Fourier Series, Fourier Transform and Their Applications to Mathematical Physics (Instructor Solution Manual, Solutions) [1 ed.] 3319652613, 9783319652610

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S OLUTION M ANUAL Fourier Series, Fourier Transform and Their Applications to Mathematical Physics

Valery Serov, Markus Harju

Preface This text presents solutions to all exercises in V. Serov: Fourier Series, Fourier Transform and Their Applications to Mathematical Physics (Springer, 2017). The solutions have been collected over several years when the content of the book was used for teaching at University of Oulu. Authors acknowledge help and support from our past and present students and colleagues. Namely we would like to thank Hanna Kiili, Jan Sandhu and Teemu Tyni.

Oulu, Finland April 2018

Valery Serov Markus Harju

5

Contents I

Fourier Series and the Discrete Fourier Transform

1

1

Introduction

1

2

Formulation of Fourier Series

7

3

Fourier Coefficients and Their Properties

19

4

Convolution and Parseval’s Equality

23

5

Fej´er Means of Fourier Series. Uniqueness of the Fourier Series.

27

6

The Riemann–Lebesgue Lemma

31

7

The Fourier Series of a Square-Integrable Function. The Riesz–Fischer Theorem. 35

8

Besov and H¨older Spaces

37

9

Absolute Convergence. Bernstein and Peetre Theorems.

43

10 Dirichlet Kernel. Pointwise and Uniform Convergence.

49

11 Formulation of the Discrete Fourier Transform and Its Properties.

59

12 Connection Between the Discrete Fourier Transform and the Fourier Transform. 61 13 Some Applications of the Discrete Fourier Transform.

65

14 Applications to Solving Some Model Equations 14.1 The One-Dimensional Heat Equation . . . . . . . . . . . . . . . . . . . . 14.2 The One-Dimensional Wave Equation . . . . . . . . . . . . . . . . . . . 14.3 The Laplace Equation in a Rectangle and in a Disk . . . . . . . . . . . .

67 67 73 79

7

II

Fourier Transform and Distributions

87

16 The Fourier Transform in Schwartz Space

89

17 The Fourier Transform in L p (Rn ), 1 ≤ p ≤ 2

95

18 Tempered Distributions

107

19 Convolutions in S and S0

119

20 Sobolev Spaces 123 20.1 Sobolev spaces on bounded domains . . . . . . . . . . . . . . . . . . . . 130 21 Homogeneous Distributions

133

22 Fundamental Solution of the Helmholtz Operator

143

23 Estimates for the Laplacian and Hamiltonian

147

III

157

Operator Theory and Integral Equations

25 Inner Product Spaces and Hilbert Spaces

159

26 Symmetric Operators in Hilbert Spaces

165

27 John von Neumann’s Spectral Theorem

171

28 Spectra of Self-Adjoint Operators

179

29 Quadratic Forms. Friedrichs Extension.

189

30 Elliptic Differential Operators

191

31 Spectral Functions

199

32 The Schr¨odinger Operator

205

33 The Magnetic Schr¨odinger Operator

211

34 Integral Operators with Weak Singularities. Integral Equations of the First and Second Kinds. 215 35 Volterra and Singular Integral Equations

221

36 Approximate Methods

223

IV

Partial Differential Equations

229

37 Introduction

231

38 Local Existence Theory

243

39 The Laplace Operator

247

40 The Dirichlet and Neumann Problems

261

41 Layer Potentials

273

42 Elliptic Boundary Value Problems

287

44 Some Inverse Scattering Problems for the Schr¨odinger Operator

291

45 The Heat Operator

297

Part I

Fourier Series and the Discrete Fourier Transform

Chapter 1

Introduction Exercise 1.1. Prove H¨older’s inequality for integrals for every 1 ≤ p < ∞. Solution. We will prove first Young’s inequality: If 1 < p < ∞ and a, b ≥ 0, then 0

ap bp ab ≤ + 0, p p where

1 p

+

1 p0

= 1.

Proof. Let’s consider the curve x = t p−1 depicted below. x = t p−1 h(t)

x

p=3 p=4

p=2

b

t

a 1

t

1

0

Then t = x p−1 = x p −1 . By the picture the rectangular area ab is less or equal to the area between x = b and the curve plus the area between the curve and t = a i.e. ab ≤

Z a

t p−1 dt +

0

Z b

0

x p −1 dx =

0

Alternative solution. Consider the function (see figure above) h(t) =

tp 1 + − t, p p0 1

0

ap bp + 0. p p

t ≥ 0.

We have h0 (t) = t p−1 − 1 = 0 if and only if t = 1 and h(1) = 1p + h0 (t) < 0,t < 1 and h0 (t) > 0,t > 1. Hence h(t) ≥ 0 for t ≥ 0. 0 Set t = ab−p /p , a, b ≥ 0. Then

1 p0

− 1 = 0. Moreover,

0

0 a p b−p 1 + 0 − ab−p /p ≥ 0 p p

or

0

0 0 ap bp + 0 − ab−p /p+p ≥ 0. p p

Since p0 =

p p−1

1 we have −p0 /p + p0 = − p−1 +

p p−1

= 1. Hence

0

ab ≤

ap bp + 0. p p

Alternative solution. Observe that the function f (x) = ex is convex. Thus 1

ab = elog(ab) = elog a+log b = e p

log a p + p10 log b p

0

= f (tx + (1 − t)y) ≤ t f (x) + (1 − t) f (y) =

1 p 1 p0 a + 0b . p p

0

Now H¨older’s inequality: Let 1 ≤ p ≤ ∞, Ω ⊆ Rn , f ∈ L p (Ω) and g ∈ L p (Ω) with 1 1 1 p + p0 = 1. Then f g ∈ L (Ω) and k f gk1 ≤ k f k p kgk p0 . Proof. If p = ∞ (or p0 = ∞), then p0 = 1 (or p = 1) and Z Ω

| f g| dx ≤ k f k∞

Z Ω

|g| dx = k f k∞ kgk1 .

Then let 1 < p < ∞. If k f k p = 0 (or kgk p0 = 0), then f = 0 a.e. and also f g = 0 a.e. and the claim holds. Then we can assume that k f k p , kgk p0 > 0. Let’s define f˜ =

f g and g˜ = . k f kp kgk p0

Now f˜ p = kgk ˜ p0 = 1. By Young’s inequality and integration 1 k f k p kgk p0

Z Ω

| f g| dx =

Z Ω

1 ≤ p

| f˜g| ˜ dx 1 | f˜| p dx + 0 p Ω

Z

2

Z Ω

0

|g| ˜ p dx =

1 1 + = 1. p p0

p !1/p f (x) ≤ ∑ j dx j=1

Z b n a

n

Z

∑

b

a

j=1

| f j (x)| p dx

1/p (1.2)

p 1/p Z d Z b 1/p Z b Z d p F(x, y)dy dx ≤ |F(x, y)| dx dy c c a a

(1.3)

Exercise 1.2. Prove (1.2) and (1.3). Solution. Let us first prove the inequality (1.3) for F(x, y) = f (x, y) ≥ 0. By Fubini’s theorem, the claim is trivial for p = 1. Let 1 < p < ∞ and denote Z d

f (x, y)dy.

J(x) := c

Then, by Fubini’s theorem and H¨older’s inequality, p Z d  Z b Z d Z b Z b p p−1 f (x, y)dy dx = J(x) dx = J(x) f (x, y)dy dx a c a a c  Z d Z b = J(x) p−1 f (x, y)dx dy c

≤

a

Z d Z b

1/p0 Z

p

b

f (x, y) dx

J(x) dx c

a

Z

b

=

1/p

p

dy

a p

1/p0 Z d Z

b

p

1/p

f (x, y) dx

J(x) dx a

c

a

It follows that b

Z

p

1−1/p0

J(x) dx a

≤

Z d Z b

p

1/p

f (x, y) dx c

dy

a

or Z b Z a

c

d

 p 1/p Z d Z f (x, y)dy dx ≤ c

b

p

f (x, y) dx

1/p dy.

a

Hence, by denoting f := |F| ≥ 0, p 1/p Z b Z d Z b Z d  p 1/p F(x, y)dy dx |F(x, y)| dy dx ≤ c a a c 1/p Z d Z b p ≤ |F(x, y)| dx dy. c

Now, (1.2) is a particular case of (1.3). 3

a

dy.

Exercise 1.3.

(1) Show that the bounded function ( x sin 1x , x ∈ (0, 1] f (x) = 0, x=0

is continuous on the interval [0, 1] but is not of bounded variation on [0, 1]. (2) Show that every piecewise constant function on [a, b] is of bounded variation. Solution. (1) Clearly f is continuous for x 6= 0 as the superposition of two continuous functions. Since limx→0 x sin 1x = 0 then f is continuous at x = 0 too. Let us prove that f is not of bounded variation. Indeed, let xj =

1 , π/2 + π j

Then n 1 1 ∑ x j sin x j − x j−1 sin x j−1 = j=1

n

>

2 π

j = 1, 2, . . . , n.

1 1 j j−1 ∑ π/2 + π j (−1) − π/2 + π( j − 1) (−1) j=1  n  1 1 =∑ + π/2 + π( j − 1) j=1 π/2 + π j  n  1 2 n 4j 2 1 + = ∑ 2 = ∑ π j=1 1 + 2 j 2 j − 1 π j=1 4 j − 1

as n → ∞. Thus

n

1 →∞ j=1 j

∑

n

1 1 =∞ sup ∑ x j sin − x j−1 sin xj x j−1 x1 ,...,xn j=1

and therefore f is not of bounded variation. (2) Let f be a piecewise constant function on the interval [a, b] i.e. f (x) = ck , when f x ∈ [xg m = b. Let k−1 , xek ] for k = 1, . . . , m for some fixed m ∈ N with xe0 = a and x x1 , . . . , xn be an arbitrary partition of [a, b]. Then n

m

∑ | f (x j ) − f (x j−1 )| ≤ ∑ |ck − ck−1 |

j=2

k=2

for any n ∈ N and for any choice of x1 , . . . , xn . But the right hand side of the latter inequality is fixed finite value which is independent on n. Therefore this function is of bounded variation. Exercise 1.4. Prove that 4

(1) Vax ( f ) is monotone increasing in x, (2) for any c ∈ (a, b), we have Vab ( f ) = Vac ( f ) +Vcb ( f ). Solution.

(1) If x ≥ y then n

Vax ( f ) −Vay ( f ) =

sup

n









sup ∑ f (x j ) − f (x j−1 ) ≥ 0. ∑ f (x j ) − f (x j−1 ) − x ,...,y=x

x0 ,...,x=xn j=1

0

n

j=1

since supremum cannot decrease if a point is added to partition. Alternatively, one can conclude from part (2) that 0 ≤ Vyx ( f ) = Vax ( f ) −Vay ( f ). (2) We follow [2, Thm 6.11]. Let f be of bounded variation on the interval [a, b]. Denote n ∑(P) = ∑ f (x j ) − f (x j−1 ) , j=1

where P = {x0 , x1 , . . . , xn } is a partition of [a, b]. Let P1 and P2 be partitions of [a, c] and [c, b], respectively. Then P0 = P1 ∪ P2 is a partition of [a, b] and

∑(P1 ) + ∑(P2 ) = ∑(P0 ) ≤ Vab ( f ). Hence f is of bounded variation on [a, c] and [c, b]. Moreover, Vac ( f ) +Vcb ( f ) ≤ Vab ( f ). To obtain the reverse inequality, denote P0 = P ∪ {c}. Since f (x j ) − f (x j−1 ) ≤ f (x j ) − f (c) + f (c) − f (x j−1 ) then

∑(P) ≤ ∑(P0 ).

Now the points of P0 in [a, c] determine a partition P1 of [a, c] and those in [c, b] determine a partition P2 of [c, b]. Thus

∑(P) ≤ ∑(P0 ) = ∑(P1 ) + ∑(P2 ) ≤ Vac ( f ) +Vcb ( f ). Therefore, Vac ( f ) +Vcb ( f ) is an upper bound for every sum ∑(P). Since this cannot be smaller than the least upper bound we must have Vab ( f ) ≤ Vac ( f ) +Vcb ( f ).

5

6

Chapter 2

Formulation of Fourier Series ∞

f (x) ∼

∑

cn einx

(2.10)

n=−∞

a bn n  n = 1, 2, . . .  2 + 2i , a0 cn ( f ) = 2 , n=0   a−n b−n 2 − 2i , n = −1, −2, . . . . 1 cn ( f ) = 2π

Z π −π

f (x)e−inx dx,

(2.11)

n = 0, ±1, ±2, . . .

cn ( f ) = c−n ( f ).

