]n (ξ )] ∨ we have
0 k{ψ(k/2B)}k k` p0 ≤ Cn,p,B kψkL p0 (Rn ) ≤ Cn,p,B kφ kL p0 (Rn ) 0
and the L p norm of φ is bounded by finite sums of Schwartz seminorms of φ . This yields that AN converge in S 0 (Rn ) to the function A, and this function satisfies for all φ ∈ S (Rn ) 0 hA, φ i ≤ Cn,p,B
1 k{ak }k k` p kφ kL p0 . (2B)n
Then A must be an L p function. Now the Fourier transforms of AN (which is are supported in [−B, B]n ) converge to that of A in S 0 (Rn ). It follows that the Fourier transform of A is also supported in [−B, B]n .
Exercise 6.6.6 Consider the function ∏nj=1 sin(πx j )/(πx j ) on Rn to show that Theorem 5.6.13 fails when ε = 0 and p ≤ 1. When 1 < p ≤ ∞ consider the function x1 + ∏nj=1 sin(πx j )/(πx j ). Solution. Consider the function
n
sin(πxi ) πxi i=1
h(x) = ∏
for 0 < p ≤ 1. Then the support of the Fourier transform of h is [−1/2, 1/2]n and kh(k)k` p (Zn ) = h(0) = 1, but kh(x)kL p
Z ∞ sin(πx1 ) p
= −∞
πx1
dx1
Z ∞ sin(πx2 ) p −∞
πx2
dx2 · · ·
Z ∞ sin(πxn ) p −∞
which is clear since p Z ∞ Z ∞ sin(πx) p sin(πx) dx = 2 dx −∞
πx
0 ∞
=2∑
πx
Z k+1 sin(πx) p
k=0 k
(πx) p
dx
πxn
dxn = ∞
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285 ∞ Z k+1 sin(πx) p
≥2∑
π p (k + 1) p
k=0 k ∞
dx
1 p k=0 (k + 1)
≥c∑ =∞
since p ≤ 1. For 1 < p ≤ ∞ let g(x) = h(x) + x1 . Then the support of the Fourier transform of g is also [−1/2, 1/2]n since the new term only contributes a derivative of a Dirac mass at the origin. But h lies in L p for 1 < p ≤ ∞ and so to show that g does not lie in L p it suffices to show that x1 is not in L p (R), which is obvious.
Exercise 6.6.7 (a) Let ψ(x) be a nonzero continuous integrable function on R that satisfies Cψ =
Z +∞ b 2 ψ(t)
t
−∞
R
R ψ(x) dx
= 0 and
dt < ∞ .
Define the wavelet transform of f in L2 (R) by setting 1 W ( f ; a, b) = p a
Z +∞
f (x)ψ
x−b
−∞
a
dx
when a 6= 0 and W ( f ; 0, b) = 0. Show that for any f ∈ L2 (R) the following inversion formula holds: f (x) =
1 Cψ
Z +∞ Z +∞ 1 −∞
a
−∞
1 2
ψ
x−b da W ( f ; a, b) db 2 . a a
n (b) State and prove an analogous wavelet transform inversion property on R . Hint: Apply Theorem 2.2.14 (5) in the bintegral and use Fourier inversion.
Solution. Part (a) We begin this problem by considering the Fourier transform with respect to b. We have Z ∞ x−b x − b −2πibt Ψ b(t) = Ψ e db a a −∞ Z −∞ y −2πi(x−y)t Ψ e dy =− a ∞ Z ∞ y 2πiyt Ψ = e−2πixt e dy a −∞ = e−2πixt
Z ∞ −∞
Ψ (z)e2πiazt adz
b (−at). = e−2πixt aΨ We now look at the inverse Fourier transform of W ( f ; a, b): x−b W ( f ; a, b) (t) = f (x)Ψ dxe2πibt db 1 a −∞ a 2 −∞ Z ∞ Z ∞ 1 x − b 2πibt Ψ = f (x) e dbdx 1 a −∞ a 2 −∞ ∨
Z ∞
1
Z ∞
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Z ∞ 1
= = = =
1 2
f (x)e2πixt
−∞
a Z 1
∞
f (x)e2πixt a
1
a 2 −∞ Z ∞ a a a a
1 2
1 2
y e−2πiyt dydx Ψ a −∞
Z ∞
Z ∞ −∞
Ψ (z)e−2πiazt dzdx
b (−at)dx f (x)e2πixtΨ
−∞
b (−at). f ∨ (t)Ψ
Turning our attention to the wavelet inversion formula, applying Theorem 2.2.14(5) and substituting in the above values, we obtain Z ∞Z ∞ 1 1 x−b da Ψ W ( f ; a, b)db 2 CΨ −∞ −∞ a 12 a a Z ∞Z ∞ 1 1 x−b da = Ψ b(t)W ( f ; a, b)∨ (t)dt 2 1 CΨ −∞ −∞ a 2 a a =
=
1 CΨ
Z ∞Z ∞
1 CΨ
Z ∞Z ∞
−∞ −∞
a
a 2 b Ψ (−at) a
−∞ −∞
2 b (y) Z ∞ Ψ 1
=
b (−at) e−2πixtΨ
1 2
a
b (−at)dt da f ∨ (t)Ψ a2 a 1 2
dae−2πixt f ∨ (t)dt
CΨ −∞ y 1 CΨ f (x) = CΨ = f (x)
Z ∞
dy
e−2πixt f ∨ (t)dt
−∞
where the third to last equality follows from a change of variables and considering t positive and t negative. Part (b). Considering the previous proof for onedimensional inversion, there appears to be only one place where the generalization to higher dimensions could be tricky. Namely translations and dilations are now in different spaces: a ∈ R but b ∈ Rn . Thus the change of variables that gave us CΨ in the third to last line of the above arguement is no longer valid. We can fix this by redefining CΨ so that it is similar in spirit, but applies to higher dimensions; let CΨ =
Z ∞ b Ψ (at)2
a
−∞
dt < ∞
for almost all t ∈ Rn . Now the wavelet transform is given by 1
W ( f ; a, b) =
Z 1
a 2
Rn
f (x)Ψ
x−b dx, a
and the inverse transform would be 1 CΨ
Z ∞Z −∞ Rn
1
1Ψ
a 2
x−b da W ( f ; a, b)db 2n . a a
The proof then is carried out exactly as before. We consider the Fourier and inverse Fourier transforms with respect to b. We perform the same changes of variables, but now in higher dimensions. This yields
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287
x−b b (−at) b(t) = e−2πix·t anΨ Ψ a and an
W ( f ; a, b) ∨ (t) =
a
1 2
b (−at). f ∨ (t)Ψ
Thus we have 1 CΨ
Z ∞Z
1
−∞ Rn a 2
=
1 CΨ
=
1 CΨ
=
1 CΨ
=
da x−b W ( f ; a, b)db 2n a a Z ∞Z 1 x−b da Ψ b(t)W ( f ; a, b) ∨ (t)dt 2n 1 a a −∞ Rn a 2 Z ∞Z an −2πix·t b da an ∨ b Ψ (−at) e 1 1 f (t)Ψ (−at)dt 2n n a −∞ R a 2 a 2
1Ψ
Z ∞Z −∞ Rn
b (−at)2 Ψ dae−2πixt f ∨ (t)dt a
Z ∞ b 1 Ψ (at)2
CΨ −∞ = f (x).
a
Z
da
Rn
e−2πix·t f ∨ (t)dt
Exercise 6.6.8 (P. Casazza) On Rn let e j be the vector whose coordinates are zero everywhere except for the jth entry, which is 1. Set q j = e j − n1 ∑nk=1 ek for 1 ≤ j ≤ n and also qn+1 = √1n ∑nk=1 ek . Prove that n+1
∑ q j · x2 = x2
j=1
for all x ∈ Rn . This provides an example of a tight frame on Rn . Solution. We need to prove that n+1
2
∑ q j · x
= x2 .
j=1
To show this, notice n+1
2 n+1 ∑ q j · x = ∑ x · q j q j · x =
j=1
j=1
n+1 ∑ (x · q j )q j · x = hSx, xi j=1
where Sx = ∑n+1 j=1 (x · q j )q j . This S is known as the frame operator, and now we must only demonstrate S = I, the indentity operator. Indeed, if we define the matrix T = q1 q2 · · · qn+1 with column vectors qi , then we see Sx = T T t x. To prove S = I, all we must show is that the rows in T are orthonormal. Let i, j ∈ {1, . . . , n + 1} so that ri and r j are any two rows from T . Then
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1 1 2 1 2 2 2 1 1 1 2 − + (n − 2) − + √ = 2 − + − 2 + =0 ri · r j = 2 1 − n n n n n n n n n when i 6= j. When i = j, we get 1 2 1 1 1 2 1 1 1 ri · ri = 1 − + (n − 1) 2 + = 1 + 2 − 2 + − 2 + = 1. n n n n n n n n 2 2 We conlcude S = I, and so ∑n+1 j=1 q j · x = x .
Chapter 7. Weighted Inequalities
Section 7.1. The A p Condition Exercise 7.1.1 Let k be a nonnegative measurable function such that k, k−1 are in L∞ (Rn ). Prove that if w is an A p weight for some 1 ≤ p < ∞, then so is kw. Solution. Since k and k−1 are in L∞ (Rn ) kL∞ , k−1 L∞ < ∞, we have
1 Q
Z
k(x)w(x)dx Q
1 Q
Z
1
(k(x)w(x))− p−1 dx
p−1
=
Q
1 Q
Z
k(x)w(x)dx Q
1 Q
Z
−1
1
k(x) p−1 w(x)− p−1 dx
p−1
Q
p−1 1 1 1 − p−1 −1 p−1 = k(x)w(x)dx (k (x)) w(x) dx Q Q Q p−1 Z Z 1 1 1 1 ≤ kkk∞ w(x)dx (k−1 ∞ ) p−1 w(x)− p−1 dx Q Q Q Q p−1 Z Z 1 1 1 − p−1 −1 ∞ ∞ = kkkL kk kL w(x)dx w(x) dx Q Q Q Q
1 Q
Z
Z
= kkkL∞ kk−1 kL∞ [w]A p < ∞. Taking supremum over all cubes Q we obtain that [kw]A p ≤ kkkL∞ kk−1 kL∞ [w]A p < ∞.
Therefore kw is an A p weight.
Exercise 7.1.2 Let w1 , w2 be two A1 weights and let 1 < p < ∞. Prove that w1 w1−p is an A p weight by showing that 2 p−1 [w1 w1−p 2 ]A p ≤ [w1 ]A1 [w2 ]A1 .
Solution. For every Q ⊆ R, we have
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1 Q
Z Q
w1 w1−p 2 dx
1 Q
Z
1
Q
− p−1 (w1 w1−p dx 2 )
p−1
1 Q
Z
1 Q
Z
≤ =
Q
p−1 w1 kw−1 2 kL∞ (Q) dx
Q
w1 dx
1 p−1 kw−1 1 k
1 Q
1 Q
Z
1
Q
p−1 w dx kw−1 2 1 k
p−1
Z Q
w2 dx
p−1
p−1 kw−1 2 kL∞ (Q) .
Taking the supremum of both sides we obtain Z Z −1 1 1 1−p p−1 p−1 sup w1 w1−p dx ( (w w ) dx) 1 2 2 Q Q Q Q Q " ≤ sup Q
1 Q
Z Q
1 p−1 kw−1 1 k
w1 dx
1 Q
p−1
Z Q
w2 dx
# p−1 kw−1 2 kL∞ (Q)
.
The right hand side satisfies " sup Q
1 Q
1 Q1 
≤ sup Q1
Z Q
w1 dx
1 p−1 kw−1 1 k
1 Q
p−1
Z Q
w2 dx
# p−1 kw−1 2 kL∞ (Q)
.
p−1 Z 1 1 p−1 p−1 k w1 dx kw−1 sup w dx kw−1 2 1 2 kL∞ (Q2 ) . L∞ (Q1 ) Q2  Q2 Q1 Q2
Z
Combining these facts we obtain p−1 [w1 w1−p 2 ]A p ≤ [w1 ]A1 [w2 ]A1 ,
which gives that w1 w1−p ∈ Ap. 2
Exercise 7.1.3 Suppose that w ∈ A p for some p ∈ [1, ∞) and 0 < δ < 1. Prove that wδ ∈ Aq , where q = δ p + 1 − δ , by showing that [wδ ]A p ≤ [w]δA p . Solution. For a given cube Q we have IQ (w) : = =
1 Q
Z
1 Q
Z
1 Q
Z
1 Q
Z
1 Q
Z
wδ
Q δ
w Q
Denote f (q) :=
q
wδ (1−p)
1 p−1
Q (1−q)
(0.0.25)
δ q−1
.
w Q
1 q
u dx
u ≥ 0, q > 0.
,
Q
We observe that f (q) is an increasing function of q. Indeed, by H¨older inequality we obtain Z
u dx ≤
Q
Let q1 > q. Set k =
q1 q
q
> 1. One has
Z
qk
1 Z k
1 dx
u dx Q
k0
Q
10 k
,
∀k > 1,
1 1 + = 1. k k0
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291
Z
uq dx ≤
Z
Q
uq1 dx
q
q1
1− qq
Q
.
1
Q
Thus, 1 Q
Z
q
u dx ≤
Q
1 Q
Z
q
q1
q1
u dx
.
Q
Take qth root on both sides, we obtain f (q) =
1 Q
Z
1
q
q
≤
u dx Q
1 Q
Z
1 Q
Z
1
q1
q1
u dx Q
= f (q1 ),
∀q1 > q
and this proves that f (q) is increasing. Using this fact, we have
1 Q
1
Z
δ
wδ dx
≤
Q
∀ 0 < δ < 1.
w dx,
(0.0.26)
Q
From (0.0.25) and (0.0.26), one has IQ (w) ≤
δ 1 δ Z q−1 1 (1−q) w w ≤ [w]δAq . Q Q Q
1 Q
Z
This implies [wδ ]A p =
sup Q cube⊂Rn
IQ (w) ≤ [w]δAq
which completes the proof.
Exercise 7.1.4 Show that if the A p characteristic constants of a weight w are uniformly bounded for all p > 1, then w ∈ A1 . Solution. As given, [w]A p are uniformly bounded for all p > 1. Thus there exists a constant M < ∞ such that [w]A p ≤ M for all p > 1. For a fixed p > 1, by definition we know that [w]A p =
sup cube Q⊂Rn
1 Q
Z
1 Q
w dx Q
Z
1
w− p−1 dx
p−1 .
Q
Therefore,
1 Q
Z
w dx Q
1 Q
Z
1
w− p−1 dx
p−1 ≤M
Q
for all cubes Q. Specifically, when p → 1+ ,the inequality holds. That is, lim
p→1+
1 Q
Z
w dx Q
1 Q
Z
1 − p−1
w
p−1
=
dx
Q
1 Q
Z Q
w dx kw−1 kL∞ (Q) ≤ M.
This is true, because lim
p→1+
1 Q
Z
1 − p−1
w Q
p−1 dx
= lim
p→1+
1 Q
p−1 lim
p→1+
= lim kw−1 k p→1+
Z
1
L p−1 (Q)
Q
1 (w−1 ) p−1 dx
p−1
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= lim kw−1 kLq (Q) , q→∞+
= kw−1 kL∞ (Q) . Take the supremum over all cubes Q, we get [w]A1 = sup Q
1 Q
Z Q
w dx kw−1 kL∞ (Q) ≤ M < ∞.
Hence w ∈ A1 from the definition of the A1 characteristic constant.
Exercise 7.1.5 Let w0 ∈ A p0 and w1 ∈ A p1 for some 1 ≤ p0 , p1 < ∞. Let 0 ≤ θ ≤ 1 and define 1 1−θ θ = + p p0 p1 Prove that
(1−θ ) pp
[w]A p ≤ [w0 ]A p
1−θ p0
1
w p = w0
and
0
0
θ p
w1 1 .
θ pp
[w1 ]A p 1 ; 1
thus w is in A p . Solution. Notice that the condition
1 1−θ θ = + p p0 p1
is equivalent to 1−θ θ 1 = + 0, 0 0 p p0 p1 that is, 1
1=
p00 (1−θ )p0
+
1 p01 θ p0
.
