Lectures on Partial Differential Equations

Lectures on Partial Differential Equations By

G.B. Folland

Tata Institute of Fundamental Research Bombay 1983

Lectures on Partial Differential Equations By

G.B. Folland Lectures delivered at the Indian Institute of Science, Bangalore under the T.I.F.R. – I.I.Sc. Programme in Applications of Mathematics

Notes by

K.T. Joseph and S. Thangavelu

Published for the

Tata Institute of Fundamental Research Bombay Springer-Verlag Berlin Heidelberg New York

1983

Author

G.B. Folland University of Washington Seattle, Washington 98175 U.S.A.

©Tata Institute of Fundamental Research, 1983

ISBN 3-540-12280-X Springer-Verlag, Berlin, Heidelberg. New York ISBN 0-387-12280-X Springer-Verlag, New York. Heidelberg. Berlin

No part of this book may be reproduced in any form by print, microfilm or any other means without written permission from the Tata Institute of Fundamental Research, Colaba, Bombay 400 005

Printed by N.S. Ray at The Book Center Limited, Sion East, Bombay 400 022 and published by H. Goetze, Springer-Verlag, Heidelberg, West Germany

Printed in India

Preface This book consists of the notes for a course I gave at the T.I.F.R. Center in Bangalore from September 20 to November 20, 1981. The purpose of the course was to introduce the students in the Programme in Application of Mathematics to the applications of Fourier analysis-by which I mean the study of convolution operators as well as the Fourier transform itself-to partial differential equations. Faced with the problem of covering a reasonably broad spectrum of material in such a short time, I had to be selective in the choice of topics. I could not develop any one subject in a really thorough manner; rather, my aim was to present the essential features of some techniques that are well worth knowing and to derive a few interesting results which are illustrative of these techniques. This does not mean that I have dealt only with general machinery; indeed, the emphasis in Chapter 2 is on very concrete calculation with distributions and Fourier transforms-because the methods of performing such calculations are also well worth knowing. If these notes suffer from the defect of incompleteness, they posses the corresponding virtue of brevity. They may therefore be of value to the reader who wishes to be introduced to some useful ideas without having to plough through a systematic treatise. More detailed accounts of the subjects discussed here can be found in the books of Folland [1], Stein [2], Taylor [3], and Treves [4]. No specific knowledge of partial differential equations or Fourier Analysis is presupposed in these notes, although some prior acquittance with the former is desirable. The main prerequisite is a familiarity with the subjects usually gathered under the rubic “real analysis”: measure v

vi and integration, and the elements of point set topology and functional analysis. In addition, the reader is expected to be acquainted with the basic facts about distributions as presented, for example, in Rudin [7]. I wish to express my gratitude to professor K.G. Ramanathan for inviting me to Bangalore, and to professor S.Raghavan and the staff of the T.I.F.R. Center for making my visit there a most enjoyable one. I also wish to thank Mr S. Thangavelu and Mr K.T. Joseph for their painstaking job of writing up the notes.

Contents Preface

v

1 Preliminaries 1 1 General Theorems About Convolutions . . . . . . . . . 1 2 The Fourier Transform . . . . . . . . . . . . . . . . . . 5 3 Some Results From the Theory of Distributions . . . . . 12 2 Partial Differential Operators with Constant Coefficients 1 Local Solvability and Fundamental Solution . . . . . . 2 Regularity Properties of Differential Operators . . . . . 3 Basic Operators in Mathematical Physics . . . . . . . 4 Laplace Operator . . . . . . . . . . . . . . . . . . . . 5 The Heat Operator . . . . . . . . . . . . . . . . . . . 6 The Wave Operator . . . . . . . . . . . . . . . . . . . 3

. . . . . .

15 15 20 22 25 41 44

L2 Sobolev Spaces 53 1 General Theory of L2 Sobolev Spaces . . . . . . . . . . 53 2 Hypoelliptic Operators With Constant Coefficients . . . 64

4 Basic Theory of Pseudo Differential Operators 1 Representation of Pseudo differential Operators . . 2 Distribution Kernels and the Pseudo Local Property 3 Asymptotic Expansions of Symbols . . . . . . . . 4 Properly Supported Operators . . . . . . . . . . . 5 ψdo′ s Defined by Multiple Symbols . . . . . . . . vii

. . . . .

. . . . .

. . . . .

69 69 73 77 79 81

Contents

viii 6 7 8 9 10 5

Products and Adjoint of ψDO′ S . . . . . . . . . . A Continuity Theorem for ψ Do on Sobolev Spaces Elliptic Pseudo Differential Operators . . . . . . . Wavefront Sets . . . . . . . . . . . . . . . . . . . Some Further Applications... . . . . . . . . . . . .

LP and Lipschitz Estimates

. . . . .

. . . . .

. . . . .

87 90 92 95 99 107

Chapter 1 Preliminaries IN THIS CHAPTER, we will study some basic results about convolu- 1 tion and the Fourier transform.

1 General Theorems About Convolutions We will begin with a theorem about integral operators. Theorem 1.1. Let K be a measurable function on Rn × Rn such that, for someR c > 0, R |K(x, y)|dy ≤ c, |K(x, y)|dx ≤ c, for every x, y in Rn .

If 1 ≤R p ≤ ∞ and f ∈ L p (Rn ), then the function T f , defined by T f (x) = K(x, y) f (y)dy for almost every x in Rn , belongs to L p (Rn ) and further, ||T f || p ≤ c|| f || p . R Proof. If p = ∞, the hypothesis |K(x, y)|dx ≤ c is superfluous and the conclusion of the theorem is obvious. If p < ∞, let q denote the conjugate exponent. Then, by Hölder ’s inequality, |T f (x)| ≤

(Z

)1/p )1/q (Z p |K(x, y)| | f (y)| dy |K(x, y)|dy 1

1. Preliminaries

2 1/q

≤c

(Z

)1/p |K(x, y)| | f (y)| dy . p

From this we have, Z x |T f (x)| p dx ≤ c p/q |K(x, y)|| f (y)| p dydx Z p 1+p/q ≤c | f (y)| p dx = c1+p/q || f || p . 2

Therefore ||T f || p ≤ c|| f || p . Next, we define the convolution of two locally integrable functions. Definition 1.2. Let f and g be two locally integrable functions. The convolution of f and g, denoted by f ∗ g, is define by Z Z ( f ∗ g)(x) = f (x − y)g(y)dy = f (y)g(x − y)dy = (g ∗ f )(x), provided that the integrals in question exist. The basic theorem on convolution is the following theorem, called Young’s inequality. Theorem 1.3 (Young’s Inequality). Let f ∈ L1 (Rn ) and g ∈ L p (Rn ), for 1 ≤ p ≤ ∞. Then f ∗ g ∈ L p (Rn ) and || f ∗ g|| p ≤ ||g|| p || f ||1 . Proof. Take K(x, y) = f (x − y) in Theorem 1.1. Then K(x, y) satisfies all the conditions of Theorem 1.1. and the conclusion follows immediately. The next theorem underlies one of the most important uses of convolution. Before coming to the theorem, let us prove the following Lemma 1.4. For a function f defined on Rn and x in Rn , we define a function f x by f x (y) = f (y − x). If f ∈ L p , 1 ≤ p < ∞, then lim || f x − f || p = 0.

x→O

1. General Theorems About Convolutions 3

3

Proof. If g is a compactly supported continuous function, then g is uniformly continuous, and so gx converges to g uniformly as x tends to 0. Further, for |x| ≤ 1, gx and g are supported in a common compact set. Therefore, lim ||gx − g|| p = 0. Given f ∈ L p , we can find a function x→0

g which is continuous and compactly supported such that || f −g|| p 0. But then ||gx − f x || p 0, choose a compact set G Rn such that Z |g(y)|dy < δ. Rn −G

5

Then Sup x∈k (| f ∗ gǫ )(x) − a f (x)| ≤ 2δ|| f ||∞ + Sup(x,y)ǫK×G | f (x − ǫy) − f (x)|

Z

|g|dy.

G

Since f is uniformly continuous on the compact set K the second term tends to 0 as ∈ to 0. Since δ is arbitrary, we see that sup x ∈ K|( f ∗ gǫ )(x) − a f (x)| → 0 as ǫ → 0. Hence the theorem is proved.

2. The Fourier Transform

5

Corollary 1.7. The space Co∞ (Rn ) is dense in L p (R⋉ ) for 1 ≤ p < ∞. Proof. Let 2

φ(x) = e−1/(1−|x| ) for |x| < 1 =0

for |x| ≤ 1

R Then φǫCo∞ (Rn ) and φ(x)dx = 1/a > 0. If f ǫL p and has compact support, then a( f ∗ φǫ )ǫCo∞ (Rn ) and by theorem 1.6, a( f ∗ φǫ ) converges to f in L p as ǫ tends to 0. Since L p functions with compact support are dense in L p , this completes the proof. Proposition 1.8. Suppose K ⊂ Rn is compact and Ω ⊃ K be an open subset of Rn . Then there exists a Co∞ function φ such that φ(x) = 1 for xǫK and Supp φ ⊂ Ω. 1 δ} where δ = d(K, Rn \Ω). Choose 6 2! R 1 a φo ǫCo∞ such that Supp φo ⊂ B 0, δ and φo (x)dx = 1. Define 2 Z φ(x) = φo (x − y)dy = (φ0 ∗ XV )(x).

Proof. Let V = {x ∈ Ω : d(x, K) ≤

V

Then φ(x) is a function with the required properties.

2 The Fourier Transform In this section, we will give a rapid introduction to the theory of the Fourier transform. For a function f ǫL1 (Rn ), the Fourier transform of the function f , denoted by fˆ, is defined by Z ˆf (ξ) = e−2πix.ξ f (x)dx, ξǫRn . Remark 1.9. Our definition of fˆ differs from some other in the placement of the factor of 2π.

1. Preliminaries

6

BASIC PROPERTIES OF THE FOURIER TRANSFORM For f ǫL1 , || fˆ||∞ ≤ || f ||1 .

(1.10)

The proof of this is trivial. For f, g ∈ L1 , ( f ∗ g) ˆ (ξ) = fˆ(ξ)ˆg(ξ).

(1.11) Indeed,

x

=

x

e−2πix.ξ f (y)g(x − y)dydx

=

Z

e−2πi(x−y)·ξ g(x − y)e−2πiY.ξ f (y)dydx Z −2πi(x−y)·ξ e g(x − y)dx f (y)e−2πiy.ξ dy

( f ∗ g)ˆ(ξ) =

= fˆ(ξ)ˆg(ξ). 7

Let us now consider the Fourier transform in the Schwartz class S = S (Rn ). Proposition 1.12. For f ∈ S , we have the following:

i) fˆǫC ∞ (Rn ) and ∂β fˆ = gˆ where g(x) = (−2πix)β f (x).

ii) (∂β f )ˆ(ξ) = (2πiξ)β fˆ(ξ). Proof.

i) Differentiation under the integral sign proves this.

ii) For this, we use integration by parts. Z

e−2πix.ξ (∂β f )(x)dx Z |β| = (−1) ∂β [e−2πix.ξ ] f (x)dx Z |β| β = (−1) (−2πiξ) e−2πix·ξ f (x)dx

(∂β f )ˆ(ξ) =

= (2πiξ)β fˆ(ξ).

2. The Fourier Transform

7

Corollary 1.13. If f ǫS , then fˆ ∈ S also. Proof. For multi-indices α and β, using proposition 1.12, we have ξ α (∂β fˆ)(ξ) = ξ α ((−2πix)β f (x))ˆ (ξ) = (2πi)−|α| [∂α ((−2πix)β f (x))ˆ(ξ)] = (−1)|β| (2πi)|β|−|α| (∂α (xβ f (x)))ˆ (ξ) Since f ǫS , ∂α (xβ f (x))ǫL1 and hence (∂α (xβ f (x)))ˆ ǫL∞ . Thus ξ α (∂β fˆ) is bounded. Since α and β are arbitrary, this proves that fˆǫS . Corollary 1.14 (RIEMANN-LEBESGUE LEMMA). If f ǫL1 , then fˆ is continuous and vanishes at ∞. Proof. By Corollary 1.13, this is true for f ǫS . Since S is dense in L1 8 and || fˆ||∞ ≤ || f ||1 , the same in true for all f ǫL1 . Let us now compute the Fourier transform of the Gaussian. 2 Theorem 1.15. Let f (x) = e−πa|x| , Re a > 0. Then, fˆ(ξ) = a−n/2 −1 e−a π|ξ|2 .

Proof. fˆ(ξ) = fˆ(ξ) =

i.e.,

Z

2

e−2πix.ξ e−aπ|x| dx ∞

n Z Y

2

e−2πix j ξ j e−aπx j dx j .

j=1 −∞

Thus it suffices to consider the case n = 1. Further, we can take a = 1 by making the change of variable x → a−1/2 x. 2

Thus we are assuming f (x) = e−πx , xǫR. Observe that f ′ (x) + 2πx f (x) = 0. Taking the Fourier transform, we obtain ′

2πiξ fˆ(ξ) + i fˆ′ (ξ) = 0. Hence

′ fˆ′ (ξ)/ f (ξ) = −2πξ

1. Preliminaries

8 2

which, on integration, gives f (ξ) = ce−πξ , c being a constant. The constant c is given by c = fˆ(0) =

Z∞

2

e−πx dx = 1.

−∞ 2

Therefore fˆ(ξ) = e−πξ , which completes the proof. We now derive the Fourier inversion formula for the Schwartz class S. 9

Let us define f V (ξ) =

R

e2πix·ξ f (x)dx = fˆ(−ξ).

Theorem 1.16 (Fouries Inversion Theorem). For f ǫ( fˆ)∨ = f . R R Proof. First, observe that for f, gǫL1 , f gˆ = fˆg. In fact, Z x fˆ(x)g(x)dx = e−2πiy·x f (y)g(x)dydx # Z "Z −2πiy·x = e g(x)dx f (y)dy Z = gˆ (y) f (y)dy. Given ǫ > 0 and x in Rn , take the function φ defined by φ(ξ) = e−2πix·ξ−πǫ

2 |ξ|2

.

Now ˆ = φ(y)

Z

e−2πiy·ξ e−2πix·ξ−πǫ

=

Z

e−2πi(y−x)·ξe −πǫ

=∈−n e−π∈

−2 |x−y|2

2 |ξ|2

2 |ξ|2

dξ

dξ

.

2

If we take g(x) = e−π|x| and define gǫ (x) = ǫ −n g(x/ǫ), then ˆ = gǫ (x − y). φ(y)

2. The Fourier Transform Therefore Z

9

e2πix·ξ fˆ(ξ)e−πǫ

2 |ξ|2

dξ =

Z

fˆ(ǫ)φ(ξ)dξ

=

Z

ˆ f (y)φ(y)dy

=

Z

f (y)gǫ (x − y)dy

= ( f ∗ gǫ )(x) But as ǫ tends to 0, ( f ∗ gǫ ) converges to f , by Theorem 1.6 and 10 clearly Z Z 2πi·xξ ˆ −πǫ 2 |ξ|2 e f (ξ)e dξ → fˆ(ξ)e2πi·x.ξdξ Therefore ( fˆ)v = f . Corollary 1.17. The Fourier transform is an isomorphism of S onto S . Next, we prove the Plancherel Theorem. Theorem 1.18. The Fourier transform uniquely extends to a unitary map of L2 (Rn ) onto itself. Proof. For f ǫS , define f˜(x) = f (−x). Then it is easily checked that fˆ˜ = f¯ˆ, so that Z 2 || f ||2 = | f (x)|2 dx Z = f (x) f˜(−x)dx = (f Z = Z = Z =

∗ f˜)(0) ( f∗ f˜)ˆ (ξ)dξ fˆ(ξ) fˆ˜(ξ)dξ fˆ(ξ) f¯ˆ(ξ)dξ = || fˆ||22 .

1. Preliminaries

10

Therefore, the Fourier transform extends continuously to an isometry of L2 . It is a unitary transformation, since its image S is dense in L2 .

11

Let us observe how the Fourier transform interacts with translations, rotations and dilations. (1.19) The Fourier transform and translation: If f x (y) f (y − x) then Z fˆx (ξ) = e2πiy·ξ f (y − x)dy Z = e2πi(z+x)·ξ f (z)dz ( by putting y − x = z) Z = e2πix·ξ fˆ(ξ). (1.20) The Fourier transform and rotations (orthogonal transformations): Let T : n → n be an orthogonal transformation. Then Z ˆ ( f ◦ T ) (ξ) = e−2πix·ξ ( f oT )(x)dx Z −1 = e−2πiT y·ξ f (y)dy ( by putting y = T x) Z = e−2πiy·T ξ f (y)dy = fˆ(T ξ) = ( fôT )(ξ). Thus, ( f oT )ˆ = fôT i.e. ˆ commutes with rotations. (1.21) The Fourier transform and dilation: Let fr (x) = r−n f (x/r). Then Z fˆr (ξ) = e−2πix·ξ r−n f (x/r)dx Z = e−2πiy·ξ f (y)dy = fˆ(rξ). The last equation suggests, roughly: the more spread out f is, the more fˆ will be concentrated at the origin and vice versa. This notion can be put in a precise form as follows.

2. The Fourier Transform (1.22)

11

HEISENBERG INEQUALITY (n = 1: For f ǫS (R), we have ||x f (x)||2 ||ξ fˆ(ξ)||2 ≥ (1/4π)|| f ||22 .

Proof. Observe that

12

df d (x f (x)) = x (x) + f (x). dx dx Thus || f ||22

=

Z

f (x) f (x)dx # " df d = (x f (x)) − x (x) dx f (x) dx dx Z Z ¯ df df = − x f (x) (x)dx − x (x) f¯(x)dx dx dx Z d f¯ = −2Re x f (x) (x)dx. dx ¯ df ≤ 2||x f (x)||2 || ||2 (by Cauchy-Schwarz) dx df 2 || f ||2 ≤ 2||x f (x)||2 || ||2 . dx Z

i.e., But

(

d f¯ ˆ ) (ξ) = 2πiξ fˆ(ξ). dx

Therefore, || f ||22 ≤ 2.2π||x f (x)||2 ||ξ fˆ(ξ)||2 or ||x f (x)||2 ||ξ f (ξ)||2 ≥ (1/4π)|| f ||22 .

12

1. Preliminaries

A GENERALISATION TO n VARIABLES ∂ d by . Also for any a j , b j ǫ, we can reWe replace x by x j , dx ∂x j ∂ ∂ place x j and by x j − a j and − b j respectively. The same proof ∂x j ∂x j then yields: (1.23)

||(x j − a j ) f (x)||2 ||(ξ j − b j ) fˆ(ξ)||2 ≥ (1/4π)|| f ||22 .

13

Let us now take || f ||2 = 1 and A = (a1 , a2 , . . . , an )ǫn . Let f be small outside a small nighbourhood of A. In this case, ||(x j − a j ) f (x)||2 will be small. Consequently, the other factor on the left in (1.23) has to be large. That is, if the mass of f is concentrated near one point, the mass of fˆ cannot be concentrated near any point. Remark 1.24. If we take Z Z 2 aj = xi | f (x)| dx, b j = ξ j | fˆ(ξ)|2 dξ, then inequality (1.23) is the mathematical formulation of the positionmomentum uncertainty relation in Quantum Mechanics.

3 Some Results From the Theory of Distributions

14

In this section, let us recall briefly some results from the theory of distributions. (For a more detailed treatment, see, for example [7] or [8]). In the sequel, D′ (Ω) will denote the space of distributions on the open set Ω ⊂ Rn which is the dual space of Co∞ (Ω). When Ω = n , we will simply write D′ instead of D′ (n ). In the same way, S ′ = S ′ (Rn ) will denote the space of tempered distributions with and E ′ = E ′ (n ) will stand for the space of distributions with compact support. The value of a distribution uǫD′ at a function φǫCo∞ will be denoted by < u, φ >. If u is a locally integrable function, then u defines a distriR bution by < u, φ >= u(x)φ(x)dx. It will sometimes be convenient to

3. Some Results From the Theory of Distributions

13

R write u(x)φ(x)dx for < u, φ >, when u is an arbitrary distribution. The convergence in D′ is the weak convergence defined by the following: un , uǫD′ , un → in means < un , φ >→< u, φ > for every φ in Co∞ . Let us now recall briefly certain operations on distributions. (1.25) We can multiply a distribution uǫD′ by a C ∞ function φ to get another distribution φu which is defined by < φu, ψ >=< u, φψ > A C ∞ function ψ is said to be tempered if, for every multiindex grows at most polynomially at ∞. We can multiply an u, ǫS ′ by a tempered function to get another tempered distribution. The definition is same as in the previous case. (1.26) If uǫD′ and f ǫCo∞ , we define the convolution u ∗ f by (u ∗ f )(x) =< u, fx > where f x (x) = f (x − y). The function u ∗ f is C ∞ and when uǫE, u ∗ f is in Co∞ . The convolution ∗ : D′ × Co∞ → C ∞ can be extended to a map from D′ × E ′ to D′ . Namely, if uǫD′ , vǫE ′R and φǫCo∞ , < u ∗ v, φ >=< u, v˜ ∗ φ > where v˜ is defined by < v˜ , ψ >= v(x)ψ(−x)dx. The associative law α, ∂α φ

u ∗ (v ∗ w) − (u ∗ w) ∗ w holds for u, v, wǫD′

provided that at most one of them does not have compact support. For 15 uǫS ′ and f ǫS , u ∗ f can also defined in the same way and u ∗ f is a tempered C ∞ function. (1.27) Since R R the Fourier transform is an isomorphism of S onto S . and f gˆ = fˆg, the Fourier transform extends by duality to an isomorphism of S ′ onto S ′ . For uǫE ′ ⊂ S ′ , we have uˆ (ξ) =< u, e−2πi(·)·ξ > which is an entire analytic function.

