Lectures on

Partial Di

fferential Equations

G.B. Folland

Tata Institute of Fundamental Research

Bombay

1983

Lectures on

Partial Di

fferential Equations

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 with-
out 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 Appli-
cation 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 cover-
ing 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 distribu-
tions 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

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 painstak-
ing job of writing up the notes.

Contents

Preface

Preliminaries

General Theorems About Convolutions

. . . . . . . . .

The Fourier Transform . . . . . . . . . . . . . . . . . .

Some Results From the Theory of Distributions . . . . . 12

Partial Di

fferential Operators with Constant Coefficients

Local Solvability and Fundamental Solution . . . . . . . 15

Regularity Properties of Differential Operators . . . . . . 20

Basic Operators in Mathematical Physics

. . . . . . . . 22

Laplace Operator . . . . . . . . . . . . . . . . . . . . . 25

The Heat Operator

. . . . . . . . . . . . . . . . . . . . 41

The Wave Operator . . . . . . . . . . . . . . . . . . . . 44

Sobolev Spaces

General Theory of L

Sobolev Spaces . . . . . . . . . . 53

Hypoelliptic Operators With Constant Coefficients . . . 64

Basic Theory of Pseudo Di

fferential Operators

Representation of Pseudo differential Operators . . . . . 69

Distribution Kernels and the Pseudo Local Property . . . 73

Asymptotic Expansions of Symbols . . . . . . . . . . . 77

Properly Supported Operators

. . . . . . . . . . . . . . 79

ψdo

′

s Defined by Multiple Symbols . . . . . . . . . . . 81

vii

viii

Contents

Products and Adjoint of ψDO

′

. . . . . . . . . . . . . 87

A Continuity Theorem for ψ Do on Sobolev Spaces . . . 90

Elliptic Pseudo Differential Operators . . . . . . . . . . 92

Wavefront Sets . . . . . . . . . . . . . . . . . . . . . . 95

Some Further Applications... . . . . . . . . . . . . . . . 99

and Lipschitz Estimates

107

Chapter 1

Preliminaries

IN THIS CHAPTER, we will study some basic results about convolu-

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 R

× R

such that, for

some c > 0,

R |K(x, y)|dy ≤ c, R |K(x, y)|dx ≤ c, for every x, y in R

If 1

≤ p ≤ ∞ and f ∈ L

), then the function T f , defined by

T f (x) =

K(x, y) f (y)dy for almost every x in R

, belongs to L

)

and further,

||T f ||

≤ c|| f ||

Proof. If p =

∞, the hypothesis

R |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 H ¨older ’s inequality,

|T f (x)| ≤

|K(x, y)|dy

)

1/q

|K(x, y)| | f (y)|

)

1/p

1. Preliminaries

≤ c

1/q

|K(x, y)| | f (y)|

)

1/p

From this we have,

|T f (x)|

≤ c

p/q

|K(x, y)|| f (y)|

dydx

≤ c

1+p/q

| f (y)|

dx = c

1+p/q

|| f ||

p
p

Therefore

||T f ||

≤ c|| f ||

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

( 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

∈ L

) and g

∈ L

), for

≤ p ≤ ∞. Then f ∗ g ∈ L

) and

|| f ∗ g||

≤ ||g||

|| f ||

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 con-

volution. Before coming to the theorem, let us prove the following

Lemma 1.4. For a function f defined on R

and x in R

, we define a

function f

by f

(y) = f (y

− x). If f ∈ L

, 1

≤ p < ∞, then lim

→O

|| f

−

= 0.

1. General Theorems About Convolutions

Proof. If g is a compactly supported continuous function, then g is uni-

formly continuous, and so g

converges to g uniformly as x tends to

Further, for

|x| ≤ 1, g

and g are supported in a common compact

set. Therefore, lim

→0

||g

− g||

= 0. Given f

∈ L

, we can find a function

g which is continuous and compactly supported such that

|| f −g||

∈ /3

for

∈> 0. But then ||g

− f

∈ /3 also holds. Therefore

|| f

− f ||

≤ || f

− g

||g

− g||

||g − f ||

≤ 2 ∈ /3 + ||g

− g||

Since lim x

→ 0||g

− g||

= 0, we can choose x close to 0 so that

||g

− g||

∈ /3. Then || f

− f || <∈ and this proves the lemme since ∈ is

arbitrary.

Remark 1.5. The above lemma is false when p =

∞. Indeed, “ f

→ f

in L

∞

” means precisely that f is uniformly continuous.

Let us now make two important observations about convolutions

which we shall use without comment later on.

i) Supp( f

∗ g) ⊂ Supp f + Supp g, where

A + B =

{x + 1 : x ∈ A, y ∈ B}.

ii) If f is of class C

and ∂

(

|α| ≤ k) and g satisfy appropriate condi-

tions so that differentiation under the integral sign is justified, then

∗ g is of class C

and ∂

( f

∗ g) = (∂

f )

∗ g.

Theorem 1.6. Let g

∈ L

) and

g(x)dx = a. Let g

∈

(x) =

∈

−n

g(x/

∈)

for

∈> 0. Then, we have the following:

i) If f

∈ L

), p <

∞, f ∗ g

∈

converges to a f in L

∈ tends to 0.

ii) If f is bounded and continuous, then f

∗g

∈

converge to a f uniformly

on compact sets as

∈ tends to 0.

1. Preliminaries

Proof. By the change of variable x

→∈ x, we see that

∈

(x)dx = a for

all

∈> 0. Now

( f

∗ g

∈

)(x)

− a f (x) =

f (x

− y)g

∈

(y)dy

−

f (x)g

∈

(y)dy

[ f (x

− y) − f (x)] g

∈

(y)dy

[ f (x

− ∈ y) − f (x)]g(y)dy

[ f

∈y

(x)

− f (x)]g(y)dy.

If f

∈ L

and p <

∞, we apply Minkoswski’s inequality for integrals

to obtain

|| f ∗ g

∈

− a f ||

≤

|| f

∈y

− f ||

|g(y)|dy.

The function y

→ || f

∈y

− f ||

is bounded by 2

|| f ||

and tends to 0 as

∈

tends to 0 for each y, by lemma 1.4. Therefore, we can apply Lebesgue
Dominated Convergence theorem to get the desired result.

On the other hand, suppose f is bounded and continuous. Let K be

any compact subset of R

. Given δ > 0, choose a compact set G R

such

that

−G

|g(y)|dy < δ.

Then

Sup

∈k

(

| f ∗ g

)(x)

− a f (x)| ≤ 2δ|| f ||

∞

+ Sup

(x,y)ǫ K

×G

| f (x − ǫy) − f (x)|

|g|dy.

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 ∗ g

)(x)

− a f (x)| → 0 as ǫ → 0.

Hence the theorem is proved.

2. The Fourier Transform

Corollary 1.7. The space C

∞

) is dense in L

⋉

) for 1

≤ p < ∞.

Proof. Let

φ(x) = e

−1/(1−|x|

)

for

|x| < 1

= 0

for

|x| ≤ 1

Then φǫC

∞

) and

φ(x)dx = 1/a > 0. If f ǫL

and has compact

support, then a( f

∗ φ

)ǫC

∞

) and by theorem 1.6, a( f

∗ φ

) converges

to f in L

as ǫ tends to 0. Since L

functions with compact support are

dense in L

, this completes the proof.

Proposition 1.8. Suppose K

⊂ R

is compact and Ω

⊃ K be an open

subset of R

. Then there exists a C

∞

function φ such that φ(x) = 1 for

xǫK and Supp φ

⊂ Ω.

Proof. Let V =

{x ∈ Ω : d(x, K) ≤

} where δ = d(K, R

\Ω). Choose

a φ

ǫC

∞

such that Supp φ

⊂ B

and

(x)dx = 1. Define

φ(x) =

− y)dy = (φ

∗ X

)(x).

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 ǫL

), the Fourier transform of the function f ,

denoted by ˆ

f , is defined by

f (ξ) =

−2πix.ξ

f (x)dx, ξǫR

Remark 1.9. Our definition of ˆ

f differs from some other in the place-

ment of the factor of 2π.

1. Preliminaries

BASIC PROPERTIES OF THE FOURIER TRANSFORM

For

(1.10)

f ǫL

|| ˆf||

∞

≤ || f ||

The proof of this is trivial.
For

(1.11)

f, g

∈ L

, ( f

∗ g)

(ξ) = ˆf(ξ)ˆg(ξ).

Indeed,

( f

∗ g)

(ξ) =

−2πix.ξ

f (y)g(x

− y)dydx

−2πi(x−y)·ξ

g(x

− y)e

−2πiY.ξ

f (y)dydx

−2πi(x−y)·ξ

g(x

− y)dx

f (y)e

−2πiy.ξ

= ˆf(ξ)ˆg(ξ).

Let us now consider the Fourier transform in the Schwartz class S =

S (R

Proposition 1.12. For f

∈ S , we have the following:

i) ˆ

f ǫC

∞

) and ∂

f = ˆg where g(x) = (

−2πix)

f (x).

ii) (∂

f )

(ξ) = (2πiξ)

f (ξ).

Proof.

i) Differentiation under the integral sign proves this.

ii) For this, we use integration by parts.

(∂

f )

(ξ) =

−2πix.ξ

(∂

f )(x)dx

= (

−1)

|β|

∂

−2πix.ξ

] f (x)dx

= (

−1)

|β|

(

−2πiξ)

−2πix·ξ

f (x)dx

= (2πiξ)

f (ξ).

2. The Fourier Transform

Corollary 1.13. If f ǫS , then ˆ

∈ S also.

Proof. For multi-indices α and β, using proposition 1.12, we have

(∂

f )(ξ) = ξ

((

−2πix)

f (x))

(ξ)

= (2πi)

−|α|

[∂

((

−2πix)

f (x))

(ξ)]

= (

−1)

|β|

(2πi)

|β|−|α|

(∂

f (x)))

(ξ)

Since f ǫS , ∂

f (x))ǫL

and hence (∂

f (x)))

ǫL

∞

. Thus ξ

(∂

f ) is bounded. Since α and β are arbitrary, this proves that ˆ

f ǫS .

Corollary 1.14 (RIEMANN-LEBESGUE LEMMA). If f ǫL

, then ˆ

f is

continuous and vanishes at

∞.

Proof. By Corollary 1.13, this is true for f ǫS . Since S is dense in L

and

|| ˆf||

∞

≤ || f ||

, the same in true for all f ǫL

Let us now compute the Fourier transform of the Gaussian.

Theorem 1.15. Let f (x) = e

−πa|x|

, Re a > 0. Then, ˆ

f (ξ) = a

−n/2

−a

−1

|ξ|

Proof.

f (ξ) =

−2πix.ξ

−aπ|x|

i.e.,

f (ξ) =

j=1

∞

−∞

−2πix

−aπx

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

Thus we are assuming f (x) = e

−πx

, xǫR. Observe that f

′

(x) +

2πx f (x) = 0. Taking the Fourier transform, we obtain

2πiξ ˆ

f (ξ) + i ˆf

′′

(ξ) = 0.

Hence

ˆf

′′

(ξ)/ f (ξ) =

−2πξ

1. Preliminaries

which, on integration, gives f (ξ) = ce

−πξ

, c being a constant.

The constant c is given by

c = ˆf(0) =

∞

−∞

−πx

dx = 1.

Therefore ˆ

f (ξ) = e

−πξ

, which completes the proof.

We now derive the Fourier inversion formula for the Schwartz class

S .

Let us define f

(ξ) =

2πix

·ξ

f (x)dx = ˆf(

−ξ).

Theorem 1.16 (Fouries Inversion Theorem). For f ǫ( ˆ

f )

∨

= f .

Proof. First, observe that for f, gǫL

f ˆg =

f g. In fact,

f (x)g(x)dx =

−2πiy·x

f (y)g(x)dydx

−2πiy·x

g(x)dx

f (y)dy

ˆg(y) f (y)dy.

Given ǫ > 0 and x in R

, take the function φ defined by

φ(ξ) = e

−2πix·ξ−πǫ

|ξ|

Now

φ(y) =

−2πiy·ξ

−2πix·ξ−πǫ

|ξ|

dξ

−2πi(y−x)·ξ

−πǫ

|ξ|

dξ

∈

−n

−π∈

−2

|x−y|

If we take g(x) = e

−π|x|

and define g

(x) = ǫ

−n

g(x/ǫ), then

φ(y) = g

− y).

2. The Fourier Transform

Therefore

2πix

·ξ

f (ξ)e

−πǫ

|ξ|

dξ =

f (ǫ)φ(ξ)dξ

f (y) ˆ

φ(y)dy

f (y)g

− y)dy

= ( f

∗ g

)(x)

But as ǫ tends to 0, ( f

∗ g

) converges to f , by Theorem 1.6 and

clearly

2πi

·xξ

f (ξ)e

−πǫ

|ξ|

dξ

→

f (ξ)e

2πi

·x.ξ

dξ

Therefore ( ˆ

f )

= 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 L

) onto itself.

Proof. For f ǫS , define ˜

f (x) = f (

−x). Then it is easily checked that

ˆ˜f = ¯ˆf, so that

|| f ||

2
2

| f (x)|

f (x) ˜

f (

−x)dx

= ( f

∗ ˜f)(0)

( f

∗

f )

(ξ)dξ

f (ξ) ˆ˜

f (ξ)dξ

f (ξ) ¯ˆ

f (ξ)dξ =

|| ˆf||

2
2

1. Preliminaries

Therefore, the Fourier transform extends continuously to an isom-

etry of L

. It is a unitary transformation, since its image S is dense in

Let us observe how the Fourier transform interacts with translations,

rotations and dilations.
(1.19)

The Fourier transform and translation: If f

(y) f (y

− x) then

(ξ) =

2πiy

·ξ

f (y

− x)dy

2πi(z+x)

·ξ

f (z)dz ( by putting y

− x = z)

2πix

·ξ

f (ξ).

(1.20)

The Fourier transform and rotations (orthogonal transfor-

mations):

Let T :

→

be an orthogonal transformation. Then

( f

◦ T )

(ξ) =

−2πix·ξ

( f oT )(x)dx

−2πiT

−1

·ξ

f (y)dy ( by putting y = T x)

−2πiy·T ξ

f (y)dy

= ˆf(T ξ) = ( ˆfoT )(ξ).

Thus, ( f oT )

= ˆfoT i.e. ˆ commutes with rotations.

(1.21)

The Fourier transform and dilation: Let f

(x) = r

−n

f (x/r).

Then

(ξ) =

−2πix·ξ

−n

f (x/r)dx

−2πiy·ξ

f (y)dy = ˆf(rξ).

The last equation suggests, roughly: the more spread out f is, the

more ˆ

f 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)

HEISENBERG INEQUALITY (n = 1: For f ǫS (R), we have

||x f (x)||

||ξ ˆf(ξ)||

≥ (1/4π)|| f ||

2
2

Proof. Observe that

(x f (x)) = x

d f

(x) + f (x).

Thus

|| f ||

2
2

f (x) f (x)dx

f (x)

" d

(x f (x))

− x

d f

(x)

−

x f (x)

d ¯

(x)dx

−

d f

(x) ¯

f (x)dx

−2Re

x f (x)

d ¯

(x)dx.

≤ 2||x f (x)||

d ¯

(by Cauchy-Schwarz)

i.e.,

|| f ||

2
2

≤ 2||x f (x)||

d f

But

(

d ¯

)

(ξ) = 2πiξ ˆf(ξ).

Therefore,

|| f ||

2
2

≤ 2.2π||x f (x)||

||ξ ˆf(ξ)||

||x f (x)||

||ξ f (ξ)||

≥ (1/4π)|| f ||

2
2

1. Preliminaries

A GENERALISATION TO n VARIABLES

We replace x by x

∂

∂x

. Also for any a

, b

ǫ, we can re-

place x

and

∂

∂x

by x

− a

and

∂

∂x

− b

respectively. The same proof

then yields:

(1.23)

||(x

− a

) f (x)

||(ξ

− b

) ˆ

f (ξ)

≥ (1/4π)|| f ||

2
2

Let us now take

|| f ||

= 1 and A = (a

, a

, . . . , a

)ǫ

. Let f be

small outside a small nighbourhood of A.

In this case,

||(x

− a

) f (x)

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 ˆ

f cannot be concentrated near

any point.

Remark 1.24. If we take

| f (x)|

dx, b

| ˆf(ξ)|

dξ,

then inequality (1.23) is the mathematical formulation of the position-
momentum uncertainty relation in Quantum Mechanics.

3 Some Results From the Theory of Distributions

In this section, let us recall briefly some results from the theory of dis-
tributions. (For a more detailed treatment, see, for example [7] or [8]).

In the sequel,

′

(Ω) will denote the space of distributions on the

open set Ω

⊂ R

which is the dual space of C

∞

(Ω). When Ω =

, we

will simply write

′

instead of

′

(

). In the same way, S

′

= S

′

)

will denote the space of tempered distributions with and E

′

= E

′

(

)

will stand for the space of distributions with compact support.

The value of a distribution uǫ

′

at a function φǫC

∞

will be denoted

by < u, φ >. If u is a locally integrable function, then u defines a distri-
bution by < u, φ >=

u(x)φ(x)dx. It will sometimes be convenient to

3. Some Results From the Theory of Distributions

write

u(x)φ(x)dx for < u, φ >, when u is an arbitrary distribution.

The convergence in

′

is the weak convergence defined by the fol-

lowing:

, uǫ

′

, u

→ in means < u

, φ >

→< u, φ > for every φ in C

∞

Let us now recall briefly certain operations on distributions.

(1.25)

We can multiply a distribution uǫ

′

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

α, ∂

φ grows at most polynomially at

∞. We can multiply an u, ǫS

′

a tempered function to get another tempered distribution. The definition
is same as in the previous case.
(1.26)

If uǫ

′

and f ǫC

∞

, we define the convolution u

∗ f by

∗ f )(x) =< u, f

> where f

(x) = f (x

− y).

The function u

∗ f is C

∞

and when uǫE, u

∗ f is in C

∞

. The convo-

lution

∗ : D

′

× C

∞

→ C

∞

can be extended to a map from

′

× E

′

Namely, if uǫ

′

, vǫE

′

and φǫC

∞

, < u

∗ v, φ >=< u, ˜v ∗ φ > where ˜v is

defined by < ˜v, ψ >=

v(x)ψ(

−x)dx. The associative law

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

′

provided that at most one of them does not have compact support. For

uǫS

′

and f ǫS , u

∗ f can also defined in the same way and u ∗ f is a

tempered C

∞

function.

(1.27)

Since the Fourier transform is an isomorphism of S onto

S . and

f ˆg =

f g, the Fourier transform extends by duality to an

isomorphism of S

′

onto S

′

For uǫE

′

⊂ S

′

, we have ˆu(ξ) =< u, e

−2πi(·)·ξ

> which is an entire

analytic function.

Chapter 2

Partial Di

fferential

Operators with Constant
Coe

fficients

1 Local Solvability and Fundamental Solution

For the sake of convenience in taking the Fourier transforms, from now

onwards, we will use the differential monomials

= (2πi)

|α|

∂

. Thus (D

)

(ξ) = ξ

f (ξ).

By a partial differential operator with constant coefficients,we mean

a differential operator L of the form

L =

|α|≤k

, a

Further, we assume that

|α|≤k

| , 0. In this case, we say that the

operator L is of order k. If we write P(ξ) =

|α|≤k

, then we have

L = P(D).

For uǫS , taking the Fourier transform, we see that

(P(D)u)

(ξ) =

|α|≤k

ˆu(ξ) = P(ξ)ˆu(ξ).

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

◦

if there is a solution of the above problem in some neighbourhood of the
point x

◦

for any f in C

∞

Remark 2.1. We can assume that f has compact support. To see this,

we can take any φ in C

∞

such that φ = 1 in some nighbourhood of the

point x

. 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 x

In the following theorem we give an affirmative answer to the ques-

tion of local solvability of L = P(D). The simple proof exhibited here is
due to L. Nirenberg.

Theorem 2.2. Let L =

|α|≤k

be a differential operator with con-

stant coefficients. If uǫC

∞

, there 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)ˆu(ξ) = ˆf(ξ). It is natural to try to define u by the formula

u(x) =

2πix

·ξ

f (ξ)/P(ξ)ξ.

In general, P will have many zeros: so there will be a problem in ap-
plying the inverse Fourier transform to ˆ

f /P. But things are not so bad.

Since f ǫC

∞

◦

, ˆ

f is an entire function of ξǫ

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

|α|≤k

0. By a rotation of coordinates, we can assume that η = (0, 0, . . . , 0, 1).
Multiplying by a constant, we can also assume that a

α0

= 1 where

= (0, 0, . . . , 0, k). Then we have P(ξ) = ξ

+(lower order terms in

). Denote ξ = (ξ

′

, ξ

) with ξ

′

= (ξ

, . . . , ξ

−1

) in

−1

1. Local Solvability and Fundamental Solution

Consider P(ξ) = P(ξ

′

, ξ

) as a polynomial in the last variable ξ

for ξ

′

in R

−1

. Let λ

(ξ

′

), . . . , λ

(ξ

′

) be the roots of P(ξ

′

, ξ

) = 0

arranged so that if i

≤ j.

ℑλ

(ξ

′

)

≤ ℑλ

(ξ

′

) and Re λ

(ξ

′

)

≤ Re λ

(ξ

′

) when

ℑλ

(ξ

′

) =

ℑλ

(ξ

′

Since the roots of a polynomial depends continuously on the coef-

ficients we see that λ

(ξ

′

) are continuous in ξ

′

. To proceed further, we

need the help of two Lemmas.

Lemma 2.3. There is a measurable function φ :

−1

→ [−k − 1, k + 1]

such that for all ξ

′

−1

min

≤ j≤k

{|φ(ξ

′

)

− Imλ

(ξ

′

)

|} ≥ 1.

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

(ξ

′

), j =

1, 2, . . . , k).

Lemma 2.4. Let P(ξ) = ξ

+( lower order terms) and let N(P) be the set

{ζǫ

: P(ζ) = 0

}. Let d(ξ, N(P)). Then, we have

|P(ξ)| ≥ (d(ξ)/2)

Proof. Take ξ in R

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 λ

, λ

, . . . , λ

be the zeros of the polynomial g. Then

g(z) = c(z

− λ

)

· · · (z − λ

)

so that

g(z)

g(0)

| =

j=1

|1 −

Since ξ + λ

η ǫ N(P)

|λ

| ≥ d(ξ) so that when |z| ≤ d(ξ), |

g(z)

g(0)

| ≤ 2

. Also

(k)

(0)

| = |

2πi

|ζ|=d(ξ)

q(ζ)ζ

−k−1

dζ

| ≤

|g(0)|

d(ξ)

k+1

2πd(ξ)

2. Partial Differential Operators with Constant Coefficients

i.e.

(k)

(0)

| ≤ k!|g(0)|2

d(ξ)

−k

. But g(0) = P(ξ) and

(k)

(0)

| =

∂

∂ξ

P(ξ) = k!

Therefore,

≤ k!|g(0)|2

d(ξ)

−k

|P(ξ)| ≥ (d(ξ)/2)

Hence the lemma is proved.

Returning to the proof of the theorem, consider the function

u(x) =

−1

IMξ

=φ(ξ

′

)

2φix

·ξ

( ˆ

f (ξ)/P(ξ))dξ

dξ.

By Lemmas 2.3 and 2.4 we have

|P(ξ)| ≥ (d(ξ)/2)

≥ 2

−k

along Imξ

= φ(ξ

′

Since f ǫC

∞

◦

, ˆ

f (ξ) is rapidly decreasing as

|Reξ| tends to ∞ as along

|Imξ| stays bounded, so the integral converges absolutely and uni-

formly together with all derivatives defining a C

∞

function u.

