# Lecture notes for pseudodifferential operators and microlocal analysis

Ma, Shiqi  Department of Mathematics and Statistics, University of Jyväskylä
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

This is a introductory course focusing some basic notions in
pseudodifferential operators ($\Psi$DOs) and microlocal analysis. We start
this lecture notes with some notations and necessary preliminaries. Then the
notion of symbols and $\Psi$DOs are introduced. In Chapter 3 we define the
oscillatory integrals of different types. Chapter 4 is devoted to the
stationary phase lemmas. One of the features of the lecture is that the
stationary phase lemmas are proved for not only compactly supported functions
but also for more general functions with certain order of smoothness and
certain order of growth at infinity. We build the results on the stationary
phase lemmas. Chapters 5, 6 and 7 covers main results in $\Psi$DOs and the
proofs are heavily built on the results in Chapter 4. Some aspects of the
semi-classical analysis are similar to that of microlocal analysis. In Chapter
8 we finally introduce the notion of wavefront, and Chapter 9 focuses on the
propagation of singularities of solution of partial differential equations.
Important results are circulated by black boxes and some key steps are marked
in red color. Exercises are provided at the end of each chapter.

Version: May, 2021.

###### Contents

  1. 1 Preliminaries
    1. 1.1 Notations
    2. 1.2 Schwartz Space and tempered distributions
    3. 1.3 Fourier transforms
  2. 2 Pseudodifferential operators
    1. 2.1 Symbols
    2. 2.2 Pseudodifferential operators
      1. 2.2.1 Some basics about the $\Psi$DOs
      2. 2.2.2 Sobolev spaces
      3. 2.2.3 Other phases
    3. 2.3 Kernels
    4. 2.4 Pseudolocal property
  3. 3 Oscillatory integrals
    1. 3.1 Oscillatory integrals - Type I
    2. 3.2 Oscillatory integrals - Type II
  4. 4 Stationary phase lemmas
    1. 4.1 A simple case
      1. 4.1.1 Preliminaries
      2. 4.1.2 A simple case
    2. 4.2 Lemma Statements
    3. 4.3 Proofs of the results
  5. 5 Symbolic calculus of $\Psi$DOs
    1. 5.1 Composition of $\Psi$DOs
    2. 5.2 Reduction of variables
    3. 5.3 The Adjoint and transpose
  6. 6 Parametrix and Boundedness of $\Psi$DOs
    1. 6.1 Parametrix
    2. 6.2 The $L^{2}$ boundedness
    3. 6.3 Gårding’s inequalities
      1. 6.3.1 Gårding’s Inequality
      2. 6.3.2 Sharp Gårding’s Inequality
  7. 7 Semi-classical $\Psi$DOs and its symbolic calculus
    1. 7.1 Semi-classical $\Psi$DOs
      1. 7.1.1 Symbol classes
      2. 7.1.2 Semiclassical pseudodifferential operators
    2. 7.2 Composition of the standard quantizations
    3. 7.3 Composition of the Weyl quantizations
      1. 7.3.1 Symplectic 2-form
      2. 7.3.2 The composition
      3. 7.3.3 Specialties of Weyl quantization
    4. 7.4 Applications in Carleman estimates
  8. 8 The wavefront set
    1. 8.1 Basic facts
    2. 8.2 Wavefront set of product of distributions
      1. 8.2.1 Direct product
      2. 8.2.2 Product
      3. 8.2.3 Convolution
    3. 8.3 The wavefront sets of Fourier integral operators
    4. 8.4 Applications
      1. 8.4.1 Microlocality of $\Psi$DOs
      2. 8.4.2 Pull-back of distributions
  9. 9 Propagation of the singularities
    1. 9.1 Microlocal parametrix
    2. 9.2 Bicharacteristics
    3. 9.3 Propagation of singularities
    4. 9.4 Cauchy problems of hyperbolic PDEs

##  Chapter 1 Preliminaries

A good reference for is [wong2014introduction, Chapters 1-5].

###  1.1. Notations

${\mathbb{R}^{n}}$: the Euclidean space. For $x\in{\mathbb{R}^{n}}$,
$|x|:=\sqrt{x_{1}^{2}+\cdots+x_{n}^{2}}$, and the inter produce $x\cdot
y:=\sum_{j=1}^{n}x_{j}y_{j}$. The notation
$\langle{x}\rangle:=(1+|x|^{2})^{1/2}$ will be frequently used throughout the
lecture. For two quantities $\mathcal{A}$ and $\mathcal{B}$, we write
$\mathcal{A}\lesssim\mathcal{B}$ to signify $\mathcal{A}\leq C\mathcal{B}$,
and write $\mathcal{A}\simeq\mathcal{B}$ to signify
$C_{1}\mathcal{B}\leq\mathcal{A}\leq C_{2}\mathcal{B}$, for some generic
positive constants $C$, $C_{1}$ and $C_{2}$. It can be checked that
$\langle{x}\rangle\simeq 1+|x|$.

######  Lemma 1.1.

For any $s\in\mathbb{R}$ and any multi-index $\alpha$, there exists a constant
$C$ independent of $x$ such that

$|\partial^{\alpha}(\langle{x}\rangle^{s})|\leq
C\langle{x}\rangle^{s-|\alpha|},\quad\text{when}\quad|x|\geq 1.$

The proof is left as an exercise.

$C^{m}({\mathbb{R}^{n}};\mathbb{C})$ is the set of complex-valued functions
that has continuous derivative up to order $m$.
$C_{c}^{\infty}({\mathbb{R}^{n}})$ is comprised of $C^{\infty}$ functions with
compact support.

The Fourier and inverse Fourier transforms of $f$ are denoted as
$\mathcal{F}f$ (also $\hat{f}$) and $\mathcal{F}^{-1}f$ (also $\check{f}$):

$\displaystyle\hat{f}(\xi)=\mathcal{F}f(\xi)$
$\displaystyle:=(2\pi)^{-n/2}\int_{{\mathbb{R}^{n}}}e^{-ix\cdot\xi}f(x)\,\mathrm{d}{x},$
$\displaystyle\check{f}(\xi)=\mathcal{F}^{-1}f(x)$
$\displaystyle:=(2\pi)^{-n/2}\int_{{\mathbb{R}^{n}}}e^{ix\cdot\xi}f(\xi)\,\mathrm{d}{\xi}.$

$\langle{f,g}\rangle:=\int_{{\mathbb{R}^{n}}}f(x)g(x)\,\mathrm{d}{x}$,
$(f,g):=\int_{{\mathbb{R}^{n}}}f(x)\overline{g(x)}\,\mathrm{d}{x}$, where
$\overline{g(x)}$ is the complex conjugation of $g(x)$.

$\partial_{j}:=\frac{\partial}{\partial x_{j}}$,
$\boxed{D_{j}:=\frac{1}{i}\partial_{j}}$, where $i$ is the imaginary unit.

Multi-index: in ${\mathbb{R}^{n}}$, a multi-index is
$\alpha=(\alpha_{1},\dots,\alpha_{n})$ where $\alpha_{j}$ are non-negative
integers. $D^{\alpha}:=D_{1}^{\alpha_{1}}\cdots D_{n}^{\alpha_{n}}$,
$\partial^{\alpha}:=\partial_{1}^{\alpha_{1}}\cdots\partial_{n}^{\alpha_{n}}$,
and $x^{\alpha}:=x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}$, and the length
of $\alpha$ is $|\alpha|:=\alpha_{1}+\cdots+\alpha_{n}$.

######  Lemma 1.2.

Assume $x\in{\mathbb{R}^{n}}$ and $\alpha$ is a multi-index. Then

$|x^{\alpha}|\leq|x|^{|\alpha|}.$

###### Proof.

We have

$\displaystyle|x^{\alpha}|$ $\displaystyle=|x_{1}^{\alpha_{1}}\cdots
x_{n}^{\alpha_{n}}|=|x_{1}^{\alpha_{1}}|\cdots|x_{n}^{\alpha_{n}}|=|x_{1}|^{\alpha_{1}}\cdots|x_{n}|^{\alpha_{n}}$
$\displaystyle\leq|x|^{\alpha_{1}}\cdots|x|^{\alpha_{n}}=|x|^{\alpha_{1}+\cdots+\alpha_{n}}=|x|^{|\alpha|}.$

∎

More on multi-index:

  * •

$\beta\leq\alpha$ means $\beta_{j}\leq\alpha_{j}$ for $j=1,\dots,n$;

  * •

the notion $\alpha-\beta$ is valid only when $\beta\leq\alpha$, and
$\alpha-\beta:=(\alpha_{1}-\beta_{1},\cdots,\alpha_{n}-\beta_{n})$;

  * •

$\alpha!:=\alpha_{1}!\cdots\alpha_{n}!$;

  * •

when $\beta\leq\alpha$,
$\binom{\alpha}{\beta}:=\frac{\alpha!}{\beta!(\alpha-\beta)!}=\binom{\alpha_{1}}{\beta_{1}}\cdots\binom{\alpha_{n}}{\beta_{n}}$,
where
$\binom{\alpha_{j}}{\beta_{j}}=\frac{\alpha_{j}!}{\beta_{j}!(\alpha_{j}-\beta_{j})!}$;

A typical form of a linear differential operator is $\sum_{|\alpha|\leq
m}a_{\alpha}(x)D^{\alpha}$. If we denote a polynomial
$p(x,\xi):=\sum_{|\alpha|\leq m}a_{\alpha}(x)\xi^{\alpha}$ where
$\xi\in{\mathbb{R}^{n}}$, then

$\sum_{|\alpha|\leq m}a_{\alpha}(x)D^{\alpha}=p(x,D).$

######  Lemma 1.3.

Assume $f,g\in C^{\infty}({\mathbb{R}^{n}})$ and $\alpha$ is a multi-index.
Then

$D^{\alpha}(fg)=\sum_{\beta\leq\alpha}\binom{\alpha}{\beta}(D^{\alpha-\beta}f)(D^{\beta}g).$

The proof is left as an exercise.

###  1.2. Schwartz Space and tempered distributions

######  Definition 1.4 (Schwartz Space).

Let $\varphi\in C^{\infty}({\mathbb{R}^{n}})$. For multi-indices $\alpha$ and
$\beta$, we define the semi-norm $|\cdot|_{\alpha,\beta}$ of $\varphi$ as

$|\varphi|_{\alpha,\beta}:=\sup_{x\in{\mathbb{R}^{n}}}|x^{\alpha}D^{\beta}\varphi(x)|<+\infty.$
(1.1)

We call $\varphi$ a _Schwartz function_ when
$|\varphi|_{\alpha,\beta}<+\infty$ for any $\alpha$ and $\beta$. The set

$\\{\varphi\in
C^{\infty}({\mathbb{R}^{n}})\,;\,|\varphi|_{\alpha,\beta}<+\infty,~{}\forall\alpha,\beta\\}$

together with the topology induced by the set of semi-norms
$|\cdot|_{\alpha,\beta}$ is call the _Schwartz space_ , denoted as
$\boxed{\mathscr{S}({\mathbb{R}^{n}})}$.

The topology $\mathcal{T}$ is induced by $\\{|\cdot|_{\alpha,\beta}\\}$ is
defined as follows. Choose

$N(\alpha,\beta;\epsilon):=\\{\varphi\in\mathscr{S}({\mathbb{R}^{n}})\,;\,|\varphi|_{\alpha,\beta}<\epsilon\\}$

to be open neighborhoods of point $0\in\mathscr{S}({\mathbb{R}^{n}})$. Choose

$\mathcal{N}:=\\{N(\alpha,\beta;\epsilon)\,;\,\alpha,\beta\text{~{}are multi-
index},\epsilon>0\\}$

to be a open neighborhood basis of $0$, and $\varphi+\mathcal{N}$ the open
neighborhood basis of $\varphi\in\mathscr{S}({\mathbb{R}^{n}})$. Then the
topology $\mathcal{T}$ is generated by these open neighborhood basis, see
[JiangFuncBook, §1.8] for more details.

######  Definition 1.5 (Convergence in Schwartz space).

A sequence of functions
$\\{\varphi_{j}\\}_{j}\subset\mathscr{S}({\mathbb{R}^{n}})$ is said to
_converge to zero in $\mathscr{S}({\mathbb{R}^{n}})$_ if

${~{}\forall\,}\alpha,\beta,~{}|\varphi_{j}|_{\alpha,\beta}\to 0\quad
j\to+\infty,$ (1.2)

denoted as _$\varphi_{j}\to 0$ in $\mathscr{S}({\mathbb{R}^{n}})$_.

######  Lemma 1.6.

We have

$\mathcal{F}\mathscr{S}({\mathbb{R}^{n}})=\mathscr{S}({\mathbb{R}^{n}}),\quad\partial^{\alpha}\mathscr{S}({\mathbb{R}^{n}})\subset\mathscr{S}({\mathbb{R}^{n}}).$

The space $\mathscr{S}({\mathbb{R}^{n}})$ are often be used as test functions
set. There is also another commonly used test functions set:
$C_{c}^{\infty}({\mathbb{R}^{n}})$. In Fourier analysis the set
$\mathscr{S}({\mathbb{R}^{n}})$ is more commonly used than
$C_{c}^{\infty}({\mathbb{R}^{n}})$, and one of the reason is that
$\mathscr{S}({\mathbb{R}^{n}})$ is closed for the Fourier transform
$\mathcal{F}$. The uncertainty principle claims that the Fourier transform of
any compactly supported function is impossible to be compactly supported,
namely, $\mathcal{F}C_{c}^{\infty}({\mathbb{R}^{n}})\neq
C_{c}^{\infty}({\mathbb{R}^{n}})$.

######  Lemma 1.7.

$C_{c}^{\infty}({\mathbb{R}^{n}})$ is dense in
$\mathscr{S}({\mathbb{R}^{n}})$.

###### Proof.

Fix a function $\phi\in C_{c}^{\infty}({\mathbb{R}^{n}})$ satisfying
$\phi\equiv 1$ when $|x|\leq 1$, and $\phi\equiv 0$ when $|x|\geq 2$, and
$0\leq\phi(x)\leq 1$.

For any $\varphi\in\mathscr{S}({\mathbb{R}^{n}})$, denote
$\varphi_{\epsilon}(x):=\varphi(x)\phi(\epsilon x)$, then
$\\{\varphi_{\epsilon}\\}_{\epsilon>0}$ is a sequence in
$C_{c}^{\infty}({\mathbb{R}^{n}})$. Then for any multi-index $\alpha$ and
$\beta$, we have

$\displaystyle|\phi-\phi_{\epsilon}|_{\alpha,\beta}$
$\displaystyle=\sup_{x\in{\mathbb{R}^{n}}}|x^{\alpha}\partial^{\beta}\big{(}\varphi(x)[1-\phi(\epsilon
x)]\big{)}|=\sup_{|x|\geq
1/\epsilon}|x^{\alpha}\partial^{\beta}\big{(}\varphi(x)[1-\phi(\epsilon
x)]\big{)}|$ $\displaystyle\leq\sup_{1/\epsilon\leq|x|\leq
2/\epsilon}|x^{\alpha}\partial^{\beta}\big{(}\varphi(x)[1-\phi(\epsilon
x)]\big{)}|+\sup_{|x|\geq 2/\epsilon}|x^{\alpha}\partial^{\beta}\varphi(x)|$
$\displaystyle=\sup_{1/\epsilon\leq|x|\leq
2/\epsilon}|x^{\alpha}\partial^{\beta}\varphi(x)||1-\phi(\epsilon
x)|+\mathcal{O}(\epsilon)+\sup_{|x|\geq
2/\epsilon}|x^{\alpha}\partial^{\beta}\varphi(x)|$
$\displaystyle\leq\sup_{1/\epsilon\leq|x|\leq
2/\epsilon}|x^{\alpha}\partial^{\beta}\varphi(x)|+\mathcal{O}(\epsilon)+\sup_{|x|\geq
2/\epsilon}|x^{\alpha}\partial^{\beta}\varphi(x)|$ $\displaystyle\leq
2\sup_{|x|\geq
1/\epsilon}|x^{\alpha}\partial^{\beta}\varphi(x)|+\mathcal{O}(\epsilon).$

Because
$\sup_{{\mathbb{R}^{n}}}|x_{j}^{2}x^{\alpha}\partial^{\beta}\varphi(x)|<+\infty$,
we have that $|x_{j}^{2}x^{\alpha}\partial^{\beta}\varphi(x)|$ is bounded in
${\mathbb{R}^{n}}$, so $|x|^{2}|x^{\alpha}\partial^{\beta}\varphi(x)|$ is
bounded in ${\mathbb{R}^{n}}$, thus
$|x^{\alpha}\partial^{\beta}\varphi(x)|\leq
C_{\alpha,\beta}\langle{x}\rangle^{-2}$ for certain constant
$C_{\alpha,\beta}$. Therefore,

$|\phi-\phi_{\epsilon}|_{\alpha,\beta}\leq 2C_{\alpha,\beta}\sup_{|x|\geq
1/\epsilon}\langle{x}\rangle^{-2}+\mathcal{O}(\epsilon)\to 0,\quad\epsilon\to
0.$

The proof is complete. ∎

######  Lemma 1.8.

Let $f\in\mathscr{S}({\mathbb{R}^{n}})$. Then ${~{}\forall\,}s\in\mathbb{R}$,
we have

$(1+|x|^{2})^{s}f(x)\in\mathscr{S}({\mathbb{R}^{n}}).$

###### Proof.

Let $\alpha$ be a multi-index. Then

$D^{\alpha}\big{[}(1+|x|^{2})^{s}f(x)\big{]}=\sum_{\delta\leq\alpha}\binom{\alpha}{\delta}D^{\delta}\big{(}(1+|x|^{2})^{s}\big{)}\cdot(D^{\alpha-\delta}f)(x).$

We should notice that $|D^{\delta}\big{(}(1+|x|^{2})^{s}\big{)}|$ can be
always controlled by $(1+|x|^{2})^{t_{\delta}}$ for $t_{\delta}\in\mathbb{R}$
large enough:

$|D^{\delta}\big{(}(1+|x|^{2})^{s}\big{)}|\leq(1+|x|^{2})^{t_{\delta}},{~{}\forall\,}x\in{\mathbb{R}^{n}}.$

So for any non-negative integer $k$ and multi-index $\alpha$, we have

$\displaystyle|(1+|x|^{2})^{k}\cdot
D^{\alpha}\big{[}(1+|x|^{2})^{s}f(x)\big{]}|$
$\displaystyle\leq\sum_{\delta\leq\alpha}\binom{\alpha}{\delta}(1+|x|^{2})^{t_{\delta}}\cdot|(D^{\alpha-\delta}f)(x)|$
$\displaystyle\leq\sum_{\delta\leq\alpha}\binom{\alpha}{\delta}|(1+|x|^{2})^{t_{\delta}+k}(D^{\alpha-\delta}f)(x)|$
$\displaystyle\leq\sum_{\delta\leq\alpha}\binom{\alpha}{\delta}C_{k,\alpha,\delta}=C_{k,\alpha}<+\infty.$

We proved the conclusion. ∎

Schwartz functions are these who decay fast enough. Now we introduce another
type of functions which grow at infinity, but with a mild speed. These
functions are called tempered functions.

######  Definition 1.9 (Tempered functions).

Let $f$ be a measurable function defined on ${\mathbb{R}^{n}}$ such that

$\sup_{x\in{\mathbb{R}^{n}}}\big{|}(1+|x|)^{-m}f(x)\big{|}<+\infty$

for some positive integer $m$. Then we call $f$ a _tempered function_. If $f$
is continuous, then we call it _continuous tempered function_.

######  Lemma 1.10.

Assume $f$ is a smooth function such that $\partial^{\alpha}f$ are tempered
functions $\forall\alpha$, then we have
$f\cdot\mathscr{S}({\mathbb{R}^{n}})\subset\mathscr{S}({\mathbb{R}^{n}}).$

The proof is left as an exercise.

######  Definition 1.11 (Tempered Distributions).

A linear functional $T$ is called a _tempered distribution_ if for any
sequence $\\{\varphi_{j}\\}_{j}$ of functions in
$\mathscr{S}({\mathbb{R}^{n}})$ converging to zero in
$\mathscr{S}({\mathbb{R}^{n}})$, we have

$T(\varphi_{j})\to 0,\quad(j\to+\infty).$

It can be checked that the set of tempered distribution, denoted as
$\boxed{\mathscr{S}^{\prime}({\mathbb{R}^{n}})}$, is the dual of Schwartz
Space $\mathscr{S}({\mathbb{R}^{n}})$.

Recall the semi-norm $|\cdot|_{\alpha,\beta}$ defined in (1.1). We define a
new norm $|\cdot|_{m}$ as

$|\varphi|_{m}:=\sum_{|\alpha|,\,|\beta|\leq m}|\varphi|_{\alpha,\beta},$

It can be seen that $|\varphi|_{m}\leq|\varphi|_{m+1}$.