Chapter 2 Partial Differential Operators with Constant Coefficients 1 Local Solvability and Fundamental Solution For the sake of convenience in taking the Fourier transforms, from now 16 onwards, we will use the differential monomials Dα = (2πi)|α| ∂α . Thus (Dα )ˆ(ξ) = ξ α fˆ(ξ). By a partial differential operator with constant coefficients,we mean a differential operator L of the form X L= aα Dα , aα ǫ |α|≤k

P

|aα | , 0. In this case, we say that the P operator L is of order k. If we write P(ξ) = aα ξ α , then we have

Further, we assume that

|α|≤k

|α|≤k

L = P(D). For uǫS , taking the Fourier transform, we see that X (P(D)u)ˆ(ξ) = aα ξ α uˆ (ξ) = P(ξ)û(ξ). |α|≤k

15

16

2. Partial Differential Operators with Constant Coefficients

Let us now consider the following problem: Given f in C ∞ , we want to find a distribution u such that P(D)u = f . We say that the differential operator L is locally solvable at x◦ ǫn , if there is a solution of the above problem in some neighbourhood of the point x◦ for any f in C ∞ . 17

Remark 2.1. We can assume that f has compact support. To see this, we can take any φ in Co∞ such that φ = 1 in some nighbourhood of the point xo . If solve the problem P(D)u = f φ near x◦ , then u is a solution of our original problem since we have f φ = f in a neighbourhood of xo . In the following theorem we give an affirmative answer to the question of local solvability of L = P(D). The simple proof exhibited here is due to L. Nirenberg. Theorem 2.2. Let L = stant coefficients. If on n .

P

aα D |α|≤k uǫCo∞ , there

α

be a differential operator with con-

is a C ∞ function u satisfying Lu = f

Proof. Taking the Fourier transform of the equation Lu = P(D)u = f , we see that P(ξ)û(ξ) = fˆ(ξ). It is natural to try to define u by the formula Z u(x) = e2πix·ξ fˆ(ξ)/P(ξ)ξ. In general, P will have many zeros: so there will be a problem in applying the inverse Fourier transform to fˆ/P. But things are not so bad. Since f ǫC◦∞ , fˆ is an entire function of ξǫn and P is obviously entire. Hence we can deform the contour of integration to avoid the zeros of P(ξ). To make this precise, let us choose a unit vector η so that

P

|α|≤k

18

aα ηα ,

0. By a rotation of coordinates, we can assume that η = (0, 0, . . . , 0, 1). Multiplying by a constant, we can also assume that aα0 = 1 where αo = (0, 0, . . . , 0, k). Then we have P(ξ) = ξnk +(lower order terms in ξn ). Denote ξ = (ξ ′ , ξn ) with ξ ′ = (ξ1 , . . . , ξn−1 ) in n−1 .

1. Local Solvability and Fundamental Solution

17

Consider P(ξ) = P(ξ ′ , ξn ) as a polynomial in the last variable ξn in C for ξ ′ in Rn−1 . Let λ1 (ξ ′ ), . . . , λk (ξ ′ ) be the roots of P(ξ ′ , ξn ) = 0 arranged so that if i ≤ j. ℑλi (ξ ′ ) ≤ ℑλ j (ξ ′ ) and Re λi (ξ ′ ) ≤ Re λi (ξ ′ ) when ℑλi (ξ ′ ) = ℑλ j (ξ ′ ). Since the roots of a polynomial depends continuously on the coefficients we see that λ j (ξ ′ ) are continuous in ξ ′ . To proceed further, we need the help of two Lemmas. Lemma 2.3. There is a measurable function φ : n−1 → [−k − 1, k + 1] such that for all ξ ′ in n−1 min {|φ(ξ ′ ) − Imλ j (ξ ′ )|} ≥ 1. 1≤ j≤k

Proof. Left as an exercise to the reader : (cf. G.B. Folland [1]). (The idea is that at least one of the k+1 intervals [−k−1, k+1], [−k− 1, k+1] . . . , [−k−1, k+1] must contain none of the k points Im λ j (ξ ′ ), j = 1, 2, . . . , k). Lemma 2.4. Let P(ξ) = ξnk +( lower order terms) and let N(P) be the set {ζǫn : P(ζ) = 0}. Let d(ξ, N(P)). Then, we have |P(ξ)| ≥ (d(ξ)/2)k . Proof. Take ξ in Rn such that P(ξ) , 0. Let η = (0, 0, . . . 0, 1) and define g(z) = P(ξ + zη) for z in C. This g is a polynomial in one complex variable z. Let λ1 , λ2 , . . . , λk be the zeros of the polynomial g. Then 19 g(z) = c(z − λ1 ) · · · (z − λk ) so that k

Y z g(z) |= |1 − |. | g(0) λ j j=1 Since ξ + λ j η ǫ N(P) |λ j | ≥ d(ξ) so that when |z| ≤ d(ξ), | k! |g (0)| = | 2πi (k)

Z

|ζ|=d(ξ)

q(ζ)ζ −k−1 dζ| ≤

g(z) | ≤ 2k . Also g(0)

k! |g(0)| k 2 2πd(ξ) 2 d(ξ)k+1


18

i.e. |g(k) (0)| ≤ k!|g(0)|2k d(ξ)−k . But g(0) = P(ξ) and |g(k) (0)| =

∂k P(ξ) = k! ∂ξnk

Therefore, k! ≤ k!|g(0)|2k d(ξ)−k or |P(ξ)| ≥ (d(ξ)/2)k Hence the lemma is proved.

Returning to the proof of the theorem, consider the function Z Z u(x) = e2φix·ξ ( fˆ(ξ)/P(ξ))dξn dξ. Rn−1 IMξn =φ(ξ′ )

By Lemmas 2.3 and 2.4 we have |P(ξ)| ≥ (d(ξ)/2)k ≥ 2−k along Imξn = φ(ξ ′ ). 20

Since f ǫC◦∞ , fˆ(ξ) is rapidly decreasing as |Reξ| tends to ∞ as along as |Imξ| stays bounded, so the integral converges absolutely and uniformly together with all derivatives defining a C ∞ function u. Finally, by Cauchy’s theorem, Z Z P(D)u(x) = e2φix·ξ ( fˆ(ξ)/P(ξ))dξn dξ ′ Rn−1 IMξn =φ(ξ′ )

=

Z

Rn

e2φix·ξ ( fˆ(ξ)dξ = f (x).

This completes the proof of Theorem 2.2. Let us now consider the local solvability of L = P(D) in the case when f is a distribution. As before, we remark that it suffices to take f ǫE ′ . Further, it is enough to consider the case where f = δ. Indeed, if K satisfies P(D)K = δ, then we have for any f ǫE ′ , P(D)(K ∗ f ) = P(D)K ∗ f = δ ∗ f = f .

1. Local Solvability and Fundamental Solution

19

Definition 2.5. A distribution K satisfying P(D)K = δ is called a fundamental solution or elementary solution of the differential operator L = P(D). A remarkable theorem due to MALGRANGE AND EHRENPREIS states that every differential operator P(D) with constant coefficients has a fundamental solution. In fact, we can prove this result by a simple extension of the preceding argument. Theorem 2.6. Every partial differential operator P(D) with constant co- 21 efficients has a fundamental solution. Proof. Proceeding as in the previous theorem, we try to define Z Z K(x) = e2Πix·ξ (p(ξ))−1 dξn dξ ′ . n−1 Imξn =φ(ξ′ )

Here, however, the integral may diverge at infinity. So, we consider the polynomial n X PN (ξ) = P(ξ)(1 + 4Π2 ξ 2j )N j=1

where N is a large positive integer. Let Z Z KN (x) = e2Πix·ξ (PN (ξ))−1 dξn dξ ′ n−1 Imξn =φ(ξ′ )

where φ is chosen appropriately for the polynomial PN .Then on the region of integration, we have |PN (ξ)| ≥ c(1 + |ξ|2 )N ; so the integral will converge when N > n/2. We claim that PN (D)KN = δ. To see this, take φǫCo∞ and observe that the transpose of PN (D) is PN (−D). Thus < PN (D)KN , φ > =< KN , PN (−D)φ >


20

Z

Z Z

(PN (−D)φ)(x) dξn dξ′ dx PN (ξ) n n−1 Imξn =φ(ξ′ ) Z Z Z (PN (ξ))−1 dξn dξ′ e2Πix·ξ PN (−D)φ(x)dx =

=

e2Πix·ξ

n−1 Imξn =φ(ξ′ )

=

Z

Z

n

′ ˆ φ(−ξ)dξ n dξ

n−1 Imξn =φ(ξ′ )

=

Z

ˆ φ(−ξ)dξ = φ(0) =< δ, φ > .

n

22

Thus, δ = PN (D)KN = P(D)(1 − ∆)N KN

so if we put K = (1 − ·∆)N KN we have P(D)K = δ. Thus K is a fundamental solution of P(D).

2 Regularity Properties of Differential Operators Definition 2.7. The singular support of a distribution f ǫD′ is defined to be the complement of the largest open set on which f is a C ∞ function. The singular support of f will be denoted by sing supp f . P Definition 2.8. Let L = aα (x)Dα where aα ǫC ∞ be a differential |α|≤K

operator. L is said to be hypoelliptic, if and only if, for any uǫD′ , sing supp u ⊂ sing supp Lu. In other words, L is hypoelliptic if and only if for any open set Ω ⊂ n and any uǫD′ (Ω), “LuǫC ∞ (Ω) implies uǫC ∞ (Ω)”. Remark 2.9. The operator L is said to be elliptic at xǫn if X aα (x)ξ α , 0 for every ξ in ǫn \{0}. |α|=k

23

L is said to be elliptic if L is elliptic at every xǫn . Elliptic operators are hypoelliptic, as we shall prove later. This accounts for the name hypoelliptic.

2. Regularity Properties of Differential Operators

21

We know that an ordinary differential operator with C ∞ coefficients is hypoelliptic as long as the top order coefficient is non-zero. But this is not the case with partial differential operators as seen from the following ∂2 in 2 . The general solution ∂x∂y of the equation Lu = 0 is given by u(x, y) = f (x) + g(y) where f and g are arbitrary C 1 functions. This shows that L is not hypoelliptic.

Example 2.10. Take the operator L =

We observe that if L is hypoelliptic, then every fundamental solution of L is a C ∞ function in n \{0}. In the case of partial differential operators with constant coefficients, this is also sufficient for hypoellipticity. Indeed, we have the following Theorem 2.11. Let L be a partial differential operator with constant coefficients. Then the following are equivalent: a) L is hypoelliptic. b) Every fundamental solution of L is C ∞ in2 \{0}. c) At least one fundamental solution of L is C ∞ in 2 \{0}. Proof. That (a) simple (b) follows from the above observation and that (b) implies (c) is completely trivial. The only nontrivial part we need to prove is that (c) implies (a). To prove this implication, we need Lemma 2.12. Suppose f ǫD′ is such that f is C ∞ in n \{0} and gǫE ′ . 24 Then sing supp( f ∗ g) ⊂ supp g. Proof. Suppose x < supp g. We will show that f ∗ g is C ∞ in a neighbourhood of x. Since x < supp g, there exists an ǫ > 0 such that B(x, ǫ)∩supp g = φ. Choose φǫC0∞ (B(0, ǫ/2)) such that φ = 1 on B(0, ǫ/4). Now, f ∗ g = (φ f ) ∗ g + (1 − φ) f ∗ g. Since (1 − φ) f is a C ∞ function, (1 − φ) f ∗ g is C ∞ . Also Supp(φ f ∗ g) ⊂ Supp φ f + Supp g

22

2. Partial Differential Operators with Constant Coefficients ⊂ {y : d(y, supp g) ≤ ǫ/2}

which does not intersect B(x, ǫ/2). Therefore, in B(x, ǫ/2), f ∗ g = (1 − φ) f ∗ G which is C ∞ . Proof of theorem 2.11. Let K be a fundamental solution of L such that K is C ∞ in n \{0}. Suppose uǫD′ and LuǫC ∞ (Ω) where Ω ⊂ n is open. ¯ ǫ) ⊂ Ω. Choose For xǫΩ, pick ǫ > 0 small enough so that B(x, ∞ φǫCo (B(x, ǫ)) so that φ = 1 on B(x, ǫ/2). Then L(φu) = φLu + v where v = 0 on B(x, ǫ/2) and also outside B(x, ǫ). We write K ∗ L(φu) = K ∗ φLu + K ∗ v. φLu is a Co∞ function so that K ∗ Lu is a C ∞ function. Also K ∗ v is a function on the ball B(x, ǫ/2) by Lemma 2.12. Therefore K ∗ L(φu) is a C ∞ function on B(x, ǫ/2). But K ∗ L(φu) = LK ∗ φu = δ ∗ φu = φu. Thus, φu is a C ∞ function on B(x, ǫ/2). Since φ = 1 on B(x, ǫ/2), u is a C ∞ function on B(x, ǫ/2). Since x is arbitrary, this completes the proof. C∞

25

3 Basic Operators in Mathematical Physics In this section, we introduce the three basic operators in Mathematical Physics. In the following sections, we shall compute fundamental solutions for these operators and show how they can be applied to solve boundary value problems and to yield other information. n ∂2 P in n . If u(x) represents 2 ∂x j=1 j electromagnetic potential (or gravitational potential ) and ρ denotes the charge (resp. mass) density, then they are connected by the equation ∆u = −4πρ. If the region contains no charge, i.e., if ρ = 0, ∆u = 0. This equation is called the homogeneous Laplace equation, and its solutions are called harmonic functions.

(i) LAPLACE OPERATOR: ∆ =

3. Basic Operators in Mathematical Physics

23

∂ − ∆ in n+1 . If u(x, t) represents the (ii) HEAT OPERATOR: L = ∂t temperature of a homogeneous body at the position x and time t, ∂u then u satisfies the heat equation − ∆u = 0 in n+1 . ∂t ∂2 − ∆ in n+1 . If u(x, t) represents the ∂t2 amplitude of an electromagnetic wave in vacuum at position x and time t, then u satisfies the wave equation @u = 0 in n+1 . 26

(iii) WAVE OPERATOR: @ =

The equation u = 0 can also be used to describe other types of wave phenomena although in most cases, it is only an approximation valid for small amplitudes. More generally, the equation u = f describes waves subject to a driving force f . The Laplace operator ∆ is an ingredient in all the above examples. The reason for this is that the basic laws of Physics are invariant under translation and rotation of coordinates which severely restricts the differential operators which can occur in them. Indeed, we have Theorem 2.13. Let L be a differential operator which is invariant under rotations and translations. Then there exists a polynomial Q in one variable with constant coefficients such that L = Q(∆). P Proof. Let L = P(D) = aα Dα . Since P(D) is invariant under trans|α|≤K

lations, aα are constants. Let T be any rotation. We have ˆ ξ) = (P(D)φ)ˆ(T ξ) P(T ξ)φ(T = (P(D)φ ◦ T )ˆ(ξ)

= (P(D)φ ◦ T ))ˆ(ξ) ˆ ξ), for every φ. = P(ξ)φ(T Thus P(T ξ) = P(ξ) so that P is rotation - invariant. Write P(ξ) =

k P

P j (ξ) where P j (ξ) is the part of P which is ho- 27

j=0

mogeneous of degree j. We claim that each P j is rotation - invariant. Indeed, if t is any real number,


24 P(tξ) =

k P

j=0

t j P j (T ξ). For a rotation T , since P(ξ) = P(T ξ), we have

P(tξ) =

k X

t j P j (T ξ)

j=0

k X

so that

j=0

t j (P j (T ξ) − P j (ξ)) = 0.

This being true for all t, it follows that P j ◦ T = P j . But the only rotation-invariant functions which are homogeneous of degree j are of the form P j (ξ) = c j |ξ| j where c j are constants. Indeed, since P j is rotation-invariant, P j (ξ) depends only on |ξ| and thus, for |ξ| , 0, ! ! ξ ξ j = |ξ| P j = c j |ξ| j . P j (ξ) = P j |ξ| |ξ| |ξ| Since |ξ| j is not a polynomial when j is odd, c j = 0 in that case. Therefore, X P(ξ) = c2 j|ξ|2 j . P Taking Q(x) = (c2 j /(−4π2 ) j )x j , we get P(D) = Q(∆).

28

Remark 2.14. This theorem applies to scalar differential operators. If one considers operators on vector or tensor functions, there are first order operators which are translation - and rotation - invariant, namely, the familiar operators grad, curl, div of 3-dimensional vector analysis and their n-dimensional generalisations. Definition 2.15. A function F(x) is said to be radial , if there is a function f of one variable such that F(x) = f (|x|) = f (r), r = |x|. When F is radial and FǫL1 (n ), we have Z

n

F(x)dx =

Z∞ Z

0 |x|=1

f (r)rn−1 dσ(x)dr

4. Laplace Operator

25

where dσ(x) is the surface measure on S n−1 , the unit sphere in n . Thus Z Z∞ F(x)dx = ωn f (r)rn−1 dr n

0

where ωn is the area of S n−1 . Let us now calculate ωn . We have Z 2 e−π|x| dx = 1. Now Z

−π|x|2

e

dx = ωn

Z∞

2

e−πr rn−1 dr

0

ωn = 2π

Z∞

e−s (s/π)n/2−1 ds, s = πr2

0

ωn = n/2 2π

Z∞

e−s sn/2−1 ds

0

= ωn Γ(n/2)/(2πn/2 ). Therefore ωn = 2πn/2 /Γ(n/2).

4 Laplace Operator First, let us find a fundamental solution of ∆, i.e., we want to find a 29 distribution K such that ∆K = δ. Since ∆ commute with rotations and δ has the same property, we observe that if u is a fundamental solution of ∆ and T is a rotation, then u ◦ T is also a fundamental solution. We therefore expect to find a fundamental solution K which is radial.

26


Let us tarry a little to compute the Laplacian of a radial function F. LetF(x) = f (r), r = |x|. Then, n X ∂ ′ ∆F(x) = [ f (r)x j /r] ∂x j j=1   2 n  X  ′′ x j f ′ (r) f ′ (r) 2   f (r) + − 3 x j  = 2 r r r j=1

= f ′′ (r) + ((n − 1)/r) f ′ (r).

Now set K(x) = f (r). If K is to be a fundamental solution of ∆, we must have f ′′ (r) + ((n − 1)/r) f ′ (r) = 0 on (0, ∞). From this equation f ′′ (r)/ f ′′ (r) = −(n − 1)/r. Integrating, we get f ′ (r) = c◦ r1−n with a constant c◦ . One more integration yields f (r) = c1 r2−n + c2 when n , 2 = c1 log r + c2 when n = 2. 30

Since constants are solutions of the homogeneous Laplace equation, we may assume that c2 = 0. Thus if we set F(x) = |x|2−n (n , 2) or F(x) = log |x|(n = 2), we expect to find that ∆F = cδ for some c , 0, and then K = c−1 F will be our fundamental solution. In fact, we have Theorem 2.16. If F(x) = |x|2−n on n (n , 2), then ∆F = (2 − n)ωn δ, where ωn is the area of the unit sphere in n ,

4. Laplace Operator

27

Proof. For ǫ > 0, we define Fǫ (x) = (ǫ 2 + |x|2 )(2−n)/2 . Then Fǫ is a C ∞ function on n , and an application of Lebesgue’s Dominated Convergence Theorem reveals that Fǫ converges to F as ǫ tends to 0, in the sense of distributions. Therefore ∆Fǫ converges to ∆F, in the same sense. Let us now compute ∆Fǫ . n X o ∂ n (2 − n)(|x|2 + ǫ 2 )−n/2 x j ∆Fǫ (x) = ∂x j j=1

=

n n X o (2 − n)(|x|2 + ǫ 2 )−n/2 + (2 − n)(−n)(|x|2 + ǫ 2 )(−n/2)−1 x2j j=1

= (2 − n)(|x|2 + ǫ 2 )(−n/2)−1 nǫ 2 . Thus we see that ∆Fǫ ǫL1 (n ). A simple computation Rshows that ∆Fǫ (x) = ǫ −n ∆F1 (x/ǫ); so, by Theorem 1.6, ∆Fǫ tends to ( ∆F1 )δ as R Rǫ tends to 0. Therefore ∆F = ( ∆F1 )δ, and we need only to compute ∆F1 . Z Z ∆F1 = n(2 − n) (1 + |x|2 )(−n/2)−1 dx = n(2 − n)ωn

Z∞

(1 + r2 )(−n/2)−1 rn−1 dr.

0

Putting r2 + 1 = s, we see that Z

31

1 ∆F1 = n(2 − n)ωn 2

Z∞

1 n(2 − n)ωn 2

Z∞

1 s−2 (1 − )(n/2)−1 ds s

1 n(2 − n)ωn 2

Z∞

(1 − σ)(n/2)−1 dσ, σ =

s(−n/2)−1 (s − 1)(n−1)/2 ds

1

=

1

=

0

1 = n(2 − n)ωn (2/n) = (2 − n)ωn 2

1 s


28

which completes the proof.

Exercise. Show that if F(x) = log |x| on 2 , then ∆F = 2πδ.   |x|2−n    (n , 2)  Corollary 2.17. Let K(x) =  (2 − n)ωn    1 log |x|(n = 2)  2π

Then K is a fundamental solution of the Laplacian.

Corollary 2.18. ∆ is hypoelliptic. Proof. Follows from Theorem 2.11.

It is also instructive to compute the fundamental solution of ∆ by the Fourier transform method. If K is a fundamental solution of ∆, we have ∆(k ∗ f ) = ∆K ∗ f = f . Since (∆g)ˆ(ξ) = −4π2 |ξ|2 gˆ (ξ), we get ˆ fˆ(ξ) = fˆ(ξ), −4π2 |ξ|2 K(ξ) 32

so that, at least formally, ˆ K(ξ) = −1/(4π2 |ξ|2 ). We observe that when n > 2, the function −1/(4π2 |ξ|2 ) is locally integrable and so defines a tempered distribution. We want to show that its inverse Fourier transform is our fundamental solution K. For that purpose, we will prove a more general theorem. Theorem 2.19. For 0 < α < n, let Fα be the locally integrable function Fα (ξ) = |ξ|−α . Then ! !! 1 1 ∨ Fα (x) = Γ (n − α) /Γ α πα−n/2 |x|α−n . 2 2 Before proving this theorem, let us pause a minute to observe that it implies F2∨ (x) = Γ((n/2) − 1)π2−n/2 |x|2−n = (Γ(n/2)/(n/2 − 1))π2−n/2 |x|2−n = (−4π2 /((2 − n)ωn ))|x|2−n , so that (−F2 /4π2 ))∨ = K

as desired.

4. Laplace Operator

29

Proof of theorem 2.19. The idea of the proof is to express Fα as a weighted average of Gaussian functions, whose Fourier transforms we can compute. To begin with, Z∞

−rt α−1

e

t

dt =

Z∞

0

e−s sα−1 r−α ds = r−α Γ(α), (r > 0, α > 0).