Finally, by Cauchy’s theorem,

P(D)u(x) =

−1

IMξ

=φ(ξ

′

)

2φix

·ξ

( ˆ

f (ξ)/P(ξ))dξ

dξ

′

2φix

·ξ

( ˆ

f (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

Definition 2.5. A distribution K satisfying P(D)K = δ is called a fun-
damental 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-

efficients has a fundamental solution.

Proof. Proceeding as in the previous theorem, we try to define

K(x) =

−1

Imξ

=φ(ξ

′

)

2Πix

·ξ

(p(ξ))

−1

dξ

′

Here, however, the integral may diverge at infinity. So, we consider

the polynomial

(ξ) = P(ξ)(1 + 4Π

j=1

where N is a large positive integer. Let

(x) =

−1

Imξ

=φ(ξ

′

)

2Πix

·ξ

(ξ))

−1

dξ

′

where φ is chosen appropriately for the polynomial P

.Then on the re-

gion of integration, we have

(ξ)

| ≥ c(1 + |ξ|

)

;

so the integral will converge when N > n/2. We claim that P

(D)K

δ.

To see this, take φǫC

∞

and observe that the transpose of P

(D) is

(

−D). Thus

< P

(D)K

, φ > =< K

, P

(

−D)φ >

2. Partial Differential Operators with Constant Coefficients

−1

Imξ

=φ(ξ

′

)

2Πix

·ξ

(

−D)φ)(x)

(ξ)

dξ

′

−1

Imξ

=φ(ξ

′

)

(ξ))

−1

dξ

′

2Πix

·ξ

(

−D)φ(x)dx

−1

Imξ

=φ(ξ

′

)

φ(

−ξ)dξ

dξ

′

φ(

−ξ)dξ = φ(0) =< δ, φ > .

Thus,

δ = P

(D)K

= P(D)(1

− ∆)

so if we put K = (1

− ·∆)

we have P(D)K = δ. Thus K is a

fundamental solution of P(D).

2 Regularity Properties of Di

fferential 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 .

Definition 2.8. Let L =

|α|≤K

(x)D

where a

ǫC

∞

be a differential

operator. L is said to be hypoelliptic, if and only if, for any uǫ

′

, sing

supp u

⊂ sing supp Lu. In other words, L is hypoelliptic if and only

if for any open set Ω

⊂

and any uǫ

′

(Ω),

“LuǫC

∞

(Ω) implies

uǫC

∞

(Ω)”.

Remark 2.9. The operator L is said to be elliptic at xǫ

|α|=k

(x)ξ

0 for every ξ in ǫ

\{0}.

L is said to be elliptic if L is elliptic at every xǫ

. Elliptic operators

are hypoelliptic, as we shall prove later. This accounts for the name

hypoelliptic.

2. Regularity Properties of Differential Operators

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

Example 2.10. Take the operator L =

∂

∂x∂y

. The general solution

of the equation Lu = 0 is given by u(x, y) = f (x) + g(y) where f and g
are arbitrary C

functions. This shows that L is not hypoelliptic.

We observe that if L is hypoelliptic, then every fundamental solution

of L is a C

∞

function in

\{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

∞

\{0}.

c) At least one fundamental solution of L is C

∞

\{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 ǫ

′

is such that f is C

∞

\{0} and gǫE

′

Then sing supp( f

∗ g) ⊂ supp g.

Proof. Suppose x < supp g. We will show that f

∗ g is C

∞

in a neigh-

bourhood of x.

Since x < supp g, there exists an ǫ > 0 such that B(x, ǫ)

∩supp g = φ.

Choose φǫC

∞

(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

∞

. Also

Supp(φ f

∗ g) ⊂ Supp φ f + Supp g

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

∞

\{0}. Suppose uǫD

′

and LuǫC

∞

(Ω) where Ω

⊂

open.

For xǫΩ, pick ǫ > 0 small enough so that ¯B(x, ǫ)

⊂ Ω. Choose

φǫC

∞

(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

∗ L(φu) = K ∗ φLu + K ∗ v.

φLu is a C

∞

function so that K

∗ Lu is a C

∞

function. Also K

∗ v is

a C

∞

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) =

∗ 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.

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 so-
lutions for these operators and show how they can be applied to solve
boundary value problems and to yield other information.

(i) LAPLACE OPERATOR: ∆ =

j=1

∂

∂x

. If u(x) represents

electromagnetic potential (or gravitational potential ) and ρ de-
notes 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.

3. Basic Operators in Mathematical Physics

(ii) HEAT OPERATOR: L =

∂

∂t

− ∆ in

n+1

. If u(x, t) represents the

temperature of a homogeneous body at the position x and time t,

then u satisfies the heat equation

∂u

∂t

− ∆u = 0 in

n+1

(iii) WAVE OPERATOR: @ =

∂

∂t

− ∆ in

n+1

. If u(x, t) represents the

amplitude of an electromagnetic wave in vacuum at position x and
time t, then u satisfies the wave equation @u = 0 in

n+1

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 un-
der 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(∆).

Proof. Let L = P(D) =

|α|≤K

. Since P(D) is invariant under trans-

lations, a

are constants. Let T be any rotation. We have

P(T ξ) ˆ

φ(T ξ) = (P(D)φ)ˆ(T ξ)

= (P(D)φ

◦ T )ˆ(ξ)

= (P(D)φ

◦ T ))ˆ(ξ)

= P(ξ) ˆφ(T ξ), for every φ.

Thus P(T ξ) = P(ξ) so that P is rotation - invariant.

Write P(ξ) =

j=0

(ξ) where P

(ξ) is the part of P which is ho-

mogeneous of degree j. We claim that each P

is rotation - invariant.

Indeed, if t is any real number,

2. Partial Differential Operators with Constant Coefficients

P(tξ) =

j=0

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

P(tξ) =

j=0

(T ξ)

so that

j=0

(T ξ)

− P

(ξ)) = 0.

This being true for all t, it follows that P

◦ T = P

. But the only

rotation-invariant functions which are homogeneous of degree j are of
the form P

(ξ) = c

|ξ|

where c

are constants.

Indeed, since P

is rotation-invariant, P

(ξ) depends only on

|ξ| and

thus, for

|ξ| , 0,

(ξ) = P

|ξ|

= c

|ξ|

Since

|ξ|

is not a polynomial when j is odd, c

= 0 in that case.

Therefore,

P(ξ) =

|ξ|

2 j

Taking Q(x) =

P(c

2 j

−4π

)

, we get P(D) = Q(∆).

Remark 2.14. This theorem applies to scalar differential operators. If
one considers operators on vector or tensor functions, there are first or-
der 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 FǫL

(

), we have

F(x)dx =

∞

|x|=1

f (r)r

−1

dσ(x)dr

4. Laplace Operator

where dσ(x) is the surface measure on S

−1

, the unit sphere in

Thus

F(x)dx = ω

∞

f (r)r

−1

where ω

is the area of S

−1

Let us now calculate ω

. We have

−π|x|

dx = 1.

Now

−π|x|

dx = ω

∞

−πr

−1

2π

∞

−s

(s/π)

n/2

−1

ds, s = πr

2π

n/2

∞

−s

n/2

−1

= ω

Γ(n/2)/(2π

n/2

Therefore

= 2π

n/2

/Γ(n/2).

4 Laplace Operator

First, let us find a fundamental solution of ∆, i.e., we want to find a

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.

2. Partial Differential Operators with Constant Coefficients

Let us tarry a little to compute the Laplacian of a radial function F.

LetF(x) = f (r), r =

|x|. Then,

∆F(x) =

j=1

∂

∂x

[ f

′

(r)x

/r]

j=1







′′

(r)

′

(r)

−

′

(r)







= 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

′′

(r) + ((n

− 1)/r) f

′

(r) = 0 on (0,

∞).

From this equation

′′

(r)/ f

′′

(r) =

−(n − 1)/r.

Integrating, we get

′

(r) = c

◦

−n

with a constant c

◦

. One more integration yields

f (r) = c

−n

+ c

when n , 2

= c

log r + c

when n = 2.

Since constants are solutions of the homogeneous Laplace equation,

we may assume that c

= 0. Thus if we set F(x) =

|x|

−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|

−n

(n , 2), then ∆F = (2

− n)ω

δ,

where ω

is the area of the unit sphere in

4. Laplace Operator

Proof. For ǫ > 0, we define F

(x) = (ǫ

|x|

)

−n)/2

. Then F

is a

∞

function on

, and an application of Lebesgue’s Dominated Con-

vergence Theorem reveals that F

converges to F as ǫ tends to 0, in

the sense of distributions. Therefore ∆F

converges to ∆F, in the same

sense. Let us now compute ∆F

∆F

(x) =

j=1

∂

∂x

− n)(|x|

+ ǫ

)

−n/2

j=1

− n)(|x|

+ ǫ

)

−n/2

+ (2

− n)(−n)(|x|

+ ǫ

)

(

−n/2)−1

= (2

− n)(|x|

+ ǫ

)

(

−n/2)−1

nǫ

Thus we see that ∆F

ǫL

(

). A simple computation shows that

∆F

(x) = ǫ

−n

∆F

(x/ǫ); so, by Theorem 1.6, ∆F

tends to (

∆F

)δ as

ǫ tends to 0. Therefore ∆F = (

∆F

)δ, and we need only to compute

∆F

= n(2

− n)

(1 +

|x|

)

(

−n/2)−1

= n(2

− n)ω

∞

(1 + r

)

(

−n/2)−1

−1

dr.

Putting r

+ 1 = s, we see that

∆F

n(2

− n)ω

∞

(

−n/2)−1

− 1)

−1)/2

n(2

− n)ω

∞

−2

−

)

(n/2)

−1

n(2

− n)ω

∞

− σ)

(n/2)

−1

dσ, σ =

n(2

− n)ω

(2/n) = (2

− n)ω

2. Partial Differential Operators with Constant Coefficients

which completes the proof.

Exercise. Show that if F(x) = log

|x| on

, then ∆F = 2πδ.

Corollary 2.17. Let K(x) =












|x|

−n

− n)ω

(n , 2)

2π

log

|x|(n = 2)

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) =

−4π

|ξ|

ˆg(ξ), we get

−4π

|ξ|

K(ξ) ˆ

f (ξ) = ˆf(ξ),

so that, at least formally,

K(ξ) =

−1/(4π

|ξ|

We observe that when n > 2, the function

−1/(4π

|ξ|

) 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 F

be the locally integrable function

(ξ) =

|ξ|

−α

. Then

∨

(x) =

− α)

/Γ

−n/2

|x|

−n

Before proving this theorem, let us pause a minute to observe that it

implies

∨

(x) = Γ((n/2)

− 1)π

−n/2

|x|

−n

= (Γ(n/2)/(n/2

− 1))π

−n/2

|x|

−n

= (

−4π

/((2

− n)ω

))

|x|

−n

, so that (

−F

/4π

))

∨

= K

as desired.

4. Laplace Operator

Proof of theorem 2.19. The idea of the proof is to express F

as a

weighted average of Gaussian functions, whose Fourier transforms we
can compute.

To begin with,

∞

−rt

−1

dt =

∞

−s

−1

−α

ds = r

−α

Γ(α), (r > 0, α > 0).

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

−α

Γ(α)

∞

−rt

−1

dt.

Taking r = π

| ξ |

and replacing α by α/2, then

| ξ |

−α

α/2

Γ(α/2)

∞

−π|ξ|

(α/2)

−1

which is the promised formula for F

as a weighted average of Gaus-

sians. Formally, we can write

2πix.ξ

| ξ |

−α

dξ =

α/2

Γ(α/2)

∞

2πix.ξ

−π|ξ|

α/2

−1

dtdξ

α/2

Γ(α/2)

∞

|ξ|

)

(x)t

(α/2)

−1

α/2

Γ(α/2)

∞

−π(|x|

)/t

−n/2

α/2

−1

α/2

Γ(α/2)

∞

−π|x|

−α)/2−1

α/2

Γ(α/2)

Γ((n

− α)/2)

−α)/2

| x |

−n

2. Partial Differential Operators with Constant Coefficients

In this computation, the change of order of the integration is unfor-

tunately not justified, because the double integral is not absolutely con-
vergent. This is not surprising, since F

is not an L

function; instead,

we must use the definition of the Fourier transform for distributions.

For every φǫS ,

< e

−π|ξ|

, ˆ

φ > =< e

−π|ξ|

)ˆ,φ >,

i.e.,

−π|ξ|

φ(ξ)dξ =

−n/2

−(π/t)|x|

φ(x)dx.

Now multiply both sides by t

(α/2)

−1

dt and integrate from 0 to

∞.

∞

−π|ξ|

(α/2)

−1

φ(ξ)dξ dt =

∞

−π/t|x|

(α

−n)/2−1

φ(x)dx dt.

The change of order in the integral is permitted now and integrating,

we obtain,

Γ(α/2)

α/2

φ(ξ)

| ξ |

−α

dξ =

Γ((n

− α)/2)

−α)/2

| x |

−n

φ(x) dx

i.e.,

| ξ |

−α

, ˆ

φ > =<

Γ((n

− α)/2)

(Γ(α/2)

−n/2

| x |

−n

, φ > .

This shows that

∨

(x) =

Γ((n

− α)/2)

Γ(α/2)

−n/2

| x |

−n

as desired.

This analysis does not suffice to explain

−(4π

| ξ |

)

−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 G

(ξ) = (2π

| ξ |)

−α

and to study the be-

haviour of G

and G

∨

as α tends to 2. This analysis can be carried out

just as easily in n dimensions where the problem is to study G

and G

as α tends to n.

4. Laplace Operator

Let R

= G

∨

for 0 < α < n. (R

called the Riesz potential of order

α). By the preceding theorem, we have

(x) =

(Γ((n

− α)/2)

n/2

Γ(α/2)

| x |

−n

Moreover, if f ǫS ,

(

−∆)

α/2

( f

∗ R

) = f,

in the sense that (2π

| ξ |)

( f

∗ R

)

(ξ) = ˆf(ξ).

As α tends to n, the Γ-function in R

blows up. However, since

(2π

| ξ |)

= 0, we see that (

−∆)

α/2

1 = 0; so we can replace R

− c, c being a constant, still having

(

−∆)

α/2

( f

∗ (R

− c)) = f.

If we choose c = c

appropriately, we can arrange that R

− c

will

have a limit as α tends to n. In fact, let us take

Γ((n

− α)/2)

n/2

Γ(α/2)

and define R

′

= R

− c

Then

′

(x) =

Γ((n

− α)/2)

n/2

Γ(a/2)

(

| x |

−n

−1)

2Γ((n

− α)/2) + 1

n/2

Γ(α/2)

(

| x |)

−n

− 1

− α

Letting α tend to n, we get in the limiting case

′

(x) =

−2

−n

n/2

Γ(n/2)

log

| x | .

when n = 2,

−R

′

(x)

2π

log

| x | and ∆( f ∗ (−R

′

)) = f , so we recover our

fundamental solution for the case n = 2.

It remains to relate the function G

(ξ) = (2π

| ξ |)

−n

to the Fourier

2. Partial Differential Operators with Constant Coefficients

transform of the tempered distribution F

′

. One way to make G

into a

distribution is the following.

Define a functional F on S by

< F, φ >=

|ξ|≤1

φ(ξ)

− φ(0)

(2π

| ξ |)

dξ +

|ξ|>1

φ(ξ)

(2π

| ξ |)

dξ.

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,
< F, φ >=

φ(ξ)G

(ξ)dξ, i.e., F agrees with G

on R

\{0}.

Just as R

′

was obtained from R

by subtracting off an infinite con-

stant, F is obtained from G

by subtracting off an infinite multiple of the

δ function. This suggests that F is essentially the Fourier transform of
R

′

. In fact, one has the following

Exercise. Show that F

− (R

′

)ˆis 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 R

with smooth boundary ∂Ω. Let v be the unit outward normal

vector on ∂Ω. Let F : R

→ R

be a C

function on ¯

Ω, the closure of

Ω. Then we have

∂Ω

F. v dσ =

Ω

div F dx =

Ω

j=1

∂F

∂x

dx.

CONSEQUENCES OF THEOREM 2.20

(2.21)

When we take F = grad u, F.v = grad u.v =

∂u

∂v

, the normal

derivative of u, and div F = div grad u = ∆u. Therefore

∂Ω

∂u

∂v

dσ =

Ω

∆u dx.

4. Laplace Operator

(2.22)

When we take F = u grad v

− v grad u, div F = u∆v − v∆u.

Therefore

∂Ω

∂v
∂v

− v

∂u

∂v

dσ =

Ω

(u∆v

− v∆u)dx.

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(x

, r). Then,for any ρ < r, we

have

u(x

) =

−1

|x−x

|=ρ

u(x)dσ(x).

That is, u(x

) is the mean value of u on any sphere centred at x

Proof. Without loss of generality, assume that x

= 0. Since ∆u =

0 and ∆ is hypoelliptic, u is a C

∞

function. Now formally,

u(0) =< δ, u >

|x|<ρ

δ(x)u(x)dx

|x|<ρ

∆K(x)u(x)dx

|x|<ρ

(u∆K

− K∆u)dx

|x|=ρ

∂K

∂v

− K

∂u

∂v

dσ, by (2.22).

Of course, we are cheating here by applying Green’s formula to the

non-smooth function K. Nonetheless the result is correct, and we leave it
as an exercise for the reader to justify it rigorously (either approximate
K by C

∞

functions as in the proof of Theorem 2.16 or apply Green’s

formula to the region ǫ <

| x |< ρ and let ǫ tend to 0.)

2. Partial Differential Operators with Constant Coefficients

On the circle

| x |= ρ, K is a constant and so

|x|=ρ

∂u

∂v

dσ =

Const

|x|=ρ

∂u

∂v

dσ = Const .

|x|≤ρ

∆ u dσ = 0.

Further on

| x |= ρ.

∂

∂v

∂

∂r

, the radial derivative. Therefore

∂K

∂v

= 1/(ω

−1

) on

| x |= ρ.

Thus we have

u(0) =

−1

|x|=ρ

u(x)dσ

as desired.

As a corollary to the mean value theorem, we can derive the maxi-

mum principle for harmonic functions.

Theorem 2.24. Suppose Ω is a connected open set in R

. Let u be a real-

valued function harmonic in Ω. If A = sup

Ω

u(x), then either u(x) < A for

all xǫΩ or u(x)

≡ A on Ω.

Proof. Suppose that u(x

) = A for some x

ǫΩ. From the mean value

theorem

u(x

) =

−1

|x−x

|=ρ

u(x)dσ(x)

where ρ is small enough so that

{x :| x − x

|≤ ρ} ⊂ Ω. By our as-

sumption, the integral does not exceed A. If u(x

) < A for some point

of B(x

, ρ) then u(x) < A in some neighbourhood of x

by continuity.

Taking r =

| x

− x

|< ρ, we have

u(x

) =

−1

|x−x

|=r

u(x)dσ(x) < A,

a contradiction. Therefore, if we set Ω

{xǫΩ : u(x) = A}, then Ω

an open subset of Ω and Ω

φ. Further Ω

{xǫΩ : u(x) < A} is also

on open subset of Ω and Ω

∪ Ω

= Ω. The connectedness of Ω and the

non-emptiness of Ω

force Ω

to be empty. Thus u(x)

≡ A on Ω.

4. Laplace Operator

Corollary 2.25. If Ω is a bounded open set in R

, uǫC( ¯

Ω) and ∆u =

0 in Ω, then

sup

Ω

| u(x) |= sup

∂Ω

| u(x) | .

Proof. Since the function u is continuous on the compact set ¯

Ω,

| u(x) |

attains its supremum on ¯

Ω at some point x

. By multiplying u by a

constant, we may assume that u(x

) =

| u(x

)

If x

ǫ∂Ω, the corollary is proved. Otherwise x

ǫΩ and by the previ-

ous theorem, on the connected component of containing x

, Re u(x) =

u(x

) and hence Im u(x) = 0. Since u is continuous on ¯

Ω, u(x) = u(x

)

for all xǫ∂Ω. Thus sup

Ω

| u(x) |= sup

∂Ω

| u(x) |.

Corollary 2.26. If u and v are in C( ¯

Ω), Ω is bounded ∆u = ∆v = 0 on

Ω and u = v on ∂Ω, then u = v everywhere.

Proof. Apply the previous corollary to u

− v.

The following boundary value problem for Laplace’s equation,

known as the Dirichlet problem, is of fundamental importance: given
a function f on ∂Ω, find a function u such that ∆u = 0 on Ω and u =

f on ∂Ω.

When Ω is bounded and f ǫC(∂Ω) the uniqueness of the solution,

if it exists at all, is assured by corollary 2.26. The problem of proving
the existence of solution of the Dirichlet problem is highly nontrivial.
We shall solve a special case, namely, when Ω is a half-space and then
indicate how similar ideas can be applied for other regions.

First of all a word about notation: we will replace R

by R

n+1

with

coordinates (x

, x

, . . . , x

, t). Our Laplacian in R

n+1

will be denoted by

∂

+ ∆ where ∂

∂

∂t

and ∆ =

j=1

∂

∂x

. Now our Dirichlet Problem is

the following :

given a function f on R

, find a function u such that

(∂

2
t

+ ∆) u(x, t) = 0, xǫR

, t > 0

(2.27)

u(x, o) = f (x), xǫR

2. Partial Differential Operators with Constant Coefficients

Since the half space is unbounded, the uniqueness argument given

above does not apply. Indeed, without some assumption on the be-
haviour of u at

∞, uniqueness does not hold. For example, if u(x, t)

is a solution, so is u(x, t) + t. However, we have

Theorem 2.28. Let u be a continuous function on R

× [0, ∞) satisfying

i) (∆ + ∂

)u = 0 on R

× (0, ∞),

ii) u(x, o) = 0 for xǫR

, and

iii) u vanishes at

∞.

Then U

≡ 0 on R

× [0, ∞).

Proof. Applying the maximum principle for u on B(0, R)

× (0, T ), we

see that maximum of

| u | on B(0, R) × (0, T ) tends to zero as R, T tend

∞. Hence u ≡ 0.

Remark 2.29. Hypothesis (iii) in Theorem 2.28 can be replaced by (iii):
u is bounded on R

× [0, ∞). But this requires a deeper argument (See

Folland [1]).

To solve the Dirichelt problem, we apply the Fourier transform in

the variable x. We denote by ˜u(ξ, t) the Fourier transform

˜u(ξ, t) =

−2πix.ξ

u(x, t)dx.

If we take Fourier transform of (2.27), we obtain

(∂

2
t

− 4π

| ξ |

)˜u(ξ, t) = o on R

× (0, ∞)

˜u(ξ, o) = ˆf(ξ) on R

The general solution of the ordinary differential equation (ODE)

(∂

− 4π

| ξ |

) ˜u(ξ, t) = 0 is given by ˜u(ξ, t) = a(ξ)e

−2π|ξ|t

+ b(ξ)e

2π

|ξ|t

and the initial condition is a(ξ) + b(ξ) = ˆf(ξ).

This formula for ˜u will define a tempered distribution in ξ, provided

that as

| ξ | tends to ∞, | a(ξ) | grows at most polynomially and | b(ξ) |

4. Laplace Operator

decreases faster than exponentially. We remove the non- uniqueness by
requiring that u should satisfy good estimates in terms of f which are
uniform in t-for example, that u should be bounded if f is bounded. This
clearly forces b(ξ) = 0, since e

2π

|ξ|t

blows up as t tends to

∞. Thus we

take ˜u(ξ, t) = ˆf(ξ)e

−2π|ξ|

t. Taking the inverse Fourier transform in the

variable ξ, u(x, t) = ( f

∗ P

)(x) where P

(x) = [e

2πt(

|.|)

]

∨

(x) is the Poisson

Kernel for R

× [0, ∞).