(0.0.27)
Given a cube Q ⊂ Rn , we apply H¨older’s inequality with the exponents in (0.0.27) to obtain (1−θ )p
θp 1 p p w dx = w0 0 w1 1 Q Q Q (1−θ )p θp Z Z p0 p1 1 1 ≤ w0 dx w1 dx , Q Q Q Q
1 Q
Z
1 Q
Z
Z
which yields w dx ≤
Q
1 Q
(1−θ )p
Z Q
w0 dx
p0
1 Q
Again, due to H¨older’s inequality with the exponents in (0.0.27), we can write
1 Q
Z
w Q
p0
0
− pp
p
dx
=
1 Q
Z Q
−
w0
(1−θ )p0 p0
! p0
0
− θpp w1 1
p
dx
θp
Z Q
w1 dx
p1
.
(0.0.28)
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293
1 Q
≤
1 Q
=
p0 − p0 0
Z Q
w0
p0
0
w0
Q
1 Q
Z
1 ≤ Q
Z
Q
1 Q
dx ! (1−θ0 )p
p0
− p0
Z
=
! (1−θ0 )p0
p0
1 Q
dx
p0 − p0 0
w0
!
! θ p0 0
p0 − p1 1
Z Q
w1
dx
Q
p1
− p1 w1 1
dx
(1−θ )p p0 p0
p00
dx
1 Q
Z
1 Q
Z
! θ 0p
p0
Z
pp0
p1
p0 − p1 1
w1
Q
! p01
θp p1
p1
dx
.
Consequently,
1 Q
0 − pp
Z
w
p0 p
dx
Q
Q
p0 − p0 0
! p00
(1−θ )p p0
p0
w0
dx
Z
p0 − p1 1
w1
Q
! p01
θp p1
p1
dx
.
(0.0.29)
Multiplying (0.0.28) and (0.0.29), we have
1 Q
Z
w dx Q
1 Q
Z
p0
0
− pp
w
p
dx
≤
Q
1 Q
(1−θ )p
Q
p0
w0 dx
1 Q
(1−θ )p p
1 Q
θp
Z
w1 dx
Q
p1
p0 − p0 0
Z Q
w0
1 Q
Z Q
! p00
(1−θ )p p0
p0
dx
p0 − p1 1
w1
! p10
θp p1
p1
dx
θp p
≤ [w0 ]A p 0 [w1 ]A1p , 0
1
and the claimed inequality follows from taking supremum over all cubes Q ⊂ Rn .
Exercise 7.1.6 ([123]) Fix 1 < p < ∞. A pair of weights (u, w) that satisfies [u, w](A p ,A p ) = sup
Q cubes in Rn
1 Q
Z
u dx Q
1 Q
Z
1
w− p−1 dx
p−1 α} is contained in the union of 3Q j and 4αn < Q1j  Q j  f (t) dt ≤ 2αn . Then u(Eα ) ≤ ∑ j u(3Q j ), and bound each u(3Q j ) by taking Q0 = 3Q j and Q = Q j in the preceding estimate. Part (c): First prove the assertion in part (b) and then derive the inequality in part (a) by adapting the idea in discussion in the beginning of Subsection 7.1.1. Solution. (a) Let f ∈ L p (w) and Q0 be a cube in Rn . p p Z Z 1 −1 1 1 0 p p  f  dx u(Q ) =  f w w dx u(Q0 ) Q0  Q0 Q0  Q0 p0 Z Z Z p0 p 1 1 1 p p ≤  f  w dx w dx u dx Q0  Q0 Q0  Q0 Q0  Q0 Z 1 ≤  f  p w dx [u, w](A p ,A p ) . Q0  Q0 (b) Replacing f by f χQ with Q ⊆ Q0 in part (a), we obtain p Z Z 1 0  f  dx u(Q ) ≤ C  f  p dx. 0 Q0  Q Q This yields R 0
Qf
0
u(Q ) ≤ C0 Q  R
p w dx
Q  f  dx
p .
By Exercise 5.3.9, there exist disjoint cubes Q j such that Eα = {x ∈ Rn : Mc ( f )(x) > α} j
∞ [
3Q j
j=1
and
1 α < 4n Q j 
Z
 f (t) dt ≤
Qj
α . 2n
∞
Then u(Eα ) ≤
∑ u(3Q j ). Taking Q0 = 3Q j and Q = Q j in the above result, we get
j=1
∞
3Q j  p p Q j  f  dx
u(Eα ) ≤ C0 ∑ R j=1 ∞
Z
 f  p w dx
Qj
C(n, p)Q j  p p k f kLp p (w,Q ) ≤ C0 ∑ R j  f  dx j=1 Qj < C0
C0 (n, p) αp
∞
∑ k f kLpp (w,Q j )
j=1
C0 (n, p) ≤ C0 k f kLp p (w) . αp
Hence, 1
1
αu(Eα ) p ≤ C0p C00 (n, p)k f kL p (w) . Now, taking supremum over α,
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295 1
kMc ( f )kL p,∞ (u) ≤ C0p C(n, p)k f kL p (w) . 1
This proves M maps L p (w) into L p,∞ (u) with norm at most C0p C(n, p). (c) Suppose that the pair of weights (u, w) is such that u
Z Mc ( f ) > λ ≤ kMc kLp p (w)→L p,∞ (u) λ −p
Rn
 f (x) p w(x)dx
for all f ∈ L p (w). Observe that for a given f ≥ 0 we have Avg f ≤ M( f χQ )(x) Q
for all x ∈ Q. Then for any λ > 0 with λ < AvgQ f we have Q j {M( f χQ ) > λ } thus u(Q) ≤ kMc kLp p (w)→L p,∞ (u) λ −p
Z
f (x) p w(x) dx
Q
which implies the inequality in part (b) with C0 = kMc kLp p (w)→L p,∞ (u) . We now take f = χS in the inequality in part (b) for some measurable subset S of Q. We obtain S p u(Q) ≤ C0 w(S) . Q 1
We choose f = w− p−1 on the set S = {w > ε} ∩ Q. Applying the inequality in part (b) we obtain p Z Z 1 1 1 w− p−1 dx u(Q) ≤ C0 w− p−1 dx . Q S S Dividing by Q and using that
1 − p−1 dx Sw
R
> 0 we obtain
1 Q
Z
u dx Q
1 Q
Z
1
w− p−1 dx
p−1
S
≤ C0
which proves that the pair (u, w) is of class (A p , A p ), letting ε → 0, and then taking the supremum over all cubes Q in Rn .
Exercise 7.1.8 ([123]) Let 1 < p < ∞ and let (u, w) be a pair of weights of class (A p , A p ). Show that for any q with p < q < ∞ there is a constant C p,q,n < ∞ such that for all f ∈ Lq (w) we have Z Rn
q
1/q
M( f )(x) u(x) dx
≤ C p,q,n
Z Rn
q
f (x) w(x) dx
Hint: Use Exercise 7.1.6 and interpolate between L p and L∞ . Solution. We have kM( f )kL∞ (u) ≤ kM( f )kL∞ ≤ k f kL∞ ≤ k f kL∞ (w) where we need to justify the first and the third inequality. Recall that
1/q .
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khkL∞ (u) = inf λ > 0 : u({h > λ }) = 0 = sup λ > 0 : u({h > λ }) > 0 . The first inequality is aR consequence of the fact that if for a measurable set E we have u(E) > 0, then we must have E > 0. This is trivial, since if E u(x) dx > 0, then E cannot have measure zero, so it must be E > 0. The third inequality is a consequence of the fact that if E > 0, then w(E) > 0. Indeed, if w(E) = 0 for some measurable set E with E > 0, then there is a bounded subset of E, say E0 such that w(E0 ) > 0. It follows from part (a) of the previous exercise that S p u(Q) ≤ C0 w(S) Q for any cube Q and any set S contained in Q. Taking S = E0 we would obtain that for any cube Q containing E0 , we would have u(Q) = 0, thus u = 0 a.e. on Rn , a contradiction, since a weight cannot vanish on a set of positive measure. In view of Exercise 7.1.6, the HardyLittlewood maximal operator M maps L p (w) → L p,∞ (u). We now interpolate between the following estimates: M : L p (w) → L p,∞ (u) M : L∞ (w) → L∞ (u) via Theorem 1.3.2 to obtain that M : Lq (w) → Lq (u) when 1 < p < q < ∞.
Exercise 7.1.8 Let k > 0. For an A1 weight w show that [min(w, k)]A1 ≤ [w]A1 . If 1 < p < ∞ and w ∈ A p , show that [min(w, k)]A p ≤ c p [w]A p , where c p = 1 if 1 < p ≤ 2 and c p = 2 p−1 if 2 < p < ∞. 1 1 1 − p−1 − p−1 1 R 1 R dx ≤ Q dx + k− p−1 and also Hint: Use the inequality Q Q min(w, k) Qw
1 R Q Q min(w, k)dx ≤ min
1 R k, Q Q w dx .
Solution. When p = 1 the above inequality also holds because: Z Z 1 1 min(w, k)dx ≤ min k, wdx Q Q Q Q ≤ min k, [w]A1 ess.inf w Q
≤ max([w]A1 , 1) min k, ess.inf w Q
= [w]A1 ess.inf min(k, w) . Q
1 1 1 Let us consider the case 1 < p ≤ 2. Since min(w, k)− p−1 = max w− p−1 , k− p−1 , we have
1 Q
Z
1
min(w, k)− p−1 dx ≤
Q
1 Q
Z
1
1
max(w− p−1 , k− p−1 ) dx ≤
Q
1 Q
Z
1
w− p−1 dx +
Q
1 Q
Z
1
k− p−1 dx .
Q
Thus,
1 Q
Z
1 − p−1
min(w, k) Q
p−1 dx
≤
1 Q
Z
1 − p−1
w Q
1 dx + Q
Z
k Q
1 − p−1
p−1 dx
≤
1 Q
Z
1 − p−1
w Q
p−1 dx
+ k−1 ,
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297
since in this case We have
Hence,
1 Q
1 p−1
≥ 1 and Minkowski’s inequality applies. 1 Q
Z
1 Q
min(w, k)dx Q
Z
h 1 Z i min(w, k)dx ≤ min k, w dx . Q Q Q
Z
1 − p−1
min(w, k)
p−1 dx
Q
p−1 Z Z 1 1 1 − p−1 −1 w dx max w ≤ min k, dx +k Q Q Q Q p−1 Z Z 1 1 1 ≤ w dx w− p−1 dx + kk−1 Q Q Q Q ≤[w]A p + 1.
This implies [min(k, w)]A p ≤ [w]A p + 1 .
For p ≥ 2, using (a + b)q ≤ 2q−1 (aq + bq ),
1 Q
Z
∀a, b ≥ 0, q ≥ 1, we have
1
1
w− p−1 dx + k− p−1
p−1
≤ 2 p−2
Q
1 Q
Z
1
w− p−1 dx
p−1
+ k−1 .
Q
Hence
1 Q
Z
1 − p−1
min(w, k) Q
p−1 dx
p−1 Z 1 1 1 − p−1 − p−1 w ≤ dx + k Q Q p−1 Z 1 1 1 − p−1 − p−1 p−2 ≤2 w dx +k . Q Q
Therefore,
1 Q
p−1 Z 1 1 − p−1 min(w, k)dx min(w, k) dx Q Q Q Z Z p−1 1 1 1 − p−1 p−2 −1 ≤ min k, w dx 2 w dx +k Q Q Q Q Z Z p−1 1 1 1 − p−1 p−2 −1 ≤2 w dx w dx + kk Q Q Q Q
Z
≤2 p−2 ([w]A p + 1). This implies [min(w, k)]A p ≤ 2 p−2 ([w]A p + 1).
Exercise 7.1.9 Suppose that w j ∈ A p j with 1 ≤ j ≤ m for some 1 ≤ p1 , . . . , pm < ∞ and let 0 < θ1 , . . . , θm < 1 be such that θ1 + · · · + θm = 1. Show that wθ11 · · · wθmm ∈ Amax{p1 ,...,pm } . Hint: First note that each weight w j lies in Amax{p1 ,...,pm } and then apply H¨older’s inequality.
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Solution. Let P = max{p1 , . . . , pm } and W = wθ11 · · · wθmm . The proof will be done as follows: (i) We prove that w j ∈ AP for all j. (ii) We show that the following inequality holds 1 Q
m
Z
θ w j j (x)dx
∏
Q j=1
m
≤∏
j=1
1 Q
θ j
Z
w j (x)dx
Q
.
(iii) We will use (ii) and H¨older’s inequality to show m
θ
[W ]AP ≤ ∏ [w j ]APj . j=1
Combining (i), (ii), and (iii) yields that W ∈ AP . (i). Since p j ≤ P for all j, from Proposition 9.1.5(6) it follows that [w j ]AP ≤ [w j ]A p j for all j. Therefore, we have proved (i). (ii). If w j = 0 for some j then the equality holds. Assuming w j 6= 0 for all j and letting x j :=
w j (x) 1 R Q Q w j (x)dx
,
one gets m
log
∏
!
m
θ xj j
=
j=1
m
∑ θ j log(x j ) ≤ log( ∑ θ j x j ).
j=1
j=1
Here we have used the concavity of the function log(x). Since log(x) is an increasing function, it follows that m
m
θ
∏ xj j ≤ j=1
∑ θ jx j.
j=1
This implies 1 Q
Z
m
θ
∏ x j j dx ≤
Q j=1
1 Q
m
Z
m
∑ θ j x j dx =
Q j=1
∑ θj
j=1
1 Q
m
Z Q
x j dx =
∑ θ j = 1.
j=1
From the above inequality (ii) follows. (iii). Let G :=
1 Q
Z
W dx Q
1 Q
Z
W
1 − P−1
p−1 .
Q
By (ii) we have " G≤
m
∏
j=1
1 Q
θ j #
Z Q
w j (x)dx
1 Q
Z
W
1 − P−1
P−1 .
dx
Q
Write W
1 − P−1
m
θ j 1 − P−1 = ∏ wj . j=1
Applying H¨older’s inequality, with exponents Z
W Q
Therefore, by (i),
1 − P−1
Z
dx =
1 θ1 ,
...,
1 θm
we obtain
m
θ j Z θ1 Z θm 1 1 1 − P−1 − P−1 − P−1 dx ≤ w1 (x)dx ··· wm (x)dx ∏ wj
Q j=1
Q
Q
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299 m
G≤∏
j=1
"
1 Q
Z Q
w j (x)dx
1 Q
Z Q
− 1 w j P−1 (x)dx
P−1 #θ j
m
θ
≤ ∏ [w]APj . j=1
Taking the sup over all cubes Q in the above inequality, we obtain that W ∈ AP .
Exercise 7.1.10 Let w1 ∈ A p1 and w2 ∈ A p2 for some 1 ≤ p1 , p2 < ∞. Prove that [w1 + w2 ]A p ≤ [w1 ]A p1 + [w2 ]A p2 , where p = max(p1 , p2 ). Solution. Let Q be a fixed cube in Rn . Throughout this solution, we put k f kLr (Q) :=
1 Q
Z
 f r dx
1 r
.
Q
Since w1 , w2 ≥ 0, it follows that 1/(w1 + w2 ) ≤ min(1/w1 , 1/w2 ), whence
(w1 + w2 )−1 1 ≤ min w−1 1
, w−1 2
.
(0.0.30)
Now, since p = max(p1 , p2 ), we have 1/(p − 1) = min (1/(p1 − 1), 1/(p2 − 1)), and therefore,
−1
w 1
1 ≤ w−1 (k = 1, 2), k k
(0.0.31)
1
L p−1 (Q)
L p−1 (Q)
1
L p−1 (Q)
L pk −1 (Q)
L p−1 (Q)
since the function t 7→ k f kLt (Q) is increasing. So, combining (0.0.30) with (0.0.31) gives
(w1 + w2 )−1
1
L p−1 (Q)
where m := min w−1 1
1 L p1 −1 (Q)
, w−1 2
1 L p2 −1 (Q)
≤ m,
. Finally, we obtain
kw1 + w2 kL1 (Q) (w1 + w2 )−1 1 L p−1 (Q) h i ≤ kw1 kL1 (Q) + kw2 kL1 (Q) m
(0.0.32)
= kw1 kL1 (Q) m + kw2 kL1 (Q) m
1 + kw2 kL1 (Q) w−1 ≤ kw1 kL1 (Q) w−1 2 1 L p1 −1 (Q)
1
L p2 −1 (Q)
≤ [w1 ]A p1 + [w2 ]A p2 . Since Q is an arbitrary cube, this implies the claim of the exercise.