######  Lemma 1.12.

“ $T\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$” is equivalent to the following
statement:

$\text{there exists a constant }C\text{ such that
}{~{}\exists\,}m\in\\{0\\}\cup\mathbb{N}^{+}\textrm{~{}s.t.~{}}\boxed{|T(\varphi)|\leq
C|\varphi|_{m}},{~{}\forall\,}\varphi\in\mathscr{S}({\mathbb{R}^{n}}).$

###### Proof.

($\Leftarrow$) Assume
${~{}\exists\,}m\in\\{0\\}\cup\mathbb{N}^{+}\textrm{~{}s.t.~{}}|T(\varphi)|\leq
C|\varphi|_{m},{~{}\forall\,}\varphi\in\mathscr{S}({\mathbb{R}^{n}})$. Then
for every sequence $\\{\varphi_{k}\\}_{k}\subset\mathscr{S}({\mathbb{R}^{n}})$
satisfying $\varphi_{k}\to 0~{}(k\to+\infty)$, we have $|\varphi_{k}|_{m}\to
0~{}(k\to+\infty)$. So $|T(\varphi_{k})|\leq C|\varphi_{k}|_{m}\to
0~{}(k\to+\infty)$. This means $T\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$.

($\Rightarrow$) Assume $T\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$. Suppose
that the claim is not true, then for every positive integer $M$, and every
$m\in\\{0\\}\cup\mathbb{N}^{+}$, there exists
$\varphi_{M,m}\in\mathscr{S}({\mathbb{R}^{n}})$ such that

$|T(\varphi_{M,m})|>M|\varphi_{M,m}|_{m},\quad\text{so}\quad|T(\varphi_{M,m}/(M|\varphi_{M,m}|_{m}))|>1.$

Let $\phi_{M,m}=\varphi_{M,m}/(M|\varphi_{M,m}|_{m})$, then
$|\phi_{M,m}|_{m}=\frac{1}{M}$ and $|T(\phi_{M,m})|\geq 1$. Further, we denote
$\phi_{M}:=\phi_{M,M}$, then $|\phi_{M}|_{M}=\frac{1}{M}$ and

$|T(\phi_{M})|\geq 1,{~{}\forall\,}M\in N^{+}.$ (1.3)

Now for every $m\in\\{0\\}\cup\mathbb{N}^{+}$, when $j$ is large enough, we
have

$|\phi_{j}|_{m}\leq|\phi_{j}|_{j}=\frac{1}{j}\to 0~{}(j\to+\infty).$

So according to the Definition 1.5, we have $\phi_{j}\to 0$ in
$\mathscr{S}({\mathbb{R}^{n}})$, so according to the definition of tempered
Distributions we shall have $|T(\phi_{j})|\to 0$. But this is contradictory
with (1.3).

The proof is complete. ∎

The notion of “tempered function” and “tempered distribution” are closely
related. Every tempered function $f$ defines a tempered distribution
$T_{f}\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$ by the following way:

$T_{f}(\varphi):=\int_{\Omega}f(x)\varphi(x)\,\mathrm{d}{x},{~{}\forall\,}\varphi\in\mathscr{S}({\mathbb{R}^{n}}).$

At the first glance, the definition of tempered distribution is not a
generalization of the definition of tempered function. But the following
theorem will characterize tempered distributions through tempered functions.

######  Theorem 1.13 (Schwartz representation Theorem).

Every $T\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$ can be represented as a sum
of certain order of derivative of continuous tempered functions in
$\mathscr{S}({\mathbb{R}^{n}})$, i.e. for every
$T\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$, there exist a finite collection
$T_{\alpha,\beta}$ of bounded continuous functions such that

$T=\sum_{|\alpha|+|\beta|\leq m}x^{\alpha}D^{\beta}T_{\alpha,\beta}$

See §1.4 in https://math.mit.edu/~rbm/iml/Chapter1.pdf

######  Theorem 1.14.

For $1\leq p\leq+\infty$, there holds
$\boxed{\mathscr{S}({\mathbb{R}^{n}})\subset
L^{p}({\mathbb{R}^{n}})\subset\mathscr{S}^{\prime}({\mathbb{R}^{n}}).}$

###  1.3. Fourier transforms

######  Definition 1.15 (Fourier Transform on
$\mathscr{S}({\mathbb{R}^{n}})$).

Let $f\in\mathscr{S}({\mathbb{R}^{n}})$, then the _Fourier transform_ of $f$
is defined as

$\boxed{(\mathcal{F}f)(\xi):=(2\pi)^{-n/2}\int_{{\mathbb{R}^{n}}}e^{-ix\cdot\xi}f(x)\,\mathrm{d}{x},{~{}\forall\,}\xi\in{\mathbb{R}^{n}},}$

where $x\cdot\xi=\sum_{i=1}^{n}x_{i}\xi_{i}$. We also denote the Fourier
transform of $f$ as $\hat{f}$.

######  Definition 1.16 (Inverse Fourier Transform).

Let $f\in\mathscr{S}({\mathbb{R}^{n}})$, then the _inverse Fourier transform_
of $f$ is defined as

$\boxed{(\mathcal{F}^{-1}f)(x):=(2\pi)^{-n/2}\int_{{\mathbb{R}^{n}}}e^{ix\cdot\xi}f(\xi)\,\mathrm{d}{\xi},{~{}\forall\,}\xi\in{\mathbb{R}^{n}}.}$

We also denote the inverse Fourier transform of $f$ as $\check{f}$.

######  Lemma 1.17.

For every $f,g\in\mathscr{S}({\mathbb{R}^{n}})$, we have:

  1. (1)

$\mathcal{F},\mathcal{F}^{-1}\colon\mathscr{S}({\mathbb{R}^{n}})\to\mathscr{S}({\mathbb{R}^{n}})$
are linear bijection;

  2. (2)

$\langle\hat{f},g\rangle=\langle f,\hat{g}\rangle$;

  3. (3)

$(f,g)=(\hat{f},\hat{g})$. (Parseval’s Relation);

  4. (4)

$\mathcal{F}(f*g)=(2\pi)^{n/2}\hat{f}\cdot\hat{g}$;

  5. (5)

$\mathcal{F}(f\cdot g)=(2\pi)^{-n/2}\hat{f}*\hat{g}$.

Let’s define an operator

$\mathcal{R}\colon
f(x)\in\mathscr{S}({\mathbb{R}^{n}})\mapsto(\mathcal{R}f)(x)=f(-x)\in\mathscr{S}({\mathbb{R}^{n}}).$

Then these four operators $\\{I,\mathcal{R},\mathcal{F},\mathcal{F}^{-1}\\}$
act very like $\\{1,-1,i,-i\\}$. Denote a multiplication operation $X_{j}$ as
$X_{j}\varphi(x):=x_{j}\varphi(x)$. We have the following relations:

######  Proposition 1.18.

(1.a) $\mathcal{R}\mathcal{F}=\mathcal{F}\mathcal{R}=\mathcal{F}^{-1};$
$\big{(}\,(-1)\cdot i=i\cdot(-1)=-i\,\big{)}$ (1.b)
$\mathcal{R}\mathcal{F}^{-1}=\mathcal{F}^{-1}\mathcal{R}=\mathcal{F};$
$\big{(}\,(-1)\cdot(-i)=(-i)\cdot(-1)=i\,\big{)}$ (1.c)
$\mathcal{F}\mathcal{F}=\mathcal{F}^{-1}\mathcal{F}^{-1}=\mathcal{R};$
$\big{(}\,i\cdot i=(-i)\cdot(-i)=-1\,\big{)}$ (1.d)
$\mathcal{R}\mathcal{R}=I.$ $\big{(}\,(-1)\cdot(-1)=1\,\big{)}$ (2.a)
$\mathcal{F}D_{j}=X_{j}\mathcal{F},\quad\mathcal{F}X_{j}=-D_{j}\mathcal{F}$,
(2.b)
$\mathcal{F}^{2}D_{j}=-D_{j}\mathcal{F}^{2},\quad\mathcal{F}^{2}X_{j}=-X_{j}\mathcal{F}^{2}$.

######  Theorem 1.19 (Plancherel Theorem).

$\mathcal{F}$ and $\mathcal{F}^{-1}$ defined on
$\mathscr{S}({\mathbb{R}^{n}})$ can be extended uniquely to a unitary operator
on $L^{2}({\mathbb{R}^{n}})$.

###### Proof.

$C_{c}^{\infty}({\mathbb{R}^{n}})\subset\mathscr{S}({\mathbb{R}^{n}})\subset
L^{2}({\mathbb{R}^{n}})$ and
$\overline{C_{c}^{\infty}({\mathbb{R}^{n}})}^{\|{\cdot}\|_{L^{2}({\mathbb{R}^{n}})}}=L^{2}({\mathbb{R}^{n}})$,
So $\mathscr{S}({\mathbb{R}^{n}})$ is dense in $L^{2}({\mathbb{R}^{n}})$ with
respect to the $L^{2}$ norm.

For any $f\in L^{2}({\mathbb{R}^{n}})$, let
$\\{\varphi_{n}\\}_{n}\subset\mathscr{S}({\mathbb{R}^{n}})$ such that
$\|{\varphi_{n}-f}\|_{L^{2}({\mathbb{R}^{n}})}\to 0~{}(n\to+\infty)$, then

$\|{\mathcal{F}\varphi_{m}-\mathcal{F}\varphi_{n}}\|_{L^{2}({\mathbb{R}^{n}})}=\|{\mathcal{F}(\varphi_{m}-\varphi_{n})}\|_{L^{2}({\mathbb{R}^{n}})}$

and by the Parseval’s Relation we can continue

$\|{\mathcal{F}\varphi_{m}-\mathcal{F}\varphi_{n}}\|_{L^{2}({\mathbb{R}^{n}})}=\|{\varphi_{m}-\varphi_{n}}\|_{L^{2}({\mathbb{R}^{n}})}\to
0,\quad(m,n\to+\infty).$

Therefore $\\{\mathcal{F}\varphi_{n}\\}_{n}$ is a Cauchy sequence in
$L^{2}({\mathbb{R}^{n}})$ and has a limit. We denote the limit as
$\mathcal{F}f$ and assign it to $f$ as the Fourier transform of $f$. ∎

The Fourier transform and inverse Fourier transform can also be uniquely
extended on $\mathscr{S}^{\prime}({\mathbb{R}^{n}})$.

######  Definition 1.20 (Fourier Transform on
$\mathscr{S}^{\prime}({\mathbb{R}^{n}})$).

Let $T\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$, then the _Fourier transform_
and _inverse Fourier transform_ of $T$ are defined to be the linear
functionals $\mathcal{F}T$ on $\mathscr{S}({\mathbb{R}^{n}})$ given by

$\boxed{\begin{aligned}
(\mathcal{F}T)(\varphi)&:=T(\hat{\varphi}),{~{}\forall\,}\varphi\in\mathscr{S}({\mathbb{R}^{n}}),\\\
(\mathcal{F}^{-1}T)(\varphi)&:=T(\check{\varphi}),{~{}\forall\,}\varphi\in\mathscr{S}({\mathbb{R}^{n}}).\end{aligned}}$

######  Theorem 1.21.

For every
$\varphi\in\mathscr{S}({\mathbb{R}^{n}}),~{}T\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$,
we have:
(1)
$\mathcal{F},\mathcal{F}^{-1}\colon\mathscr{S}^{\prime}({\mathbb{R}^{n}})\to\mathscr{S}^{\prime}({\mathbb{R}^{n}})$
are linear continuous bijection.
(2) $\mathcal{F}(\varphi*T)=(2\pi)^{n/2}\hat{\varphi}\cdot\hat{T}$;
(3) $\mathcal{F}(\varphi\cdot T)=(2\pi)^{-n/2}\hat{\varphi}*\hat{T}$.

The Proposition 1.18 also holds on $\mathscr{S}^{\prime}({\mathbb{R}^{n}})$.
The operator $\mathcal{R}$ for $T\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$ is
defined as:

$(\mathcal{R}T)(\varphi):=T(\mathcal{R}\varphi),{~{}\forall\,}\varphi\in\mathscr{S}({\mathbb{R}^{n}}).$

The Fourier transform are both $(1,+\infty)$-type and $(2,2)$-type bounded.
And we have $L^{p}({\mathbb{R}^{n}})\subset
L^{1}({\mathbb{R}^{n}})+L^{2}({\mathbb{R}^{n}})$ when $1<p<2$. Therefore we
can define the Fourier transform $\mathcal{F}$ on
$L^{1}({\mathbb{R}^{n}})+L^{2}({\mathbb{R}^{n}})$ by (1.4) and then study the
boundedness of Fourier transform on $L^{p}({\mathbb{R}^{n}})$ with $1<p<2$.
For details about these $L^{1}({\mathbb{R}^{n}})+L^{2}({\mathbb{R}^{n}})$
things, please Google “Riesz–Thorin theorem”. Therefore according to the
Marcinkiewicz interpolation theorem (see [stein2016singular, Appendix B]), we
have the following result.

$\begin{cases}f=f_{1}+f_{2}\in L^{p}({\mathbb{R}^{n}}),\\\ f_{1}\in
L^{1}({\mathbb{R}^{n}}),~{}f_{2}\in L^{2}({\mathbb{R}^{n}}),\\\
\mathcal{F}_{1}\colon\text{Fourier transform from
}L^{1}({\mathbb{R}^{n}})\text{ to }L^{1}({\mathbb{R}^{n}}),\\\
\mathcal{F}_{2}\colon\text{Fourier transform from
}L^{2}({\mathbb{R}^{n}})\text{ to }L^{2}({\mathbb{R}^{n}}),\\\
\mathcal{F}(f):=\mathcal{F}_{1}(f_{1})+\mathcal{F}_{2}(f_{2}).\end{cases}$
(1.4)

######  Theorem 1.22 (Hausdorff-Young inequality).

Define the Fourier transform on
$L^{1}({\mathbb{R}^{n}})+L^{2}({\mathbb{R}^{n}})$ by (1.4), then there exists
a constant $C_{p}$ such that for all $f\in L^{p}({\mathbb{R}^{n}}),~{}(1\leq
p\leq 2)$, we have

$\boxed{\|{\hat{f}}\|_{L^{p^{\prime}}({\mathbb{R}^{n}})}\leq
C\|{f}\|_{L^{p}({\mathbb{R}^{n}})},}$

where $1\leq p\leq 2$ and $1/p+1/p^{\prime}=1$.

###### Proof.

We know that $\mathcal{F}$ and $\mathcal{F}^{-1}$ are bounded from
$L^{1}({\mathbb{R}^{n}})$ to $L^{\infty}({\mathbb{R}^{n}})$ and
$L^{2}({\mathbb{R}^{n}})$ to $L^{2}({\mathbb{R}^{n}})$. So according to
Marcinkiewicz interpolation theorem, $\forall t\in[0,1]$, $\mathcal{F}$ and
$\mathcal{F}^{-1}$ are bounded from $L^{t_{1}}({\mathbb{R}^{n}})$ to
$L^{t_{2}}({\mathbb{R}^{n}})$, where

$\left\\{\begin{aligned} t_{1}&=\left(t\cdot\frac{1}{2}+(1-t)\cdot
1\right)^{-1}=\frac{2}{2-t}\\\ t_{2}&=\left(t\cdot\frac{1}{2}+(1-t)\cdot
0\right)^{-1}=\frac{2}{t}\\\ \end{aligned}\right..$

Let $p=t_{1}$ $p^{\prime}=t_{2}$, then we proved the theorem. ∎

### Exercise

######  Exercise 1.1.

Prove Lemma 1.1.

######  Exercise 1.2.

Prove Lemma 1.3.

######  Exercise 1.3.

Prove Lemma 1.10.

##  Chapter 2 Pseudodifferential operators

In this chapter we introduce the pseudodifferential operators, and in most of
the place we abbreviate it as $\Psi$DOs. First, we introduce symbols and its
asymptotics. Then the $\Psi$DOs are its kernels are defined. Finally, we prove
an important property of $\Psi$DOs–the pseudolocal property. Other references
are [wong2014introduction, Chapter 6], [grigis94mic, §1 & §3].

###  2.1. Symbols

Recall the general form $p(x,D)=\sum_{|\alpha|\leq m}a_{\alpha}(x)D^{\alpha}$
of the linear differential operators mentioned is §1.1. For a test function
$\varphi$, we have

$\displaystyle p(x,D)\varphi(x)$ $\displaystyle=\sum_{|\alpha|\leq
m}a_{\alpha}(x)D^{\alpha}\varphi(x)=\sum_{|\alpha|\leq
m}a_{\alpha}(x)\mathcal{F}^{-1}\\{\widehat{D^{\alpha}\varphi}\\}(x)$
$\displaystyle=\sum_{|\alpha|\leq
m}a_{\alpha}(x)\mathcal{F}^{-1}\\{\xi^{\alpha}\hat{\varphi}\\}(x)$
$\displaystyle=\sum_{|\alpha|\leq
m}a_{\alpha}(x)(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}e^{ix\cdot\xi}\xi^{\alpha}\hat{\varphi}(\xi)\,\mathrm{d}{\xi}$
$\displaystyle=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}e^{ix\cdot\xi}\sum_{|\alpha|\leq
m}a_{\alpha}(x)\xi^{\alpha}\hat{\varphi}(\xi)\,\mathrm{d}{\xi}$
$\displaystyle=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}e^{ix\cdot\xi}p(x,\xi)\hat{\varphi}(\xi)\,\mathrm{d}{\xi}.$

This observation encourages us to define operators by functions $p(x,\xi)$.

######  Definition 2.1 (Kohn-Nirenberg symbol).

Let $m\in(-\infty,+\infty)$. Then we define $S^{m}$ to be the set of all
functions $\sigma(x,\xi)\in
C^{\infty}({\mathbb{R}^{n}}\times{\mathbb{R}^{n}};\mathbb{C})$ such that for
any two multi-indices $\alpha$ and $\beta$, there is a positive constant
$C_{\alpha,\beta}$, independent of $(x,\xi)$, such that

$\boxed{|(D_{x}^{\alpha}D_{\xi}^{\beta}\sigma)(x,\xi)|\leq
C_{\alpha,\beta}\langle{\xi}\rangle^{m-|\beta|},\quad\forall
x,\xi\in{\mathbb{R}^{n}}}$

holds. We call any function $\sigma$ in $S^{m}$ a _symbol_ of order $m$. We
write $S^{-\infty}=\cap_{m\in\mathbb{R}}S^{m}$ and
$S^{+\infty}=\cup_{m\in\mathbb{R}}S^{m}$.

######  Example 2.2.

Here we give some examples of symbols.

  * •

$\sum_{|\alpha|\leq m}a_{\alpha}(x)\xi^{\alpha}$ is a symbol of order $m$ when
$a_{\alpha}\in\mathscr{S}({\mathbb{R}^{n}})$;

  * •

$\mathscr{S}({\mathbb{R}^{n}})\subset S^{-\infty}$;

  * •

Fix a bounded $\psi\in C^{\infty}({\mathbb{R}^{n}})$, then
$\psi(x)\langle{\xi}\rangle^{m}$ is a symbol of order $m$;

  * •

Fix a $\phi\in C_{c}^{\infty}({\mathbb{R}^{n}})$ with $\phi(0)=1$, then
$(1-\phi(\xi))(1+|\xi|)^{m}$ is a symbol of order $m$.

######  Lemma 2.3.

Assume $\sigma_{j}\in S^{m_{j}}~{}(j=1,2)$, then $\sigma_{1}\sigma_{2}\in
S^{m_{1}+m_{2}}$. $\partial^{\alpha}\sigma_{1}\in S^{m_{j}-|\alpha|}$.

The proof is left as an exercise.

One can also define a more general symbol which the effect of $x$ is taken
into consideration, and the dimension of $x$ variable and $\xi$ variable can
be different.

######  Definition 2.4.

Let $m\in(-\infty,+\infty)$ and $0\leq\delta<\rho\leq 1$. Then we define
$S^{m}_{\rho,\delta}$ to be the set of all functions $\sigma(x,\xi)\in
C^{\infty}(\mathbb{R}^{n_{1}}\times\mathbb{R}^{n_{2}};\mathbb{C})$ such that
for any two multi-indices $\alpha$ and $\beta$, there is a positive constant
$C_{\alpha,\beta}$, depending on $\alpha$ and $\beta$ only, for which

$\boxed{\big{|}(D_{x}^{\alpha}D_{\xi}^{\beta}\sigma)(x,\xi)\big{|}\leq
C_{\alpha,\beta}\langle{\xi}\rangle^{m-\rho|\beta|+\delta|\alpha|},\quad\forall
x\in\mathbb{R}^{n_{1}},\xi\in\mathbb{R}^{n_{2}}}$

holds. We also call any function $\sigma$ in $S^{m}_{\rho,\delta}$ a _symbol_.