We now compute P

explicitly. When n = 1, this is easy:

(x) =

−∞

2πixξ

2πξt

dξ +

∞

2πixξ

−2πξt

dξ

2π

[(t + ix)

−1

+ (t

− ix)

−1

] =

+ t

)

−1

To compute P

(x) for the case n > 1, as in the proof of Theorem

2.19, the idea is to express e

−2π|ξ|t

as a weighted average of Gaussians.

Here’s how!

Lemma 2.30. When β > 0, e

−β

∞

−s

−β

/4s

√

(πs)

ds.

Proof. First, observe that

−β

∞

−∞

iβτ

+ 1

dτ.

This can be proved by using the residue theorem or by applying the

Fourier inversion theorem to P

(x) =

(1 + x

)

−1

on R

. We also have

+ 1

∞

−(τ

+1)s

ds.

2. Partial Differential Operators with Constant Coefficients

Therefore,

−β

′

∞

−∞

∞

i2πτ

−(τ

+1)s

ds dτ.

Putting τ = 2πσ and changing the order of integration,

=β

∞

−∞

i2πσβ

−s

−4π

dσ ds

i.e.,

−β

= 2

∞

−2

(4πs)

−1/2

/4s

∞

−s

/4s

√

(πs)

which is the required expression.

Returning to the computation of P

(x), we have

(x) =

2πix.ξ

−2π|ξ|t

dξ.

Taking β = 2π

| ξ | t in the lemma, we get

(x) =

∞

2πix.ξ

−s

√

πs

(4π

|ξ|

)/4s

dsdξ

∞

−s

√

πs

2πixξ

−(π

|ξ|

)/S

dξ.ds

∞

−s

√

πs

πt

− n/2e

−(s|x|

)/t

= π

−(n+1)/2

−n

∞

−s(1+(|x|

))

−1)/2

4. Laplace Operator

−(n+1)/2

−n

Γ(n + 1)/2

1 +

|x|

(n + 1/2)

Thus we have

(2.31)

(x) = Γ((n + 1)/2)π

(n+1)/2

t/(t

| x |

)

(n+1)/2

In particular, we see that P

ǫL

∩ L

∞

, so that f

∗ P

is well defined if,

for example, f ǫL

, 1

≤ p ≤ ∞. That (∆ + ∂

= 0 follows by taking

the Fourier transform, and hence

(∆ + ∂

2
t

)( f

∗ P

) = f

∗ (∆ + ∂

2
t

= 0 for any f.

Further note that

(x) = t

−n

(x/t) and

(x)dx = ˆ

(0) = 1.

Therefore, by Theorem 1.6, when f ǫL

, 1

≤ p < ∞, f ∗ P

tends to

f in the L

norm and when f is bounded and continuous f

∗ P

tends to

f uniformly on compact sets as t tends to 0.

If we take f continuous and bounded, then for u(x, t) = P

∗ f , we

have lim

→0

u(x, t) = u(x, 0) = f (x). Thus the function u(x, t) = ( f

∗ P

)(x)

is a solution of the Dirichlet Problem for the half space.

Remark 2.32. The Poisson Kernel P

is closely related to the fundamen-

tal solution K

n+1

of the Laplacian in R

n+1

. Indeed

n+1

(x, t) =

− n)ω

n+1

(

|x|

+ t

)

−((n−1)/2)

and hence

(x) = 2∂

n+1

(x, t).

Exercise. Check the above equation using the Fourier transform. (Start
with (∂

+ ∆)K

n+1

= δ(x)δ(t) and take the Fourier transform in the vari-

able x to obtain

(∂

2
t

− 4π

|ξ|

) ˜

n+1

= δ(t).

2. Partial Differential Operators with Constant Coefficients

Solve this equation to obtain

n+1

(ξ, t) =

2π

|ξ||t|

4π

|ξ|

Then

∂

n+1

−2π|ξ|t

, t > 0, so that ∂

n+1

Our formula (2.31) for P

makes sense even when t < 0 and we have

−t

(x) =

−P

(x), so that lim

→o±

∗ P

± f . We further observe that

∗

(x)

= f (x)δ(t)

∗

(x,t)

∂

n+1

(x, t)

where

∗

(

x) and

∗

(x,t)

mean convolution on R

and R

n+1

respectively.

Therefore,

∗

(x)

= f (x)δ

′

(t)

∗

(x,t)

n+1

(x, t)

i.e.,

u(x, t) = 2 f (x)δ

′

(t)

∗

(

x, t)K

n+1

(x, t).

Form this, we

(δ

2
t

+ ∆)u(x, t) = 2 f (x)δ

′

(t)

∗ δ(x)δ(t) = 2 f (x)δ

′

(t).

Exercise. Show directly that if

i) (∆ + ∂

)u = 0 on R

n+1

× {0}

ii) u(x,

−t) = −u(x, t)

iii) lim

→0±

u(x, t) =

± f (x), then u is a distribution solution of (∂

+ ∆)u =

2 f (x)δ

′

(t).

[To avoid technicalities, assume f is sufficiently smooth so that

lim

→0±

∂u

∂t

exists; e.g. f ǫC

is sufficient ].

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 R

. We

assume that the boundary ∂Ω of Ω is of class C

5. The Heat Operator

We know that the solution of the Dirichlet Problem in the case when

Ω = R

−1

× (0, ∞) is given by u = f ∗ 2

∂K

∂x

(the convolution is in R

−1

we note that

∂K

∂x

is the inward normal derivative of K. For bounded Ω

with C

boundary let us consider

u(x) = 2

∂Ω

f (y)

∂

∂v

K(x

− y)dσ(y).

Here σ(y) is the surface measure on the boundary, ν is the unit out-

ward normal on ∂Ω, and

∂φ

∂ν

(x, y) = Σν

∂φ

∂y

. The minus sign in K(x

−y)

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

uǫC ¯

(Ω) and u

∂Ω

= f + T f

where T is a compact integral operator on L

(∂Ω) or C(∂Ω). Hence if

we can find φ in C(∂Ω) such that φ + T φ = f , and we set

u(x) = 2

∂Ω

φ(y)

∂

∂ν

K(x

− y)dσ(y)

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 (pro-
vided, as we have assumed, that ∂Ω is of class C

). See Folland [1] for

a detailed treatment.

5 The Heat Operator

The Heat operator is given by ∂

− ∆. We want to find a distribution

K such that (∂

− ∆)K = δ(x)δ(t). Taking the Fourier transform in both

variables we have

(2.33)

K(ξ, τ) = (2πi + 4π

|ξ|

)

−1

2. Partial Differential Operators with Constant Coefficients

Exercise . Show that ˆ

K is locally integrable near the origin, and hence

defines a tempered distribution.

K is not globally integrable, however; so computing its inverse Fou-

rier transform directly is a bit tricky. Instead, if we take the Fourier
transform in the variable x only,

(∂

+ 4π

|ξ|

) ˜

K(ξ, t) = δ(t).

we can solve this ODE get

K(ξ, t) =












a(ξ)e

−4π

|ξ|

t > 0

b(ξ)e

−4π

|ξ|

t < 0

with a(ξ)

− b(ξ) = 1.

As in the previous section, there is some latitude in the choice of a

and b, but the most natural choice is the one which makes ˜

K tempered

in t as well as ξ, namely a = 1, b = 0. So,

K(ξ, t) =












−4π

|ξ|

, t > 0

0 otherwise .

Taking the inverse Fourier transform,

K(x, t) =












(4πt)

−n/2

(

|x|

/4t)

t > 0

t > 0.

This is a fundamental solution of the heat operator.

Remark 2.34. If we take the Fourier transform of ˜

K(ξ, t) in the t vari-

able, we obtain

K(ξ, τ) =

∞

−∞

K(ξ, t)e

2π i t τ

∞

−t(4π

|ξ|

+π i τ)

= (4π

|ξ|

+ 2πiτ)

−1

thus recovering formula (2.33).

5. The Heat Operator

Exercise . Show that this really works, i.e., the iterated Fourier trans-
form 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 R

n+1

\{(0, 0)}. Therefore by Theorem 2.11, ∂

− ∆ is

hypoelliptic.

(2) K(x, t) = t

−n/2

, 1

. Thus if we set K(x, 1) = φ(x), then K(x, ǫ

)

= ǫ

−n

φ(x/ǫ) = φ

(x); so, as t tends to 0, K(x, t) tends to δ, Theorem

1.6. From this, we infer

(3) If f ǫL

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 ||

converges to 0. Thus we have

solved the initial value problem for the homogeneous heat equation,
when f ǫL

. Actually, since K(x, t) decreases so rapidly as

|x| tends

∞, this works for much wider classes of f

′

s. The convolution

∗

(x)

K(., t) make sense, for example, if

| f (x)| < Ce

|x|

−ǫ

. 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:

(∂

− ∆)u = F on R

× (0, ∞),

u(x, 0) = f (x) on R

If we take u

= F

∗

(x,t)

K and

= ( f

− u

(., 0))

∗

(x)

2. Partial Differential Operators with Constant Coefficients

then we see that

(∂

− ∆)u

= F on R

× (0, ∞), (∂

− ∆)u

= 0

on R

× (0, ∞), and (u

+ u

)(x, 0) = f (x). Thus u = u

+ u

solves the

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 R

, then, for any compact Ω

⊂ R

, there exists a sequence

) of polynomials such that P

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) =

f (y)(4πt)

−n/2

−

j=1

−y

)

/4t

is an entire holomorphic function of x

∈

. 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

(∂

2
t

− ∆)K = δ(x)δ(t)

in both variables x and t, we have, formally

(2.36)

K(ξ, τ) = (4π

|ξ|

− 4π

)

−1

This function ˆ

K is not locally integrable, so it is not clear how to

interpret it as a distribution. Again, it is better to take the Fourier trans-
form in the x variable obtaining

(∂

2
t

+ 4π

|ξ|

) ˜

K(ξ, t) = δ(t).

6. The Wave Operator

Solving this ODE,

K(ξ, t) = a

(ξ)e

2πi

|ξ|t

+ b

(ξ)e

−2πi|ξ|t

for t/

|t| = ±1

and the coefficients a

, b

must be determined so that

K(ξ, 0+) = ˜

K(ξ, 0

−), ∂

K(ξ, 0+)

− ∂

K(ξ, 0 =) = 1.

This gives two equations in four unknowns. In contrast to the situa-

tion 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 opera-

tor 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 sup-

ported 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

(ξ, t) = H(t)

sin 2π

|ξ|t

2π

|ξ|

−

(ξ, t) =

−H(−t)

sin 2π

|ξ|t

2π

|ξ|

= ˜

(ξ,

−t)

where H is the Heaviside function, i.e., the characteristic function of

[0,

∞). Let us compute the Fourier transforms of ˜

and ˜

−

in t to see

how to make sense out of (2.36). ˜

and ˜

−

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

(ξ, t) = e

−2πǫt

H(t)

sin 2π

|ξ|t

2π

|ξ|

, ǫ > 0,

then ˜

is an integrable function and ˜

converges to ˜

is S

′

as ǫ tends

to 0. Therefore ˜

= lim

→0

where

(ξ, τ) =

∞

2πǫt

−2π it τ

sin 2π

|ξ|t

2π

|ξ|

2. Partial Differential Operators with Constant Coefficients

∞

2πǫt

−2π it τ

πi

|ξ|t

− e

πi

|ξ|t

)

4πi

|ξ|

= (4π

)

−1

(

|ξ|

− (τ − iǫ)

)

−1

Exercise. Prove that ˆ

−

= lim

→0

−

in S

′

where

−

(ξ, τ) = (4π

)

−1

(

|ξ|

− (τ + iǫ)

)

−1

Thus we have two distinct ways of making the function [4π

(

|ξ|

−

)]

−1

into a tempered distribution. The difference ˜

− ˜

−

is of course

a distribution supported on the cone

|ξ| = |τ|.

We now propose to use the fundamental solutions K

and K

−

solve the Initial Value Problem or Cauchy Problem, for the operator:

(2.37)

(∂

2
t

− ∆)u = f on R

n+1

u(x, o) = u

(x),

∂

u(x, 0) = u

(x),

where u

, u

, f are given functions.

For the moment, we assume that u

, u

ǫS and f ǫC

∞

)) ( That is,

→ f (., t) is a C

∞

function with values in S (R

)).

Taking the Fourier transform in the variable x in (2.37),

(∂

2
t

+ 4π

|ξ|

) ˜u(ξ, t) = ˜f(ξ, t)

˜u(ξ, 0) = ˜u

(ξ)

∂

˜u(ξ, 0) = ˜u

(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,

A(ξ) =

ˆu

(ξ)

2π

|ξ|

and B(ξ) = ˜u

(ξ).

when u

= u

= 0, the solution is

˜u(ξ, t) =

f (ξ, s)

sin(2π

|ξ|(t − s))

2π

|ξ|

6. The Wave Operator

(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) = ˆu

(ξ) cos 2π

|ξ|t +

ˆu

(ξ)

2π

|ξ|

sin 2π

|ξ|t+

f (ξ, s)

sin(2π

|ξ|(t − s))

2π

|ξ|

ds.

For t > 0, we can rewrite this as

˜u(ξ, t) = ˆu

(ξ) ˜

(ξ, t) + ˆu

(ξ)∂

(ξ, t) +

f (ξ, s) ˜

f (ξ, s)K

(ξ, t

− s)ds.

Since K

(ξ, t

− s) = 0 for t < s,

˜u(ξ, t) = ˆu

(ξ) ˜

(ξ, t) + ˆu

(ξ)∂

(ξ, t) +

f (ξ, s) ˜

f (ξ, s)K

(ξ, t

− s)ds.

Take the inverse Fourier transform:

u(x, t) = (u

∗

(x)

) + u

∗

(x)

∂

) + (H(t) f

∗

(x,t)

Likewise, for t < 0

u(x, t) =

−(u

∗

(x)

−

)

− u

∗

(x)

∂

−

) + ((H(

−t) f ) ∗

(x,t)

−

So, for arbitrary t, our solution u can be expressed as

u(x, t) = (u

∗

(x)

− K

)

+ u

∗

(x)

∂

− K

−

)

+ (H(t) f

∗

(x,t)

) + (H(

−t) f ∗

(x,t)

)

(2.38)

So far, we have avoided the question of computing K

and K

−

ex-

plicitly. Indeed, since ˜

and ˜

−

are not L

functions, it is not an easy

matter to calculate their inverse Fourier transform,

2. Partial Differential Operators with Constant Coefficients

However, for the case n = 1, we can find K

and K

−

by solving the

wave wave equation directly. We have

∂

2
t

− ∆ = ∂

2
t

− ∂

If we make the change of variables ξ = x + t, n = x

− t, then the wave

operator becomes ∂

− ∂

−4

∂

∂ξ∂η

. The general solution of

∂

∂ξ∂η

= 0

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 (∂

− ∂

)u = 0, u(x, 0) = u

(x), ∂

u(x, 0) = u

(x) we must

have

(x) = f (x) + g(x)

(x) = f

′

(x)

− g

′

(x).

From these equations, we have

′

(x) =

′

(x) + u

(x))

′

(x) =

′

(x)

− u

(x)).

Thus u(x, t) is given by

u(x, t) =

(x + t) + u

− t)) +

x+t

−t

(s)ds.

Comparing with the previous formula (2.38), we find that

(x, t) =

H(t

− |x|), K

−

(x, t) =

−t − |x|).

Exercise. Compute ˜

directly from these formulas.

6. The Wave Operator

It turns out that, for n = 2,

(x, t) =

2π

√

− x

±t − |x|),

and, for n = 3,

(x, t) =

±1

4πt

δ(

±t − |x|).

For n = 1, 2, K

, K

−

are functions; for n = 3, K

, K

−

are not func-

tions 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.

Theorem 2.39. Suppose u is a C

function on

{(x, t) : t ≥ o} such that

(∂

−△)u = 0 for t > 0 and u = ∂

u = 0 on the set B

{(x, 0) : |x− x

| ≤

}. Them u vanishes on Ω = {(x, t) : 0 ≤ t ≤ t

|x − x

| ≤ t

− t}.

Proof. Assume u is real valued; otherwise, we can consider the real and
imaginary parts separately. Let

{x : |x − x

| ≤ t

− t} and E(t) =

| grad

x,t

i.e.,

E(t) =







(∂

∂u

∂x







dx.

Then







∂u

∂t

∂

∂u

∂x

∂

∂x

∂t







−

∂B







∂u

∂t

∂u

∂x







dσ(x).

2. Partial Differential Operators with Constant Coefficients

(The second term comes from the change in the region B

. If this is

not clear, write the derivative as a limit of difference quotients and work
it our).

Since

∂u

∂x

∂

∂x

∂t

∂

∂u

∂t

= div

∂u

∂t

grad

applying the divergence theorem and using (∂

− △)u = 0, we obtain

∂B

" ∂u

∂t

∂u

∂ν

−

| grad

x,t

dσ

where ν is the unit normal to B

in R

. Now

∂u

∂t

∂u

∂ν

| <

∂u

∂t

∂u

∂ν

∂u

∂t

| grad

x,t

Thus we see that the integrand is non-positive, and hence

≤ 0.

Also E(0) = 0 since u = ∂

u = 0 on B

, so E(t)

≤ 0. But E(t) ≥ 0 by

definition, so E(t) = 0. This implies that grad

x,t

u = 0 on Ω =

≤t

and

since u = 0 on B

, we conclude that u = 0 on Ω.

Corollary 2.40. Suppose uǫC

on R

× [0, ∞), (∂

− △)u = 0 for t >

0, u(x, 0) = u

(x), ∂

u(x, 0) = u

(x). If Ω

= (supp u

)

∪ (supp u

), then

supp u

⊂ Ω = {(x, t) : d(x, Ω

)

≤ t}.

(The set Ω is the union of the forward light cones with vertices in

Ω

Proof. Suppose (x, t

) < Ω. Then for some ǫ > 0, the set

{x : d(x, x

)

≤ t

+ ǫ

} is disjoint from Ω

6. The Wave Operator

Therefore, by Theorem 2.39, u = 0 on the cone

{(x, t) : |x − x

| ≤ t

+ ǫ

− t, 0 ≤ t ≤ t

+ ǫ

In particular, u = 0 on a neighbourhood of (x

, t

), i.e., (x

, t

) is not in

the support of u.

Corollary 2.41.

SuppK

⊂ {(x, t) : t ≥ |x|}

SuppK

⊂ {(x, t) : −t ≥ |x|}.

Proof. Pick aφǫC

α
0

(B(0, 1)) such that

φ = 1. Put

(x) = ǫ

−n

φ(ǫ

−1

x). Let u

(x, t) = φ

∗

(x)

, t > 0.

Then (∂

− △)u

= 0, u(x, 0) = 0, ∂

u(x, 0) = φ

(x) and u

is C

By the previous corollary

supp u

⊂ {(x, t) : |x| ≤ t + ǫ}.

Now u

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 ˜

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].

2. Partial Differential Operators with Constant Coefficients

(iii) The distributions K

are smooth functions of t (except at t = 0)

with values in E

′

). Therefore, we now see that our formula

(2.38) for the solution of the Cauchy problem makes sense even
when u

, u

, ǫD

′

) and f ǫC(R

, D

′

)), and it is easily checked

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 (∂

−△)u = f on R

n+1

, u(x, 0) = u

(x), and ∂

u(x, 0)

= u

(x), figure out how supp u is related to supp f, supp u

and supp u

Chapter 3

Sobolev Spaces

THERE ARE MANY ways of measuring smoothness properties of func-

tions in terms of various norms. Often it is convenient to use L

norms,

since L

interacts nicely with the Fourier transforms. In this chapter, we

set up a precise theory of L

differentiability and use it to prove Horman-

der’s theorem on the hypoellipticity of constant coefficient differential
operators.

1 General Theory of L

Sobolev Spaces

Definition 3.1. For a non-negative integer k, the Sobolev space H

defined to be the space of all tempered distributions all of whose deriva-
tives of order less than or equal to k are in L

Thus

{ f ǫS

′

: D

f ǫL

) for 0

≤ |α| ≤ k}.

From the definition, we note that f ǫH

if and only if ξ

f (ξ)ǫL

for

≤ |α| ≤ k.

Proposition 3.2. f ǫH

if and only if (1 +

|ξ|

)

k/2

f ǫL

Proof. First assume that (1 +

|ξ|

)

k/2

f ǫL

. Since, for

|ξ| ≥ 1,

|ξ

| ≤ |ξ|

≤ (1 + |ξ|

)

k/2

for all

|α| ≤ k

3. L

Sobolev Spaces

and for

|ξ| < 1,

|ξ

| ≤ 1 ≤ (1 + |ξ|

)

k/2

for all

|α| ≤ k,

we have

|ξ

f (ξ)

| ≤ (1 + |ξ|

)

k/2

f (ξ)

and hence

f ǫL

for

|α| ≤ k which implies f ǫH

Conversely, assume that f ǫH

. Since

|ξ|

and

j=1

|ξ

are homoge-

neous of degree k and nonvanishing for ξ , 0, we have

(1 +

|ξ|

)

k/2

≤ c

(1 +

|ξ|

)

≤ c







1 + c

j=1

|ξ







so that

||(1 + |ξ|

)

k/2

≤ c

|| f ||

+ c

j=1

||∂

which shows that (1 +

|ξ|

)

k/2

f ǫL

The characterisation of H

given in the above proposition immedi-

ately 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 by

(Λ

f )ˆ(ξ) = (1 +

|ξ|

)

s/2

f (ξ).

In other words, Λ

−

△

4π

s/2

. Clearly Λ

maps S continuously

onto itself. We can therefore extend Λ

continuously from S

′

onto itself.

The Sobolev space of order s is defined by

{ f ǫS

′

: Λ

f ǫL

We equip H

with the norm

|| f ||

(s)

||Λ

. If s is a positive integer,

the proof of Proposition 3.2 shows that

|| ||

(s)

is equivalent to the norm

|| f || = Σ

≤|α|≤s

||D

PROPERTIES OF H

1. General Theory of L

Sobolev Spaces

(i) H

is a Hilbert space with the scalar product defined by (u, v)

(s)

(Λ

u, Λ

v). Here the scalar product on the right is that in L

. The

Fourier transform is a unitary isomorphism between H

and the

space of functions which are square integrable with respect to the
measure (1 +

|ξ|

)

dξ.

(ii) For every sǫR, S is dense in H

(iii) If s > t, H

⊂ H

with continuous imbedding. In fact, for uǫH

||u||

(t)

≤ ||u||

(s)

. In particular, H

⊂ L

for s > 0.

(iv) D

is a bounded operator from H

into H

−|α|

sǫR.

(v) If f ǫH

−s

, then f as a linear functional on S extends continuously

to H

and

|| f ||

(

−s)

is the norm of f in (H

)

∗

. So we can identify

)

∗

with H

(

−s)

. For f ǫH

−s

, gǫH

the pairing is given by

< f, g >=< Λ

−s

f, Λ

g >=

f ˆg.

(If s = 0, this identification of H

= L

with its dual is the complex

conjugate of the usual one).