(2.12) (2.13)

Exercise 2.1. Prove formulas (2.10), (2.11), (2.12) and (2.13). Solution. We consider the Fourier series in the form f (x) ∼

∞ a0 + ∑ (an cos(nx) + bn sin(nx)) 2 n=1

with 1 a0 = π

Z π

f (x)dx, −π

1 an = π

Z π

f (x) cos(nx)dx, −π

1 bn = π

Z π

f (x) sin(nx)dx. −π

It follows from eix = cos x + i sin x that 1 cos x = (eix + e−ix ), 2

sin x = 7

1 ix (e − e−ix ). 2i

Thus ∞ a0 + ∑ (an cos(nx) + bn sin(nx)) 2 n=1  ∞  a0 1 inx 1 inx −inx −inx = + ∑ an (e + e ) + bn (e − e ) 2 n=1 2 2i    ∞  an bn inx ∞ an bn −inx a0 + − = +∑ e +∑ e 2 n=1 2 2i 2 2i n=1    ∞  an bn inx −∞ a−n b−n inx a0 e + ∑ e . + − = +∑ 2 n=1 2 2i 2 2i n=−1

f (x) ∼

This proves (2.10)-(2.11). For (2.12), if n > 0 then π an bn 1 π 1 cn ( f ) = + = f (x) cos(nx)dx + f (x) sin(nx)dx 2 2i 2π −π 2πi −π Z π Z π 1 1 = f (x) (cos(nx)dx − i sin(nx)) dx = f (x)e−inx dx. 2π −π 2π −π

Z

Z

If n < 0 then π 1 π 1 a−n b−n − = f (x) cos(−nx)dx − f (x) sin(−nx)dx 2 2i 2π −π 2πi −π Z Z 1 π 1 π = f (x) (cos(nx)dx − i sin(nx)) dx = f (x)e−inx dx. 2π −π 2π −π

Z

Z

cn ( f ) =

Moreover, a0 1 c0 = = 2 2π

Z π −π

1 f (x)dx = 2π

Z π −π

f (x)e−i0x dx.

Finally, (2.13) is clear from c−n ( f ) =

1 2π

Z π

f (x)einx dx =

−π

1 2π

Exercise 2.2. Find the Fourier series of   −1, −π ≤ x < 0 (1) sgn(x) = 0, x=0   1, 0 < x ≤ π. (2) |x|, −1 ≤ x ≤ 1. (3) x, −1 ≤ x ≤ 1. ( 0, −L ≤ x ≤ 0 (4) f (x) = L, 0 < x ≤ L. 8

Z π −π

f (x)e−inx dx = cn ( f ).

(5) f (x) = sin x, |x| ≤ 2. Solution. (1) Since L = π and sgn(x) is odd function we have a Fourier sine series with   Z Z 1 π 2 π 2 cos(mx) π bm = sgn(x) sin(mx)dx = sin(mx)dx = − π −π π 0 π m 0     cos(mπ) 1 2 1 − (−1)m 2 − + = = π m m π m ( 0, m = 2k, k = 1, 2, . . . = 4 πm , m = 2k − 1, k = 1, 2, . . . . That’s why ∞

sgn(x) =

4

∑ π(2k − 1) sin((2k − 1)x).

k=1

In particular, putting x = π/2 gives ∞ ∞ π sin((k − 1/2)π) (−1)k+1 =∑ =∑ . 2 k=1 k − 1/2 k=1 k − 1/2

(2) In this case L = 1 and f (x) is even. Hence we will have a Fourier cosine series, where Z 1

a0 =

−1

|x|dx = 2

Z 1

xdx = 1 0

and, by integration by parts,  Z 1 sin(mπx) sin(mπx) 1 dx − 2 am = 2 x cos(mπx)dx = 2 x mπ mπ 0 0 0     cos(mπx) 1 cos(mπ) 1 =2 = 2 − (mπ)2 (mπ)2 (mπ)2 0 ( 0, m = 2k, k = 1, 2, . . . 2((−1)m − 1) = = 4 2 (mπ) − (mπ)2 , m = 2k − 1, k = 1, 2, . . . . Z 1



So we have |x| =

1 4 ∞ cos((2k − 1)πx) − 2∑ . 2 π k=1 (2k − 1)2

In particular, putting x = 0 yields ∞ π2 1 =∑ . 2 8 k=1 (2k − 1)

9

(3) Since L = 1 and f (x) is odd we will have a Fourier sine series ∞

f (x) =

∑ bm sin(mπx),

m=1

where Z 1

Z 1

bm =

x sin(mπx)dx = 2 x sin(mπx)dx −1 0   Z 1 cos(mπx) − cos(mπx) 1 + 2 dx =2 x mπ mπ 0 0   cos(mπ) 2 sin(mπx) 1 2(−1)m+1 = −2 + = mπ mπ mπ mπ 0

by integration by parts. Thus x=

2 ∞ (−1)m+1 sin(mπx) . ∑ π m=1 m

This representation actually holds on the interval (−1, 1]. For the periodic extension of f it receives the mean value 0 at x = ±1, ±2, . . .. (4) Let f be defined outside this interval so that f (x + 2L) = f (x) for all x, except at the points x = 0, ±L, ±2L, . . .. We will temporarily leave open the definition of f at these points. The Fourier coefficients are a0 = 1 am = L

Z L −L

1 L

Z L

f (x)dx = −L

mπx f (x) cos dx = L

Z L 0

1 L

Z L

Ldx = L, 0

L sin mπx mπx L cos dx = mπ = 0 L L 0

and L Z Z L − cos mπx 1 L mπx mπx L f (x) sin dx = sin dx = bm = mπ L −L L L 0 L 0 L L = (1 − cos(mπ)) = (1 − (−1)m ) mπ mπ ( 0, m = 2k, k = 1, 2, . . . = 2L mπ , m = 2k − 1, k = 1, 2, . . . . Hence (2k−1)πx

L 2L ∞ sin f (x) ∼ + ∑ 2k −L1 2 π k=1 10

.

(5) Now L = 2 and f is odd so we have the Fourier sine series with the coefficients 1 2 (cos(mπ/2 − 1)x − cos(mπ/2 + 1)x)dx bm = sin x sin(mπx/2)dx = 2 0 0 sin(mπ/2 − 1)x sin(mπ/2 + 1)x 2 = − mπ − 2 mπ + 2 0 sin(mπ − 2) sin(mπ + 2) − sin 2 cos(mπ) sin 2 cos(mπ) − = − = mπ − 2  mπ + 2 mπ + 2  mπ − 2 −1 2mπ 1 = (−1)m+1 sin 2 2 2 . = (−1)m sin 2 − mπ − 2 mπ + 2 m π −4 Z 2

Z

So

∞

sin x =

2mπ

∑ (−1)m+1 sin 2 m2 π 2 − 4 sin(mπx/2).

m=1

Exercise 2.3. Prove, using Part (2) of Exercise 2.2, that ∞ π2 1 =∑ 2 8 k=1 (2k − 1)

and

∞ π2 1 = ∑ 2. 6 k=1 k

Solution. The first part is proved in Exercise 2.2. For the second part we observe that ∞ ∞ 1 1 π2 1 ∞ 1 1 = + = ∑ 2 ∑ (2k)2 ∑ (2k − 1)2 8 + 4 ∑ k2 . k=1 k=1 k=1 k k=1 ∞

This gives the claim after rearranging. Exercise 2.4. Suppose that ( 1 − x, 0 ≤ x ≤ 1 f (x) = 0, 1 < x ≤ 2. Find the Fourier cosine and sine series of f (x). Solution. As discussed in the book, we can represent f either by a cosine series or sine series. For cosine series we define an even extension of f as follows:   1 − x, 0 ≤ x ≤ 1    0, 1 0 and α > 1. Prove that f is constant almost everywhere. Hint. First show that ω p,2δ ( f ) ≤ 2ω p,δ ( f ), then iterate this to obtain a contradiction. Solution. Let us first prove that ω p,2δ ( f ) ≤ 2ω p,δ ( f ). Indeed, Z ω p,2δ ( f ) = sup

π

| f (x + h) − f (x)| dx

−π

|h|≤2δ

Z

≤ sup

π

p

Z

π

| f (x + h/2) − f (x)| p dx

+ sup −π

|h|≤2δ

Z

≤ sup

π+h/2

−π+h/2

|h|≤2δ

Z

π

= 2 sup −π

Z

1/p

| f (x + h/2) − f (x)| dx

Z

π

= 2 sup ˜ |h|≤δ

1/p | f (y + h/2) − f (y)| dy p

−π

|h|/2≤δ

1/p

p

π

+ sup |h|≤2δ

1/p

| f (x + h) − f (x + h/2)| dx

−π

|h|≤2δ

1/p

p

−π

1/p | f (y + h/2) − f (y)| dy p

1/p ˜ − f (y)| p dy | f (y + h) = 2ω p,δ ( f ).

By iterating this we obtain or, by denoting δ = δ0 2−n ,

ω p,2n δ ( f ) ≤ 2n ω p,δ ( f )

ω p,δ0 ( f ) ≤ 2n ω p,δ0 2−n ( f ) ≤ 2nC(δ0 2−n )α = Cδ0α (2n )1−α , α > 1. Letting n → ∞ yields ω p,δ0 ( f ) = 0 for all δ0 > 0. It means that f (x + h) − f (x) = 0 almost everywhere i.e. f is constant almost everywhere.

33

34

Chapter 7

The Fourier Series of a Square-Integrable Function. The Riesz–Fischer Theorem. 1 2π

2 ∑ cn ( f )einx − f (x) dx = ∑ |cn ( f )|2 . −π |n|≤N |n|>N

Z π

(7.5)

Exercise 7.1. Prove (7.5). Solution. Denote the difference on the left by g(x). By Parseval’s equality 1 2π

2 Z ∞ 1 π |g(x)|2 dx = ∑ |cn (g)|2 ∑ cn ( f )einx − f (x) dx = 2π −π −π |n|≤N n=−∞

Z π

so it remains to compute cn (g). We have 1 cn (g) = 2π

1 g(x)e−inx dx = 2π −π

Z π

= −cn ( f ) +

∑

|k|≤N

Z π −π

" e−inx

#

∑

|k|≤N

ck ( f )eikx − f (x) dx

Z 1 π −inx ikx ck ( f ) e e dx

2π

−π

( ( 0, |n| ≤ N 1 2π, k = n = −cn ( f ) + ∑ ck ( f ) = . 2π 0, k= 6 n −cn ( f ), |n| > N |k|≤N

Exercise 7.2. Prove that 2 (1) ∑M n=1 (2n − 1) = M −2α ) = O(M 2−2α ) for 0 < α < 1. (2) ∑M n=2 O((2n − 1)(n − 1)

35

Solution.

(1) Compute directly M

M

M

∑ (2n − 1) = 2 ∑ n − ∑ 1 = M(M + 1) − M = M2 .

n=1

n=1

n=1

Alternatively one can use induction. (2) If 1 − 2α ≥ 0 then we may argue that !

M

O

−2α

∑ (2n − 1)(n − 1)

=O

−2α

∑ (2n + 1)n

n=2

n=1

M−1

=O

!

M−1

M−1

∑n

1−2α

+

n=1

! −2α

∑n

!

M

=O

∑n

n=1

1−2α


= O M 2−2α .

n=1

If 1 − 2α < 0 then we notice (by comparing areas under the curve t 1−2α ) that M

M

∑ n1−2α ≤

n=1

Z n

Z M

t 1−2α dt =

∑

t 1−2α dt =

0

n=1 n−1

M 2−2α = O(M 2−2α ). 2 − 2α

Exercise 7.3. Suppose that a periodic function f ∈ L2 (−π, π) satisfies the condition Z π −π

| f (x + h) − f (x)|2 dx ≤ Ch2

with some C > 0. Prove that ∞

∑

n=−∞

|n|2 |cn ( f )|2 < ∞

and therefore f ∈ W21 (−π, π). Solution. Since Ch

2

≥ ≥

1 2π

Z π −π

∑

|nh|≤π

≥ 4

∑

| f (x + h) − f (x)|2 dx =

|einh − 1|2 |cn ( f )|2 = 4

|nh|≤π

2

2

∞

∑

n=−∞

∑

|nh|≤π

(nh/π) |cn ( f )|

then

∑

|n|≤π/|h|

n2 |cn ( f )|2 ≤ C

and letting h → 0 gives the claim. 36

|cn ( f )|2 |einh − 1|2

sin2 (nh/2)|cn ( f )|2

Chapter 8

Besov and H¨older Spaces (1) Bα2,2 (−π, π) = W2α (−π, π), α ≥ 0. (2) W20 (−π, π) = L2 (−π, π). (3) Bα2,1 (−π, π) ⊂ Bα2,θ (−π, π) ⊂ H2α (−π, π), α ≥ 0, 1 ≤ θ < ∞. (4) B02,θ (−π, π) ⊂ L2 (−π, π), 1 ≤ θ ≤ 2 and L2 (−π, π) ⊂ B02,θ (−π, π), 2 ≤ θ < ∞. (5) L2 (−π, π) ⊂ H20 (−π, π). Exercise 8.1. Prove embeddings (3), (4) and (5). Solution.