Exercise 7.1.11 Show that the function
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u(x) =
( 1 log x 1
when x < 1e , otherwise,
in is an A1 weight on Rn . Example 7.1.8 balls Hint: Use [u]A1 instead of [u]A1 and consider balls of type I and II as in Example 7.1.7. Solution. We note that
1 u(x) = max 1, log . x
We divide all balls B(x0 , R) in Rn into two categories: balls of type I that satisfy x0  ≥ 3R and type II that satisfy x0  < 3R. If B = B(x0 , R) is a ball of type I, then for x satisfying x − x0  ≤ R we must have 2 4 x0  ≤ x0  − R ≤ x ≤ x0  + R ≤ x0  . 3 3 This implies that for x − x0  ≤ we have 1 1 3 1 3 ≤ log ≤ log + log . log + log 4 x0  x 2 x0  hence
log 1 − log 1 ≤ log 1 − log 1 ≤ log 3 ≈ 0.4 . x0  x x0  x 2
Consequently, for x − x0  ≤ R we have u(x) ≥ but also
1 1 1 1 3 1 1 + log + log ≥ − log 2 2 x 2 2 2 x0  3 1 u(x) ≤ log + log . 2 x0 
These two facts together imply that
1 B(x0 , R)
log 23 + log x1  0 u(x) dx ess.sup u(y)−1  ≤ 1 − log 3 + 1 log B(x0 ,R) y∈B(x ,R)
Z
0
2
2
2
≤ c.
1 x0 
Taking the supremum over all balls B(x0 , R) we obtain that balls of type I are OK. If B(x0 , R) is a ball of type II, then B(0, 4R) contains B(x0 , R). We have Z Z 1 1 −1 u(x) dx ess.sup u(y)  ≤ u(x) dx ess.sup u(y)−1  . B(x0 , R) B(x0 ,R) vn Rn B(0,4R) y∈B(x0 ,R) y∈B(0,4R) If 4R ≥ 1/e, then the preceding expression is bounded by Z Z c + vn (4R)n 1 1 log dy + dy ≤ n vn R y vn Rn 1/e≤y≤4R y≤1/e which is bounded by a constant. We consider the case where 4R < 1/e. Then Z 4R Z Z 1 1 1 1 1 1 1 −1 n−1 u(x) dx ess.sup u(y)  ≤ log dy = r log dr . 1 1 n n n vn R B(0,4R) vn R y≤4R y vn R r log 4R log 4R 0 y∈B(0,4R) But we have
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Z 4R 0
(4R)n 1 (4R)n (4R)n 1 1 log + ≤2 log rn−1 log dr = r n 4R n n 4R
and the desired conclusion follows.
Exercise 7.1.12 p,∞ Let 1 < p < ∞ and w ∈ A1 . Show that the uncentered HardyLittlewood maximal function M maps L (w) to itself. Hint: Prove first the inequality 2Z 3n ([w]balls A1 ) w({M(g) > λ }) ≤ g w dx λ {M(g)>λ }
and then use the characterization of L p,∞ given in Exercise 1.1.12.] Solution. We fix w ∈ A1 . We work with [w]balls A1 which is comparable to [w]A1 . The weight w is doubling, precisely w(3B) ≤ 3n [w]balls A1 w(B) for all balls B. We may extend the definition of M on any measurable function, allowing values in [0, ∞] via 1 B3x B
Mw (g)(x) = sup
Z
g(x) dx
B
We claim that for and any measurable function g, we have w({M(g) > λ }) ≤
2Z 3n ([w]balls A1 )
λ
g w dx .
(0.0.33)
{M(g)>λ }
This is proved by adapting the proof of Theorem 2.1.6 in which the measure wdx is replaced by Lebesgue measure. Details are given later. We write the above inequality as 2 1/p−1 λ w({M(g) > λ })1/p ≤ 3n ([w]balls A1 ) w({M(g) > λ })
Z
g w dx .
{M(g)>λ }
Using Exercise 1.1.12 we obtain that 2 n balls 2 λ w({M(g) > λ })1/p ≤ 3n ([w]balls A1 ) gL p,∞ (w) ≤ 3 ([w]A1 )
p kgkL p,∞ (w) , p−1
and after taking the supremum in λ , we obtain that M maps L p,∞ (w) to itself. This proves the claim of the exercise. We now prove (0.0.33). First we note that the inner regularity of Lebesgue measure implies the inner regularity of the measure w dx; indeed if U is open and K j compact that increase to U, we have that χK j w ↑ χU w and the Lebesgue monotone theorem implies that w(K j ) → w(U). We claim that the set Eλ = {x ∈ Rn : M(g)(x) > λ } is open. Indeed, for x ∈ Eλ , there is an open ball Bx that contains x such that the average of g over Bx is strictly bigger than λ . Then the uncentered maximal function Mw (g) of any other point in Bx is also bigger than λ , and thus Bx is contained in Eλ . This proves that Eλ is open. Let K be a compact subset of Eλ . For each x ∈ K there exists an open ball Bx containing the point x such that Z Bx
g(y) dy > λ Bx  .
(0.0.34)
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Observe that Bx j Eλ for all x. By compactness there exists a finite subcover {Bx1 , . . . , Bxk } of K. Using Lemma 2.1.5 we find a subcollection of pairwise disjoint balls Bx j1 , . . . , Bx jl such that each ball in {Bx1 , . . . , Bxk } is contained in the triple of some ball in the set {Bx j1 , . . . , Bx jl }. Using this property and (0.0.34) we obtain k [ w(K) ≤ w Bxi i=1 l [ ≤w 3Bx ji i=1 l
≤
3Bx ji
∑w
i=1
l
≤ 3n [w]balls A1 ∑ w(Bx ji ) i=1 l
≤ 3n [w]balls A1 ∑
w(Bx ji )
i=1
≤ ≤ ≤
≤
3n [w]balls A1
l
∑
λ
i=1
3n [w]balls A1
l
λ
∑
Bx ji 
w(Bx ji ) Z Bx ji 
∑
λ
Bx ji  w(Bx ji ) Bx ji 
i=1
2Z 3n ([w]balls A1 )
α
g(y) dy
Bx j
i
w(Bx ji ) Z
i=1
l 3n [w]balls A1
Bx ji 
g(y) w w−1 dy
Bx j
i
sup w−1 Bx j
Z
g(y) w dy
Bx j
i
i
g(y) w(y) dy ,
Eα
since all the balls Bx ji are disjoint and contained in Eλ . Taking the supremum over all compact K ⊆ Eα and using the inner regularity of w measure, we deduce (0.0.33).
Section 7.2. Reverse H¨older Inequality and Consequences Exercise 7.2.1 Let w ∈ A p for some 1 < p < ∞ and let 1 ≤ q < ∞. Prove that the sublinear operator S( f ) = M( f q w)w−1
1
q
0
is bounded on L p q (w). Solution. 0 First we show that S is indeed sublinear. Let f , g ∈ L p q (w). For any cube Q ⊂ Rn , Minkowski’s Inequality gives
1 Q
Z Q
 f + gq w dx
1
q
≤
1 Q
Z
 f q w dx
Q
1
q
+
1 Q
As the preceding inequality holds for any cube Q, it follows that M ( f + gq w) ≤ M ( f q w) + M (gq w) ,
Z Q
gq w dx
1 q
.
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303
and therefore, M ( f + gq w) w−1 ≤ M ( f q w) w−1 + M (gq w) w−1 , since w−1 is nonnegative. Now, if we raise both sides of the last displayed inequality to the 1/q power and invoking the fact 1
1
1
that (a + b) q ≤ a q + b q , whenever a, b ≥ 0, because 1 ≤ q < ∞, we find that S( f + g) ≤ S( f ) + S(g); i.e. S is subadditive. So, to see sublinearity, it remains to compute 1 q −1 q S(λ f ) = [M (λ f  w)] w 1 q q −1 q    = [M (λ f w)] w 1 = λ q [M (λ q  f q w)] w−1 q 1 = λ  [M (λ q  f q w)] w−1 q = λ  S( f ) . 0
0
0
Moving on, since w ∈ A p , we have w−p /p ∈ A p0 , and consequently, the maximal operator M carries L p (w−p /p ) → 0 0 L p (w−p /p ), say with bound C. Thus, Z
0
Rn
S( f ) p q w dx =
Z Rn
Z
=
Rn
≤C
0
0
p0
[M ( f q w)] p w− p dx
Z
p0
0
Rn
Z
( f q w) p w− p dx 0
=C ZR
=C
0
[M ( f q w)] p w−p w dx
n
Rn
0
p0
 f  p q w p − p dx 0
 f  p q w dx.
0
This shows that S is bounded on L p q (w) with bound at most C.
Exercise 7.2.2 Let v be a realvalued locally integrable function on Rn and let 1 < p < ∞. For a cube Q, let νQ be the average of ν over Q. (a) If ev is an A p weight, show that 1 Q cubes Q
Z
sup
1 Q cubes Q
Z
sup
Q
Q
ev(t)−vQ dt ≤ [eν ]A p , 1
e−(v(t)−vQ ) p−1 dt ≤ [eν ]A p .
(b) Conversely, if the preceding inequalities hold with some constant C in place of [ν]A p , then ν lies in A p with [ν]A p ≤ C. Hint: Part (a): If ev ∈ A p , use that 1 Q
Z
p−1 v ev(t)−vQ dt ≤ Avg e− p−1 Avg ev
Q
and obtain a similar estimate for the second quantity.
Q
Q
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Solution. (a) Let 1 supQ cubes Q
(i)
v(t)−vQ dt Qe
R
≤ C,
1 −(v(t)−vQ ) p−1 1 R supQ cubes Q dt ≤ C . Qe
(ii) Suppose eν ∈ A p , 1 < p < ∞. Then, 1 Q
Z
eν(t)−νQ dt = e−νQ
Q
1 eν(t) dt Q Q Z νQ p−1 1 Z
= e− p−1
Q
1 Z ν(t) e− p−1 dt Q Q ≤ [eν ]A p < ∞.
≤
eν(t) dt.
Q p−1
1 Q
Z
eν(t) dt
Q
This proves (i). Again, 1 Q
Z Q
1
νQ ν(t) 1 e− p−1 e p−1 dt Q Q 1 Z ν(t) 1 Z 1 p−1 ≤ e p−1 dt eν(t) dt Q Q Q Q ≤ [eν ]A p < ∞.
Z
e−(ν(t)−νQ ) p−1 dt =
This proves (ii). (b) suppose that the conditions (i) and (ii) hold with a constant C in place of [eν ]A p . We have 1 Z 1 Z − 1 p−1 1 Z ν(t)−ν 1 Z − ν(t)−νQ p−1 Q dt eν(t) dt e e p−1 dt eν(t) p−1 dt = ≤ CC p−1 = C p , Q Q Q Q Q Q Q Q hence eν lies in A p with [eν ]A p ≤ C.
Exercise 7.2.3 This exercise assumes familiarity with the space BMO. (a) Show that if ϕ ∈ A2 , then log ϕ ∈ BMO and k log ϕkBMO ≤ [ϕ]A2 . (b) Prove that every BMO function is equal to a constant multiple of the logarithm of an A2 weight. Precisely, given f ∈ BMO show that cf e A ≤ 1 + 2e , 2
= 1/(2n+1 k f k
where c BMO ). (c) Prove that if ϕ is in A p for some 1 < p < ∞, then log ϕ is in BMO by showing that [ϕ]A p when 1 < p ≤ 2,
log ϕ 1 ≤ BMO (p − 1)[ϕ] p−1 when 2 < p < ∞ . Ap Hint: Part (a): Use Exercise 7.2.2 with p = 2. Part (b): Use Exercise 7.2.2 and Corollary 3.1.7 in [132]. Use Part (c): Use that 1 ϕ − p−1 ∈ A p0 when p > 2. Solution.
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(a) Suppose that ϕ = eν ∈ A2 , then from Exercise 7.2.2 with p = 2 we obtain for any cube Q 1 Q
Z
1 Q
Z
Q
eν−νQ  dt ≤ [eν ]A2 .
Consequently Q
ν − νQ  dt ≤ [ϕ]A2 .
This says that for any cube Q we have 1 Q
Z Q
 log ϕ − (log ϕ)Q  dt ≤ [ϕ]A2 ,
i.e., log ϕ lies in BMO and k log ϕkBMO ≤ [ϕ]A2 . (b) It follows from Corollary 3.1.7 in [132] that for any γ < 1/(2n e), for all f ∈ BMO(Rn ), and all cubes Q we have 1 Q Choosing γ =
1 1 2 2n e
Z
eγ f (x)−AvgQ f /k f kBMO dx ≤ 1 +
Q
2n e2 γ . 1 − 2n eγ
we obtain that 1 Q
Z
ec f (x)−AvgQ f  dx ≤ 1 + 2e .
Q
Exercise 7.2.2 with p = 2 now implies that ec f lies in A2 and we have cf e A ≤ 1 + 2e , 2
where c = γ/(2k f kBMO ). (c) We already know that k log ϕkBMO ≤ [ϕ]A2 and when 1 < p ≤ 2, [ϕ]A p = sup Q
1 ≥ sup Q Q
1 Q
Z
ϕ(x)dx Q
Z
ϕ(x)dx Q
1 Q
1 Q
Z
1
ϕ(x)− p−1 dx
p−1
Q
Z
ϕ
−1
Q
(x)dx = [ϕ]A2 .
Therefore, when 1 < p ≤ 2 k log ϕkBMO ≤ [ϕ]A p . Now fix p > 2. We have 1
1
k log ϕ − p−1 kBMO ≤ [ϕ − p−1 ]A p0 , that is 1 k log ϕkBMO ≤ sup p−1 Q
"
1 Q
Z
ϕ
1 − p−1
Q
1 Q
Z
ϕ
1 (p−1)(p0 −1)
p0 −1 #
Q
which implies 1 k log ϕkBMO ≤ sup p−1 Q
"
1 Q
Z
ϕ Q
1 Q
Z
ϕ
1 − p−1
Q
1 p−1 # p−1
1
= [ϕ]Ap−1 p
that is, 1
k log ϕkBMO ≤ (p − 1)[ϕ]Ap−1 p when p > 2.
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Exercise 7.2.4 Prove the following quantitative versions of Theorem 7.2.2 and Corollary 7.2.6. (a) For any 1 ≤ p < ∞ and B > 1, there exists a positive constant C3 (n, p, B) and γ = γ(n, p, B) such that for all w ∈ A p satisfying [w]A p ≤ B, (7.2.4) holds for every cube Q with C3 (n, p, B) in place of C. (b) Given any 1 < p < ∞ and B > 1 there exists a constant C4 (n, p, B) and δ = δ (n, p, B) such that for all w ∈ A p we have [w]A p ≤ B =⇒ [w]A p−δ ≤ C4 (n, p, B) . Solution. (a) Let us fix a cube Q and set α0 =
1 R Q Q w(x)dx.
We also fix an 0 < α < 1. We define an increasing sequence of scalars
α0 < α1 < α2 < ... < αk < ... for k ≥ 0 by setting αk = 2n α −1 αk or αk = (2n α −1 )k α0 , and for each k ≥ 1 we apply the Calder´onZygmund decomposition to w at height αk . Precisely, for dyadic subcubes R of Q, we let 1 R
Z R
w(x)dx > αk
(0.0.35)
be the selection criterion. Since Q does not satisfy the criterion, it is not selected. We divide the cube Q into a mesh of 2n subcubes of equal side length and among these cubes we select those that satisfy (0.0.35). We subdivide each non selected subcube into 2n subcubes of equal side length and we continue in this way infinitely. We denote by {Qk, j } j the collection of all selected subcubes of Q. We observe that the following properties are satisfied: (1) αk < Q1  Qk, j w(t)dt ≤ 2n αk . k, j S (2) For almost all x ∈ / Uk we have w(x) ≤ αk , where Uk = j Qk, j . (3) Each Qk+1, j is contained in some Qk,l . R
Property (1) is satisfied since the unique dyadic parent of Qk, j was not chosen in the selection procedure, while (2) follows from the Lebesgue differentiation theorem using the fact that for almost all x ∈ / Uk there exists a sequence of nonselected cubes of decreasing lengths whose intersection is {x}. Property (3) is satisfied since each Qk, j is the maximal subcube of Q satisfying (0.0.35). And since the average of w over Qk+1, j is also bigger then αk , it follows that Qk+1, j must be contained in some maximal cube that possesses this property. We now compute the portion of Qk,l that is covered by cubes of the form Qk+1, j for some j. We have 1 Qk,l  1 = Qk,l 
2n αk ≥
>
Z
w(t)dt Qk,l ∩Uk+1
∑
j:Qk+1 ⊆Qk,l
Qk+1, j 
1
Z
Qk+1, j 
w(t)dt Qk+1, j
Qk,l ∩Uk+1  Qk,l ∩Uk+1  n −1 αk+1 = 2 α αk . Qk,l  Qk,l 
It follows that Qk,l ∩Uk+1  ≤ αQk,l ; thus, applying Lemma 9.2.1, we obtain w(Qk,l ∩Uk+1 ) (1 − α) p < β = 1− w(Qk,l ) [w]A p from which, summing over all l, we obtain w(Uk+1 ) ≤ β w(Uk ). The latter gives w(Uk+1 ) ≤ β k w(U0 ). We also have Uk+1  ≤ αUk ; hence Uk  → 0 as k → ∞. Therefore, the intersection of the Uk0 s is a set of Lebesgue measure zero. We can therefore write
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307
!