The Kohn-Nirenberg symbol $S^{m}=S^{m}_{1,0}$. In what follows, we only focus
on $S^{m}$, and the situations for $S^{m}_{\rho,\delta}$ shall be followed in
similar manners.

Now we introduce an important notion: the asymptotic expansion of symbols.

######  Definition 2.5 (Asymptotics).

Let symbol $a\in S^{m}$ and $a_{j}\in S^{m_{j}}$ $(j=0,1,\cdots)$ where the
orders $m_{j}$ satisfies

$m=m_{0}>m_{1}>\cdots>m_{j}>m_{j+1}\to-\infty,\quad j\to\infty.$

If

$a-\sum_{j=0}^{N}a_{j}\in S^{m_{N+1}},$

holds for every integer $N$, we write

$a\sim\sum_{j}a_{j}\text{~{}in~{}}S^{m},$

and we call $\\{a_{j}\\}$ an asymptotics of $a$. The $a_{0}$ is called the
_principal symbol_ of $a$.

We often write $a=b+S^{m}$ as a shorthand of $a=b+r$ for some $r\in S^{m}$.
Then we can summarize Definition 2.5 as follows,

$\boxed{a\sim\sum_{j}a_{j}\ \text{in}\ S^{m_{N+1}}\quad\Leftrightarrow\quad
a=\sum_{j=0}^{N}a_{j}+S^{m_{N+1}}.}$

Now let’s randomly pick up some $m$, $m_{j}$ that satisfy the requirement in
Definition 2.5, and randomly pick up $a_{j}\in S^{m^{j}}$. A natural question
is to ask, does there exist $a\in S^{m}$ such that
$a\sim\sum_{j}a_{j}\text{~{}in~{}}S^{m}$? The answer is yes.

######  Theorem 2.6.

For any $m$ and $m_{j}$ satisfying

$m=m_{0}>m_{1}>\cdots>m_{j}>m_{j+1}\to-\infty,\quad j\to\infty,$

and for any $a_{j}\in S^{m_{j}}$, there exists a symbol (not unique) $a\in
S^{m}$ such that $a\sim\sum_{j}a_{j}\text{~{}in~{}}S^{m}$.

When $x\to+\infty$,
$\frac{1}{1-1/\langle{x}\rangle}=1+1/\langle{x}\rangle+1/\langle{x}\rangle^{2}+\mathcal{O}(1/\langle{x}\rangle^{3})$.
Arbitrarily pick up $\alpha_{j}$, is there a function $f(x)$ such that in
$[1,+\infty)$,

$f(x)=\sum_{0\leq j\leq
N}\alpha_{j}/\langle{x}\rangle^{j}+\mathcal{O}(1/\langle{x}\rangle^{N+1}),\quad
x\to+\infty,$ (2.1)

holds for all $N\in\mathbb{N}$? The answer is no and an example is
$\alpha_{j}:=j!$ (the convergence radius goes to infinity as $N$ grows). The
problem is that $\alpha_{j}/\langle{x=1}\rangle^{j}$ will be too big when
$j\to+\infty$. However, we can fix this problem by cutoff, so that there exist
a function $f$ (not unique!) such that (2.1) holds on intervals
$[A_{j},+\infty)$ where the $A_{j}$ is in accordance with $a_{j}$,
$A_{1}<A_{2}<\cdots$. The key step is to choose a cutoff function $\chi_{j}$
to cutoff term $\alpha_{j}/\langle{x}\rangle^{j}$ such that

$\chi_{j}(x)\alpha_{j}/\langle{x}\rangle^{j}\leq
1/2^{j},\text{~{}or~{}}1/3^{j}\text{~{}etc}.$ (2.2)

The following function satisfies the requirement:

$\left\\{\begin{aligned} &\chi_{j}\in C^{\infty}(\mathbb{R}),\\\
&\chi_{j}\equiv 0,|x|\leq 2(\alpha_{j})^{1/j},\\\ &\chi_{j}\equiv 1,|x|\geq
1+2(\alpha_{j})^{1/j},\\\ &0\leq\chi_{j}\leq
1,\text{otherwise}.\end{aligned}\right.$ (2.3)

The requirement (2.3) can also be realized by fix some $\chi$ satisfying

$\left\\{\begin{aligned} &\chi\in C^{\infty}(\mathbb{R}),\ 0\leq\chi(x)\leq
1\\\ &\chi\equiv 0\text{~{}when~{}}|x|\leq 1,\ \chi_{j}\equiv
1\text{~{}when~{}}|x|\geq 2.\end{aligned}\right.$ (2.4)

and then set $\chi_{j}(x):=\chi(\epsilon_{j}x)$, where the $\epsilon_{j}$
shall be chosen according to (2.3).

###### Sketch of the proof of Theorem 2.6.

Choose suitable coefficients $\epsilon_{j}$ and define

$a(x,\xi):=\sum_{j\geq 0}\chi(\epsilon_{j}\xi)a_{j}(x,\xi),\quad
x,\xi\in{\mathbb{R}^{n}}.$

It can be checked that $\chi(\epsilon_{j}\xi)\in S^{0}$. For any fixed
$(x_{0},\xi_{0})$, there is only finitely many terms in $\sum_{j\geq
0}\chi(\epsilon_{j}\xi_{0})a_{j}(x_{0},\xi_{0})$ which are non-zero, so
$a(x,\xi)\in C^{\infty}({\mathbb{R}^{n}}\times{\mathbb{R}^{n}};\mathbb{C})$.
Moreover, we need to show first $a$ is a symbol, and second $a$ is an
asymptotics of $a_{j}$.

First, we show that $a\in S^{m}$. It can be checked that for any multi-index
$\beta$,

$|D_{\xi}^{\beta}(\chi(\epsilon_{j}\xi))|\leq
C_{\beta}\langle{\xi}\rangle^{-\beta},\quad\forall\xi\in{\mathbb{R}^{n}},$

where the constant $C_{\beta}$ is independent of $\epsilon_{j}$. we notice
that every term $\chi(\epsilon_{j}\xi)a_{j}(x,\xi)$ is in
$S^{0+m_{j}}=S^{m_{j}}$, so

$\displaystyle\
D_{x}^{\alpha}D_{\xi}^{\beta}(\chi(\epsilon_{j}\xi)a_{j}(x,\xi))={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\chi(2\epsilon_{j}\xi)}D_{x}^{\alpha}D_{\xi}^{\beta}(\chi(\epsilon_{j}\xi)a_{j}(x,\xi))$
$\displaystyle=$ $\displaystyle\
\chi(2\epsilon_{j}\xi)\sum_{\beta^{\prime}\leq\beta}\binom{\beta}{\beta^{\prime}}D_{\xi}^{\beta^{\prime}}(\chi(\epsilon_{j}\xi))\cdot(D_{x}^{\alpha}D_{\xi}^{\beta-\beta^{\prime}}a_{j})(x,\xi)$
$\displaystyle\leq$ $\displaystyle\
\chi(2\epsilon_{j}\xi)\sum_{\beta^{\prime}\leq\beta}\binom{\beta}{\beta^{\prime}}C_{\beta^{\prime}}\langle{\xi}\rangle^{-|\beta^{\prime}|}\cdot
C_{\alpha,\beta,\beta^{\prime}}\langle{\xi}\rangle^{m_{j}-|\beta|+|\beta^{\prime}|}$
$\displaystyle=$ $\displaystyle\
\chi(2\epsilon_{j}\xi)\sum_{\beta^{\prime}\leq\beta}\binom{\beta}{\beta^{\prime}}C_{\beta^{\prime}}C_{\alpha,\beta,\beta^{\prime}}\langle{\xi}\rangle^{m_{j}-|\beta|}=\chi(2\epsilon_{j}\xi)C_{\alpha,\beta}\langle{\xi}\rangle^{m_{j}-|\beta|}$
$\displaystyle=$ $\displaystyle\
C_{\alpha,\beta}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\langle{\xi}\rangle^{m_{j}-m}}\chi(2\epsilon_{j}\xi)\cdot\langle{\xi}\rangle^{m-|\beta|}$
$\displaystyle\leq$ $\displaystyle\
C_{\alpha,\beta}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}(2\epsilon_{j})^{m-m_{j}}}\chi(2\epsilon_{j}\xi)\cdot\langle{\xi}\rangle^{m-|\beta|}$
$\displaystyle\leq$ $\displaystyle\
C_{\alpha,\beta}(2\epsilon_{j})^{m-m_{j}}\cdot\langle{\xi}\rangle^{m-|\beta|},$

where the change “$\langle{\xi}\rangle^{m_{j}-m}\to(2\epsilon_{j})^{m-m_{j}}$”
is due to the presence of $\chi(2\epsilon_{j}\xi)$. Hence,

$|D_{x}^{\alpha}D_{\xi}^{\beta}a(x,\xi)|\leq\langle{\xi}\rangle^{m-|\beta|}\cdot\sum_{j\geq
0}C_{\alpha,\beta}(2\epsilon_{j})^{m-m_{j}}.$

We choose $\epsilon_{j}$ to decrease fast enough such that $\sum_{j\geq
0}C_{\alpha,\beta}(2\epsilon_{j})^{m-m_{j}}$ is finite for every
$\alpha,\beta$ (see [wong2014introduction, Theorem 6.10] for details). We
proved $a\in S^{m}$.

Second, to show $a\sim\sum_{j}a_{j}$ in $S^{m}$, we see

$\displaystyle a-\sum_{0\leq j\leq N}a_{j}$ $\displaystyle=\sum_{0\leq j\leq
N}[\chi(\epsilon_{j}\xi)-1]a_{j}(x,\xi)+\sum_{j\geq
N+1}\chi(\epsilon_{j}\xi)a_{j}(x,\xi)$ $\displaystyle\in
S^{-\infty}+S^{m_{N+1}}=S^{m_{N+1}}.$

The proof is complete. ∎

###  2.2. Pseudodifferential operators

####  2.2.1. Some basics about the $\Psi$DOs

Based on the notion of symbols, we introduce the pseudodifferential operators.

######  Definition 2.7 (Pseudodifferential operator, $\Psi$DO).

Let $\sigma$ be a symbol. Then the _pseudo-differential operator_
$T_{\sigma}$, defined on $\mathscr{S}({\mathbb{R}^{n}})$ and associated with
$\sigma$, is defined as

$\displaystyle(T_{\sigma}\varphi)(x)$
$\displaystyle:=(2\pi)^{-n/2}\int_{{\mathbb{R}^{n}}}e^{ix\cdot\xi}\sigma(x,\xi)\hat{\varphi}(\xi)\,\mathrm{d}{\xi}$
$\displaystyle\
=\boxed{(2\pi)^{-n}\iint_{{\mathbb{R}^{n}}\times{\mathbb{R}^{n}}}e^{i(x-y)\cdot\xi}\sigma(x,\xi)\varphi(y)\,\mathrm{d}{y}\,\mathrm{d}{\xi}},\quad\forall\varphi\in\mathscr{S}({\mathbb{R}^{n}}).$

We denote the set of $\Psi$DOs of order $m$ as $\Psi^{m}$ We write
$\Psi^{-\infty}=\cap_{m\in\mathbb{R}}\Psi^{m}$ and
$\Psi^{+\infty}=\cup_{m\in\mathbb{R}}\Psi^{m}$.

######  Example 2.8.

Here we give some examples of $\Psi$DOs:

  * •

$-\Delta\in\Psi^{2}$, with symbol $|\xi|^{2}$;

  * •

$\sum_{|\alpha|\leq m}a_{\alpha}(x)D^{\alpha}\in\Psi^{m}$, with symbol
$\sum_{|\alpha|\leq m}a_{\alpha}(x)\xi^{\alpha}$;

  * •

$(I-\Delta)^{m/2}\in\Psi^{m}$, which is defined by the symbol
$\langle{\xi}\rangle^{m}=(I+|\xi|^{2})^{m/2}$;

  * •

The DtN map of the Calderón problem is a $\Psi$DO living on the boundary, see
[MR1029119].

It is an interesting question to ask for the symbol when given a certain
$\Psi$DO.

######  Example 2.9.

Some simple $\Psi$DOs whose symbol are also simple:

  * •

$D\mapsto\xi$;

  * •

$-\Delta=D\cdot D$, so $-\Delta\mapsto|\xi|^{2}$.

Similar to Lemma 2.3, we have the following claim, whose proof will be
provided in Theorem 5.3.

######  Lemma 2.10.

Assume $\sigma_{j}\in S^{m_{j}}~{}(j=1,2)$, then $T_{\sigma_{1}}\circ
T_{\sigma_{2}}\in\Psi^{m_{1}+m_{2}}$.

We show that the map $\sigma\mapsto T_{\sigma}$ is a bijection.

######  Lemma 2.11.

map $\sigma\mapsto
T_{\sigma}\in\mathcal{L}(\mathscr{S}({\mathbb{R}^{n}}),\mathscr{S}({\mathbb{R}^{n}}))$
is a bijection.

###### Proof.

The $\Psi$DO $T_{\sigma}$ is defined by $\sigma$, so the surjectivity is
obvious. The injectivity amounts to prove
$T_{\sigma}=T_{\tau}\Rightarrow\sigma=\tau$.

Let’s assume $\sigma$ and $\tau$ are two symbols and $T_{\sigma}=T_{\tau}$,
then

$\int
e^{ix\cdot\xi}[\sigma(x,\xi)-\tau(x,\xi)]\hat{\varphi}(\xi)\,\mathrm{d}{\xi}=0$

holds for any $x\in{\mathbb{R}^{n}}$ and any
$\varphi\in\mathscr{S}({\mathbb{R}^{n}})$. Replace $\varphi$ by its inverse
Fourier transform, and fix $x$ to some $x_{0}$, we can see

$\int
e^{ix_{0}\cdot\xi}[\sigma(x_{0},\xi)-\tau(x_{0},\xi)]\varphi(\xi)\,\mathrm{d}{\xi}=0.$

The arbitrary of $\varphi$ gives

$e^{ix_{0}\cdot\xi}[\sigma(x_{0},\xi)-\tau(x_{0},\xi)]=0,\quad\forall
x_{0}\in{\mathbb{R}^{n}}.$

And the arbitrary of $x_{0}$ gives $\sigma(x_{0},\xi)=\tau(x_{0},\xi)$ for
$\forall x_{0}\in{\mathbb{R}^{n}}$. The injectivity is proved. We arrive at
the conclusion. ∎

######  Remark 2.12.

For people who the first time encounter the form

$T_{\sigma}\varphi(x)\simeq\int_{{\mathbb{R}^{n}}}e^{ix\cdot\xi}\sigma(x,\xi)\hat{\varphi}(\xi)\,\mathrm{d}{\xi},$

one may think that $T_{\sigma}\varphi(x)$ is just the inverse Fourier
transform of $\sigma(x,\xi)\hat{\varphi}(\xi)$, and consequently,
$\sigma(x,\xi)\hat{\varphi}(\xi)$ can be recovered by taking the Fourier
transform of $T_{\sigma}\varphi(x)$. Unfortunately this is not true. The
function $\sigma(x,\xi)\hat{\varphi}(\xi)$ depends not only $\xi$ but also
$x$, so that is not a Fourier transform anymore. When a symbol $a$ is
independent of $x$, we have

$\mathcal{F}\\{T_{\sigma}\varphi\\}(\xi)=a(\xi)\hat{\varphi}(\xi).$

But when $a$ depends on $x$, the Fourier transform of $(T_{\sigma}\varphi)(x)$
is generally _NOT_ $a(x,\xi)\hat{\varphi}(\xi)$. In generally, we cannot use
the expression above to get the symbol $a$:

$\boxed{a(x,\xi)\hat{\varphi}(\xi)\neq\mathcal{F}\big{\\{}(T_{\sigma}\varphi)(\cdot)\big{\\}}(\xi).}$

Similar to generalizing the Fourier transform from functions to distributions,
the notion of $\Psi$DOs can also extend to
$\mathscr{S}^{\prime}({\mathbb{R}^{n}})$ by using duality arguments. Formally
speaking, we have the following computation,

$\displaystyle(T_{\sigma}u,\varphi)$
$\displaystyle=\int(T_{\sigma}u)\overline{\varphi}\,\mathrm{d}{x}=\int\big{[}(2\pi)^{-n/2}\int
e^{ix\cdot\xi}\sigma(x,\xi)\hat{u}(\xi)\,\mathrm{d}{\xi}\big{]}\overline{\varphi(x)}\,\mathrm{d}{x}$
$\displaystyle=\int\big{[}(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}\sigma(x,\xi)u(y)\,\mathrm{d}{y}\,\mathrm{d}{\xi}\big{]}\overline{\varphi(x)}\,\mathrm{d}{x}$
$\displaystyle=\int u(y)\big{[}\overline{(2\pi)^{-n}\int
e^{i(y-x)\cdot\xi}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\overline{\sigma}(x,\xi)}\varphi(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi}}\big{]}\,\mathrm{d}{y}$
$\displaystyle\sim\int u(y)\big{[}\overline{(2\pi)^{-n}\int
e^{i(y-x)\cdot\xi}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sum_{\alpha}\frac{(x-y)^{\alpha}}{\alpha!}\partial_{y}^{\alpha}\overline{\sigma}(y,\xi)}\varphi(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi}}\big{]}\,\mathrm{d}{y}\quad\text{(Taylor's)}$
$\displaystyle\sim\int
u(y)\big{[}\overline{(2\pi)^{-n}\int{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}(-D)_{\xi}^{\alpha}}(e^{i(y-x)\cdot\xi}){\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sum_{\alpha}\frac{1}{\alpha!}\partial_{y}^{\alpha}\overline{\sigma}(y,\xi)}\varphi(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi}}\big{]}\,\mathrm{d}{y}$
$\displaystyle\sim\int u(y)\big{[}\overline{(2\pi)^{-n}\int
e^{i(y-x)\cdot\xi}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sum_{\alpha}\frac{1}{\alpha!}\partial_{y}^{\alpha}D_{\xi}^{\alpha}\overline{\sigma}(y,\xi)}\varphi(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi}}\big{]}\,\mathrm{d}{y}$
$\displaystyle=\int u(y)\big{[}\overline{(2\pi)^{-n/2}\int
e^{iy\cdot\xi}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\sum_{\alpha}\frac{1}{\alpha!}\partial_{y}^{\alpha}D_{\xi}^{\alpha}\overline{\sigma}(y,\xi)}\hat{\varphi}(\xi)\,\mathrm{d}{\xi}}\big{]}\,\mathrm{d}{y}$
$\displaystyle=\int
u(y)\overline{T_{\sigma^{*}}\varphi(y)}\,\mathrm{d}{y},\qquad\sigma^{*}(y,\xi):=\sum_{\alpha}\frac{1}{\alpha!}\partial_{y}^{\alpha}D_{\xi}^{\alpha}\overline{\sigma}(y,\xi)$
$\displaystyle=(u,T_{\sigma^{*}}\varphi).$ (2.5)

The computation (2.5) implies the existence of the adjoint of $T_{\sigma}$
(denoted as $T_{\sigma}^{*}$), and we leave the rigorous proof of the
existence of $T_{\sigma}^{*}$ to §5.3. Now, by assuming the existence of
$T_{\sigma}^{*}$, we extend the domain of $T_{\sigma}$ from
$\mathscr{S}({\mathbb{R}^{n}})$ to $\mathscr{S}^{\prime}({\mathbb{R}^{n}})$ as
follows.

######  Definition 2.13 (Pseudodifferential operators in
$\mathscr{S}^{\prime}$).

Let $\sigma$ be a symbol. For every
$u\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$, we can define the _pseudo-
differential operator_ $T_{\sigma}$ acting on $u$ as

$\boxed{(T_{\sigma}u,\varphi):=(u,T_{\sigma}^{*}\varphi),{~{}\forall\,}\varphi\in\mathscr{S}({\mathbb{R}^{n}}).}$

where $T_{\sigma}^{*}$ is the adjoint of $T_{\sigma}$ and the bracket
$(\cdot,\cdot)$ signifies the pair of distributions with test functions.

######  Lemma 2.14.

Let $\sigma$ be a symbol, and denote its corresponding $\Psi$DO as
$T_{\sigma}$. Then
$T_{\sigma}(\mathscr{S}({\mathbb{R}^{n}}))\subset\mathscr{S}({\mathbb{R}^{n}})$.
And also,
$T_{\sigma}(\mathscr{S}^{\prime}({\mathbb{R}^{n}}))\subset\mathscr{S}^{\prime}({\mathbb{R}^{n}})$.