0

In other words, for any r > 0 and α > 0, r

−α

1 = Γ(α)

Z∞

33

e−rt tα−1 dt.

0

Taking r = π | ξ |2 and replacing α by α/2, then πα/2 | ξ |−α = Γ(α/2)

Z∞

2t

e−π|ξ| t(α/2)−1 dt

0

which is the promised formula for Fα as a weighted average of Gaussians. Formally, we can write Z

Rn

2πix.ξ

e

−α

|ξ|

πα/2 dξ = Γ(α/2) =

=

= =

Z Z∞

2t

e2πix.ξe −π|ξ| tα/2−1 dtdξ

πα/2 Γ(α/2)

Rn 0 Z∞

πα/2 Γ(α/2)

Z∞

e−π(|x|

πα/2 Γ(α/2)

Z∞

e−π|x|

πα/2 Γ(α/2)

2t

(eπ|ξ| )v (x)t(α/2)−1 dt

0

2 )/t

t−n/2 tα/2−1 dt

0

2S

S (n−α)/2−1

ds

0

Γ((n − α)/2) | x |α−n . n(n−α)/2


30

In this computation, the change of order of the integration is unfortunately not justified, because the double integral is not absolutely convergent. This is not surprising, since Fα is not an L1 function; instead, we must use the definition of the Fourier transform for distributions. For every φǫS , 2t 2t < e−π|ξ| , φˆ > =< e−π|ξ| )ˆ,φ >, Z Z 2 −π|ξ|2t ˆ φ(ξ)dξ = e t−n/2 e−(π/t)|x| φ(x)dx.

i.e.,

Rn

Rn

34

Now multiply both sides by t(α/2)−1 dt and integrate from 0 to ∞. Z∞ Z

−π|ξ|2t (α/2)−1 ˆ

e

φ(ξ)dξ dt =

t

0 Rn

Z∞ Z

2

e−π/t|x| t(α−n)/2−1 φ(x)dx dt.

0 Rn

The change of order in the integral is permitted now and integrating, we obtain, Z Z Γ(α/2) Γ((n − α)/2) −α ˆ | ξ | dξ = | x |α−n φ(x) dx φ(ξ) πα/2 π(n−α)/2 Rn

Rn

Γ((n − α)/2) α−n/2 =< π | x |α−n , φ > . (Γ(α/2)

i.e., This shows that

Fα∨ (x) =

Γ((n − α)/2) α−n/2 π | x |α−n Γ(α/2)

as desired. This analysis does not suffice to explain −(4π2 | ξ |2 )−1 as (in some sense) the Fourier transform of (2π)−1 log | x | in the case n = 2. One way to proceed is to define Gα (ξ) = (2π | ξ |)−α and to study the behaviour of Gα and G∨α as α tends to 2. This analysis can be carried out just as easily in n dimensions where the problem is to study Gα and Gvα as α tends to n.

4. Laplace Operator

31

Let Rα = G∨α for 0 < α < n. (Rα called the Riesz potential of order α). By the preceding theorem, we have Rα (x) =

(Γ((n − α)/2) | x |α−n . 2α πn/2 Γ(α/2) 35

Moreover, if f ǫS , (−∆)α/2 ( f ∗ Rα ) = f, in the sense that (2π | ξ |)α ( f ∗ Rα )ˆ (ξ) = fˆ(ξ). As α tends to n, the Γ-function in Rα blows up. However, since (2π | ξ |)αδ = 0, we see that (−∆)α/2 1 = 0; so we can replace Rα by Rα − c, c being a constant, still having (−∆)α/2 ( f ∗ (Rα − c)) = f. If we choose c = cα appropriately, we can arrange that Rα − cα will have a limit as α tends to n. In fact, let us take cα =

Γ((n − α)/2) and define R′α = Rα − cα . 2α πn/2 Γ(α/2)

Then Γ((n − α)/2) (| x |α−n −1) 2α nn/2 Γ(a/2) 2Γ((n − α)/2) + 1 (| x |)α−n − 1 = . n−α 2α πn/2 Γ(α/2)

R′α (x) =

Letting α tend to n, we get in the limiting case R′n (x) =

−21−n log | x | . πn/2 Γ(n/2)

1 when n = 2, −R′2 (x) log | x | and ∆( f ∗ (−R′2 )) = f , so we recover our 2π fundamental solution for the case n = 2. It remains to relate the function Gn (ξ) = (2π | ξ |)−n to the Fourier 36

32


transform of the tempered distribution Fn′ . One way to make Gn into a distribution is the following. Define a functional F on S by Z Z φ(ξ) − φ(0) φ(ξ) < F, φ >= dξ + dξ. n (2π | ξ |) (2π | ξ |)n |ξ|≤1

|ξ|>1

Note that the first integral converges, since, in view of the mean value theorem, | φ(ξ) − φ(0) |≤ c | ξ |. It is easy to see that F is indeed a tempered distribution. Further, we observe that when φ(0) = 0, R < F, φ >= φ(ξ)Gn (ξ)dξ, i.e., F agrees with Gn on Rn \{0}. Just as R′n was obtained from Rn by subtracting off an infinite constant, F is obtained from Gn by subtracting off an infinite multiple of the δ function. This suggests that F is essentially the Fourier transform of R′n . In fact, one has the following Exercise. Show that F − (R′n )îs a multiple of δ. We now examine a few of the basic properties of harmonic functions, i.e., functions satisfying ∆u = 0. We shall need the following results from advanced calculus. Theorem 2.20 (DIVERGENCE THEOREM). Let Ω be a bounded open set in Rn with smooth boundary ∂Ω. Let v be the unit outward normal ¯ the closure of vector on ∂Ω. Let F : Rn → Rn be a C 1 function on Ω, Ω. Then we have Z Z X Z n ∂F j dx. F. v dσ = div F dx = ∂x j j=1 Ω

∂Ω

Ω

37

CONSEQUENCES OF THEOREM 2.20 ∂u (2.21) When we take F = grad u, F.v = grad u.v = , the normal ∂v derivative of u, and div F = div grad u = ∆u. Therefore Z Z ∂u ∆u dx. dσ = ∂v Ω ∂Ω

4. Laplace Operator

33

(2.22) When we take F = u grad v − v grad u, div F = u∆v − v∆u. Therefore ! Z Z ∂u ∂v dσ = (u∆v − v∆u)dx. u −v ∂v ∂v ∂Ω

Ω

This formula is known as Green’s formula. For harmonic functions, we have the following mean value theorem. Theorem 2.23. Let u be harmonic in B(x0 , r). Then,for any ρ < r, we have Z 1 u(x)dσ(x). u(x0 ) = ωn ρn−1 |x−x0 |=ρ

That is, u(x0 ) is the mean value of u on any sphere centred at x0 Proof. Without loss of generality, assume that x0 = 0. Since ∆u = 0 and ∆ is hypoelliptic, u is a C ∞ function. Now formally, u(0) =< δ, u > Z δ(x)u(x)dx = |x| 0, so that ∂t K˜ n+1 = Pˆ t ). 2 2 Our formula (2.31) for Pt makes sense even when t < 0 and we have P−t (x) = −Pt (x), so that lim f ∗ Pt = ± f . We further observe that t→o±

f ∗(x) Pt = f (x)δ(t) ∗(x,t)2 ∂t Kn+1 (x, t) 46

where ∗( x) and ∗(x,t) mean convolution on Rn and Rn+1 respectively. Therefore, f ∗(x) Pt = f (x)δ′ (t) ∗(x,t) 2Kn+1 (x, t)

u(x, t) = 2 f (x)δ′ (t) ∗( x, t)Kn+1 (x, t).

i.e., Form this, we

(δ2t + ∆)u(x, t) = 2 f (x)δ′ (t) ∗ δ(x)δ(t) = 2 f (x)δ′ (t). Exercise. Show directly that if i) (∆ + ∂2t )u = 0 on Rn+1 \Rn × {0} ii) u(x, −t) = −u(x, t) iii) lim u(x, t) = ± f (x), then u is a distribution solution of (∂2t + ∆)u = t→0±

2 f (x)δ′ (t).

[To avoid technicalities, assume f is sufficiently smooth so that ∂u exists; e.g. f ǫC 2 is sufficient ]. lim t→0± ∂t We now indicate, without giving any proof, how these ideas can be used to solve the Dirichlet Problem in a bounded open set Ω in R2 . We assume that the boundary ∂Ω of Ω is of class C 2 .

5. The Heat Operator

41

We know that the solution of the Dirichlet Problem in the case when ∂K (the convolution is in Rn−1 ). Ω = Rn−1 × (0, ∞) is given by u = f ∗ 2 ∂xn ∂K is the inward normal derivative of K. For bounded Ω we note that ∂xn with C 2 boundary let us consider Z ∂ f (y) K(x − y)dσ(y). u(x) = 2 ∂vy ∂Ω

Here σ(y) is the surface measure on the boundary, ν is the unit out- 47 ∂φ ∂φ ward normal on ∂Ω, and (x, y) = Σν j . The minus sign in K(x−y) ∂νy ∂y j compensates the switch from inward to outward normal. Since ∆K(x − y) = δ(x − y) we see that ∆u = 0 in Ω. What about the behaviour of u on ∂Ω? It turns out that if u is defined as above, then ¯ and u|∂Ω = f + T f uǫC (Ω) where T is a compact integral operator on L2 (∂Ω) or C(∂Ω). Hence if we can find φ in C(∂Ω) such that φ + T φ = f , and we set Z ∂ K(x − y)dσ(y) u(x) = 2 φ(y) ∂νy ∂Ω

then u satisfies ∆u = 0 in Ω and further u|∂Ω = φ + T φ = f . Hence the Dirichlet problem is reduced to solving the equation φ + T φ = f , and for this purpose, the classical Fredholm - Riesz theory is available. The upshot is that a solution to the Dirichlet problem always exists (provided, as we have assumed, that ∂Ω is of class C 2 ). See Folland [1] for a detailed treatment.

5 The Heat Operator The Heat operator is given by ∂t − ∆. We want to find a distribution 48 K such that (∂t − ∆)K = δ(x)δ(t). Taking the Fourier transform in both variables we have (2.33)

ˆ τ) = (2πi + 4π2 |ξ|2 )−1 . K(ξ,

42


Exercise . Show that Kˆ is locally integrable near the origin, and hence defines a tempered distribution. Kˆ is not globally integrable, however; so computing its inverse Fourier transform directly is a bit tricky. Instead, if we take the Fourier transform in the variable x only, ˜ t) = δ(t). (∂t + 4π2 |ξ|2 ) K(ξ, we can solve this ODE get  2 2   a(ξ)e−4π |ξ| t , ˜ K(ξ, t) =   b(ξ)e−4π2 |ξ|2 t ,

49

t>0 t 0 ˜ K(ξ, t) =   0 otherwise . Taking the inverse Fourier transform,  2   (4πt)−n/2 e(|x| /4t) , K(x, t) =   0,

t>0 t > 0.

This is a fundamental solution of the heat operator.

˜ t) in the t variRemark 2.34. If we take the Fourier transform of K(ξ, able, we obtain Z ∞ ˜ t)e2π i t τ dt ˜ K(ξ, K(ξ, τ) = −∞ Z ∞ 2 2 = e−t(4π |ξ| +π i τ) dt 0

= (4π2 |ξ|2 + 2πiτ)−1 , thus recovering formula (2.33).

5. The Heat Operator

43

Exercise . Show that this really works, i.e., the iterated Fourier transform of K first in x, then in t, is the distribution Fourier transform of K in all variables. OBSERVATIONS: From the formula for K, we get (1) K(x, t) vanishes to infinite order as t tends to 0 when x , 0 and hence is C ∞ on Rn+1 \{(0, 0)}. Therefore by Theorem 2.11, ∂t − ∆ is hypoelliptic. x (2) K(x, t) = t−n/2 K , 1 . Thus if we set K(x, 1) = φ(x), then K(x, ǫ 2 ) t = ǫ −n φ(x/ǫ) = φǫ (x); so, as t tends to 0, K(x, t) tends to δ, Theorem 1.6. From this, we infer 50 (3) If f ǫL p and if we set u(x, t) = f ∗(x) K(x, t), then (∂ − ∆)u = 0 for t > 0 u(x, 0) = f (x).

i.e., as t tends to 0, ||u(., t) − f ||P converges to 0. Thus we have solved the initial value problem for the homogeneous heat equation, when f ǫL p . Actually, since K(x, t) decreases so rapidly as |x| tends to ∞, this works for much wider classes of f ′ s. The convolution 2−ǫ f ∗(x) K(., t) make sense, for example, if | f (x)| < Ce|x| . If f is also continuous, it is not hard to see that f ∗(x) K(., t) converges to f uniformly on compact sets as t tends to 0. It is now a simple matter to solve the inhomogeneous initial value problem: (∂t − ∆)u = F on Rn × (0, ∞), u(x, 0) = f (x) on Rn .

If we take u1 = F ∗(x,t) K and u2 = ( f − u1 (., 0)) ∗(x) K,

44


then we see that (∂t − ∆)u1 = F on Rn × (0, ∞), (∂t − ∆)u2 = 0 on Rn × (0, ∞), and (u1 + u2 )(x, 0) = f (x). Thus u = u1 + u2 solves the 51 problem. As another application of the fundamental solution K, we can derive the Weierstrass approximation theorem. Theorem 2.35. (WEIERSTRASS) If f is continuous with compact support on Rn , then, for any compact Ω ⊂ Rn , there exists a sequence (Pm ) of polynomials such that Pm converges to f uniformly on Ω Proof. Let u(x, t) = ( f ∗(x) K(., t))(x). Then u(x, t) converges to f (x) uniformly as t tends to 0. Moreover, for any t > 0, u(x, t) =

Z

−

f (y)(4πt)−n/2 e

n P

(x j −y j )2 /4t

j=1

dy

is an entire holomorphic function of x ∈ n . So u(., t) can be uniformly approximated on any compact set by partial sums of its Taylor series.

6 The Wave Operator If we take the Fourier transform of the equation (∂2t − ∆)K = δ(x)δ(t) in both variables x and t, we have, formally (2.36) 52

ˆ τ) = (4π2 |ξ|2 − 4π2 τ2 )−1 . K(ξ,

This function Kˆ is not locally integrable, so it is not clear how to interpret it as a distribution. Again, it is better to take the Fourier transform in the x variable obtaining ˜ t) = δ(t). (∂2t + 4π2 |ξ|2 )K(ξ,

6. The Wave Operator

45

Solving this ODE, ˜ t) = a± (ξ)e2πi|ξ|t + b± (ξ)e−2πi|ξ|t for t/|t| = ±1 K(ξ, and the coefficients a± , b± must be determined so that ˜ 0+) = K(ξ, ˜ 0−), ∂t K(ξ, ˜ 0+) − ∂t K(ξ, 0 =) = 1. K(ξ, This gives two equations in four unknowns. In contrast to the situation with the Dirichlet problem and the heat operator, there is no way to narrow down the choices further by imposing growth restrictions on K as |t| tends to ∞. Rather, it is a characteristic feature of the wave operator that one can adapt the choice of fundamental solution to the problem at hand. The two which we shall use, called K+ and K, are the ones supported in the half - space t ≥ 0 and t ≤ 0. K+ and K− are thus determined by the requirements a− = b− = 0 and a+ = b+ = 0 respectively, from which one easily sees that sin 2π|ξ|t K˜ + (ξ, t) = H(t) 2π|ξ| sin 2π|ξ|t K˜ − (ξ, t) = −H(−t) = K˜ + (ξ, −t) 2π|ξ| where H is the Heaviside function, i.e., the characteristic function of 53 [0, ∞). Let us compute the Fourier transforms of K˜ + and K˜ − in t to see how to make sense out of (2.36). K˜ + and K˜ − are not integrable on t, but it is easy to approximate them in the distribution topology by integrable functions whose Fourier transforms we can calculate.Indeed, if we set sin 2π|ξ|t , ǫ > 0, K˜ +ǫ (ξ, t) = e−2πǫt H(t) 2π|ξ| then K˜ +ǫ is an integrable function and K˜ +ǫ converges to K˜ + is S ′ as ǫ tends to 0. Therefore K˜ + = lim K˜ +ǫ where ǫ→0

Kˆ + (ξ, τ) =

Z∞ 0

e2πǫt−2π it τ

sin 2π|ξ|t dt 2π|ξ|

46


=

Z∞

e2πǫt−2π it τ

(eπi|ξ|t − eπi|ξ|t ) dt 4πi|ξ|

0

= (4π2 )−1 (|ξ|2 − (τ − iǫ)2 )−1 . Exercise. Prove that Kˆ − = lim Kˆ −ǫ in S ′ where ǫ→0

Kˆ −ǫ (ξ, τ)

= (4π2 )−1 (|ξ|2 − (τ + iǫ)2 )−1 .

Thus we have two distinct ways of making the function [4π2 (|ξ|2 − into a tempered distribution. The difference K˜ + − K˜ − is of course a distribution supported on the cone |ξ| = |τ|. We now propose to use the fundamental solutions K+ and K− to solve the Initial Value Problem or Cauchy Problem, for the operator: τ2 )]−1

54

(2.37)

(∂2t − ∆)u = f on Rn+1 u(x, o) = uo (x),

∂t u(x, 0) = u1 (x), where u0 , u1 , f are given functions. For the moment, we assume that u0 , u1 ǫS and f ǫC ∞ (Rnx )) ( That is, t → f (., t) is a C ∞ function with values in S (Rn )). Taking the Fourier transform in the variable x in (2.37), (∂2t + 4π2 |ξ|2 ) u˜ (ξ, t) = f˜(ξ, t) u˜ (ξ, 0) = u˜ 0 (ξ)

∂t u˜ (ξ, 0) = u˜ 0 (ξ). When f = 0, the general solution of the ODE is u˜ (ξ, t) = A(ξ) sin 2π|ξ|t + B(ξ) cos 2π|ξ|t, uˆ 1 (ξ) and B(ξ) = u˜ 0 (ξ). A(ξ) = 2π|ξ| when u0 = u1 = 0, the solution is u˜ (ξ, t) =

Zt 0

sin(2π|ξ|(t − s)) ds f˜(ξ, s) 2π|ξ|

6. The Wave Operator 55

47

(This may be derived by variation of parameters; in any case it is easy to check that this u is in fact the solution). Therefore, the solution for the general case is given by u˜ (ξ, t) = uˆ 0 (ξ) cos 2π|ξ|t + +

Zt

uˆ 1 (ξ) sin 2π|ξ|t+ 2π|ξ|

sin(2π|ξ|(t − s)) f˜(ξ, s) ds. 2π|ξ|

0

For t > 0, we can rewrite this as u˜ (ξ, t) = uˆ 1 (ξ)K˜ + (ξ, t) + uˆ 0 (ξ)∂t K˜ + (ξ, t) +

Zt

f˜(ξ, s) f˜(ξ, s)K+ (ξ, t − s)ds.

Zt

f˜(ξ, s) f˜(ξ, s)K+ (ξ, t − s)ds.

0

Since K+ (ξ, t − s) = 0 for t < s, u˜ (ξ, t) = uˆ 1 (ξ)K˜ + (ξ, t) + uˆ 0 (ξ)∂t K˜ + (ξ, t) +

0

Take the inverse Fourier transform: u(x, t) = (u1 ∗(x) K+ ) + u0 ∗(x) ∂t K+ ) + (H(t) f ∗(x,t) K+ ). Likewise, for t < 0 u(x, t) = −(u1 ∗(x) K− ) − u0 ∗(x) ∂t K− ) + ((H(−t) f ) ∗(x,t) K− ). So, for arbitrary t, our solution u can be expressed as

(2.38)

u(x, t) = (u1 ∗(x) (K+ − K) + u0 ∗(x) ∂t (K+ − K− )

+ (H(t) f ∗(x,t) K+ ) + (H(−t) f ∗(x,t) K) ).

So far, we have avoided the question of computing K+ and K− ex- 56 plicitly. Indeed, since K˜ + and K˜ − are not L1 functions, it is not an easy matter to calculate their inverse Fourier transform,

48


However, for the case n = 1, we can find K+ and K− by solving the wave wave equation directly. We have ∂2t − ∆ = ∂2t − ∂2x . If we make the change of variables ξ = x + t, n = x − t, then the wave ∂2 ∂2 u operator becomes ∂2t − ∂2x = −4 . The general solution of =0 ∂ξ∂η ∂ξ∂η is given by u(ξ, n) = f (ξ) + g(η) where f and g are arbitrary functions. Therefore u(x, t) = f (x + t) + g(x − t). To solve (∂2t − ∂2x )u = 0, u(x, 0) = u0 (x), ∂t u(x, 0) = u1 (x) we must have u0 (x) = f (x) + g(x) u1 (x) = f ′ (x) − g′ (x). From these equations, we have 1 f ′ (x) = (u′0 (x) + u1 (x)) 2 1 ′ g (x) = (u′0 (x) − u1 (x)). 2 57

Thus u(x, t) is given by 1 1 u(x, t) = (u0 (x + t) + u0 (x − t)) + 2 2

Zx+t u1 (s)ds.

x−t

Comparing with the previous formula (2.38), we find that K+ (x, t) =

1 1 H(t − |x|), K− (x, t) = H(−t − |x|). 2 2

Exercise. Compute K˜ ± directly from these formulas.