∞ [
Q = (Q\U0 ) ∪
Uk \Uk+1
k=0
modulo a set of Lebesgue measure zero. Let us now find a γ > 0 so that the reverse H¨older inequality (7.2.4) holds. We have w(x) ≤ αk for almost all x in Q\Uk and therefore Z
w(t)1+γ dt =
∞
Z
Q
Q\U0
w(t)γ w(t)dt + ∑
Z
w(t)γ w(t)dt
k=0 Uk \Uk+1
∞ γ
γ
≤ α0 w (Q\U0 ) + ∑ αk+1 w(Uk ) k=0 ∞ h
≤ α0 w (Q\U0 ) + ∑ (2n α −1 )k+1 α0 γ
iγ
β k w(U0 )
k=0
" n
γ ≤ α0
1 + (2 α
−1 γ
)
#
∞
n
2 α
∑
−1 γk
β
k
w(Q)
k=0
1 Q
=
Z
w(t)dt
γ 1+
Q
(2n α −1 )γ 1 − (2n α −1 )γ β
Z (0.0.36)
w(t)dt, Q
provided γ > 0 is chosen small enough so that (2n α −1 )γ β < 1. Keeping track of the constants, we need to have log [w]A p − log [w]A p − (1 − α) p − log β γ< = log 2n − log α log 2n − log α
(0.0.37)
and to achieve this we set 1 2
γ = γ(n, p, B) =
log B − log [B − (1 − α) p ] log 2n − log α
=−
h i p log 1 − (1−α) B n
2 log 2α
.
Then γ decreases as a function of [w]A p and satisfies (0.0.37) as [w]A p ≤ B. Notice that if γ = γ(n, p, B), then γ 2n α −1 1 1 = 1+ = 1+ 1+ 1 . −γ (1−α) p p n −1 1 − (2n α −1 )γ β 2 (1−α) p (2 α ) − 1 − [w] 1 − (1−α) − 1 − Ap [w] [w] Ap
(0.0.38)
Ap
We now pick 0 < α < 1 such that (1 − α) p = 34 . Then the constant in (0.0.38) increases as [w]A p increases. We define " C3 (n, p, B) = 1 +
#
1 3 1 − 4B
12
3 − 1 − 4B
1 γ(n,p,B)+1
and we notice that C3 (n, p, B) is an increasing function of B. Then, by (0.0.36), we have Z
1+γ
w(t) Q
C3 (n, p, B)γ+1 dt = Qγ
and hence
1 Q
Z
1+γ
w(t) Q
dt
1 1+γ
≤
1+γ
Z w(t)dt
,
Q
C3 (n, p, B) Q
Z
w(t)dt. Q
which is (7.2.4) with C = C3 (n, p, B). (b) Given w ∈ A p , let γ = min(γ1 , γ2 ), C1 and C2 be as in the proof of Theorem 7.2.5. Then
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1
), C2 (n, p, [w]A p ) = C3 (n, p0 , [w− p−1 ]A p0 ) = C3 (n, p0 , [w]Ap−1 p
C1 (n, p, [w]A p ) = C3 (n, p, [w]A p ),
where C3 is defined in part (a). By Theorem 7.2.5 we know that w1+γ ∈ A p , so apply the result in Exercise 7.1.3 to get 1
w = (w1+γ ) 1+γ ∈ Aq , where q= p and
1 1 p+γ +1− = , 1+γ 1+γ 1+γ
i h 1 1 ≤ C1C2p−1 [w]A p . [w]Aq ≤ (w1+γ ) 1+γ w1+γ A1+γ p Aq
p+γ 1+γ
Since 1 < q = < p, we can take δ = p−q = increase as [w]A p increases, so [w]A p ≤ B implies
γ γ+1 (p−1) > 0, then q =
p−δ . Observe that the C1 (n, p, [w]A p ),C2 (n, p, [w]A p )
[w]A p−δ ≤ C1 (n, p, [w]A p )C2 (n, p, [w]A p ) p−1 [w]A p ≤ C1 (n, p, B)C2 (n, p, B) p−1 B = C4 (n, p, B).
Exercise 7.2.5 Given a positive doubling measure µ on Rn , define the characteristic constant [w]A p (µ) and the class A p (µ) for 1 < p < ∞. (a) Show that statement (8) of Proposition 7.1.5 remains valid if Lebesgue measure is replaced by µ. (b) Obtain as a consequence that if w ∈ A p (µ), then for all cubes Q and all µmeasurable subsets A of Q we have w(A) µ(A) p ≤ [w]A p (µ) . µ(Q) w(Q) Conclude that if Lebesgue measure is replaced by µ in Lemma 7.2.1, then the lemma is valid for w ∈ A p (µ). (c) Use Corollary 7.2.4 to obtain that weights in A p (µ) satisfy a reverse H¨older condition. (d) Prove that given a weight w ∈ A p (µ), there exists 1 < q < p, which depends on [w]A p (µ) , such that w ∈ Aq (µ). Solution. In this exercise the characteristic constant [w]A p (µ) and the class A p (µ) are defined as follows
1 [w]A p (µ) := sup µ(Q) Q
Z
wdµ Q
1 µ(Q)
Z
w Q
and A p (µ) := {w : [w]A p (µ) < ∞}. (a) Applying H¨older’s inequality with
1 p
+
1 p0
= 1 and p > 1, one gets
p0
0
− pp
p
dµ
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309
k f pk
1 L p (µ)
p 1 − 1p p  f (x)w(x) w(x) dµ(x)
1 µ(Q) Q Z Z p0 0 p 1 − pp p  f (x) w(x)dµ(x) w(x) dµ(x) ≤ µ(Q) p Q Q Z 1  f (x) p w(x)dµ(x) kwkL1 (µ) kw−1 k 1 = wµ (Q) Q L p−1 (µ) Z 1 ≤ [w]A p (µ)  f (x) p w(x)dµ(x) . wµ (Q) Q Z
=
(0.0.39)
Thus, the right hand side F of (8) of Proposition 7.1.5 satisfies F ≤ [w]A p (µ) . Let us show the reverse inequality, i.e., we want to show that there exists f > 0 in L p (Q, wdµ) such that [w]A p (µ) ≤ F. Let 0
f = (w + ε)−p /p , ε > 0. Then
p 1 R  f (t)dµ(t) Q µ(Q) 1 R p wµ (Q) Q  f (t) w(t)dµ(t)
= kwkL1 (µ)
p 1 R −p0 /p dµ(t) (w(t) + ε) Q µ(Q) 1 R −p0 w(t)dµ(t) wµ (Q) Q (w(t) + ε)
=
p−1 R 0 ( Q (w(t) + ε)−p /p dµ(t))
1 R −p0 /p dµ(t) µ(Q) Q (w(t) + ε)
−p0 w(t)dµ(t) Q (w(t) + ε) p−1 R 0 1 R −p0 /p dµ(t) (w(t) + ε) ( Q (w(t) + ε)−p /p dµ(t)) µ(Q) Q R ≥ kwkL1 (µ) −p0 (w(t) + ε)dµ(t) Q (w(t) + ε) p−1 Z 0 1 (w(t) + ε)−p /p dµ(t) . = kwkL1 (µ) µ(Q) Q
R
(0.0.40)
Letting ε → 0, one gets F ≥ [w]A p (µ) . (b) Let us take f = χA in part (a). Then one obtains
µ(A) µ(Q)
p
wµ (A) wµ (Q)
≤ [w]A p (µ) ,
(0.0.41)
which gives the desired inequality. Let w ∈ A p (µ) and µ(S) ≤ α µ(Q), where 0 < α < 1. We want to show that there exists β < 1 such that wµ (S) ≤ β wµ (Q). Using (0.0.41) and letting S = Q \ A, we obtain wµ (A) wµ (S) µ(S) p µ(A) p 1− = ≤ [w]A p (µ) = [w]A p (µ) 1 − . µ(Q) µ(Q) wµ (Q) wµ (Q) From the above inequality, one gets wµ (S) ≤ 1− wµ (Q) Setting α :=
µ(S) µ(Q)
µ(S) p 1 − µ(Q) [w]A p (µ)
.
p
(1−α) and β := 1 − [w] , one can find β < 1 such that wµ (S) ≤ β wµ (Q) for a given 0 < α < 1. A p (µ)
(c) From Part (b) there exist 0 < α, β < 1, such that if µ(S) ≤ α µ(Q) then wµ (S) ≤ β wµ (Q). Since µ is also a doubling measure, one may use Corollary 7.2.4 to conclude that weights in A p (µ) satisfy a reverse H¨older condition, i.e. there exist 0 < C, γ < ∞ such that for every cube Q ∈ Rn we have
1 µ(Q)
Z
1+γ
w(t) Q
1 1+γ dµ(t) ≤
C µ(Q)
Z
w(t)dµ(t). Q
(0.0.42)
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(d) From Part (c) for w ∈ A p (µ) and w− p−1 ∈ A p0 (µ) there exist γ1 , γ2 > 0 and C1 ,C2 > 0 such that
1 µ(Q)
Z
1+γ1
w(t) Q
1 Z 1+γ1 C1 dµ(t) ≤ w(t)dµ(t) µ(Q) Q
and
1 µ(Q)
Z
1 (1+γ ) − p−1 2
w(t) Q
1 Z 1+γ2 1 C2 ≤ w(t)− p−1 dµ(t). dµ(t) µ(Q) Q
Take γ = min{γ1 , γ2 } and p > 1 to obtain 1+γ
[w1+γ ]A p (µ) ≤ (C1C2p−1 )1+γ [w]A p (µ) . 1 1 Let q = p 1+γ + 1 − 1+γ =
p+γ 1+γ .
Since
1 1+γ
∈ (0, 1), it follows that 1 < q < p. We have
kwkL1 (µ) kwk
(0.0.43)
1 L q−1 (µ)
= ≤
1 µ(Q)
Z
1 µ(Q)
Z
1 µ(Q) 1
wdµ Q 1+γ
w
Z
1+γ
dµ
Q
− 1+γ p−1
w
p−1 1+γ
dµ
Q
1 µ(Q)
Z
− 1+γ p−1
w
p−1 1+γ
dµ
Q
1 1+γ
≤ [w1+γ ]A p (µ) . Taking the sup in the above inequality and applying the inequality (0.0.43), one gets [w]Aq (µ) ≤ C1C2p−1 [w]A p (µ) .
Exercise 7.2.6 Let 1 < q < ∞ and µ a positive measure on Rn . We say that a positive function K on Rn satisfies a reverse H¨older condition of order q with respect to µ, symbolically K ∈ RHq (µ) , if 1
[K]RHq (µ) =
sup Q cubes in Rn
1 R q q µ(Q) Q K dµ R 1 µ(Q) Q K dµ
< ∞.
For positive functions u, v on Rn and 1 < p < ∞, show that 1
[vu−1 ]RHp0 (u dx) = [uv−1 ]Ap p (v dx) , that is, vu−1 satisfies a reverse H¨older condition of order p0 with respect to u dx if and only if uv−1 is in A p (v dx). Conclude that w ∈ RH p0 (dx) ⇐⇒ w−1 ∈ A p (w dx) , w ∈ A p (dx) ⇐⇒ w−1 ∈ RH p0 (w dx) . Solution. We have p0 = 1 −
1 , 1− p
1 − p0 =
1 , 1− p
1 p−1 = . 0 p p
(0.0.44)
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Denote
Z
Z
u(Q) :=
v(Q) :=
u dx, Q
v dx Q
We have [vu−1 ]RHp0 (udx) = sup
1 R −1 p0 u(Q) Q (vu ) udx
10 p
1 R −1 u(Q) Q (vu )udx
Q
v(Q) u(Q)
10 p
= sup
1 R p0 1−p0 dx v(Q) Q v u
10 p
1 R u(Q) Q vdx
Q
v(Q) u(Q)
p−1 p
1
1
1− 1−p 1−p 1 R u dx v(Q) Q v
= sup
p−1 p
v(Q) u(Q)
Q
= sup Q
v(Q) u(Q)
− 1 p
1 v(Q)
Z
1
(uv−1 ) 1−p vdx
p−1 p
Q
p−1 Z Z p p 1 1 1 −1 −1 1−p (uv )vdx (uv ) vdx = sup v(Q) Q v(Q) Q Q p−1 1 Z Z p 1 1 1 (uv−1 )vdx (uv−1 ) 1−p vdx = sup v(Q) Q v(Q) Q Q 1
1
= [uv−1 ]Ap p (vdx) . We deduce that [vu−1 ]RHp0 (udx) =
I(Q) =
sup Q cubes in Rn
sup Q cubes in Rn
1
1
J(Q) p = [uv−1 ]Ap p (vdx) .
Setting u = 1, v = w and v = 1, u = w, we have 1
1
[w]RHp0 (dx) = [w−1 ]Ap p (wdx) ,
[w−1 ]RHp0 (wdx) = [w]Ap p (dx) .
This implies w ∈ RH p0 (dx) ⇐⇒ w−1 ∈ A p (w dx), −1
w ∈ A p (dx) ⇐⇒ w
and
∈ RH p0 (w dx).
Exercise 7.2.7 ([126]) Suppose that a positive function K on Rn lies in RH p (dx) for some 1 < p < ∞. Show that there exists a δ > 0 such that K lies in RH p+δ (dx). Hint: By Exercise 7.2.6, K ∈ RH p (dx) is equivalent to the fact that K −1 ∈ A p0 (K dx), and the index p0 can be improved by Exercise 7.2.5 (d). Solution.
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We know from Exercise 7.2.6 that K ∈ RH p (dx) is equivalent to the fact that K −1 ∈ A p0 (Kdx). Then Exercise 7.2.5 (d), there exists q0 = q0 ([K −1 ]A p0 (kdx) ) with 1 < q0 < p0 such that K −1 ∈ Aq0 (Kdx). This, again by Exercise 7.2.6, is equivalent to the fact that K ∈ RHq (dx). But, q0 < p0 ⇒ p0 q0 − p0 < p0 q0 − q0 ⇒ p0 (q0 − 1) < q0 (p0 − 1) q0 p0 < p0 − 1 q0 − 1 ⇒ p < q = p+δ,
⇒
for some δ > 0.
Exercise 7.2.8 (a) Show that for any w ∈ A1 and any cube Q in Rn and a > 1 we have ess.inf w ≤ an [w]A1 ess.inf w . Q
aQ
(b) Prove that there is a constant Cn such that for all locally integrable functions f on Rn and all cubes Q in Rn we have ess.inf M( f ) ≤ Cn ess.inf M( f ) , 3Q
Q
and an analogous statement is valid for Mc . 1 Hint: Part (a): Use (7.1.18). Part (b): Apply part (a) to M( f ) 2 , which is an A1 weight in view of Theorem 7.2.7. Solution. (a) Since we obviously have ess.inf w ≤ w(x) a.e. x ∈ Q, Q
it follows that 1 w(x)dx Q Q Z an = w(x)dx aQ Q Z an ≤ w(x)dx aQ aQ ≤ an [w]A1 ess.inf w(x).
ess.inf w ≤ Q
Z
aQ
(b) Using Theorem 7.2.7, we have M( f )ε ∈ A1 for 0 < ε < 1. Now, taking w = M( f )ε in part (a), we have ess.inf M( f )ε ≤ 3n [M( f )ε ]A1 ess.inf M( f )ε . aQ
Q
Since [M( f )ε ]A1 is independent of f , we get
ess.inf M( f )ε Q
1
1 ε
1 ε ≤ 3n [M( f )ε ]A1 ess.inf M( f )ε .