(3) Let f ∈ Bα2,1 (−π, π) i.e. !1/2

∞

∑

j=0

∑

2 j ≤|n|0

and Z ∞ sin(ax) 0

x

dx =

Z ∞ sin(−bx)

x

0

dx = −

Z ∞ sin(ay)

y

0

π dy = − , 2

−b = a < 0

Alternatively, one may perform complex integration. Indeed, since sin x/x is even and cos x/x is odd then Z ∞ ix e

or

−∞

x

dx =

Z ∞ cos x

x

−∞

dx + i

x

−∞

Z ∞ sin x

x

0

Z ∞ sin x

dx =

1 2i

dx = 2i

Z ∞ sin x 0

Z ∞ ix e

x

−∞

x

dx

dx.

Im z CR Cρ b

ρ

−R −ρ

R

Re z

Figure 18.1: Contour C of integration. iz

Let us integrate the function F(z) = ez over the contour C illustrated in Figure 18.1. Since F(z) has a simple pole at z = 0 then eiz dz = 0. C z Now we turn to the four parts of this integral. Z

On CR we have z = Reiθ so that dz = iReiθ and dzz = idθ . Hence, using sin θ ≥ 2θ /π for 0 ≤ θ ≤ π/2, we have Zπ Z Z π iz iz e iz iθ C z dz = 0 e idθ ≤ 0 e dθ , z = Re R

Z π

=

e−R sin θ dθ = 2

0

≤2

Z π/2

e−R sin θ dθ

0

Z π/2

e−2Rθ /π dθ =

0

113

π π −R − e →0 R R

as R → ∞. Hence

Z CR

On Cρ we have z = ρeiθ so that Z Cρ

eiz dz = z

Z Cρ

eiz −1 z

dz z

1 dz + z

= −πi + since

eiz dz → 0, z

Z

R → ∞.

= idθ . Hence eiz − 1 dz = z

Z

Cρ eiz − 1

z

Cρ

Z 0

Z

idθ + Cρ

π

dz → −πi,

eiz − 1 dz z

ρ →0

is analytic and bounded at z = 0.

These considerations allos us to conclude that p. v.

Z ∞ ix e −∞

or p. v.

x

dx = πi

Z ∞ sin x

x

−∞

dx = π

after taking imaginary parts. Actually, we do not need p. v. here. Indeed, not have any singularities near x = 0. Moreover, if Ik =

Z (k+1)π sin x

x

kπ

sin x x

does

k = 0, 1, 2, . . .

dx,

then |Ik | is a monotone decreasing sequence with Ik → 0 as k → ∞. As Ik alternates in sign then I0 + I1 + · · · converges. The sum of the series is clearly Z ∞ sin x

x

0

dx.

Similarly for Z 0 sin x

x

−∞

Thus p. v.

Z ∞ sin x

x

−∞

Hence

dx =

0

x

−∞

Z ∞ sin xξ

x

Z ∞ sin x

dx =

dx.

dx = 2

0

Z ∞ sin y 1 0

Z ∞ sin x

y/ξ ξ

dy =

π , 2

x

dx = π.

ξ >0

and Z ∞ sin xξ 0

x

dx =

Z −∞ sin y 1 0

y/ξ ξ

dy = − 114

Z ∞ sin(−y) 0

−y

π dy = − , 2

ξ < 0.

(2) By (2.1) and Exercises 18.2 and 18.3 we obtain   1 1 1 0 1 d b b = hp. v. , ϕ(x)i b hF(p. v. 2 ), ϕi = hp. v. 2 , ϕi = −h p. v. , ϕi x x x x dx 1 \ = hp. v. , −iξ ϕ(ξ )(x)i x Z ∞ \ ξ ϕ(ξ )(x) − ξ\ ϕ(ξ )(−x) = −i dx x 0 Z ∞ Z ∞ 1 1 √ = −i ξ ϕ(ξ )[e−ixξ − eixξ ]dξ dx x 0 −∞ 2π Z ∞ Z −2i 1 ∞ √ = −i ξ ϕ(ξ ) sin(xξ )dξ dx 0 2π x −∞ Z ∞ Z ∞ 2 sin(xξ ) ξ ϕ(ξ ) dxdξ = −√ x 0 2π −∞ r Z ∞ π 2 π |ξ |, ϕi, ϕ ∈ S. = −√ ξ ϕ(ξ ) sgn ξ dξ = h− 2 2 2π −∞

Exercise 18.7. Prove that (1) \ √ 1 = − 2πξ H(ξ ) and 2 (x + i0)

\ √ 1 = 2πξ H(−ξ ) 2 (x − i0)

(2) 1 1 1 + = 2 p. v. 2 2 2 (x + i0) (x − i0) x

and

1 1 − = −2πiδ 0 2 (x − i0) (x + i0)2

1 1 = p. v. 2 + πiδ 0 2 (x + i0) x

and

1 1 = p. v. 2 − πiδ 0 2 (x − i0) x

(3)

(4) r π 1 \ log |x| = − p. v. 2 |ξ | (5) xbβ = (2π)n/2 i|β | ∂ β δ . Solution.

(1) We know from Example 17.10 that \ √ 1 (ξ ) = − 2πξ H(ξ )e−εξ , 2 (x + iε) 115

\ √ 1 (ξ ) = 2πξ H(−ξ )eεξ . 2 (x − iε)

Hence

and

\ \ √ 1 1 (ξ ) = lim (ξ ) = − 2πξ H(ξ ) 2 2 ε→0 (x + iε) (x + i0) \ \ √ 1 1 (ξ ) = lim (ξ ) = 2πξ H(−ξ ). ε→0 (x − iε)2 (x − i0)2

(2) We prove that Fourier transforms coincide. Part (1) implies that \ \ √ 1 1 (ξ ) + (ξ ) = 2πξ (H(−ξ ) − H(ξ )) 2 2 (x + i0) (x − i0) √  √2πξ (1 − 0), ξ < 0 = 2πξ (1 − 1), ξ = 0  √ 2πξ (0 − 1), ξ > 0 r √ √ 2 1 = − 2π|ξ | = 2π F(p. v. 2 ) π x 1 = 2F(p. v. 2 ), x where we have used Exercise 18.6. Similarly, \ \ √ √ 1 1 (ξ ) − (ξ ) = 2πξ (H(−ξ ) + H(ξ )) = 2πξ = −2πiδb0 (x − i0)2 (x + i0)2 since

1 δb0 = √ iξ 2π

by Example 4.10. (3) Follows by subtracting and adding the identities from part (2). (4) We know from Example 18.12 that 1 (log |x|)0 = p. v. . x Hence r \ 1 π 0 \ sgn ξ (log |x|) = p. v. = −i x 2 by Exercise 18.6 or

by formula (2.2). Thus

1 \ ξ log |x| = −i −i

r

π sgn ξ 2

r π \ ξ log |x| = − sgn ξ 2 116

and so (at least formally, in spirit) r r r π sgn ξ π 1 π 1 1 \ log |x| = − =− =− ∈ / Lloc (R). 2 ξ 2 ξ sgn ξ 2 |ξ | We can make this proof more precise when we define (introduce) p. v. |ξ1 | in Chapter 21. \ β f from (16.1). Applying it to f = 1 and using Example 4.8 5) Recall ∂ β fb = (−ix) implies that \β = ∂ β b (−ix) 1 = ∂ β (2π)n/2 δ or xbβ = (2π)n/2 i|β | ∂ β δ .

Exercise 18.8. Prove that (1) b ) = − √i · 1 H(ξ 2π ξ − i0 (2) i 1 c ) = − p π p. v. . sgn(ξ ξ 2 Solution.

(1) Consider the reflection operator R f (x) := f (−x). Then 1 F 2 f (x) = √ 2π

Z ∞ −∞

F f (ξ )e−ixξ dξ = F −1 F f (−x) = f (−x) = R f (x)

holds for functions. One can extend this property to distributions too. Indeed, hF 2 T, ϕi = hT, F 2 ϕi = hT, Rϕi := hRT, ϕi,

ϕ ∈ S, T ∈ S0 .

Example 4.9 implies now that d √ 1 −i 2πH(ξ ) = x + i0 or H(ξ ) =

d 1 1 √ . −i 2π x + i0

Taking the Fourier transform yields     i 1 i 1 i 1 2 b )= √ F H(ξ =√ R =√ . x + i0 x + i0 2π 2π 2π −ξ + i0 117

(2) It follows from Exercise 18.6 that r sgn ξ = i

  1 2 . F p. v. π x

As above, this implies that r [ sgn ξ =i

r   1 1 2 2 R p. v. = −i p. v. . π x π ξ

118

Chapter 19

Convolutions in S and S0 Exercise 19.1. Prove Lemma 19.6 and find a counterexample showing that the first part fails for p = ∞. Solution. We start by writing ( jε ∗ f )(x) − f (x) =

Z n

ZR

= Rn

jε (y) f (x − y)dy − f (x)

Z Rn

jε (y)dy

jε (y) ( f (x − y) − f (x)) dy.

H¨older inequality implies that |( jε ∗ f )(x) − f (x)| ≤ ≤

Z

1

1

Rn

| jε (y)| p | f (x − y) − f (x)| | jε (y)| p0 dy

Z

 1p Z | jε (y)| | f (x − y) − f (x)| dy p

Rn

Rn

 10 p | jε (y)|dy

for 1 ≤ p < ∞. Let δ > 0. Take L p -norms in x and use Fubini to get Z Rn

|( jε ∗ f )(x) − f (x)| p dx  Z  p0 Z Z p p | jε (y)|dy ≤ | jε (y)| | f (x − y) − f (x)| dy dx Rn Rn Rn   Z p Z 0 = k jkLp1 | jε (y)| | f (x − y) − f (x)| p dy dx Rn Rn Z  p Z p p0 | f (x − y) − f (x)| dx dy = k jkL1 | jε (y)| Rn Rn Z  Z p δ δ p0 = k jkL1 dy + dy < + 2 2 |y|r

since Z |y|>r

| jε (y)|dy =

Z |y|>r

ε −n | j(y/ε)|dy = 119

Z |x|>r/ε

| j(x)|dx → 0,

ε →0

and (this is done first actually for r small enough) Z Rn

| f (x − y) − f (x)| p dx → 0,

|y| → 0

Indeed, denote τy f (x) = f (x + y). Take g ∈ C0 such that k f − gkL p
2. Consider the function 1 2 f (x) = n/2−ε ∈ Lloc . |x| Then, for bounded Ω, Z Ω

if and only if

q

|f| =

1

Z

Ω |x|qn/2−qε

qn − qε < n 2

or q
0 so small that q≥ q i.e. f ∈ / Lloc for q > 2. Let now f ∈ Ls2 i.e.

Z Rn

n >2 n/2 − ε

(1 + |x|)2s | f |2 dx < ∞.