49

It turns out that, for n = 2, K± (x, t) =

√

1

2π t2 − x2

H(±t − |x|),

and, for n = 3,

±1 δ(±t − |x|). 4πt For n = 1, 2, K+ , K− are functions; for n = 3, K+ , K− are not functions but they are measures. For n > 3, K+ , K− are neither functions nor measures; they are more singular distributions. The exact formula for K± is rather messy and we shall not write it out; it may be read off from Theorem 5.13 and 5.14 of Folland [1] in view of our formula (2.38). The most important qualitative feature of K± , however, is that it is always supported in the light cone {(x, t) : ±t > |x|} and we shall now prove this as a consequence of the following result. K± (x, t) =

Theorem 2.39. Suppose u is a C 2 function on {(x, t) : t ≥ o} such that 58 (∂2t −△)u = 0 for t > 0 and u = ∂t u = 0 on the set B0 = {(x, 0) : |x− x0 | ≤ t0 }. Them u vanishes on Ω = {(x, t) : 0 ≤ t ≤ t0 , |x − x0 | ≤ t0 − t}. Proof. Assume u is real valued; otherwise, we can consider the real and imaginary parts separately. Let Z 1 | grad x,t u|2 dx Bt = {x : |x − x0 | ≤ t0 − t} and E(t) = 2 Bt !2  Z  n X 1 ∂u   2 2 i.e., E(t) =  dx. (∂t u) + 2 Bt  ∂x j  1 Then

dE 1 = dt 2

Z Bt

  n 2u   ∂u ∂2 u X ∂u ∂  2  . 2 +  dx− ∂t ∂t ∂x ∂x ∂t j j 1 !2  !2 X Z  n 1 ∂u   ∂u − +  dσ(x).  2 ∂t ∂x j  1 ∂Bt

50


(The second term comes from the change in the region Bt . If this is not clear, write the derivative as a limit of difference quotients and work it our). Since ! X ∂2 u ∂u X ∂u ∂2 u ∂u + = div gradx u , ∂x j ∂x j ∂t ∂ x2 ∂t ∂t j

applying the divergence theorem and using (∂2t − △)u = 0, we obtain dE = dt

Z "

# ∂u ∂u 1 2 − | grad x,t u| dσ ∂t ∂ν 2

∂Bt

59

where ν is the unit normal to Bt in Rn . Now " # 1 ∂u 2 ∂u 2 ∂u ∂u |< | | +| | | ∂t ∂ν 2 ∂t ∂ν " # 1 1 ∂u 2 2 2 | | + | grad x u| = | grad x,t u| . < 2 ∂t 2 dE Thus we see that the integrand is non-positive, and hence ≤ 0. dt Also E(0) = 0 since u = ∂t u = 0 on B0 , so E(t) ≤ 0. But E(t) ≥ 0 by S definition, so E(t) = 0. This implies that grad x,t u = 0 on Ω = Bt and t≤t0

since u = 0 on B0 , we conclude that u = 0 on Ω.

Corollary 2.40. Suppose uǫC 2 on Rn × [0, ∞), (∂2t − △)u = 0 for t > 0, u(x, 0) = u0 (x), ∂t u(x, 0) = u1 (x). If Ω0 = (supp u0 ) ∪ (supp u1 ), then supp u ⊂ Ω = {(x, t) : d(x, Ω0 ) ≤ t}. (The set Ω is the union of the forward light cones with vertices in Ω0 ). Proof. Suppose (x, t0 ) < Ω. Then for some ǫ > 0, the set B0 = {x : d(x, x0 ) ≤ t0 + ǫ} is disjoint from Ω0 .


51

Therefore, by Theorem 2.39, u = 0 on the cone {(x, t) : |x − x0 | ≤ t0 + ǫ − t, 0 ≤ t ≤ t0 + ǫ}. In particular, u = 0 on a neighbourhood of (x0 , t0 ), i.e., (x0 , t0 ) is not in the support of u. Corollary 2.41. SuppK+ ⊂ {(x, t) : t ≥ |x|}

SuppK+ ⊂ {(x, t) : −t ≥ |x|}. 60

Proof. Pick aφǫC0α (B(0, 1)) such that

R

φ = 1. Put

φǫ (x) = ǫ −n φ(ǫ −1 x). Let uǫ (x, t) = φǫ ∗(x) K+ , t > 0. Then (∂2t − △)uǫ = 0, u(x, 0) = 0, ∂t u(x, 0) = φǫ (x) and uǫ is C α . By the previous corollary supp uǫ ⊂ {(x, t) : |x| ≤ t + ǫ}. Now uǫ converges to K+ in S ′ , as ǫ tends to 0. Therefore, supp K+ ⊂ {(x, t) : |x| ≤ t}. The result for K− follows then since K− (x, t) = K+ (x, −t). Remarks 2.42. (i) One could also deduce the above result from our formulas for K˜ ± by using the Paley-Wiener theorem. (ii) Actually for n = 3, 5, 7, . . ., it turns out that supp K± = {(x, t) : |x| = ±t}. This is known as the Huygens principle. See Folland [1].

52


(iii) The distributions K± are smooth functions of t (except at t = 0) with values in E ′ (Rn ). Therefore, we now see that our formula (2.38) for the solution of the Cauchy problem makes sense even when u0 , u1 , ǫD′ (Rn ) and f ǫC(Rt , D′ (Rnx )), and it is easily checked 61 that u thus defined still solves the Cauchy problem in the sense of distributions. Corollary 2.40 remains valid in this more general setting, as can seen by an approximation argument as in the proof of Corollary 2.41. EXERCISE If (∂2t −△)u = f on Rn+1 , u(x, 0) = u0 (x), and ∂t u(x, 0) = u1 (x), figure out how supp u is related to supp f, supp u0 and supp u1 .

Chapter 3 L2 Sobolev Spaces THERE ARE MANY ways of measuring smoothness properties of func- 62 tions in terms of various norms. Often it is convenient to use L2 norms, since L2 interacts nicely with the Fourier transforms. In this chapter, we set up a precise theory of L2 differentiability and use it to prove Hormander’s theorem on the hypoellipticity of constant coefficient differential operators.

1 General Theory of L2 Sobolev Spaces Definition 3.1. For a non-negative integer k, the Sobolev space Hk is defined to be the space of all tempered distributions all of whose derivatives of order less than or equal to k are in L2 . Thus Hk = { f ǫS ′ : Dα f ǫL2 (Rn ) for 0 ≤ |α| ≤ k}.

From the definition, we note that f ǫHk if and only if ξ α fˆ(ξ)ǫL2 for 0 ≤ |α| ≤ k. Proposition 3.2. f ǫHk if and only if (1 + |ξ|2 )k/2 fˆǫL2 . Proof. First assume that (1 + |ξ|2 )k/2 fˆǫL2 . Since, for |ξ| ≥ 1, |ξ α | ≤ |ξ|k ≤ (1 + |ξ|2 )k/2 for all |α| ≤ k 53

3. L2 Sobolev Spaces

54 and for |ξ| < 1, we have and hence

63

|ξ α | ≤ 1 ≤ (1 + |ξ|2 )k/2 for all |α| ≤ k, |ξ α fˆ(ξ)| ≤ (1 + |ξ|2 )k/2 fˆ(ξ)

ξ α fˆǫL2 for |α| ≤ k which implies f ǫHk .

2

Conversely, assume that f ǫHk . Since |ξ|k and

n P

j=1

|ξ j |k are homoge-

neous of degree k and nonvanishing for ξ , 0, we have   n   X |ξ j |k  (1 + |ξ|2 )k/2 ≤ c0 (1 + |ξ|k ) ≤ c0 1 + c j=1

so that

||(1 + |ξ|2 )k/2 fˆ||2 ≤ c0 || f ||2 + c0 c

n X j=1

||∂kj f ||2

which shows that (1 + |ξ|2 )k/2 fˆǫL2 . The characterisation of Hk given in the above proposition immediately suggests a generalisation to non-integral values of k which turns out to be very useful. Definition 3.3. For sǫR, we define the operator Λ s : S → S by

64

(Λ s f )ˆ(ξ) = (1 + |ξ|2 ) s/2 fˆ(ξ). △ s/2 In other words, Λ s = I − 2 . Clearly Λ s maps S continuously 4π onto itself. We can therefore extend Λ s continuously from S ′ onto itself. The Sobolev space of order s is defined by H s = { f ǫS ′ : Λ s f ǫL2 }. We equip H s with the norm || f ||(s) = ||Λ s f ||2 . If s is a positive integer, the proof of Proposition 3.2 shows that || ||(s) is equivalent to the norm || f || = Σ0≤|α|≤s ||Dα f ||2 . PROPERTIES OF H s

1. General Theory of L2 Sobolev Spaces

55

(i) H s is a Hilbert space with the scalar product defined by (u, v)(s) = (Λ s u, Λ s v). Here the scalar product on the right is that in L2 . The Fourier transform is a unitary isomorphism between H s and the space of functions which are square integrable with respect to the measure (1 + |ξ|2 ) s dξ. (ii) For every sǫR, S is dense in H s . (iii) If s > t, H s ⊂ Ht with continuous imbedding. In fact, for uǫH s , ||u||(t) ≤ ||u||(s) . In particular, H s ⊂ L2 for s > 0. (iv) Dα is a bounded operator from H s into H s−|α| sǫR. (v) If f ǫH−s , then f as a linear functional on S extends continuously to H s and || f ||(−s) is the norm of f in (H s )∗ . So we can identify (H s )∗ with H(−s) . For f ǫH−s , gǫH s the pairing is given by −s

s

< f, g >=< Λ f, Λ g >=

Z

fˆgˆ .

(If s = 0, this identification of H0 = L2 with its dual is the complex conjugate of the usual one). (vi) The norm ||.||(s) is translation invariant. Indeed, if g(x) = f (x − x0 ), then gˆ (ξ) = e2πix0 .ξ fˆ(ξ) and hence || f ||(s) = ||g||(s) . Proposition 3.4. For s > n/2, we have δǫH−s . Proof. (1 + |ξ|2 )−s/2 δˆ = (1 + |ξ|2 )−s/2 ǫL2 , whenever s > n/2. This is a consequence of the following observation : R

(1 + |ξ|2 )−s dξ ∼ 1 +

R∞ 1

r−2s rn−1 dr < ∞ if and only if s > n/2.

As an immediate consequence of this proposition, we have Corollary 3.5. For s > n/2 + |α|, Dα δǫH−s .

65


56

Theorem 3.6 (SOBOLEV IMBEDDING THEOREM). For s > k, H s ⊂ C k . Further, we have X (3.7) sup |Dα f | ≤ C sk || f ||(s) .

n + 66 2

N |α|≤k R

Proof. Since S is dense in H s , it suffices to prove (3.7) for f ǫS . Let δx n denote the Dirac measure at x. For |α| ≤ k, since s > + |α|, Dα δx ǫH−s . 2 Since < Dα δx , f >= (−1)|α| < δx , Dα f >= (−1)|α| (Dα f )(x) and ||Dα δx ||(−s) is independent of x, X X sup |Dα f (x)| = sup |Dα δx , f > | n |α|≤k R

n |α|≤k R

≤

X

sup |Dα δx ||(−s) || f ||(s) = C sk || f ||(s)

n |α|≤k R

Now, given uǫH s , choose a sequence (u j ) in S such that ||u − u j ||(s) converges to 0, as j tends to ∞. The above inequality with f = ui − u j shows that (Dα u j ) is a Cauchy sequence in the uniform norm for |α| ≤ k; so its limit Dα u is continuous. T Corollary 3.8. If uǫH s for all sǫR, then uǫC ∞ i.e., H s ⊂ C ∞ . s

67

n + k, then This argument can be extended to show that if s > 2 elements of H s and their derivatives of order less than or equal to k are not merely continuous but actually Hölder continuous. n Proposition 3.9. If 0 < α < 1 and s = + α, then ||δx − δy ||(−s) ≤ 2 Cα |x − y|α Proof. We have ||δx − δy ||2(−s) =

Z

|e−2πix.ξ − e−2πiy.ξ |2 (1 + |ξ|2 )−s dξ.


57

Let R be a positive number, to be fixed later. When |ξ| ≤ R we use the estimate |e−2πix.ξ − e−2πiy.ξ | ≤ 2π|ξ||x− y| (by the mean value theorem) and when |ξ| > R, we use |e−2πix.ξ − e−2πiy.ξ| ≤ 2. Then we have Z Z 2 2 −s 2 2 2 (1 + |ξ|2 )−s dξ |ξ| (1 + |ξ| ) dξ + 4 ||δ x − δy ||(−s) ≤ 4π |x − y| |ξ|>R

|ξ|≤R

  ZR ZR     ≤ c |x − y|2 (1 + r2 )−s rn+1 dr + r−2s+n−1 dr   0

0

h i n n ≤ c′ |x − y|2 R−2s+n+2 + R−2s+n as < s< +1 2 2 h i = c′ |x − y|2 R2−2α + R−2α

When we take R = |x − y|−1 we get our result.

Exercise . Show that the above argument does not work when α = 1. Instead, we get ||δx − δy ||(−s) ≤ c|x − y|| log |x − y||1/2 when x is near y. What happens when α > 1? Corollary 3.10. Let 0 < α < 1 and Λα = bounded functions g : sup 68 x,y

n |g(x) − g(y)| < ∞. If s = + α + l and f ǫH s , then Dβ f ǫΛα for |β| ≤ k. α |x − y| 2 Remark 3.11. We shall obtain an analogue of this result for L p norms in Chapter 5. The following lemma will be used in several arguments hereafter. Lemma 3.12. For all ξ, nǫRn and sǫR, we have " #s 1 + |ξ|2 < 2|s| (1 + |ξ − η|2 )|s| . 1 + |η|2 Proof. |ξ| ≤ |η| + |ξ − η| gives |ξ|2 < 2(|η|2 + |ξ − η|2 ) so that

(1 + |ξ|2 ) < 2(1 + |η|2 )(1 + |ξ − η|2 ).


58

If s > 0, then raise both sides to the sth power. If s < 0, interchange ξ and η and raise to the −sth power. Proposition 3.13. If φǫS , then the operator f → φ f is bounded for all s. Proof. The operator f → φ f is bounded on H s if and only if the operator of g → Λ s φΛ−s g is bounded on L2 , as one sees by 69

setting g = ∧s f . But (∧s φ−s g)∧ (ξ) = (1 + |ξ|2 )s/2 (φ ∧−s g)ˆ(ξ) i h ˆ ∗ (∧−s g)ˆ(ξ) = (1 + |ξ|2 )s/2 φ(ξ) Z ˆ − η)(1 + |η|2 )ˆg(η)dη = (1 + |ξ|2 )s/2 φ(ξ Z = gˆ (η)K(ξ, η)dη where ˆ − η). K(ξ, η) = (1 + η|2 )−s/2 (1 + |ξ|2 ) s/2 φ(ξ By lemma 3.12, ˆ − η)|. |K(ξ, η)| < 2|s|/2 (1 + |ξ − η|2 )|s|/2 |φ(ξ Therefore, since φˆ is rapidly decreasing at ∞, Z |K(ξ, η)|dξ ≤ c for every ηǫRn , Z |K(ξ, η)|dη ≤ c for every ηǫRn . Thus from Theorem 1.1, the operator with kernel K is bounded on L2 . Hence our proposition is proved. The spaces H s are defined on Rn globally by means of the Fourier transform. Frequently, it is more appropriate to consider the following versions of these spaces.


59

Definition 3.14. If Ω ⊂ Rn is open and sǫR, we define H sloc Ω = { f ǫD′ (Ω) : ∀Ω′ ⋐ Ω, ∃ gΩ , ǫH s such that gΩ′ = f on Ω′ }. Proposition 3.15. f ǫH sloc (Ω) if and only if φ f ǫH s for φǫC0∞ (Ω).

70

Proof. If f ǫH sloc (Ω) and φǫC0∞ (Ω), then there exists gǫH s such that f = g on supp φ. Therefore φ f = φgǫH s by proposition 3.13. Conversely, if φ f ǫH s for all φǫC0∞ (Ω′ ), and Ω′ ⋐ Ω, choose φǫC0∞ (∞) with φ ≡ 1 on Ω′ . Then φ f ǫH s and f = φ f on Ω′ . P Corollary 3.16. If L = a∝ (x)D∝ with a∝ ǫC ∞ (Ω), then L maps |∝|≤k

loc (Ω) for all sǫR. H sloc (Ω) into H s−k

It is a consequence of the Arzela-Ascoli theorem that if (u j ) is a sequence of C k functions such that |u j | and |∂α u j |(|α| ≤ k) are bounded on compact set uniformly in j, there exists a subsequence (v j ) of (u j ) such that (∂α v j ) converges uniformly on compact set for |α| ≤ k − 1. In particular, if the u′j s are supported in a common compact set, then (∂α v j ) converges uniformly. There is an analogue of this result for H s spaces. Lemma 3.17. Suppose (uk ) is a sequence of C ∞ functions supported in a fixed compact set Ω such that sup ||uk ||(s) < ∞. Then there exists a k

subsequence which converges in the Ht norm for all t < s. Proof. Pick a φǫC0∞ such that φ = 1 on Ω so that uk = φuk and hence 71 uˆ k = φˆ ∗ uˆ k . Then 2 s/2

(1 + |ξ| )

2 s/2

Z

ˆ − η)uˆ k (η)dη| |uˆ k (ξ)| = (1 + |ξ| ) | φ(ξ Z ˆ − η)||uˆ k (η)|2|s|/2(1 + |η|2 ) s/2 (1 + |ξ − η|2 )|s|/2dη ≤ |φ(ξ ≤ 2|s|/2||φ||(|s|)||uk ||(s) ≤ c −1 independent of k.

Likewise, we have (1 + |ξ|2 )s/2 |∂ j uˆ k (ξ)| ≤ 2|s|/2 ||2πx j φ(x)|||s| ||uk ||(s) ≤ c2


60

independently of k. Therefore, by the Arzela-Ascoli theorem there exists a subsequence (ˆvk ) of (ûk ) which converges uniformly on compact sets. For t ≤ s, Z 2 ||v j − vk ||(t) = (1 + |ξ|2 )t |ˆv j − vˆ k |2 dξ Z Z 2 t 2 (1 + |ξ|2 )t |ˆv j − vˆ k |2 dξ (1 + |ξ| ) |ˆv j − vˆ k | dξ + = |ξ|>R

|ξ|≤R

2 max(t,0)

< (1 + R )

2

sup |ˆv j (ξ) − vˆ k (ξ)|

|ξ|≤R

+ (1 + R2 )t−s

Z

|ξ|>R

Z

dξ+

|ξ|≤R

[(1 + |ξ|2 ) s |ˆv j (ξ) − vˆ k (ξ)|2 dξ]

≤ c(l + R2 )n+|t| sup |ˆv j (ξ) − vˆ j (ξ)|2 + (1 + R2 )t−s ||v j − vk ||2(s) . |ξ|≤R

72

Given ǫ > 0, choose R large enough so that the second term is less than ǫ/2 for all j and k. This is possible since ||v j − vk ||(s) ≤ c and t − s < 0. Then for j and k large enough the first term is less than ǫ/2, since (ˆvk ) converges uniformly on compact sets. Thus we see that (vk ) is a Cauchy sequence in Ht′ and since Ht is complete we are done. Remark 3.12. Lemma 3.17 is false, if we do not assume that all the u′k s have support in a fixed compact set. For example, for uǫC0∞ and xk ǫRn with |xk | tending to ∞, define uk (x) = u(x−xk ). Then the invariance of H s norms shows that ||uk ||(s) = ||u||(s) . But no subsequence of (uk ) converges, in any Ht . For, if a subsequence (vk ) of (uk ) converges, it must converge to 0, since uk converges to 0 in S ′ . But then lim ||vk ||(t) = 0 which is not the case. Theorem 3.20 (RELLICH THEOREM). Let H s0 (Ω) be the closure of C0∞ (Ω) in H s . If Ω is bounded and t < s, the inclusion H s0 (Ω) ֒→ Ht is compact, i.e., bounded sets in H s0 (Ω) are relatively compact in Ht . Proof. Let (uk ) be a sequence in H s0 (Ω). To each k, find a vk ǫC0∞ (Ω)


61

1 such that ||uk − vk ||(s) ≤ . Then we have ||vk ||(s) ≤ ||uk ||(s) + | ≤ c k (independent of k.) Therefore, by lemma 3.17, a subsequence (wk ) of (vk ) exists such that (wk ) converges in Ht . If (u′k ) is the subsequence of (uk ) corresponding to the sequence (wk ), we have 73 ||u′i − u′j ||(t) < ||u′i − wi ||(t) + ||wi − w j ||(t) + ||w j − u′j ||(t) 1 1 < + + ||wi − w j ||(t) → 0 as ||i, j → ∞. i j Hence (u′k ) converges in Ht .

I the proof of the next theorem, we will use the technique of complex interpolation, which is based on the following result from elementary complex analysis called ‘Three lines lemma’. Lemma 3.20. Suppose F(z) is analytic in o < ReZ < 1, continuous and bounded on 0 ≤ ReZ ≤ 1. If |F(1 + it)| ≤ c0 and |F(l + it)| ≤ c1 , then s F(s + it) ≤ c1−s 0 c1 , for 0 < s < 1. Proof. If ǫ > 0, the function gǫ (z) = c0z−1 c1z−1 eǫ(z

2 −z)

f (z)

satisfies the hypotheses with c0 and c1 replaced by 1, and also |gǫ (z)| converges to 0 as |Imz| → ∞ for 0 ≤ ReZ ≤ 1. From the maximum modulus principle, it follows that |g(z)| ≤ 1 for 0 ≤ Rez ≤ 1 and letting ǫ tend to 0, we obtain the desired result. Theorem 3.21. Suppose that −∞ < s0 < s1 < ∞ and T is a bounded linear operator H s0 such that T |H s1 is bounded on H s1 . Then T |H s is bounded on H s for all s with s0 ≤ s ≤ s1 . Proof. Our hypothesis means that

74

∧s0 T ∧−s0 and ∧s1 T ∧−s1 are bounded operators on L2 . For 0 ≤ Rez ≤ 1 we define sz = (l − z)s0 + zs1 and T z = ∧sz T ∧−sz .


62

Then what we wish to prove is that T z is bounded on L2 for 0 ≤ z ≤ 1. Observe that when w = x + iy, ∧w , ∧x ∧iy and (∧iy f )ˆ(ξ) = (1 + |ξ|2 )iy/2 fˆ2 (ξ) so that |(∧iy f )ˆ(ξ) = | fˆ(ξ)|. Thus ∧iy is unitary on H s for all s. For φ, ψǫS , we define Z F(z) = (T z φ)ψ =< ∧sz T ∧−sz φ, ψ > . Then |F(z)| = | < ∧sz T ∧−sz φ, ψ > | = | < T ∧sz φ, ∧−sz ψ > |

≤ ||T ∧−sz φ||( s0 )|| ∧−sz ψ||(−s0 )

≤ c|| ∧−sz φ|| s0 || ∧sz ||(−s0 )

≤ c||φ||(s0 −s1 )Re z ||ψ||(s1 −s0 )Rez

≤ c||φ||(0) ||ψ|| s1 −s0 75

F(z) is clearly an analytic function of z for 0 < Re z < 1. Further, by our hypothesis on T , when Rez = 0, we have |F(z)| ≤ c0 ||φ||(0) ||ψ||(0) and when Rez = 1, we have |F(z)| ≤ c1 ||φ||(0) ||ψ||(0) . Therefore, by the Three lines lemma z |F(z)| ≤ c1−z 0 c0 ||φ||(0) ||ψ||(0) for 0 < z < 1.