###### Sketch of the proof.

Let $\varphi\in\mathscr{S}({\mathbb{R}^{n}})$, we need to show
$x^{\alpha}D^{\beta}(T_{\sigma}\varphi)$ are bounded in ${\mathbb{R}^{n}}$. We
can show

$x^{\alpha}D^{\beta}(T_{\sigma}\varphi)(x)\simeq\sum_{\alpha^{\prime},\beta^{\prime}}\int
e^{ix\cdot\xi}\tilde{\sigma}_{\alpha^{\prime},\beta^{\prime}}(x,\xi)\mathcal{F}\\{x^{\alpha^{\prime}}D^{\beta^{\prime}}\varphi\\}(\xi)\,\mathrm{d}{\xi}$

where $\tilde{\sigma}_{\alpha^{\prime},\beta^{\prime}}$ are also symbols of
certain orders, say, $m$. Because $\varphi\in\mathscr{S}({\mathbb{R}^{n}})$,
we know that the Fourier transform
$\mathcal{F}\\{x^{\alpha^{\prime}}D^{\beta^{\prime}}\varphi\\}\in\mathscr{S}({\mathbb{R}^{n}})$,
so

$|\mathcal{F}\\{x^{\alpha^{\prime}}D^{\beta^{\prime}}\varphi\\}(\xi)|\lesssim\langle{\xi}\rangle^{-M}$

for any positive integer $M$. Hence,

$|x^{\alpha}D^{\beta}(T_{\sigma}\varphi)(x)|\lesssim\sum_{\alpha^{\prime},\beta^{\prime}}\int\langle{\xi}\rangle^{m}\langle{\xi}\rangle^{-M}\,\mathrm{d}{\xi}<+\infty$

when we take $M$ to be large enough.

For the second conclusion, from
$(T_{\sigma}u,\varphi):=(u,T_{\sigma}^{*}\varphi),{~{}\forall\,}\varphi\in\mathscr{S}({\mathbb{R}^{n}})$
we have

$|(T_{\sigma}u,\varphi)|\leq\|{u}\|\|{T_{\sigma}^{*}\varphi}\|\lesssim\|{u}\||\varphi|_{m}$

where $m$ is the order of $\sigma$. Then by Lemma 1.12 we can conclude
$T_{\sigma}u\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})$. ∎

In conclusion, there holds

$T_{\sigma}\colon\left\\{\begin{aligned} \mathscr{S}&\to\mathscr{S},\\\
\mathscr{S}^{\prime}&\to\mathscr{S}^{\prime},\end{aligned}\right.$ (2.6)

where $\mathscr{S}$ is a shorthand for $\mathscr{S}({\mathbb{R}^{n}})$. Space
$\mathscr{S}$ represents functions which are extremely smooth (good), while
$\mathscr{S}^{\prime}$ represents “functions” which are extremely rough (bad).
To quantize the goodness and the badness, we introduce the potential spaces.

####  2.2.2. Sobolev spaces

######  Definition 2.15 (Sobolev spaces).

We denote

$\boxed{H^{s,p}({\mathbb{R}^{n}}):=\\{f\in\mathscr{S}^{\prime}({\mathbb{R}^{n}})\,;\,(I-\Delta)^{s/2}f\in
L^{p}({\mathbb{R}^{n}})\\}},$

and define the norm
$\|{f}\|_{H^{s,p}}:=\|{(I-\Delta)^{s/2}f}\|_{L^{p}({\mathbb{R}^{n}})}$. Write
$H^{s}({\mathbb{R}^{n}}):=H^{s,2}({\mathbb{R}^{n}})$.

######  Lemma 2.16.

The normed vector space $(H^{s,p}({\mathbb{R}^{n}}),\|{\cdot}\|_{H^{s,p}})$ is
a Banach space, and $(H^{s}({\mathbb{R}^{n}}),\|{\cdot}\|_{H^{s}})$ is a
Hilbert space.

######  Theorem 2.17.

Let $\sigma\in S^{m}$ and denote its corresponding $\Psi$DO as $T_{\sigma}$.
Then the mapping $T_{\sigma}\colon H^{s}({\mathbb{R}^{n}})\to
H^{s-m}({\mathbb{R}^{n}})$ is bounded.

The proof of Theorem 2.17 is based on the $L^{2}$ boundedness of $\Psi$DOs of
order 0. Formally speaking,

$\displaystyle\|{T_{\sigma}f}\|_{H^{s-m}}$
$\displaystyle=\|{(I-\Delta)^{(s-m)/2}\circ
T_{\sigma}f}\|_{L^{2}}\lesssim\|{(I-\Delta)^{s/2}f}\|_{L^{2}}=\|{f}\|_{H^{s}}.$

######  Theorem 2.18.

For a fixed constant $s\in\mathbb{R}$, ${~{}\forall\,}r,t\colon r\leq s\leq
t,{~{}\forall\,}C>0,{~{}\forall\,}\varphi\in\mathscr{S}({\mathbb{R}^{n}})$, we
have:

$\boxed{\|{\varphi}\|_{H^{s}}^{2}\leq\frac{1}{C^{t-s}}\|{\varphi}\|_{H^{t}}^{2}+C^{s-r}\|{\varphi}\|_{H^{r}}^{2}.}$
(2.7)

######  Remark 2.19.

We know that when $t>s$, $\|{\varphi}\|_{H^{s}}$ can be controlled by
$\|{\varphi}\|_{H^{t}}$:

$\|{\varphi}\|_{H^{s}}\leq 1\cdot\|{\varphi}\|_{H^{t}}.$

The key point of Theorem 2.18 is that $\|{\varphi}\|_{H^{s}}$ can even be
“controlled” by $c\cdot\|{\varphi}\|_{H^{t}}$ with $0<c<1$. But we need to pay
for this: being dominated only by $c\cdot\|{\varphi}\|_{H^{t}}$ is not enough.
Due to the fact that $c$ is less than 1, certain “byproduct” should cost to
compensate the advantage, and this so-called “byproduct” is
$\|{\varphi}\|_{H^{r}}$.

###### Proof of Theorem 2.18.

When $0<C\leq 1$, it is trivial. When $C\geq 1$, we have:

$\displaystyle\|{\varphi}\|_{H^{s}}^{2}$
$\displaystyle=\|{\mathcal{F}^{-1}\sigma_{-s}\mathcal{F}\varphi}\|_{2}^{2}=\|{\sigma_{-s}\hat{\varphi}}\|_{2}^{2}=\int_{{\mathbb{R}^{n}}}\langle{\xi}\rangle^{2s}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle=\int_{\\{\langle{\xi}\rangle\leq\sqrt{C}\\}}\langle{\xi}\rangle^{2s}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}+\int_{\\{\langle{\xi}\rangle>\sqrt{C}\\}}\langle{\xi}\rangle^{2s}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle=\int_{\\{\langle{\xi}\rangle\leq\sqrt{C}\\}}\langle{\xi}\rangle^{2s-2r}\cdot\langle{\xi}\rangle^{2r}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}+\int_{\\{\langle{\xi}\rangle^{-1}<\frac{1}{\sqrt{C}}\\}}(\langle{\xi}\rangle^{-1})^{2t-2s}\cdot\langle{\xi}\rangle^{2t}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle\leq\int_{\\{\langle{\xi}\rangle\leq\sqrt{C}\\}}\sqrt{C}^{2s-2r}\cdot\langle{\xi}\rangle^{2r}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}+\int_{\\{\langle{\xi}\rangle^{-1}<\frac{1}{\sqrt{C}}\\}}(\frac{1}{\sqrt{C}})^{2t-2s}\cdot\langle{\xi}\rangle^{2t}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle\leq\sqrt{C}^{2s-2r}\int_{{\mathbb{R}^{n}}}\langle{\xi}\rangle^{2r}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}+(\frac{1}{\sqrt{C}})^{2t-2s}\int_{{\mathbb{R}^{n}}}\langle{\xi}\rangle^{2t}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle=C^{s-r}\|{\varphi}\|_{H^{r}}^{2}+C^{s-t}\|{\varphi}\|_{H^{t}}^{2}.$

This completes the proof. ∎

######  Remark 2.20.

In the proof of Theorem 2.18, when $\varphi$ is compactly supported and $s=0$,
and if we replace $\langle{\xi}\rangle$ by $|\xi|$ and choose $r=0$ and $C$ to
be small enough and use the fact that

$\int_{\\{|\xi|\leq\sqrt{C}\\}}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}\lesssim
C^{n/2}\sup_{|\xi|\leq\sqrt{C}}|\hat{\varphi}(\xi)|^{2}\lesssim
C^{n/2}\|{\varphi}\|_{L^{1}}^{2}\leq C^{n/2}\sqrt{|\mathop{\rm
supp}\varphi|}\|{\varphi}\|_{L^{2}},$

we can prove the Poincare’s inequality
$\|{\varphi}\|_{L^{2}}\lesssim\|{(-\Delta)^{t/2}\varphi}\|_{L^{2}}$ for the
set of functions with uniformly compact support.

Noticing that $\langle{\xi}\rangle\geq 1$, we can further extend Theorem 2.18
to more generalized situation.

######  Theorem 2.21.

For a fixed constant $s\in\mathbb{R}$, ${~{}\forall\,}t\geq s$ and
${~{}\forall\,}r\in\mathbb{R}^{1}$,
${~{}\forall\,}\epsilon>0,{~{}\forall\,}\varphi\in\mathscr{S}({\mathbb{R}^{n}})$,
there exists a constant $C_{r,s,t,\epsilon}$ such that:

$\boxed{\|{\varphi}\|_{H^{s}}^{2}\leq\epsilon\|{\varphi}\|_{H^{t}}^{2}+C_{r,s,t,\epsilon}\|{\varphi}\|_{H^{r}}^{2}.}$
(2.8)

As mentioned in Remark 2.19, Theorem 2.21 expresses the same information, in
addition that the “byproduct” can be $\|{\varphi}\|_{H^{r}}$ with any
$r\in\mathbb{R}^{1}$.

###### Proof of Theorem 2.21.

We pick up some constant $C>1$ first, and then we decide its value later.

$\displaystyle\|{\varphi}\|_{H^{s}}^{2}$
$\displaystyle=\|{\mathcal{F}^{-1}\sigma_{-s}\mathcal{F}\varphi}\|_{2}^{2}=\|{\sigma_{-s}\hat{\varphi}}\|_{2}^{2}=\int_{{\mathbb{R}^{n}}}\langle{\xi}\rangle^{2s}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle=\int_{\\{1\leq\langle{\xi}\rangle\leq\sqrt{C}\\}}\langle{\xi}\rangle^{2s}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}+\int_{\\{\langle{\xi}\rangle>\sqrt{C}\\}}\langle{\xi}\rangle^{2s}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle=\int_{\\{1\leq\langle{\xi}\rangle\leq\sqrt{C}\\}}\langle{\xi}\rangle^{2s-2r}\cdot\langle{\xi}\rangle^{2r}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}+\int_{\\{\langle{\xi}\rangle^{-1}<\frac{1}{\sqrt{C}}\\}}(\langle{\xi}\rangle^{-1})^{2t-2s}\cdot\langle{\xi}\rangle^{2t}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle\leq\max\\{1,\sqrt{C}^{2s-2r}\\}\cdot\int_{\\{1\leq\langle{\xi}\rangle\leq\sqrt{C}\\}}\langle{\xi}\rangle^{2r}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle\quad+\int_{\\{\langle{\xi}\rangle^{-1}<\frac{1}{\sqrt{C}}\\}}(\frac{1}{\sqrt{C}})^{2t-2s}\cdot\langle{\xi}\rangle^{2t}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle\leq\max\\{1,\sqrt{C}^{2s-2r}\\}\cdot\int_{{\mathbb{R}^{n}}}\langle{\xi}\rangle^{2r}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}+(\frac{1}{\sqrt{C}})^{2t-2s}\int_{{\mathbb{R}^{n}}}\langle{\xi}\rangle^{2t}|\hat{\varphi}(\xi)|^{2}\,\mathrm{d}{\xi}$
$\displaystyle=\max\\{1,\sqrt{C}^{2s-2r}\\}\cdot\|{\varphi}\|_{H^{r}}^{2}+C^{s-t}\|{\varphi}\|_{H^{t}}^{2}.$

Now let $C=\epsilon^{t-s}$ and let
$C_{r,s,t,\epsilon}=\max\\{1,\sqrt{C}^{2s-2r}\\}$, then we completes the
proof. ∎

####  2.2.3. Other phases

Beside the phase $(x-y)\cdot\xi$ in the expression

$(T_{\sigma}\varphi)(x)=(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}\sigma(x,y,\xi)\varphi(y)\,\mathrm{d}{y}\,\mathrm{d}{\xi}$

in Definition 2.7, it is possible to use more general functions as the phase
functions and the corresponding operators are still $\Psi$DOs, i.e.,

$P\varphi(x)=(2\pi)^{-n}\int
e^{i\phi(x,y,\xi)}\sigma(x,y,\xi)\varphi(y)\,\mathrm{d}{y}\,\mathrm{d}{\xi}$

will still be a $\Psi$DO is the $\phi$ satisfies certain conditions. See
[so17fo, §3.2] for details.

###  2.3. Kernels

The expression in Definition 2.7 can also be represented as

$\boxed{(T_{\sigma}\varphi)(x)=\int_{{\mathbb{R}^{n}}}K(x,y)\varphi(y)\,\mathrm{d}{y},}$
(2.9)

where $K(x,y)$ is called the _kernel_ of $T_{\sigma}$,

$K(x,y):=(2\pi)^{-n}\int e^{i(x-y)\cdot\xi}\sigma(x,\xi)\,\mathrm{d}{\xi},$

and the integration shall be understood as an oscillatory integral (see
Definition 3.8).

Differential operators such as $P=\sum_{j=1}^{n}x_{j}\partial_{j}$ maps
$\mathscr{S}({\mathbb{R}^{n}})\to\mathscr{S}({\mathbb{R}^{n}})$,
$C_{c}^{\infty}({\mathbb{R}^{n}})\to C_{c}^{\infty}({\mathbb{R}^{n}})$, and
$\mathcal{E}({\mathbb{R}^{n}})\to\mathcal{E}({\mathbb{R}^{n}})$, and so by
duality argument, we know differential operators $P$ maps
$\mathscr{S}^{\prime}({\mathbb{R}^{n}})\to\mathscr{S}^{\prime}({\mathbb{R}^{n}})$,
$\mathcal{D}^{\prime}({\mathbb{R}^{n}})\to\mathcal{D}^{\prime}({\mathbb{R}^{n}})$,
and
$\mathcal{E}^{\prime}({\mathbb{R}^{n}})\to\mathcal{E}^{\prime}({\mathbb{R}^{n}})$:

$P\colon\left\\{\begin{aligned}
\mathcal{E}^{\prime}({\mathbb{R}^{n}})&\to\mathcal{E}^{\prime}({\mathbb{R}^{n}})\\\
\mathcal{D}^{\prime}({\mathbb{R}^{n}})&\to\mathcal{D}^{\prime}({\mathbb{R}^{n}})\end{aligned}\right.$

But for pseudo-differential operators $T_{\sigma}$, generally speaking, we
only have
$T_{\sigma}\colon\mathcal{E}^{\prime}({\mathbb{R}^{n}})\to\mathcal{D}^{\prime}({\mathbb{R}^{n}})$.

$T_{\sigma}\colon\left\\{\begin{aligned}
\mathcal{E}^{\prime}({\mathbb{R}^{n}})&\to\mathcal{D}^{\prime}({\mathbb{R}^{n}})\\\
\mathcal{E}^{\prime}({\mathbb{R}^{n}})&\not\to\mathcal{E}^{\prime}({\mathbb{R}^{n}})\\\
\mathcal{D}^{\prime}({\mathbb{R}^{n}})&\not\to\mathcal{D}^{\prime}({\mathbb{R}^{n}}).\end{aligned}\right.$

A $\Psi$DO which maps $\mathcal{E}^{\prime}$ to $\mathcal{E}^{\prime}$ is
called _properly supported_. In fact any $\Psi$DO can be divided into a
properly supported part and a $C^{\infty}$-smooth part.

######  Lemma 2.22.

Assume $m\in\mathbb{R}$ and $\sigma\in S^{m}$ is a symbol, and $K(x,y)$ is the
kernel of $T_{\sigma}$. Then for any $\epsilon>0$, there exists two symbols
$\sigma_{1}\in S^{m}$ and $\sigma_{2}\in S^{-\infty}$ such that
$\sigma=\sigma_{1}+\sigma_{2}$, $T_{\sigma_{1}}$ is properly supported,
$T_{\sigma_{2}}$ is smooth, and their kernels $K_{1},K_{2}$ has the following
properties:

$\displaystyle\mathop{\rm supp}K_{1}$
$\displaystyle\subset\\{(x,y)\in\mathbb{R}^{2n}\,;\,|x-y|\leq\epsilon\\},\quad\text{(properly
supported)}$ $\displaystyle\mathop{\rm supp}K_{2}$
$\displaystyle\subset\\{(x,y)\in\mathbb{R}^{2n}\,;\,|x-y|\geq\epsilon/2\\}.\quad\text{(smooth)}$

###### Idea of the proof.

The proof needs Theorem 5.6.

Fix a cutoff function $\chi\in C_{c}^{\infty}(\mathbb{R})$ such that
$\chi(t)=1$ when $|t|\leq\epsilon/2$ and $\chi(t)=0$ when $|t|\geq\epsilon$.
We have

$T_{\sigma}\varphi(x)=(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}\sigma(x,\xi)\varphi(y)\,\mathrm{d}{y}\,\mathrm{d}{\xi}\\\
=T^{\prime}\varphi(x)+T^{\prime\prime}\varphi(x),$

where

$\left\\{\begin{aligned} T^{\prime}\varphi(x)&=(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}\chi(|x-y|^{2})\sigma(x,\xi)\varphi(y)\,\mathrm{d}{y}\,\mathrm{d}{\xi},\\\
T^{\prime\prime}\varphi(x)&=(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}(1-\chi(|x-y|^{2}))\sigma(x,\xi)\varphi(y)\,\mathrm{d}{y}\,\mathrm{d}{\xi}.\end{aligned}\right.$
(2.10)

By Theorem 5.6 we see that there exist $\sigma_{1}$, $\sigma_{2}\in S^{m}$
such that $T^{\prime}=T_{\sigma_{1}}$ and $T^{\prime\prime}=T_{\sigma_{2}}$,
so $T_{\sigma}=T_{\sigma_{1}}+T_{\sigma_{2}}=T_{\sigma_{1}+\sigma_{2}}$. By
Lemma 2.11 we know $\sigma=\sigma_{1}+\sigma_{2}$.

The fact $\sigma_{2}\in S^{-\infty}$ can be seen when using the asymptotics in
Theorem 5.6, namely,

$\sigma_{2}(x,\xi)=\sum_{|\alpha|\leq
N}\frac{1}{\alpha!}D_{y}^{\alpha}\partial_{\eta}^{\alpha}\big{(}(1-\chi(|x-y|^{2}))\sigma(x,\eta)\big{)}|_{(y,\eta)=(x,\xi)}+S^{m-N-1}=S^{m-N-1}$

holds for $\forall N\in\mathbb{N}$, so $\sigma_{2}\in S^{-\infty}$.

From (2.10) we can see

$\left\\{\begin{aligned} K_{1}(x,y)&=(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}\chi(|x-y|^{2})\sigma(x,\xi)\,\mathrm{d}{\xi},\\\
K_{2}(x,y)&=(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}(1-\chi(|x-y|^{2}))\sigma(x,\xi)\,\mathrm{d}{\xi}.\end{aligned}\right.$

which implies $T_{1}$ is properly supported. And the requirements for the
$\mathop{\rm supp}K_{1}$ and $\mathop{\rm supp}K_{1}$ can be seen from the
expression above. The proof is complete. ∎

###  2.4. Pseudolocal property

We talk about singular support and pseudolocal property.

######  Definition 2.23 (Singular support).

For a distribution $u\in\mathscr{D}^{\prime}$, we define its _singular
support_ to be the complement of the set
$\bigcap\\{O\subset{\mathbb{R}^{n}}\,;\,O\text{~{}is open and~{}}A\subset
O\\}$ where

$A=\\{x\in{\mathbb{R}^{n}}\,;\,u\text{~{}is~{}}C^{\infty}\text{~{}at~{}}x\\}.$

We denote the singular support of a distribution $u\in\mathscr{D}^{\prime}$ as
$\mathop{\rm sing\,supp}u$.

It is obvious that $\mathop{\rm sing\,supp}u$ is as closed set and

$\mathop{\rm sing\,supp}u\subset\mathop{\rm supp}u.$

We know a differential operator doesn’t increase the support of a
distribution, but this is not true for a $\Psi$DO. More specifically, if a
distribution $u$ is supported in $\Omega$, then $Tu$ might not be supported in
a domain $\Omega$ anymore. Instead, $\Psi$DOs have another property, called
pseudolocal property, which means $\Psi$DOs don’t increase the singular
support of a distribution.