1) If q = 2 then Z Rn

| f |q dx ≤

Z Rn

(1 + |x|)2s | f |2 dx < ∞

for any s ≥ 0. 2) If 1 ≤ q < 2 then it follows from H¨older inequality that Z Rn

| f |q dx = ≤

Z Rn

(1 + |x|)−t (1 + |x|)t | f |q dx

Z Rn

2 −t 2−q

(1 + |x|)

Z  2−q 2 dx Rn

(1 + |x|)2t/q | f |2 dx

q/2 n 2−q   1 1 2−q =n − . s>n 2q q 2 t

or

Conversely, if  s=n then f :=

1 1 − q 2



1 (1 + |x|)n/q (log(2 + |x|))1/q

Indeed, (put t := log r, dt =

dr r

∈ Ls2 ∩ (Lq )c .

in the last step)

(1 + |x|)2s dx dx = (1 + |x|)−n 2n/q 2/q (log(2 + |x|)) (log(2 + |x|))2/q |x|>1 (1 + |x|) |x|>1 Z ∞ Z Z ∞ ∞ rn−1 dr dr dt   = < ∞, n β β c (1 + r) (log(2 + r)) c r(log r) c tβ

Z

Z

where β = 2/q > 1. But (put again t := log r) Z Rn

For s < n

q

| f | dx = 

1 q

Z Rn

1  n (1 + |x|) (log(2 + |x|))

 − 12 the space Ls2 is even wider. 127

Z ∞ c

dr = r(log r)

Z ∞ dt c

t

=∞

Exercise 20.6. Prove that (1) χ[0,1] ∈ H s (R) if and only if s < 1/2. (2) χ[0,1]×[0,1] ∈ H s (R2 ) if and only if s < 1/2.   1 ∈ H s (Rn ) if and only if s < 2 − n/2. (3) K(x) := F −1 1+|ξ 2 | (4) Let f (x) = χ(x) log log |x|−1 in R2 , where χ(x) ∈ C0∞ (|x| < 1/3). Prove that f ∈ H 1 (R2 ) but f ∈ / L∞ (R2 ). Solution. Recall from Exercise 20.2 that f ∈ H s if and only if fb ∈ Ls2 . s (1) Since we need χd [0,1] for the H norm let us calculate it first. We have 1 χd [0,1] (ξ ) = √ 2π 1 =√ 2π

 ξ =0 1 1, 1 e−ixξ dx = √ −ixξ 0 2π  e−iξ , ξ 6= 0. 0 ( ( 1, ξ =0 1, ξ =0 1 =√ e−iξ −1 e−iξ /2 −eiξ /2 −iξ /2 , ξ 6= 0. , ξ= 6 0. 2π e −iξ −iξ Z 1

ξ

sin 2 1 = √ 2e−iξ /2 ξ 2π for all ξ . Thus χ[0,1] ∈ H s (R) if and only if Z ∞

C −∞

(1 + |ξ |2 )s

sin2 (ξ /2) dξ < ∞. ξ2

Near ξ = 0 we have no problems since ξ = ∞ requires 2 − 2s > 1 or s < 1/2.

sint t

→ 1 as t → 0. The integrability at

(2) It follows from part (1) that χ[ [0,1]2 (ξ ) =

1 2π

Z 1 0

Z 1

dx1

0

ξ

e−i(x1 ξ1 +x2 ξ2 ) dx2 =

Hence

2

χ[0,1]2 s

H (R2 )

Z

=C R2

=C

2 ξ1 2 ξ12

2 s sin

(1 + |ξ | )

sin2 ξ22 dξ1 dξ2 ξ22

Z ∞ Z ∞ 2 ξ1 sin2 ξ22 2 s sin 2 dξ (1 + |ξ | ) dξ1 < ∞ 2 2 2 −∞

ξ2

if and only if 2 − 2s > 1 or s < 1/2. 128

−∞

ξ

4 −iξ1 /2 sin 21 −iξ2 /2 sin 22 e e . 2π ξ1 ξ2

ξ1

(3) We have that K ∈ H s if and only if Z Rn

(1 + |ξ |2 )s



1 1 + |ξ |2

2 dξ < ∞

or 4 − 2s > n or s < 2 − n2 . (4) In dimension n = 2 the Sobolev embedding theorem reads H s (R2 ) ⊂ C˙ k (R2 ),

s > k + 1.

Here we construct a counterexample when k = 0 that shows that ˙ 2 ). H 1 (R2 ) 6⊂ C(R Consider the function f (x) = χ(x) log log(1/|x|),

χ(x) ∈ C0∞ (|x| < 1/3).

Then d d 1 d 1 1 f (x) = log(1/|x|) = |x|−1 dx j log(1/|x|) dx j log(1/|x|) 1/|x| dx j xj 1 −1 1 |x| (x12 + x22 )−3/2 2x j = − . = log(1/|x|) 2 log(1/|x|) |x|2 It follows that 2 d 1 1 dx j f (x) ≤ (log(1/|x|))2 |x|2 , and thus 2 Z 1/3 Z d 1 1 f (x) ≤ C ρdρ, 2 (log(1/ρ)) ρ 2 |x| |γ|,

|α| > |β | + m + n,

|α| > |γ| + m + n.

For example, α = (α1 , 0, . . . , 0) with α1 = max{|γ|, |β | + m + n, |γ| + m + n} + 1 will do. Exercise 21.5. Prove that, Z

hp. v. T−n , ϕi = lim

ε→0+ |x|≥ε

where T−n = |x|−n ω

T−n (x)ϕ(x)dx,

  R x |x| , Sn−1 ω(θ )dθ = 0.

Solution. Recall that hp. v. T−n , ϕi :=

Z Rn

T−n (x) [ϕ(x) − ϕ(0)ψ(x)] dx,

where ϕ ∈ S and ψ(x) = ψ(|x|) ∈ S with ψ(0) = 1. We have that hp. v. T−n , ϕi =

Z Rn

T−n (x) [ϕ(x) − ϕ(0)ψ(x)] dx Z

= lim ε→0 |x|≥ε

Z

= lim ε→0 |x|≥ε

T−n (x) [ϕ(x) − ϕ(0)ψ(x)] dx T−n (x)ϕ(x)dx − lim

ε→0 |x|≥ε

Z

= lim ε→0 |x|≥ε

Z

T−n (x)ϕ(0)ψ(x)dx

T−n (x)ϕ(x)dx

since Z |x|≥ε

Z

T−n (x)ψ(x)dx =

|x|≥ε

−n

|x| ω

Z ∞Z

= Zε ∞

=



Sn−1

r−n ω(θ )ψ(r)rn−1 dθ dr

r−1 ψ(r)

ε

137

 x ψ(|x|)dx |x|

Z Sn−1

ω(θ )dθ dr = 0.

Exercise 21.6. Prove that (1) kH f kL2 (R) = k f kL2 (R) . (2) Hilbert transform has an extension to functions from L2 (R). (3) H 2 = −I i.e. H −1 = −H.

0

(4) (H f1 , H f2 )L2 = ( f1 , f2 )L2 for f1 ∈ L p and f2 ∈ L p , where (5) H : L p (R) → L p (R), 1 < p < ∞ i.e.



1 Z f (t)dt

π x−t

|x−t|≥ε

1 p

+

1 p0

= 1, 1 < p < ∞.

≤ c k f kL p

Lp

for all ε > 0 where c does not depend on ε. Solution. Since

1 Hf = π

  1 p. v. ∗ f x

then it follows from Exercise 18.6 that √  r  \   \ 2 π 1 1 1b 1/2 1 c Hf = p. v. ∗ f = (2π) p. v. f = √ sgn ξ fb = −i sgn ξ fb. −i π x π x 2 π (1) Hence



c kH f kL2 (R) = H f

L2 (R)



= fb

L2 (R)

= k f kL2 (R)

by Parseval equality. (2) Follows from part (1). (3) Since cf = −i sgn ξ fb, H

  H f = F −1 −i sgn ξ fb

then     cf = F −1 −i sgn ξ (−i sgn ξ fb) = − f . HH f = F −1 −i sgn ξ H The zero measure set ξ = 0 where sgn ξ sgn ξ = 0 can be ignored in Lebesgue integration. (4) We consider only p = 2. Recall from Exercise 17.5, that (F f1 , F f2 )L2 = ( f1 , f2 )L2 ,

f1 , f2 ∈ L2 (Rn ).

Hence (H f1 , H f2 )L2 = (F H f1 , F H f2 )L2 = (−i sgn ξ F f1 , −i sgn ξ F f2 )L2 = −i

Z

Rn

sgn ξ fb−i sgn ξ fbdξ = (F f1 , F f2 )L2 = ( f1 , f2 )L2 . 138

(5) See part 1) for p = 2. Alternatively, we can proceed as in [5, Theorem 4.1.7]. Let us start by proving that (H f )2 = f 2 + 2H( f H f ) cf (ξ ) = −i sgn ξ fb(ξ ) := m(ξ ) fb(ξ ) we have, for any real-valued f ∈ S(R). Since H by taking the Fourier transform of the right hand, that \ cf ) fb2 + 2H( f H f ) = cn ( fb∗ fb) + 2cn m(ξ )( fb∗ H Z

= cn

fb(η) fb(ξ − η)dη + 2cn m(ξ )

RZ

= cn R

Z

fb(η) fb(ξ − η)m(η)dη

RZ

fb(η) fb(ξ − η)dη + 2cn m(ξ )

R

fb(η) fb(ξ − η)m(ξ − η)dη.

Here we have changed variables in the last step and used the fact that c f g = (2π)−n/2 fb∗ gb = cn fb∗ gb. It follows (by averaging, a = b = c implies a = b = b/2 + b/2 = b/2 + c/2) that \ fb2 + 2H( f H f ) = cn

Z R

Z

= cn R

fb(η) fb(ξ − η)(1 + m(ξ )(m(η) + m(ξ − η)))dη

\ cf ∗ H cf = (H fb(η) fb(ξ − η)m(η)m(ξ − η)dη = cn H f )2 .

Here we have used the fact that 1 + m(ξ )m(η) + m(ξ )m(ξ − η) = m(η)m(ξ − η) which is easy to verify by investigating the signs of ξ and η. Indeed, 1) ξ > η: 1 + m(ξ )m(η) + m(ξ )m(ξ − η) = m(η)m(ξ − η)

1 − sgn ξ sgn η − sgn ξ sgn(ξ − η) = − sgn(η) sgn(ξ − η) 1 − sgn ξ sgn η − sgn ξ = − sgn η

1 − sgn ξ sgn η = − sgn η + sgn ξ    0, ξ > 0, η > 0  0, ξ > 0, η > 0 0, ξ < 0, η < 0 = 0, ξ < 0, η < 0     2, ξ > 0, η < 0 2, ξ > 0, η < 0

2) ξ < η: 1 − sgn ξ sgn η + sgn ξ = sgn η

1 − sgn ξ sgn η = sgn η − sgn ξ    0, ξ > 0, η > 0  0, ξ > 0, η > 0 = 0, ξ < 0, η < 0 0, ξ < 0, η < 0     2, ξ < 0, η > 0 2, ξ < 0, η > 0 139

3) ξ = η 6= 0: 1 + m(ξ )m(η) + m(ξ )m(ξ − η) = m(η)m(ξ − η)

1 − sgn ξ sgn η − sgn ξ sgn(ξ − η) = − sgn(η) sgn(ξ − η) 1 − (sgn ξ )2 = 0

4) ξ = η = 0 may be ignored as a set of measure zero. The identity just obtained is used to prove L p boundedness for p = 2k by induction. The case k = 1 (p = 2) is already known. Suppose that H is bounded on L p with norm c p for p = 2k . Then

kH f k2L2p = (H f )2 L p ≤ f 2 L p + k2H( f H f )kL p ≤ k f k2L2p + 2c p k f H f kL p ≤ k f k2L2p + 2c p k f kL2p kH f kL2p .

It follows that kH f k2L2p − 2c p k f kL2p kH f kL2p − k f k2L2p ≤ 0 or

2

kH f kL2p − 2c p − 1 ≤ 0. k f kL2p q Hence (x2 − 2c p x − 1 = (x − (c p + c2p + 1))2 ≤ 0. ) 

kH f kL2p k f kL2p

q kH f kL2p ≤ c p + c2p + 1 k f kL2p

and so H is bounded on L2p with bound

c2p ≤ c p +

q c2p + 1.

Interpolation gives that H maps L p to L p for p ≥ 2. Parts 3) and 4) imply that h f , Hgi = h−H 2 f , Hgi = h−H f , gi

so that H ∗ = −H. By duality H is bounded for 1 < p ≤ 2.

I2

µ→+∞

→ =

= =

Z |x|>1

∂ i ∂ξj −

e−i(x,ξ ) x j ∂ dx = i |x|n+1 ∂ξj Z |x|>1

iξ j |ξ |

Z |x|>1

Z |x|>1

e−i(x,ξ ) dx |x|n+1

|x|−n−1 cos(|ξ | · x1 )dx x1 sin(|ξ |x1 ) dx |x|n+1

−iξ j · const,

|ξ | = 1.

140

Exercise 21.7. Prove the convergence of the last integral. Solution. We have, by setting x = rθ , that x1 sin(|ξ |x1 ) dx = |x|n+1 |x|>1

Z

Z ∞ rθ1 sin(|ξ |rθ1 ) n−1 r drdθ n+1

Z

r

Sn−1 1

Z

= Sn−1

θ1

Z

= Sn−1

θ1

Z ∞ sin(rθ1 )

drdθ , |ξ | = 1 r  Z 1 ∞ sin(rθ ) sin(rθ1 ) 1 dr − dr dθ . r r 0 0

1Z

Hence it suffices to check for the finiteness of Z ∞ sin(rθ1 ) 1

We know that

r

dr =

r

0

Z ∞ sin(rθ1 ) 0

and

Z ∞ sin(rθ1 )

r

dr =

dr −

Z 1 sin(rθ1 ) 0

r

dr.