Finally, by the self duality of H0 = L2 , this gives z ||T z φ|| ≤ c1−z 0 c0 ||φ||(0)

which completes the proof.


63

Remark 3.22. The same proof also yields the following more general result: Suppose ∞ < s0 < s1 < ∞ < t0 < t1 < ∞. If T is a bounded linear operator from H s0 to Ht0 whose restriction to H s1 to Ht1 , then the restriction of T to Htθ is bounded from Htθ to Htθ for 0 < θ < 1 where sθ = (1 − θ)s0 + θs1 and tθ = (1 − θ)t0 + θt1 . As a consequence of this result, we obtain an easy proof that H sloc is invariant under smooth coordinate changes. Theorem 3.23. Suppose Ω and Ω′ are open subsets of Rn and φ : Ω → Ω′ is a C ∞ diffeomorphism. Then the mapping f → f oφ maps H sloc (Ω′ ). 76 continuously onto H sloc (Ω). Proof. The statement of the theorem is equivalent to the assertion that for any φǫC0∞ (Ω′ ), the map T f = (φ f ) ◦ φ is bounded on H s for sǫR. If s = 0, 1, 2, . . . , this follows from the chain rule and the fact that H s = { f : D∝ f ǫL2 for | ∝ | ≤ s}. By Theorem 3.21, it is true for all s ≥ 0. But the adjoint of T is another map of the same form : T ∗ g = (ψg)◦ ψ where ψ = θ−1 and ψ = φ|J| ◦ Φ, J being the Jacobian determinant of Φ−1 . Hence T ∗ is bounded on H s for all s ≥ 0 and by duality of H s and H−s , this yields the boundedness of T on H s for s < 0. Finally, we ask to what extend the H s spaces include all distributions. Globally they do not, since, if f ǫH s , then f is tempered and fˆ is a function. But locally they do, as we see from the following result. Proposition 3.24. Every distribution with compact support lies in some S Hs . H s : i.e., E ′ ⊂ s∈R

Proof. If f ǫE ′ , then it is a continuous linear functional on C ∞ . Therefore, there exists a constant c > 0, a compact set K, and a nonnegative integer k such that X | < f, φ > | ≤ c sup |D∝ φ| for all φǫC ∞ , |∝|≤k k

i.e., | < f, φ > | ≤ c

P

sup |Dα φ| for all φǫC ∞ .

|α|≤|k Rn

77


64

By the Sobolev imbedding theorem, X sup |Dα φ| ≤ c′ ||φ||(k+ n2 +ǫ) for ǫ > 0. n |∝|≤k R

Therefore, | < f, φ > | ≤ c′′ ||(k+ n +ǫ ) for all φǫS since S is dense in 2 H(k+ n2 +ǫ) , this shows that f is a continuous linear functional on H(k+ n +ǫ ) . 2 Hence f ǫH− n2 −k−ǫ . Corollary 3.25. If f ǫD′ (Ω) and Ω′ has compact closure in Ω then there exists s in R such that f ǫH sloc (Ω′ ).

2 Hypoelliptic Operators With Constant Coefficients We now apply the machinery of Sobolev spaces to derive a criterion for the hypoellipticity of constant coefficient differential operators. First, we have a few preliminaries. Definition 3.26. Let P be a polynomial in n variables. For a multi-index ∂ ∝, P(∝) will ne defined by P(∝) (ξ) = ( )∝ P(ξ). ∂ξ Proposition 3.27. LEIBNIZ RULE When f ǫC ∞ , gǫD′ and P(D) is a constant-coefficient partial differential operator of order k, we have X 1 (P(∝) (D)g)D∝ f ). P(D)( f g) = ∝! |∝|≤k The proof of this proposition is left as an exercise to the reader. 78

Definition 3.28. We say that a polynomial P satisfies condition (H) if there exists a δ > 0 such that |P(∝) (ξ)| = 0 (|ξ|−δ|α| ) as |δ| → ∞, ∀ ∝ . |P(ξ)| ¨ Theorem 3.29 (HORMANDER). If P satisfies condition (H), then P(D) loc (Ω), is hypoelliptic. More precisely, if f is in D′ Ω)and P(D) f H s+kδ where δ is as in condition (H) and k is the degree of P.

2. Hypoelliptic Operators With Constant Coefficients

65

Proof. We first observe that the second assertion implies the first since T loc H s (Ω) by Corollary 3.8. C ∞ (Ω) = sǫR

Suppose therefore that P(D) f ǫH sloc (Ω), f ǫD′ (Ω). Given φǫC◦∞ (Ω) we have to prove that φ f ǫH s+kδ . Let Ω′ , be an open set such that Ω′ ⋐ Ω and supp φ ⊂ Ω′ . By Corollary 3.25, therefore exists t in R such that f ǫHtloc (Ω′ ). By decreasing t, we can some that t = s + k − 1 − mδ for assume positive integer m. Set φm = φ and then choose φm−1 , φm−2 , . . . , φ0 , φ−1 in C0∞ (Ω′ ) such that φ j = 1 on supp φ j+1 . Then φ j P(D) f ǫHS ⊂ Ht−k+l+ jδ for 0 ≤ j ≤ m and φ−1 f ǫHt . Now P(D)(φ0 f ) = φ0 P(D) f +

X 1 P∝ (D)(φ−1 f )D∝φ0 ∝! ∝,0

since θ−1 = 1 on the support of φ0 . So, P(D)(φ0 f )ǫHr−k+1 . This means that Z (1 + |ξ|2 )t−k+1 |P(ξ)(φ0 f )ˆ(ξ)|2 dξ < ∞. By condition (H) Z (1 + |ξ|2 )t−k+l+δ|δ| |P∝| (ξ)(φ0 f )( ξ)|2 dξ < ∞ This implies that P(α) (D)(φ◦ f )ǫHt − k + l + δ|δ|. Next, P(D)(φ1 f ) = φ1 P(D) f +

X 1 p(∝) (D)(φo f )D∝ (φ1 f ) ∝! ∝,0

since φ0 = 1 on the support of φ1 , so P(D)(φ1 f )ǫHt−k+1+δ . By the same argument same argument as above, P(D) (φ1 f )ǫHt+k+l+δ(1+|∝|) . Continuing inductively, we obtain P(D)(φ j f )ǫHt−k+l+ jδ , which implies that P(∝) (D)(φ j f )ǫHt−k+l+δ( j+|∝|) .

79


66 For j = m, we above

P(∝) (D)(φm f )ǫHt−k+l+δ(m+(δ)) = H s+δ|∝| . If P(ξ) =

X

|∝|≤k

a∝ ξ ∝ ,

choose ∝ with | ∝ | = k such that a∝ , 0. Then P∝ (ξ) =∝!a∝ , 0. whence φ f = φm f ǫH s+kδ and we are done. 80

Remark 3.30. The condition (H) is equivalent to the following apparently weaker condition (H ′ ) :

|p(∝)(ξ) | → 0 as |ξ| → ∞ for ∝, 0. |P(ξ)|

Condition (H ′ ) is in turn equivalent to (H ′′ ) : |Imζ| → ∞, in the set {ζǫn : P(ζ) = 0}. The converse of Hörmander’s theorem is also true, i.e., hypoellipticity implies condition (H). The proofs of these assertions can be found in Hörmander [6]. The logical order of the proofs is (H) ⇒ hypoellipticity ⇒ (H ′′ ) ⇒ (H ′ ) ⇒ (H). The implication (H ′ ) ⇒ (H) requires the use of some results from (semi) algebraic geometry. P P Definition 3.31. P(ξ) = a∝ ξ ∝ is called elliptic if a∝ ξ ∝ , 0 for every ξ , 0.

|∝|≤k

|∝|=k

EXERCISES 1. Prove that P is elliptic if and only if |P(ξ)| ≥ c◦ |ξ|k for large |ξ|. 2. Prove that P is elliptic if and only if P satisfies condition (H) with δ = 1.

2. Hypoelliptic Operators With Constant Coefficients

67

3. Prove that no P satisfies condition (H) with δ > 1. (Hint: If | ∝ | = k, P(∝) is a constant). 81

4. Let P be elliptic and real valued. Define Q on Rn+1 by Q(ξ, τ) = 2πiτ + P(ξ), so that Q(D x , Dt ) = ∂t + P(D x ). Show that Q satisfies condition (H) with δ = 1/k where k is the degree of P and that 1/k is the best possible value of δ. (Hint : Consider the regions |ξ|k ≤ |τ| and |τ| ≤ |ξ|k separately).

Chapter 4 Basic Theory of Pseudo Differential Operators 1 Representation of Pseudo differential Operators Let L =

P

|α|≤k

aα (x)Dα be a partial differential operator with C ∞ coeffi- 82

cients on Ω. Using the Fourier transform, we can write Z X (Lu)(x) = aα (x) e2πix.ξ uˆ (ξ)ξ α dξ |α|≤k

= where p(x, ξ) =

P

Z

e2πix.ξ p(x, ξ)û(ξ)dξ

aα (x)ξ α . This representation suggests that p(x, ξ)

|α|≤k

can be replaced by more general functions. So, we make the following definition. Definition 4.1. For an open set Ω ⊂ Rn and a real number m we define S m (Ω), the class of symbols of order m on Ω, by S m (Ω) = {pǫC ∞ (Ω × Rn ) : ∀α, β, VΩ′ ⊂ Ω, c = cαβΩ′ β

such that sup xǫΩ′ |D x Dαξ p(x, ξ)| ≤ c(1 + |ξ|)m−|α| }. 69

4. Basic Theory of Pseudo Differential Operators

70

T

We note that S m (Ω) ⊂ S m whenever m < m′ , and we set S −∞ (Ω) = S m (Ω).

mǫR

Examples (i) Let p(x, ξ) = 83

P

|α|≤k

(ii) Let p(x, ξ) =

aα (x)ξ α with aα ǫC ∞ (Ω). Then pǫS k (Ω).

N P

j=1

a j (x) f j (ξ) where a f ǫC ∞ (Ω) and f j ǫC ∞ (Rn ) is

homogeneous of degree m j large ξ : that is, fi (rξ) = rm j f j (ξ) for |ξ| ≥ c, r ≥ 1. In this case, pǫS m (Ω), where m = max {m j }. 1≤ j≤N

(iii) Let p(x, ξ) = (1 + |ξ|2 )s/2 . This p belongs to S s (Rn ). (iv) Let p(x, ξ) = φ(ξ) sin log |ξ| with φǫC ∞ , φ = 0 near the origin and φ = 1 when |ξ| ≥ 1. Then pǫS o (Rn ). β

Remark 4.2. We observe that when pǫS m (Ω), D x Dαξ pǫS m−|α| (Ω). Further, if pǫS m1 (Ω) and qǫS m2 (Ω),then p + qǫS m (Ω), where m = max{m1 , m2 } and pqǫS m1 +m2 (Ω). Remark 4.3. Our symbol classes S m (Ω) are special cases of Hormanm (Ω). Namely, for 0 ≤ δ ≤ ρ ≤ 1, and mǫR, der’s classes S ρ,δ m S ρ,δ (Ω) = {pǫC ∞ (Ω × Rn ) : ∀α, β, VΩ′ ⋐ Ω, c = cαβΩ′

such that

β

sup |D x Dαξ p(x, ξ)| ≤ c(1 + |ξ|)m−ρ|α|+δ|β| }. xǫΩ′

m (Ω). In this terminology, S m (Ω) = S 1,0

Definition 4.4. For pǫS m (Ω), we define the operator p(x.D) on the domain Co∞ (Ω) by Z p(x, D) u(x) = e2πix.ξ p(x, ξ) uˆ (ξ)dξ, uǫCo∞ (Ω).

1. Representation of Pseudo differential Operators

84

71

(Sometimes, we shall denote p(x, D) by p). Operators of the form p(x, D) with pǫS m (Ω) are called pseudo differential operators of order m on Ω. The set of all pseudo differential operators of order m on Ω will denoted by Ψm (Ω). For brevity, we will sometimes write “ΨDO′′ instead of “pseudo differential operators”. The next theorem states that p(x, D) is a continuous linear map of ∞ Co (Ω) into C ∞ (Ω which extends to E ′ (Ω). For the proof, we need a result which depends on Lemma 4.5. Let pǫS m (Ω) and φǫCo∞ (Ω). Then, for each positive integer N, there exists cN > 0 such that for all ξ, η in Rn , Z | e2πix.η p(x, ξ)φ(x)dx| ≤ cN (1 + |ξ|)m (1 + |η|)−N . Proof. RFor any ξ and η in Rn , |ηα e2πix.η p(x, ξ)φ(x)dx| =|

Z

Dαx e2πix.η p(x, ξ)φ(x)dx|.

Integrating by parts, we have |ηα

Z

e2πix.ξ p(x, ξ)φ(x)dx| = |

Z

e2πix.η Dαx (p(x, ξ)φ(x))dx|

≤ Cα (1 + |ξ|)m for all α. Therefore X

|α|≤N

α

|η

Z

e2πix.η p(x, ξ)φ(x)dx| ≤ c′N (1 + |ξ|)m for all N.

Since (1 + |η)N ≤ c

P

|α|≤N

|ηα |, the required result follows.

85


72

Corollary 4.6. If pǫS m (Ω) and φǫCo∞ (Ω), then the function Z g(ξ) = e2πix.ξ p(x, ξ)φ(x)dx is rapidly decreasing as ξ tends to ∞. Proof. Set ξ = η in the lemma.

Theorem 4.7. If pǫS m (Ω), then p(x, D) is a continuous linear map of Co∞ (Ω) into C ∞ (Ω) which can be extended as a linear map from E ′ (Ω) into D′ (Ω). Proof. For uǫCo∞ (Ω), p(x, D)u(x) =

Z

e2πix.ξ p(x, ξ)û(ξ)dξ.

The integral converges absolutely and uniformly on compact sets, as do the integrals Z Dαx (e2πix.ξ p(x, ξ) uˆ (ξ)dξ for all α, since pǫS m and uˆ ǫS.

This proves that p(x, D)uǫC ∞ (Ω), and continuity of p(x, D) from to C ∞ (Ω) is an easy exercise. To prove the rest of the theorem, we will make use of Corollary 4.6. For uǫE ′ (Ω), we define p(x, D)u as a functional on Co∞ (Ω) as follows: x < p(x, D)u, φ > = p(x, ξ)û(ξ)e2πix.ξ φ(x)dxdξ Z = g(ξ)û(ξ)dξ

Co∞ (Ω)

86

where g(ξ) =

Z

e2πix.ξ p(x, ξ)φ(x)dx, for φǫCo∞ (Ω).

2. Distribution Kernels and the Pseudo Local Property

73

By Corollary 4.6, g(ξ) is rapidly decreasing, while uˆ is of polynomial growth; so the last integral is absolutely convergent and the functional on Co∞ thus defined is easily seen to be continuous. Moreover, if uǫCo∞ , the double integral is absolutely convergent, and by interchanging the order of integration, we see that this definition of p(x, D)u coincides with the original one. Hence we have extended p(x, D) to a map from E ′ (Ω) to D(Ω). Remark 4.8. It follows easily from the above argument that p(x, D) is sequentially from E ′ (Ω) to D(Ω), i.e., if uk converges to u in E ′ (Ω), then p(x, D)uk converges to p(x, D)u in D′ (Ω). Actually, p(x, D) is continuous from E ′ (Ω) to D′ (Ω) : this follows from the fact (which we shall prove later) that the transpose of a pseudo differential operator is again a pseudo differential operator. Thus p(x, D)t : Co∞ (Ω) → C ∞ (Ω) is continuous, and so, by duality, p(x, D) = (p(x, D)t )t : E ′ (Ω) → D′ (Ω) is continuous.

2 Distribution Kernels and the Pseudo Local Property Definition 4.9. Let T be an operator from Co∞ (Ω) to C ∞ (Ω). If there exists a distribution K on Ω × Ω such that < T u, v >=< k, v ⊗ u > for u, vǫCo∞ (Ω),

we say that K is the distribution kernel of the operator T . In this definition, v⊗u is defined by (v⊗u)(x, y) = v(x)u(y). Formally, this definition says that. Z T u(x) = K(x.y)u(y)dy. K is uniquely determined since linear combinations of functions of the 87 form v ⊗ u are dense in Co∞ (Ω × Ω). If pǫS m (Ω), it is easy to compute the distribution kernel of p(x, D). In fact, x < p(x, D)u, v > = p(x, ξ)û(ξ)e2πix.ξ v(x)dξdx


74

=

x

p(x, ξ)e2πix.ξ (v ⊗ u)ˆ2 (x, ξ)dξdx

where (v ⊗ u)ˆ2 (x, ξ) means the Fourier transform in the second variable. It follows immediately from the definition of K that x < K, w > = e2πix.ξ p(x, ξ)wˆ2 (x, ξ)dξdx, ∀wǫCo∞ (Ω × Ω). y or < K, w > = e2πi(x−y).ξ p(x, ξ)w(x, y)dy dξ dx. From this it is easy to see that K(x, y) = pv2 (x, x − y) where pv2 (x, .) is the inverse Fourier transform of the tempered distribution p(x, .). In particular, this shows that p is uniquely determined by the operator p(x, D). If PǫΨm (Ω), we shall sometimes denote the unique pǫS m (Ω) such that p = p(x, D) by σ p . The following theorem gives precise results on the kernel K of p(x, D). Theorem 4.10. The distribution kernel K of p(x, D) with pǫS m (Ω) is in C ∞ on (Ω × Ω)/∆ where ∆{(x, x) : xǫΩ} is the diagonal. More precisely, if |α| > M + n + j for a positive integer j, then (x − y)α K(x, y)ǫC j (Ω × Ω). 88

Proof. For wǫCo∞ (Ω × Ω), let us compute < (x − y)α K, w >. < (x − y)α K, w > =< K, (x − y)α w > x = e2πix.ξ p(x, ξ)(x + Dξ )α wˆ2 (x, ξ)dξ dx x = wˆ2 (x, ξ)(x − Dξ )α {e2πix.ξ p(x, ξ)}dξ dx. Using the Leibniz formula, it is easily seen that (x − Dξ )α {e2πix.ξ p(x, ξ)} = (−Dξ )α p(x, ξ).e2πix.ξ Therefore < (x − y)α K, w > =

x

wˆ2 (x, ξ)e2πix.ξ (−Dξ )α p(x, ξ)dξ dx y = w(x, y)e2πi(x−y).ξ (−Dξ )α p(x, ξ)dy dξ dx.

2. Distribution Kernels and the Pseudo Local Property

75

From the above expression, we infer that Z (x − y)α K(x, y) = e2πi(x−y).ξ (−Dξ )α p(x, ξ)dξ. Since |(−Dξ )α p(x, ξ)| ≤ c(1 + |ξ|)m−|α| , the integral converges absolutely and uniformly on compact sets whenever m − |α| < −n. Also we can differentiate with respect to x and y j times under the integral sign provided m − |α| + j < −n or |α| > m + n + j. Thus, we see that (x − y)α KǫC j (Ω × Ω). Corollary 4.11 (PSEUDO LOCAL PROPERTY OF ΨDO). If PǫΨm (Ω), then, for all uǫE ′ (Ω), sing supp Pu is contained in sing supp u. Proof. Let uǫE ′ and let V be an arbitrary neighbourhood of sing supp u. Take a φǫCo∞ (V) such that φ = 1 on sing supp u. Then u = φu+(1−φ)u = u1 + u2 ; u2 is a Co∞ function and supp u1 ǫV. Therefore, Pu = Pu1 + Pu2 and Pu2 is a C ∞ function. Moreover, when xo < V, 89 Z K(x, y)u1 (y)dy

Pu1 (x) =

V

is also a C ∞ function in a neighbourhood of xo since u1 (y) = 0 for y near xo . This implies that sing supp Pu ⊂ V. Since V is any neighbourhood of sing supp u, the corollary is proved. Corollary 4.12. If pǫS −∞ (Ω), then the distribution kernel K of p(x, D) is in C ∞ (Ω × Ω). Proof. Follows from the theorem if we take |α| = 0.

Corollary 4.13. If pǫS −∞ (Ω), then p(x, D), maps E ′ (Ω) continuously into C ∞ (Ω). Proof. If uǫE ′ (Ω), in view of Corollary 4.12, it is easily seen that p(x, D)u is a smooth function defined by Z P(x, D)u(x) = K(x, y)u(y)dy =< u, K(x, .) >, whence the result follows.

76


Definition 4.14. A smoothing operator is a linear operator T which maps E ′ (Ω) continuously into C ∞ (Ω). If KǫC ∞ (Ω × Ω), then the operator T defined by (T f )(x) =< K(x, .), f >=

Z

K(x, y) f (y)dy

is a smoothing operator. Conversely, every smoothing operator T can be given in the above form with K(x, y) = (T δy )(x). As we have already remarked if pǫS −∞ , then the corresponding ΨDO is smoothing. However, not every smoothing operator is a ΨDO. For example, we have 90

Proposition 4.15. Suppose p(x, .) is a C ∞ function of x with values in E ′ . Then p(x, D) (defined in the same way as in the case of pǫS m (Ω)) is a smoothing operator. Proof. The distribution kernel K of p(x, D) is given by K(x, y) =

Z

e2πi(x−y).ξ p(x, ξ)dξ

=< p(x, .), e2πi(x−y).(.) > and hence K(x, y)ǫC ∞ . Thus K defines a smoothing operator.

Remark 4.16. Sometimes, it is convenient to enlarge the class of pseudo-differential operators of order m by including operators of the form P + S where PǫΨm (Ω) and S is smoothing. However, the general philosophy is the following: 1. On the level of operators, smoothing operators are negligible. 2. On the level of symbols, what counts is the asymptotic behaviour at ∞, so that symbols of order −∞ are negligible.