######  Theorem 2.24 (Pseudolocal property).

Assume $T$ is a $\Psi$DO, then

$\boxed{\mathop{\rm sing\,supp}(Tu)\subset\mathop{\rm sing\,supp}u.}$

###### Proof.

Assume $x_{0}\notin\mathop{\rm sing\,supp}u$. Because $\mathop{\rm
sing\,supp}u$ is closed, we can find $\epsilon>0$ such that $u$ is
$C^{\infty}$ in $B(x_{0},\epsilon)$. According to Lemma 2.22, we can divided
$T$ into $T_{1}$ and $T_{2}$ such that $T_{2}$ is $C^{\infty}$-smooth and the
kernel $K_{1}$ of $T_{1}$ satisfies

$\mathop{\rm
supp}K_{1}\subset\\{(x,y)\in\mathbb{R}^{2n}\,;\,|x-y|\leq\epsilon/4\\}.$

Hence, for $\forall x\in B(x_{0},\epsilon/4)$, we have $K(x,y)=0$ when
$|y-x_{0}|\geq\epsilon/2$.

Fix a function $\chi\in C_{c}^{\infty}$ such that $\chi(y)=1$ when
$|y-x_{0}|\leq\epsilon/2$ and $\mathop{\rm supp}\chi=B(x_{0},\epsilon)$, then

$\forall x\in B(x_{0},\epsilon/4),\quad T_{1}u(x)=\int
K(x,y)u(y)\,\mathrm{d}{y}=\int K(x,y)\chi(y)u(y)\,\mathrm{d}{y}=T_{1}(\chi
u)(x).$

Note that $\chi u\in C_{c}^{\infty}\subset\mathscr{S}$ due to the fact that
$u$ is $C^{\infty}$ in $\mathop{\rm supp}\chi$, so $T_{1}(\chi
u)\in\mathscr{S}$. Because $T_{1}u=T_{1}(\chi u)\in\mathscr{S}$ on
$B(x_{0},\epsilon/4)$, we conclude that $T_{1}u\in
C^{\infty}(B(x_{0},\epsilon/4))$. Also, $T_{2}u\in C^{\infty}$ because $T_{2}$
is $C^{\infty}$-smooth. In total, $Tu$ is $C^{\infty}$-smooth in a small
neighborhood of $x_{0}$, so $x_{0}\notin\mathop{\rm sing\,supp}(Tu)$.

We obtain $(\mathop{\rm sing\,supp}u)^{c}\subset(\mathop{\rm
sing\,supp}(Tu))^{c}$. The proof is complete. ∎

### Exercise

######  Exercise 2.1.

Prove $\mathscr{S}({\mathbb{R}^{n}})\subset S^{-\infty}$, namely,
$\forall\varphi\in\mathscr{S}({\mathbb{R}^{n}})$, $\varphi(\xi)\in
S^{-\infty}$.

######  Exercise 2.2.

Prove Lemma 2.3.

######  Exercise 2.3.

Prove Lemma 2.14. See [wong2014introduction, Prop. 6.7] for reference.

##  Chapter 3 Oscillatory integrals

In §2.1 we encountered the notion of kernel of a $\Psi$DO,

$K(x,y):=(2\pi)^{-n}\int e^{i(x-y)\cdot\xi}\sigma(x,\xi)\,\mathrm{d}{\xi}.$

which might not be integral in the Lebesgue sense (e.g. when
$\sigma(x,\xi)=1$). However, if we look back to the original definition of a
$\Psi$DO,

$(T_{\sigma}\varphi)(x)=(2\pi)^{-n/2}\int_{{\mathbb{R}^{n}}}e^{ix\cdot\xi}\sigma(x,\xi)\hat{\varphi}(\xi)\,\mathrm{d}{\xi},$

the integral above is always well-defined in the Lebesgue sense, because
$\hat{\varphi}$ is rapidly decaying. Specifically, for any $m\in\mathbb{R}$
and any $\sigma\in S^{m}$, we have

$|(T_{\sigma}\varphi)(x)|\lesssim\int|\sigma(x,\xi)\hat{\varphi}(\xi)|\,\mathrm{d}{\xi}\lesssim\int_{{\mathbb{R}^{n}}}\langle{\xi}\rangle^{m}\langle{\xi}\rangle^{-m-n-1}\,\mathrm{d}{\xi}<+\infty.$

The problems emerges when we expand the Fourier transform $\hat{\varphi}$ (by
a variable $y$) and exchange the integration order of $y$ and $\xi$:

$\displaystyle(T_{\sigma}\varphi)(x)$
$\displaystyle=(2\pi)^{-n}\int\big{(}\int
e^{i(x-y)\cdot\xi}\sigma(x,\xi)\varphi(y)\,\mathrm{d}{y}\big{)}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\,\mathrm{d}{\xi}},$
(3.1) $\displaystyle\int_{{\mathbb{R}^{n}}}K(x,y)\varphi(y)\,\mathrm{d}{y}$
$\displaystyle=(2\pi)^{-n}\int\big{(}\int
e^{i(x-y)\cdot\xi}\sigma(x,\xi){\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\,\mathrm{d}{\xi}}\big{)}\varphi(y)\,\mathrm{d}{y}.$

According to Fubini’s theorem, this exchange is valid only when all of the
integrals involved are absolutely integrable. From time to time we will
encounter integrals of the form (3.1), but also more general than that. A
rigorous framework is appealing for making the integrals of these type always
well-defined.

###  3.1. Oscillatory integrals - Type I

Generally speaking, for any $u\in\mathscr{S}({\mathbb{R}^{n}})$ and $\sigma\in
S^{m}(\mathbb{R}_{x}^{n}\times R_{\xi}^{N})$, one is interested in the
following integral

$I(u):=\int
e^{i\varphi(x,\xi)}\sigma(x,\xi)u(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi}$ (3.2)

where $\varphi$ is a phase function defined as follows.

######  Definition 3.1 (Phase function).

Function $\varphi$ is called a _phase function (of order $\mu$)_ if it
satisfies

  1. (1)

$\varphi\in
C^{\infty}(\mathbb{R}_{x}^{n}\times(\mathbb{R}_{\xi}^{N}\backslash\\{0\\});\mathbb{R})$
is real-valued;

  2. (2)

$\varphi$ is homogeneous w.r.t. $\xi$ of order $\mu>0$, i.e.
$\varphi(x,t\xi)=t^{\mu}\varphi(x,\xi)$;

  3. (3)

$|\nabla_{(x,\xi)}\varphi(x,\xi)|\neq 0$ for
$\forall(x,\xi)\in\mathbb{R}_{x}^{n}\times(\mathbb{R}_{\xi}^{N}\backslash\\{0\\})$.

There are different ways to define the notion of phase functions, see
[horm2003IIV, §7.8], and we don’t pursue diversity here. Note that $n$ might
not equal $N$, and most of the results in §2.1 holds also for the case $n\neq
N$. Here we consider phases of order $\mu$, instead of just order 1, because
in §5 and §7 we do encounter phases of order 2. The condition $\mu>0$ is
indispensable.

In contrast with (3.1), in (3.2) it is not sure that integrating first w.r.t
$x$ (or $\xi$) can guarantee it’s integrable. Instead, we study

$\lim_{\epsilon\to 0^{+}}I_{\epsilon}(u):=\lim_{\epsilon\to 0^{+}}\int
e^{i\varphi(x,\xi)}\sigma(x,\xi){\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\chi(\epsilon\xi)}u(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi},$
(3.3)

where $\chi$ is a function in $C_{c}^{\infty}({\mathbb{R}^{n}})$ with
$\chi(0)=1$. We show that the limit (3.3) exists and its value is independent
of the choice of $\chi$.

######  Theorem 3.2.

Assume $m\in\mathbb{R}$, $\sigma\in S^{m}(\mathbb{R}_{x}^{n}\times
R_{\xi}^{N})$ and $\varphi$ is a phase function of order $\mu$. Fix a function
$\chi\in C_{c}^{\infty}({\mathbb{R}^{n}})$ with $\chi(0)=1$. Assume either
$u\in C_{c}^{\infty}({\mathbb{R}^{n}})$, or
$u\in\mathscr{S}({\mathbb{R}^{n}})$ and $\partial_{x}^{\alpha}\varphi(x,\xi)$
is tempered w.r.t. $x$ for any $\alpha$. Then the limit (3.3) exists and its
value is independent of the choice of $\chi$, and it equals to

$\int
e^{i\varphi(x,\xi)}L^{T}\big{(}\sigma(x,\xi)u(x)\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi},$

when integer $T>(m+N)/\mu$ where $L$ is given in Lemma 3.4.

The proof Theorem 3.2, we fist do some preparation.

######  Lemma 3.3.

Assume $\chi\in C_{c}^{\infty}({\mathbb{R}^{n}})$ and let
$\epsilon\in\mathbb{R}$. There exists a constant $C$ independent of $\epsilon$
such that

$\boxed{|\partial_{\xi}^{\alpha}\big{(}\chi(\epsilon\xi)\big{)}|\leq
C\langle{\xi}\rangle^{-|\alpha|}.}$

###### Proof.

Because $\chi\in C_{c}^{\infty}({\mathbb{R}^{n}})$, there exists a fixed
constant $C$ such that $\chi(\epsilon\xi)\equiv 0$ when
$|\epsilon\langle{\xi}\rangle|\geq C$. When $|\epsilon\langle{\xi}\rangle|\geq
C$, $\chi(\epsilon\xi)=0$ so
$\partial_{\xi}^{\alpha}\big{(}\chi(\epsilon\xi)\big{)}=0$; when
$|\epsilon\langle{\xi}\rangle|\leq C$, we have

$|\partial_{\xi}^{\alpha}\big{(}\chi(\epsilon\xi)\big{)}|=|\epsilon|^{|\alpha|}|\partial_{\xi}^{\alpha}\chi(\epsilon\xi)|\leq(C\langle{\xi}\rangle)^{-|\alpha|}\sup_{\mathbb{R}^{n}}|\partial_{\xi}^{\alpha}\chi|\simeq
C\langle{\xi}\rangle^{-|\alpha|}.$

We arrive at the conclusion. ∎

######  Lemma 3.4.

Assume $\varphi$ is a phase function $\varphi$ of order $\mu$, and
$\partial_{x}^{\alpha}\varphi(x,\xi)$ is tempered w.r.t. $x$ for any $\alpha$.
Then there exists an first order linear differential operator

$L=a_{j}(x,\xi)\partial_{x_{j}}+b_{j}(x,\xi)\partial_{\xi_{j}}+c(x,\xi)$

such that ${}^{t}L(e^{i\varphi(x,\xi)})=e^{i\varphi(x,\xi)}$, and for any
fixed $x_{0}$, $a_{j}(x_{0},\cdot)\in S^{-\mu}$, $b_{j}(x_{0},\cdot)\in
S^{1-\mu}$, $c(x_{0},\cdot)\in S^{-\mu}$, and $a_{j},b_{j},c$ are tempered
functions of $x$ variable.

Here $\langle{{}^{t}Lf,g}\rangle:=\langle{f,Lg}\rangle$, where the integral is
w.r.t. $(x,\xi)$, and $f,g\in C_{0}^{\infty}$. ${}^{t}L$ is call the
_transpose_ of $L$, e.g. ${}^{t}(\nabla_{\xi})=-\nabla_{\xi}$.

###### Proof of Lemma 3.4.

We write $\nabla_{x}\varphi=\varphi_{x}$ and
$\nabla_{\xi}\varphi=\varphi_{\xi}$ for short. Fix a $\chi\in
C_{c}^{\infty}(\mathbb{R}^{N})$ with $\chi\equiv 1$ in a neighborhood of 0.
Construct

$M:=(1-\chi(\xi))\frac{\varphi_{x}\cdot
D_{x}+\langle{\xi}\rangle^{2}\varphi_{\xi}\cdot
D_{\xi}}{|\varphi_{x}|^{2}+\langle{\xi}\rangle^{2}|\varphi_{\xi}|^{2}}+\chi(\xi).$

We mention several facts about $M$:

  * •

First, $M$ is well-defined. Note that the denominator
$|\varphi_{x}|^{2}+\langle{\xi}\rangle|\varphi_{\xi}|^{2}\geq|\varphi_{x}|^{2}+|\varphi_{\xi}|^{2}\neq
0$ when
$(x,\xi)\in\mathbb{R}_{x}^{n}\times(\mathbb{R}_{\xi}^{N}\backslash\\{0\\})$,
and the point $\xi=0$ has been cutoff by $1-\chi$, so $M$ is always well-
defined;

  * •

Second, away from $\xi=0$, $\varphi_{x}(x_{0},\cdot)\in S^{\mu}$,
$\varphi_{\xi}(x_{0},\cdot)\in S^{\mu-1}$, the denominator$(x_{0},\cdot)\in
S^{2\mu}$;

  * •

Third, $Me^{i\varphi(x,\xi)}=(1-\chi)e^{i\varphi(x,\xi)}+\chi
e^{i\varphi(x,\xi)}=e^{i\varphi(x,\xi)}.$

The transpose of $M$ is the desired operator. Indeed, it can be checked that,
when $x$ is fixed,

${}^{t}M=(1-\chi)S^{-\mu}\partial_{x}+(1-\chi)S^{1-\mu}\partial_{\xi}+(1-\chi)S^{-\mu}+\chi(\xi).$
(3.4)

The proof is complete. ∎

######  Remark 3.5.

Here $S^{-\mu}\partial_{x}$ is a shorthand of $a\in C^{\infty}$ such that
$a_{j}\partial_{x_{j}}$ for some $a(x_{0},\cdot)\in S^{-\mu}$. Readers should
note that (3.4) is somehow misleading because the coefficients might not be
bounded w.r.t. $x$, e.g. ${}^{t}M=x_{1}\partial_{x_{1}}$. However, they must
be tempered, and these tempered growth will be balanced by the rapid decay of
$u$, The notations in (3.4) wouldn’t hurt.

###### Proof of Theorem 3.2.

Choose $L$ according to Lemma 3.4, then

$\displaystyle I_{\epsilon}(u)$
$\displaystyle=\int({}^{t}L)^{T}(e^{i\varphi(x,\xi)})\,\sigma(x,\xi)\chi(\epsilon\xi)u(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi}$
$\displaystyle=\int
e^{i\varphi(x,\xi)}L^{T}\big{(}\sigma(x,\xi)\chi(\epsilon\xi)u(x)\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi}.$
(3.5)

Readers should note that the transpose of $L$ is realized by the classical
integration by parts (nothing fancy here), and it is the presence of
$\chi(\epsilon\xi)$ that cancels the boundary terms and makes the integration
by parts applicable.

The conditions “$a_{j}\in S^{-\mu}$, $b_{j}\in S^{1-\mu}$, $c\in S^{-\mu}$” in
Lemma 3.4 give us

$\displaystyle\ L^{T}\big{(}\sigma(x,\xi)\chi(\epsilon\xi)u(x)\big{)}$
$\displaystyle=$ $\displaystyle\
\big{(}(a_{j}(x,\xi)\partial_{x_{j}}+b_{j}(x,\xi)\partial_{\xi_{j}}+c(x,\xi)\big{)}^{T}\big{(}\sigma(x,\xi)\chi(\epsilon\xi)u(x)\big{)}$
$\displaystyle=$ $\displaystyle\ \sum_{|\alpha+\beta|\leq
T}S^{-\mu|\alpha|+(1-\mu)|\beta|-\mu(T-|\alpha|-|\beta|)}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\big{(}\sigma(x,\xi)\chi(\epsilon\xi)u(x)\big{)}$
$\displaystyle=$ $\displaystyle\ \sum_{|\alpha+\beta|\leq T}S^{|\beta|-\mu
T}\partial_{x}^{\alpha}\partial_{\xi}^{\beta}\big{(}\sigma(x,\xi)\chi(\epsilon\xi)u(x)\big{)}$
$\displaystyle=$ $\displaystyle\ \sum_{|\alpha+\beta|\leq
T}\sum_{\beta^{\prime}+\beta^{\prime\prime}=\beta}C_{\beta^{\prime},\beta^{\prime\prime}}S^{|\beta|-\mu
T}\partial_{x}^{\alpha}\big{(}\partial_{\xi}^{\beta^{\prime}}\sigma(x,\xi)\partial_{\xi}^{\beta^{\prime\prime}}[\chi(\epsilon\xi)]u(x)\big{)}$
$\displaystyle=$ $\displaystyle\ \sum_{|\alpha+\beta|\leq
T}\sum_{\beta^{\prime}+\beta^{\prime\prime}=\beta}CS^{|\beta|-\mu
T}\partial_{x}^{\alpha}\big{(}\partial_{\xi}^{\beta^{\prime}}\sigma(x,\xi)\partial_{\xi}^{\beta^{\prime\prime}}[\chi(\epsilon\xi)]u(x)\big{)}$
$\displaystyle=$ $\displaystyle\ \sum_{|\alpha+\beta|\leq
T}\sum_{\beta^{\prime}+\beta^{\prime\prime}=\beta}\sum_{\alpha^{\prime}+\alpha^{\prime\prime}=\alpha}CS^{|\beta|-\mu
T}\partial_{x}^{\alpha^{\prime}}\partial_{\xi}^{\beta^{\prime}}\sigma(x,\xi){\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\partial_{\xi}^{\beta^{\prime\prime}}[\chi(\epsilon\xi)]}\partial_{x}^{\alpha^{\prime\prime}}u(x).$

The term $\partial_{\xi}^{\beta^{\prime\prime}}[\chi(\epsilon\xi)]$ is the
only term that depends on $\epsilon$. Hence, by Lemma 3.3 we can have

$\displaystyle\ |L^{T}\big{(}\sigma(x,\xi)\chi(\epsilon\xi)u(x)\big{)}|$
$\displaystyle\leq$ $\displaystyle\ \sum_{|\alpha+\beta|\leq
T}\sum_{\beta^{\prime}+\beta^{\prime\prime}=\beta}\sum_{\alpha^{\prime}+\alpha^{\prime\prime}=\alpha}C\langle{\xi}\rangle^{|\beta|-\mu
T}C\langle{\xi}\rangle^{m-|\beta^{\prime}|}C\langle{\xi}\rangle^{-|\beta^{\prime\prime}|}\mathscr{S}(\mathbb{R}_{x}^{n})$
$\displaystyle\leq$ $\displaystyle\ C\langle{\xi}\rangle^{m-\mu
T}\mathscr{S}(\mathbb{R}_{x}^{n}),$

where the constant $C$ is independent of $\epsilon$. Then $T$ is chosen to be
larger than $(m+N)/\mu$, the integrand in (3.5) is bounded by a absolutely
integral function. Therefore, according to LDCT, the limit $\lim_{\epsilon\to
0^{+}}I_{\epsilon}(u)$ exists. Readers may think where we used the condition
$\mu>0$.

We also show that the limit is independent of $\chi$. Fix a $T>(m+N)/\mu$,
then by (3.5) we have

$\displaystyle\lim_{\epsilon\to 0^{+}}I_{\epsilon}(u)$
$\displaystyle=\lim_{\epsilon\to 0^{+}}\int
e^{i\varphi(x,\xi)}L^{T}\big{(}\sigma(x,\xi)\chi(\epsilon\xi)u(x)\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi}$
$\displaystyle=\int e^{i\varphi(x,\xi)}\lim_{\epsilon\to
0^{+}}L^{T}\big{(}\sigma(x,\xi)\chi(\epsilon\xi)u(x)\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi}\quad\text{(thanks
to LDCT)}$ $\displaystyle=\int
e^{i\varphi(x,\xi)}L^{T}\big{(}\sigma(x,\xi)u(x)\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi},$
(3.6)

which implies $\lim_{\epsilon\to 0^{+}}I_{\epsilon}(u)$ is independent of
$\chi$. The proof is complete. ∎

Readers may think about if the framework can be generalized to symbols in
$S_{\rho,\delta}^{m}$.

Now let’s summarize the definition of oscillatory integrals.

######  Definition 3.6 (Oscillatory integral).

For any $m\in\mathbb{R}$, any $\sigma\in S^{m}(\mathbb{R}_{x}^{n}\times
R_{\xi}^{N})$, and any phase function $\varphi$ of order $\mu$, and either
$u\in C_{c}^{\infty}({\mathbb{R}^{n}})$, or
$u\in\mathscr{S}({\mathbb{R}^{n}})$ and $\partial_{x}^{\alpha}\varphi(x,\xi)$
is tempered w.r.t. $x$ for any $\alpha$, the integral

$I(u)=\int
e^{i\varphi(x,\xi)}\sigma(x,\xi)u(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi}$

is defined as

$\boxed{I(u):=\lim_{\epsilon\to 0^{+}}\int
e^{i\varphi(x,\xi)}\sigma(x,\xi)\chi(\epsilon\xi)u(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi},}$
(3.7)

where the result is independent of $\chi$, as long as $\chi\in C_{c}^{\infty}$
and $\chi(0)=1$. The limit equals

$\boxed{I(u)=\int
e^{i\varphi(x,\xi)}L^{T}\big{(}\sigma(x,\xi)u(x)\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi}}$

when integer $T>(m+N)/\mu$, where $L$ is given in Lemma 3.4.

In many cases we will meet oscillatory integrals involving parameters.