(vi) The norm

||.||

(s)

is translation invariant. Indeed, if g(x) = f (x

− x

then ˆg(ξ) = e

2πix

.ξ

f (ξ) and hence

|| f ||

(s)

||g||

(s)

Proposition 3.4. For s > n/2, we have δǫH

−s

Proof. (1 +

|ξ|

)

−s/2

ˆδ = (1 + |ξ|

)

−s/2

ǫL

, whenever s > n/2. This is a

consequence of the following observation :

(1 +

|ξ|

)

−s

dξ

∼ 1 +

∞

−2s

−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|, D

δǫH

−s

3. L

Sobolev Spaces

Theorem 3.6 (SOBOLEV IMBEDDING THEOREM). For s >

k, H

⊂ C

. Further, we have

(3.7)

|α|≤k

sup

| ≤ C

|| f ||

(s)

Proof. Since S is dense in H

, it suffices to prove (3.7) for f ǫS . Let δ

denote the Dirac measure at x. For

|α| ≤ k, since s >

|α|, D

ǫH

−s

Since

< D

, f >= (

−1)

|α|

< δ

, D

f >= (

−1)

|α|

f )(x)

and

||D

(

−s)

is independent of x,

|α|≤k

sup

f (x)

| =

|α|≤k

sup

, f >

≤

|α|≤k

sup

(

−s)

|| f ||

(s)

= C

|| f ||

(s)

Now, given uǫH

, choose a sequence (u

) in S such that

||u − u

(s)

converges to 0, as j tends to

∞. The above inequality with f = u

− u

shows that (D

) is a Cauchy sequence in the uniform norm for

|α| ≤ k;

so its limit D

u is continuous.

Corollary 3.8. If uǫH

for all sǫR, then uǫC

∞

i.e.,

⊂ C

∞

This argument can be extended to show that if s >

+ k, then

elements of H

and their derivatives of order less than or equal to k are

not merely continuous but actually H ¨older continuous.

Proposition 3.9. If 0 < α < 1 and s =

+ α, then

||δ

− δ

(

−s)

≤

|x − y|

Proof. We have

||δ

− δ

2
(

−s)

−2πix.ξ

− e

−2πiy.ξ

(1 +

|ξ|

)

−s

dξ.

1. General Theory of L

Sobolev Spaces

Let R be a positive number, to be fixed later. When

|ξ| ≤ R we use

the estimate

−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

||δ

− δ

2
(

−s)

≤ 4π

|x − y|

|ξ|≤R

|ξ|

(1 +

|ξ|

)

−s

dξ + 4

|ξ|>R

(1 +

|ξ|

)

−s

dξ

≤ c







|x − y|

(1 + r

)

−s

n+1

dr +

−2s+n−1







≤ c

′

|x − y|

−2s+n+2

+ R

−2s+n

< s <

+ 1

= c

′

|x − y|

−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

||δ

− δ

(

−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 Λ

= bounded functions g : sup

x,y

|g(x) − g(y)|

|x − y|

∞. If s =

+ α + l and f ǫH

, then D

f ǫΛ

for

|β| ≤ k.

Remark 3.11. We shall obtain an analogue of this result for L

norms

in Chapter 5.

The following lemma will be used in several arguments hereafter.

Lemma 3.12. For all ξ, nǫR

and sǫR, we have

" 1 + |ξ|

1 +

|η|

< 2

|s|

(1 +

|ξ − η|

)

|s|

Proof.

|ξ| ≤ |η| + |ξ − η| gives

|ξ|

< 2(

|η|

|ξ − η|

) so that

(1 +

|ξ|

) < 2(1 +

|η|

)(1 +

|ξ − η|

3. L

Sobolev Spaces

If s > 0, then raise both sides to the s

power. If s < 0, interchange

ξ and η and raise to the

−s

power.

Proposition 3.13. If φǫS , then the operator f

→ φ f is bounded for all

Proof. The operator f

→ φ f is bounded on H

if and only if the opera-

tor of g

→ Λ

φΛ

−s

g is bounded on L

, as one sees by

setting g =

∧

f . But

(

∧

−s

∧

(ξ) = (1 +

|ξ|

)

s/2

(φ

∧

−s

g)ˆ(ξ)

= (1 +

|ξ|

)

s/2

h ˆφ(ξ) ∗ (∧

−s

(ξ)

= (1 +

|ξ|

)

s/2

φ(ξ

− η)(1 + |η|

)ˆg(η)dη

ˆg(η)K(ξ, η)dη

where

K(ξ, η) = (1 + η

)

−s/2

(1 +

|ξ|

)

s/2

φ(ξ

− η).

By lemma 3.12,

|K(ξ, η)| < 2

|s|/2

(1 +

|ξ − η|

)

|s|/2

| ˆφ(ξ − η)|.

Therefore, since ˆ

φ is rapidly decreasing at

∞,

|K(ξ, η)|dξ ≤ c for every ηǫR

|K(ξ, η)|dη ≤ c for every ηǫR

Thus from Theorem 1.1, the operator with kernel K is bounded on

. Hence our proposition is proved.

The spaces H

are defined on R

globally by means of the Fourier

transform. Frequently, it is more appropriate to consider the following
versions of these spaces.

1. General Theory of L

Sobolev Spaces

Definition 3.14. If Ω

⊂ R

is open and sǫR, we define H

loc

Ω =

{ f ǫD

′

(Ω) :

∀Ω

′

⋐ Ω,

∃ g

Ω

, ǫH

such that

Ω

′

= f on Ω

′

Proposition 3.15. f ǫH

loc

(Ω) if and only if φ f ǫH

for φǫC

∞

(Ω).

Proof. If f ǫH

loc

(Ω) and φǫC

∞

(Ω), then there exists gǫH

such that f =

g on supp φ. Therefore φ f = φgǫH

by proposition 3.13.

Conversely, if φ f ǫH

for all φǫC

∞

(Ω

′

), and Ω

′

⋐ Ω, choose

φǫC

∞

(

∞) with φ ≡ 1 on Ω

′

. Then φ f ǫH

and f = φ f on Ω

′

Corollary 3.16. If L =

|∝|≤k

∝

(x)D

∝

with a

∝

ǫC

∞

(Ω), then L maps

loc

(Ω) into H

loc
s

−k

(Ω) for all sǫR.

It is a consequence of the Arzela-Ascoli theorem that if (u

) is a

sequence of C

functions such that

| and |∂

|(|α| ≤ k) are bounded

on compact set uniformly in j, there exists a subsequence (v

) of (u

)

such that (∂

) converges uniformly on compact set for

|α| ≤ k − 1. In

particular, if the u

′

s are supported in a common compact set, then (∂

)

converges uniformly.

There is an analogue of this result for H

spaces.

Lemma 3.17. Suppose (u

) is a sequence of C

∞

functions supported in

a fixed compact set Ω such that sup

||u

(s)

∞. Then there exists a

subsequence which converges in the H

norm for all t < s.

Proof. Pick a φǫC

∞

such that φ = 1 on Ω so that u

= φu

and hence

ˆu

= ˆφ

∗

ˆu

. Then

(1 +

|ξ|

)

s/2

|ˆu

(ξ)

| = (1 + |ξ|

)

s/2

φ(ξ

− η)ˆu

(η)dη

≤

| ˆφ(ξ − η)||ˆu

(η)

|s|/2

(1 +

|η|

)

s/2

(1 +

|ξ − η|

)

|s|/2

dη

≤ 2

|s|/2

||φ||

(

|s|)

||u

(s)

≤ c −

independent of k.

Likewise, we have

(1 +

|ξ|

)

s/2

|∂

ˆu

(ξ)

| ≤ 2

|s|/2

||2πx

φ(x)

|s|

||u

(s)

≤ c

3. L

Sobolev Spaces

independently of k. Therefore, by the Arzela-Ascoli theorem there ex-
ists a subsequence (ˆv

) of (ˆu

) which converges uniformly on compact

sets. For t

≤ s,

||v

− v

2
(t)

(1 +

|ξ|

)

|ˆv

− ˆv

dξ

|ξ|≤R

(1 +

|ξ|

)

|ˆv

− ˆv

dξ +

|ξ|>R

(1 +

|ξ|

)

|ˆv

− ˆv

dξ

< (1 + R

)

max(t,0)

sup

|ξ|≤R

|ˆv

(ξ)

− ˆv

(ξ)

|ξ|≤R

dξ+

+ (1 + R

)

−s

|ξ|>R

[(1 +

|ξ|

)

|ˆv

(ξ)

− ˆv

(ξ)

dξ]

≤ c(l + R

)

|t|

sup

|ξ|≤R

|ˆv

(ξ)

− ˆv

(ξ)

+ (1 + R

)

−s

||v

− v

2
(s)

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

− v

(s)

≤ c and

− s < 0. Then for j and k large enough the first term is less than ǫ/2,

since (ˆv

) converges uniformly on compact sets. Thus we see that (v

)

is a Cauchy sequence in H

′

and since H

is complete we are done.

Remark 3.12. Lemma 3.17 is false, if we do not assume that all the u

′

have support in a fixed compact set. For example, for uǫC

∞

and x

ǫR

with

| tending to ∞, define u

(x) = u(x

−x

). Then the invariance of H

norms shows that

||u

(s)

||u||

(s)

. But no subsequence of (u

) converges,

in any H

. For, if a subsequence (v

) of (u

) converges, it must converge

to 0, since u

converges to 0 in S

′

. But then lim

||v

(t)

= 0 which is not

the case.

Theorem 3.20 (RELLICH THEOREM). Let H

(Ω) be the closure of

∞

(Ω) in H

. If Ω is bounded and t < s, the inclusion H

(Ω) ֒

→ H

compact, i.e., bounded sets in H

(Ω) are relatively compact in H

Proof. Let (u

) be a sequence in H

(Ω). To each k, find a v

ǫC

∞

(Ω)

1. General Theory of L

Sobolev Spaces

such that

||u

− v

(s)

≤

. Then we have

||v

(s)

≤ ||u

(s)

| ≤ c

(independent of k.) Therefore, by lemma 3.17, a subsequence (w

) of

) exists such that (w

) converges in H

. If (u

′

) is the subsequence of

) corresponding to the sequence (w

), we have

||u

′

− u

′

(t)

||u

′

− w

(t)

||w

− w

(t)

||w

− u

′

(t)

||w

− w

(t)

→ 0 as ||i, j → ∞.

Hence (u

′

) converges in H

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)| ≤ c

and

|F(l + it)| ≤ c

, then

F(s + it)

≤ c

−s

, for 0 < s < 1.

Proof. If ǫ > 0, the function

(z) = c

−1

ǫ(z

−z)

f (z)

satisfies the hypotheses with c

and c

replaced by 1, and also

(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

−∞ < s

< s

∞ and T is a bounded

linear operator H

such that T

is bounded on H

. Then T

bounded on H

for all s with s

≤ s ≤ s

Proof. Our hypothesis means that

∧

−s

and

∧

−s

are bounded operators on L

. For 0

≤ Rez ≤ 1 we define

= (l

− z)s

+ zs

and T

∧

−s

3. L

Sobolev Spaces

Then what we wish to prove is that T

is bounded on L

for 0

≤ z ≤

1. Observe that when w = x + iy,

∧

and

(

∧

f )ˆ(ξ) = (1 +

|ξ|

)

iy/2

(ξ) so that

|(∧

f )ˆ(ξ) =

| ˆf(ξ)|.

Thus

∧

is unitary on H

for all s.

For φ, ψǫS , we define

F(z) =

φ)ψ =<

∧

−s

φ, ψ > .

Then

|F(z)| = | < ∧

∧

−s

φ, ψ >

| < T ∧

φ,

∧

−s

ψ >

≤ ||T ∧

−s

(

)

|| ∧

−s

(

−s

)

≤ c|| ∧

−s

|| ∧

(

−s

)

≤ c||φ||

−s

)Re z

||ψ||

−s

)Rez

≤ c||φ||

(0)

||ψ||

−s

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)| ≤ c

||φ||

(0)

||ψ||

(0)

and when Rez = 1, we have

|F(z)| ≤ c

||φ||

(0)

||ψ||

(0)

Therefore, by the Three lines lemma

|F(z)| ≤ c

−z

z
0

||φ||

(0)

||ψ||

(0)

for 0 < z < 1.

Finally, by the self duality of H

= L

, this gives

||T

|| ≤ c

−z

z
0

||φ||

(0)

which completes the proof.

1. General Theory of L

Sobolev Spaces

Remark 3.22. The same proof also yields the following more general
result:

Suppose

∞ < s

< s

∞ < t

< t

∞. If T is a bounded

linear operator from H

to H

whose restriction to H

to H

, then the

restriction of T to H

is bounded from H

to H

for 0 < θ < 1 where

= (1

− θ)s

+ θs

and t

= (1

− θ)t

+ θt

As a consequence of this result, we obtain an easy proof that H

loc

invariant under smooth coordinate changes.

Theorem 3.23. Suppose Ω and Ω

′

are open subsets of R

and φ : Ω

→

Ω

′

is a C

∞

diffeomorphism. Then the mapping f

→ f oφ maps H

loc

(Ω

′

continuously onto H

loc

(Ω).

Proof. The statement of the theorem is equivalent to the assertion that
for any φǫC

∞

(Ω

′

), the map T f = (φ f )

◦ φ is bounded on H

for sǫR. If

s = 0, 1, 2, . . . , this follows from the chain rule and the fact that H

{ f : D

∝

f ǫL

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 :

∗

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

for all s

≥ 0 and by

duality of H

and H

−s

, this yields the boundedness of T on H

for s < 0.

Finally, we ask to what extend the H

spaces include all distribu-

tions. Globally they do not, since, if f ǫH

, then f is tempered and ˆ

f 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
H

: i.e., E

′

⊂

∈R

Proof. If f ǫE

′

, then it is a continuous linear functional on C

∞

. There-

fore, there exists a constant c > 0, a compact set K, and a nonnegative
integer k such that

| < f, φ > | ≤ c

|∝|≤k

sup

∝

| for all φǫC

∞

i.e.,

| < f, φ > | ≤ c P

|α|≤|k

sup

| for all φǫC

∞

3. L

Sobolev Spaces

By the Sobolev imbedding theorem,

|∝|≤k

sup

| ≤ c

′

||φ||

(k+

n
2

+ǫ)

for ǫ > 0.

Therefore,

| < f, φ > | ≤ c

′′

||(

n
2

+ǫ

) for all φǫS since S is dense in

(k+

n
2

+ǫ)

, this shows that f is a continuous linear functional on H(

n
2

+ǫ

Hence f ǫH

−

n
2

−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

loc

(Ω

′

2 Hypoelliptic Operators With Constant

Coe

fficients

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) = (

∂

∂ξ

)

∝

P(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

P(D)( f g) =

|∝|≤k

∝!

(

∝)

(D)g)D

∝

f ).

The proof of this proposition is left as an exercise to the reader.

Definition 3.28. We say that a polynomial P satisfies condition (H) if

there exists a δ > 0 such that

(

∝)

(ξ)

|P(ξ)|

= 0 (

|ξ|

−δ|α|

) as

|δ| → ∞, ∀ ∝ .

Theorem 3.29 (H ¨

ORMANDER). If P satisfies condition (H), then P(D)

is hypoelliptic. More precisely, if f is in D

′

Ω)and P(D) f H

loc
s+kδ

(Ω),

where δ is as in condition (H) and k is the degree of P.

2. Hypoelliptic Operators With Constant Coefficients

Proof. We first observe that the second assertion implies the first since
C

∞

(Ω) =

sǫR

loc

(Ω) by Corollary 3.8.

Suppose therefore that P(D) f ǫH

loc

(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 ǫH

loc

(Ω

′

). By decreasing t, we can some that t = s + k

−

− mδ for assume positive integer m. Set φ

= φ and then choose

−1

, φ

−2

, . . . , φ

, φ

−1 in C

∞

(Ω

′

) such that φ

= 1 on supp φ

j+1

Then φ

P(D) f ǫH

⊂ H

−k+l+ jδ

for 0

≤ j ≤ m and φ

−1

f ǫH

. Now

P(D)(φ

f ) = φ

P(D) f +

∝,0

∝!

∝

(D)(φ

−1

f )D

∝

φ0

since θ

−1

= 1 on the support of φ

. So, P(D)(φ

f )ǫH

−k+1

. This means

that

(1 +

|ξ|

)

−k+1

|P(ξ)(φ

f )ˆ(ξ)

dξ <

∞.

By condition (H)

(1 +

|ξ|

)

−k+l+δ|δ|

∝|

(ξ)(φ

f )

(

ξ)

dξ <

∞

This implies that P

(α)

(D)(φ

◦

f )ǫH

− k + l + δ|δ|.

Next,

P(D)(φ

f ) = φ

P(D) f +

∝,0

∝!

(

∝)

(D)(φ

f )D

∝

(φ

f )

since φ

= 1 on the support of φ

, so P(D)(φ

f )ǫH

−k+1+δ

. By the same

argument same argument as above,

(D)

(φ

f )ǫH

t+k+l+δ(1+

|∝|)

Continuing inductively, we obtain P(D)(φ

f )ǫH

−k+l+ jδ

, which im-

plies that

(

∝)

(D)(φ

f )ǫH

−k+l+δ( j+|∝|)

3. L

Sobolev Spaces

For j = m, we above

(

∝)

(D)(φ

f )ǫH

−k+l+δ(m+(δ))

= H

s+δ

|∝|

P(ξ) =

|∝|≤k

∝

choose

∝ with | ∝ | = k such that a

∝

0. Then P

∝

(ξ) =

∝!a

∝

whence φ f = φ

f ǫH

s+kδ

and we are done.

Remark 3.30. The condition (H) is equivalent to the following appar-

ently weaker condition

′

) :

(

∝)(ξ)

|P(ξ)|

→ 0 as |ξ| → ∞ for ∝, 0.

Condition (H

′

) is in turn equivalent to

′′

) :

|Imζ| → ∞, in the set {ζǫ

: P(ζ) = 0

The converse of H ¨ormander’s theorem is also true, i.e., hypoelliptic-

ity implies condition (H).

The proofs of these assertions can be found in H ¨ormander [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.

Definition 3.31. P(ξ) =

|∝|≤k

∝

is called elliptic if

|∝|=k

∝

0 for

every ξ , 0.

EXERCISES

1. Prove that P is elliptic if and only if

|P(ξ)| ≥ c

◦

|ξ|

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

3. Prove that no P satisfies condition (H) with δ > 1. (Hint: If

| ∝ | = k, P

(

∝)

is a constant).

4. Let P be elliptic and real valued. Define Q on R

n+1

by Q(ξ, τ) =

2πiτ + P(ξ), so that Q(D

, D

) = ∂

+ P(D

). 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|

≤ |τ| and |τ| ≤ |ξ|

separately).

Chapter 4

Basic Theory of Pseudo
Di

fferential Operators

1 Representation of Pseudo di

fferential Operators

Let L =

|α|≤k

(x)D

be a partial differential operator with C

∞

coeffi-

cients on Ω. Using the Fourier transform, we can write

(Lu)(x) =

|α|≤k

(x)

2πix.ξ

ˆu(ξ)ξ

dξ

2πix.ξ

p(x, ξ)ˆu(ξ)dξ

where p(x, ξ) =

|α|≤k

(x)ξ

. This representation suggests that p(x, ξ)

can be replaced by more general functions. So, we make the following
definition.

Definition 4.1. For an open set Ω

⊂ R

and a real number m we define

(Ω), the class of symbols of order m on Ω, by

(Ω) =

{pǫC

∞

(Ω

× R

) :

∀α, β, VΩ

′

⊂ Ω, c = c

αβΩ

′

such that sup

xǫΩ

′

α
ξ

p(x, ξ)

| ≤ c(1 + |ξ|)

−|α|

4. Basic Theory of Pseudo Differential Operators

We note that S

(Ω)

⊂ S

whenever m < m

′

, and we set S

−∞

(Ω) =

mǫR

(Ω).

Examples

(i) Let p(x, ξ) =

|α|≤k

(x)ξ

with a

ǫC

∞

(Ω). Then pǫS

(Ω).

(ii) Let p(x, ξ) =

j=1

(x) f

(ξ) where a

ǫC

∞

(Ω) and f

ǫC

∞

) is

homogeneous of degree m

large ξ : that is,

(rξ) = r

(ξ) for

|ξ| ≥ c, r ≥ 1.

In this case, pǫS

(Ω), where m = max

≤ j≤N

(iii) Let p(x, ξ) = (1 +

|ξ|

)

s/2

. This p belongs to S

(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

Remark 4.2. We observe that when pǫS

(Ω), D

α
ξ

pǫS

−|α|

(Ω).

Further, if pǫS

(Ω) and qǫS

(Ω),then p + qǫS

(Ω), where m =

max

, m

} and pqǫS

(Ω).

Remark 4.3. Our symbol classes S

(Ω) are special cases of Horman-

der’s classes S

m
ρ,δ

(Ω). Namely, for 0

≤ δ ≤ ρ ≤ 1, and mǫR,

m
ρ,δ

(Ω) =

{pǫC

∞

(Ω

× R

) :

∀α, β, VΩ

′

⋐ Ω, c = c

αβΩ

′

such that

sup

xǫΩ

′

α
ξ

p(x, ξ)

| ≤ c(1 + |ξ|)

−ρ|α|+δ|β|

In this terminology, S

(Ω) = S

m
1,0

(Ω).

Definition 4.4. For pǫS

(Ω), we define the operator p(x.D) on the do-

main C

∞

(Ω) by

p(x, D) u(x) =

2πix.ξ

p(x, ξ) ˆu(ξ)dξ, uǫC

∞

(Ω).

1. Representation of Pseudo differential Operators

(Sometimes, we shall denote p(x, D) by p). Operators of the form

p(x, D) with pǫS

(Ω) are called pseudo differential operators of order

m on Ω.

The set of all pseudo differential operators of order m on Ω will de-

noted by Ψ

(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

∞

(Ω) into C

∞

(Ω which extends to E

′

(Ω). For the proof, we need a

result which depends on

Lemma 4.5. Let pǫS

(Ω) and φǫC

∞

(Ω). Then, for each positive inte-

ger N, there exists c

> 0 such that for all ξ, η in R

2πix.η

p(x, ξ)φ(x)dx

| ≤ c

(1 +

|ξ|)

(1 +

|η|)

−N

Proof. For any ξ and η in R

|η

2πix.η

p(x, ξ)φ(x)dx

2πix.η

p(x, ξ)φ(x)dx

Integrating by parts, we have

|η

2πix.ξ

p(x, ξ)φ(x)dx

| = |

2πix.η

(p(x, ξ)φ(x))dx

≤ C

(1 +

|ξ|)

for all α.

Therefore

|α|≤N

|η

2πix.η

p(x, ξ)φ(x)dx

| ≤ c

′

(1 +

|ξ|)

for all N.

Since (1 +

|η)

≤ c P

|α|≤N

|η

|, the required result follows.

4. Basic Theory of Pseudo Differential Operators

Corollary 4.6. If pǫS

(Ω) and φǫC

∞

(Ω), then the function

g(ξ) =

2πix.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

(Ω), then p(x, D) is a continuous linear map of

∞

(Ω) into C

∞

(Ω) which can be extended as a linear map from E

′

(Ω)

into D

′

(Ω).

Proof. For uǫC

∞

(Ω),

p(x, D)u(x) =

2πix.ξ

p(x, ξ)ˆu(ξ)dξ.

The integral converges absolutely and uniformly on compact sets, as

do the integrals

2πix.ξ

p(x, ξ) ˆu(ξ)dξ for all α,

since pǫS

and ˆuǫ

This proves that p(x, D)uǫC

∞

(Ω), and continuity of p(x, D) from

∞

(Ω) 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 C

∞

(Ω) as fol-

lows:

< p(x, D)u, φ > =

p(x, ξ)ˆu(ξ)e

2πix.ξ

φ(x)dxdξ

g(ξ)ˆu(ξ)dξ

where

g(ξ) =

2πix.ξ

p(x, ξ)φ(x)dx, for φǫC

∞

(Ω).