Using the fact that 0 < ε < 1, the function t ε is increasing, we can write
3Q
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313 1
n
ess.inf M( f ) ≤ 3 ε [M( f )ε ]Aε 1 ess.inf M( f ). 3Q
Q
n
1
Letting cn = 3 ε [M( f )ε ]Aε 1 , we have
ess.inf M( f ) ≤ cn ess.inf M( f ). 3Q
Q
Now to get the analogous statement for Mc , we note the following relation between Mc and M, AMc ( f )(x) ≤ M( f )(x) ≤ BMc ( f )(x). for suitable constants A and B. This gives, 1 ess.inf M( f ) A Q cn ≤ ess.inf M( f ) A 3Q cn ≤ B ess.inf Mc ( f ) 3Q A = c˜ ess.inf Mc ( f )
ess.inf Mc ( f ) ≤ Q
3Q
where c˜ =
cn A B.
Exercise 7.2.9 ([224]) For a weight w ∈ A1 (Rn ) define a quantity r = 1 + 2n+11[w] . Show that A1
1
Mc (wr ) r ≤ 2 [w]A1 w
a.e. Hint: Fix a cube Q and consider the family FQ of all cubes obtained by subdividing Q into a mesh of (2n )m subcubes R of side length 2−m `(Q) for all m = 1, 2, . . . . Define MQd ( f )(x) = supR∈FQ ,R3x R−1 R  f  dy. Using Corollary 2.1.21 obtain R
d (w)>λ } w(x) dx Q∩{MQ
≤ 2n λ {x ∈ Q : MQd (w)(x) > λ } for λ > wQ = Z Q
MQd (w)δ w dx ≤ (wQ )δ
Replace w by wk = min(k, w) and select δ =
using [wk ]A1
1 Q ≤ [w]A1 . Then let k → ∞.
Z Q
1 2n+1 [w]A1
wδk +1 dx ≤
Z
w dx + Q
1 R Q Q w dt.
2n δ δ +1
Z Q
Multiply by λ δ −1 and integrate to obtain
MQd (w)δ +1 dx .
to deduce 1 Q
Z Q
MQd (wk )δ wk dx ≤ 2(wQ )δ +1 ,
Solution. Fix a cube Q and consider the family FQ of all cubes obtained by subdividing Q into a mesh of (2n )m subcubes of side length 2−m `(Q) for all m = 1, 2, . . . . Define a maximal operator MQd adapted to Q by setting MQd ( f )(x) =
−1
R
sup R∈FQ ,R3x
Z
 f  dy .
R
1 Applying Corollary 2.1.21 to the function w and λ > wQ = Q Q w dt, we obtain the existence of disjoint subcubes Q j of Q S such that Q j ∈ FQ and such that for almost all x ∈ Q \ j Q j we have w(x) ≤ λ and
R
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1 Q j 
λ
λ } j
where the last inequality is due to the fact that on every Q j , the average of w over Q j is bigger than λ . Next we claim that [ Q ∩ {MQd (w) > λ } j Q j . j
Indeed, suppose that y ∈ Q is such that MQd (w)(y) > λ . Then there is a maximal cube (in terms of size) from FQ such that y lies in it and the average of w over this cube is bigger than λ . But this cube must be a selected one according to the selection procedure of Proposition 2.1.20, which is also adopted by Corollary 2.1.21. So this maximal cube must be one of the Q j ’s, and the claim follows. We have therefore proved that Z d (w)>λ } Q∩{MQ
for λ > wQ =
1 R Q Q w dt.
w(x) dx ≤ 2n λ {x ∈ Q : MQd (w)(x) > λ }
Multiply by δ λ δ −1 and integrate from wQ to ∞ to obtain
Z ∞
δ λ δ −1
Z d (w)>λ } Q∩{MQ
wQ
w(x) dxdλ ≤ 2n
Z ∞ wQ
δ λ δ {x ∈ Q : MQd (w)(x) > λ } dλ
which implies Z Z M d (w) Q
δ λ δ −1 dλ w(x)dx ≤
Q wQ
2n δ δ +1
This gives Z Q
Setting δ =
1 , 2n+1 [w]A1
MQd (w)δ w dx ≤ (wQ )δ
Z
1 Q
Z
w dx + Q
Z Q
2n δ δ +1
MQd (w)δ +1 dx .
Z Q
MQd (w)δ +1 dx .
we obtain that 1 Q
Z
MQd (w)δ w dx ≤ (wQ )δ
Q
w dx + Q
2n δ 1 δ + 1 Q
Z Q
MQd (w)δ [w]A1 w dx .
Up until this point w could have been any A1 weight. Replacing w by wk = min(k, w) we obtain 1 Q
Z Q
MQd (wk )δ wk dx ≤ ((wk )Q )δ
1 Q
Z Q
wk dx +
2n δ 1 δ + 1 Q
Z Q
MQd (wk )δ [wk ]A1 wk dx
but since [wk ]A1 ≤ [w]A1 (Exercise 7.1.8) and (wk )Q ≤ wQ we have 1 Q
Z Q
MQd (wk )δ wk dx ≤ (wQ )δ +1 +
1 1 2δ + 2 Q
Z Q
MQd (wk )δ wk dx .
The right hand side is finite, so we obtain 1 1 2 Q and from this we deduce
Z Q
MQd (wk )δ wk dx 1 Q
Z Q
Z 1 1 1 ≤ 2− M d (wk )δ wk dx ≤ (wQ )δ +1 2 δ + 1 Q Q Q
wδk +1 dx ≤
1 Q
Z Q
MQd (wk )δ wk dx ≤ 2(wQ )δ +1
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315
which implies
1 Q
Z Q
wδk +1 dx
1 δ +1
1
≤ 2 δ +1 wQ ≤ 2wQ ≤ 2[w]A1 w
a.e.
Then we let k → ∞ and use the Lebesgue monotone convergence theorem to deduce the conclusion with r = 1 + δ .
Exercise 7.2.10 Let 1 < p < ∞. Recall that a pair of weights (u, w) that satisfies [u, w](A p ,A p ) = sup
Q cubes in Rn
1 Q
Z
u dx Q
1 Q
Z
1
w− p−1 dx
p−1 1 such that 1
−p0 −q0 = εp q 1 0 (p − 1) = q0 − 1 ε 1 q0 = (p0 − 1) + 1 > 1 ε
if
p0 > 1,
where we used, Z 1 Z ε 1 f (y)dy ≤ Mc ( f )ε (x)dx. Q Q Q Q
which is a consequence of the fact that
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Contents
ε 1 Z f (y)dy ≤ Mc ( f )(x)ε Q Q
∀x ∈ Q .
(b) Let g ≥ 0, integrable and nontrivial (so that M(g) ∈ / L1 ). In addition to this, take g bounded so that M(g)(x) is always finite. Then, p (g, M(g)) ∈ (A1 , A1 ) ⊆ (A p0 , A p0 ), p0 = . p−1 This implies that, − 1 − 1 M(g) p0 −1 , g p0 −1 = M(g) p−1 , g1−p ∈ (A p , A p ) Now take u = M(g)1−p , w = g1−p . Then, Z Rn
M(g)(x) p u(x)dx = Z Rn
Z Rn
M(g)(x) p M(g)(x)1−p dx =
g(x) p w(x)dx =
Z Rn
Z Rn
g p g1−p dx =
M(g)(x)dx = ∞,
Z Rn
g(x)dx < ∞ ,
where we picked g ∈ L1 (Rn ) such that M(g) ∈ / L1 (Rn ). Thus the inequality Z Rn
M( f )(x) p u(x)dx ≤ c
Z Rn
 f (x) p w(x)dx
cannot hold, in particular, for f = g. (c) The weak type estimate that M maps L p (M(g)dx) to L p,∞ (gdx) is a consequence of Exercise 7.1.6. The strong type estimate that M maps Lq (M(g)dx) to Lq (gdx) is a consequence of Exercise 7.1.7.
Section 7.3. The A∞ Condition Exercise 7.3.1 Let λ > 0, Q be a cube in Rn , and w ∈ A∞ (Rn ). (a) Show that property (6) in Proposition 7.3.2 can be improved to n
w(λ Q) ≤ min ε>0
(b) Prove that
n
(1 + ε)λ [w]λA∞ − 1 w(Q) . ε n (1+log [w] ) A∞ 2
w(λ Q) ≤ (2λ )2
w(Q) . n 1/λ Hint: Part (a): = (1 + ε)[w]A∞ . Part (b): Use the estimate in property (6) of Proposition 7.3.2 Take c in (7.3.4) such that c with λ = 2. Solution. (a) We prove the inequality by applying property (5) of Proposition 7.3.2 to the cube λ Q in place of Q and to the function c, on Q; f= 1, on Rn \ Q, where c is a constant which will be chosen later. We have log c w(λ Q) R e λ n ≤ [w]A∞ . Q cw(t)dt + λ Q\Q w(t)dt
R 1
Choosing c so that c λ n = (1 + ε)[w]A∞ , one gets
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317
w(λ Q) (1 + ε)[w]A∞ ≤ [w]A∞ . n n (1 + ε)λ [w]λA∞ w(Q) + w(λ Q \ Q) Since w(λ Q \ Q) = w(λ Q) − w(Q), it follows that n
n
(1 + ε)λ [w]λA∞ w(Q) − w(Q) w(λ Q) ≤ ε for ε > 0. Taking the minimum over all ε > 0 we obtain the conclusion. (b) This is proved in the following way. Property (6) in Proposition 7.3.2 gives n
n
w(λ Q) ≤ 2λ [w]λA∞ w(Q).
(0.0.45)
Letting λ = 2 in (0.0.45), we obtain n
n
n
w(2Q) ≤ 22 [w]2A∞ w(Q) = 22 22
n log [w] A∞ 2
n (1+log
w(Q) = 22
2 [w]A∞ )
w(Q).
(0.0.46)
For 1 < λ ≤ 2 we argue in the following way: We obtain from (0.0.45) n (1+log [w] ) A∞ 2
w(λ Q) ≤ w(2Q) ≤ 22
n (1+log
w(Q) ≤ (2λ )2
2 [w]A∞ )
w(Q) .
(0.0.47)
since by Proposition 7.3.2 (4), one has log2 [w]A∞ ≥ 0. Now let us prove the inequality for λ > 2. For any fixed λ > 2 there exist k ∈ N such that λ < 2k . Let k0 := min{k ∈ N : λ < 2k }.
(0.0.48)
Then n (1+log
w(λ Q) ≤ w(2k0 Q) ≤ (2k0 )2
2 [w]A∞ )
w(Q) = (2 · 2k0 −1 )2
n (1+log
2 [w]A∞ )
w(Q).
(0.0.49)
By (0.0.48), one obtains 2k0 −1 < λ . Therefore, from (0.0.49) the desired inequality follows, i.e. , w(λ Q) ≤ (2λ )2
n (1+log
2 [w]A∞ )
w(Q).
Exercise 7.3.2 Suppose that µ is a positive Borel measure on Rn with the property that for all cubes Q and all measurable subsets A of Q we have A < αQ =⇒ µ(A) < β µ(Q) for some fixed 0 < α, β < 1. Show that µ is doubling [i.e., it satisfies (7.2.9)]. Hint: Use that S > (1 − α)Q ⇒ µ(S) > (1 − β )µ(Q) when S j Q. Solution. Given a cube Q, take any measurable subset S j Q and set A = Q \ S. Then, by assumption, we have Q − S < αQ ⇒ µ(Q) − µ(S) < β ω(Q), which yields S > (1 − α)Q ⇒ µ(S) > (1 − β )µ(Q) . Now, choose λ ∈ (0, 1) such that
λn
> 1 − α. Since λ < 1 we have S = λ Q ⊂ Q. Then, since S = λ n Q > (1 − α)Q,
(0.0.50)
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by (0.0.50) we must have for any cube Q in Rn 1 µ(λ Q) 1−β
µ(Q)
1 we obtain that µ is a doubling measure, since the double of a cube is contained in a finite union of 1/λ dilates. Precisely, pick q ∈ Z+ such that 3 < (1/λ )q and we obtain that 1 µ(R) . (1 − β )q
µ(3R) ≤
Exercise 7.3.3 1
− p−1 are in A∞ . Prove that a weight w is in A p if and only if both w and w Hint: You may want to use the result of Exercise 7.2.2.
Solution. By Exercise 7.2.2, w ∈ A p if and only if both of the following hold: 1 elog w−log wQ dt ≤ C Q Q Q Z 1 1 sup e−(log w−log wQ ) p−1 dt ≤ C . Q Q Q Z
(i)
sup
(ii) Inequality (i) is equivalent to sup Q
1 1 Q elog wQ
Z
elog w dt ≤ C.
Q
But, sup Q
1 1 Q elog wQ
Z
1 R log(w−1 )dt Q
elog w dt = sup e Q
Q
Q
1 Q
Z Q
w(t)dt = [w]A∞
and thus inequality (i) is equivalent to w ∈ A∞ . Inequality (ii) is equivalent to Z 1 R log(w p−1 )dt 1 1 Q sup e−(log w) p−1 dt ≤ C. e Q Q Q But we have sup Q
1 R log(w p−1 )dt 1 Q e Q
Z Q
1
1
e−(log w) p−1 dt = [w− p−1 ]A∞
1
which means that inequality (ii) is equivalent to w− p−1 ∈ A∞ . 1 − p−1
Therefore w ∈ A p if and only if w ∈ A∞ and w
∈ A∞ .
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Exercise 7.3.4 ([33], [344]) Prove that if P(x) is a polynomial of degree k in Rn , then log P(x) is in BMO with norm depending only on k and n and not on the coefficients of the polynomial. Hint: Use that all norms on the finitedimensional space of polynomials of degree at most k are equivalent to show that P(x) satisfies a reverse H¨older inequality. Therefore, P(x) is an A∞ weight and thus Exercise 7.2.3 (c) is applicable. Solution. Let Pk,n = {polynomials of degree k or less in Rn } which is a vector space of dimension k. Let P(x) be a polynomial of degree k in Rn . Then P(x) ∈ Pk,n . Denote B = B(0, 1). For 1 ≤ q < ∞, 1 Z q 1 P(x)q dx kPkLq = B B defines a norm on Pk,n ∼ = Rk . But in Rk , all norms are equivalent. Hence, in particular, k · kL1 and k · kL2 are equivalent norms in Pk,n . Hence, there exists c1 , c2 > 0 (independent of the coefficients) such that c1 (k, n)kPkL1 ≤ kPk2 ≤ c2 (k, n)kPkL1 , i.e., Z
C1 (k, n)
P(x) dx ≤
Z
2
1
P(x) dx
B
B
2
≤ C2 (k, n)
Z
P(x) dx.
B
Given any ball B0 = B(x0 , r), x0 ∈ Rn , r > 0, define T : Rn → Rn by T (y) = ry + x0 . Now,
1 B0 
Z B0
2
1
P(x) dx
2
= = ≤ = = =
1 B0 
Z
2
1 2
P(x) dx
T (B)
1 P(T (y))2  det JacT  dy αrn B Z c(k, n) P(T (y)) dy α B Z c(k, n) P(T (y))rn dy α B Z c(k, n) P(T (y)) det JacT  dy B0  B Z c(k, n) P(x) dx. B0  B0 Z
1 2
This is a reverse H¨older’s inequality. Hence, P(x) ∈ RH1 with a constant c(k, n). By Theorem 7.3.3 (c), this implies that P(x) ∈ A∞ with [P]A∞ depending only on k and n. Hence, P ∈ A p for some 1 < p < ∞ with [P]A p depending only on k and n. Then by Exercise 7.2.3 (c), log P ∈ BMO with k log PkBMO depending only on k and n and not on the coefficients of the polynomial P(x).
Exercise 7.3.5 Show that the product of two A1 weights may not be an A∞ weight. Solution. We note the following results from Example 7.1.7:
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(a) (b)
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xa is an A p weight if and only if −n < a < n(p − 1). xa is an A1 weight if and only if −n < a ≤ 0.