π sgn θ1 2

Z1 Z 1 Z 1 Z 1 sin(rθ1 ) | sin(rθ1 )| r|θ1 | dr ≤ dr ≤ dr = |θ1 |dr = |θ1 |. 0 r r r 0 0 0

141

142

Chapter 22

Fundamental Solution of the Helmholtz Operator Exercise 22.1. Let L be a differential operator with constant coefficients. Prove that u = q ∗ E = E ∗ q solves the inhomogeneous equation Lu = q in D0 . Solution. If L=

∑

aα D α ,

|α|≤m

aα ∈ C

and u = q ∗ E with LE = δ then Lu = L(q ∗ E) =

∑

|α|≤m

aα Dα (q ∗ E) =

∑

|α|≤m

aα (q ∗ Dα E) = q ∗

∑

aα D α E

|α|≤m

= q ∗ LE = q ∗ δ = q. Note that there may be other solutions as well. If Lv = 0 and Lu = q then L(u + v) = q too. Definition. Denote by a0 (x, ξ ) the main (or principal) symbol a0 (x, ξ ) =

∑

aα (x)ξ α ,

|α|=m

ξ ∈ Rn

of L(x, ξ ). Assume that aα (x) are ”smooth”. Operator L(x, D) is said to be elliptic in Ω if for any x ∈ Ω and ξ ∈ Rn \{0} it follows that a0 (x, ξ ) 6= 0. Exercise 22.2. Let aα (x) be real for |α| = m. Prove that the previous definition is equivalent to 143

(1) m is even, (2) a0 (x, ξ ) ≥ CK |ξ |m (or −a0 (x, ξ ) ≥ CK |ξ |m ), CK > 0, for every compact set K ⊂ Ω and for all ξ ∈ Rn and x ∈ K. Solution. Let us note first that since a0 (x, ξ ) ∈ R is smooth with respect to x and ξ then it cannot change sign without passing through origin. Hence there are two possibilities: either a0 (x, ξ ) > 0, x ∈ Ω, ξ ∈ Rn \ {0} or a0 (x, ξ ) < 0,

ξ ∈ Rn \ {0}.

x ∈ Ω,

Let us consider the first case. Since a0 (x, λ ξ ) =

∑

aα (x)(λ ξ )α = λ m

|α|=m

∑

aα (x)ξ α = λ m a0 (x, ξ )

|α|=m

then a0 is homogeneous of order m with respect to ξ . If L is elliptic then     ξ ξ m a0 (x, ξ ) = a0 x, |ξ | = |ξ | a0 x, ≥ |ξ |mCK , ξ 6= 0, |ξ | |ξ |

x∈K

since a0 (x, η) ≥ CK > 0 for x ∈ K and η ∈ Rn \ {0} as a smooth function on a compact set (we cannot conclude that a0 ≥ CΩ > 0 for x ∈ Ω since a0 might attain 0 at ∂ Ω; in such case there exists x0 ∈ Ω such that a0 (x0 , ξ ) < CΩ , no matter how small CΩ is). What is more, a0 (x, 0) = 0 ≥ CK 0 = 0. This proves (2). To see (1) note that a0 (x, −ξ ) =

∑

aα (x)(−ξ )α = (−1)m

∑

aα (x)ξ α = (−1)m a0 (x, ξ ) > 0

|α|=m

|α|=m

for ξ ∈ Rn \ {0} forces m to be even since a0 (x, ξ ) > 0. Conversely, a0 (x, ξ ) ≥ |ξ |mCK , CK > 0 implies that a0 (x, ξ ) > 0,

ξ ∈ Rn \ {0}

i.e. L is elliptic. It also implies that a0 (x, −ξ ) = (−1)m a0 (x, ξ ) ≥ |ξ |mCK i.e. that m is even. In the second case a0 (x, ξ ) < 0 we can consider −a0 (x, ξ ) and obtain that a0 (x, ξ ) ≤ −|ξ |mCK .

144

Exercise 22.3. Let us define a fundamental solution Γ(x,t) of

∂ ∂t

+ L(D) as a solution of

( ∂ ( ∂t + L)Γ(x,t) = δ (x)δ (t) Γ(x, 0) = 0. Prove that Z ∞

F(x, λ ) :=

e−λt Γ(x,t)dt

0

is a fundamental solution of the operator L(D) + λ I, λ > 0. Solution. Let ϕ = ϕ(x) ∈ S(Rn ). Then Z ∞

he−λt L(D)Γ(x,t), ϕidt   Z ∞ Z ∞ ∂ ∂ −λt Γ(x,t), ϕidt − = he−λt Γ(x,t), ϕidt he L(D) + ∂t ∂t 0 0   Z ∞ ∞ ∂ = h L(D) + Γ(x,t), e−λt ϕi − he−λt Γ(x,t), ϕi − λ he−λt Γ(x,t), ϕi ∂t 0 0

hL(D)F(x, λ ), ϕi =

0

= hδ (x)δ (t), e−λt ϕi − 0 − λ hF, ϕi. Note that in the last two equalities brackets involve both x and t integration. Hence h(L(D) + λ )F(x, λ ), ϕi = hδ (x)δ (t), e−λt ϕi = ϕ(0) = hδ , ϕi.

Exercise 22.4. Prove that Z

lim ε→0+ |x|=ε

∂ En dσ (x) = 1 ∂r

or lim rn−1 ωn

r→0

∂ En = 1, ∂r

where ωn = |Sn−1 | is the area (measure) of the unit sphere Sn−1 . Solution. Consider a fundamental solution En of the Helmholtz operator −∆ − λ : h−∆En − λ En , ϕi = hδ , ϕi = ϕ(0), If we take a test function ϕ ∈ C0∞ (Rn ) with ( 1, |x| ≤ ε ϕ(x) = 0, |x| ≥ 2ε 145

ϕ ∈ C0∞ (Rn ).

then, using ∆En = −λ En , x 6= 0, −1 = h∆En + λ En , ϕi = hEn , ∆ϕi + λ hEn , ϕi Z

Z

=

En ∆ϕdx + λ ε≤|x|≤2ε

Z

|x|≤2ε

En ϕdx

Z

=

Z

En ∆ϕdx + λ ε≤|x|≤2ε

Z

= ε≤|x|≤2ε

En ϕdx + λ ε≤|x|≤2ε

En ∆ϕdx −

Z

Z

∆En ϕdx + λ ε≤|x|≤2ε

|x| 0,

δ2 > 0.

2 -norm of Let f ∈ Lδ22 (R3 ). Let us estimate the L−δ 1

(−∆)−1 f =

Z R3

f (y)dy . 4π|x − y|

We have that

(−∆)−1 f 2 2 L

3 −δ1 (R )

Z

= R3

Z f (y)dy 2 R3 4π|x − y| dx.

−2δ1

(1 + |x|)

Here Z Z  Z  f (y)dy 2 dy 2δ2 2 R3 |x − y| ≤ R3 (1 + |y|) | f (y)| dy R3 (1 + |y|)2δ2 |x − y|2  Z dy 2 = k f kL2 (R3 ) . δ2 R3 (1 + |y|)2δ2 |x − y|2 Next we estimate dy (1 + |y|)2δ2 |x − y|2 Z Z dy dy = + =: I1 + I2 . 2δ 2 |y|1 (1 + |y|)2δ2 |x − y|2 Z

I :=

R3

We have ( C , |x| > 2 dy C C |x|2 I1 ≤ ≤ ≤ ≤ . 2 2 (1 + |x|)2 |y|1 |y|2δ2 |x − y|2 R3 |x|2δ2 |z|2δ2 |e − z|2 |x|2 Z dz 1 C = 2δ −1 ≤ 2δ −1 , 1/2 < δ2 < 3/2. 2δ 2 2 2 3 |x| |x| 2 R |z| |e − z|

I2 ≤

Z

Z

Indeed: 1) If |z| < 1/2 then |e − z| ≥ |e| − |z| = 1 − |z| > 1/2 and dz ≤C 2δ |z| 2. In particular, choosing 3/2 > δ1 = δ2 = δ > 1 we get 2 (−∆)−1 : Lδ2 (R3 ) → L−δ (R3 ), 3/2 > δ > 1. Finally, if δ ≥ 3/2 then the result just proved implies that 2 2 (R3 ) ⊂ L−δ (R3 ), (−∆)−1 : Lδ2 (R3 ) ⊂ Lδ20 (R3 ) → L−δ 0

where δ0 < 3/2. Exercise 23.3 (Sobolev inequality). Let 0 < α < n, 1 < p < q < ∞ and that

Z

f (y)dy

Rn |x − y|n−α q ≤ Ck f kL p . L

1 q

= 1p − αn . Prove

Hint: For K := |x|−n+α use the representation K = K1 + K2 , where ( ( K, |x| < µ 0, |x| < µ K1 = and K2 = 0, |x| > µ K, |x| > µ.

Solution. Our first observation is that we can assume without loss of generality that k f k p = 1. Indeed, suppose we have proved kK ∗ gkq ≤ C for g ∈ L p (Rn ) with kgk p = 1. Then, setting g := kK ∗ gkq =

f k f kp ,

we have

1 kK ∗ f kq ≤ C k f kp 149

or kK ∗ f kq ≤ C k f k p

L p (Rn ).

for any f ∈ Next, we are going to make use of the Marcinkiewicz interpolation theorem which says the following: if T is a sublinear operator that is weak type (p1 , q1 ) and (p2 , q2 ) with p1 < p2 , q1 6= q2 then it is strong type (p, q) where p and q are related according to 1 1−θ θ = + , p p1 p2

1 1−θ θ = + q q1 q2

with 0 < θ < 1. Recall that an operator T is said to be of weak type (p, q) if   k f kp q m{x : |T f (x)| > λ } ≤ Cp,q λ for f ∈ L p (Rn ) and for any λ > 0. Here m denotes the Lebesgue measure. Let us check that our particular operator f 7→ K ∗ f is of weak type. Indeed, with the notation K = K1 + K2 it is always true that m{x : |K ∗ f | > λ } ≤ m{x : |K1 ∗ f | > λ /2} + m{x : |K2 ∗ f | > λ /2}. Here Z

kK1 ∗ f k pp =

Z

|K1 ∗ f | p dx ≥ |K1 ∗ f | p dx Rn {x:|K1 ∗ f |>λ /2}  p λ m{x : |K1 ∗ f | > λ /2} ≥ 2

or

kK1 ∗ f k pp kK1 k1p k f k pp m{x : |K1 ∗ f | > λ /2} ≤   p ≤   p λ 2

λ 2

for 1 ≤ p < ∞ by Young’s inequality for convolution. Note that Z p p −n+α kK1 k1 = |x| dx |x|≤µ p  p  Z µ 1 = c1 µ α p . = ωn−1 r−n+α rn−1 dr = ωn−1 µ α α 0 On the other hand, H¨older’s inequality implies that kK2 ∗ f k∞ ≤ kK2 k p0 k f k p , where kK2 k p0 =

Z

(−n+α)p0

|x|≥µ

|x|

1/p0 dx

 Z = ωn−1

∞

r µ

0

= c2 µ n/p −n+α = c2 µ −n/p+α = c2 µ −n/q 150

(−n+α)p0 n−1

r

1/p0 dr

if 1 < p < ∞. Now we choose µ so that c2 µ −n/q = λ /2 i.e. µ = c3 λ −q/n . If p = 1 then kK2 k∞ = µ −n+α and our choice for µ becomes µ = c3 λ n−α . In either case we get kK2 ∗ f k∞ ≤

λ λ k f kp = . 2 2

It follows that m{x : |K2 ∗ f | > λ /2} = 0. These considerations allow us to conclude that m{x : |K ∗ f | > λ } ≤ m{x : |K1 ∗ f | > λ /2} + m{x : |K2 ∗ f | > λ /2}
α p kK1 k1p c1 µ α p c1 c3 λ −q/n  p ≤  p =  p = = c4 λ −α pq/n−p = c4 λ −q . λ 2

λ 2

λ 2

Therefore the operator f 7→ K ∗ f is weak type (p, q) if 1 ≤ p < q < ∞ and 1q = 1p − αn . The interpolation theorem of Marcinkiewicz implies now that the operator is strong type (p, q) with (p1 = 1, p2 > 1) 1 1−θ θ = + , p 1 p2 where