2. Distribution Kernels and the Pseudo Local Property

By Corollary 4.6, g(ξ) is rapidly decreasing, while ˆu is of polyno-

mial growth; so the last integral is absolutely convergent and the func-
tional on C

∞

thus defined is easily seen to be continuous. Moreover, if

uǫC

∞

, the double integral is absolutely convergent, and by interchang-

ing the order of integration, we see that this definition of p(x, D)u coin-
cides 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 u

converges to u in E

′

(Ω), then

p(x, D)u

converges to p(x, D)u in

′

(Ω). Actually, p(x, D) is contin-

uous from E

′

(Ω) to

′

(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)

: C

∞

(Ω)

→ C

∞

(Ω) is con-

tinuous, and so, by duality, p(x, D) = (p(x, D)

)

: E

′

(Ω)

→ D

′

(Ω) is

continuous.

2 Distribution Kernels and the Pseudo Local Prop-

erty

Definition 4.9. Let T be an operator from C

∞

(Ω) to C

∞

(Ω). If there

exists a distribution K on Ω

× Ω such that

< T u, v >=< k, v

⊗ u > for u, vǫC

∞

(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.

T u(x) =

K(x.y)u(y)dy.

K is uniquely determined since linear combinations of functions of the

form v

⊗ u are dense in C

∞

(Ω

× Ω).

If pǫS

(Ω), it is easy to compute the distribution kernel of p(x, D).

In fact,

< p(x, D)u, v > =

p(x, ξ)ˆu(ξ)e

2πix.ξ

v(x)dξdx

4. Basic Theory of Pseudo Differential Operators

p(x, ξ)e

2πix.ξ

⊗ u)

(x, ξ)dξdx

where (v

⊗ u)

(x, ξ) means the Fourier transform in the second variable.

It follows immediately from the definition of K that

< K, w > =

2πix.ξ

p(x, ξ)w

(x, ξ)dξdx,

∀wǫC

∞

(Ω

× Ω).

or < K, w > =

2πi(x

−y).ξ

p(x, ξ)w(x, y)dy dξ dx.

From this it is easy to see that K(x, y) = p

v
2

(x, x

− y) where p

v
2

(x, .) is

the inverse Fourier transform of the tempered distribution p(x, .). In par-
ticular, this shows that p is uniquely determined by the operator p(x, D).
If PǫΨ

(Ω), we shall sometimes denote the unique pǫS

(Ω) such that

p = p(x, D) by σ

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

(Ω) is in

∞

on (Ω

× Ω)/∆ where ∆{(x, x) : xǫΩ} is the diagonal. More precisely,

|α| > M + n + j for a positive integer j, then (x − y)

K(x, y)ǫC

(Ω

× Ω).

Proof. For wǫC

∞

(Ω

× Ω), let us compute < (x − y)

K, w >.

< (x

− y)

K, w > =< K, (x

− y)

w >

2πix.ξ

p(x, ξ)(x + D

)

(x, ξ)dξ dx

(x, ξ)(x

− D

)

2πix.ξ

p(x, ξ)

}dξ dx.

Using the Leibniz formula, it is easily seen that

− D

)

2πix.ξ

p(x, ξ)

} = (−D

)

p(x, ξ).e

2πix.ξ

Therefore

< (x

− y)

K, w > =

(x, ξ)e

2πix.ξ

(

−D

)

p(x, ξ)dξ dx

w(x, y)e

2πi(x

−y).ξ

(

−D

)

p(x, ξ)dy dξ dx.

2. Distribution Kernels and the Pseudo Local Property

From the above expression, we infer that

− y)

K(x, y) =

2πi(x

−y).ξ

(

−D

)

p(x, ξ)dξ.

Since

|(−D

)

p(x, ξ)

| ≤ c(1 + |ξ|)

−|α|

, the integral converges abso-

lutely 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

− y)

KǫC

(Ω

× Ω).

Corollary 4.11 (PSEUDO LOCAL PROPERTY OF ΨDO). If PǫΨ

(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 φǫC

∞

(V) such that φ = 1 on sing supp u. Then u = φu+(1

−φ)u =

+ u

; u

is a C

∞

function and supp u

ǫV.

Therefore, Pu = Pu

+ Pu

and Pu

is a C

∞

function. Moreover,

when x

(x) =

K(x, y)u

(y)dy

is also a C

∞

function in a neighbourhood of x

since u

(y) = 0 for y near

. 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

P(x, D)u(x) =

K(x, y)u(y)dy =< u, K(x, .) >,

whence the result follows.

4. Basic Theory of Pseudo Differential Operators

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 >=

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 δ

)(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

Proposition 4.15. Suppose p(x, .) is a C

∞

function of x with values in

′

. Then p(x, D) (defined in the same way as in the case of pǫS

(Ω)) is

a smoothing operator.

Proof. The distribution kernel K of p(x, D) is given by

K(x, y) =

2πi(x

−y).ξ

p(x, ξ)dξ

=< p(x, .), e

2πi(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 PǫΨ

(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

∞, so that symbols of order −∞ are negligible.

3. Asymptotic Expansions of Symbols

3 Asymptotic Expansions of Symbols

Definition 4.17. Suppose m

> m

· · · m

ǫR, m

tends to

−∞,

and p

ǫS

(Ω), pǫS

(Ω). We say that p

∼

∞

j=0

if p

−

j<k

ǫS

(Ω)

for all k.

Proposition 4.18. Suppose m

tends to

−∞ and p

ǫS

m j

(Ω). Then there

exists a p in S

(Ω) such that p

∼

∞

P j = 0p

. This p is unique modulo

−∞

(Ω).

Proof. Let (Ω

) be an increasing sequence of compact subsets of Ω

whose union is Ω. Fix φǫC

∞

with φ = 1 for

|ξ| ≥ 1 > and φ = 0

for

|ξ| ≤

Claim

There exists a sequence (t

),t

≥ 0 and t

tending to

∞ so

rapidly that we have
(4.19)
|D

α
ξ

(φ(ξ/t

(x, ξ))

| ≤ 2

− j

(1+

|ξ|)

−1

−|α|

for xǫΩ

, and

|α|+|β|+i ≤ j.

Granted this, we define

p(x, ξ) =

∞

j=0

φ(ξ/t

(x, ξ).

Note that for each x and ξ, the sum is finite. Using (4.19), it is

straightforward to show that pǫS

(Ω) and p

∼

∞

j=0

. Moreover, sup-

pose that qǫS

(Ω) and q

∼

∞

j=0

. Then

− q = (p −

j<k

)

− (q −

j<k

)ǫS

(Ω) for all k.

Hence p

− qǫS

−∞

we now briefly indicate the steps involved in proving the claim :

i) First observe that

ν
ξ

φ(ξ/t

)

| ≤ c|ξ|

−|ν|

uniformly in j.

4. Basic Theory of Pseudo Differential Operators

ii) Hence we have

α
ξ

(φ(ξ/t

(x, ξ))

| ≤ c

(1 +

|ξ|)

−|α|

for xǫΩ

|α| + |β| + i ≤ j.

iii) Finally, pick t

so large that “

|ξ| ≥ t

≥ c

(1 +

|ξ|)

−m

−1

≤ 2

− j

”

Details are left to the reader.

The following theorem provides a useful criterion for the asymptotic

relation p

∼

∞

j=0

to hold.

Theorem 4.20. Suppose p

ǫS

, m

tends to

−∞, and pǫC

∞

(Ω

× R

)

satisfies following conditions:

i) for all α and β and all Ω

′

⋐ Ω, there exists c > 0, µǫRsuch that

α
ξ

p(x, ξ)

| ≤ c(1 + |ξ|)

, xǫΩ, and

ii) there exists a sequence (µ

), µ

tending to

−∞ so that |p(x, ξ) −

j<k

(x, ξ)

| ≤ C

Ω

, (1 +

|ξ|)

for xǫΩ

′

. Then pǫS

(Ω) and p

∼

∞

j=0

To prove this theorem, we need the following

Lemma 4.21. Let Ω

and Ω

be two compact subsets of R

such that Ω

is contained in the interior of Ω

. Then there exists constants c

> 0 and

> 0 such that for all f ǫC

(Ω

Sup

Ω

|∂

≤ c

(Sup

Ω

| f |

) + c

(Sup

Ω

| f |)(Sup

Ω

|∂

|).

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

′′

∞

are both small, then so is

k f

′

∞

is small and

| f

′

)

| is large, then

| f

′

| will be large in some sizable interval [a, b] containing x

. 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

Proof of the theorem . By Proposition 4.18, there exists q in S

such

that q

∼

∞

j=0

; so, it will suffice to show that p

− qǫS

−∞

. First,

|p(x, ξ) − q(x, ξ)| = |(p(x, ξ) −

j<k

(x, ξ))

− (q(x, ξ) −

j<k

(x, ξ))

For

Ω

′

⊂⊂ Ω, |p(x, ξ) −

j<k

(x, ξ)

| ≤ c

Ω

, (1 +

|ξ|)

, xǫΩ

′

Also q

∼

∞

j=0

. These show that, for any N,

|p(x, ξ)| − q(x, ξ)| ≤ C

NΩ

′

(1 +

|ξ|)

−N

, xǫΩ

′

We want to prove such an estimate for D

α
ξ

− q) also. To this

end, we will apply Lemma 4.21 to the function (x, η)

→ (p − q)(x, ξ +

η) considering ξ as a parameter, and taking Ω

= Ω

′

× 0 and Ω

small neighbourhood of Ω

. If

|α| + |β| = 1, we use the estimate just

established for p

− q, together with the hypothesis (i) on the second

order derivatives of p; the lemma implies that D

α
ξ

− q) is rapidly

decreasing for

|α| + |β| = 1. Combining this with the hypothesis (i) on

the third order derivatives of p, we see that D

α
ξ

is rapidly decreasing

for

|α| + |β| = 2. Proceeding by induction on |α| + |β| we get the required

result.

4 Properly Supported Operators

Since pseudo differential operators map C

∞

to C

∞

rather then C

∞

, it

is generally not possible to compose two of them. The problem may
be remedied by considering a more restricted class of operators, the so
called “properly supported” ones.

Definition 4.22. A subset K of Ω

× Ω is said to be proper if, for any

compact set Ω

′

⊂ Ω, both π

−1

(Ω

′

)

∩ K and π

−1

(Ω

′

)

∩ K are compact.

4. Basic Theory of Pseudo Differential Operators

Here π

and π

are projections of Ω

× Ω onto the first and second

factors.

For example the diagonal ∆ =

{(x, x) : xǫΩ} is proper. Most of the

proper subsets we will be considering are neighbourhoods of subsets of
the diagonal.

Figure 4.1: A proper set

Definition 4.23. An operator T : C

∞

(Ω)

→ C

∞

(Ω) is said to be prop-

erly supported, if its distribution kernel K has proper support.

Exercise. Let T =

P a

(x)D

be a differential operator on Ω. Computer

the distribution kernel K of T and show that supp K is a subset of the
diagonal. Thus T is properly supported.

If T is properly supported then it maps C

∞

into itself, since supp

T u

⊂ π

(π

−1

(supp u)

∩ supp K), as is easily seen from the formula

T u(x) =

K(x, y)u(y)dy. More generally, for any Ω

′

⊂⊂ Ω, there exists

Ω

′′

⊂⊂ Ω such that the values of T u on Ω

′

depend only on the values of

u on Ω

′′

namely, Ω

′′

= π

(π

−1

(Ω

′

)

∩ supp K). From this it follows that

T can be extended to a map from C

∞

(Ω) to itself. In fact, if uǫC

∞

(Ω)

and Ω

′

, Ω

′′

are as above, we define Tu on Ω

′

T u

Ω

′

= T (φu)

Ω

where φǫC

∞

(Ω) and φ = 1 on Ω

′′

. This definition is independent of the

choice of φ and gives the same result on the intersection of two Ω

′

s; so

T u is well defined on all of Ω.

If T is a properly supported pseudo differential operator, so that T

extends to a map from E

′

(Ω) to

′

(Ω), the same arguments show that T

maps E

′

(Ω) into itself and extends further to a map of

′

(Ω) to itself.

Suppose T and S are properly supported operators on C

∞

(Ω) with

distribution kernels K and L which are C

∞

off the diagonal. Then T S

5. ψdo

′

Defined by Multiple Symbols

is an operator on C

∞

(Ω) with distribution kernel M formally given by

M(x, y) =

K(x, z)L(z, y)dz. In fact, if x , y, since K(x, .) and L(., Y)

are smooth except at x and y, the product K(x, .)L(., Y) is a well defined
element of E

′

, and the formula M(x, y) =< K(x, .)L(., y), 1 > displays M

as a C

∞

function off the diagonal.

Proposition 4.24. Supp M is proper. Thus, T S is properly supported.

Proof. Clearly,

Supp M

⊂ {(x, y) : π

(π

−1

(x)

∩ supp K) ∩ π

(π

−1

(y)

∩ supp L) , φ}

Suppose A

⊂ Ω is compact and set B = π

(π

−1

(A)

∩ supp K), then B is

compact and

−1

(A)

∩ supp M ⊂ A ×

y : B

∩ π

(π

−1

(y)

∩ supp L) , φ

= A

y : π

−1

(B)

∩ π

−1

(y)

∩ supp L , φ

= A

× π

(π

−1

(B)

∩ supp L)

which is compact. Likewise π

−1

(A)

∩ supp M is also compact.

Exercise . Suppose A

⊂ (Ω × Ω) is proper. Show that there exists a

properly supported φǫC

∞

(Ω

×Ω) such that φ = 1 on A. (Hint: Let {φ

} ⊂

∞

(Ω

× Ω) be a partition of unity on Ω × Ω and let φ =

∩supp φ

= φ

5 ψdo

′

s Defined by Multiple Symbols

We have arranged our definition of pseudo differential operators to agree
with the usual convention for differential operators, according to which
differentiations are performed first, followed by multiplication by the
coefficients. However, in some situations (for example, computing ad-
joints), it is convenient to have a more flexible setup which allows mul-
tiplication operators both before and after differentiations. We there-
fore, introduce the following apparently more general class of operators
(which, however, turns out to coincide with the class of ψDO, modulo
smoothing operators).

4. Basic Theory of Pseudo Differential Operators

Definition 4.25. For Ω open in R

and m real, we define the class of

multiple symbols of order m on Ω,

(Ω

× Ω) = {aǫC

∞

(Ω

× R

× Ω) : ∀α, β, ∀Ω

′

⊂⊂ Ω, ∃c = c

αβΩ

′

such that sup

x,yǫΩ

′

x,y

α
ξ

a(x, ξ, y)

| ≤ c(1 + |ξ|)

−|α|

}

When aǫS

(Ω

× Ω) we define the operator A = a(x, D, y) by

Au(x) =

2πi(x

−y).ξ

a(x, ξ, y)u(y)dydξ

Here the integral must be interpreted as an iterated integral with in-

tegration performed first in y then in ξ as it is not absolutely convergent
as a double integral.

We observe that if a(x, ξ, y) = a(x, ξ) is independent of y, then

A = a(x, D); thus this class of operators include the ψDO

′

s. We also

observe that different a

′

s may give rise to same operator. For example,

if a(x, ξ, y) = φ(x)ψ(y) with ψ, φǫC

∞

(Ω) and supp ψ

∩ supp φ = φ, then

aǫ s

(Ω

× Ω) and a(x, D, y) = 0.

Definition 4.26. Given aǫS

(Ω

× Ω) we define

{(x, y)ǫΩ × Ω : (x, ξ, y)ǫ supp a for some ξǫR

Proposition 4.27. Let aǫS

(Ω

× Ω) and let K be the distribution kernel

of A = a(x, D, y). Then

i) Supp K

⊂ P

, and

ii) if the support of K is proper, then there exists a

′

ǫS

(Ω

×Ω) such that

′

(x, ξ, y) = a(x, ξ, y) when (x, y) is near the diagonal in Ω

× Ω, P

is proper and a(x, D, y) = a

′

(x, D, y).

Proof. The kernel K is given by

< K, w >=

2πi(x

−y).ξ

a(x, ξ, y)w(x, y)dydξdx.

From this, we see that < K, w > . = 0 whenever supp w

∩ P

= φ.

Therefore, supp K

⊂ P

. This proves (i).

5. ψdo

′

Defined by Multiple Symbols

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).

Then

⊂ supp φ and hence P

, 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) >

2πi(x

−y)·ξ

a(x, ξ, y)φ(x, y)

× v(x)u(y)dydξdx

2πi(x

−y).ξ

′

(x, ξ, y)u(y)dydξdx

=< a

′

(x, D, y)u, v > .

Since v is arbitrary, we have a(x, D, y) = a

′

(x, D, y).

Theorem 4.28. Suppose aǫS

(Ω

× Ω) and A = a(x, D, y) is prop-

erly supported. Let p(x, ξ) = e

−2πix.ξ

A(e

2πi(.).ξ

)(x). Then pǫS

(Ω) and

p(x, D) = A. Further,

p(x, ξ)

∼

α!

∂

α
ξ

α
y

a(x, ξ, y)

y=x

Proof. For u in C

∞

(Ω), we have

u(x) =

2πix.ξ

ˆu(ξ)dξ.

Therefore, by linearity and continuity of A,

Au(x) =

A(e

2πi(.).ξ

)(x)ˆu(ξ)dξ

2πix.ξ

p(x, ξ)ˆu(ξ)dξ = p(x, D)u(x).

4. Basic Theory of Pseudo Differential Operators

By the proposition above, we can modify a so that

is proper with-

out 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,

for fixed x and ξ, if yǫ

′

: b(x, ξ, y

′

) , 0

} then x + yǫπ

(π

−1

(x)

∩ P

)

which is compact. Now,

p(x, η) = e

−2πix.η

2πi(x

−y).ξ

b(x, ξ, y

− x)e

2πiy.η

dydξ

= e

−2πix.η

2πiz.ξ

b(x, ξz)e

2πi(z+x).η

dzdξ

ˆb

(x, ξ, ξ

− η)dξ =

ˆb

(x, ξ + η, ξ)dξ,

where ˆb

is the Fourier transform of b in the third variable. Since

b(x, ξ, .) is in C

∞

, this calculation is justified and ˆb

(x, η, ξ) is rapidly

decreasing in the variable ξ.

More precisely, since aǫS

(Ω

× Ω), we have

(4.29)

α
η

ˆb

(x, η, ξ)

| ≤ c

αβN

(1 +

|η|)

−|α|

(1 +

|ξ|)

−N

Since

1 + |η|

1 +

|ξ|

≤ (1 + |ξ − η|)

|s|

, taking η + ξ instead of η

we have (1 +

|ξ + η|)

≤ (1 + |ξ|)

(1 +

|η|)

|s|

. If we use this in

α
η

ˆb

(x, ξ + η, ξ)

| ≤ c

αβN

(1 +

|ξ + η|)

−|α|

(1 +

|ξ|)

−N

we get

α
η

ˆb

(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

α
η

p(x, η)

| = |

α
η

ˆb

(x, ξ + η, ξ)dξ

(4.30)

≤ c

αβ

(1 +

|η|)

|m|+|α|

5. ψdo

′

Defined by Multiple Symbols

On the other hand, taking the Taylor expansion of ˆb

(x, ξ + η, ξ) in

100

the middle argument about the point η, (4.29) gives

|ˆb

(x, ξ + η, ξ)

−

|α|<k

∂

α
η

ˆb

(x, η, ξ)

α!

≤ C|ξ|

sup

|α|=k

≤t≤1

|∂

α
η

ˆb

(x, η + tξ, ξ)

≤ C

|ξ|

sup

≤t≤1

(1 +

|η + tξ|)

−k

(1 +

|ξ|)

−N

When

|ξ| <

1
2

|η|, taking N = k we get

|ˆb

(x, ξ + η, ξ)

−

|α|<k

∂

α
η

ˆb

(x, η, ξ)

α!

| ≤ c

(1 +

|η|)

−k

When

|ξ| ≥

|η|, we have

|ˆb

(x, ξ + η, ξ)

−

|α|<k

∂

α
η

ˆb

(x, η, ξ)

α!

| ≤ C

(1 +

|ξ|)

m+k

−N

(Actually, the exponent of (1 +

|ξ|)) can be taken as m − N when

− k ≥ 0 and when m − k < 0). Also we have

∂

α
η

ˆb

(x, η, ξ)ξ

dξ = D

α
Y

2πiY

·ξ

∂

α
η

ˆb

(x, η, ξ)dξ

Y=0

= D

α
Y

∂

α
η

b(x, η, y)

Y=0

= D

α
Y

∂

α
η

a(x, η, y)

y=x

So finally

|p(x, η) −

|α|<k

α
y

∂

α
η

a(x, η, y)

α!

y=x

ˆb

(x, ξ + η, ξ)dξ

−

|α|<k

α!

∂

α
η

ˆb

(x, η, ξ)ξ

dξ

4. Basic Theory of Pseudo Differential Operators

≤ C

|ξ|<

|η|

(1 +

|η|)

−k

dξ + C

|ξ|≥

|η|

(1 +

|ξ|)

m+k

−N

dξ.

Now

101

|ξ|<

|η|

(1 +

|η|)

−k

dξ = C

′

(1 +

|η|)

−k

|η|

≤ C

′

(1 +

|η|)

−k+n

In the second integral, we take N > n + m + k so that

|ξ|

≥ |η|

(1 +

|ξ|)

−N

dξ

≤ C

′

(1 +

|η|)

−N+n

by the usual integration in polar coordinates. Therefore, since N

≥ k we

obtain

|p(x, η) −

|α|<k

α!

α
y

α
η

a(x, η, y)

x=y

| ≤ C

(1 +

|η|)|

with µ

= m

− k + n.

Combining this with (4.30), we see that all the conditions of Theo-

rem 4.20 are satisfied, so we are done.

Corollary 4.31. If Pǫψ

, there exists a Qǫψ

such that Q is properly

supported and P

− Q is smoothing.

Proof. If P = p(x, D) choose φǫC

∞

(Ω

× Ω) which is properly sup-

ported and φ = 1 near the diagonal. Set a(x, ξ, y) = φ(x, y)p(x, ξ). Then
aǫS

(Ω

× Ω) and by construction

is proper. So a(x, D, y) = Qǫψ

If K

and K

denote the distribution kernels of P and Q then it is eas-

ily seen that K

= φK

. This shows that K

− K

vanishes near the

diagonal: so, by Theorem 4.10, K

− K

is C

∞

. Therefore, Q

− p is

smoothing.

Remark 4.32. The idea of multiple symbols can clearly be generalised.
For example, if aǫC

∞

(Ω

× R

× Ω × R

) satisfies estimates of the form

x,y

α
ξ

β
η

a(x, ξ, y, η)

| ≤ C(1 + η|)

−|β|

(1 +

|ξ|)

−|α|

, one can define an

102

operator A = a(x, D, y, D) by

6. Products and Adjoint of ψDO

′

Au(x) =

2πi(x.ξ

−y.ξ+y.η)

a(x, ξ, y, η)ˆu(η)dηdydξ

which agree with our previous definitions, if a is independent of the last
one or two variables. As one would expect, under reasonable restriction
on supp a, it turns out that Aǫψ

(Ω). The reader may wish to amuse

himself by working this out and computing the symbol of A.

6 Products and Adjoint of ψDO

′

We are now in a position to compute products and adjoints of ψDO

′

S .