We start with the weight w(x) = x−n+ε which gives w2 (x) = x−2n+2ε . For w(x) to be in A1 , we must have −n < −2n + 2ε < 0, 1 1 that is, n < 2ε. Choose ε = 41 . Then, w(x) = x−n+ 4 ∈ A1 . However, w2 (x) = x−2n+ 4 ∈ / A1 . Moreover, w2 (x) ∈ / A p for any 1 < p < ∞. Indeed, since [ Ap, A∞ = 1≤p 0, since w cannot vanish on a set of positive measure. (2) 1 ≤ p < ∞. In this case apply H¨older’s inequality to obtain Z Z 1 1 gdx = gw p w− p dx B
B
≤
Z
1
(gw p ) p dx
1 Z p
B
Z = 1 p
+
1 p0
0
10 p
B
g p wdx
1 Z p
p0
w− p dx
10 p
,
B
B
where
1
(w− p ) p dx
= 1.
p0 p0 R − p0 10 R 1 (Rn ) and hence p dx p < ∞. Therefore By Proposition 7.1.5 (4), w− p ∈ A p0 as w ∈ A p . Thus w− p ∈ Lloc Bw B gdx < 1 (Rn ). ∞, which shows that g ∈ Lloc S (3) p = ∞. From Corollary 9.3.4, since A∞ = 1≤p 0 such that for all cubes Q we have w x ∈ Q : w(x) > λ ≤ C λ x ∈ Q : w(x) > γλ for all λ > AvgQ w. Hint: The displayed condition easily implies that 1 Q
Z Q
w(Q) ε+1
w1+ε dx ≤ k
Q
+
C0 δ 1 γ 1+ε Q
Z Q
wk1+ε dx ,
where k > 0, wk = min(w, k) and δ = ε/(1 + ε). Take ε > 0 small enough to obtain the reverse H¨older condition (c ) in Theorem 7.3.3 for wk . Let k → ∞ to obtain the same conclusion for w. Conversely, find constants γ, δ ∈ (0, 1) as in condition (a) of Theorem 7.3.3 and for λ > AvgQ w write the set {w > λ } ∩ Q as a union of maximal dyadic cubes Q j such that λ < 2n λ AvgQ j w ≤ 2n λ for all j. Then w(Q j ) ≤ 2n λ Q j  ≤ 1−δ Q j ∩ {w > γλ } and the required conclusion follows by summing on j. Solution. Suppose w ∈ A∞ . Then, by Theorem 7.3.3 (a), there exist 0 < γ, δ < 1 such that for all cubes Q in Rn {x ∈ Q : w(x) ≤ γ Avg w} ≤ δ Q . Q
Equivalently, this can be written as {x ∈ Q : w(x) > γ Avg w} ≥ (1 − δ )Q Q 1 R Q Q w(x)dx
For λ > AvgQ w = we may choose a collection Q j of maximal (hence disjoint) subcubes of Q (obtained by a dyadic decomposition of Q) such that \ [ {w > λ } Q = Q j j
and such that λ < Avg w = Qj
1 Q j 
Z
w dx ≤ 2n λ
∀ j.
Qj
This implies that Z
w(Q j ) =
Qj
w(x)dx ≤ 2n λ Q j  ≤
\ \ 2n λ 2n λ Q j {w > γ Avg w} ≤ Q j {w > γλ }. 1−δ 1−δ Qj
Summing over all j yields one direction of the desired equivalence. Conversely, suppose that there exists γ,C > 0 such that for all cubes Q in Rn we have w({x ∈ Q : w(x) > λ }) ≤ cλ {x ∈ Q : w(x) > γλ }, for every λ > Avg w. Let wk = min{w, k} where k is a positive constant. We wish to prove the following inequality: Q
1 Q
Z Q
w1+ε k dx ≤
w(Q) Q
Now we start with, Z Q
w1+ε k dx =
Z Q
wεk wk dx
1+ε +
c0 δ 1+ε γ Q
Z Q
w1+ε k .
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dλ . ελ ε wk {x ∈ Q : wk (x) > λ } λ 0 = I + II Z ∞
=
where, wk (Q) Q
Z
I= 0
dλ ελ ε wk {x ∈ Q : wk (x) > λ } λ
≤ εwk (Q) =
wk
wk (Q) Q
Z
0 (Q)1+ε
Qε
λ ε−1 dλ
,
and, Z ∞
II =
wk (Q) Q
≤C ≤C
dλ ελ ε wk {x ∈ Q : wk (x) > λ } λ
Z ∞ wk (Q) Q
ελ ε+1 {x ∈ Q : wk (x) > γλ }
Z ∞ γwk (Q) Q
ε
dλ λ
1 ε dt t {x ∈ Q : wk (x) > t} ε γ t
∞ 1 dt CQ ε (ε + 1)t ε+1 {x ∈ Q : wk (x) > t} wk (Q) γ ε+1 ε + 1 0 t Z CQ ε 1 = wk (x)ε+1 dx. wk (Q) ε + 1 γ ε+1 Q
Z
≤
Set δ =
ε 1+ε .
Then, 1 Q
Z Q
wε+1 k dx ≤
That is, 1 Q
w(Q) 1+ε Q
+
C0 δ 1 γ ε+1 Q
Z Q
wε+1 k dx
C0 δ wk (Q) ε+1 wε+1 . k dx 1 − γ ε+1 ≤ Q Q
Z
0
δ Take ε > 0 such that 1 − γCε+1 > 0. Then,
Z 1 1 Z C0 1+ε wε+1 (x)dx ≤ wk (x)dx. Q Q k Q Q
Therefore by Theorem 7.3.3 (c), wk ∈ A∞ . Now let k → ∞. Then by the same theorem it follows that w ∈ A∞ .
Section 7.4. Weighted Norm Inequalities for Singular Integrals Exercise 7.4.1 1 (Rn ) satisfy w > 0 a.e. Show that C ∞ (Rn ) is dense in L p (w). In particular this assertion holds Let 1 ≤ p < ∞ and let w ∈ Lloc 0 for any w ∈ A∞ .
Solution. Fix w a weight and f in L p (w). For M, N ∈ Z+ define
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fN (x) = f (x)χx≤N χ f ≤N and wM (x) = w(x)χw≤M . Then fN is a compactly supported bounded function that converges to f in L p (w) by the Lebesgue dominated convergence theorem. Indeed,  fN − f  p w ≤  f  p w ∈ L1 and  fN − f  p w → 0 a.e. as N → ∞. Thus, for an arbitrary small ε > 0, there exists N such that k f − fN kL p (w) ≤
ε . 3
For the given N and ε > 0 we find M such that Z
1
p
M
(w − w ) dx
N x≤N
ε ≤ . 6
This is possible again by the Lebesgue dominated convergence theorem, since the integrand tends to zero pointwise a.e. as M → ∞ and is bounded by w which is an L1 function over the ball x ≤ N. Finally, given N and M there is a smooth function with compact support φ supported in B(0, N) such that φ  ≤ N and k f N − φ kL p ≤
ε , 3M
since fN ∈ L p and C0∞ is dense in L p for p < ∞. (This can be done via the intermediate step of approximating with step functions). Combining these three inequalities we obtain k f − φ kL p (w) ≤ k f − fN kL p (w) + k fN − φ kL p (wM ) + k fN − φ kL p (w−wM ) ε ≤ + Mk fN − φ kL p + 2NkχB(0,N) kL p (w−wM ) 3 ε ε ε +2 ≤ +M 3 3M 6 = ε. This implies that C0∞ is dense in L p (w).
Exercise 7.4.2 ([74]) Let T be in CZO(δ , A, B). Show that for all ε > 0 and all 1 < p < ∞ there exists a constant Cn,p,ε,δ such that for all 1 (Rn ) and M(u1+ε ) < ∞ a.e. we have f ∈ L p (Rn ) and for all measurable nonnegative functions u with u1+ε ∈ Lloc Z Rn
T
(∗)
p
( f ) u dx ≤ Cn,p,ε,δ (A + B)
p
Z Rn
1
 f  p M(u1+ε ) 1+ε dx .
Hint: Obtain this result as a consequence of Theorems 7.4.6 and 7.2.7. Solution. By Theorem 7.4.6 we know that there is a constant C p = C p (n, δ , [w]A∞ ) such that 1
kT ∗ ( f )kL p (w) ≤ C p (A + B) [w]Ap−1 k f kL p (w) p whenever w ∈ A p and f ∈ L p (w). 1
1 (Rn ) we have u ≤ M(u1+ε ) 1+ε for all ε > 0. So we have Then, for any function such that u1+ε ∈ Lloc
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Z Rn
T ∗ ( f ) p udx ≤
Z
1
Rn
T ∗ ( f ) p M(u1+ε ) 1+ε dx.
Since by assumption (M(u1+ε ) < ∞ a.e., applying Theorem 7.2.7 with k ≡ 1, we obtain that 1 w := M(u1+ε ) 1+ε is an A1 weight with [w]A1 characteristic constant bounded by a constant that depends only on n and ε. Consequently, Z Rn
T ∗ ( f ) p u dx ≤
p
Z Rn
T ∗ ( f ) p w dx = kT ∗ ( f )kLp p (w) ≤ C p (n, δ , [w]A∞ ) p (A + B) p [w]Ap−1 k f kLp p (w) . p
Since w ∈ A1 ⊂ A p , thus [w]A∞ ≤ [w]A p ≤ [w]A1 ≤ c(n, ε) is also bounded by a constant that depends on ε and n. Thus we have p
C p (n, δ , [w]A∞ ) p (A + B) p [w]Ap−1 ≤ p
h
Also we have k f kLp p (w) =
ip p C p (n, δ ,t) c(n, ε) p−1 (A + B) p = Cn,p,ε,δ (A + B) p .
sup 1≤t≤c(n,ε)
Z Rn
 f  p wdx =
Z Rn
1
 f  p M(u1+ε ) 1+ε dx.
Therefore Z Rn
p
k f kLp p (w) ≤ Cn,p,ε,δ (A + B) p T ∗ ( f ) p u dx ≤ C p (n, δ , [w]A∞ ) p (A + B) p [w]Ap−1 p
Z Rn
1
 f  p M(u1+ε ) 1+ε dx
and the constant in front of the second integral is independent of u.
Exercise 7.4.3 Use the idea of the proof of Theorem 7.4.6 to prove the following result. Suppose that for some fixed A, B > 0 the nonnegative µmeasurable functions F and G on a σ finite measure space (X, µ) satisfy the distributional inequality µ {G > α} ∩ {F ≤ cα} ≤ A µ {G > Bα} for all α > 0. Given 0 < p < ∞, if A < B p and kGkL p (µ) < ∞, show that kGkL p (µ) ≤
B (B p − A)1/p
1 kFkL p (µ) . c
Solution. We have, for 0 < p < ∞, kGkLp p (µ) =
Z ∞
pα p−1 µ({G > α})dα
0
Z ∞
=
pα
p−1
µ({G > α} ∩ {F ≤ cα})dα +
Z ∞
pα p−1 Aµ({G > Bα})dα +
Z ∞
0
Z ∞
=
pα p−1 µ({G > α} ∩ {F > cα})dα
0
0
≤
Z ∞
pα p−1 µ({F > cα})dα
0
pα p−1 Aµ({G > Bα})dα +
0
A ∞ p−1 1 pt µ({G > t})dt + p Bp 0 c A 1 = p kGkLp p (µ) + p kFkLp p (µ) . B c Z
1 cp
Z ∞
Z ∞
=
0
pt p−1 µ({F > t})dt
0
pt p−1 µ({F > t})dt
Contents A Bp
Since
325
< 1 and kGkLp p (µ) < ∞, it follows that B
kGkL p (µ) ≤
1
c(B p − A) p
kFkL p (µ) .
Exercise 7.4.4 Let α > 0, w ∈ A1 , and f ∈ L1 (Rn , w) ∩ L1 (Rn ). Let f = g + b be the Calder´on–Zygmund decomposition of f at height α > 0 R given in Theorem 5.3.1, such that b = ∑ j b j , where each b j is supported in a dyadic cube Q j , Q j b j (x)dx = 0, and Q j and Qk have disjoint interiors when j 6= k. Prove that (a) kgkL1 (w) ≤ [w]A1 k f kL1 (w) and kgkL∞ (w) = kgkL∞ ≤ 2n α, (b) kb j kL1 (w) ≤ (1 + [w]A1 )k f kL1 (Q j ,w) and kbkL1 (w) ≤ (1 + [w]A1 )k f kL1 (w) , (c) ∑ j w(Q j ) ≤
[w]A1 α k f kL1 (w) .
Solution. According to the proof of Theorem 5.3.1 we have (1)
f = g + b.
(2)
kgkL1 ≤ k f kL1 and kgkL∞ ≤ 2n α.
(3)
b = ∑ j b j , where each b j is supported in a dyadic cube Q j . Furthermore, the cubes Qk and Q j are disjoint when j 6= k. Z
(4) Qj
b j (x) dx = 0.
b j 1 ≤ 2n+1 αQ j . L
∑ j Q j  ≤ α −1 f L1 ,
(5) (6)
where
bj =
and g=
f
1 f− Q j 
Z Qj
f dx χQ j ,
on Rn \
Q1j 
S
j Q j,
Z
on Q j .
f dx Qj
To prove (c), simply observe that
∑ w(Q j ) ≤ ∑ j
j
≤∑ j
≤ ≤
1 α
w(Q j ) Q j  Q j  w(Q j ) 1 Q j  α Z
∑ j
[w]A1 α
Qj
j
Qj
 f  dx
Qj
1 Q j  Z
∑
Z
Z
w(t)dt f (x) dx Qj
 f (x) w(x) dx
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=
[w]A1
f 1 . L (w) α
We have Z Qj
b j  w dx ≤
Z
 f  w dx +
Qj
Z
Qj
Z
 f  w dx +
=
Z
Qj
≤
1  f (t) dtw(x) dx Q j  Q j Z 1 w(x) dx f (t) dt Q j  Q j
Z
Qj
Z
 f  w dx +
Z
Qj
[w]A1 w(t) f (t) dt
Qj
Z
= (1 + [w]A1 )
 f  w dx ,
Qj
which proves the first claim in (b). The second claim in (b) follows by summing over j. S It remains to prove (a). Since g = f a.e. on ( j Q j )c , we obviously have Z ( j Q j )c
g w dx ≤
Z ( j Q j )c S
S
 f  w dx ≤ [w]A1
Z ( j Q j )c
 f  w dx .
S
Then (a) would follow if we had Z Qj
g w dx ≤ [w]A1
Z
 f  w dx
Qj
for each j. But for each j we have Z
g w dx =
Qj
Z Qj
1 Q j 
Z
 f (y) dy w dx
Qj
1 =  f (y) Q Qj j Z
≤
Z Qj
Z
w dx dy Qj
 f (y) [w]A1 w(y) dy Z
= [w]A1
 f  w dx .
Qj
Finally we notice that {g > t} = 0 ⇐⇒ w({g > t}) = 0 , since a weight cannot vanish on a set of positive Lebesgue measure. This implies that kgkL∞ = kgkL∞ (w) and we know that this is at most 2n α.
Exercise 7.4.5 Assume that T is an operator associated with a kernel in SK(δ , A). Suppose that T maps L2 (w) to L2 (w) for all w ∈ A1 with bound Bw . Prove that there is a constant Cn,δ such that kT kL1 (w)→L1,∞ (w) ≤ Cn,δ (A + Bw ) [w]2A1 for all w ∈ A1 . Hint: Apply the idea of the proof of Theorem 5.3.3 using the Calder´onZygmund decomposition f = g + b of Exercise 7.4.4 at height γα for a suitable γ. To estimate T (g) use an L2 (w) estimate and Exercise 7.4.4. To estimate T (b) use the mean value property, the fact that
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327
y − c j δ w(x) dx ≤ Cδ ,n M(w)(y) ≤ Cδ0 ,n [w]A1 w(y) , Rn \Q∗j x − c j n+δ and Exercise 7.4.4 to obtain the required estimate. Z
Solution. We have, for K ∈ SK(δ , A), Z
T ( f )(x) =
Rn
K(x, y) f (y) dy.