α 1 = 1− , q1 n

1 1−θ θ = + , q q1 q2 1 1 α = − . q2 p2 n

It follows that 1 α θ α θ α α − = 1−θ + − = 1−θ + +θ − p n p2 n q2 n n  α θ 1−θ θ 1 = (1 − θ ) 1 − + = + = . n q2 q1 q2 q c+ ◦ q for all k > 0 is a compact operator Exercise 23.4. Show that the integral operator G k in L∞ (Rn ), where q satisfies the conditions of Theorem 23.6. Solution. The function G+ k (x)

i = 4



|k| 2π|x|

(n−2)/2

(1)

H(n−2)/2 (|k||x|)

satisfies the estimates 151

(1) for |x| < 1,

( |x|2−n , n≥3 |G+ (x)| ≤ C k | log |x||, n = 2

(2) for |x| > 1,

|G+ k (x)| ≤

(3) for n ≥ 3 and |x − x0 | small enough,

C , |x|1/2

n≥2

+ 0 |G+ k (x − y) − Gk (x − y)|

( 2+ j−n |x0 − y|− j−1 , |x − y| < 1 ∑n−3 j=0 |x − y| ≤ C|x − x | |x − y|−(n+1)/2 + |x0 − y|−(n+1)/2 , |x − y| > 1 0

(4) for n = 2, ( |x − x0 |ε−1 |x0 − y|−ε , |x − y| < 1 + 0 0 − y)| ≤ C|x − x | |G+ (x − y) − G (x k k −1/2 0 −1/2 |x − y| |x − y| , |x − y| > 1. + d ∞ n First we show that G k q maps L (R ) into itself. Indeed, if n ≥ 3 then by (1) and (2) we get,

Z

|G+ k (x − y)||q(y)||u(y)|dy Z Z 2−n ≤ C kukL∞ |x − y| |q(y)|dy +

+ d |G k qu(x)| ≤

Rn

|x−y| 0. Next we show using Ascoli–Arzel`a theorem (Theorem 34.7) that this operator is compact in L∞ (Rn ). The previous calculations show that if u ∈ U ⊂ L∞ (Rn ) and there is a constant M > 0 such that kuk∞ ≤ M then

d

qu

G+

≤ MC, k ∞

+ d where the constant C is independent on u. Thus the image of U under G k q is uniformly bounded. Next we observe that

152

(1) for n ≥ 3, 00 0 + + d d |G k qu(x ) − Gk qu(x )| ≤ M

Z Rn

0 + 00 |G+ k (x − y) − Gk (x − y)||q(y)|dy

n−3 Z

≤ M c1 |x − x0 | ∑

l=0 |x−y|1

Z

(2+l−n)p0

|x−y| 1 this inequality shows that the image of U under G k q is uniformly continuous. (2) for n = 2, see (4) above, 0 00 + + d d |G k qu(x ) − Gk qu(x )|  Z ≤ M c1 |x − x0 |ε

≤ M0

 Z |q(y)|dy |q(y)|dy 0 + c2 |x − x | 0 ε |x−y|1 |x − y|1/2 |x0 − y|1/2 ! Z 1/p0 Z 0 ε 0 −ε p0 0 kqkL p + |x − x | |x − x | |x − y| dy |q(y)|dy Rn

|x−y|1 |x − y|1/2−ε

Z

Z

154

|x−y|>1

|q(y)|dy < ∞

ε > 0.

and |q(y)|dy ≤ |x−y| 2. Solution. We make use of the representation formula (Theorem 39.20) Z

u(x) = S


u(y)∂νy K(x − y) − K(x − y)∂ν u(y) dσ (y), 275

x ∈ Ω,

where Ω is a bounded domain with boundary S. Let first n > 2 and let x ∈ Rn \ Ω. Consider BR (x) such that Ω ⊂ BR (x). Denote DR := BR (x) \ Ω. Then Z

u(x) = −

Z

|x−y|=R


u(y)∂νy K(x − y) − K(x − y)∂ν u(y) dσ (y)


u(y)∂νy K(x − y) − K(x − y)∂ν u(y) dσ (y),

S

x ∈ DR .

We claim that the first integral tends to 0 as R tends to ∞. Indeed, on the sphere |y−x| = R we have that (x − y, ν(y)) = cn |x − y|1−n |∂νy K(x − y)| = ωn |x − y|n since ν(y) = (y − x)/|y − x|. Therefore Z |x−y|=R

u(y)∂νy K(x − y)dσ (y) = cn Z

= cn

Z

u(y)R1−n dσ (y)

|x−y|=R 1−n

|z|=R

u(x − z)R

Z

dσ (z) = cn

|θ |=1

u(x − Rθ )dθ → 0, R → ∞

since u(x) is o(1) for large x. On the other hand, since u is harmonic in the annulus R < |x| < R1 then by Exercise 39.2 we have that Z

Z

|x−y|=R

∂νy u(y)dσ (y) =

Hence

|x−y|=R1

∂νy u(y)dσ (y).

Z |x−y|=R

∂νy u(y)dσ (y)

is constant with respect to R. That’s why Z |x−y|=R

K(x − y)∂νy u(y))dσ (y) = cn R2−n

Z |x−y|=R

∂νy u(y))dσ (y) = c0n R2−n → 0

as R → ∞. This proves that u(x) = −

Z S


u(y)∂νy K(x − y) − K(x − y)∂ν u(y) dσ (y),

x ∈ Rn \ Ω.

Here K(x − y) = cn |x − y|2−n = O(|x|2−n ), ∂νy K(x − y) = −

(x − y, ν(y)) = O(|x|1−n ) ωn |x − y|n

for large x and bounded y. Indeed,   (x, y) |y|2 |x − y| = |x| − 2(x, y) + |y| = |x| 1 − 2 2 + 2 |x| |x| 2

2

2

276

2

implies that |x − y|2 /|x|2 = 1 − 2

(x, y) |y|2 |y| |y|2 + ≤ 1 + 2 + ≤C |x|2 |x|2 |x| |x|2

or |x − y| = O(|x|) for large x and bounded y. Thus u(x) = O(|x|2−n ),

|x| → ∞.

In particular, u is bounded at infinity. Moreover, ∂r u(x) = ∑ ∂x j u j

xj ∂xj = ∑ ∂x j u. ∂r j r

Observe that !(2−n)/2 2

∂x j K(x − y) = cn ∂x j =

c0n

∑(x j − y j ) j

2

∑(x j − y j )

!−n/2

j

2(x j − y j ) = c00n |x − y|−n (x j − y j ) = O(|x|1−n )

as |x| → ∞. Similarly, ∂x j ∂νy K(x − y) = O(|x|−n ),

|x| → ∞.

Therefore, ∂r u(x) = O(|x|1−n ),

|x| → ∞.

It remains to consider the case n = 2. Here we have only that u = o(log |x|) for large x. Start as above and write Z


u(y)∂νy K(x − y) − K(x − y)∂ν u(y) dσ (y)

u(x) = −

Z

|x−y|=R

S


u(y)∂νy K(x − y) − K(x − y)∂ν u(y) dσ (y)

1 = o(log R) − log R 2π −

Z S

Z |x−y|=R

∂ν u(y)dσ (y)


u(y)∂νy K(x − y) − K(x − y)∂ν u(y) dσ (y).

Since u = o(log R) then Z |x−y|=R

∂ν u(y)dσ (y)

must be zero. Hence u(x) = o(log R) −

Z S


u(y)∂νy K(x − y) − K(x − y)∂ν u(y) dσ (y). 277

Since the left hand side and the integral do not depend on R then letting R → ∞ yields u(x) = c0 −

Z


u(y)∂νy K(x − y) − K(x − y)∂ν u(y) dσ (y),

S

where c0 is the limit of the first term on the right hand side. The behaviour for large x can now be studied by applying this result outside some ball. So write u(x) = c0 − = c0 − Z

Z

Z |y−x0 |=R0

|y−x0 |=R0

+ |y−x0 |=R0


u(y)∂νy K(x − y) − K(x − y)∂ν u(y) dσ (y)

u(y)∂νy K(x − y)dσ (y)

∂νy u(y)(K(x − y) − K(x))dσ (y) + K(x)

Z |y−x0 |=R0

∂νy u(y)dσ (y).

Here the last integral is again zero and thus Z

u(x) = c0 − +

1 2π

|y−x0 |=R0

Z

|y−x0 |=R0

u(y)∂νy K(x − y)dσ (y)

∂νy u(y)(log |x − y| − log |x|)dσ (y).

But log |x − y| − log |x| → 0 as |x| → ∞1 . Since ∂νy K(x − y) = O(|x|−1 ) then u(x) is bounded at infinity. Moreover, ∂r u(x) = − Z

Z |y−x0 |=R0

+ |y−x0 |=R0

u(y)∂νy ∂r K(x − y)dσ (y)

∂νy u(y)∂r (log |x − y| − log |x|)dσ (y).

Here ∂νy ∂r K(x − y) = O(|x|−2 ) (at least) and 2

∂r (log |x − y| − log |x|) = 1 This

xj

∑ ∂x (log |x − y| − log |x|) r . j

j=1

can be seen as follows: If |x| < |x − y| then   |x − y| |x − y| − |x| |x − y| − |x| |y| log = log 1 + < ≤ |x| |x| |x| |x|

Here we have used the fact that log(1 + a) < a, ex − 1

a > 0.

This is true since > 0, x > 0 and hence the function ex − x is increasing for x > 0. This implies that x x e − x > 1 or e > 1 + x or x > log(1 + x) for x > 0. If |x| > |x − y| we get   log |x − y| = − log |x| = log |x| = log 1 + |x| − |x − y| < |x| − |x − y| ≤ |y| . |x| |x − y| |x − y| |x − y| |x − y| |x − y|

278

Here x j /r is only O(1) but, for j = 1, 2 it is true that, 1 xj −yj 1 xj − |x − y| |x − y| |x| |x| xj xj yj = − 2− 2 |x − y| |x| |x − y|2 yj |x|2 − |x − y|2 − = xj |x − y|2 |x|2 |x − y|2 yj 2(x, y) − |y|2 = xj − = O(|x|−2 ), |x| → ∞. 2 2 |x − y| |x| |x − y|2

∂x j (log |x − y| − log |x|) =

Thus finally ∂r u(x) = O(|x|−2 ), |x| → ∞. Exercise 41.3. Prove that a Hilbert-Schmidt operator i.e. an integral operator whose kernel I(x, y) satisfies Z Z S S

|I(x, y)|2 dxdy < ∞

is compact in L2 (S). Solution. We use the following facts about compact operators. Definition. A bounded linear operator on a Banach space X is compact if it maps bounded sequences (in X) into sequences having a convergent subsequence. Theorem. A bounded operator T is compact if there is a sequence Tm of operators of finite rank (i.e. R(Tm ) is finite dimensional) such that kT − Tm k → 0. Let

Z

Ibf (x) = be such that

I(x, y) f (y)dy, S

Z Z S S

f ∈ L2 (S)

|I(x, y)|2 dxdy < ∞.