######  Lemma 3.7.

For any $\sigma\in
S^{m}(\mathbb{R}_{x}^{n_{1}}\times\mathbb{R}_{y}^{n_{2}}\times R_{\xi}^{N})$,
and any phase function $\varphi$ of order $\mu$, and for either $u\in
C_{c}^{\infty}(\mathbb{R}_{x}^{n_{1}}\times\mathbb{R}_{y}^{n_{2}})$, or
$u\in\mathscr{S}(\mathbb{R}_{x}^{n_{1}}\times\mathbb{R}_{y}^{n_{2}})$ and
$\partial_{(x,y)}^{\alpha}\varphi(x,\xi)$ is tempered w.r.t. $(x,y)$ for any
$\alpha$, the integral

$I(u)(y):=\int
e^{i\varphi(x,y,\xi)}\sigma(x,y,\xi)u(x,y)\,\mathrm{d}{x}\,\mathrm{d}{\xi}$
(3.8)

is a well-defined oscillatory integral, and
$I\colon\mathscr{S}(\mathbb{R}^{n_{1}})\to\mathscr{S}(\mathbb{R}^{n_{2}})$
bounded. Moreover, we have

$\displaystyle\frac{\partial}{\partial y}\big{(}I(u)(y)\big{)}$
$\displaystyle=\int\frac{\partial}{\partial
y}\big{(}e^{i\varphi(x,y,\xi)}\sigma(x,y,\xi)u(x,y)\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi},$
$\displaystyle\int I(u)(y)\,\mathrm{d}{y}$ $\displaystyle=\int
e^{i\varphi(x,y,\xi)}\sigma(x,y,\xi)u(x,y)\,\mathrm{d}{x}\,\mathrm{d}{y}\,\mathrm{d}{\xi}.$

We omit the proof. The take-home message of Lemma 3.7 is that oscillatory
integrals can have parameters, and there are much freedom to put operations
w.r.t. $y$ inside the integration $I(u)(y)$.

Now we go back to $\Psi$DOs and its kernel. We have intuitively claimed that
the kernel of $T_{\sigma}$ is of the form

$K(x,y)=\int e^{i(x-y)\cdot\xi}\sigma(x,\xi)\,\mathrm{d}{\xi}.$

Note that this object has variables $(x,y)$, so a proper candidate of test
functions should be $w(x,y)\in\mathscr{S}$. We choose $w(x,y)=u(y)v(x)$ where
$u,v\in\mathscr{S}$, then formally we should have

$\langle{K,v\otimes u}\rangle\simeq\int
e^{i(x-y)\cdot\xi}\sigma(x,\xi)u(y)v(x)\,\mathrm{d}{x}\,\mathrm{d}{y}\,\mathrm{d}{\xi}\simeq\langle{T_{\sigma}u,v}\rangle.$

The integral above is exactly an example of Lemma 3.7, so it is a well-defined
oscillatory integral. Now the kernel of a $\Psi$DO can be defined.

######  Definition 3.8 (Kernel).

Assume $m\in\mathbb{R}$ and $\sigma\in S^{m}$, and $T_{\sigma}$ is the
corresponding $\Psi$DO. The _kernel_ of $T_{\sigma}$ is defined as a map:

$K_{\sigma}\colon w\in\mathscr{S}(\mathbb{R}^{2n})\ \mapsto\
\boxed{\langle{K_{\sigma},w}\rangle:=(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}\sigma(x,\xi)w(x,y)\,\mathrm{d}{x}\,\mathrm{d}{y}\,\mathrm{d}{\xi}}\in\mathbb{C}.$

When $x\neq y$, we write $K_{\sigma}(x,y)$ as

$\boxed{K_{\sigma}(x,y)=(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}\sigma(x,\xi)\,\mathrm{d}{\xi}.}$ (3.9)

The well-definedness of Definition 3.8 is guaranteed by Lemma 3.7.

######  Remark 3.9.

From (3.9), we know that when $m<-N$, $K_{\sigma}$ is a well-defined bounded
function for any $(x,y)$ because

$|K_{\sigma}|\lesssim\int|\sigma(x,\xi)|\,\mathrm{d}{\xi}\lesssim\int\langle{\xi}\rangle^{m}\,\mathrm{d}{\xi}\leq
C<+\infty.$

This implies when the order of $\sigma$ is small enough, there should hold
some type of boundedness for $T_{\sigma}$, and we will cover this in §6.
However when $m\geq-N$, only when $x\neq y$ the kernel $K_{\sigma}$ can be
expressed as (3.9).

######  Lemma 3.10.

Under the assumption of Definition 3.8, we have

$\langle{K_{\sigma},u(y)v(x)}\rangle=\langle{T_{\sigma}u,v}\rangle.$

We omit the proof.

######  Lemma 3.11.

Under the assumption of Definition 3.8, when $x\neq y$, $K_{\sigma}$ is
$C^{\infty}$ smooth Moreover, for $T$ large enough, we have

$\boxed{|K_{\sigma}(x,y)|\leq C_{T}|x-y|^{-T},\quad|x-y|\geq 1.}$

###### Proof.

For any fixed point $(x,y)$ with $x\neq y$, we show that $K_{\sigma}$ is
$C^{\infty}$ at this point. Fix a function $\chi\in C^{\infty}$ satisfying
$\chi\equiv 0$ in a small neighborhood $U$ of the diagonal $\\{x=y\\}$ and
$\chi\equiv 1$ in the interior of the complement of $U$. We can shrink $U$
such that for any $x\neq y$, $\chi(x,y)=1$. For any $w\in\mathscr{S}$, we have

$\langle{\partial_{x}^{\alpha}(\chi K_{\sigma}),w}\rangle\simeq\int
e^{i(x-y)\cdot\xi}\xi^{\alpha}\sigma(x,\xi)w(x,y)\,\mathrm{d}{x}\,\mathrm{d}{y}\,\mathrm{d}{\xi}.$

Apply the operator $L$ to $e^{i(x-y)\cdot\xi}$ and integrate by parts, we
obtain

$\langle{\partial_{x}^{\alpha}\chi
K_{\sigma},w}\rangle=\langle{(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}\big{(}\frac{-(x-y)\cdot
D_{\xi}}{|x-y|^{2}}\big{)}^{T}\big{(}\xi^{\alpha}\sigma(x,\xi)\big{)}\,\mathrm{d}{\xi},w\chi}\rangle,$

which implies

$\partial_{x}^{\alpha}(\chi K_{\sigma})(x,y)\simeq\int
e^{i(x-y)\cdot\xi}\big{(}\frac{-(x-y)\cdot
D_{\xi}}{|x-y|^{2}}\big{)}^{T}\big{(}\xi^{\alpha}\sigma(x,\xi)\big{)}\,\mathrm{d}{\xi}.$

It can be checked that when $T$ is large enough, the integral above will be
absolutely integrable, and is of the order $|x-y|^{-T}$ for $T$ large enough.
Therefore, $K_{\sigma}\in C^{\infty}(\mathbb{R}^{2n}\backslash\\{x=y\\})$, and
$K$ satisfies the desired decay. ∎

By Lemma 3.11, we see that $K(x,y)$ behave nicely when off the diagonal,
thanks to the notion of oscillatory integrals. However, on the diagonal,
$K(x,y)$ might still be ill-defined. See Example 3.12.

######  Example 3.12.

The kernel corresponding to the identity operator (symbol $=1$) is the
distribution $\delta(x-y)$. This is because

$\displaystyle\langle{K,v\otimes u}\rangle$
$\displaystyle=\langle{T_{1}u,v}\rangle=\langle{u,v}\rangle=\int
u(x)v(x)\,\mathrm{d}{x}$
$\displaystyle=\int\delta(x-y)u(y)v(x)\,\mathrm{d}{x}\,\mathrm{d}{y}$
$\displaystyle=\langle{\delta(x-y),v\otimes u}\rangle.$

which gives $K(x,y)=\delta(x-y)$.

######  Example 3.13.

The kernel corresponding to $D_{1}$ (symbol $=\xi_{1}$) is $D_{1}\delta(x-y)$.
This is because

$\displaystyle\langle{K,v\otimes u}\rangle$
$\displaystyle=\langle{D_{1}u,v}\rangle=\int
D_{1}u(x)v(x)\,\mathrm{d}{x}=\int\delta(x-y)D_{1}u(y)v(x)\,\mathrm{d}{x}\,\mathrm{d}{y}$
$\displaystyle=-\int
D_{y_{1}}\big{(}\delta(x-y)\big{)}u(y)v(x)\,\mathrm{d}{x}\,\mathrm{d}{y}=\int
D_{1}\delta(x-y)u(y)v(x)\,\mathrm{d}{x}\,\mathrm{d}{y}$
$\displaystyle=\langle{D_{1}\delta(x-y),v\otimes u}\rangle.$

which gives $K(x,y)=D_{1}\delta(x-y)$. Besides, readers may also have tried
another way to compute the kernel and get a zero result: when $x\neq y$

$\displaystyle K(x,y)$ $\displaystyle=(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}\xi_{1}\,\mathrm{d}{\xi}=(2\pi)^{-n}\int\big{(}\frac{(x-y)\cdot
D_{\xi}}{|x-y|^{2}}\big{)}^{2}\big{(}e^{i(x-y)\cdot\xi}\big{)}\xi_{1}\,\mathrm{d}{\xi}$
$\displaystyle=(2\pi)^{-n}\int e^{i(x-y)\cdot\xi}\big{(}\frac{-(x-y)\cdot
D_{\xi}}{|x-y|^{2}}\big{)}^{2}(\xi_{1})\,\mathrm{d}{\xi}=0.$

This result is technically correct (because $D_{1}\delta(x-y)=0$ when $x\neq
y$), but is not complete: it cannot speak about the behavior of $K$ on the
diagonal. This example told us, none of the methods is the best one to get
most accurate expression for a kernel, sometimes we need to do complicated and
delicate computations.

###  3.2. Oscillatory integrals - Type II

We know the Fourier transform of a constant function is the $\delta$ function
and so the inverse Fourier transform of the $\delta$ function should be the
constant, namely,

$\int e^{i(x-y)\cdot\xi}\,\mathrm{d}{x}\,\mathrm{d}{\xi}\simeq 1.$ (3.10)

This integral can be regarded as the $I(u)$ defined in Lemma 3.7 where the
symbol and the $u$ are both constant 1. However, this is not covered by Lemma
3.7 because $1\notin\mathscr{S}$. The map $I$ in Lemma 3.7 is defined on
$\mathscr{S}$. Now by using duality arguments we shall generalize it from
$\mathscr{S}$ to $S^{+\infty}$.

Let $\sigma\in S^{m}$ and $u\in\mathscr{S}$, so $T_{\sigma}u\in\mathscr{S}$.
Let $f\in S^{+\infty}$, then $f$ is a smooth tempered function, so
$\langle{T_{\sigma}u,f}\rangle$ is meaningful and

$\langle{T_{\sigma}u,f}\rangle=\lim_{\epsilon\to
0^{+}}\langle{T_{\sigma}u,\chi(\epsilon\cdot)f}\rangle$

holds for any $\chi\in C_{c}^{\infty}$. Expanding the integral, we have

$\displaystyle\langle{T_{\sigma}u,f}\rangle$ $\displaystyle=\lim_{\epsilon\to
0^{+}}\int e^{i(x-y)\cdot\xi}\sigma(x,\xi)u(y)f(x)\chi(\epsilon
x)\,\mathrm{d}{x}\,\mathrm{d}{y}\,\mathrm{d}{\xi}$
$\displaystyle=\lim_{\epsilon\to 0^{+}}\int
e^{i(x-y)\cdot\xi}\sigma(x,\xi){\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\chi(\epsilon\xi)}u(y)f(x)\chi(\epsilon
x)\,\mathrm{d}{x}\,\mathrm{d}{y}\,\mathrm{d}{\xi}$
$\displaystyle=\lim_{\epsilon\to 0^{+}}\int u(y)\big{(}\int
e^{i(x-y)\cdot\xi}\sigma(x,\xi)f(x){\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\chi(\epsilon
x)\chi(\epsilon\xi)}\,\mathrm{d}{x}\,\mathrm{d}{\xi}\big{)}\,\mathrm{d}{y}.$

This inspired us to define (3.10) as

$\lim_{\epsilon\to 0^{+}}\int e^{i(x-y)\cdot\xi}\chi(\epsilon
x)\chi(\epsilon\xi)\,\mathrm{d}{x}\,\mathrm{d}{\xi}.$

More generally, we can generalize Definition 3.6 as follows.

######  Definition 3.14 (Oscillatory integral).

For any $m\in\mathbb{R}$ and $\rho>0$, any $\sigma\in S_{\rho}^{m}(R^{N})$,
and any phase function $\varphi\in
C^{\infty}(\mathbb{R}^{N}\backslash\\{0\\})$ of order $\mu$ (real-valued,
$\varphi(\theta)=t^{\mu}\varphi(\theta)$, and
$\nabla_{\theta}\varphi(\theta)\neq 0$ when $\theta\neq 0$) satisfying

$\boxed{\rho+\mu>1,}$ (3.11)

the integral

$I=\int e^{i\varphi(\theta)}\sigma(\theta)\,\mathrm{d}{\theta}$

is defined as

$\boxed{I:=\lim_{\epsilon\to 0^{+}}\int
e^{i\varphi(\theta)}\sigma(\theta)\chi(\epsilon\theta)\,\mathrm{d}{\theta},}$
(3.12)

where the result is independent of $\chi$, as long as $\chi\in C_{c}^{\infty}$
and $\chi(0)=1$. The limit equals

$\boxed{I=\int
e^{i\varphi(\theta)}L^{T}\big{(}\sigma(\theta)\big{)}\,\mathrm{d}{\theta}}$
(3.13)

when $T>(m+N)/(\rho+\mu-1)$, where $L$ is given in Lemma 3.16 below.

######  Remark 3.15.

The space $S_{\rho}^{m}(R^{N})$ is defined as

$S_{\rho}^{m}(R^{N}):=\\{\varphi\in
C^{\infty}(\mathbb{R}^{N})\,;\,|\partial^{\alpha}\varphi(\theta)|\lesssim\langle{\theta}\rangle^{m-\rho|\alpha|}\\}.$

See [alinhac2007pseudo, §I.8.1] for more details. Also, the formula (3.9) in
the definition of the kernel is meaningful now.

######  Lemma 3.16.

Under the condition in Definition 3.14, there exists an first order linear
differential operator

$L=b_{j}(\theta)\partial_{\theta_{j}}+c(\theta)$

such that ${}^{t}L(e^{i\varphi(\theta)})=e^{i\varphi(\theta)}$,
$0\notin\mathop{\rm supp}b_{j}$, and $b_{j}\in S^{1-\mu}$, $c\in S^{-\mu}$,
and in a small neighborhood of $\theta=0$ there holds $b\equiv 0$ and $c\equiv
1$.

###### Proof.

We write $\nabla_{\theta}\varphi=\varphi_{\theta}$ for short. Fix a $\chi\in
C_{c}^{\infty}({\mathbb{R}^{n}})$ with $\chi\equiv 1$ in a neighborhood of 0.
Construct

$M:=(1-\chi(\theta))\frac{\varphi_{\theta}\cdot
D_{\theta}}{|\varphi_{\theta}|^{2}}+\chi(\theta).$

We mention several facts about $M$:

  * •

First, $M$ is well-defined. Note that the denominator
$|\varphi_{\theta}|^{2}\neq 0$ when
$\theta\in\mathbb{R}^{N}\backslash\\{0\\}$, and the point $\theta=0$ has been
cutoff by $1-\chi$, so $M$ is always well-defined;

  * •

Second, away from $\theta=0$, $\varphi_{\theta}\in S^{\mu-1}$;

  * •

Third, $Me^{i\varphi(\theta)}=(1-\chi)e^{i\varphi(\theta)}+\chi
e^{i\varphi(\theta)}=e^{i\varphi(\theta)}.$

The transpose of $M$ is the desired operator. Indeed, it can be checked that,
when $x$ is fixed,

${}^{t}M=(1-\chi)S^{1-\mu}\partial_{\theta}+(1-\chi)S^{-\mu}+\chi(\theta).$

The proof is complete. ∎

Readers may compare Lemma 3.16 with Lemma 3.4.

######  Lemma 3.17.

Assume $L$ is chosen as in Lemma 3.16, and $\sigma\in S_{\rho}^{m}(R^{N})$,
then

$|L^{T}\big{(}\sigma(\theta)\big{)}|\lesssim\langle{\theta}\rangle^{m-(\rho+\mu-1)T}.$

The proof is left as an exercise. Combining Lemmas 3.16 and 3.17, we obtain a
result similar to Theorem 3.2.

######  Theorem 3.18.

Assume $m\in\mathbb{R}$ and $\rho>0$, $\sigma\in S_{\rho}^{m}(R^{N})$ and
$\varphi\in C^{\infty}(\mathbb{R}^{N}\backslash\\{0\\})$ is a phase function
of order $\mu$. Fix a function $\chi\in C_{c}^{\infty}({\mathbb{R}^{n}})$ with
$\chi(0)=1$. Then the limit (3.12) exists and its value is independent of the
choice of $\chi$, and it equals to

$\int
e^{i\varphi(\theta)}L^{T}\big{(}\sigma(\theta)\big{)}\,\mathrm{d}{\theta},$

when $T>(m+N)/(\rho+\mu-1)$ where $T$ is given in Lemma 3.16.

The proof is similar to that of Theorem 3.2. The generalized definition of the
oscillatory integral can handle more cases. One of the examples is as follows.

######  Lemma 3.19.

The following equality holds in oscillatory sense,

$\int_{\mathbb{R}^{2n}}e^{\pm
ix\cdot\xi}\,\mathrm{d}{x}\,\mathrm{d}{\xi}=(2\pi)^{n}.$

###### Proof.

We shall regard $(x,\xi)$ as the $\theta$ in Definition 3.14, then this phase
function $x\cdot\xi$ is of order 2. One can also check that
$\nabla_{(x,\xi)}(x\cdot\xi)\neq 0$ when $(x,\xi)\neq 0$. We choose the cutoff
function as $\chi(\epsilon x)\chi(\epsilon\xi)$ where $\chi\in C_{c}^{\infty}$
with $\chi(0)=1$, then

$\displaystyle\int e^{ix\cdot\xi}\,\mathrm{d}{x}\,\mathrm{d}{\xi}$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}:=}\lim_{\epsilon\to
0^{+}}\int e^{ix\cdot\xi}\chi(\epsilon
x)\chi(\epsilon\xi)\,\mathrm{d}{x}\,\mathrm{d}{\xi}=\lim_{\epsilon\to
0^{+}}\int(\int
e^{ix\cdot\xi}\chi(x)\,\mathrm{d}{x})\cdot\chi(\epsilon^{2}\xi)\,\mathrm{d}{\xi}$
$\displaystyle=(2\pi)^{n/2}\lim_{\epsilon\to
0^{+}}\int\widehat{\chi}(-\xi)\chi(\epsilon^{2}\xi)\,\mathrm{d}{\xi}=(2\pi)^{n/2}\int\widehat{\chi}(-\xi)\,\mathrm{d}{\xi}\qquad\text{(LDCT)}$
$\displaystyle=(2\pi)^{n}\chi(0)=(2\pi)^{n}.$

The case for $e^{-ix\cdot\xi}$ is similar. Note that all of the integrals
above are usual integral besides the first one on the LHS. ∎

The following result will be useful.