Fix a function f ∈ L1 (Rn , w) ∩ L1 (Rn ) and write f = g + b = g + ∑ j b j in terms of its Calder´onZygmund decomposition at height αγ. Since, T ( f ) = T (g) + T (b), we have α α w {x ∈ Rn : T ( f )(x) > α} ≤ w {x ∈ Rn : T (g)(x) > } + w {x ∈ Rn : T (b)(x) > } 2 2 =: I + II We start by estimating I. α I = w {x ∈ Rn : T (g)(x) > } 2 Z 4 2 ≤ 2 T (g)(x) w(x)dx α Rn 4B2 ≤ 2w kgk2L2 (w) α 4B2w ≤ 2 kgkL1 (w) kgkL∞ (w) α 4B2w n ≤ 2 2 γα[w]A1 k f kL1 (w) α 2n+2 B2w = γ [w]A1 k f kL1 (w) . α Therefore, choosing γ = (A + Bw )−1 we obtain 2n+2 [w]A1 (A + Bw )k f kL1 (w) . α
I≤
(0.0.52)
We now turn to term II. Let Q∗j = 2Q j and Ω ∗ = ∪ j Q∗j . Then, since w is doubling with constant at most a multiple of 2n [w]A1 [Proposition 7.1.5 (9)], we obtain that w(Ω ∗ ) ≤ ∑ w(Q∗j ) ≤ 2n [w]A1 ∑ w(Q j ) ≤ 2n [w]A1 j
j
[w]A1 1 k f kL1 (w) = 2n [w]2A1 k f kL1 (w) . αγ αγ
Now, since T (b)(x) = T (∑ j b j )(x) ≤ ∑ j T (b j )(x), we have w {x ∈ / Ω ∗ : T (b)(x) >
Z α 2 2 } ≤ T (b)(x)w(x)dx ≤ 2 α Rn \Ω ∗ α
Z
∑ j
Rn \Q∗j
T (b j )(x)w(x)dx.
Hence, α 2 w x ∈ Rn : T (b)(x) > ≤ Ω ∗  + 2 α ≤ 2n [w]2A1 Let c j be the center of Q j . We have
Z
∑ j
Rn \Q j ∗
T (b)(x)w(x)dx
1 2 k f kL1 (w) + αγ α
Z w(x)dx. K(x, y)b (y)dy j Q Rn \Q∗
Z
∑ j
j
j
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Z
Z w(x)dx T (b)(x)w(x)dx = K(x, y)b (y)dy j ∗ Rn \Q∗ Q Z
Rn \Q j
j
j
Z
= ≤
Z
Rn \Q∗j Q j
Z Qj
≤A
Z b j (y)
Z Qj
≤A
K(x, y) − K(x, c j )b j (y)dyw(x)dx
Z Qj
Rn \Q∗j
Z b j (y)
Rn \Q∗j
Z Rn \Q∗j
y − c j δ w(x)dx dy x − c j n+δ
 b j (y)  Cδ0 ,n [w]A1 w(y) dy
= Cδ0 ,n [w]A1 A
Therefore,
K(x, y) − K(x, c j )w(x)dx dy
Z
b j (y) w(y) dy.
Qj
T (b)(x)w(x)dx = Cδ0 ,n [w]A1 A
Z Qj
b j (y) w(y) dy.
(0.0.53)
This estimate is based on the following inequality Z Rn \Q∗j
y − c j δ w(x)dx ≤ Cδ ,n M(w)(y) ≤ Cδ0 ,n [w]A1 w(y). x − c j n+δ
(0.0.54)
which will be proved later. Using this fact we obtain Z α 1 2A w x ∈ Rn : T (b)(x) > ≤ 2n [w]2A1 Cδ0 ,n [w]A1 k f kL1 (w) + b j (y)w(y)dy ∑ 2 αγ α j Qj
≤ 2n [w]2A1
2A 1 k f kL1 (w) + Cδ0 ,n [w]A1 kbkL1 (w) . αγ α
Therefore, II ≤ 2n [w]2A1
2A 1 k f kL1 (w) + Cδ0 ,n [w]A1 2[w]A1 k f kL1 (w) . αγ α
Recalling the choice of γ = (A + Bw )−1 yields that kT ( f )kL1,∞ (w) ≤ Cn,δ (A + Bw ) [w]2A1 k f kL1 (w) . Therefore, the result holds, but it remains to prove inequality (0.0.54) to complete the proof. Proof: We have x − c j  ≈ x − y for y ∈ Q j , x ∈ / Q∗j . Let `(Q j ) denote the side length of the cube Q j . Then, Z Rn \Q∗j
y − c j δ w(x)dx ≤ Cn,δ x − c j n+δ
Z Rn \Q∗j
≤ `(Q∗j )δ
y − c j δ dx x − yn+δ
Z Rn \Q∗j
= Cn,δ `(Q∗j )δ
w(x) dx x − yn+δ
Z Rn
h j (y − x)w(x)dx
= Cn,δ `(Q∗j )δ (h j ∗ w)(y),
where h j (y − x) =
1 x−yn+δ
if x ∈ / Q∗j and h j (y − x) = 0 otherwise. Passing to polar coordinates we find that,
(0.0.55)
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kh j kL1 (Rn ) ≤ cn,δ `(Q∗j )−δ On the other hand, by Theorem 2.1.10, we obtain Cn,δ `(Q∗j )(h j ∗ w)(y) ≤ Cn,δ `(Q∗j )δ `(Q∗j )−δ M(w)(y) ≤ Cn,δ [w]A1 w(y).
and this proves (0.0.54).
Exercise 7.4.6 Recall that the transpose T t of a linear operator T is defined by
T ( f ), g = f , T t (g) for all suitable f and g. Suppose that T is a linear operator that maps L p (Rn , vdx) to itself for some 1 < p < ∞ and some v ∈ A p . 0 0 Show that the transpose operator T t maps L p (Rn , wdx) to itself with the same norm, where w = v1−p ∈ A p0 . Solution. 0 Fix f ∈ L p (w). Given g ∈ L p (w) with kgkL p (w) ≤ 1, observe that kgwkLp p (v) =
Z Rn
gw p v =
Z
0
Rn
g p (v1−p ) p v =
Z
0
Rn
g p v1−p = kgkLp p (w) ≤ 1,
whence kgwkL p (v) ≤ 1.
(0.0.56)
Thus, Z Z t Rn T ( f )gw dx = Rn f T (gw) dx ≤
Z ZR
= ≤
n
(definition of T t )
 f  T (gw) dx − p10
1
Rn
 f  w p0 T (gw) w
Z Rn
p0
 f  w dx
= k f kL p0 (w)
Z Rn
10 Z p
p
Rn
T (gw)
dx
p
− pp0
T (gw) w
0 − p (v1−p ) p0 dx
1
p
(H¨older)
dx
1
p
(definition of w)
= k f kL p0 (w) kT (gw)kL p (v)
(algebra)
≤ kT kL p (v)→L p (v) k f kL p0 (w) kgwkL p (v)
(boundedness of T )
≤ kT kL p (v)→L p (v) k f kL p0 (w)
(inequality (0.0.56)).
In summary, we have shown that Z t ≤ kT k p T ( f )gw dx L (v)→L p (v) k f kL p0 (w) Rn 0
holds for every f ∈ L p (w) and g ∈ L p (w) with kgkL p (w) ≤ 1. Taking the supremum over all g ∈ L p (w) with kgkL p (w) yields that T t ( f ) ∈ L p (v) and that
t
T ( f ) p0 ≤ kT k p k f k p0 . (0.0.57) p L (w)
p0
L (v)→L (v)
L (w)
0
Now, since (0.0.57) holds for every f ∈ L (w) we may further deduce that T t is a bounded linear operator on L p (w), and moreover,
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t
T p0 0 ≤ kT kL p (v)→L p (v) . L (w)→L p (w)
(0.0.58)
That we actually have equality in (0.0.58) follows from (0.0.58) applied to T = (T t )t .
Exercise 7.4.7 Suppose that T is a linear operator that maps L2 (Rn , vdx) to itself for all v such that v−1 ∈ A1 . Show that the transpose operator T t of T maps L2 (Rn , wdx) to itself for all w ∈ A1 . Solution. Fix w ∈ A1 and set v = w−1 , then v−1 = w ∈ A1 . By assumption T : L2 (v) ⇒ L2 (v). Fix g ∈ L2 (w) and f ∈ L2 (w). Then f w ∈ L2 (w−2 w) = L2 (v). We have Z Z t = gT ( f w) dx T (g) f w dx Rn
Rn
≤
Z
1
Z =
1
gw 2 v 2 T ( f w) dx
Rn
g2 w dx
1 Z 2
Rn
≤ kgkL2 (w) kT kL2 (v)→L2 (v) ≤ kT kL2 (v)→L2 (v) kgkL2 (w)
T ( f w)2 v dx Z Rn
Z Rn
1
 f w2 v dx
2
1 2
 f 2 w2 w−1 dx
1 2
≤ kT kL2 (v)→L2 (v) kgkL2 (w) k f kL2 (w) . Taking supremum over f ∈ L2 (w) with k f kL2 (w) ≤ 1 we obtain that kT t (g)kL2 (w) ≤ kT kL2 (v)→L2 (v) kgkL2 (w) . Consequently, kT t kL2 (w)→L2 (w) ≤ kT kL2 (v)→L2 (v) which says that T t maps L2 (w) to L2 (w), for all w ∈ A1 .
Exercise 7.4.8 Let 1 < p < ∞. Suppose that T is a linear operator that maps L p (v) to itself for all v satisfying v−1 ∈ A p . Show that the transpose 0 operator T t of T maps L p (w) to itself for all w satisfying w−1 ∈ A p0 . Solution. p −p Let w be a weight such that w−1 ∈ A p0 . Set v = w p0 ∈ A p , hence w p0 is a weight whose reciprocal lies in A p and the hypothesis can be applied for this weight. Using H¨older’s inequality we obtain Z t t k T (g) kL p0 (w) = sup T (g) f wdx f ∈L p (w) k f kL p (w) ≤1
=
sup k f kL p (w) ≤1
Rn
Z Rn gT ( f w)dx
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331
=
sup k f kL p (w) ≤1
≤
Z 1 −1 p0 w p0 gT ( f w)dx w Rn Z
sup
Rn
k f kL p (w) ≤1
≤
sup k f kL p (w) ≤1
=
sup k f kL p (w) ≤1
p0
g wdx
10 Z p
Rn
kT kL p (v)→L p (v) kgkL p0 (w)
−p T ( f w) p w p0 dx
Z Rn
1
−p  f w p w p0 dx
p
1
p
kT kL p (v)→L p (v) kgkL p0 (w) k f kL p (w)
≤ kT kL p (v)→L p (v) kgkL p0 (w) . Therefore, kT t kL p0 (w)→L p0 (w) ≤ kT kL p (v)→L p (v) < ∞.
Section 7.5. Further Properties of A p Weights Exercise 7.5.1 Let (X, µ) be a measure space, 0 < s < 1, and f ∈ Ls (X, µ). Show that Z
f s = inf  f u−1 dµ : kuk L X
s L 1−s
≤1
and is attained. that the infimum1−s Hint: Try u = c  f  for a suitable constant c. Solution. s Let u ∈ L 1−s (X, µ) be such that kuk
s
L 1−s
Z
≤ 1. By H¨older’s Inequality, with p = 1/s > 1, we obtain
 f s dµ =
X
Z
( f  /u)s us dµ
(0.0.59)
X
≤
Z
1 Z p
s p
[( f  /u) ] dµ
(u ) dµ
( f  /u) dµ
=
s Z u
X
( f  /u) dµ
=
kuks
1−s
s
L 1−s
X
≤
s 1−s
X
s
Z
p
X
X
Z
10
s p0
s
Z
( f  /u) dµ
.
X
Taking the pth root in (0.0.59), shows that k f kLs ≤ s
As u ∈ L 1−s (X, µ) with kuk
s
L 1−s
Z
( f  /u) dµ.
(0.0.60)
X
≤ 1 was chosen arbitrarily, we may take the infimum over all such u in (0.0.60) to obtain k f kLs ≤ inf
Z X
( f  /u) dµ : kuk
s
L 1−s
≤1 .
(0.0.61)
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Now, put u := c  f 1−s , where ( k f ks−1 6 0 L s , k f kL s = c= . k f kL s = 0 1, s
Then u ∈ L 1−s , kuk
s
L 1−s
≤ 1 and Z X
( f  /u) dµ = k f kLs .
This shows that the infimum in (0.0.61) is achieved, and consequently, the claimed identity follows.
Exercise 7.5.2 p 1 (Rn ). (K. Yabuta) Let w ∈ A p for some 1 < p < ∞ and let f be in Lloc (Rn , wdx). Show that f lies in Lloc p−1 Hint: Write w = w1 /w2 via Theorem 7.5.1.
Solution. Using the factorization theorem (Theorem 7.5.1) write w = w1 /w2p−1 , where w1 , w2 lie in A1 . Given a cube Q write 1
Z
 f (x) dx =
Q
Z
 f (x)
Q
≤ ≤
1
dx
w2 (x) p0
Z
1  f (x) dx p−1 w (x) Q 2 p
1 Z
w1 (x)w1  f (x) p w2 (x) p−1 Q
[w1 ]A1 AvgQ w1
10
p
Q
(x)−1
Z
≤
w2 (x) p0
w2 (x) dx
p
dx Q
p
1 p
− 1p
= k f kL p (Q,w) [w]A1 w1 (Q)
10
1 Z
w1 (x) dx  f (x) p−1 w Q 2 (x)
Z
p
thus f is locally integrable on Rn .
p
1 Z
1− 1p
w2 (Q)
w2 (x) dx
10
p
Q
w2 (x) dx
p
1 p
Q < ∞ ,
Exercise 7.5.3 Use the same idea of the proof of Theorem 7.5.1 to prove the following general result: Let µ be a positive measure on a measure space X and let T be a bounded sublinear operator on L p (X, µ) for some 1 ≤ p < ∞. Suppose that T ( f ) ≥ 0 for all f in L p (X, µ). Prove that for all f0 ∈ L p (X, µ), there exists an f ∈ L p (X, µ) such that (a) f0 (x) ≤ f (x) for µalmost all x ∈ X. (b) k f kL p (X) ≤ 2 k f0 kL p (X) . (c) T ( f )(x) ≤ 2 kT kL p →L p f (x) for µalmost all x ∈ X. Hint: Try the expression in (7.5.2) starting the sum at j = 0. Solution. Following the hint, we start with, ∞
f=
∑ (2kT k)− j T j ( f0 ).
j=0
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333
Clearly this series converges in norm since f0 ∈ L p (X, µ) and T is bounded. Therefore, ∞
∑ (2T )− j kT j ( f0 )kL p
k f kL p ≤
j=0 ∞
∑ (2T )− j kT k j k f0 kL p
≤
j=0 ∞
=
∑ 2− j k f0 kL p
j=0
= 2k f0 kL p . This proves (b). We also have f0 (x) ≤ f (x) because all the partial sums are greater than or equal to f0 as a consequence of the fact that T ( f ) ≥ 0 for every f (x) and this gives (a). Finally, ∞
f=
∑ (2T )− j T j ( f0 )
j=0
yields T(f) = T
∑ (2kT k) T ( f0 ) ∞
−j
j
j=0 ∞
≤
∑ (2kT k)− j T j+1 ( f0 ))
j=0
∞ − j−1 j+1 T ( f0 ) = 2kT k ∑ (2kT k) j=0
≤ 2kT k f
which yields (c).
Exercise 7.5.4 ([101]) Suppose that T is an operator defined on 1 1. Equivalently, kT f kLr (w) ≤ Nr (B)k f kLr (w) for any f ∈ Lr (w). Hence, Z Rn
T f r vM( f )−(r−q) dx
Also, Z Rn
vM( f )q dx
1 r
= kT f kLr (w) ≤ Nr (B)k f kLr (w) .
r−q
r−q
r−q
rq
= kM f kLqr(v) ≤ Cq k f kLqr(v) .
Therefore,
1− q
r kT f kLq (v) ≤ Nr (B)Cq k f kLr (w) k f kLq (v)
and k f kLr (w) =
Z Rn
 f r w dx
1
Z
r
=
Rn
(0.0.62)
 f r vM( f )−(r−q) dx
1 r
.
(0.0.63)
Since  f  ≤ M( f ) a.e. and −(r − q) < 0, thus M( f )−(r−q) ≤  f −(r−q) a.e. Hence from (0.0.63) we have k f kLr (w) ≤
Z
−(r−q)
r
Rn
 f  v f 
1
Z
r
=
dx
Together with (0.0.62) we get
q
q
Rn
 f  v dx
1
q
r
= k f kLr q (v) .
1− q
r kT f kLq (v) ≤ (Nr (B)Cq ) k f kLr q (v) k f kLq (v) = Ck f kLq (v) .