3. Asymptotic Expansions of Symbols

77

3 Asymptotic Expansions of Symbols Definition 4.17. Suppose m0 > m1 > m2 > · · · m j ǫR, m j tends to −∞, ∞ P P and p j ǫS m j (Ω), pǫS mo (Ω). We say that p ∼ p j if p − p j ǫS mk (Ω) j=0

j and φ = 0 1 for |ξ| ≤ . 2 Claim There exists a sequence (t j ),t j ≥ 0 and t j tending to ∞ so 91 rapidly that we have (4.19) β |D x Dαξ (φ(ξ/t j )p j (x, ξ))| ≤ 2− j (1+|ξ|)m j−1 −|α| for xǫΩi , and |α|+|β|+i ≤ j. Granted this, we define p(x, ξ) =

∞ X

φ(ξ/t j )p j (x, ξ).

j=0

Note that for each x and ξ, the sum is finite. Using (4.19), it is ∞ P straightforward to show that pǫS mo (Ω) and p ∼ p j . Moreover, supj=0

pose that qǫS mo (Ω) and q ∼ p − q = (p −

X j 0 such that for all f ǫC 2 (Ω2 ), SupΩ1 |∂ j f |2 ≤ c1 (SupΩ2 | f |2 ) + c2 (SupΩ2 | f |)(SupΩ2 |∂2j f |). Proof. It suffices to assume that n = 1 and f is real valued. With this reduction, the proof becomes an exercise in elementary calculus. This idea is roughly as follows: We wish to show that if k f k∞ and k f ′′ k∞ are both small, then so is k f ′ k∞ is small and | f ′ (xo )| is large, then | f ′ | will be large in some sizable interval [a, b] containing xo . But then | f (b) − f (a)| and hence k f k∞ is large. We leave it to the reader to work out the quantitative details of this argument.

4. Properly Supported Operators 93

79

Proof of the theorem . By Proposition 4.18, there exists q in S mo such ∞ P that q ∼ p j ; so, it will suffice to show that p − qǫS −∞ . First, j=0

|p(x, ξ) − q(x, ξ)| = |(p(x, ξ) −

X j . = 0 whenever supp w ∩ P Therefore, supp K ⊂ a . This proves (i).

P

a

= φ.

5. ψdo′ s Defined by Multiple Symbols

83

To prove (ii), choose a properly supported φǫC ∞ (Ω × Ω) with φ = 1 on ∆ ∪ supp K, ∆ being the diagonal. Set a(x, ξ, y) = φ(x, y)a(x, ξ, y). P P Then a , ⊂ supp φ and hence a , is proper. Also a′ (x, ξ, y) = a(x, ξ, y) when (x, y) is near the diagonal. Now, < a(x, D, y)u, v > =< K, v ⊗ u >

=< φK, v ⊗ u >

=< K, φ(v ⊗ u) > y = e2πi(x−y)·ξ a(x, ξ, y)φ(x, y) × v(x)u(y)dydξdx y = e2πi(x−y).ξ a′ (x, ξ, y)u(y)dydξdx =< a′ (x, D, y)u, v > . 98

Since v is arbitrary, we have a(x, D, y) = a′ (x, D, y). Theorem 4.28. Suppose aǫS m (Ω × Ω) and A = a(x, D, y) is properly supported. Let p(x, ξ) = e−2πix.ξ A(e2πi(.).ξ )(x). Then pǫS m (Ω) and p(x, D) = A. Further, p(x, ξ) ∼

X 1 ∂αξ Dαy a(x, ξ, y)|y=x . α! a

Proof. For u in Co∞ (Ω), we have u(x) =

Z

e2πix.ξ uˆ (ξ)dξ.

Therefore, by linearity and continuity of A, Z Au(x) = A(e2πi(.).ξ )(x)û(ξ)dξ Z = e2πix.ξ p(x, ξ)û(ξ)dξ = p(x, D)u(x).


84

P By the proposition above, we can modify a so that a is proper without affecting A or the behaviour of a along the diagonal x = y (which is what enters into the asymptotic expansion of p). Set b(x, ξ, y) = a(x, ξ, x + y). Then as a function of y, b has compact support. Indeed, 99 P for fixed x and ξ, if yǫ{y′ : b(x, ξ, y′ ) , 0} then x + yǫπ2 (π−1 1 (x) ∩ a ) which is compact. Now, x p(x, η) = e−2πix.η e2πi(x−y).ξ b(x, ξ, y − x)e2πiy.η dydξ x = e−2πix.η e2πiz.ξ b(x, ξz)e2πi(z+x).η dzdξ Z Z ˆ = b3 (x, ξ, ξ − η)dξ = bˆ 3 (x, ξ + η, ξ)dξ, where bˆ 3 is the Fourier transform of b in the third variable. Since b(x, ξ, .) is in Co∞ , this calculation is justified and bˆ 3 (x, η, ξ) is rapidly decreasing in the variable ξ. More precisely, since aǫS m (Ω × Ω), we have β |D x Dαη bˆ 3 (x, η, ξ)| ≤ cαβN (1 + |η|)m−|α| (1 + |ξ|)−N .

(4.29) Since

1 + |η| 1 + |ξ|

!s

≤ (1 + |ξ − η|)|s| , taking η + ξ instead of η

we have (1 + |ξ + η|)s ≤ (1 + |ξ|)s (1 + |η|)|s| . If we use this in β |D x Dαη bˆ 3 (x, ξ + η, ξ)| ≤ cαβN (1 + |ξ + η|)m−|α| (1 + |ξ|)−N

we get β |D x Dαη bˆ 3 (x, ξ + η, ξ)| ≤ c′αβN (1 + |ξ|)−N+m−|α| (1 + |η|)|m|+|α| .

If we take N so that −N + m − |α| < −n we have Z β α β (4.30) |D x Dη p(x, η)| = | D x Dαη bˆ 3 (x, ξ + η, ξ)dξ| ≤ cαβ (1 + |η|)|m|+|α|

5. ψdo′ s Defined by Multiple Symbols 100

85

On the other hand, taking the Taylor expansion of bˆ 3 (x, ξ + η, ξ) in the middle argument about the point η, (4.29) gives |bˆ 3 (x, ξ + η, ξ) −

X

∂αη bˆ 3 (x, η, ξ)

|α| = e2πix,ξ p(x, ξ)û(ξ)v(x)dξdx ! Z Z 2πix,ξ = e p(x, ξ)v(x)dx uˆ (ξ)dξ

103


88

=

Z

g(ξ)û(ξ)dξ

where g(ξ) =

Z

e2πix,ξ p(x, ξ)v(x)dx.

By Corollary 4.6, g us a rapidly decreasing function. Thus Z Z < Pu, v >= gû = uˆg =< u, Ptv > so that Pt v(y) = gˆ (y) =

(4.35)

=

x

x

e2πi(x−y).ξ p(x, ξ)v(x)dxdξ e2πi(y−x).ξ p(x, −ξ)v(x)dxdξ

= a(x, D, y)v(y) where a(x, ξy) = p(y. − ξ). Therefore, by Theorem 4.28, Pt ǫψm (Ω) and σPt (x, ξ) ∼

X (−1)|α| a

α!

∂αξ Dαx σP (x, −ξ).

The assertions about P∗ follows along similar lines Theorem 4.36. If Pǫψm1 (Ω) and Qǫψm2 (Ω) are properly supported, then QPǫψm1 +m2 (Ω) and σQP(x, ξ) ∼ 104

X 1 ∂αξ σQ (x, ξ)Dαx σ p (x, ξ). α! α

Proof. Since P = (Pt )t , we have x (Pu)(x) = e2πi(x−y)·ξ σtP (y, −ξ)u(y)dydξ by (4.35).

6. Products and Adjoint of ψDO′ S

89

In other words (Pu)ˆ(ξ) =

Z

e−2πiy·ξ σPt (y, −xi)u(y)dy.

Therefore, (QP)u(x) =

x

e2πi(x−y)·ξ σQ (x, ξ)σPt (y, −xi)u(y)dydξ

= a(x, D, y)u(x) where a(x, ξ, y) = σQ (x, ξ)σPt (y, −ξ). Clearly aǫS m1 +m2 (Ω × Ω) which implies QPǫψm1 +m2 (Ω). Moreover X 1 ∂αξ Dαy (σQ (x, ξ)σ pt (y, −ξ))|y=x α! α X 1 X α β ∼ ∂ξ σQ (x, ξ)∂νξ DαY σPt (x, −ξ) α! β!ν! α β+ν=α

σQP (x, ξ) ∼

∼

X X (−1)|δ| α,ν

δ

ν!β!δ!

β

ν+δ ∂ξ σQ (x, ξ)Dβ+ν+δ ∂ν+δ x ∂ξ σ P (x, −ξ)

  X  X (−1)|δ|  1   ∂β σ (x, ξ)Dβ+λ ∂λ σ (x, ξ) ∼  x  β! ξ Q ξ p ν!β!δ! β,λ ν+δ=λ

But

P

v+δ=λ

(−1)|δ| v!δ!

= (x0 − x0 )λ with x0 = (1, 1, . . . , 1)    1 if λ = 0 =  0 if λ , 0

Therefore σQP (x, ξ) ∼

P 1 β β ∂ξ σQ (x, ξ)D x σ p (x, ξ). β β!

Corollary 4.37 (PRODUCT RULES FOR ψDO′ S ). If Q = q(x, D)ǫψm (λ) and f ǫC ∞ (Ω), then for any positive integer N, 105 P 1 α D f [(∂αξ q)(x, D)u] + T N u where T N ǫψm−N (Ω). q(x, D)( f u) = |α| x = e2πi(ξ−η).x p(x, ξ)û(ξ)dxdξ Z = pˆ 1 (η − ξ, ξ)û(ξ)dξ and hence, if VǫS , < Pu, v¯ > = where

Z

(Pu)ˆvˆ =

x

K(η, ξ) f (ξ)¯g(η)dξdη

f (ξ) = (1 + |ξ|2 ) s/2 uˆ (ξ)

g(η) = (1 + |η|2 )(m−s)/2 vˆ (η) and K(η, ξ) = pˆ 1 (η − ξ, ξ)(1 + |ξ|2 )−s/2 (1 + |η|2 )(s−m)/2 . 107 We wish to estimate K(η, ξ). For any multi-index α, since p(., ξ)ǫCo∞ , we have Z α |ζ pˆ 1 (ζ, ξ)| = | (Dαx e−2πix.ζ) p(x, ξ)dx| Z = | e−2πix.ζ) (Dαx p(x, ξ)dx| ≤ Cα (1 + |ξ|)m , so that for any positive integer N, | pˆ 1 (ζ, ξ)| ≤ C N (1 + |ξ|)m (1 + |ζ|)−N


92 and hence

|K(η, ξ)| ≤ C N (1 + |ξ|)m−s (1 + |η|)s−m (1 + |ξ − η|)N ≤ C N (1 + |ξ|)|m−s|−N

If we take N > n + |m − s|, we see that Z Z |K(η, ξ)|dη ≤ C, |K(η, ξ)|dξ ≤ C. Therefore by Theorem 1.1 and the Schwarz inequality, | < Pu, v¯ > | ≤ C|| f ||2 ||g||2 = C||u||(s) ||v||(m−s) .

108

From this, it follows that ||Pu||(s−m) ≤ C||u||(s) for uǫH s which establishes (4.41) and hence (i). Now suppose P is properly supported and uǫH sloc (Ω). If φǫCo∞ (Ω) there exists Ω′ ⋐ Ω such that the values of Pu on supp φ depend only the values of u on Ω. Thus if we pick φ′ ǫCo∞ (Ω) with φ′ = 1 on Ω′ , we have φPu = φP(φ′ u). But φ′ uǫH s and (4.41) φP is bounded from H s to H s−m . This establishes (ii) and completes the proof.

8 Elliptic Pseudo Differential Operators Definition 4.42. Pǫψm (Ω) is said to be elliptic of order m if |σP (x, ξ)| ≥ CΩ , |ξ|m for large |ξ|, for all xǫΩ′ , Ω′ ⋐ Ω. Definition 4.43. If Pǫψm (Ω), a left (resp. right ) parametrix of P is a ψDOQ such that QP − I(resp. PQ -I) is smoothing. Theorem 4.44. If Pǫψm is elliptic of order m and properly supported, then there exists a properly supported Qǫψ−m which is a two-sided parametrix for P. Proof. Let P = p(x, D). We will obtain Q = q(x, D) with q ∼

∞ P

j=0

qj

where the q′j s are defined recursively. Let ζ(x, ξ) be a C ∞ function such that ζ(x, ξ) = 1 for large ξ and ζ(x, ξ) = 0 in a neighbourhood of the

8. Elliptic Pseudo Differential Operators zeros of p. Define qo (x, ξ) =

93

ζ(x, ξ) . Then qo ǫ s−m . Let Qo = qo (x, D). p(x, ξ)

We have p

mod S −1 = 1 mod S −1 = 1 + r1 with r1 ǫS −1 .

σQ p = q0 O

Let q1 =

−r1 ζ , Q1 = q1 (x, D). Then p

σ(Q0 +Q1 )P = σQ p + σQ p = 1 + r1 − r1 ζ O

1

mod S −2

= 1 + r2 with r2 ǫS −2 .

Let q1 =

set

−r 2ζ

p

etc. Having determined q1 , q2 , . . . q j so that

σ(Q0 +Q1 +···Q j )P = 1 + r j+1 , r j+1 ǫS −( j+1) −r j+1 ζ . q j+1 = p Let Q be a properly supported operator with symbol q ∼

∞ P

109

q j.

j=0

Then σQP = 1 mod S −∞ , i.e., QP − I is smoothing. Thus Q is a left parametrix. In the same way, we can construct a right parametrix Q′ . Then if S = QP − I and S ′ = PQ′ − I, we have QPQ′ = S Q′ + Q′ = QS ′ + Q. So Q − Q′ = S Q′ − QS ’ is smoothing. Hence Q is a two sided parametrix. Exercise. What happens if P is not properly supported? (cf. the remarks at the end of §6). The left parametrices are used to prove regularity theorems and the right parametrices are used to prove existence theorems. Indeed, we have:


94

Corollary 4.45 (ELLIPTIC REGUIARITY THEOREM). If P is elliptic loc (Ω). In particular, P is of order m, uǫD′ (Ω), PuǫH sloc (Ω) implies uǫH s+m hypoelliptic. Proof. Let Q be a left parametrix and set S = QP − I. Then u = QPu − S u. SincePuǫH sloc (Ω) and Q is properly supported, Qǫψ−m we have loc (Ω) by Theorem 4.40. Also S uǫC ∞ since S is smoothing. QPuǫH s+m loc (Ω). Hence uǫH s+m 110

Theorem 4.46. Every elliptic differential operator is locally solvable. In fact, if P is elliptic on Ω, f ǫD′ (Ω) and xo ǫΩ, there exists uǫD′ (Ω) such that Pu = f in a neighbourhood of xo . (Of course, if f ǫC ∞ , then uǫC ∞ near xo , by the previous corollary). Proof. BY cutting f off away from xo , we can assume that f ǫE ′ and hence f ǫH s for some s. Let Q be a properly supported parametrix for P and let S = PQ − I. Then S is also properly supported. If Ω1 ⊂⊂ Ω is a neighbourhood of xo then there exists Ω2 ⊂⊂ Ω such that the values of Su on Ω1 depend only on the values of u on Ω2 . Pick φ1 , φ2 ǫCo∞ (Ω) such that φ j ≡ 1 on Ω j , j = 1, 2 and set T u = φ1 S (φ2 u). Now observe the following : i) T u = su on Ω1 . ii) T : H s → Co∞ (suup φ1 ) continuously. From (ii) and the Arzela -Ascoli theorem, it follows that if (uk ) is a bounded sequence in H s′ (T uk ) has a convergent subsequence in Co∞ (suup φ1 ) and hence in H s . Therefore, T is compact on H s . So the equation (T + I)u = f can be solved if f is orthogonal (with respect to the pairing of H s and H−s ) to the space N = {g : (T ∗ + I)g = 0}. This space N is a finite dimensional space of C ∞ functions so we can always make this happen by modifying f outside a small neighbourhood of xo . Indeed, pick a basis g1 , g2 , . . . , gν for N and pick a neighbourhood U of xo so small that g1 , . . . , gν are linearly independent as functionals on Co∞ (Ω\U). (Such a U exists; otherwise, by a limiting argument using the local compactness of N, we could find a nontrivial linear combination of

9. Wavefront Sets

95

g1 , . . . , gν supported at {xo }, which is absurd). Then we can make f ⊥ N by adding to f a function in Co∞ (Ω\U). But then PQu = (I + S )u = (I + T )u on Ω1 and (I + T )u = f in a neighbourhood of xo . So Qu solves the problem.

111

9 Wavefront Sets We now introduce the notion of wavefront sets, which provides a precise way of describing the singularities of distributions: it specifies not only the points at which a distribution is not smooth but the directions in which it is not smooth. All pseudo differential operators encountered in this section will be presumed to be properly supported, and by “ψDO” we shall always mean “properly supported ψDO”. Definition 4.47. (i) Let Ω be an open set in Rn . Then we define T o Ω = Ω × (Rn \{0}). (In coordinate - invariant terms, T o Ω is the cotangent bundle of Ω with the zero section removed). (ii) A set S ⊂ T o Ω is called conic if (x, ξ)ǫS ⇒ (x, rξ)ǫS , ∀r > 0 (iii) Suppose P = p(x, D)ǫψm (Ω). Then (xo , ξo )ǫT o Ω is said to be noncharacteristic for P if |p(x, ξ)| ≥ c|ξ|m for |ξ| large and (x, ξ) in some conic neighbourhood of (xo , ξo ). (iv) The characteristic variety of P, denoted by char P, is defined by char P = {(x, ξ)ǫT o Ω : (x, ξ) is characteristic for p }. (v) Let uǫD′ (Ω). The wavefront set of u, denoted by W F(u) is defined by W F(U) = ∩{ charP : Pǫψ(Ω), PuǫC ∞ (Ω)}.

The restriction Pǫψ0 (Ω) in the definition of W F(u) is merely a con- 112 venient normalisation. We could allow ψDO of arbitrary order without changing anything, for if Pǫψm (Ω) and Qǫψ−m (Ω) is elliptic, then QPǫψ0 (Ω), char (QP) = char (P)(by Theorem 4.36), and QPuǫC ∞ if and only if PuǫC ∞ (by Corollaries 4.11 and 4.45).


96 Exercise. When p =

P

|α|=≤m

aα (x)ξ α show that the characteristic variety

of the differential operator p = P(x, D) is X char P = {(x, ξ) : aα (x)ξ α = 0}. |α|=m

The motivation for W F(u) is as follows. If uǫE ′ , to say that u is not smooth in the direction ξo should mean that uˆ is not rapidly decreasing on the through ξo . On the other hand, if PuǫCo∞ then (Pu)ˆ should be rapidly decreasing everywhere. If these conditions both hold, then ξo must be characteristic for P. Localising these ideas, we arrive at W F(u). Thus (xo ξo )ǫW F(u) means roughly that u fails to be C ∞ at xo in the direction ξo . For another interpretation of this statement, see Theorem 4.56 below. For the present, we show that W F(u) is related to sing supp u as it should be. We denote the projection T o Ω onto Ω by π. Theorem 4.48. For uǫD′ (Ω), sing supp u = π(W F(u)). 113

Proof. Suppose x0 < sin g supp u. Then there exists φǫCo∞ with φ = 1 near xo and φuǫCo∞ . But multiplication by φ is a ψDO of order 0. Call it Pφ . Then char Pφ = π−1 (φ−1 (0)). Since φ = 1 near xo , π−1 (φ−1 (0)) is disjoint from π−1 (xo ). Therefore, π−1 (xo ) ∩ W F(u) = φ, i.e., xo < π(W F(u)). Conversely, suppose xo < π(W F(u)). Then for each ξ with |ξ| = 1, there exists Pǫψ0 (Ω) with PuǫC ∞ (Ω) and (xo , ξ) < char P. Each char P is a closed conic set; so, by the compactness of the unit sphere, there exists a finite number of ψDO′ s say, P1 , . . . PN ǫψ0 (Ω) with P j uǫC ∞ (Ω) and   N N \  X  ∩ π−1 (x ) = φ. Set P =  char P P∗j P j . j o   j=1

j=1

Then P is elliptic near xo and PuǫC ∞ (Ω). Therefore by Corollary 4.45, u is C ∞ near xo , i.e., xo < sin g supp u.

9. Wavefront Sets

97

Definition 4.49. Let P = p(x, D)ǫψm (Ω) and U be an open conic set in T o Ω. We say that P has order −∞ on U, if, for all closed conic sets K ⊂ U with π(K) compact for every positive integer N and multi-indices α and β, there exist constants CαβKN such that β

|D x Dαξ p(x, ξ)| ≤ CαβKN (1 + |ξ|)−N , (x, ξ)ǫK. The essential support of P is defined to be the smallest closed conic set outside of which P has order −∞. Exercise . If P is a differential operator whose coefficients do not all vanish at any point of Ω, show that the essential support of P is T Ω. Proposition 4.50. Ess. supp PQ ⊂ Ess. supp P∩ Ess. supp Q.

114

Proof. This follows immediately from the expansion 2

σPQ ∼

X 1 ∂α σP Dαx σC . α! ξ

Lemma 4.51. Let (xo , ξo )ǫT o Ω and U be any conic neighbourhood of (xo , ξo ). Then there exists a Pǫψ0 (Ω) such that (xo , ξo ) < char P but Ess. supp P ⊂ U. Proof. Choose po (ξ)ǫC ∞ with the following properties: i) po (ξ) = 1 for ξ near ξo , po (ξ) = 0 outside {ξ : (xo ξ)ǫU} and ii) po is homogeneous of degree 0 for large |ξ|. Then take φǫC ∞ (π(U)) with φ = 1 near xo and put p(x, ξ) = po (ξ) φ(x). This will do the job. The following theorem and its corollary are refinements of Corollary 4.11 (pseudo local property of ψDO) and Corollary 4.45 (elliptic regularity theorem). Theorem 4.52. If Pǫψm (Ω) and uǫD′ (Ω), then W F(Pu) ⊂ W F(u)∩ Ess. supp P.