######  Lemma 3.20.

The following equality holds in oscillatory sense,

$\int_{\mathbb{R}^{2n}}e^{\pm
ix\cdot\xi}x^{\alpha}\xi^{\beta}\,\mathrm{d}{x}\,\mathrm{d}{\xi}=(\pm
i)^{|\alpha|}(2\pi)^{n}\alpha!\delta^{\alpha\beta}.$

###### Proof.

We have

$\displaystyle\int e^{\pm
ix\cdot\xi}x^{\alpha}\xi^{\beta}\,\mathrm{d}{x}\,\mathrm{d}{\xi}=\int(\pm
D_{\xi})^{\alpha}\big{(}e^{\pm
ix\cdot\xi}\big{)}\xi^{\beta}\,\mathrm{d}{x}\,\mathrm{d}{\xi}$
$\displaystyle=$ $\displaystyle\int e^{\pm ix\cdot\xi}(\mp
D_{\xi})^{\alpha}\big{(}\xi^{\beta}\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi}=(\pm
i)^{|\alpha|}\int e^{\pm
ix\cdot\xi}\partial_{\xi}^{\alpha}(\xi^{\beta})\,\mathrm{d}{x}\,\mathrm{d}{\xi}.$

It can be check that
$\partial_{\xi}^{\alpha}(\xi^{\beta})=\beta!/(\beta-\alpha)!\,\xi^{\beta-\alpha}$
when $\alpha\leq\beta$, and $=0$ otherwise. Hence, when $\alpha\leq\beta$ we
have

$\int e^{\pm
ix\cdot\xi}x^{\alpha}\xi^{\beta}\,\mathrm{d}{x}\,\mathrm{d}{\xi}=(\pm
i)^{|\alpha|}\beta!/(\beta-\alpha)!\int e^{\pm
ix\cdot\xi}\xi^{\beta-\alpha}\,\mathrm{d}{x}\,\mathrm{d}{\xi}.$ (3.14)

When $\alpha\neq\beta$, we can continue

$\int e^{\pm
ix\cdot\xi}\xi^{\beta-\alpha}\,\mathrm{d}{x}\,\mathrm{d}{\xi}\simeq\int
D_{x}^{\beta-\alpha}\big{(}e^{\pm
ix\cdot\xi}\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi}=0\simeq\int e^{\pm
ix\cdot\xi}D_{x}^{\beta-\alpha}\big{(}1\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi}=0.$

Therefore, when $\alpha\neq\beta$, we have

$\int_{\mathbb{R}^{2n}}e^{\pm
ix\cdot\xi}x^{\alpha}\xi^{\beta}\,\mathrm{d}{x}\,\mathrm{d}{\xi}=0.$ (3.15)

When $\alpha=\beta$, by Lemma 3.19 we can continue (3.14) as follows,

$\int_{\mathbb{R}^{2n}}e^{\pm
ix\cdot\xi}x^{\alpha}\xi^{\beta}\,\mathrm{d}{x}\,\mathrm{d}{\xi}=(\pm
i)^{|\alpha|}\alpha!\int e^{\pm
ix\cdot\xi}\,\mathrm{d}{x}\,\mathrm{d}{\xi}=(\pm
i)^{|\alpha|}\alpha!(2\pi)^{n}.$

The proof is complete. ∎

######  Remark 3.21.

The space $S_{\rho}^{m}(R^{N})$ is defined as

$S_{\rho}^{m}(R^{N}):=\\{\varphi\in
C^{\infty}(\mathbb{R}^{N})\,;\,|\partial^{\alpha}\varphi(\theta)|\lesssim\langle{\theta}\rangle^{m-\rho|\alpha|}\\}.$

See [alinhac2007pseudo, §I.8.1] for more details. Also, the formula (3.9) in
the definition of the kernel is meaningful now.

We can generalize Lemma 3.19

######  Lemma 3.22.

Assume $m\in\mathbb{R}$ and $f\in S_{\rho}^{m}({\mathbb{R}^{n}})$ (see Remark
3.15) with $\rho+2>1$. The following equality holds in oscillatory sense,

$\int_{\mathbb{R}^{2n}}e^{\pm
ix\cdot\xi}f(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi}=(2\pi)^{n}f(0).$

######  Remark 3.23.

Lemma 3.22 indicates the the “inverse Fourier transform” is indeed the inverse
of “Fourier transform”.

###### Proof.

We only show the case $+ix\cdot\xi$. The condition on $f$ guarantees the
integral is well-defined, see Definition 3.14. By Taylor’s expansion we have

$f(x)=f(0)+\sum_{j=1}^{n}x_{j}g_{j}(x),\quad\text{where}\quad
g_{j}(x):=\int_{0}^{1}\partial_{x_{j}}f(tx),\,\mathrm{d}{t}.$

so

$\displaystyle\int e^{ix\cdot\xi}f(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi}$
$\displaystyle=\int
e^{ix\cdot\xi}[f(0)+\sum_{j=1}^{n}x_{j}g_{j}(x)]\,\mathrm{d}{x}\,\mathrm{d}{\xi}$
$\displaystyle=f(0)\int
e^{ix\cdot\xi}\,\mathrm{d}{x}\,\mathrm{d}{\xi}+\sum_{j=1}^{n}\int
D_{\xi_{j}}(e^{ix\cdot\xi})g_{j}(x)\,\mathrm{d}{x}\,\mathrm{d}{\xi}$
$\displaystyle=(2\pi)^{n}f(0)-\sum_{j=1}^{n}\int
e^{ix\cdot\xi}D_{\xi_{j}}\big{(}g_{j}(x)\big{)}\,\mathrm{d}{x}\,\mathrm{d}{\xi}$
$\displaystyle=(2\pi)^{n}f(0).$

The proof is complete. ∎

### Exercise

######  Exercise 3.1.

Check that the ${}^{t}M$ given in (3.4) satisfies (3.4). Hint: utilize the
second fact about $M$ to facilitate the derivation.

######  Exercise 3.2.

Prove Lemma 3.17.

##  Chapter 4 Stationary phase lemmas

The stationary phase lemmas is a useful tools for computing certain
asymptotics. Some useful references are [dim1999spe, §5], [eskin2011lectures,
§19.3], [horm2003IIV, §7.7], [zw2012semi, §3].

From time to time we will encounter oscillatory integrals of the form

$I(\lambda)=\int e^{i\lambda\varphi(x)}a(x)\,\mathrm{d}{x}$ (4.1)

where $\varphi$ is a phase function of some order and $a$ is a symbol
($|\partial^{\beta}a(\xi)|\lesssim\langle{\xi}\rangle^{m-|\beta|}$). In §3 we
have introduced schemes to make $I(\lambda)$ well-defined. Now we focus on the
asymptotics of $I(\lambda)$ w.r.t. $\lambda\to+\infty$ when $\varphi$
satisfies certain conditions.

When $\varphi$ is linear, i.e. $\varphi(x)=p\cdot x$ for certain fixed vector
$p\neq 0$, there is no critical point of $\varphi$ ($|\nabla\varphi|=|p|\neq
0$). In this case we call $\varphi$ _non-stationary_. For the non-stationary
case, the asymptotics of $I$ is straightforward:

$\displaystyle I(\lambda)$ $\displaystyle=\int e^{i\lambda p\cdot
x}a(x)\,\mathrm{d}{x}=\int\big{(}\frac{p\cdot
D_{x}}{\lambda|p|^{2}}\big{)}^{N}(e^{i\lambda p\cdot x})\,a(x)\,\mathrm{d}{x}$
$\displaystyle=\lambda^{-N}\int e^{i\lambda p\cdot x}\big{(}\frac{-p\cdot
D_{x}}{|p|^{2}}\big{)}^{N}(a(x))\,\mathrm{d}{x}\lesssim|p|^{-N}\lambda^{-N}\int\sum_{|\beta|=N}C_{\beta}\partial^{\beta}a(x)\,\mathrm{d}{x}$
$\displaystyle\lesssim|p|^{-N}\lambda^{-N}\int\langle{x}\rangle^{m-N}\,\mathrm{d}{x}\lesssim\lambda^{-N},$

provided $N$ is large enough. This means that $\int e^{i\lambda p\cdot
x}a(x)\,\mathrm{d}{x}$ is of rapid decay w.r.t. $\lambda$.

The gradient of $\varphi$ has been put in the denominator in the derivation
above, so the method will not be applicable when the phase function contains
critical points. In this case we call the phase _stationary_. In this chapter
we devote ourselves into the stationary case.

###  4.1. A simple case

To help readers understand the method of stationary phase, we start with a
simple case where the phase is stationary. To that end, we need to do some
preparations.

####  4.1.1. Preliminaries

We need the Taylor’s expansion. Suppose $f\in
C^{N+1}({\mathbb{R}^{n}};\mathbb{C})$, then we have that

$\displaystyle f(x)$ $\displaystyle=\sum_{|\delta|\leq
N}\frac{1}{\delta!}\big{(}\partial^{\delta}f\big{)}(x_{0})\cdot(x-x_{0})^{\delta}$
$\displaystyle\quad+(N+1)\sum_{|\delta|=N+1}\frac{(x-x_{0})^{\delta}}{\delta!}\int_{0}^{1}(1-t)^{N}\big{(}\partial^{\delta}f\big{)}(x_{0}+t(x-x_{0}))\,\mathrm{d}{t}.$
(4.2)

The proof of (4.2) can be found in most of the calculus textbook and we omit
it here.

Secondly, for a measurable function $u$ in ${\mathbb{R}^{n}}$, as long as
$\partial^{\alpha}u\in L^{1}({\mathbb{R}^{n}})$ for $|\alpha|\geq n+1$, then
$\hat{u}$ exists and there exists a constant $C$ such that

$\|{\hat{u}}\|_{L^{1}({\mathbb{R}^{n}})}\leq C\sum_{|\alpha|\leq
n+1}\|{\partial^{\alpha}u}\|_{L^{1}({\mathbb{R}^{n}})}.$ (4.3)

###### Proof.

We have

$\displaystyle\int|\hat{u}(\xi)|\,\mathrm{d}{\xi}$
$\displaystyle=\int\langle{\xi}\rangle^{-n-1}|\langle{\xi}\rangle^{n+1}\hat{u}(\xi)|\,\mathrm{d}{\xi}\leq
C\sup_{{\mathbb{R}^{n}}}|\langle{\xi}\rangle^{n+1}\hat{u}(\xi)|$
$\displaystyle\leq C\sum_{|\alpha|\leq
n+1}C_{\alpha}\sup_{{\mathbb{R}^{n}}}|\xi^{\alpha}\hat{u}(\xi)|\leq\sum_{|\alpha|\leq
n+1}C_{\alpha}\sup_{{\mathbb{R}^{n}}}|\mathcal{F}\\{\partial^{\alpha}u\\}(\xi)|$
$\displaystyle\leq\sum_{|\alpha|\leq
n+1}C_{\alpha}\|{\partial^{\alpha}u}\|_{L^{1}({\mathbb{R}^{n}})}.$

We arrive at the conclusion. ∎

We also need the following transformation. For a fixed non-degenerate,
symmetric, real-valued square matrix $Q$, we have

$\mathcal{F}\\{e^{\pm i\,\langle{Q\cdot,\cdot}\rangle/2}\\}(\xi)=\frac{e^{\pm
i\frac{\pi}{4}{\rm sgn}\,Q}}{|\det Q|^{1/2}}e^{\mp
i\langle{Q^{-1}\xi,\xi}\rangle/2}.$ (4.4)

Here the non-degeneracy condition of $Q$ means $\det Q\neq 0$.

###### Proof.

We have

$\displaystyle\mathcal{F}\\{e^{\pm i\,|\cdot|^{2}/2}\\}(\xi)$
$\displaystyle=(2\pi)^{-n/2}\int_{{\mathbb{R}^{n}}}e^{-ix\cdot\xi}e^{\pm
i\,|x|^{2}/2}\,\mathrm{d}{x}=(2\pi)^{-n/2}e^{\mp
i|\xi|^{2}/2}\int_{{\mathbb{R}^{n}}}e^{\pm i|x\mp\xi|^{2}/2}\,\mathrm{d}{x}$
$\displaystyle=\pi^{-n/2}e^{\mp i|\xi|^{2}/2}\int_{{\mathbb{R}^{n}}}e^{\pm
i|x|^{2}}\,\mathrm{d}{x}.$

By standard Cauchy’s integral theorem we can have

$\int_{-\infty}^{+\infty}e^{\pm ix^{2}}\,\mathrm{d}{x}=\sqrt{\pi}e^{\pm
i\pi/4},$

so we can continue,

$\mathcal{F}\\{e^{\pm i\,|\cdot|^{2}/2}\\}(\xi)=\pi^{-n/2}e^{\mp
i|\xi|^{2}/2}(\sqrt{\pi}e^{\pm i\frac{\pi}{4}})^{n}=e^{\pm
i\frac{\pi}{4}n}e^{\mp i|\xi|^{2}/2}.$ (4.5)

We left the computation from (4.5) to (4.4) as an exercise. ∎

####  4.1.2. A simple case

We study the quadratic case in $\mathbb{R}^{1}$.

######  Lemma 4.1.

Assume $a\in C_{c}^{\infty}(\mathbb{R})$ with $a(0)\neq 0$. Fix an arbitrary
integer $N\in\mathbb{N}$. Then for the integral $I(\lambda)$:

$I(\lambda)=\int_{\mathbb{R}}e^{i\lambda x^{2}/2}a(x)\,\mathrm{d}{x},$

there holds

$I(\lambda)=\left(\frac{2\pi}{\lambda}\right)^{1/2}e^{i\frac{\pi}{4}}\sum_{0\leq
j\leq
N}\frac{\lambda^{-j}}{j!}\left(\frac{i}{2}\right)^{j}a^{(2j)}(0)+\mathcal{O}(\lambda^{-\frac{1}{2}-N-1}\sum_{j\leq
2N+4}\sup|a^{(j)}|),$ (4.6)

where $a^{(j)}$ signifies $\frac{\mathrm{d}^{j}a}{\mathrm{d}x^{j}}$.

###### Proof.

By the Plancherel theorem (which claims $(f,g)=(\hat{f},\hat{g})$) we have

$\displaystyle I(\lambda)$ $\displaystyle=\int\overline{e^{-i\lambda
x^{2}/2}}a(x)\,\mathrm{d}{x}=\int\overline{(\lambda)^{-1/2}e^{-\frac{i\pi}{4}}e^{\frac{i}{2\lambda}\xi^{2}}}\cdot\hat{a}(\xi)\,\mathrm{d}{\xi}$
$\displaystyle=\lambda^{-1/2}e^{\frac{i\pi}{4}}\int
e^{-\frac{i\xi^{2}}{2}h}\hat{a}(\xi)\,\mathrm{d}{\xi},\qquad h:=\lambda^{-1}.$

Using (4.2) to expand $I(\lambda)$ w.r.t. $h$ at $h=0$, we obtain

$\displaystyle\int e^{-\frac{i\xi^{2}}{2}h}\hat{a}(\xi)\,\mathrm{d}{\xi}$
$\displaystyle=\sum_{j=0}^{N}\int
h^{j}(-\frac{i\xi^{2}}{2})^{j}/j!\,\hat{a}(\xi)\,\mathrm{d}{\xi}$
$\displaystyle\quad+|\frac{h^{N+1}}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}N!}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\int_{0}^{1}}\int{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}(1-t)^{N}}(-\frac{i\xi^{2}}{2})^{N+1}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}e^{-\frac{i\xi^{2}}{2}th}}\hat{a}(\xi)\,\mathrm{d}{\xi}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\,\mathrm{d}{t}}|$
$\displaystyle=\sum_{j=0}^{N}h^{j}\frac{(\frac{i}{2})^{j}}{j!}\int\mathcal{F}\\{a^{(2j)}\\}(\xi)\,\mathrm{d}{\xi}+\mathcal{O}(h^{N+1}\int|\xi^{2N+2}\hat{a}(\xi)|\,\mathrm{d}{\xi})$
$\displaystyle=\sum_{j=0}^{N}\frac{(2\pi)^{1/2}(\frac{i}{2})^{j}h^{j}}{j!}a^{(2j)}(0)+\mathcal{O}(h^{N+1}\|{\mathcal{F}\\{a^{(2N+2)}\\}}\|_{L^{1}})$
(4.7)
$\displaystyle=\sum_{j=0}^{N}\frac{(2\pi)^{1/2}(\frac{i}{2})^{j}h^{j}}{j!}a^{(2j)}(0)+\mathcal{O}(h^{N+1}\sum_{j\leq
2N+4}\|{a^{(j)}}\|_{L^{1}({\mathbb{R}^{n}})}).$

Note that we used (4.3). Combining the computations and changing $h$ back to
$\lambda^{-1}$, we arrive at

$I(\lambda)=\left(\frac{2\pi}{\lambda}\right)^{1/2}e^{\frac{i\pi}{4}}\sum_{j=0}^{N}\frac{\lambda^{-j}}{j!}\left(\frac{i}{2}\right)^{j}a^{(2j)}(0)+\mathcal{O}(\lambda^{-\frac{1}{2}-N-1}\sum_{j\leq
2N+4}\sup|a^{(j)}|).$

The proof is complete. ∎

From this short proof, we see extract several main steps:

  1. (1)

use Plancherel theorem to turn $\lambda$ into $\lambda^{-1}$ in the exponent;

  2. (2)

expand the integral w.r.t. $h:=\lambda^{-1}$ at $h=0$ with integral remainder;

  3. (3)

estimate the remainder with (4.3).

###  4.2. Lemma Statements

######  Theorem 4.2.

Let $n\in\mathbb{N}^{+}$ be the dimension. We consider the oscillatory
integral $I(\lambda)$:

$I(\lambda)=\int_{{\mathbb{R}^{n}}}e^{i\lambda\langle{Q(x-x_{0}),x-x_{0}}\rangle/2}a(x;\lambda)\,\mathrm{d}{x},$

where $\langle{Qx,y}\rangle$ signifies $(Qx)^{T}y$ as matrix multiplication.
Fix two arbitrary integers $M$, $N\in\mathbb{N}$, and we assume

  * •

$Q$ is a non-degenerate, symmetric, real-valued matrix;

  * •

for each $\lambda$, $a(\cdot;\lambda)\in
C^{n+2N+3}({\mathbb{R}^{n}};\mathbb{C})$;

  * •

for each $\lambda$, $a(\cdot;\lambda)\in C^{M}({\mathbb{R}^{n}};\mathbb{C})$,
and $\forall\alpha:|\alpha|\leq M$, there exists $\lambda$-dependent constants
$C_{M,\alpha}(\lambda)>0$ such that $\forall x\in{\mathbb{R}^{n}}$ there holds

$|\partial_{x}^{\alpha}a(x-x_{0};\lambda)|<C_{M,\alpha}(\lambda)\langle{x}\rangle^{2M-n-1-|\alpha|}.$

Then the integral $I(\lambda)$ is well-defined in the oscillatory integral
sense, and as $\lambda\to+\infty$ we have

$\displaystyle I(\lambda)$
$\displaystyle=\left(\frac{2\pi}{\lambda}\right)^{n/2}\frac{e^{i\frac{\pi}{4}\mathop{\rm
sgn}Q}}{|\det Q|^{1/2}}\sum_{0\leq j\leq
N}\frac{\lambda^{-j}}{j!}\left(\frac{\langle{Q^{-1}D,D}\rangle}{2i}\right)^{j}\big{(}a(x;\lambda)\big{)}|_{x=x_{0}}$
$\displaystyle\qquad+\mathcal{O}\big{(}\lambda^{-\frac{n}{2}-N-1}\times\sum_{|\alpha|\leq
n+2N+3}\sup_{B(x_{0},1)}|\partial^{\alpha}a(\cdot;\lambda)|\big{)}$ (4.8)
$\displaystyle\qquad+\mathcal{O}\big{(}\lambda^{-M}\times\sum_{|\alpha|\leq
M}\sup_{{\mathbb{R}^{n}}}\frac{|\partial^{\alpha}a(x-x_{0};\lambda)|}{\langle{x}\rangle^{2M-n-1-|\alpha|}}\big{)},$

where $B(x_{0},1)$ stands for the ball centered at $x_{0}$ with radius 1, and
$\mathop{\rm sgn}Q$ stands for the difference between the number of positive
eigenvalues and the number of negative eigenvalues of the matrix $Q$.

######  Remark 4.3.

In contrast to many other versions of the stationary phase lemma, here we
don’t require $a$ to be compactly supported. Instead, some other boundedness
conditions are required, which makes the oscillatory integral well-defined.

######  Remark 4.4.

Eq. (4.8) is not an asymptotics w.r.t. $j$, but is rather w.r.t. $\lambda$. To
get enough terms w.r.t. $j$, one could choose $M$ be large enough first, and
then check if $a$ satisfies the requirements.

######  Remark 4.5.

The integers $M$ and $N$ in Theorem 4.2 shall be chosen properly to serve for
your own purposes. For example, if one cares more about the decaying behavior
w.r.t. $\lambda$, then the $M$ can be set to $\lceil n/2\rceil+N+1$. However,
if one is dealing with these functions $a$ which doesn’t have good decaying
behavior at the infinity, then one could set $M$ to be large enough such that
$\langle{x}\rangle^{2M-n-1-|\alpha|}$ can dominate $\partial^{\alpha}a$, with
the cost that we should demand more smoothness of $a(x)$.

######  Remark 4.6.

The unit ball $B(x_{0},1)$ involved in the term
$\sup_{B(0,1)}|\partial^{\alpha}a|$ can be changed to other bounded domain
containing $x_{0}$. But one should be careful that when the domain used has a
very small radius, the underlying coefficients of the $\mathcal{O}(\cdot)$
term will be very large accordingly.

######  Remark 4.7.

The function $a$ is allowed to be dependent on $\lambda$, hence the expression
(4.8) is an asymptotic expansion only when $\partial_{x}^{\alpha}a(x;\lambda)$
doesn’t increase significantly when $\lambda\to+\infty$.

If chosen $M=n+2N+3$, Theorem 4.2 will be simplified as follows.