Here C = Nr (B)Cq is a constant depending on q, r, n and [v]A1 . We conclude that kT kLq (v)→Lq (v) ≤ C (q, r, n, [v]A1 ) .
Exercise 7.5.5 Let T be a sublinear operator defined on
S
2≤q 1 one has Z sup
( 2p )0
Rn
kuk ( p )0 ≤1 L 2
M(u)
dx
1 p 2( 2 )0
1
≤ kMk 2( p )0
p 0
L 2 →L( 2 )
,
(0.0.67)
since p > 2. From (0.0.66) and (0.0.67), we have 1
kT ( f )kL p ≤ k f kL p kMk 2( p )0
p 0
L 2 →L( 2 )
.
This means that T maps L p (Rn ) into itself for all 2 < p < ∞.
Exercise 7.5.6 (X. C. Li ) Let T be a sublinear operator defined on 1 λ } has finite measure, otherwise there is nothing to prove. Then for each x ∈ Ωλ there is a maximal dyadic cube Qx that contains x such that 1 Qx 
Z Qx
 f (y)dy > λ ;
(0.0.68)
otherwise Ωλ would have infinite measure. Let Q j be the collection of all such maximal dyadic cubes containing all x in Ωλ (i.e., {Q j } j = {Qx : x ∈ Ωλ }). Maximal dyadic cubes are disjoint hence any two different Q j ’s are disjoint; Moreover, we note that if x, y ∈ Q j , then Q j = Qx = Qy . It follows that [ Ωλ = Q j . j
First, let us prove that for all Q j we have the estimate {x ∈ Q j : Md ( f )(x) > 2λ , M ] ( f )(x) ≤ γλ } ≤ 2n γQ j .
(0.0.69)
We fix j and x ∈ Q j such that Md ( f )(x) > 2λ . Then the following supremum 1 R3x R
Md ( f )(x) = sup
Z
 f (y)dy
(0.0.70)
R
is taken over all dyadic cubes R that either contain Q j or are contained in Q j (since Q j ∩R 6= 0). If R ⊃6= Q j , the maximality of Q j implies that (0.0.68) does not hold for R; thus the average of  f  over R is at most λ . Thus if Md ( f )(x) > 2λ , then the supremum in (0.0.70) is attained for some dyadic cube R contained (not properly) in Q j . Therefore, if x ∈ Q j and Md ( f )(x) > 2λ , then we can replace f by f χQ j in (0.0.70) and we must have Md ( f χQ j )(x) > 2λ . We let Q0j be the unique dyadic cube of twice the side length of Q j . Therefore, for x ∈ Q j we have Md (( f − Avg f )χQ j )(x) ≥ Md ( f χQ j )(x) −  Avg f  > 2λ − λ = λ Q0j
Q0j
since  AvgQ0 f  ≤ AvgQ0  f  ≤ λ because of the maximality of Q j . We conclude that j
j
{x ∈ Q j : Md ( f )(x) > 2λ } ≤ {x ∈ Q j : Md (( f − Avg f )χQ j )(x) > λ }
(0.0.71)
Q0j
and, using the fact that Md is of weak type (1, 1) with constant 1, we can control the last expression in (0.0.71) by 1 λ
Z Qj
 f (y) − Avg f dy ≤ Q0j
2n Q j  1 λ Q0j 
Z Q0j
 f (y) − Avg f dy
2n Q j  ] ≤ M ( f )(ξ j ) λ
Q0j
(0.0.72)
for all ξ j ∈ Q j . In proving (0.0.69) we may assume that for some ξ j ∈ Q j we have M ] ( f )(ξ j ) ≤ γλ ; otherwise there is nothing to prove. For this ξ j , using (0.0.71) and (0.0.72) we obtain (0.0.69).
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Denote A j,λ := {x ∈ Q j : Md ( f )(x) > 2λ , M ] ( f )(x) ≤ γλ }. Inequality (0.0.69) implies A j,λ  ≤ 2n γ, Q j 
if
Q j  6= 0.
By property (d) in Theorem 9.3.3, there exist C2 , ε0 > 0 such that A j,λ  ε0 w(A j,λ ) ≤ C2 (2n γ)ε0 . ≤ C2 w(Q j ) Q j 
(0.0.73)
(0.0.74)
We have used inequality (0.0.73). Inequality (0.0.74) implies w({x ∈ Q j : Md ( f )(x) > 2λ , M ] ( f )(x) ≤ γλ }) ≤ C2 (2n γ)ε0 w(Q j ). This implies w({x ∈ Rn : Md ( f )(x) > 2λ ,M ] ( f )(x) ≤ γλ }) ≤ C2 (2n γ)ε0 w({x ∈ Rn : Md ( f )(x) > λ }).
(0.0.75)
For a positive real number N we set Z N
IN =
0
pλ p−1 w({x ∈ Rn : Md ( f )(x) > λ })dλ .
We note that IN is finite since it is bounded by pN p−p0 p0
Z N 0
p0 λ p0 −1 w({x ∈ Rn : Md ( f )(x) > λ })dλ ≤
since p ≥ p0 . We now write IN = 2 p
N 2
Z 0
pN p−p0 p kMd ( f )kL0p0 (w) < ∞, p0
pλ p−1 w({x ∈ Rn : Md ( f )(x) > 2λ })dλ
and use (0.0.75) to obtain the following sequence of inequalities: IN ≤2 p
N 2
Z
+2
pλ p−1 w({x ∈ Rn : Md ( f )(x) > 2λ , M ] ( f )(x) ≤ γλ })dλ
0 p
N 2
Z
pλ p−1 w({x ∈ Rn : M ] ( f )(x) > γλ })dλ
0 p
n
ε0
≤2 C2 (2 γ) + 2p
Z
N 2
Z
N 2
0
pλ p−1 w({x ∈ Rn : Md ( f )(x) > λ })dλ
pλ p−1 w({x ∈ Rn : M ] ( f )(x) > γλ })dλ
0
≤2 pC2 (2n γ)ε0 IN +
2p γp
Z
γN 2
pλ p−1 w({x ∈ Rn : M ] ( f )(x) > λ })dλ .
0
At this point we pick a γ such that 2 pC2 (2n γ)ε0 = 12 . Since IN is finite, we can subtract from both sides of the inequality the quantity 12 IN to obtain IN ≤ C(n, p, [w]A∞ )
Z
γN 2
pλ p−1 w({x ∈ Rn : M ] ( f )(x) > λ })dλ .
0
Let N → ∞ we obtain kMd ( f )kL p (w) ≤ C(p, n, [w]A∞ )kM ] ( f )kL p (w) .
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339
Exercise 9.5.7 (First edition) Let 1 < r < ∞ be fixed. Let T j be a sequence of linear operators on L p (Rn ) with the property that for each B > 1, there is a constant Cr (B) > 0 such that for all w in Ar , with [w]Ar ≤ B, we have sup kT j ( f )kLr,∞ (w) ≤ Cr (B)k f kLr (w) . j
Then the vectorvalued inequality 1
1
k(∑ T j ( f j ) p ) p kL p ≤ Ck(∑  f j  p ) p kL p j
j
holds for all 1 < p, q < ∞, p 6= q. Solution. By the proof of Theorem 9.5.10, it suffices to show the inequality when p = r. q −1 s σ 1/σ , σ = s+1 Consider the cases r < q and r > q. When 1 < r < q let s = q−r 2 , for a given u ≥ 0 in L (w) set U = kMkLs M(u ) for some function M. Then U satisfies kUkLs ≤ kukLs and U pointwise controls u. Then since sup j kT j ( f )kLr,∞ (w) ≤ Cr (B)k f kLr (w) for all w, Z Z sup T j ( f )(ν)r u(ν)dν ≤ Cr  f (µ)rU(µ)dµ j
Then
1
(∑ T j ( f j )r ) 1r r = ∑ T j ( f j )r r 1 L L j
j
Z = sup kukLs ≤1
≤ sup C kukLs ≤1
∑ Tj ( f j )(ν)r u(ν)dν
1 r
Rn j
Z Rn
1 r
r
∑( f j )(µ) U(µ)dµ j
1 ≤C (∑  f j r ) r Lq (µ) j
When q < r, denote by T t the transpose of a linear operation T defined by hT ( f ), gi = h f , T t (g)i
by the proof of Theorem 9.5.10, we obtain Z Rn 0
0
0
T jt ( f )(x)r wr −1 (x)dx ≤ (Cr (B))r
0
Z Rn
0
0
( f (x))r wr −1 (x)dx
wr −1 ∈ Ar0 , [wr −1 ]A0r ≤ B, we have r0 < q0 applying the result to previous case, we obtain the conclusion.
Exercise 9.5.8 (First edition) Prove part (b) of Lemma 9.5.4 using only the conclusion in part (a). Solution.
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We have 1 < p < r < ∞ and w ∈ A p . Let p0 and r0 be the conjugate of p and r, respectively, i.e.
1 p
+
1 p0
= 1 and
1 r
+ r10 = 1.
1
Then 1 < r0 < p0 < ∞ and v := w− p−1 ∈ A p0 . By part (a) there exists a constant C1 such that for every nonnegative function 0 0 0 g ∈ L(p /r ) (v), there is a function G(g) such that g ≤ G(g) (0.0.76) kG(g)k
≤ 2kgk
p0 0 0
L p −r (v)
,
p0 0 0
(0.0.77)
L p −r (v)
[G(g)v]Ar0 ≤ C1 . By direct calculation, one obtains
p0 r0
0
and
= (r − 1)p(r − p)−1
p0 p0 − r0 0
r−p
0 0
(0.0.78)
=
p(r − 1) . r− p
p
Claim 1: g ∈ L(p /r ) (v) if and only if gr−1 w− p−1 ∈ L r−p (w). Proof. We have kgkL(p0 /r0 )0 (v) =
Z g
(r−1)p r−p
1 − p−1
w
r−p p(r−1)
dx
r−p p(r−1) Z p p r−p 1 − r−1 − p−1 r−p w p−1 w p−1 dx ⇔ g w
r−p ⇔ kgr−1 w− p−1 k
(0.0.79)
1 r−1
.
p L r−p (w)
Thus, Claim 1 is proved. Claim 2: G(g)v ∈ Ar0 if and only if 1 − r0 −1
(G(g)v)
r−p
= (G(g)r−1 w− p−1 )−1 w ∈ Ar .
Proof. We have !r0 −1 − 0 1 Z Z 1 1 r −1 1 1 − (p−1) − (p−1) G(g)w dx dx G(g)w Q Q Q Q r0 −1 Z Z r−p 1 1 − 1 = G(g)w (p−1) dx (G(g)r−1 w− p−1 )−1 wdx Q Q Q Q !r0 −1 r−p 1 r−p r−p 1 Z Z r−1 w− p−1 ) r−1 w (p−1)(r−1) w− (p−1) r−1 w− p−1 )−1 w (G(g) (G(g) = dx dx Q Q Q Q
= =
1 Q
Z
r−p
1
Q 1 i− r−1 r−p (G(g)r−1 w− p−1 )−1 w dx
Z h 1
Q
1
(G(g)r−1 w− p−1 ) r−1 w− r−1 dx
1 Q !
Q
Z
r−p
(G(g)r−1 w− p−1 )−1 wdx
Q
1 Q
Z Q
(G(g)r−1 w− p−1 )−1 w ∈ Ar . Thus, Claim 2 is proved.
r−p
(G(g)r−1 w− p−1 )−1 wdx
From (0.0.80) it follows that G(g)v ∈ Ar0 if and only if r−p
r0 −1
1 r−1
.
(0.0.80)
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341 r−p
r−p
p
Let h = gr−1 w− p−1 and H(h) = G(g)r−1 w− p−1 . By Claim 1, we have h ∈ L r−p (w). Since g ≤ G(g) and r > 1, it follows that r−p
r−p
h = gr−1 w− p−1 ≤ G(g)r−1 w− p−1 = H(h).
(0.0.81)
We have kG(g)k
Z p0 0 0
=
G(g)
p0 p0 −r0
1 − p−1
w
p0 −r0 0
Z
p
=
dx
G(g)
p(r−1) r−p
1 − p−1
w
r−p p(r−1)
dx
L p −r (v)
Z =
G(g)r−1
p r−p
1
w− p−1 dx
r−p p(r−1)
r−p p(r−1) Z p r−p r−p 1 r−1 − p−1 p−1 r−p − p−1 w dx = G(g) w w
(0.0.82)
r−p Z p(r−1) p r−p r−1 − p−1 r−p = G(g) w wdx
Z =
r−p p(r−1)
= kH(h)k
H(h)wdx
1 r−1
.
p
L r−p (w)
By the same calculations as above, we obtain = khk
kgk
p0 0 0 L p −r
(v)
1 r−1
.
p L r−p (w)
(0.0.83)
Applying the equalities (0.0.82) and (0.0.83) to the inequality (0.0.77), we get kH(h)k
1 r−1
p
L r−p (w)
≤ 2 khk
p
≤ 2r−1 khk
p
L r−p (w)
1 r−1
.
Therefore, kH(h)k
p
L r−p (w)
L r−p (w)
.
(0.0.84)
By (0.0.78), we have G(g)v ∈ Ar0 . Applying Claim 2, we get H(h)−1 w ∈ Ar , i.e., [H(h)−1 w]Ar ≤ C2 .
(0.0.85)
p
Thus, for every h ∈ L r−p (w) there is a function H(h) satisfying inequalities (0.0.81) , (0.0.84) and (0.0.85).
Exercise 9.5.4 (first edition) Give an alternate proof of the extrapolation theorem (Theorem (9.5.3)) stated below: [ [ Lq (w) and takes values in the space of all measurable complexvalued functions. Fix Suppose that T is defined on 1≤ 1 there exists a constant Nr (B) > 0 such that for all weights ν in Ar with [ν]Ar ≤ B, we have T Lr (ν)→Lr (ν) ≤ Nr (B). Then for any 1 < p < ∞ and all B > 1, there exists N p (B) > 0 such that for all weights w in A p satisfying [w]A p ≤ B, we have T L p (w)→L p (w) ≤ N p (B). Solution. Let 1 < p < ∞ , B > 1 and w ∈ A p such that [w]A p ≤ B. Since w ∈ A p , by Exercise 9.2.4 (b), ∃ δ > 0 such that p w ∈ A p−δ . Further, we can choose δ small enough and B0 = B0 (n, p, B) in such a way that 1 < p−δ < r and [w]A p−δ ≤ B0 . Now
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by Proposition (9.1.5)(4), w−(p−δ ) (p − δ ) ∈ A(p−δ )0 . Then again by Exercise (9.2.4)(b), ∃ θ > 0 such that w
−(p−δ )0 (p−δ )
Further, we can choose θ small enough and B” = B”(n, (p − δ )0 , B0 ) in such a way that θ < (p − δ )0 − 1 and [w p(1+η) (p−δ )0
∈ A(p−δ )0 −θ .
−(p−δ )0 (p−δ )
]A(p−δ )0 −θ .
(p−δ )0
1 1+η
p where 1 + η = (p−δ )0 −θ . By Theorem 9.2.7, ν ∈ A1 . Then Let f ∈ L p (w) Put q = p−δ and ν = {M(T f  w1+η )(x)} ˜ r, n, [ν]A1 ) such that T f Lq (ν) ≤ B ˜ f Lq (ν) . by Exercise 9.5.3 with this choice of q and ν, ∃ B˜ = B(q,
Z
T f  p w = ≤
p
Z
T f q T f  (p−δ )0 w
Z
T f q {M(T f  (p−δ )0 w1+η )} 1+η
p(1+η)
1
≤ B˜ q 1
Z
p
Z
p
≤ B˜ q  f Lp−δ p (w) N1 1
p(1+η)
q
1
 f q w p {M(T f  (p−δ )0 w1+η )} 1+η w
≤ B˜ q  f Lp−δ p (w) 1
1
p
≤ B˜ q  f Lp−δ p (w) N1
{M T f 
Z Z
p (p−δ )0 −θ
p (p−δ )0
T f  w
w w
−q p
(p−δ )0 (p−δ )0 −θ
−(p−δ )0 (p−δ )
(p−δ )0 −θ −(p−δ )0 } w (p−δ )
1 (p−δ )0
1 (p−δ )0
1 0 (p−δ ) T f  w . p
p 1 1 R So, T f  p w (p−δ ) ≤ B˜ q N1  f Lp−δ p (w) . Consequently, T f L p (w) ≤ N(p, B) f L p (w) , and T L p (w)→L p (w) ≤ N(p, B).
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