98


Proof. If (xo , ξo ) < Ess. supp P, by Lemma 4.51, we can find a Qǫψ0 such that (xo , ξo ) < char Q and Ess. supp P∩ Ess. supp Q = φ Then QPǫψ−∞ by Proposition 4.50, so that QPǫC ∞ . But this implies that (xo , ξo ) < W F(Pu). Suppose now that (xo , ξo ) < W F(u). Then there exists Aǫψ0 with AuǫC ∞ and (xo , ξo ) < char A. 115

Claim . There exist operators B, Cǫψ0 with (xo , ξo ) < char B and BP = CA mod ψ−∞ . Granted this, BPu = CAu + (C ∞ function) ǫC ∞ which implies that (xo , ξo ) < W F(Pu). To prove the claim, let Ao be elliptic with σA = σA on a conic neighbourhood U of (xo , ξo ). Then Ess. supp(A − Ao ) ∩ U = φ. By Lemma 4.51, choose B with (xo , ξo ) < char B and Ess. supp B ⊂ U. Let Eo be a parametrix for Ao and set C = BPEo .Then CA = BPEo A = BPEo (A − Ao ) + BPEo Ao . Since Eo Ao = I mod ψ−∞ and BPEo (A − Ao ) also belongs to ψ−∞ (by Proposition 4.50 again), CA = BP mod ψ−∞ . This completes the proof. Corollary 4.53. If P is elliptic, then W F(u) = W F(Pu)). Proof. By the theorem, W F(Pu) ⊂ W F(u). If E is a parametrix for P, then W F(u) = W F(EPu + C ∞ function) = W F(EPu) ⊂ W F(Pu), by the theorem again Hence W F(u) = W F(Pu).

2

Theorem 4.56. Let uǫD′ (D). Then (xo , ξo ) < W F(u) if and only if there exists φǫCo∞ (Rn ) such that φ(xo ) = 1 and (φu)ˆ is rapidly decreasing on a conic neighbourhood of ξo .

116

Proof. (Sufficiency) If such a φ exists, choose pǫS 0 , p(x, ξ) = p(ξ) with p(ξ) = 1 near ξo and p(ξ) = 0 outside the region where (φu)ˆ is rapidly decreasing. Then p(φu)ˆǫS and hence p(D)(φu)ǫS where p(D) = p(x, D). The operator Pu = φp(D)(φu) is a pseudo differential operator of order 0 with symbol φ(x)2 p(ξ) modulo S −1 , so (xo , ξo ) < char P. This implies that (xo , ξo ) < W F(u).

10. Some Further Applications...

99

(Necessity) Suppose (xo , ξo ) < W F(u). Then there exists a neighbourhood U of xo in Ω such that (xo , ξo ) < W F(u) for all xǫU. Choose φǫCo∞ (U) such that φ(xo ) = 1. Then (xo , ξo ) < W F(φu) for all xǫRn . Let X = {ξ : (x, ξ)ǫW F(φu) for somex}. P This does not contain ξo and is a closed conic set. There exists p(ξ)ǫS 0 , p = 1 on a conic neighbourhood of ξo and p = 0 on a neighP P P bourhood of , say o . Since p(ξ) = 0 on o , Ess. supp p(D) ∩ (Rn × P c o ) and hence Ess. supp p(D) ∩ W F(φu) = φ. But this gives Ess. supp(p(D)φu) = φ. So p(D)φuǫC ∞ . We now claim that p(D)φuǫS . Accepting this, we see that p(φu)ˆǫS and in particular, (φu)ˆ is rapidly decreasing near ξo . To prove the claim, observe that n Dβ (ξ α p(ξ)) = 0(1 + |ξ|)|α|−|β| ǫL2 if |α| − |β| < − . 2 So if we put K(x) = p˜ (x), xβ Dα K(x)ǫL2 if |α| − |β| < − n2 . An application of Leibniz’s rule and the the Sobolev imbedding theorem shows that xβ Dα K(x)ǫL∞ if |β| − |α| > − 3n 2 + 2. Hence K and all its derivatives are rapidly decreasing at ∞ and since φuǫE ′ , the same is true of p(D)φu = φu ∗ K.

10 Some Further Applications of Pseudo Defferential Operators We conclude our discussion of pseudo differential operators by giving 117 brief and informal descriptions of some further applications. We recall that an mth order ordinary differential equation u(m) = F(x, u, u(1) , . . . u(m−1) ) can be reduced to the first order system     u2  u1        u3 d  u2     ..  =  ..  dx  .   .     um F(x, u1 , . . . , um )

simply by introducing the derivatives of u of order < m as new variables, and this reduction is frequently a useful technical device. To do

100


something similar for mth order partial differential equations, however, is more problematical. If one simply introduces all partial derivatives of u of order < m as new variables and writes down the first order differential relations they satisfy, one usually obtains more equations than there are unknowns, because of the equality of the mixed partials. Moreover, such a reduction usually does not preserve the character of the original equation. For example take the Laplace equation in two variables : ∂2 u ∂2 u + = f. ∂x2 ∂y2 putting u1 = u, u2 =

∂ ∂u , u3 = we have a 3 × 3 system given by ∂x ∂y ∂u1 − u2 = 0 ∂x ∂u1 − u3 = 0 ∂y ∂u2 ∂u3 + = f. ∂x ∂y

118

The original equation is elliptic. But consider the matrix of the top order symbols of the 3 × 3 system obtained above. The matrix is given by   ξ 0 0   2πi η 0 0   0 ξ η

which is not invertible. This means that the first order system is not elliptic. There is, however, a method, due to A.P. CALDERON, of reducing an mth order linear partial differential equation to an m × m system of first order pseudo differential equations which preserves the characteristic variety of the equation in a sense which we shall make precise below. In this method, one of the variables is singled out to play a special role, so we shall suppose that we are working on Rn+1 with coordinates (x1 , x2 , . . . , xn , t).


101

Let L be a partial differential operator of order m on Rn+1 such that the coefficient of ∂m t is nowhere vanishing. Dividing throughout by this coefficient, we can assume that L is of the following form: L=

∂m t

−

m−1 X

j

Am− j (x, t, D x )∂t .

j=0

119

Here Am− j is a differential operator of order ≤ m − j in the x variable with coefficients depending on t. We want to reduce the equation Lv = f to a first order system. To this end we proceed as follows: Using the familiar operators Λ s with symbol (1 + |ξ|)s/2 (acting in the x variables ) we put u1 = Λm−1 v u2 = Λm−2 ∂tv u3 = Λm−3 ∂2t v .. . um = ∂m−1 v. t For j < m, we observe that ∂t u j = Λu j+1 . Therefore, ∂t um = f +

m−1 X

Am− j (x, t, D x )∂t v

= f+

m−1 X

Am− j (x, t, D x )Λ j−m+1 u j+1

j

j=0

= f+

j=0 m X

Am− j (x, t, D x )Λ j−m u j

j=1

Set B j(x, t, D x ) = Am− j+1 (x, t, D x )Λ j−m . This B j is a ΨDO of order 1 in the variable x. Then the equation Lv = f is equivalent to ∂t u = Ku + f˜


102

where u = (u1 , . . . , um )t , f˜ = (0, 0, . . . , f )t . (Here (· · · )t denotes the 120 transpose of the row vector (· · · )) and K is the matrix given by   0 · · · 0   0 Λ   0 0 Λ · · · 0     . . · · ·   K =   . . · · ·    0 · · · Λ   0 0  B1 B2 B3 · · · Bm This K is a matrix of first order pseudo differential operators in x, with coefficients depending on t. Let us now make a little digression to fix on some notations which will be used in the further development. For a pǫS m , a function pm (x, ξ) homogeneous of degree m in ξ is said to be the Principal symbol of p(x, D) if p − pm agrees with an element of S m−1 for large |ξ|. We remark that not all ΨDO′ s have a principal symbol; but most of them that arise in practice do. Moreover, the principal symbol, if it exists, is clearly unique. Examples (i) If p is a polynomial, p(x, ξ) =

X

aα (x)ξ α ,

X

aα (x)ξ α .

|α|≤m

then pm (x, ξ) =

|α|=m

121

(ii) If p(x, ξ) = (1 + |ξ|2 )s/2 , ps (x, ξ) = |ξ| s . Returning to our discussion, let ak (x, t, ξ) be the principal symbol of Ak (x, t, dx ) (Here we are considering Ak as an operator of order k, so if it happens to be of lower order, its principal symbol is zero). Then


103

the principal symbol of B j is b j (x, t ξ) = am− j+1 (x, t, ξ)|ξ| j−m and the principal symbol of k is the matrix K1    0 |ξ| 0 0 · · · 0     0 0 |ξ| 0 · · · 0   0 0 0 |ξ| · · · 0      . . . ···  K1 (x, t, ξ) =  . . ···   .   . . . · · ·    0 0 0 0 · · · |ξ|    b1 b2 b3 b4 · · · bm The principal symbol of L, on the other hand, is Lm (x, t, ξ, τ) = (2πiτ)m −

m−1 X

am− j (x, t, ξ)(2πiτ) j

j=0

The characteristic variety of L, i.e., the set of zeros of Lm , can easily be read off from the matrix K as follows: Proposition 4.57. The eigenvalues of K1 (x, t, ξ) are precisely 2πiτ1 , . . ., 2πiτm where τ1 , τ2 , . . . , τm are the roots of the polynomial Lm (x, t, ξ, .). Proof. The characteristic polynomial p(λ) of K1 is the determinant of   0   λ −|ξ| 0 · · ·   0 λ −|ξ| · · · 0     ..   .    0 0 · · · −|ξ|  0  ˙ −b1 −b2 −b3 λ − bm Expanding in minors along the last row, we get p(λ) =

m−1 X j=1

(−1)m+ j (−b j )λ j−1 (−|ξ|)m− j + λm−1 (λ − bm )

= λm − λm−1 bm −

m−1 X j=1

b j |ξ|m−1 λ j−1

122


104 = λm −

m X j=1

am− j+1 λ j−1 = λm −

m−1 X

am− j λ j

j=0

−1

= Lm (x, t, ξ, (2πi) λλ. This proves the proposition.

It is also easy to incorporate boundary conditions into this scheme. Suppose we want to solve the equation Lv = f with the boundary conditions B j v|t=0 = g j , 1, 2, . . . , ν where Bj =

mj X k=0

bkj (x, t, D x )∂kt , bkj is of order m j − k and m j ≤ m − 1.

If we set k−m Bkj = Λm−m j −1 bk−1 , j (x, 0, D x )Λ

φ j = Λm−m j −1 g j j−1

then with u j = Λm− j ∂t v as above and u0j (x) = u j (x, 0), the boundary conditions will become m j +1 X

Bkj u0k = φ j , j = 1, 2, . . . ν.

k=1

123

This is a system of zeroth order pseudo differential equations. The above method of reduction is useful, for example, in the following two important problems. (1) Cauchy Problem for Hyperbolic Equations Lv = f, ∂tj v|t=0 = g j , j = 0, 1, . . . , m − 1. Here the equation is said to be hyperbolic when the eigenvalues of the matrix K1 (i.e., the principal symbol of K occurring in the linear system corresponding to Lv = f ) are purely imaginary.Equivalently, the roots of Lm (x, t, ξ, .) are real.


105

(2) Elliptic Boundary Value Prob Lems Lu = f on Ω, B ju = g j on ∂Ω where L is elliptic of order 2m and j = 1, 2, . . . m. Here one works locally near a point x0 ǫ∂Ω and makes a change of coordinates so that ∂Ω becomes a hyperplane near xo . One then can apply Calderon’s reduction technique, taking t as the variable normal to ∂Ω. (See Michael Taylor [3]). We shall now sketch an example of a somewhat different technique for applying ψDO′ S to elliptic boundary value problems. Let Ω be a bounded open set of Rn with C ∞ boundary ∂Ω Consider the problem ∆u = 0 in Ω, ∂X u + au = g on ∂Ω, aǫC ∞ (∂Ω). Here X is a real a vector field on the boundary which is nowhere 124 tangent to the boundary and ∂X u = grad u.χ. If ν is the unit outward normal to ∂Ω, by normalising we can assume that χ = ν + τ where τ is tangent to ∂Ω. We want to use ΨDO to reduce this to the Dirichlet problem. Pretend for the moment that ∂Ω = ∂ . Rn−1 x{0} and Ω = {x : xn < 0} so that ∂ν = ∂xn n−1 P ∂2 u = −∆b u, ∆b being the Laplacian on If ∆u = 0 then ∂2ν u = − 2 j=1 ∂x j √ the boundary. So formally ∂ν u = ± (∆b u). Indeed, if we compare this with our discussion of the Poisson kernel in Chapter 2(taking account of the fact that Ω is now the lower rather than the upper half space), we √ see that the equation ∂ν u = (−∆b )u is correct if interpreted in terms of the Fourier transform in the variables x′ = (x1 , . . . xn−1 ); ∂ν u˜ (ξ ′ , xn ) = 2π|ξ ′ |˜u(ξ ′ , xn ). It turns out that something similar works for our original domain Ω. ¯ and ∆u = 0 on Ω then ∂ν u|∂Ω = P(u|∂Ω ) Namely, if u is smooth on Ω where P is a pseudo differential operator of order 1 on ∂Ω which equals √ (−∆b ) modulo terms of order ≤ 0 where ∆b is the Laplace -Beltrami operator on ∂Ω. In particular, P is elliptic and has real principal symbol.

106

125


The boundary conditions become Pv + |∂t v + av = g, v = u|∂Ω . This is a first order pseudo differential equation for ν. It is elliptic (because ∂τ has imaginary symbol). So it can be solved modulo smoothing operators. Since ∂Ω is compact. smoothing operators on ∂Ω are compact, so our boundary equation can be solved provided g is orthogonal to some finite dimensional space of smooth functions. Having done this, we have reduced our original problem to the familiar Dirichlet problem ∆u = 0 in Ω and u|∂Ω = ν which has a unique solution.

Chapter 5 LP and Lipschitz Estimates OUR AIM IN this chapter is to study how to measure the smoothing 126 properties of pseudo differential operators of non positive order in terms of various important function spaces. Most of the interesting results are obtained by considering operators of order −λ with 0 ≤ λ ≤ n. Indeed, if PǫΨ−λ with λ > n and K(x, y) is the distribution kernel of P then, by Theorem 4.10 we know that KǫC j Ω × Ω) when j < λ − n. So these operators can be studied by elementary methods. What is more, when PǫΨ−λ , Dα PǫΨ−λ+|α| . So, by a proper choice of α, we can make 0 ≤ λ − |α| ≤ n and then study Dα P rather than P itself. Actually, we shall restrict attention to operators of order −λ where 0 ≤ λ < n. The transitional case λ = n requires a separate treatment to obtain sharp results; however, for many purpose, it suffices to make the trite observation that an operator of order −n can be regarded as an operator of order −n + ǫ. Further, we restrict our attention to ΨDO′ s whose symbols have asymptotic expansions p(x, ξ) ∼

∞ X

p j (x, ξ)

j=0

where P j is homogeneous of degree −λ− j. (These p′j s no longer belong to our symbol classes, being singular at ξ = 0, but we can still consider the operators p j (x, D).) By the preceding remarks, it will suffice to consider the operators corresponding to the individual terms in the series 127 107

5. LP and Lipschitz Estimates

108 P

p j whose degrees of homogeneity are between 0 and −n. Thus we are looking at p(x, D) where p(x, ξ) is C ∞ on Rn × Rn \{0} and homogeneous of degree −λ in ξ where 0 ≤ λ < n. Since λ < n, for each x, p(x, .) is locally integrable at the origin and hence defines a tempered distribution. Denoting the inverse Fourier transform of this distribution p∨2 (x, .) we then have p(x, D)u(x) =

Z

e2πix.ξ p(x, ξ)û(ξ)dξ

= (p∨2 (x, .) ∗ u)(x) Z = p∨2 (x, x − y)u(y)dy. Thus we see that the distribution kernel of p(x, D) is given by K(x, y) = pv2 (x, x − y). Let us digress a little to make some remarks on homogeneous distributions. Definition 5.1. A distribution f ǫS ′ is said to be homogeneous of degree µ, i f < f, φr >= rµ < f, φ > for all φǫS where φr is the function defined by φr (x) = r−n φ(x/r). Exercise. 1. Show that the above definition agrees with the usual definition of homogeneity when f is a locally integrable function. 2. Show that if f ǫS ′ is homogeneous of degree µ, then Dα f is homogeneous of degree µ − |α|. 128

3. Show that Dα δ is homogeneous of degree −n − |α| (where δ is the Dirac measure at 0). Let us now prove a proposition concerning the Fourier transform of homogeneous distributions. It turns out that the Fourier transform of such a distribution is also a homogeneous one. Precisely, we have the following

109 Proposition 5.2. If f is a tempered distribution, homogeneous of degree µ, then fˆ is homogeneous of degree −µ − n. If f is also C ∞ away from the origin, then the same is true of fˆ. Proof If f is homogeneous of degree µ, then for φǫS

2

< fˆ, φr > =< f, (φr )ˆ >=< f, r−n φˆ 1/r >= r−n−µ < f, φˆ > = r−µ−n < fˆ, φ > .

This proves the first assertion. To prove the second assertion, choose φǫC0∞ with φ = 1 in a neighbourhood of the origin and write f = φ f + (1 − φ) f . Since φ f ǫE ′ , (φ f )ˆ is C ∞ everywhere. On the other hand, (1 − φ) f is C ∞ and homogeneous of degree µ for large ξ, and hence lies in S µ (Rn ). Let p(x, ξ) = (1 − φ)(ξ) f (ξ). The distribution kernel of p(x, D) is given by Z K(x, y) = e−2πi(y−x),ξ (1 − φ)(ξ) f (ξ)dξ = ((1 − φ) f )ˆ(y − x).

Since K is C ∞ away from the diagonal, we see that ((1 − φ) f )îs C ∞ away from the origin. Hence fˆ is C ∞ on Rn \{0}. Returning to our operators p(x, D) with p(x, ξ) homogeneous of de- 129 gree −λ in ξ, 0 ≤ λ < n, by the proposition above, we have Z p(x, D)u = K(x, x − y)u(y)dy where K = P∨2 is C ∞ on Rn × Rn \{0} and homogeneous of degree λ − n in the second variable. Finally, we restrict attention to constant coefficient case, i.e., K(x, x− y) = K(x − y). The essential ideas are already present in this case and the results we shall obtain can be generalised to the variable coefficient case in a rather routine fashion. Let us give a name to the objects we are finally going to study. Definition 5.3. A tempered distribution K which is homogeneous of degree λ − n and C ∞ away from the origin is called a kernel of type λ. If K is a kernel of type λ, the operator T f = K ∗ f is called an operator of type λ.


110

We now classify the kernels of type λ ≥ 0. Proposition 5.4. Suppose λ > 0 and f ǫC ∞ (Rn \{0}) is homogeneous of degree λ − n in the sense of functions. Then f is locally integrable and defines a distribution F which is a kernel of type λ. Conversely, if F is a kernel of type R λ and f is the function with which F agrees on Rn \{0},then < F, φ >= f φ for every φ in S .

130

Proof. Since λ > 0, f is locally integrable and of (at most ) polynomial growth at ∞, so f defines an F in S ′ . It is easy to check that F is homogeneous in the sense defined above, i.e.< F, φr >= rλ−n < F, φ >, and so F is a kernel of type λ. R For the converse, define G by < G, φ >=< F, φ > − f φ, φ ∈ S . P Then G is a distribution supported at 0 and hence we have G = cα Dα δ. Therefore, X < G, φr >= cα r−n−|α| (Dα φ)(0) = O(r−n ), as r → ∞. On the other hand, < G, φr >= r−n−λ < G, φ > and this is not O(r−n ) as r tends to ∞ unless < G, φ >= 0. Hence G = 0 and this completes the proof. Suppose that f ǫC ∞ (Rn \{0}) is homogeneous of degree −n. Then f is not locally integrable near 0 and so does not define a distribution in a trivial way. However, let us define Z f (x)dσ(x). µf = |x|=1

If µ f = 0 there is a canonical distribution associated with f which is called principal value of f, PV( f ), defined by Z f (X)dx. < PV( f ), φ >= lim ǫ→0 |x|>ǫ

111 To see that this limit exists, we observe that Z

f (x)dx = µ f

Z1

r−1 dr = 0.

ǫ

ǫ= |x|>1

|x|≤1

By the same argument as above, G is a tempered distribution and P G = F on Rn /{0}. Therefore, G − F = cα Dα δ. Now, < F, φr >= r−n < P, φ > so that < G, φr > −r−n < G, φ >=< G − Fφr > −r−n < G − F, φ >= 0(r−n ) 132

as r tends to ∞. On the other hand, for r > 1, Z −n < G, φr > − r < G, φ >= r−n f (x)(φ(x) − φ(0))dx +

Z

|x|≤1/r

r

−n

f (x)φ(x)dx − r

−n

Z

|x|≤1

|x|>1/r

= r−n φ(0)

Z

f (x)(φ(x) − φ(0))dx − r−n Z

f (x)φ(x)dx

|x|>1

f (x)dx = r−n φ(0) log rµ f ′ for every φ.

1r≤|x|≤1

This is not 0(r−n ) as r tends to ∞ unless µ f = 0. Finally, F − PV( f ) is a kernel of type 0 which is supported at the origin and hence is a multiple of δ. To study the boundedness of operators of type λ on L p spaces, we need some concepts from measure theory. Definition 5.6. Let F be a measurable function defined on Rn . Then the distribution function of f is the function δ f : (0, ∞) → [0, ∞] given by δ f (t) = |Et |, where Et = {x : | f (x)| > t}. From the distribution function of f , we can get a large amount of information regarding f . For example, we have Z Z∞ p | f (x)| dx = − t p dδ f (t) 0

113 (To see this, observe that the Riemann sums for the Stieltjes integral on the right are approximating sums for the Lebesque integral on the left). If t p δ f (t) converges to 0 as t tends to 0 and t tends to ∞, we can R∞ R integrate by parts to obtain | f (x)| p dx = p t p−1 δ f (t)dt. 0

Using the concept of distribution functions, we will now define weak 133 L p spaces.

Definition 5.7. For 1 ≤ p < ∞ we define weak L p as { f : δ f (t) ≤ (c/t) p for some constant c}. For f ǫ weak L p , the smallest such constant c will be denoted by [ f ] p . Thus, if f ǫweak L p , we have δ f (t) ≤ ([ f ] p /t) p . For p = ∞, we set weak L∞ = L∞ . Proposition 5.8. CHEBYSHEV ’S INEQUALITY) L p ⊂ weak L p and [ f ] p ≤ || f || p . Proof. For f ǫL p , if Et = {x : | f (x)| > t}, Z Z p || f || p = | f (x)| p dx ≥ | f (x)| p dx ≥ t p |Et | Rn

i.e.,

Et

|Et | ≤ (|| f || p /t)p. From this, it follows that f ǫ weak L p and [ f ] p ≤ || f || p .