First we clarify our terminology. Let T S be linear operators from
C

∞

(Ω) into C

∞

(Ω). We say that S = T

if < T u, v >=< u, S u >

for u, vǫC

∞

(Ω) and we say that S = T

∗

if < T u, ¯v >=< u, ¯

S v > for

u, vǫC

∞

(Ω).

Remark 4.33. If T is properly supported, then T

and T

∗

are also prop-

erly supported. Indeed, if K is the distribution kernel of T , then the dis-
tribution kernel of T

is K

and that of T

∗

is K

∗

where K

(x, y) = K(y, x)

and K

∗

(x, y) = K(y, x).

We recall that if Pǫψ

(Ω), we denote the corresponding symbol in

(Ω) by σ

Theorem 4.34. If Pǫψ

(Ω) is properly supported, then P

, P

∗

ǫψ

(Ω)

and σ

t(x, ξ)

∼

(

−1)

|α|

α!

∂

α
ξ

(x,

−ξ),

∗

(x, ξ)

∼

α!

∂

α
ξ

(x, ξ).

Proof. For u, vǫC

∞

(Ω), we have

103

< Pu, v > =

2πix,ξ

p(x, ξ)ˆu(ξ)v(x)dξdx

2πix,ξ

p(x, ξ)v(x)dx

ˆu(ξ)dξ

4. Basic Theory of Pseudo Differential Operators

g(ξ)ˆu(ξ)dξ

where

g(ξ) =

2πix,ξ

p(x, ξ)v(x)dx.

By Corollary 4.6, g us a rapidly decreasing function. Thus

< Pu, v >=

gˆu =

uˆg =< u, P

t
v

so that

v(y) = ˆg(y) =

2πi(x

−y).ξ

p(x, ξ)v(x)dxdξ

(4.35)

2πi(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,

ǫψ

(Ω) and σ

(x, ξ)

∼

(

−1)

|α|

α!

∂

α
ξ

(x,

−ξ).

The assertions about P

∗

follows along similar lines

Theorem 4.36. If Pǫψ

(Ω) and Qǫψ

(Ω) are properly supported,

then QPǫψ

(Ω) and

(x, ξ)

∼

α!

∂

α
ξ

(x, ξ)D

(x, ξ).

Proof. Since P = (P

)

, we have

104

(Pu)(x) =

2πi(x

−y)·ξ

t
P

(y,

−ξ)u(y)dydξ

by (4.35).

6. Products and Adjoint of ψDO

′

In other words

(Pu)ˆ(ξ) =

−2πiy·ξ

(y,

−xi)u(y)dy.

Therefore,

(QP)u(x) =

2πi(x

−y)·ξ

(x, ξ)σ

(y,

−xi)u(y)dydξ

= a(x, D, y)u(x)

where a(x, ξ, y) = σ

(x, ξ)σ

(y,

−ξ). Clearly aǫS

(Ω

× Ω) which

implies QPǫψ

(Ω). Moreover

(x, ξ)

∼

α!

∂

α
ξ

α
y

(σ

(x, ξ)σ

(y,

−ξ))|

y=x

∼

α!

β+ν=α

β!ν!

∂

β
ξ

(x, ξ)∂

ν
ξ

α
Y

(x,

−ξ)

∼

α,ν

(

−1)

|δ|

ν!β!δ!

∂

β
ξ

(x, ξ)D

β+ν+δ

∂

ν+δ

∂

ν+δ

(x,

−ξ)

∼

β,λ







ν+δ=λ

(

−1)

|δ|

ν!β!δ!







β!

∂

β
ξ

(x, ξ)D

β+λ

∂

λ
ξ

(x, ξ)

But

v+δ=λ

(

−1)

|δ|

v!δ!

= (x

− x

)

with x

= (1, 1, . . . , 1)












1 if λ = 0
0 if λ , 0

Therefore σ

(x, ξ)

∼

β!

∂

β
ξ

(x, ξ)D

(x, ξ).

Corollary 4.37 (PRODUCT RULES FOR ψDO

′

S ).

If Q = q(x, D)ǫψ

(λ) and f ǫC

∞

(Ω), then for any positive integer N,

105

q(x, D)( f u) =

|α|<N

α!

f [(∂

α
ξ

q)(x, D)u] + T

u where T

ǫψ

−N

(Ω).

4. Basic Theory of Pseudo Differential Operators

Corollary 4.38. The correspondence p

→ p(x, D) and P → σ

are *

homomorphisms modulo lower order term i.e., if pǫS

(Ω) and qǫS

(Ω), then (pq)(x, D)

−p(x, D)q(x, D)ǫψ

−1

(Ω) and p(x, D)

∗

− ¯p(x, D)

ǫψ

−1

and also, if Pǫψ

(Ω) and Qǫψ

(Ω), then σ

− σ

ǫS

−1

(Ω), σ

∗

− ¯σ + pǫS

−1

(Ω).

Corollary 4.39. If Pǫψ

(Ω) and Qǫψ

(Ω), then [P, Q] = PQ

− QP

ǫ ψ

−1

(Ω) and σ

[P,Q]

−

2πi

{σ

, σ

}ǫS

−1

(Ω) where

{ f, q} stands

for the Poisson bracket defined by

{ f, q} =

∂ f

∂ξ

∂g

∂x

−

∂ f

∂x

∂g

∂ξ

Following the philosophy that smoothing operators are negligible

it is a trivial matter to extend these results to non-properly supported
operators.

Suppose Pǫψ

(Ω) is not properly supported. By corollary 4.31, we

can write P = P

+ s where P

is properly supported and S is smoothing.

Then P

= P

t
1

+ S where P

t
1

is given by Theorem 4.34 and S

is again

smoothing (c.f. Remark 4.33). Likewise for P

∗

. If Q is a properly

supported ψDO, the products PQ and QP are well defined as operators
from C

∞

(Ω) to C

∞

(Ω). Again, we have PQ = P

Q + S Q and QP =

+ Qs; P

Q and QP

are described by Theorem 4.36, while S Q and

QS are smoothing.

7 A Continuity Theorem for ψ Do on Sobolev

Spaces

We now state and prove a continuity theorem for pseudo defferential

106

operators acting on Sobolev spaces. We are indebted to Dr P.N. Srikanth
for simplifying our original argument.

Theorem 4.40. Suppose P = p(x, D)ǫψ

(Ω). Then

i) P : H

→ H

loc

−m

(Ω) continuous for all sǫR.

7. A Continuity Theorem for ψ Do on Sobolev Spaces

ii) If P is properly supported, P : H

loc

(Ω)

→ H

loc

−m

(Ω) continuously

for all sǫR

Proof. To established (i), we must show that the map u

→ φPu is

bounded from H

→ H

−m

for any φǫC

∞

(Ω). Replacing p(x, ξ) by

φ(x)p(x, ξ) we must show:

(4.41)

if pǫ s

(R) and p(x, ξ) = 0 for x outside a compact set, then

p = p(x, D) is bounded from H

and H

−m

for every s in R.

Suppose then that uǫH

. Then puǫE

′

, and from the definition of P

on distributions, we see that

(Pu)

(η) =< Pu, e

−2πiη.(.)

2πi(ξ

−η).x

p(x, ξ)ˆu(ξ)dxdξ

ˆp

(η

− ξ, ξ)ˆu(ξ)dξ

and hence, if VǫS ,

< Pu, ¯v > =

(Pu)

ˆv =

K(η, ξ) f (ξ)¯g(η)dξdη

where

f (ξ) = (1 +

|ξ|

)

s/2

ˆu(ξ)

g(η) = (1 +

|η|

)

−s)/2

ˆv(η)

and K(η, ξ) = ˆp

(η

− ξ, ξ)(1 + |ξ|

)

−s/2

(1 +

|η|

)

−m)/2

107

We wish to estimate K(η, ξ).

For any multi-index α, since

p(., ξ)ǫC

∞

, we have

|ζ

ˆp

(ζ, ξ)

| = |

−2πix.ζ)

p(x, ξ)dx

−2πix.ζ)

p(x, ξ)dx

≤ C

(1 +

|ξ|)

so that for any positive integer N,

| ˆp

(ζ, ξ)

| ≤ C

(1 +

|ξ|)

(1 +

|ζ|)

−N

4. Basic Theory of Pseudo Differential Operators

and hence

|K(η, ξ)| ≤ C

(1 +

|ξ|)

−s

(1 +

|η|)

−m

(1 +

|ξ − η|)

≤ C

(1 +

|ξ|)

|m−s|−N

If we take N > n +

|m − s|, we see that

|K(η, ξ)|dη ≤ C,

|K(η, ξ)|dξ ≤ C.

Therefore by Theorem 1.1 and the Schwarz inequality,

| < Pu, ¯v > | ≤ C|| f ||

||g||

= C

||u||

(s)

||v||

−s)

From this, it follows that

||Pu||

−m)

≤ C||u||

(s)

for uǫH

which estab-

lishes (4.41) and hence (i).

Now suppose P is properly supported and uǫH

loc

(Ω). If φǫC

∞

(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

′

ǫC

∞

(Ω) with φ

′

= 1 on Ω

′

, we

have φPu = φP(φ

′

u). But φ

′

uǫH

and (4.41) φP is bounded from H

108

−m

. This establishes (ii) and completes the proof.

8 Elliptic Pseudo Di

fferential Operators

Definition 4.42. Pǫψ

(Ω) is said to be elliptic of order m if

|σ

(x, ξ)

| ≥

Ω

|ξ|

for large

|ξ|, for all xǫΩ

′

, Ω

′

⋐ Ω.

Definition 4.43. If Pǫψ

(Ω), a left (resp. right ) parametrix of P is a

ψDOQ such that QP

− I(resp. PQ -I) is smoothing.

Theorem 4.44. If Pǫψ

is elliptic of order m and properly supported,

then there exists a properly supported Qǫψ

−m

which is a two-sided para-

metrix for P.

Proof. Let P = p(x, D). We will obtain Q = q(x, D) with q

∼

∞

j=0

where the q

′

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 q

(x, ξ) =

ζ(x, ξ)

p(x, ξ)

. Then q

ǫ s

−m

. Let Q

= q

(x, D).

We have

= q

mod S

−1

= 1 mod S

−1

= 1 + r

with r

ǫS

−1

Let q

−r

, Q

= q

(x, D). Then

= σ

+ σ

= 1 + r

− r

mod S

−2

= 1 + r

with r

ǫS

−2

Let q

−r

etc. Having determined q

, q

, . . . q

so that

···Q

= 1 + r

j+1

, r

j+1

ǫS

−( j+1)

set

j+1

−r

j+1

109

Let Q be a properly supported operator with symbol q

∼

∞

j=0

Then σ

= 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:

4. Basic Theory of Pseudo Differential Operators

Corollary 4.45 (ELLIPTIC REGUIARITY THEOREM). If P is elliptic
of order m, uǫ

′

(Ω), PuǫH

loc

(Ω) implies uǫH

loc

s+m

(Ω). In particular, P is

hypoelliptic.

Proof. Let Q be a left parametrix and set S = QP

− I. Then u = QPu −

S u. SincePuǫH

loc

(Ω) and Q is properly supported, Qǫψ

−m

we have

QPuǫH

loc

s+m

(Ω) by Theorem 4.40. Also S uǫC

∞

since S is smoothing.

Hence uǫH

loc

s+m

(Ω).

Theorem 4.46. Every elliptic differential operator is locally solvable.
In fact, if P is elliptic on Ω, f ǫ

′

(Ω) and x

ǫΩ, there exists uǫ

′

(Ω)

110

such that Pu = f in a neighbourhood of x

. (Of course, if f ǫC

∞

, then

uǫC

∞

near x

, by the previous corollary).

Proof. BY cutting f off away from x

, we can assume that f ǫE

′

and

hence f ǫH

for some s. Let Q be a properly supported parametrix for P

and let S = PQ

− I. Then S is also properly supported. If Ω

⊂⊂ Ω is

a neighbourhood of x

then there exists Ω

⊂⊂ Ω such that the values

of Su on Ω

depend only on the values of u on Ω

. Pick φ

, φ

ǫC

∞

(Ω)

such that φ

≡ 1 on Ω

, j = 1, 2 and set T u = φ

S (φ

u).

Now observe the following :

i) T u = su on Ω

ii) T : H

→ C

∞

(suup φ

) continuously.

From (ii) and the Arzela -Ascoli theorem, it follows that if (u

)

is a bounded sequence in H

′

(T u

) has a convergent subsequence in

∞

(suup φ

) and hence in H

. Therefore, T is compact on H

. So

the equation (T + I)u = f can be solved if f is orthogonal (with respect
to the pairing of H

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 x

Indeed, pick a basis g

, g

, . . . , g

for N and pick a neighbourhood U

of x

so small that g

, . . . , g

are linearly independent as functionals on

∞

(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

, . . . , g

supported at

}, which is absurd). Then we can make f ⊥ N

by adding to f a function in C

∞

(Ω

\U). But then

PQu = (I + S )u = (I + T )u on Ω

and (I + T )u = f

in a neighbourhood of x

. 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 R

. Then we define

Ω = Ω

× (R

\{0}). (In coordinate - invariant terms, T

Ω is

the cotangent bundle of Ω with the zero section removed).

(ii) A set S

⊂ T

Ω is called conic if (x, ξ)ǫS

⇒ (x, rξ)ǫS, ∀r > 0

(iii) Suppose P = p(x, D)ǫψ

(Ω). Then (x

, ξ

)ǫT

Ω is said to be non-

characteristic for P if

|p(x, ξ)| ≥ c|ξ|

for

|ξ| large and (x, ξ) in

some conic neighbourhood of (x

, ξ

(iv) The characteristic variety of P, denoted by char P, is defined by

char P =

{(x, ξ)ǫT

Ω : (x, ξ) is characteristic for p

(v) Let uǫ

′

(Ω). The wavefront set of u, denoted by WF(u) is defined

W F(U) =

∩{ charP : Pǫψ(Ω), PuǫC

∞

(Ω)

The restriction Pǫψ

(Ω) in the definition of WF(u) is merely a con-

112

venient normalisation. We could allow ψDO of arbitrary order with-
out changing anything, for if Pǫψ

(Ω) and Qǫψ

−m

(Ω) is elliptic, then

QPǫψ

(Ω), char (QP) = char (P)(by Theorem 4.36), and QPuǫC

∞

and only if PuǫC

∞

(by Corollaries 4.11 and 4.45).

4. Basic Theory of Pseudo Differential Operators

Exercise . When p =

|α|=≤m

(x)ξ

show that the characteristic variety

of the differential operator p = P(x, D) is

char P =

{(x, ξ) :

|α|=m

(x)ξ

= 0

The motivation for W F(u) is as follows. If uǫE

′

, to say that u is not

smooth in the direction ξ

should mean that ˆu is not rapidly decreasing

on the through ξ

. On the other hand, if PuǫC

∞

then (Pu)

should be

rapidly decreasing everywhere.

If these conditions both hold, then ξ

must be characteristic for P.

Localising these ideas, we arrive at W F(u).

Thus (x

)ǫW F(u) means roughly that u fails to be C

∞

at x

in the

direction ξ

. 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

Ω onto Ω by π.

Theorem 4.48. For uǫ

′

(Ω), sing supp u = π(WF(u)).

Proof. Suppose x

sin g supp u. Then there exists φǫC

∞

with φ = 1

near x

and φuǫC

∞

. But multiplication by φ is a ψDO of order 0. Call

113

it P

. Then char P

= π

−1

(φ

−1

(0)). Since φ = 1 near x

, π

−1

(φ

−1

(0))

is disjoint from π

−1

). Therefore, π

−1

)

∩ WF(u) = φ, i.e., x

π(W F(u)).

Conversely, suppose x

π(W F(u)). Then for each ξ with

|ξ| = 1,

there exists Pǫψ

(Ω) with PuǫC

∞

(Ω) and (x

, 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, P

, . . . P

ǫψ

(Ω) with

uǫC

∞

(Ω) and







j=1

char P







∩ π

−1

) = φ. Set P =

j=1

∗

Then P is elliptic near x

and PuǫC

∞

(Ω). Therefore by Corollary

4.45, u is C

∞

near x

, i.e., x

sin g supp u.

9. Wavefront Sets

Definition 4.49. Let P = p(x, D)ǫψ

(Ω) and U be an open conic set in

Ω. We say that P has order

−∞ on U, if, for all closed conic sets

⊂ U with π(K) compact for every positive integer N and multi-indices

α and β, there exist constants C

αβKN

such that

α
ξ

p(x, ξ)

| ≤ C

αβKN

(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

∼

α!

∂

α
ξ

Lemma 4.51. Let (x

, ξ

)ǫT

Ω and U be any conic neighbourhood of

, ξ

). Then there exists a Pǫψ

(Ω) such that (x

, ξ

) < char P but Ess.

supp P

⊂ U.

Proof. Choose p

(ξ)ǫC

∞

with the following properties:

i) p

(ξ) = 1 for ξ near ξ

, p

(ξ) = 0 outside

{ξ : (x

ξ)ǫU

} and

ii) p

is homogeneous of degree 0 for large

|ξ|.

Then take φǫC

∞

(π(U)) with φ = 1 near x

and put p(x, ξ) = p

(ξ)

φ(x). This will do the job.

The following theorem and its corollary are refinements of Corol-

lary 4.11 (pseudo local property of ψDO) and Corollary 4.45 (elliptic
regularity theorem).

Theorem 4.52. If Pǫψ

(Ω) and uǫD

′

(Ω), then WF(Pu)

⊂ WF(u)∩ Ess.

supp P.

4. Basic Theory of Pseudo Differential Operators

Proof. If (x

, ξ

) < Ess. supp P, by Lemma 4.51, we can find a Qǫψ

such that (x

, ξ

) < char Q and Ess. supp P

∩ Ess. supp Q = φ Then

QPǫψ

−∞

by Proposition 4.50, so that QPǫC

∞

. But this implies that

, ξ

) < W F(Pu).

Suppose now that (x

, ξ

) < W F(u). Then there exists Aǫψ

with

AuǫC

∞

and (x

, ξ

) < char A.

Claim . There exist operators B, Cǫψ

with (x

, ξ

) < char B and BP =

115

CA mod ψ

−∞

Granted this, BPu = CAu + (C

∞

function) ǫC

∞

which implies that

, ξ

) < W F(Pu).

To prove the claim, let A

be elliptic with σ

= σ

on a conic

neighbourhood U of (x

, ξ

). Then Ess. supp(A

− A

)

∩ U = φ. By

Lemma 4.51, choose B with (x

, ξ

) < char B and Ess. supp B

⊂ U.

Let E

be a parametrix for A

and set C = BPE

.Then CA = BPE

A =

BPE

− A

) + BPE

. Since E

= I mod ψ

−∞

and BPE

− A

)

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) = WF(Pu)).

Proof. By the theorem, W F(Pu)

⊂ WF(u). If E is a parametrix for

P, then W F(u) = WF(EPu + C

∞

function)

= WF(EPu)

⊂ WF(Pu), by the theorem again

Hence

W F(u) = WF(Pu).

Theorem 4.56. Let uǫ

′

(D). Then (x

, ξ

) < W F(u) if and only if there

exists φǫC

∞

) such that φ(x

) = 1 and (φu)

is rapidly decreasing on

a conic neighbourhood of ξ

Proof. (Sufficiency) If such a φ exists, choose pǫS

, p(x, ξ) = p(ξ)

with p(ξ) = 1 near ξ

and p(ξ) = 0 outside the region where (φu)

rapidly decreasing. Then p(φu)

ǫS and hence p(D)(φu)ǫS where p(D) =

p(x, D). The operator Pu = φp(D)(φu) is a pseudo differential operator

116

of order 0 with symbol φ(x)

p(ξ) modulo S

−1

, so (x

, ξ

) < char P. This

implies that (x

, ξ

) < W F(u).

10. Some Further Applications...

(Necessity) Suppose (x

, ξ

) < W F(u). Then there exists a neigh-

bourhood U of x

in Ω such that (x

, ξ

) < W F(u) for all xǫU. Choose

φǫC

∞

(U) such that φ(x

) = 1. Then (x

, ξ

) < W F(φu) for all xǫR

. Let

{ξ : (x, ξ)ǫWF(φu) for somex}.

This

P does not contain ξ

and is a closed conic set. There exists

p(ξ)ǫS

, p = 1 on a conic neighbourhood of ξ

and p = 0 on a neigh-

bourhood of

P, say P

. Since p(ξ) = 0 on

, Ess. supp p(D)

∩ (R

)

and hence Ess. supp p(D)

∩ WF(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)

ǫS and in particular, (φu)

is rapidly

decreasing near ξ

. To prove the claim, observe that

(ξ

p(ξ)) = 0(1 +

|ξ|)

|α|−|β|

ǫL

|α| − |β| < −

So if we put K(x) = ˜p(x), x

K(x)ǫL

|α| − |β| < −

n
2

. An application

of Leibniz’s rule and the the Sobolev imbedding theorem shows that

K(x)ǫL

∞

|β| − |α| > −

+ 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 De

fferen-

tial Operators

We conclude our discussion of pseudo differential operators by giving

117

brief and informal descriptions of some further applications. We recall
that an m

order ordinary differential equation u

(m)

= F(x, u, u

(1)

, . . .

−1)

) can be reduced to the first order system



















F(x, u

, . . . , u

)







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

100

4. Basic Theory of Pseudo Differential Operators

something similar for m

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 differen-
tial 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 :

∂

∂x

∂

∂y

= f.

putting u

= u, u

∂u

∂x

, u

∂

∂y

we have a 3

× 3 system given by

∂u

∂x

− u

= 0

∂u

∂y

− u

= 0

∂u

∂x

∂u

∂y

= f.

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

2πi













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 m

order linear partial differential equation to an m

× m system of

first order pseudo differential equations which preserves the character-
istic variety of the equation in a sense which we shall make precise be-
low. In this method, one of the variables is singled out to play a special
role, so we shall suppose that we are working on R

n+1

with coordinates

, x

, . . . , x

, t).

10. Some Further Applications...

101

Let L be a partial differential operator of order m on R

n+1

such that

the coefficient of ∂

is nowhere vanishing. Dividing throughout by this

coefficient, we can assume that L is of the following form:

L = ∂

m
t

−

−1

j=0

− j

(x, t, D

)∂

119

Here A

− 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

with symbol (1 +

|ξ|)

s/2

(acting in

the x variables ) we put

= Λ

−1

= Λ

−2

∂

= Λ

−3

∂

2
t

= ∂

−1

For j < m, we observe that ∂

= Λu

j+1

. Therefore,

∂

= f +

−1

j=0

− j

(x, t, D

)∂

= f +

−1

j=0

− j

(x, t, D

)Λ

−m+1

j+1

= f +

j=1

− j

(x, t, D

)Λ

−m

Set B

(x, t, D

) = A

− j+1

(x, t, D

)Λ

−m

. This B

is a ΨDO of order

1 in the variable x. Then the equation Lv = f is equivalent to

∂

u = Ku + ˜f

102

4. Basic Theory of Pseudo Differential Operators

where u = (u

, . . . , u

)

, ˜

f = (0, 0, . . . , f )

. (Here (

· · · )

denotes the

120

transpose of the row vector (

· · · )) and K is the matrix given by

K =







· · · 0

· · ·

· · · Λ

· · · B







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

, a function p

(x, ξ)

homogeneous of degree m in ξ is said to be the Principal symbol of

p(x, D) if p

− p

agrees with an element of S

−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) =

|α|≤m

(x)ξ

then

(x, ξ) =

|α|=m

(x)ξ

(ii) If p(x, ξ) = (1 +

|ξ|

)

s/2

, p

(x, ξ) =

|ξ|

121

Returning to our discussion, let a

(x, t, ξ) be the principal symbol

of A

(x, t, d

) (Here we are considering A

as an operator of order k, so

if it happens to be of lower order, its principal symbol is zero). Then

10. Some Further Applications...

103

the principal symbol of B

is b

(x, t ξ) = a

− j+1

(x, t, ξ)

|ξ|

−m

and the

principal symbol of k is the matrix K

(x, t, ξ) =







|ξ|

· · ·

|ξ|

· · ·

|ξ| · · ·

· · ·

· · · |ξ|

· · · b







The principal symbol of L, on the other hand, is

(x, t, ξ, τ) = (2πiτ)

−

−1

j=0

− j

(x, t, ξ)(2πiτ)

The characteristic variety of L, i.e., the set of zeros of L

, can easily

be read off from the matrix K as follows:

Proposition 4.57. The eigenvalues of K

(x, t, ξ) are precisely 2πiτ

, . . .,

2πiτ

where τ

, τ

, . . . , τ

are the roots of the polynomial L

(x, t, ξ, .).