Then Z 1/2 Z 1/2 Z b 2 2 |I(x, y)| dy | f (y)| dy I f (x) ≤ |I(x, y)| | f (y)| dy ≤ S

S

Z = S

S

1/2 2 k f k2 . |I(x, y)| dy

Hence  Z Z

2 Z 2

b b 2 |I(x, y)| dy k f k22 dx

I f = I f (x) dx ≤ 2

S

Z Z

= S S

S

S

|I(x, y)|2 dydx k f k22 = kIk22 k f k22 . 279

This shows that Ibf is finite almost everywhere and that the integral operator Ib is bounded:





b

I = sup Ibf ≤ kIk2 . 2

2

k f k2 =1

 ∞ We will now show that Ib is a limit of finite rank operators. To that end, let φ j 1 be an  ∞  ∞ orthonormal basis for L2 (S). Then ψi j i, j=1 = φi φ j i, j=1 is an orthonormal basis for L2 (S × S). Hence ∞

I(x, y) =

∑

ai j ψi j ,

ai j = (I, ψi j )L2 .

i, j=1

Set IN (x, y) =

∑

ai j ψi j ,

N = 1, 2, . . . .

i+ j≤N

Then R(IbN ) ⊂ span {φi , . . . , φN } so IbN has finite rank. Moreover, I − IN =

∑

ai j ψi j

i+ j>N

so that kI − IN k22 =

∑

i+ j>N

|ai j |2 → 0,

N→∞

2

since IN → I (L2 is Hilbert space). This implies that Ib− IbN ≤ kI − IN k22 → 0 as N → 2 ∞

∂νy K(x − y) = −

(x − y, ν(y)) . ωn |x − y|n

Exercise 41.4. Prove that (41.4) holds for all n ≥ 1. Solution. If n = 2 then K(x − y) =

1 log |x − y| 2π

and ∂ 1 1 ∂ K(x − y) = |x − y| ∂yj 2π |x − y| ∂ y j
1/2 1 1 ∂ = (x1 − y1 )2 + (x2 − y2 )2 2π |x − y| ∂ y j
−1/2 1 1 1 =− (x1 − y1 )2 + (x2 − y2 )2 2(x j − y j ) 2π |x − y| 2 1 yj −xj = . 2π |x − y|2 280

(41.4)

Hence ∇y K(x − y) = If n 6= 2 then

1 y−x y−x = . 2π |x − y|2 ωn |x − y|n

K(x − y) = It follows that

|x − y|2−n . (2 − n)ωn

∂ 1 ∂ K(x − y) = |x − y|2−n ∂yj (2 − n)ωn ∂ y j
(2−n)/2 1 ∂ = (x1 − y1 )2 + · · · + (xn − yn )2 (2 − n)ωn ∂ y j
−n/2 2−n 1 =− (x1 − y1 )2 + · · · + (xn − yn )2 2(x j − y j ) (2 − n)ωn 2 yj −xj = ωn |x − y|n the same as above. Hence again ∇y K(x − y) =

y−x . ωn |x − y|n

In any case ∂νy K(x − y) = ν(y) · ∇y K(x − y) = −

(x − y, ν(y)) . ωn |x − y|n

Exercise 41.5. Prove that (41.4) defines a harmonic function at infinity. Solution. Recall that a function u is harmonic at infinity if the function   x 2−n ue(x) = |x| u |x|2 has the behaviour

( o(|x|2−n ), n ≥ 3 ue(x) = o(log |x|), n = 2

as x → 0. It is enough to consider the function u(x) =

(x − y, ν(y)) . |x − y|n

Then  
2−n −2 x 2 , ν(y) − y, ν(y) 2−n |x|2 |x| |x| x − y|x| n = |e u(x)| = |x| −2n |x − y|x|2 |n x |x| 2 − y |x|

≤ |x|n

|x − y|x|2 | n 2 1−n ≤ C|x|n |x|1−n = C|x| n = |x| x − y|x| 2 |x − y|x| | 281

since x − y|x|2 ≤ |x| + |y||x|2 ≤ |x| + |y||x| = (1 + |y|)|x| = C|x| for |x| < 1. If n ≥ 3 it follows that |e u(x)| ≤ C|x|n−1 → 0, |x|2−n If n = 2 then

x → 0.

C|x| |e u(x)| ≤ → 0, |log |x|| |log |x||

x → 0.

Exercise 41.6. Prove that ( δ, n>2 (|K(x − y)| + |K(x0 − y)|) |ϕ(y)|dσ (y) ≤ c kϕk∞ 1 Bδ δ log δ , n = 2.

Z

Solution. Here x, x0 ∈ S and ϕ ∈ C(S). We prove that if |x − x0 | < δ then ( Z δ, n>2 I := (|K(x − y)| + |K(x0 − y)|) dσ (y) ≤ c 1 Bδ δ log δ , n = 2, where Bδ = {y ∈ S : |x0 − y| < δ }. Let first n > 2. Then |I| ≤

1 (n − 2)ωn

Z Bδ


|y − x|2−n + |y − x0 |2−n dσ (y).

(1) If |y − x| ≥ |y − x0 | then |y − x|2−n ≤ |y − x0 |2−n . Hence 2 |I| ≤ |y − x0 |2−n dσ (y), y − x0 = rθ (n − 2)ωn Bδ Z δ Z 2ωn−1 2 = dr r2−n rn−2 dθ = δ = cδ . (n − 2)ωn 0 (n − 2)ωn |θ |=1 Z

(2) If |y − x| ≤ |y − x0 | then |y − x|2−n ≥ |y − x0 |2−n . Since |y − x| ≤ |y − x0 | + |x − x0 | < 2δ then Bδ = {y : |y − x0 | < δ } ⊂ {y : |y − x| < 2δ } . Hence Z Bδ

2−n

|x − y|

2−n

+ |x0 − y|




dσ (y) ≤ 2

after using polar coordinates. 282

Z {y:|y−x| 0. Exercise 42.2. Show that Theorem 42.6 does not hold for L = ∆2 on Ω ⊂ Rn for any u ∈ H 2 (Ω). 287

Solution. Let us assume on the contrary that Z

Re Ω

|∆u|2 dx ≥ ν1 kuk22 − ν2 kuk20

for any u ∈ H 2 (Ω) with some positive constants ν1 and ν2 . If u ∈ H 2 (Ω) is harmonic on Ω then the latter inequality implies that ν2 kuk20 ≥ ν1 kuk22 . This contradiction proves the claim. Note that this proof does not hold for harmonic function u ∈ H02 (Ω) since in that case u ≡ 0 due to homogeneous boundary condition. Exercise 42.3. Prove that the range of the operator (−∆)m + µI, µ > 0 considered on H0m (B), B the unit ball in Rn , is complete in H l (B) for all l = 0, 1, 2, . . .. Hint: Prove the solvability of the equation ((−∆)m + µI)u = (|x|2 − 1)P(x) for a polynomial P(x). Solution. Let f ∈ C0∞ (B). Then for m = 1, 2, . . . we have ((−∆)m f + µ f , f )L2 (B) = µ k f k2L2 ((B) ( ((−∆)m/2 f , (−∆)m/2 f )L2 (B) , m is even + (m−1)/2 (m−1)/2 ((−∆) (∇ f ), (−∆) (∇ f ))L2 (B) , m is odd Using Parseval equality we obtain ((−∆)m f + µ f , f )L2 (B) =

Z Rn

(|ξ |2m + µ)| fb|2 dξ ≥ min(1, µ)

Z Rn

(|ξ |2m + 1)| fb|2 dξ .

Using the elementary inequality |ξ |2m + 1 ≥ 2−m (|ξ |2 + 1)m we get m

((−∆) f + µ f , f )L2 (B) ≥ cµ,m

Z Rn

(1 + |ξ |2 )m | fb|2 dξ = cµ,m k f k2H m (B) ,

This implies that (−∆)m + µI is invertible on H0m (B) and it maps as (−∆)m + µI : H0m (B) → H −m (B), where H −m (B) denotes the dual space of H0m (B). It is also true that L2 (B) ⊂ Range((−∆)m + µI) defined on H0m (B). 288

cµ,m > 0.

Next, any function h ∈ H l (B), l = 0, 1, . . . can be approximated by C∞ (B) functions with respect to the norm of the space H l (B) (see [1]) i.e. Hl

C∞ (B) = H l (B). On the other hand any function g ∈ C∞ (B) can be approximated with any of its derivative in L∞ norm by the polynomial P(x) (Taylor expansion) and therefore by the polynomials of the form (1 − |x|2 )P(x). Hence we may conclude that any function h ∈ H l (B) can be approximated in the norm of H l (B) by the polynomials of the form (1 − |x|2 )P(x). Finally, it is easy to see that (1 − |x|2 )P(x) ∈ L2 (B) ⊂ Range((−∆)m + µI). Exercise 42.4. Prove that if {ϕ j }∞j=1 is an orthonormal basis in H0k (Ω), k = 1, 2, . . . then {ϕ j }∞j=1 is a basis in H0l (Ω) for every integer 0 ≤ l < k. Show that the converse is not true. Solution. Since C0∞ (Ω) ⊂ H0k (Ω) ⊂ H0l (Ω),

0≤l 0 and any g ∈ H0l (Ω) there is u ∈ H0k (Ω) such that kg − ukH l (Ω) < ε/2. Let {ϕ j }∞j=1 be an orthonormal basis in H0k (Ω) i.e. ∞

u(x) =

∑ (u, ϕ j )H (Ω) ϕ j (x). k

j=0

So for any ε > 0 there is Nε ∈ N such that



Nε

u − ∑ (u, ϕ j )H k (Ω) ϕ j

j=1

< ε/2.

H k (Ω)

Since l < k then for any g ∈ H0l (Ω) we have





Nε Nε



kg u − (u, ϕ ) ϕ ≤ − uk + g − (u, ϕ ) ϕ k k

∑ j H (Ω) j

∑ j H (Ω) j

Hl

j=1 j=1 H l (Ω) H l (Ω)



Nε

< ε/2 + u − ∑ (u, ϕ j )H k (Ω) ϕ j < ε.

k j=1 H (Ω)

Thus {ϕ j }∞j=1 is also a basis in H0l (Ω). 289

290

Chapter 44

Some Inverse Scattering Problems for the Schr¨odinger Operator k=

|ξ | , 2(θ , ξb)

ξ ξb = . |ξ |

θ 0 = θ − 2(θ , ξb)ξb,

Exercise 44.1. Let uθ (k, θ 0 ) be the coordinate mapping M0 → Rn given as uθ (k, θ 0 ) = k(θ − θ 0 ), where θ is considered a fixed parameter. Prove that (1) the formulas (44.1) for k and θ 0 hold (2) the following is true: Z

0

M0

0

ϕ ◦ uθ (k, θ )dµθ (k, θ ) =

Z Rn

ϕ(x)dx

if ϕ ∈ S is even and Z M

ϕ ◦ uθ (k, θ 0 )dµ(k, θ 0 , θ ) =

Z Rn

ϕ(x)dx

if ϕ ∈ S. (3) in addition, FM−10 (ϕ ◦ uθ ) = F −1 ϕ if ϕ ∈ S is even and

FM−1 (ϕ ◦ uθ ) = F −1 ϕ

if ϕ ∈ S. Here F −1 is the usual inverse Fourier transform in Rn . 291

(44.1)

(1) If ξ = k(θ − θ 0 ) then

Solution.

(θ , ξ ) = k(θ , θ − θ 0 ) = k(1 − (θ , θ 0 ))). Hence |ξ |2 = k2 |θ − θ 0 |2 = k2 (2 − 2(θ , θ 0 )) = 2k(θ , ξ ) or k=

|ξ |

2(θ , ξˆ )

,

ξ ξˆ = . |ξ |

Moreover 2(θ , ξˆ )ξˆ =

|ξ | ξ = θ − θ 0. k |ξ |

(2) Recall that M0 = R × Sn−1 and M = M0 × Sn−1 with 1 dµθ (k, θ 0 ) = |k|n−1 dk|θ − θ 0 |2 dθ 0 , 4

dµ(k, θ 0 , θ ) =

1 |Sn−1 |

dθ dµθ (k, θ 0 ).

If now ϕ is even then Z M0

1 ϕ(k(θ − θ 0 ))|k|n−1 |θ − θ 0 |2 dθ 0 dk 4 R Sn−1 Z Z 1 0 = ϕ(k(θ − θ 0 ))(−k)n−1 |θ − θ 0 |2 dθ 0 dk 4 −∞ Sn−1 Z Z 1 ∞ + ϕ(k(θ − θ 0 ))kn−1 |θ − θ 0 |2 dθ 0 dk 4 0 Sn−1 Z Z 1 0 =− ϕ(−k(θ − θ 0 ))kn−1 |θ − θ 0 |2 dθ 0 dk 4 ∞ Sn−1 Z Z 1 ∞ + ϕ(k(θ − θ 0 ))kn−1 |θ − θ 0 |2 dθ 0 dk 4 0 Sn−1 Z Z 1 ∞ = ϕ(k(θ − θ 0 ))kn−1 |θ − θ 0 |2 dθ 0 dk n−1 2 Z 0 S

ϕ ◦ uθ (k, θ 0 )dµθ (k, θ 0 ) =

Z Z

= Rn

ϕ(ξ )dξ .