Remark 5.9. It is not true that L p = weak L p for 1 ≤ p < ∞. For example, f (x) = |x|−n/p belongs to weak L p but not L p . Remark 5.10. The function f → [ f ] p satisfies [c f ] p = |c|[ f ] p . But it fails to satisfy triangle inequality and hence is not a norm. However, since {| f + g| > t} ⊂ {| f |t/2} ∪ {|g| > t/2}, we have δ f +g (t) ≤ δ f (t/2) + δg (t/2)


114

which gives, when f and g are in weak L p , p

p

δ f +g (t) ≤ (2 p [ f ] p + 2 p [g] p )/t p so that

[ f + g] p ≤ 2([ f ] pp + [g] pp )1/p ≤ 2([ f ] p + [g] p ). The functional [.] p thus defines a topology on weak L p which will turn it into a (non-locally convex) topological vector space. Definition 5.11. A linear operator T defined on a space of functions is said to be of weak type (p, q), 1 ≤ p, q ≤ ∞, if T is a bounded linear operator from L p into weak Lq , i.e. for every f ǫL p , [T f ]q ≤ c|| f || p for some constant c independent of f . We will now state the Marcinkiewicz interpolation theorem and use it to prove generalisations of Young’s inequality (Theorem 1.3). Theorem 5.12 (J. Marcinkiewicz). Suppose T is of weak types (po , qo ) and (p1 , q1 ) with 1 ≤ pi ≤ qi ≤ ∞, p0 < p1 , q0 , q1 , i.e., [′ Γ f ]qi ≤ ci || f || pi for i = 0, 1. Then, if 1/pθ = (1 − θ)/p0 + (θ/p1 ), 1/qθ = (1 − θ)/q0 + (θ/q1 ), 0 < θ < 1, T is bounded from L pθ to Lqθ i.e., ||T f ||qθ ≤ cθ || f || pθ where the constant cθ depends only on p0 , q0 , p1 , q1 , c0 , c1 and θ. For the proof of this theorem, see A. Zygmund [5] or E.M.Stein [2]. Theorem 5.13 (GENERAL FROM OF YOUNG’S INEQUALITY). If (1/p) + (1/q) − 1 = 1/r, 1 ≤ p, q, r < ∞, Then, for f ǫL p , gǫLq f ∗ gǫLr and we have || f ∗ g||r ≤ c pq || f || p ||g||q . 135

(In fact, c pq ≤ 1 for all p, q, although our proof does not yield this estimate). Proof. Fixing f ǫL p , consider the convolution operator g → T q = f ∗ g. We know that for gǫL1 , T gǫL p and ||T g|| p ≤ || f || p ||g||1 . Further if p′ is the conjugate of p, then, by Hölder’s inequality ||T g||∞ ≤ ||g|| p , || f || p ′ for gǫL p . Thus we see that the operator T is of weak types (1, p) and

134

115 (p′ , ∞). Therefore, by the Marcinkiewicz interpolation theorem, T maps L pθ boundedly into Lqθ for every 0 < θ < 1 with (1/pθ ) = 1 − θ + (θ/p′ ) = 1 − (θ/p) and 1/qθ = (1/p) − θ/p. Given r with (1/p) + (1/q) − 1 = (1/r), set qθ=r . Then (1/pθ ) = 1 + (1/r) − (1/p) = (1/q). Hence we get the required result. Theorem 5.14 (WEAK TYPE YOUNG’S INEQUALITY). Let 1 ≤ p < q < ∞, (1/p) + (1/q) > 1 and (1/p) + (1/q) − 1 = (1/r). Suppose f ǫL p and gǫ weak Lq . Then, we have a) f ∗ g exists a.e. and is in weak Lr , and [ f ∗ g]r ≤ c pq || f || p [g]q b) If p > 1, then f ∗ gǫLr and || f ∗ g||r ≤ c′pq || f || p [g]q . Proof. a) It suffices to assume that || f || p = [g] p = 1 and to show that [ f ∗ g]r ≤ c pq . Given α > 0, let ′

′

′

M = (α/2) p /(p −q) (p′ /(p′ − q)−1/(p −q) where p′ is as usual the conjugate of p. Define g1 (x) = g(x), if |g(x)| > M = 0 otherwise

and

g2 (x) = g(x) − g1 (x).

Then 136

δ f ∗g (α) ≤ δ f ∗g1 (α/2) + δ f ∗g2 (α/2) and we shall estimate the quantities on the right separately. By Holder’s inequality, || f ∗ g2 ||∞ ≤ || f || p ||g2 || p′ = ||g2 || p′ .


116

Since (1/q) − (1/p′ ) = 1/r > 0, we see that p′ − q > 0, and hence p′ ||g2 || p′

′

= p

Z∞

= p′

ZM

t p −1 (δg (t) − δg (M))dt

0 ZM

t p −1 t−qdt = (p′ /(p′ − q))M p −q = (α/2) p .

′

t p −1 δg2 (t)dt

0

≤ p′

′

′

′

′

0

Thus ( f ∗ g2 )(x) exists at all points and || f ∗ g2 ||∞ ≤ α/2.

Consequently δ f ∗g2 (α/2) = 0. Next consider ||g1 ||1 =

Z∞

≤

ZM

δg1 (t)dt =

0

R∞

δg (M)dt +

M −q dt +

Z∞

Z∞

δg (t)dt

M

0

0

The integral

ZM

t−q dt.

M

t−q dt converges since q > 1, and we obtain

M

||g1 ||1 ≤ M 1−q + M 1−q /(q − 1) = (q/(q − 1))M 1−q . Also, by Chebyshev inequality, ! α || f ∗ g1 || ≤ p p δ f ∗g1 1 α/2

p′

≤ (2 p /α p )(q/(q − 1)) p (α/2) p′ −q

= c pq α−r 137

Hence (a) is proved.

p(1−q)

− p(1−q) p′ −q

(p′ /(p′ − q))

117 b) The operator f → f ∗ g is of weak type (1, q) by (a). Also, if we choose p¯ > p with (1/ p) ¯ + (1/q) > 1 and put (1¯r) = (1/ p) ¯ + (1/q) − 1 then by (a), f → f ∗ g is of weak type ( p¯ , r¯). Therefore, by Marcinkiewicz, f → f ∗ g is bounded from L p into Lqθ with 1/pθ = 1 − θ + (θ/ p¯ ), (1/qθ ) = ((1 − θ)/q) + (θ/¯r). Putting qθ = r we get pθ = p. Hence T maps L p continuously to Lr and || f ∗ g||r ≤ c′pq || f || p [g]q . Corollary 5.15. If T is an operator of type λ, 0 < λ < n, then T is bounded from L p into Lq , whenever 1 < p < q < ∞ and 1/q = 1/p−λ/n. Also, T is of weak type (1, n/(n − λ)). Proof. If K is a kernel of type λ, we have |K(x)| = |x|λ−n K(x/|x|)| ≤ c|x|λ−n , which implies that Kǫ weak Ln/(n−λ) . Therefore, by Theorem 5.14(b), if f ǫL p and T f = K ∗ f then T f ǫLq where (1/q) = (1/p) + ((n − λ(/n) − 1 = (1/p) − (λ/n) and ||T f ||q ≤ c[ f ] p ≤ c|| f || p .

(a).

Also we see that T is of weak type (1, n/(n − λ)) by Theorem 5.14

The limiting case of this result with λ = 0 is also true, but this is a 138 much deeper theorem: Theorem 5.16. (Calderon - Zygmund) Operators of type 0 are bounded on L p , 1 < p < ∞. Proof. Let T be an operator of type 0 with kernel K i.e., T f = K ∗ f . To ∞ begin with, we can regard T as a map from c∞ o to c and we shall show that ||T f || p ≤ c p || f || p for f ǫc∞ o , 1 < p < ∞, so that T extends uniquely to a bounded operator on L p . We have K = PV(k) + cδ so that T f = PV(k) ∗ f + c f . Since the identity operator is continuous, we shall assume that c = 0 and also we shall identify K with k. The proof now proceed in several steps.


118

Step 1. T is bounded on L2 . Indeed, we have (T f )ˆt = kˆ fˆ.kˆ is smooth away from 0 and homogeneous of degree 0, hence is bounded on Rn Therefore, ˆ ∞ || fˆ||2 = ||k|| ˆ ∞ || f ||2 . ||T f ||2 = ||(T f )ˆ||2 ≤ ||k|| 1 Step 2. Fix a radial function φǫc∞ and φ(x) = 0 o with φ(x) = 1 for |x| ≤ 2 for |x| ≥ 1. For ǫ > 0, we define kǫ (x) = k(x)(1−φ(x/ǫ)) and T ǫ f = kǫ ∗ f . Then we claim that T ǫ is bounded on L2 uniformly in ǫ. To see this, we observe that (T ǫ f )ˆ = fˆkˆ ǫ = fˆ(k − kφ(x/ǫ))ˆ = fˆkˆ − fˆ(kˆ ∗ (φ(x/ǫ))ˆ) which gives ˆ ∞ {1 + ǫ n ||T ǫ f ||2 = ||(T ǫ f )ˆ||2 ≤ || fˆ||2 ||k||

Z

ˆ |φ(ǫξ)|dξ}

ˆ ∞ {1 + ||φ|| ˆ 1} = || f ||2 ||k|| 139

Step 3. T ǫ is of weak type (1,1) uniformly in ǫ. The proof of this is more involved and will be given later. Step 4. By steps 2 and 3, using Marcinkiewicz, we get that T ǫ is bounded on L p for 1 < p < 2 uniformly in ǫ. Step 5. T ǫ is bounded on L p , for 2 < p < ∞, uniformly in ǫ. Indeed, for f, gǫCo∞ , Z Z (T ǫ f ) (x) g(x) dx = f (x) (T˜ǫ g) (x) dx with T˜ǫ g = k˜ ǫ ∗ g, k˜ ǫ = kǫ (−x) = k(−x)(1 − φ(x/ǫ)). Since k˜ ǫ satisfies the same conditions as kǫ , we see that T˜ ǫ is bounded on Lq for 1 < q < 2 uniformly in ǫ. So if (1/p) + (1/q) = 1, 1 < q < 2, R | (T ǫ f )g| ||T˜ǫ g||q ≤ sup || f || p = c|| f || p . ||T ǫ f || p = sup ||g||q gǫCo∞ ||g||q g ǫ Co∞

119 Step 6. If f ǫCo∞ , T ǫ f converges to T f in the L p norm, 1 ≤ p ≤ ∞ as ǫ tends to 0. Since φ is radial and µk = 0, Z Z k(x)dσ(x) = 0; φ(x/ǫ)k(x)dσ(x) = c |x|=r

|x|=r

so, for ǫ ≤ 1, we have Z Z (T ǫ f )(x) = ( f (x − y) − f (x)) k(y)(1 − φ(y/ǫ))dy + f (x − y) k(y)dy |y|≤1

and hence

(T ǫ f )(x) − (T f )(x) =

Z

|y|>1

f (x − y) − f (x)k(y)φ(y/ǫ)dy.

|y|≤ǫ

Now supp(T ǫ f − T f ) ⊂ {x : d(x, supp f ) ≤ ǫ} ⊂ A, a fixed compact set. Since, on compact sets, the uniform norm dominates all L p norms, it suffices to show that (T ǫ f − T f ) converges to 0 uniformly on A. But Z c|y||y|−n dy ≤ c′ || grad f ||∞ ||T ǫ f − f ||∞ ≤ || grad f ||∞ |y|≤ǫ

Zǫ

r1−n rn−1 dr = c′′ ǫ → 0, as ǫ → 0.

0

Step 7. If f ǫL p and η > 0, choose gǫCo∞ with ||g − f || p < η. Then ||T ǫ f − T δ f || p ≤ ||T ǫ ( f − g)|| p + ||T ǫ g − T δ g|| p + ||T δ (g − f )|| p ≤ 2cη + ||T ǫ g − T δ g|| p . Since η is arbitrary and ||T ǫ g − T δ g|| p converges to 0 as ǫ, δ tend to 0 by step 6, we see that (T ǫ f ) is Cauchy in the L p norm. Setting T f = lim T ǫ f, ||T f || p ≤ c|| f || p . ǫ→0

Thus the theorem is proved modulo Step 3. Let us now proceed to the proof of Step 3. First, we need a lemma which will be used in the proof.

140


120

Lemma 5.17. Suppose FǫL1 , F ≥ 0 and α > 0. Then there exists a sequence (Qk ) of closed cubes with sides parallel to the coordinate axes 141 and disjoint interiors such that a) α
0, let (Qk ) be the sequence of cubes as in the lemma with F = | f |. We write f = g+

∞ X

bk with

k=1

 R   f (x) − |Q1k | f (y)dy, for xǫQk    Qk bk (x) =     0, otherwise  1 R      |Qk | f (y)dy, for xǫQk g(x) =  Qk     f (x), for x < Ω.

and

Now me make the following observations : R a) Supp bk ⊂ Qk , bk = 0 and (5.18)

∞ X 1

Z ∞ ∞ X X ||bk ||1 ≤ 2 |f| ≤ 2 2n α|Qk | ≤ 2n+1 || f ||1 1

Qk

1

143


122

b) |g(x)| ≤ 2n α for xǫΩ and |g(x)| ≤ α for a.e. x < Ω. Therefore, |g(x)| ≤ 2n α a.e. and Z Z (5.19) |g|2 |g|2 + ||g||22 = Rn −Ω

Ω

≤ (2n α)2 |Ω| + α 2n

Z

Rn −Ω

|| f ||

≤ (2 + 1)α|| f ||1 . We put

∞ P 1

bk = b so that T ǫ f = T ǫ g + T ǫ b and

|{|T ǫ f | > α}| ≤ |{|T ǫ g| > α/2}| + |{|T ǫ b| > α/2}|. We shall show that both terms on the right are dominated by || f ||1/α uniformly in ǫ. To estimate the first term on the right, we use Chebyshev inequality, Step 2 and the estimate (5.19) obtaining |{|T ǫ g| > α/2}| ≤ (2/α)||T ǫ g||2 )2 ≤ c(||g||2 /α)2 ≤ c1 (|| f ||1 α). To estimate the second term, let yk be the center of the cube Qk and √ Q˜ k be the cube centred at yk but with length side 2 n times that of Qk . ∞ S ˜ Then We put Q˜ k = Ω. 1

˜ ≤ |Ω| So

144

∞ X 1

∞ √ n X √ n ˜ | Qk | = 2 n |Qk | ≤ 2 n || f ||1 /α = c(|| f ||1 /α). 1

˜ + |{|T ǫ b| > α/2}\Ω| ˜ |{|T ǫ b| > α/2} ≤ |Ω|

˜ ≤ c(|| f ||1 /α) + |{|T ǫ b| > α/2}\Ω|, R ˜ Since b = 0, and it suffices to estimate |{|T ǫ | > α/2}/Ω. Qk

T ǫ b(x) =

Z

kǫ (x − y)b(y)dy

123

=

=

∞ Z X

1 Q k ∞ XZ 1 Q k

kǫ (x − y)b(y)dy (kǫ (x − y) − kǫ (x − yk ))b(y)dy

Therefore, we have ˜ |{|T ǫ b| > α/2}\Ω| Z ≤ (2/α) |T ǫ b|dx ˜ Rn \Ω ∞ X

≤ (2/α)

≤ (2/α)

1

Z Z

|(kǫ (x − y) − kǫ (x − yk ))||b(y)|dy dx

˜ Qk Rn \Ω

∞ Z X

Z

1 Q n ˜ k R \ Qk

|(kǫ (x − y) − kǫ (x − yk ))||b(y)|dx dy

We R now claim that |(kǫ (x − y) − kǫ (x − yk ))|dx ≤ c independent of k and ǫ for yǫQk .

Rn \Q˜ k

Accepting this claim, by the estimate (5.18), we have ∞ Z X ˜ ≤ (2/α)c |{|T ǫ | > α/2}\Ω| |b(y)|dy ≤ 2n+2 c(|| f ||1 /α). 1 Q k

3.

Thus |{|T ǫ b| > α}| ≤ co (|| f ||1 /α) which completes the proof of Step

Returning to the claim, we observe that if xǫRn \Q˜ k and yǫQk , then |x − yk | ≥ 2|y − yk |. So, if we set z = x − yk , w = y − yk we must show that 145 Z |(kǫ (z − w) − kǫ (z))|dz ≤ c |z|>2|w|

independent of w and ǫ.


124

First consider the case ǫ = 1. Now k1 (z) is a C ∞ function which is homogeneous of degree −n for large z. Therefore, |gradk1 (z)| ≤ c′′ |z|−n−1 . By the mean value theorem, |k1 (z − w) − k1 (z)| ≤ c′ |w| sup |z − tw|−n−1 02|w|

Z

|w||z|−n−1 dz

≤ c′′′ |w|

r−2 dr = c.

′′

≤c

|z|>2|w| Z∞ 2|w|

Now for general ǫ, kǫ (z) = (1 − φ(z/ǫ))k(z/ǫ)ǫ −n by the homogeneity of k. Therefore, if we set z′ = ǫ −1 z and w′ = ǫ −1 w, we see that Z Z |(k1 (z′ − w′ ) − k1 (z′ ))|dz′ |(kǫ (z − w) − kǫ (z))|dz = |z|>2|w|

146

|z′ |>2|w′ |

which is bounded by a constant, by the result for ǫ = 1. Hence the claim above is proved. To complete the picture, we should observe that operators of type

125 0 are not bounded on L1 (and hence, by duality, not bounded on L∞ ). Indeed, if T is an operator of type 0 with kernel k, (T f )ˆ = kˆ fˆ. Since fˆ is homogeneous of degree 0, it has a discontinuity at 0 (unless k = cδ). Thus if f ǫL1 , (T f )ˆ is not continuous at 0 whenever fˆ(0) , 0 and this implies that T f is not in L1 . R This reflects the fact that if f = fˆ(0) , 0, then T f will not be integrable near ∞, because k is itself not integrable at ∞. However, there are also problems with the local integrability of T f caused by the singularity of k at the origin. In fact, let φǫCo∞ be a radial function such that φ = 1 near 0, and set S f = f ∗ (φk). Then the argument used to prove Theorem 5.16 shows that S is bounded on L p for 1 < p < ∞. In this case (φk)ˆ = φˆ ∗ kˆ is in C ∞ , but still S is not bounded on L1 . This follows from the following general fact. Proposition 5.20. If kǫS ′ and the operator f → k ∗ f is bounded on L1 , then k is necessarily a finite Borel measure. R Proof. Choose φǫCo∞ with φ = 1 and put φǫ (x) = ǫ −n φ(x/ǫ). Then ||φǫ ||1 is independent of ǫ, so ||φǫ ∗ k||1 ≤ c. Therefore there exists a sequence ǫk tending to 0 such that φǫk ∗ k converges to a finite Borel measure µ in the weak* topology of measures and hence φǫk ∗k converges to µ in S ′ also. On the other hand, since (φǫk ) is an approximate identity, φǫk ∗ k converges to k in S ′ . Hence µ = k and the proposition is proved. 147 It is easy to see that kernels of type 0 are not measures even when truncated away from the origin, as their total variation in any neighbourhood of 0 is infinite. One can also see directly that they do not define bounded functionals on Co . Exercise . Let k(x) = 1/x on R and let f be a continuous compactly 1 supported function such that f (x) = (log x)−1 for 0 < x < and f (x) = 2 0 for x ≤ 0. Show that lim ( f ∗ PV(k))(x) = ∞.

x→0−


126

Generalise this to kernels of type 0 on Rn , n > 1. This example also shows that operators of type 0 do not map continuous functions into continuous functions. However, they do preserve Lipschitz or Hölder continuity, as we shall now see. Definition 5.21. For 0 < α < 1, we define | f |α = sup x,y

| f (x + y) − f (x)| |y|α

Λα = { f : || f ||Λα =de f || f ||∞ + | f |α < ∞}. Λα is called Lipschitz class of order α. Remark 5.22. The definition makes perfectly good sense for α = 1 (When α > 1 it is an easy exercise to show that if | f |α < ∞ then f is constant ). However, we shall not use this definition for α = 1, because the theorems we wish to prove are false in this case.

148

We are going to prove, essentially, that operators of type 0 are bounded on Λα . However, if k is a kernel of type 0 and f ǫΛα , the integral defining k ∗ f will usually diverge f need not decay at ∞. Consequently, we shall work instead with Λα ∩ L p (1 < p < ∞), concerning which we have the following useful result. Proposition 5.23. If f ǫL p , 1 ≤ p < ∞ and | f |α < ∞, then f ǫΛα and || f ||∞ ≤ c(|| f || p + | f |α ). Consequently, L p ∩ Λα is a Banach space with norm || f || p + | f |α . Proof. Let Ax = (

| f (x)| 1/α ) for xǫRn . 2| f |α

Then | f (y)| ≥ | f (x)|/2 for all y such that |x − y| ≤ A x . So Z Z p | f (y)| p dy ≥ | f (x)| p 2−p c′ Anx | f (y)| dy ≥ |x−y|≤A x

= c′′ | f (x)| p+(n/α) | f |−n/α α or

n/alphap

| f (x)|1+(n/αp) ≤ c′′′ || f || p | f |α

.

2

127 Since this is true for all x, setting θ = n/pα we have || f ||∞ ≤ c′′′ || f ||1/(1+θ) | f |θ/(1+θ) p α ≤ c(|| f || p + | f |α ).

Theorem 5.24. Operators of type 0 are bounded on Λα ∩ L p (0 < α < 1, < 1 < p < ∞). Proof. Let T : f → K ∗ f be an operator of type 0. Since we know that ||T f || p ≤ c p || f || p , by proposition 5.23, it will suffice to show |T f |α ≤ cα | f |α for 0 < α < 1, f ǫL p ∩ Λalpha . As in the proof of Theorem 5.16, 149 we may assume that K = PV(k) and identify K with k. Given yǫRn \{0} and f ǫL p ∩ Λα define Z k(z) f (x − z)dz g(x) = |z|≤3|y|

Z

h(x) =

k(z) f (x − z)dz

|z|≤3|y|

so that T f = g + h. Since µk = 0, we have Z |g(x)| = | k(z)( f (x − z) − f (x))dz| |z|≤3|y|

≤

Z

c|z|−n | f |α |z|α dz ≤ c1 | f |α |y|α .

|z|≤3|y|

Since this is true for all x, |g(x + y) − g(x)| ≤ 2c1 | f |α |y|α . Next h(x + y) = lim

Z

η→∞ 3|y|