Proof. The characteristic polynomial p(λ) of K

is the determinant of







−|ξ|

· · ·

−|ξ| · · ·

· · ·

−|ξ|

−b

˙˙

− b







122

Expanding in minors along the last row, we get

p(λ) =

−1

j=1

(

−1)

m+ j

(

−b

)λ

−1

(

−|ξ|)

− j

+ λ

−1

(λ

− b

)

= λ

− λ

−1

−

−1

j=1

|ξ|

−1

104

4. Basic Theory of Pseudo Differential Operators

= λ

−

j=1

− j+1

−1

= λ

−

−1

j=0

− j

= L

(x, t, ξ, (2πi)

−1

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 con-
ditions B

t=0

= g

, 1, 2, . . . , ν where

k=0

(x, t, D

)∂

k
t

, b

is of order m

− k and m

≤ m − 1.

If we set

= Λ

−m

−1

(x, 0, D

)Λ

−m

= Λ

−m

−1

then with u

= Λ

− j

∂

−1

v as above and u

(x) = u

(x, 0), the boundary

conditions will become

k=1

0
k

= φ

, j = 1, 2, . . . ν.

123

This is a system of zeroth order pseudo differential equations.
The above method of reduction is useful, for example, in the follow-

ing two important problems.

(1) Cauchy Problem for Hyperbolic Equations

Lv = f, ∂

t=0

= g

, j = 0, 1, . . . , m

− 1.

Here the equation is said to be hyperbolic when the eigenvalues of the
matrix K

(i.e., the principal symbol of K occurring in the linear system

corresponding to Lv = f ) are purely imaginary.Equivalently, the roots
of L

(x, t, ξ, .) are real.

10. Some Further Applications...

105

(2) Elliptic Boundary Value Prob Lems

Lu = f on Ω, B

u = g

on ∂Ω

where L is elliptic of order 2m and j = 1, 2, . . . m. Here one works
locally near a point x

ǫ∂Ω and makes a change of coordinates so that ∂Ω

becomes a hyperplane near x

. 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 R

with C

∞

boundary ∂Ω Consider

the problem

∆u = 0 in Ω, ∂

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 ∂

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 =

−1

{0} and Ω = {x : x

< 0

} so that ∂

∂

∂x

If ∆u = 0 then ∂

u =

−

−1

j=1

∂

∂x

−∆

u, ∆

being the Laplacian on

the boundary. So formally ∂

u =

√

(∆

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 ∂

u =

√

(

−∆

)u is correct if interpreted in terms of

the Fourier transform in the variables x

′

= (x

, . . . x

−1

);

∂

˜u(ξ

′

, x

) = 2π

|ξ

′

|˜u(ξ

′

, x

It turns out that something similar works for our original domain Ω.

Namely, if u is smooth on ¯

Ω and ∆u = 0 on Ω then ∂

∂Ω

= P(u

∂Ω

)

where P is a pseudo differential operator of order 1 on ∂Ω which equals

√

(

−∆

) modulo terms of order

≤ 0 where ∆

is the Laplace -Beltrami

operator on ∂Ω. In particular, P is elliptic and has real principal symbol.

106

4. Basic Theory of Pseudo Differential Operators

The boundary conditions become Pv +

|∂

v + av = g, v = u

∂Ω

. This

is a first order pseudo differential equation for ν. It is elliptic (because
∂

has imaginary symbol). So it can be solved modulo smoothing oper-

ators. Since ∂Ω is compact. smoothing operators on ∂Ω are compact, so

125

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 fa-

miliar Dirichlet problem ∆u = 0 in Ω and u

∂Ω

= ν which has a unique

solution.

Chapter 5

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

Ω

× Ω) when j < λ −

n. So these operators can be studied by elementary methods. What
is more, when PǫΨ

−λ

, D

PǫΨ

−λ+|α|

. So, by a proper choice of α, we

can make 0

≤ λ − |α| ≤ n and then study D

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)

∼

∞

j=0

(x, ξ)

where P

is homogeneous of degree

−λ− j. (These p

′

s no longer belong

to our symbol classes, being singular at ξ = 0, but we can still consider
the operators p

(x, D).) By the preceding remarks, it will suffice to con-

sider the operators corresponding to the individual terms in the series

127

107

108

5. L

and Lipschitz Estimates

P p

whose degrees of homogeneity are between 0 and

−n. Thus we are

looking at p(x, D) where p(x, ξ) is C

∞

on R

×R

\{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

∨

(x, .) we then have

p(x, D)u(x) =

2πix.ξ

p(x, ξ)ˆu(ξ)dξ

= (p

∨

(x, .)

∗ u)(x)

∨

(x, x

− y)u(y)dy.

Thus we see that the distribution kernel of p(x, D) is given by K(x, y)

= p

v
2

(x, x

− y).

Let us digress a little to make some remarks on homogeneous dis-

tributions.

Definition 5.1. A distribution f ǫS

′

is said to be homogeneous of degree

µ, i f < f, φ

>= r

< f, φ > for all φǫS where φ

is the function defined

by φ

(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 D

f is ho-

mogeneous of degree µ

− |α|.

3. Show that D

δ is homogeneous of degree

−n − |α| (where δ is the

128

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 ˆ

f is homogeneous of degree

−µ − n. If f is also C

∞

away from

the origin, then the same is true of ˆ

f .

Proof If f is homogeneous of degree µ, then for φǫS

< ˆ

f , φ

> =< f, (φ

)ˆ >=< f, r

−n

1/r

>= r

−n−µ

< f, ˆ

φ >

= r

−µ−n

< ˆ

f , φ > .

This proves the first assertion. To prove the second assertion, choose

φǫC

∞

with φ = 1 in a neighbourhood of the origin and write f = φ f +

− φ) f . Since φ f ǫE

′

, (φ f )ˆ is C

∞

everywhere. On the other hand,

− φ) f is C

∞

and homogeneous of degree µ for large ξ, and hence

lies in S

). Let p(x, ξ) = (1

− φ)(ξ) f (ξ). The distribution kernel of

p(x, D) is given by

K(x, y) =

−2πi(y−x),ξ

− 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 )ˆis C

∞

away from the origin. Hence ˆ

f is C

∞

on R

\{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

p(x, D)u =

K(x, x

− y)u(y)dy

where K = P

∨

is C

∞

on R

× R

\{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 de-
gree 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

5. L

and Lipschitz Estimates

We now classify the kernels of type λ

≥ 0.

Proposition 5.4. Suppose λ > 0 and f ǫC

∞

\{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 λ and f is the function with which

F agrees on R

\{0},then < F, φ >=

f φ for every φ in S .

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, φ

>= r

−n

< F, φ >,

130

and so F is a kernel of type λ.

For the converse, define G by < G, φ >=< F, φ >

−

f φ, φ

∈ S .

Then G is a distribution supported at 0 and hence we have G =

P c

δ.

Therefore,

< G, φ

−n−|α|

φ)(0) = O(r

−n

), as r

→ ∞.

On the other hand,

< G, φ

>= 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

∞

\{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

|x|=1

f (x)dσ(x).

If µ

= 0 there is a canonical distribution associated with f which is

called principal value of f, PV( f ), defined by

< PV( f ), φ >= lim

→0

|x|>ǫ

f (X)dx.

111

To see that this limit exists, we observe that

ǫ<

|x|<1

f (x)dx = µ

−1

dr = 0.

Hence

< PV( f ), φ >= lim

→0

ǫ<

|x|<1

f (x)(φ(x)

− φ(0))dx +

|x|≥1

f (x)φ(x)dx,

where the last integrals are absolutely convergent, since

131

|φ(x) − φ(0)| ≤ c|x|,

so that

|x|<1

| f (x)||φ(x) − φ(0)|dx ≤ c

|x|<1

|x|

−n+1

dx <

∞.

Further, the estimate on φ(x)

− φ(0) depends only on the first deriva-

tives of φ via the mean value theorem, so it is easily verified that the
functional PV( f ) is continuous on S . Finally, we observe that

< PV( f ), φ

> = lim

→0

|x|>ǫ

f (x)r

−n

φ(x/r)dx

= lim

→0

|y|>(ǫ/r)

f (y)r

−n

φ(y)dy

= r

−n

< PV( f ), φ > .

Thus PV( f ) is homogeneous of degree -n and so is a kernel of type

The following theorem gives a sort of converse to the above results.

Theorem 5.5. Suppose F is a kernel of type 0 and f is the function with
which F agrees on R

\{0}. Then µ

= 0 and F = PV( f ) + cδ, for some

constant c.

112

5. L

and Lipschitz Estimates

Proof. Define the functional G on S by

< G, φ >=

|x|≤1

f (x)(φ(x)

− (φ(0))dx +

|x|>1

f (x)φ(x)dx.

By the same argument as above, G is a tempered distribution and

G = F on R

{0}. Therefore, G − F =

P c

δ. Now, < F, φ

>= r

−n

P, φ > so that

< G, φ

−r

−n

< G, φ >=< G

− Fφ

−r

−n

< G

− F, φ >= 0(r

−n

)

as r tends to

∞. On the other hand, for r > 1,

132

< G, φ

− r

−n

< G, φ >=

|x|≤1/r

−n

f (x)(φ(x)

− φ(0))dx

|x|>1/r

−n

f (x)φ(x)dx

− r

−n

|x|≤1

f (x)(φ(x)

− φ(0))dx − r

−n

|x|>1

f (x)φ(x)dx

= r

−n

φ(0)

≤|x|≤1

f (x)dx = r

−n

φ(0) log rµ

′

for every φ.

This is not 0(r

−n

) as r tends to

∞ unless µ

= 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

spaces, we

need some concepts from measure theory.

Definition 5.6. Let F be a measurable function defined on R

. Then the

distribution function of f is the function δ

: (0,

∞) → [0, ∞] given by

(t) =

|, where E

{x : | f (x)| > t}.

From the distribution function of f , we can get a large amount of

information regarding f . For example, we have

| f (x)|

dx =

−

∞

dδ

(t)

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

(t) converges to 0 as t tends to 0 and t tends to

∞, we can

integrate by parts to obtain

R | f (x)|

dx = p

∞

−1

(t)dt.

Using the concept of distribution functions, we will now define weak

133

spaces.

Definition 5.7. For 1

≤ p < ∞ we define weak L

{ f : δ

(t)

≤ (c/t)

for some constant c

}. For f ǫ weak L

, the smallest such constant c will

be denoted by [ f ]

. Thus, if f ǫweak L

, we have

(t)

≤ ([ f ]

/t)

For p =

∞, we set weak L

∞

= L

∞

Proposition 5.8. CHEBYSHEV ’S INEQUALITY) L

⊂ weak L

and

[ f ]

≤ || f ||

Proof. For f ǫL

, if E

{x : | f (x)| > t},

|| f ||

p
p

| f (x)|

≥

| f (x)|

≥ t

i.e.,

| ≤ (|| f ||

/t)p.

From this, it follows that f ǫ weak L

and [ f ]

≤ || f ||

Remark 5.9. It is not true that L

= weak L

for 1

≤ p < ∞. For

example, f (x) =

|x|

−n/p

belongs to weak L

but not L

Remark 5.10. The function f

→ [ f ]

satisfies [c f ]

|c|[ f ]

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

f +g

(t)

≤ δ

(t/2) + δ

(t/2)

114

5. L

and Lipschitz Estimates

which gives, when f and g are in weak L

134

f +g

(t)

≤ (2

[ f ]

p
p

+ 2

[g]

p
p

)/t

so that

[ f + g]

≤ 2([ f ]

p
p

+ [g]

p
p

)

1/p

≤ 2([ f ]

+ [g]

The functional [.]

thus defines a topology on weak L

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

into weak L

, i.e. for every f ǫL

, [T f ]

≤ c|| f ||

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 (p

, q

)

and (p

, q

) with 1

≤ p

≤ q

≤ ∞, p

< p

, q

, i.e., [

′

Γ f ]

≤

|| f ||

for i = 0, 1. Then, if

1/p

= (1

− θ)/p

+ (θ/p

), 1/q

= (1

− θ)/q

+ (θ/q

), 0 < θ < 1,

T is bounded from L

to L

i.e.,

||T f ||

≤ c

|| f ||

where the constant

depends only on p

, q

, p

, q

, c

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

, gǫL

∗ gǫL

and we have

|| f ∗ g||

≤ c

|| f ||

||g||

(In fact, c

≤ 1 for all p, q, although our proof does not yield this

135

estimate).

Proof. Fixing f ǫL

, consider the convolution operator g

→ T

= f

∗ g.

We know that for gǫL

, T gǫL

and

||T g||

≤ || f ||

||g||

. Further if p

′

is the conjugate of p, then, by H ¨older’s inequality

||T g||

∞

≤ ||g||

|| f ||

for gǫL

′

. Thus we see that the operator T is of weak types (1, p) and

115

′

∞). Therefore, by the Marcinkiewicz interpolation theorem, T maps

boundedly into L

for every 0 < θ < 1 with

(1/p

) = 1

− θ + (θ/p

′

) = 1

− (θ/p) and 1/q

= (1/p)

− θ/p.

Given r with (1/p) + (1/q)

− 1 = (1/r), set q

θ=r

. Then (1/p

) =

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

and gǫ weak L

. Then, we have

a) f

∗ g exists a.e. and is in weak L

, and

[ f

∗ g]

≤ c

|| f ||

[g]

b) If p > 1, then f

∗ gǫL

and

|| f ∗ g||

≤ c

′

|| f ||

[g]

Proof. a) It suffices to assume that

|| f ||

= [g]

= 1 and to show that

[ f

∗ g]

≤ c

. Given α > 0, let

M = (α/2)

′

/(p

′

−q)

′

/(p

′

− q)

−1/(p

′

−q)

where p

′

is as usual the conjugate of p. Define

(x) = g(x), if

|g(x)| > M

= 0 otherwise

and

(x) = g(x)

− g

(x).

Then

136

∗g

(α)

≤ δ

∗g

(α/2) + δ

∗g

(α/2)

and we shall estimate the quantities on the right separately.

Holder’s inequality,

|| f ∗ g

∞

≤ || f ||

||g

′

||g

′

116

5. L

and Lipschitz Estimates

Since (1/q)

− (1/p

′

) = 1/r > 0, we see that p

′

− q > 0, and hence

||g

′

= p

′

∞

′

−1

(t)dt

= p

′

−1

(δ

(t)

− δ

(M))dt

≤ p

′

−1

−q

= (p

′

/(p

′

− q))M

′

−q

= (α/2)

′

Thus ( f

∗ g

)(x) exists at all points and

|| f ∗ g

∞

≤ α/2.

Consequently δ

∗g

(α/2) = 0. Next consider

||g

∞

(t)dt =

(M)dt +

∞

(t)dt

≤

−q

dt +

∞

−q

dt.

The integral

∞

−q

dt converges since q > 1, and we obtain

||g

≤ M

−q

+ M

−q

/(q

− 1) = (q/(q − 1))M

−q

Also, by Chebyshev inequality,

∗g

≤

|| f ∗ g

α/2

≤ (2

/α

)(q/(q

− 1))

(α/2)

p′

−q

p(1

−q)

′

/(p

′

− q))

−

p(1

−q)

p′

−q

= c

−r

137

Hence (a) is proved.

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

into L

with

1/p

= 1

− θ + (θ/ ¯p), (1/q

) = ((1

− θ)/q) + (θ/¯r).

Putting q

= r we get p

= p. Hence T maps L

continuously to L

and

|| f ∗ g||

≤ c

′

|| f ||

[g]

Corollary 5.15. If T is an operator of type λ, 0 < λ < n, then T is
bounded from L

into L

, 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|

−n

K(x/

|x|)| ≤ c|x|

−n

which implies that Kǫ weak L

n/(n

−λ)

. Therefore, by Theorem 5.14(b), if

f ǫL

and T f = K

∗ f then T f ǫL

where

(1/q) = (1/p) + ((n

− λ(/n) − 1 = (1/p) − (λ/n) and ||T f ||

≤ c[ f ]

≤ c|| f ||

Also we see that T is of weak type (1, n/(n

− λ)) by Theorem 5.14

(a).

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

, 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

∞

to c

∞

and we shall show

that

||T f ||

≤ c

|| f ||

for f ǫc

∞

, 1 < p <

∞, so that T extends uniquely

to a bounded operator on L

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

5. L

and Lipschitz Estimates

Step 1. T is bounded on L

. Indeed, we have (T f )ˆt = ˆk ˆf.ˆk is smooth

away from 0 and homogeneous of degree 0, hence is bounded on R

Therefore,

||T f ||

||(T f )ˆ||

≤ ||ˆk||

∞

|| ˆf||

||ˆk||

∞

|| f ||

Step 2. Fix a radial function φǫc

∞

with φ(x) = 1 for

|x| ≤

and φ(x) = 0

for

|x| ≥ 1. For ǫ > 0, we define k

(x) = k(x)(1

−φ(x/ǫ)) and T

f = k

∗ f .

Then we claim that T

is bounded on L

uniformly in ǫ. To see this, we

observe that

f )ˆ = ˆfˆk

= ˆf(k

− kφ(x/ǫ))ˆ

= ˆfˆk

− ˆf(ˆk ∗ (φ(x/ǫ))ˆ)

which gives

||T

||(T

f )ˆ

≤ || ˆf||

||ˆk||

∞

{1 + ǫ

| ˆφ(ǫξ)|dξ}

|| f ||

||ˆk||

∞

{1 + || ˆφ||

}

Step 3. T

is of weak type (1,1) uniformly in ǫ. The proof of this is more

139

involved and will be given later.

Step 4. By steps 2 and 3, using Marcinkiewicz, we get that T

is bounded

on L

for 1 < p < 2 uniformly in ǫ.

Step 5. T

is bounded on L

, for 2 < p <

∞, uniformly in ǫ. Indeed, for

f, gǫC

∞

f ) (x) g(x) dx =

f (x) ( ˜

g) (x) dx

with ˜

g = ˜k

∗ g, ˜k

= k

(

−x) = k(−x)(1 − φ(x/ǫ)). Since ˜k

satisfies

the same conditions as k

, we see that ˜

is bounded on L

for 1 < q < 2

uniformly in ǫ. So if (1/p) + (1/q) = 1, 1 < q < 2,

||T

sup

g ǫ C

∞

f )g

||g||

≤ sup

gǫC

∞

|| ˜T

||g||

|| f ||

= c

|| f ||

119

Step 6. If f ǫC

∞

, T

f converges to T f in the L

norm, 1

≤ p ≤ ∞ as ǫ

tends to 0. Since φ is radial and µ

= 0,

|x|=r

φ(x/ǫ)k(x)dσ(x) = c

|x|=r

k(x)dσ(x) = 0;

so, for ǫ

≤ 1, we have

f )(x) =

|y|≤1

( f (x

− y) − f (x)) k(y)(1 − φ(y/ǫ))dy +

|y|>1

f (x

− y) k(y)dy

and hence

f )(x)

− (T f )(x) =

|y|≤ǫ

f (x

− y) − f (x)k(y)φ(y/ǫ)dy.

140

Now supp(T

− T f ) ⊂ {x : d(x, supp f ) ≤ ǫ} ⊂ A, a fixed compact

set. Since, on compact sets, the uniform norm dominates all L

norms,

it suffices to show that (T

− T f ) converges to 0 uniformly on A. But

||T

− f ||

∞

≤ || grad f ||

∞

|y|≤ǫ

|y||y|

−n

≤ c

′

|| grad f ||

∞

−n

−1

dr = c

′′

→ 0, as ǫ → 0.

Step 7. If f ǫL

and η > 0, choose gǫC

∞

with

||g − f ||

< η.

Then

||T

− T

≤ ||T

( f

− g)||

||T

− T

||T

− f )||

≤

2cη +

||T

− T

Since η is arbitrary and

||T

− T

converges to 0 as ǫ, δ tend to 0

by step 6, we see that (T

f ) is Cauchy in the L

norm. Setting

T f = lim

→0

||T f ||

≤ c|| f ||

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.

120

5. L

and Lipschitz Estimates

Lemma 5.17. Suppose FǫL

, F

≥ 0 and α > 0. Then there exists a

sequence (Q

) of closed cubes with sides parallel to the coordinate axes

141

and disjoint interiors such that

a) α <

≤ 2

α for all k,

b) If Ω =

∞

, then

|Ω| ≤ (1/α)||F||

c) F(x)

≤ α for a.e. x < Ω.

Proof. Let r = (

||F||

/α)

1/n

and for j = 1, 2, . . . let Q

be the collection

of closed cubes of side length r/2

and vertices in (r/2

. Our se-

quence will be constructed in the following way. Put those cubes QǫQ

in the sequence which satisfy α <

|Q|

F. Then

|Q|

≤

|Q|

||F||

= (2

α/

||F||

)

||F||

= 2

so that the first condition is satisfied.

Put those cubes QǫQ

into the sequence which are not contained in

one of the previously accepted cubes and satisfy

α <

|Q|

Inductively, put those QǫQ

which are not contained in one of the

previously accepted cubes and satisfy α <

|Q|

F. If QǫQ

is in the

sequence and Q

′

is the cube in Q

−1

containing Q then

|Q|

≤

|Q|

, F =

′

Since Q

⊂ Q

′

, Q

′

cannot be in the sequence and hence

142

121

′

≤ α which then yields

|Q|

≤ 2

α.

Thus the condition (a) is satisfied. Also,

|Ω| =

∞

| ≤

≤ (1/α)||F||

; so (b) follows .

Finally, by the Lebesgue differentiation theorem,

lim

xǫ QǫQ

→∞

|Q|

F = F(x) for a.e. x.

So, if x < Ω,

|Q|

≤ α for all those Q

′

s and hence F(x)

≤ α a.e.

on R

\Ω.

Coming back to the proof of Step 3, given f ǫL

and α > 0, let (Q

)

be the sequence of cubes as in the lemma with F =

| f |. We write

f = g +

∞

k=1

with

(x) =












f (x)

−

f (y)dy, for xǫQ

0, otherwise

and

g(x) =












f (y)dy, for xǫQ

f (x), for x < Ω.

Now me make the following observations :

a) Supp b

⊂ Q

= 0 and

(5.18)

∞

||b

≤

∞

| f | ≤ 2

∞

| ≤ 2

n+1

|| f ||

143

122

5. L

and Lipschitz Estimates

|g(x)| ≤ 2

α for xǫΩ and

|g(x)| ≤ α for a.e. x < Ω.