Here we have used the fact that the Jacobian of the transformation is ∂ ξ 1 n−1 0 2 ∂ (k, θ ) = 2 |k| |θ − θ | . 292

Similarly, Z M

ϕ ◦ uθ (k, θ 0 )dµ(k, θ 0 , θ ) 1 = ϕ(k(θ − θ 0 ))|k|n−1 |θ − θ 0 |2 dθ 0 dθ dk 4|Sn−1 | R Sn−1 Sn−1 Z 0 Z Z 1 ϕ(k(θ − θ 0 ))(−k)n−1 |θ − θ 0 |2 dθ 0 dθ dk = 4|Sn−1 | −∞ Sn−1 Sn−1 Z ∞Z Z 1 ϕ(k(θ − θ 0 ))kn−1 |θ − θ 0 |2 dθ 0 dθ dk + 4|Sn−1 | 0 Sn−1 Sn−1 Z ∞Z Z 1 = ϕ(−k(θ − θ 0 ))kn−1 |θ − θ 0 |2 dθ 0 dθ dk 4|Sn−1 | 0 Sn−1 Sn−1 Z ∞Z Z 1 + ϕ(k(θ − θ 0 ))kn−1 |θ − θ 0 |2 dθ 0 dθ dk 4|Sn−1 | 0 Sn−1 Sn−1 Z ∞Z Z 1 = ϕ(k(θ − θ 0 ))kn−1 |θ − θ 0 |2 dθ 0 dθ dk 2|Sn−1 | 0 Sn−1 Sn−1 Z Z Z 1 ϕ(ξ )dξ dθ = ϕ(ξ )dξ . = n−1 |S | Rn Sn−1 Rn Z Z

Z

(3) Since (FM−10 ϕ1 )(x)

1 = (2π)n/2

Z

0

M0

e−ik(θ −θ ,x) ϕ1 (k, θ 0 )dµθ ,

then 0 1 e−ik(θ −θ ,x) (ϕ ◦ uθ )(k, θ 0 )dµθ n/2 (2π) M0 Z 1 = (e−i(·,x) ◦ uθ )(k, θ 0 )(ϕ ◦ uθ )(k, θ 0 )dµθ (2π)n/2 M0 Z 1 e−i(ξ ,x) ϕ(ξ )dξ = F −1 ϕ(x) = (2π)n/2 Rn

Z

(FM−10 ϕ ◦ uθ )(x) =

by part (2) for even ϕ. Similarly, (FM−1 ϕ ◦ uθ )(x) =

1 (2π)n/2

Z M

0

e−ik(θ −θ ,x) (ϕ ◦ uθ )(k, θ 0 , θ )dµ.

Exercise 44.2. Prove that (1) A(−k, θ 0 , θ ) = A(k, θ 0 , θ ) (2) A(k, θ 0 , θ ) = A(k, −θ , −θ 0 ). Solution.

(1) Since 0

A(k, θ , θ ) =

Z

0

Rn

e−ik(θ ,y) q(y)u(y, k, θ )dy

293

and q is real then A(k, θ 0 , θ )

Z

0

= Rn

Z

eik(θ ,y) q(y)u(y, k, θ )dy 0

= Rn

eik(θ ,y) q(y)u(y, −k, θ )dy = A(−k, θ 0 , θ ).

Here we extended u to negative k by u(y, k, θ ) = u(y, −k, θ ). (2) This is the reciprocity relation (see [7, p. 185]). Since Hu = k2 u,

u = u0 + usc

and (−∆ − k2 )u0 = 0 then (H − k2 )usc = −qu0 . Therefore A(k, θ 0 , θ ) = −A−qu0 (k, θ 0 ) =−

Z

Rn

Z

0

= n

ZR

Rn

Z Rn

cq (q(·)u0 (·, k, θ ))(y))dy e−ik(θ ,y) q(y)(u0 (y, k, θ ) − G 0

= −

0 cq (q(·)u0 (·, k, θ ))(y))dy e−ik(θ ,y) (−q(y)u0 (y, k, θ ) + q(y)G

e−ik(θ ,y) q(y)u0 (y, k, θ )dy 0

cq (q(·)u0 (·, k, θ ))(y))dy e−ik(θ ,y) q(y)G

The first term is alright. For the second term we note that Z   + c Gk f , g 2 = L

c+ f (y)g(y)dy G k

n ZR Z

= n n ZR R

=

G+ k (|y − z|) f (z)dzg(y)dy Z

f (z) Rn

Z

=

Rn



Z

f (z) Rn

G+ k (|y − z|)g(y)dydz

G+ k (|y − z|)g(y)dydz Rn

cq via The same is true for G c+ − G c+ qG cq = G cq G k k 294

=

 c+ f , Gk g . L2

since q is real. Hence Z

  0 cq (q(·)u0 (·, k, θ ))(y))dy = G cq (qu0 (θ )), eik(θ 0 ,y) q e−ik(θ ,y) q(y)G L2 Rn   c u (θ 0 )q = qu0 (θ ), G q 0 2 L

Z

ik(θ ,y)

=

e R3

Z

= R3

c u (θ 0 )qdy q(y)G q 0

cq u0 (−θ 0 )qdy eik(θ ,y) q(y)G

and thus A(k, θ 0 , θ ) =

Z Rn

Z

= Rn

0

e−ik(θ ,y) q(y)u0 (y, k, θ )dy −

Z

eik(θ ,y) q(y)u0 (y, k, −θ 0 )dy −

= A(k, −θ , −θ 0 ).

0

Rn

cq (qu0 (θ ))(y))dy e−ik(θ ,y) q(y)G

Z R3

cq u0 (−θ 0 )qdy eik(θ ,y) q(y)G

Alternative solution. We follow [6, footnote 7]. Firstly, Z

Z

0

eik(θ ,y) q(y)u(y, k, θ )dy = u(y, k, θ )q(y)(u(y, k, θ 0 ) − usc (y, k, θ 0 ))dy Rn Rn   Z Z 0 + 0 = u(y, k, θ )q(y) u(y, k, θ ) + Gk (|y − z|)q(z)u(z, k, θ )dz dy Rn

Rn

Z

= Rn

eik(θ ,y) q(y)u(y, k, θ 0 )dy.

The last step holds since the right hand side is symmetric in θ and θ 0 . Hence A(k, −θ 0 , θ ) =

Z ZR

0

n

= Rn

eik(θ ,y) q(y)u(y, k, θ )dy eik(θ ,y) q(y)u(y, k, θ 0 )dy = A(k, −θ , θ 0 )

or A(k, θ 0 , θ ) = A(k, −θ , −θ 0 ).

295

296

Chapter 45

The Heat Operator   |x|2 2 Kt (x) := (2π)−n/2 F −1 e−|ξ | t = (4πt)−n/2 e− 4t ,

t > 0.

Exercise 45.1. Prove (45.2). Solution. We have proved in Exercise 37.7 that   |ξ |2 2 F e−a|x| = (2a)−n/2 e− 4a ,

a > 0.

But Z   2 F −1 e−a|ξ | (x) = (2π)−n/2

= (2π)−n/2

2

ZR

eix·ξ e−a|ξ | dξ

n

  2 2 e−ix·η e−a|η| dη = F e−a|ξ | (x).

Rn

Hence     2 2 Kt (x) = (2π)−n/2 F −1 e−t|ξ | (x) = (2π)−n/2 F e−t|ξ | (x) = (2π)−n/2 (2t)−n/2 e−

|x|2 4t

= (4πt)−n/2 e−

|x|2 4t

.

Exercise 45.2. Prove Theorem 45.6. Solution. First we observe that Z Rn

|u(x,t)|dx ≤ ≤

Z t

Z

dx Rn Z t

−∞

Z

ds −∞

Z

ds

Rn

Rn

Kt−s (x − y)| f (y, s)|dy

| f (y, s)|dy 297

Z Rn

Kt−s (x − y)dx ≤ k f kL1 (Rn+1 )

(45.2)

so that u(·,t) ∈ L1 (Rn ). Now write

Z ∞

u(x,t) =

Z

ds Rn

−∞

et−s (x − y) f (y, s)dy, K

et−s = 0 for s > t. Let ϕ ∈ C∞ (Rn+1 ). Then where K 0 h∂t u − ∆x u, ϕi = hu, −∂t ϕ − ∆x ϕi = Z ∞

=

dt

=

dx Rn

dy Rn

Z−∞ ∞

=

Z ∞

I=

Z

dt −∞

Rn

Rn

Z

dt n Z R∞

−∞

Z

dy f (y, s)

ds Rn

−∞

Here

ds Z−∞ ∞

Z

ds

Z

dt Z−∞

Z ∞

Z

Z−∞ ∞

Z ∞

Rn

u(x,t)(−∂t − ∆x )ϕ(x,t)dx

et−s (x − y) f (y, s)(−∂t − ∆x )ϕ dyK et−s (x − y) f (y, s)(−∂t − ∆x )ϕ dxK Z

dt Rn

−∞

et−s (x − y)(−∂t − ∆x )ϕ. dxK

et−s (x − y)(−∂t − ∆x )ϕ(x,t) dxK

et−s (x − y), (−∂t − ∆x )ϕ(x,t)i = h(∂t − ∆x )K et−s (x − y), ϕ(x,t)i = hK = hδ (t − s)δ (x − y), ϕ(x,t)i = ϕ(y, s),

since Kt is a fundamental solution of the heat operator. Hence h∂t u − ∆x u, ϕi = h f , ϕi.

∞

u(x,t) =

2 j

∑ f j Fj (x)e−λ t ,

(45.7)

j=1

Exercise 45.3. Prove that u(x,t) of the form (45.7) is a distributional solution of the heat equation in Ω × (0, ∞). Solution. Recall from the book that Fj is an orthonormal basis in L2 (Ω) and ( ∆Fj = −λ j2 Fj , in Ω Fj = 0, on ∂ Ω. Moreover f j = ( f , Fj )L2 . Let ϕ = ϕ(x,t) ∈ C0∞ (Ω × (0, ∞)). Then h∂t u − ∆x u, ϕi = hu, (−∂t − ∆x )ϕi = ∞

=

∑

Z Z ∞

fj

j=1 ∞

=

∑ fj

j=1 ∞

+ ∑ fj j=1

Ω 0

Z Z ∞ Ω 0

Z Z ∞ Ω 0

Z Z ∞ Ω 0

u(x,t)(−∂t ϕ − ∆ϕ)dtdx

2

Fj (x)e−λ j t (−∂t ϕ − ∆ϕ)dtdx 2

Fj (x)e−λ j t (−∂t ϕ)dtdx 2

Fj (x)e−λ j t (−∆ϕ)dtdx =

∞

∑ (I1 + I2 ).

j=1

298

Perform integration by parts in both I1 and I2 . So  Z  Z ∞ ∞ −λ j2 t 2 −λ j2 t I1 = − Fj (x)λ j e ϕ(x,t)dt dx Fj (x)e ϕ + t=0 Ω 0  Z Z ∞ 2 =− Fj (x)λ j2 e−λ j t ϕ(x,t)dt dx 0

Ω

and Z Z ∞

I2 =

Ω 0 n

= −∑

i=1

Z Z ∞

i=1 Ω 0 n Z Z ∞

= −∑

i=1 Ω 0

Z Z ∞

= Ω 0

Hence

Z Z ∞

I1 + I2 =

0

Ω

n

2

Fj (x)e−λ j t (− ∑ ∂i2 ϕ)dtdx 2

Fj (x)e−λ j t (∂i2 ϕ)dtdx 2

(∂i2 Fj (x))e−λ j t ϕdtdx 2

(−∆Fj (x))e−λ j t ϕdtdx.

−Fj (x)λ j2 − ∆Fj (x)




−λ j2 t

e

 ϕ(x,t)dt dx = 0.

Exercise 45.4. Show that 0π |u(x,t)|2 dx is a decreasing function of t > 0, where u(x,t) is the solution of ( ut − uxx = 0, 0 < x < π,t > 0 u(0,t) = u(π,t) = 0, t > 0. R

Solution. Denote Z π

G(t) = 0

2

|u(x,t)| dx =

Z π

u(x,t)u(x,t)dx. 0

We have G0 (t) =

Z π 0

Z π

(ut (x,t)u(x,t) + u(x,t)ut (x,t))dx =

= ux u|π0 −

Z π 0

ux ux dx + uux |π0 −

Z π 0

0

(uxx u + uuxx )dx

ux ux dx = −2

Z π 0

|ux |2 dx ≤ 0

since the boundary terms vanish due to zero boundary conditions for u.

299

300

Bibliography [1] R. A. Adams, Sobolev Spaces, Academic Press, 1975 [2] T. Apostol, Mathematical Analysis, Addison-Wesley, 1974 [3] G. B. Folland, Introduction to Partial Differential Equations, Princeton University Press, 1995 [4] I. M. Gel’fand, G.E. Shilov, Generalized functions Vol 1, Properties and operations, Academic Press, 1964 [5] L. Grafakos, Classical and Modern Fourier Analysis, Pearson, 2004 [6] T. Ikebe, On the phase-shift formula for the scattering operator, Pacific J. Math. 15, 1965, 511–523 [7] A. Kirsch, An Introduction to the Mathematical Theory of Inverse Problems, Springer, 1996 [8] R. Kress, Linear Integral Equations, Springer, 1999 [9] C. S. Kubrusly, The Elements of Operator Theory, Springer, 2011 [10] N. N. Lebedev, Special Functions and Their Applications, Dover, 1972 [11] M. Pinsky, Introduction to Fourier Analysis and Wavelets ,Brooks/Cole,2002 [12] R. S. Strichartz, A guide to distribution theory and Fourier transforms, World Scientific, 2003 [13] E. Zeidler, Applied Functional Analysis, Springer, 1995

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