Therefore,

|g(x)| ≤ 2

α a.e. and

||g||

2
2

Ω

|g|

−Ω

|g|

(5.19)

≤ (2

α)

|Ω| + α

−Ω

|| f ||

≤ (2

+ 1)α

|| f ||

We put

∞

= b so that T

f = T

g + T

b and

|{|T

| > α}| ≤ |{|T

| > α/2}| + |{|T

| > 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

| > α/2}| ≤ (2/α)||T

)

≤ c(||g||

/α)

≤ c

(

|| f ||

α).

To estimate the second term, let y

be the center of the cube Q

and

be the cube centred at y

but with length side 2

√

n times that of Q

We put

∞

= ˜

Ω. Then

| ˜Ω| ≤

∞

| ˜

| =

√

∞

| ≤

√

|| f ||

/α = c(

|| f ||

/α).

|{|T

| > α/2} ≤ | ˜Ω| + |{|T

| > α/2}\ ˜Ω|

≤ c(|| f ||

/α) +

|{|T

| > α/2}\ ˜Ω|,

and it suffices to estimate

|{|T

| > α/2}/ ˜Ω. Since

b = 0,

144

b(x) =

− y)b(y)dy

123

∞

− y)b(y)dy

∞

− y) − k

− y

))b(y)dy

Therefore, we have

|{|T

| > α/2}\ ˜Ω|

≤ (2/α)

\ ˜Ω

|dx

≤ (2/α)

∞

\ ˜Ω

|(k

− y) − k

− y

))

||b(y)|dy dx

≤ (2/α)

∞

\ ˜

|(k

− y) − k

− y

))

||b(y)|dx dy

We now claim that

\ ˜

|(k

− y) − k

− y

))

|dx ≤ c independent of k and ǫ for yǫQ

Accepting this claim, by the estimate (5.18), we have

|{|T

| > α/2}\ ˜Ω| ≤ (2/α)c

∞

|b(y)|dy ≤ 2

n+2

|| f ||

/α).

Thus

|{|T

| > α}| ≤ c

(

|| f ||

/α) which completes the proof of Step

Returning to the claim, we observe that if xǫR

\ ˜

and yǫQ

, then

|x − y

| ≥ 2|y − y

So, if we set z = x

− y

, w = y

− y

we must show that

145

|z|>2|w|

|(k

− w) − k

(z))

|dz ≤ c

independent of w and ǫ.

124

5. L

and Lipschitz Estimates

First consider the case ǫ = 1. Now k

(z) is a C

∞

function which

is homogeneous of degree

−n for large z. Therefore, |gradk

(z)

| ≤

′′

|z|

−n−1

. By the mean value theorem,

− w) − k

(z)

| ≤ c

′

|w| sup

0<t<1

|z − tw|

−n−1

≤ c

′′

|w||z|

−n−1

for

|z| > 2|w|.

|z|>2|w|

|(k

− w) − k

(z))

|dz

≤ c

′′

|z|>2|w|

|w||z|

−n−1

≤ c

′′′

|w|

∞

|w|

−2

dr = c.

Now for general ǫ,

(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|>2|w|

|(k

− w) − k

(z))

|dz =

′

|>2|w

′

|(k

′

− w

′

)

− k

′

))

|dz

′

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

146

125

0 are not bounded on L

(and hence, by duality, not bounded on L

∞

Indeed, if T is an operator of type 0 with kernel k, (T f )ˆ = ˆk ˆf. Since ˆf
is homogeneous of degree 0, it has a discontinuity at 0 (unless k = cδ).
Thus if f ǫL

, (T f )ˆ is not continuous at 0 whenever ˆ

f (0) , 0 and this

implies that T f is not in L

This reflects the fact that if

f = ˆf(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 φǫC

∞

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

for 1 < p <

∞. In

this case (φk)ˆ = ˆφ

∗ ˆk is in C

∞

, but still S is not bounded on L

This follows from the following general fact.

Proposition 5.20. If kǫS

′

and the operator f

→ k ∗ f is bounded on L

then k is necessarily a finite Borel measure.

Proof. Choose φǫC

∞

with

φ = 1 and put φ

(x) = ǫ

−n

φ(x/ǫ). Then

||φ

is independent of ǫ, so

||φ

∗ k||

≤ c. Therefore there exists a

sequence ǫ

tending to 0 such that φ

∗ k converges to a finite Borel

measure µ in the weak* topology of measures and hence φ

∗k converges

to µ in S

′

also. On the other hand, since (φ

) is an approximate identity,

∗ 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 neighbour-
hood of 0 is infinite. One can also see directly that they do not define
bounded functionals on C

Exercise . Let k(x) = 1/x on R and let f be a continuous compactly

supported function such that f (x) = (log x)

−1

for 0 < x <

and f (x) =

0 for x

≤ 0. Show that

lim

→0−

( f

∗ PV(k))(x) = ∞.

126

5. L

and Lipschitz Estimates

Generalise this to kernels of type 0 on R

, n > 1.

This example also shows that operators of type 0 do not map con-

tinuous functions into continuous functions. However, they do preserve
Lipschitz or H ¨older continuity, as we shall now see.

Definition 5.21. For 0 < α < 1, we define

| f |

= sup

x,y

| f (x + y) − f (x)|

|y|

{ f : || f ||

de f

|| f ||

∞

| f |

∞}.

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 |

∞ 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.

We are going to prove, essentially, that operators of type 0 are boun-

ded on Λ

. However, if k is a kernel of type 0 and f ǫΛ

, the integral

148

defining k

∗ f will usually diverge f need not decay at ∞. Consequently,

we shall work instead with Λ

∩ L

(1 < p <

∞), concerning which we

have the following useful result.

Proposition 5.23. If f ǫL

, 1

≤ p < ∞ and | f |

∞, then f ǫΛ

and

|| f ||

∞

≤ c(|| f ||

| f |

). Consequently, L

∩ Λ

is a Banach space with

norm

|| f ||

| f |

Proof. Let

= (

| f (x)|

| f |

)

1/α

for xǫR

Then

| f (y)| ≥ | f (x)|/2 for all y such that |x − y| ≤ A

. So

| f (y)|

≥

|x−y|≤A

| f (y)|

≥ | f (x)|

−p

′

= c

′′

| f (x)|

p+(n/α)

| f |

−n/α

| f (x)|

1+(n/αp)

≤ c

′′′

|| f ||

| f |

n/alphap
α

127

Since this is true for all x, setting θ = n/pα we have

|| f ||

∞

≤ c

′′′

|| f ||

1/(1+θ)

| f |

θ/(1+θ)

≤ c(|| f ||

| f |

Theorem 5.24. Operators of type 0 are bounded on Λ

∩ L

(0 < α <

1, < 1 < p <

∞).

Proof. Let T : f

→ K ∗ f be an operator of type 0. Since we know that

||T f ||

≤ c

|| f ||

, by proposition 5.23, it will suffice to show

|T f |

≤

| f |

for 0 < α < 1, f ǫL

∩ Λ

alpha

. As in the proof of Theorem 5.16,

149

we may assume that K = PV(k) and identify K with k.

Given yǫR

\{0} and f ǫL

∩ Λ

define

g(x) =

|z|≤3|y|

k(z) f (x

− z)dz

h(x) =

|z|≤3|y|

k(z) f (x

− z)dz

so that T f = g + h. Since µ

= 0, we have

|g(x)| = |

|z|≤3|y|

k(z)( f (x

− z) − f (x))dz|

≤

|z|≤3|y|

|z|

−n

| f |

|z|

≤ c

| f |

|y|

Since this is true for all x,

|g(x + y) − g(x)| ≤ 2c

| f |

|y|

h(x + y) = lim

→∞

|y|<|z|<η

k(z)( f (x + y

− z) − f (x))dz

128

5. L

and Lipschitz Estimates

= lim

→∞

|y|<|z+y|<η

k(z + y)( f (x

− z) − f (x))dz.

Therefore,

h(x + y)

− h(x) = lim

→∞

|y|<|z|<η

(k(z + y)

− k(z))( f (x − z) − f (x))dz + ǫ

+ ǫ

where ǫ

and ǫ

are errors coming from difference between the regions

of integration.

is the error coming from the difference between the regions

|z| < η

and

|z + y| < η.

The symmetric difference between these two regions is contained in

150

the annulus η

− |y| < |z| < η + |y|.

If z is in this region and η

≫ |y| then |z| ≈ |z + y| ≈ η so that

|ǫ

| ≤ c

′

−|y|<|z|<η+|y|

2η

−n

|| f ||

∞

≤ c

′′

−n

|| f ||

∞

((η +

|y|)

− (η − |y|)

)

= 0(η

−1

)

→ 0 as η → ∞.

The term ǫ

comes from the symmetric difference of the regions

|z| >

|y| and |z + y| > 3|y| which is contained in the annulus 2|y| < |z| < 4|y|.

In this region

|z + y| ≈ |z| ≈ |y|.

Therefore

|ǫ

| ≤ c

′′′

|y|<|z|<4|y|

|y|

−n+α

| f |

dz =

= c

′

| f |

|y|

−n+α

((4

|y|)

− (2|y|)

)

= c

| f |

|y|

129

Finally coming to the main term, we have

|k(z + y) − k(z)| ≤ |y| sup

0<t<1

| grad k(z + ty)|

≤ |y| sup

0<t<1

|z + ty|

−n−1

≤ c

|y||z|

−n−1

for

|z| ≥ 3|y|.

Hence, since α < 1 so that

−n − 1 + α < −n,

151

|y|<|z|<η

(k(z + y)

− k(z))( f (x − z) − f (x))dz|

≤ c

|y|<|z|<η

|y||z|

−n−1

| f |

|z|

≤ c

|y|<|z|

|y||z|

−n−1+α

| f |

≤ c

|y|| f |

|y|

−1

= c

|y|

| f |

Therefore,

|T f (x + y) − T f (x)|

|y|

≤ c| f |

and consequently

|T f |

≤ c| f |

For kernels of positive type, we have the following result.

Theorem 5.25. Suppose 0 < λ < n, 1 < p < n/λ < q < Q where
Q =

∞ if λ ≤ 1, Q = n/(λ − 1), if λ < 1. Let 1/r = (1/p) − (λ/n) and

α = λ

− (n/q).(Thus r < ∞ and 0 < α < 1). Thus operators of type λ

are bounded from L

∩ L

into L

∩ Λ

Proof. Let T f = k

∗ f be an operator of type λ. By Corollary 5.15, T is

bounded from L

to L

, so by proposition 5.23, it is enough to show that

|T f |

≤ c|| f ||

We have (T f )(x) =

f (x

− z)k(z)dz

130

5. L

and Lipschitz Estimates

(T f )(x + y) =

f (x + y

− z)k(z)dz

f (x

− z)k(z + y)dz,

so that

(T f )(x + y)

− T f (x) =

f (x

− z)(k(z + y) − k(z))dz

|z|≤2|y|

f (x

− z)(k(z + y) − k(z))dz

|z|>2|y|

f (x

− z)(k(z + y) − k(z))dz.

152

If q

′

is the conjugate exponent of q,

|z|≤2|y|

f (x

− z)(k(z + y) − k(z))dz|

≤ || f ||


















|w|≤3|y|

|k(w)

′







1/q

′







|z|≤2|y|

|k(z)

′







1/q

′












≤ 2|| f ||







|z|≤3|y|

|k(z)|

′







1/q

′

≤ c

|| f ||







|z|≤3|y|

|z|

(λ

−n)q

′







1/q

′

≤ c

′

|| f ||

|y|

−n+λ+(n/q

′

)

= c

′

|| f ||

|y|

by the definition of α

For the second integral, we estimate k(z +y)

−k(z) by the mean value

theorem:

|k(z + y) − k(z)| ≤ |y| sup

0<t<1

|gradk(z + ty)|

131

≤ c|y||z|

−n−1

for

|z| ≥ 2|y|.

Thus

|z|>2|y|

f (x

− z)(k(z + y) − k(z))dz|

≤ c|| f ||












|z|>2|y|

|y||z|

−n−1

′












1/q

′

≤ c

′

|| f ||

|y||y|

−n−1+(n/q

′

)

since (λ

− n − 1)q

′

−n

= c

′

|| f ||

|y|

Hence

153

|T f (x + y) − T f (x)|

|y|

≤ c|| f ||

which gives

|T f |

≤ c|| f ||

Remark 5.26. As in the preceding theorem, the reason for taking the
domain of T to be L

∩ L

instead of just L

is that the integral defining

T f will usually diverge when f is merely in L

Nonetheless, the point of these theorems is that operators of type 0

are in essence, bounded from L

to Λ

(for appropriate q and α). To

make this precise without losing simplicity, one can observe that opera-
tors of type 0 map Λ

∩ E

′

into Λ

while operators of type λ > 0 map

∩ E

′

into Λ

We now introduce spaces of functions whose derivatives upto a cer-

tain order are in L

or Λ

Definition 5.27. Suppose 1

≤ p ≤ ∞ and k is a positive integer. We

define

{ f : D

f ǫL

for0

≤ |β| ≤ k}.

We equip L

with the norm

|| f ||

k,p

|β|≤k

||D

. (Thus L

2
k

= H

the notation of Chapter 3).

132

5. L

and Lipschitz Estimates

Definition 5.28. Suppose k is a positive integer and k < α < k + 1. We
define

{ f : D

f ǫΛ

−k

for0

≤ |β| ≤ k}.

We equip Λ

with the norm

|| f ||

|β|≤k

||D

−k

Remarks 5.29.
(i) f ǫΛ

if and only if D

f is bounded and continuous for 0

≤ |β| ≤ k

154

and D

f ǫΛ

−k

for

|β| = k. Indeed, if |β| < k, for |y| ≤ 1,

f (x + y)

− D

f (x)

|y|

−k

≤

f (x + y)

− D

f (x)

|y|

≤ c

|v|=|β|+1

||D

∞

(by the mean value theorem) and for

|y| > 1,

f (x + y)

− D

f (x)

|y|

−k

≤ 2||D

∞

This shows that D

f ǫΛ

−k

for 0

≤ |β| ≤ k i.e., f ǫΛ

(ii) If k < α < k + 1 then L

∩ Λ

is a Banach space with the norm

|β|≤k

(

||D

−k

This follows from the corresponding fact that L

∩ Λ

is a Banach space

(Proposition 5.23).

Theorem 5.30. Suppose 0

≤ λ < n, 1 < p < n/λ, 1/r = (1/p) − (λ/n)

and k = 0, 1, 2, . . .. Then we have:

a) Operators of type λ are bounded from L

into L

r
k

b) If λ is an integer, λ = 0, 1, 2, . . . and k < α < k + 1, then operators of

type λ are bounded from L

∩ Λ

to L

r
k+λ

∩ Λ

α+λ

Proof. a) This is an easy consequence of Corollary 5.15 and Theorem
5.16, since convolution commutes with differentiation. In the same way
(b) follows Theorem 5.24 when λ = 0.

133

We now proceed by induction on λ. Therefore, assume that λ

≥ 1.

Suppose f ǫL

∩ Λ

. Then ∂

( f

∗ k) = f ∗ ∂

k and ∂

k is a kernel of

155

type λ

− 1. So, if 1/s = (1/p) − (λ − 1)/n, by our induction hypothesis

∂

( f

∗ k)ǫL

k+λ

−1

∩ Λ

α+λ

−1

. Since s < r <

∞, L

∩ L

∞

⊂ L

, so ∂

( f

∗

k)ǫL

r
k+λ

−1

∩ Λ

α+λ

−1

. Also f

∗ kǫL

∩ C

, hence in Λ

α+λ

provided it is

bounded. But ∂

( f

∗ k) ∈ Λ

α+λ

−1

implies that ∂

( f

∗ k) is bounded. By

the mean value theorem, we then have

|( f ∗ k)(x + y) − ( f ∗ k)(x)|

|y|

≤ c.

This together with f

∗ kǫL

implies that f is bounded (by definition

5.21). Hence f

∗ kǫL

r
k+λ

∩ Λ

α+λ

The above theorem can be generalised. For example, if 0

≤ λ < n

one can show that operators of type λ map Λ

∩ E

′

in to Λ

α+λ

Also generalisations of the L

Sobolev spaces L

can be given for

non-integral values of k. In fact, a theorem due to CALDERON says
that for 1 < p <

∞, f ǫL

if and only if Λ

f ǫL

. (Here Λ = (1

− ∆)

1/2

Therefore, we can define L

p
s

for any real s by

p
s

{ f : Λ

f ǫL

} with the norm || f ||

s,p

||Λ

Then part (b) of the above theorem is still true for 0

≤ λ < n, λ not

necessarily an integer in this case.

Refer to E.M. Stein [2].
We will now prove the Sobolev imbedding theorem for L

Sobolev

spaces L

with positive integral k. This theorem can also be generalised

156

to L

p
s

for s in R.

Theorem 5.31 (SOBOLEV IMBEDDING THEOREM). Suppose 1 <

p <

∞ and k a positive integer. If k < n/p, then L

⊂ L

for 1/r =

(1/p)

− (k/n) (Hence also L

k+ j

⊂ L

for any j). If k > n/p, and α =

− n/p is not integer, then L

⊂ Λ

Proof. Let N be the fundamental solution of ∆ given by

N(x =)












− n)

−1

|x|

−n

for n , 2

(2π)

−1

log

|x| for n = 2.

Then K

(x) = ∂

N(x) = ω

|x|

−n

( true for all n) is a kernel of type

134

5. L

and Lipschitz Estimates

Now, if f

∈ L

∩ E

′

f = f

∗ δ = f ∗ ∆N = f ∗

j=1

∂

j=1

(∂

∗ k

Suppose k = 1. If 1 < n/p, ∂

f ǫL

⇒ ∂

∗ k

ǫL

where 1/r =

(1/p)

− (1/n), by Theorem 5.25 and hence f ǫL

. If 1 > n/p, ∂

∗

ǫΛ

−(n/p)

by Theorem 5.25 which implies that f ǫΛ

1(n/p)

. Thus the

theorem is true for k = 1.

For k > 1, we proceed by induction. Now

f ǫL

∩ E

′

⇒ f and ∂

f are in L

−1

∩ E

′

Therefore, if p < n/(k

− 1) and 1/q = (1/p) − (k − 1)/n, we have f ,

∂

f ǫL

, i.e., f

∈ L

q
1

, while if p > n/(k

− 1), we have f , ∂

f ǫΛ

−1−(n/p)

i.e., f ǫΛ

−(n/p)

. In the second case, we are done, and in the first case, we

apply the result for k = 1 to see that f is in the required space. Finally,

157

it is easy to check that if we keep track of the norm inequalities that are
implicit in the above arguments, we obtain

|| f ||

≤ c|| f ||

k,p

|| f ||

≤ c|| f ||

k,p

as appropriate, for f ǫL

∩ E

′

. Since L

∩ E

′

is clearly dense in L

, the

desired result follows immediately.

We can summarise these theorems in an elegant way using the fol-

lowing picture.

For

−n < α < 0 we define x

= L

|α|

and when α > 0, α not an

integer, we define x

= Λ

(The small circle represent missing spaces)
In this terminology, we have

Theorem 5.32. Operators of type λ map x

∩ E

′

into x

α+λ

≤ λ < n.

Theorem 5.33 (SOBOLEV IMBEDDING THEOREM). If D

f ǫ x

for

≤ |β| ≤ k, then f ∈ x

α+k

135

We now indicate how to fill the gaps in this picture at α = 0, 1, 2, . . .
For α = 1, we define
Λ

{ f : f is continuous, bounded and

sup

x,y

| f (x + y) + f (x − y) − 2 f (x)|

|y|

∞}.

For k = 2, 3, 4, . . . we define Λ

{ f : D

f ǫΛ

for

|β| ≤ k − 1}. The

158

sudden jump from first differences in the definition of Λ

for α < 1 to

the second differences at α = 1 is less mysterious than it seems at first,
because it can be shown that if 0 < α < 2 then f ǫΛ

if and only if f is

bounded, continuous and satisfies

sup

x,y

| f (x + y) + f (x − y) − 2 f (x)|

|y|

∞.

To fill the gap at α = 0 we use the space BMO (“ bounded mean

oscillation”) first introduced by F. JOHN and L. NIRENBERG in 1961,
which is defined as follows.

For f ǫL

1
loc

), we denote by m

f mean value of f over a measur-

able set E

⊂ R

, that is,

f =

|E|

Let

Q denote the collection of all cubes in R

with sides parallel to

the axes.

Definition 5.34. BMO =

{ f ǫL

1
loc

) : sup

QǫQ

(

| f − m

|) < ∞}.

Clearly L

∞

⊂ BMO, for, f ǫL

∞

⇒ |m

| ≤ || f ||

∞

for every QǫQ

and consequently

(

| f − m

|) ≤ 2|| f ||

∞

It can be shown that BMO

⊂ L

q
loc

for every q <

∞.

If we define x

= Λ

for α = 1, 2, . . . and X

= BMO then Theorem

5.32 remains valid for all α

(

−n, ∞), except that the Sobolev imbedding

theorem for α = 0 must be slightly modified as follows:

If D

f is in the closure of BMO

∩E

′

in BMO for

|β| ≤ k then f ǫΛ

159

136

5. L

and Lipschitz Estimates

WARNING. BMO is not an interpolation space between L

and Λ

i.e.,

it is not true that if T is a linear operator which is bounded on L

for

some p <

∞ and on Λ

some α > 0, then T is bounded on BMO.

For proofs of the foregoing assertions, see E.M. Stein [2] and also

the following papers :

(i) E.M Stein and A. Zygmund: Boundedness of translation invari-

ant operators on Holder spaces and L

spaces, Ann. Math 85

(11967)337-349, and

(ii) C. Fefferman and E.M. Stein : H

spaces of several variables Acta

Math. 129, (1972), 137-193.

As we indicated at the beginning of this chapter, the arguments we

have developed can be extended in a fairly straightforward way to give
estimates for ψDO with variable coefficients. We conclude by sum-
marising the result in

Theorem 5.35. [Let] P = p(x, D) ba a properly supported ψDO of

order

−λ on Ω, where λ ≥ 0 and p ∼

∞

j=0

with p

(x, ξ) homogeneous

of degree λ

− j for large |ξ|. Then P maps L

(Ω, loc) into L

k+λ

(Ω, loc) for

1 < p <

∞, and in the terminology of Theorem 5.32 P maps x

(Ω, loc)

into x

α+λ

(Ω, loc) for

−n < α < ∞.

Bibliography

[1] G. B. FOLLAND: Introduction to partial differential equations

160

Princeton University press, Princeton, N. J., 1975.

[2] E.M. STEIN : Singular integrals and differentiability properties of

functions, Priceton University press, Princeton, N.J., 1970.

[3] M. TAYLOR: (i) Pseudo differential operators,

Lecture Notes in Math # 416, Springer-Verlag, New York, 1974.
(ii) Pseudo differential operators, Princeton University press,
Princeton, N,J., 1981

[4] F. TREVES: Basic linear partial differential equations, Academic

press, New York, 1975.

[5] A. ZYGMUND : Trigonometric series, Cambridge University

press Cambridge, U.K. 1959.

[6] L. HORMANDER : Linear partial differential operators, Springer-

Verlag, New York, 1963.

[7] W. RUDIN : Functional Analysis, McGraw-Hill, New York, 1973.

[8] F. TREVES : Topological Vectors spaces, Distributions and Ker-

nels, Academic press, New York, 1967.

137

Document Outline

Preface
Preliminaries
Partial Differential Operators with Constant Coefficients
L2 Sobolev Spaces
- General Theory of L2 Sobolev Spaces
- Hypoelliptic Operators With Constant Coefficients
Basic Theory of Pseudo Differential Operators
LP and Lipschitz Estimates

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