######  Proposition 4.8.

Let $n\in\mathbb{N}^{+}$ be the dimension. We consider the oscillatory
integral $I(\lambda)$:

$I(\lambda)=\int_{{\mathbb{R}^{n}}}e^{i\lambda\langle{Q(x-x_{0}),x-x_{0}}\rangle/2}a(x;\lambda)\,\mathrm{d}{x},$

where $\langle{Qx,y}\rangle$ signifies $(Qx)^{T}y$ as matrix multiplication.
Fix an integer $N\in\mathbb{N}$, and we assume

  * •

$Q$ is a non-degenerate, symmetric, real-valued matrix;

  * •

for each $\lambda$, $a(\cdot;\lambda)\in
C^{n+2N+3}({\mathbb{R}^{n}};\mathbb{C})$, and $\forall\alpha:|\alpha|\leq
n+2N+3$, there exists $\lambda$-dependent constants
$C_{N,n,\alpha}(\lambda)>0$ such that $\forall x\in{\mathbb{R}^{n}}$ there
holds

$|\partial_{x}^{\alpha}a(x-x_{0};\lambda)|<C_{N,n,\alpha}(\lambda)\langle{x}\rangle^{2N+2}.$
(4.9)

Then the integral $I(\lambda)$ is well-defined in the oscillatory integral
sense, and as $\lambda\to+\infty$ we have

$\displaystyle I(\lambda)$
$\displaystyle=\left(\frac{2\pi}{\lambda}\right)^{n/2}\frac{e^{i\frac{\pi}{4}\mathop{\rm
sgn}Q}}{|\det Q|^{1/2}}\sum_{0\leq j\leq
N}\frac{\lambda^{-j}}{j!}\left(\frac{\langle{Q^{-1}D,D}\rangle}{2i}\right)^{j}\big{(}a(x;\lambda)\big{)}|_{x=x_{0}}$
$\displaystyle\qquad+\mathcal{O}\big{(}\lambda^{-\frac{n}{2}-N-1}\sum_{|\alpha|\leq
n+2N+3}\sup_{x\in{\mathbb{R}^{n}}}\frac{|\partial^{\alpha}a(x-x_{0};\lambda)|}{\langle{x}\rangle^{n+4N+5-|\alpha|}}\big{)}.$
(4.10)

Proposition 4.8 can be extend to a more general case where the phase function
is not quadratic.

######  Theorem 4.9 (Stationary phase lemma [eskin2011lectures]).

We consider the oscillatory integral $I(\lambda)$:

$I(\lambda)=\int_{{\mathbb{R}^{n}}}e^{i\lambda\varphi(x)}a(x)\,\mathrm{d}{x}.$

For an arbitrary integer $N\in\mathbb{N}$, assume

  * •

$a\in C^{n+2N+3}({\mathbb{R}^{n}};\mathbb{C})$ with $\sum_{|\alpha|\leq
n+2N+3}\sup_{{\mathbb{R}^{n}}}|\partial^{\alpha}a|<+\infty$;

  * •

$\varphi\in C^{n+2N+6}({\mathbb{R}^{n}};\mathbb{R})$ with $\sum_{|\alpha|\leq
n+2N+6}\sup_{{\mathbb{R}^{n}}}|\partial^{\alpha}\varphi|<+\infty$;

  * •

$x_{0}$ is the only critical point of $\varphi(x)$ on $\mathop{\rm supp}a(x)$,
i.e., $\varphi(x_{0})=\nabla\varphi(x_{0})=0$, $\varphi_{x}(x)\neq 0$ for
$x\neq x_{0}$;

  * •

the Hessian $\varphi_{xx}(x_{0}):=[\frac{\partial^{2}\varphi}{\partial
x_{j}\partial x_{k}}(x_{0})]_{j,k=1}^{n}$ satisfies
$\det\varphi_{xx}(x_{0})\neq 0$.

Then the integral $I(\lambda)$ is well-defined in the oscillatory integral
sense, and as $\lambda\to+\infty$ we have

$\displaystyle I(\lambda)$
$\displaystyle=\left(\frac{2\pi}{\lambda}\right)^{n/2}\frac{e^{i\lambda\varphi(x_{0})+i\frac{\pi}{4}\mathop{\rm
sgn}\varphi_{xx}(x_{0})}}{|\det\varphi_{xx}(x_{0})|^{1/2}}\Big{(}a(x_{0})+\sum_{j=1}^{N}a_{j}(x_{0})\lambda^{-j}\Big{)}$
$\displaystyle\quad+\mathcal{O}\big{(}\lambda^{-\frac{n}{2}-N-1}\times\sum_{|\alpha|\leq
n+2N+3}\sup_{{\mathbb{R}^{n}}}|\partial^{\alpha}a|\times\sum_{|\alpha|\leq
n+2N+6}\sup_{{\mathbb{R}^{n}}}|\partial^{\alpha}\varphi|\big{)},$ (4.11)

for some functions $a_{j},~{}1\leq j\leq N$.

######  Remark 4.10.

Proposition 4.8 is a special case of Theorem 4.9 where

$\varphi(x)=\langle{Q(x-x_{0}),x-x_{0}}\rangle/2,$

which guarantees $\varphi_{xx}(x_{0})=Q$. Theorem 4.9 is not a generalization
of Theorem 4.2 because unlike the quadratic phase function in Theorem 4.2, the
phase function $\varphi$ in Theorem 4.9 is not assumed to possess the property
that $|\nabla\varphi(x)|\simeq|x|$ as $|x|$ is large. However if $\varphi$ is
a homeomorphism of ${\mathbb{R}^{n}}$, it is possible to generalize Theorem
4.2.

In one-dimensional case, explicit expressions for these $a_{j}(x_{0})$ are
given in [zw2012semi, (3.4.11)], and the details are given in [zw2012semi,
Second proof of Theorem 3.11]. For explicit expressions for these
$a_{j}(x_{0})$ in higher dimension, readers may refer to [horm2003IIV, Theorem
7.7.5] for details. In [eskin2011lectures, Lemma 19.3] there is also another
routine to prove the stationary phase lemma. [dim1999spe, Chapter 5] by
Dimassi and Sjöstrand is also a good reference. See also [won89asy, §2.3 &
§6.4].

###  4.3. Proofs of the results

We first prove the quadratic case.

###### Proof of Theorem 4.2.

We omit notationally the dependence of $a$ on $\lambda$ until related
clarifications are needed. Without loss of generality we assume $x_{0}=0$, and
$a(0)=1$. For readers’ convenient we rewrite the expression of $I$ here:
$I(\lambda)=\int_{{\mathbb{R}^{n}}}e^{i\lambda\langle{Qx,x}\rangle/2}a(x)\,\mathrm{d}{x}$.

Step 1: cutoff singularity of the phase function. According to the assumption
on $Q$, we know there exists a decomposition $Q=P\Lambda P^{T}$ where $P$ is
an orthogonal matrix and $\Lambda:=(\alpha_{j})_{j=1,\cdots,n}$ is a diagonal
matrix. Make the change of variable $y=P^{T}x$, we can have

$\displaystyle I$
$\displaystyle=\int_{\mathbb{R}^{n}}a(Py)e^{i\lambda\sum_{j=1}^{n}\alpha_{j}y_{j}^{2}/2}\,\mathrm{d}{y}$
$\displaystyle=\int_{\mathbb{R}^{n}}e^{i\lambda\sum_{j=1}^{n}\alpha_{j}y_{j}^{2}/2}(1-\chi(y))f(y)\,\mathrm{d}{y}+\int_{\mathbb{R}^{n}}e^{i\lambda\sum_{j=1}^{n}\alpha_{j}y_{j}^{2}/2}\chi(y)f(y)\,\mathrm{d}{y}\quad(\det
P=1)$ $\displaystyle=:J_{1}+J_{2},$ (4.12)

where $f(y):=a(Py)$ and $\chi\in C_{c}^{\infty}({\mathbb{R}^{n}})$ is a cutoff
function satisfying $0\leq\chi\leq 1$ and $\chi\equiv 1$ in a neighborhood of
the origin. We will see:
$J_{1}$ is rapidly decaying and $J_{2}$ gives the desired asymptotics.

Step 2: $J_{1}$ is rapidly decaying. Noting that neighborhoods of the origin
is not included in the support of the integrand in $J_{1}$, we can estimate
$J_{1}$ by using integration by parts (in the oscillatory integral sense). For
any integer $M\in\mathbb{N}$ we have

$\displaystyle|J_{1}|$
$\displaystyle=|\int_{\mathbb{R}^{n}}\big{(}\frac{\sum_{j}\alpha_{j}^{-1}y_{j}\partial_{j}}{i\lambda|y|^{2}}\big{)}^{M}(e^{i\lambda\sum_{j=1}^{n}\alpha_{j}y_{j}^{2}/2})\big{[}(1-\chi(y))f(y)\big{]}\,\mathrm{d}{y}|$
$\displaystyle\lesssim\lambda^{-M}\int_{\mathbb{R}^{n}}|\big{(}\sum_{j}\partial_{j}\circ(y_{j}|y|^{-2})\big{)}^{M}\big{[}(1-\chi(y))f(y)\big{]}|\,\mathrm{d}{y}$
$\displaystyle\lesssim\lambda^{-M}\int_{\mathop{\rm
supp}(1-\chi)}\sum_{|\alpha|\leq
M}C_{M;\alpha}|y|^{|\alpha|-2M}|\partial^{\alpha}((1-\chi)f(y))|\,\mathrm{d}{y}$
(4.13) $\displaystyle\lesssim\lambda^{-M}\big{(}C\sum_{|\alpha|\leq
M}\sup_{\\{0<\chi<1\\}}|\partial^{\alpha}f|+\sum_{|\alpha|\leq
M}C_{M;\alpha}\int_{\\{\chi=0\\}}|y|^{|\alpha|-2M}|\partial^{\alpha}f(y)|\,\mathrm{d}{y}\big{)}$
(4.14) $\displaystyle\lesssim\lambda^{-M}\sum_{|\alpha|\leq
M}\big{(}\sup_{\\{0<\chi<1\\}}|\partial^{\alpha}a|+\int_{\\{\chi=0\\}}|y|^{-n-1}\langle{y}\rangle^{|\alpha|-2M+n+1}|\partial^{\alpha}a(y)|\,\mathrm{d}{y}\big{)}$
$\displaystyle\lesssim\lambda^{-M}\sum_{|\alpha|\leq
M}\big{(}\sup_{\\{0<\chi<1\\}}|\partial^{\alpha}a|+\sup_{{\mathbb{R}^{n}}}\frac{|\partial^{\alpha}a(y)|}{\langle{y}\rangle^{2M-n-1-|\alpha|}}\big{)}$
$\displaystyle\lesssim\lambda^{-M}\sum_{|\alpha|\leq
M}\sup_{{\mathbb{R}^{n}}}\frac{|\partial^{\alpha}a(y)|}{\langle{y}\rangle^{2M-n-1-|\alpha|}}.$
(4.15)

The inequality (4.13) is due to the fact that

$\big{(}\sum_{1\leq j\leq
n}\partial_{j}\circ(y_{j}|y|^{-2})\big{)}^{M}\varphi=\sum_{|\alpha|\leq
M}C_{M;\alpha}|y|^{|\alpha|-2M}\partial^{\alpha}\varphi$

which can be derived by induction and we omit the details. Inequality (4.14)
is due to the fact that $\partial^{\alpha}((1-\chi)f)=\partial^{\alpha}f$ in
$\\{\chi=0\\}$.

Step 3: $J_{2}$ and Plancherel theorem. We turn to $J_{2}$. Keep in mind that
$f(y)=a(Py)$ and $\chi f$ is compactly support and in
$C_{c}^{n+2N+3}({\mathbb{R}^{n}})$. Here we analyze $I$ by borrowing idea from
[zw2012semi, First proof of Theorem 3.11]. First we use Plancherel theorem
(which states $(f,g)=(\hat{f},\hat{g})$),

$\displaystyle J_{2}$
$\displaystyle=\int_{\mathbb{R}^{n}}\overline{e^{-i\lambda\sum_{j=1}^{n}\alpha_{j}y_{j}^{2}/2}}\chi
f(y)\,\mathrm{d}{y}=\int_{\mathbb{R}^{n}}\overline{\mathcal{F}\\{e^{-i\lambda\sum_{j=1}^{n}\alpha_{j}(\cdot)^{2}/2}\\}(\xi)}\cdot\widehat{\chi
f}(\xi)\,\mathrm{d}{\xi}$
$\displaystyle=\int_{\mathbb{R}^{n}}\overline{(\lambda)^{-n/2}\frac{e^{-\frac{i\pi}{4}{\rm
sgn}\,Q}}{|\det
Q|^{1/2}}e^{\frac{i}{2\lambda}\alpha_{j}^{-1}\xi_{j}^{2}}}\cdot\widehat{\chi
f}(\xi)\,\mathrm{d}{\xi}\qquad(\text{by~{}}\eqref{eq:FrGauiQ-PM2021})$
$\displaystyle=:\big{(}\frac{2\pi}{\lambda}\big{)}^{n/2}\frac{e^{\frac{i\pi}{4}{\rm
sgn}\,Q}}{|\det Q|^{1/2}}J(1/\lambda,1/\lambda,\chi f),$ (4.16)

where we ignored the summation notation over $j$ and the function $J$ is
defined by

$J(h_{1},h_{2},\chi
f):=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}e^{\frac{\xi_{j}^{2}h_{1}}{i2\alpha_{j}}}\cdot\widehat{\chi
f}(\xi;1/h_{2})\,\mathrm{d}{\xi}.$ (4.17)

Note that in (4.17) we put emphasize on the dependence of $f$ on $h_{2}$ (i.e.
dependence of $a$ on $\lambda$). The smoothness of $J$ w.r.t. $h_{1}$ is
guaranteed by the $L^{1}$ of derivatives of $f$, namely, we have the following
claim whose justification will be clear in (4.20),

$\forall m\in\mathbb{N},\,\max_{|\alpha|\leq
n+2m+1}\|{\partial^{\alpha}f}\|_{L^{1}({\mathbb{R}^{n}})}<+\infty\
\Rightarrow\ J(\cdot,h_{2},f)\in C^{m}(\mathbb{R}).$

Step 4: Taylor’s expansion. We abbreviate $\partial_{h_{1}}J$ as
$\partial_{1}J$. Expand $J$ w.r.t. $h_{1}$,

$\displaystyle\partial_{1}^{k}J(0,h_{2},\chi f)$
$\displaystyle=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}\partial_{1}^{k}(e^{\frac{\xi_{j}^{2}h_{1}}{i2\alpha_{j}}})|_{h_{1}=0}\cdot\widehat{\chi
f}(\xi)\,\mathrm{d}{\xi}$
$\displaystyle=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}(\sum_{j}\frac{\xi_{j}^{2}}{i2\alpha_{j}})^{k}\cdot\widehat{\chi
f}(\xi)\,\mathrm{d}{\xi}$
$\displaystyle=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}\widehat{T^{k}\chi
f}(\xi)\,\mathrm{d}{\xi}=T^{k}f(0)=T^{k}\big{(}a(Py)\big{)}|_{y=0}$
$\displaystyle=(\frac{i}{2}P^{lj}P^{kj}\alpha_{j}^{-1}\partial_{kl})^{k}a(0)=A^{k}a(0;1/h_{2}),$

where $T=\frac{i}{2}\sum_{j}\frac{\partial_{j}^{2}}{\alpha_{j}}$ and
$A=\frac{i}{2}(Q^{-1})^{jl}\partial_{jl}=\frac{1}{2i}\langle{Q^{-1}D,D}\rangle$
(recall that $D=\frac{1}{i}\nabla$ is vertical). We expand $J$ using Taylor
series (i.e. (4.2)),

$\displaystyle J(h,h_{2},\chi f)$ $\displaystyle=\sum_{k\leq
N}\frac{h^{k}}{k!}\partial_{h}^{k}J(0,h_{2},\chi
f)+\frac{h^{N+1}}{N!}\int_{0}^{1}(1-t)^{N}\cdot\partial_{1}^{N+1}J(th,h_{2},\chi
f)\,\mathrm{d}{t}$ $\displaystyle=\sum_{0\leq k\leq
N}\frac{(hA)^{k}}{k!}a(0;h_{2})+\frac{h^{N+1}}{N!}\int_{0}^{1}(1-t)^{N}\cdot\partial_{1}^{N+1}J(th,h_{2},\chi
f)\,\mathrm{d}{t}.$ (4.18)

Step 5: The remainder term. By invoking (4.17), the remainder term in (4.18)
can be estimated as

$\displaystyle\
|\frac{h^{N+1}}{N!}\int_{0}^{1}(1-t)^{N}\cdot\partial_{1}^{N+1}J(th,h_{2},\chi
f)\,\mathrm{d}{t}|$ $\displaystyle\leq$ $\displaystyle
C_{N}h^{N+1}\int_{\mathbb{R}^{n}}|(\frac{-i}{4\alpha_{j}}\xi_{j}^{2})^{N+1}\cdot\widehat{\chi
f}(\xi;1/h_{2})|\,\mathrm{d}{\xi}$ $\displaystyle\leq$ $\displaystyle\
C_{N}h^{N+1}\sum_{|\beta|\leq 2N+2}C_{\beta}\|{(\partial_{x}^{\beta}(\chi
f(\cdot;1/h_{2})))^{\wedge}}\|_{L^{1}({\mathbb{R}^{n}})}.$

By using (4.17) and (4.3), we can continue

$\displaystyle\
|\frac{h^{N+1}}{N!}\int_{0}^{1}(1-t)^{N}\cdot\partial_{1}^{N+1}J(th,h_{2},\chi
f)\,\mathrm{d}{t}|$ $\displaystyle\leq$ $\displaystyle\
C_{N}h^{N+1}\sum_{\begin{subarray}{c}|\alpha|\leq n+1\\\ |\beta|\leq
2N+2\end{subarray}}\|{\partial_{x}^{\alpha+\beta}(\chi
f(\cdot;1/h_{2}))}\|_{L^{1}({\mathbb{R}^{n}})}$ (4.19) $\displaystyle\leq$
$\displaystyle\ C_{N}h^{N+1}\sum_{|\alpha|\leq
n+2N+3}\|{\partial_{x}^{\alpha}(\chi
f(\cdot;1/h_{2}))}\|_{L^{1}({\mathbb{R}^{n}})}$ $\displaystyle\leq$
$\displaystyle\ C_{N}h^{N+1}\sum_{|\alpha|\leq n+2N+3}\sup_{\mathop{\rm
supp}\chi}|\partial^{\alpha}a(\cdot;1/h_{2})|.$ (4.20)

Letting $h=h_{2}=1/\lambda$ and combining (4.16), (4.17), (4.18) and (4.20),
we obtain

$\displaystyle J_{2}$
$\displaystyle=\left(\frac{2\pi}{\lambda}\right)^{n/2}\frac{e^{\frac{i\pi}{4}{\rm
sgn}\,Q}}{|\det Q|^{1/2}}\sum_{j\leq
N}\frac{\lambda^{-j}}{j!}\left(\frac{\langle{Q^{-1}D,D}\rangle}{2i}\right)^{j}a(0;\lambda)$
$\displaystyle\quad+C_{N}\lambda^{-\frac{n}{2}-N-1}\sum_{|\alpha|\leq
n+2N+3}\sup_{\mathop{\rm supp}\chi}|\partial^{\alpha}a(\cdot;\lambda)|.$
(4.21)

Combining (4.21) with (4.12), (4.15), we have

$\displaystyle I$
$\displaystyle=\left(\frac{2\pi}{\lambda}\right)^{n/2}\frac{e^{\frac{i\pi}{4}{\rm
sgn}\,Q}}{|\det Q|^{1/2}}\sum_{j\leq
N}\frac{\lambda^{-j}}{j!}\left(\frac{\langle{Q^{-1}D,D}\rangle}{2i}\right)^{j}a(0;\lambda)$
$\displaystyle\qquad+\mathcal{O}\big{(}\lambda^{-\frac{n}{2}-N-1}\sum_{|\alpha|\leq
n+2N+3}\sup_{\mathop{\rm
supp}\chi}|\partial^{\alpha}a(\cdot;\lambda)|\big{)}+\mathcal{O}\big{(}\lambda^{-M}\sum_{|\alpha|\leq
M}\sup_{{\mathbb{R}^{n}}}\frac{|\partial^{\alpha}a(y;\lambda)|}{\langle{y}\rangle^{2M-n-1-|\alpha|}}\big{)},$

which is (4.8). The proof is complete. ∎

###### Proof of Proposition 4.8.

The statement of Proposition 4.8 is almost the same as Theorem (4.2), except
that $M$ is set to be $n+2N+3$. Hence, we set $M:=n+2N+3$, then we have
$-M\leq-n/2-N-1$, so

$\displaystyle\lambda^{-M}\sum_{|\alpha|\leq
M}\sup_{{\mathbb{R}^{n}}}\frac{|\partial^{\alpha}a(y;\lambda)|}{\langle{y}\rangle^{2M-n-1-|\alpha|}}$
$\displaystyle=\lambda^{-M}\sum_{|\alpha|\leq
n+2N+3}\sup_{{\mathbb{R}^{n}}}\frac{|\partial^{\alpha}a(y;\lambda)|}{\langle{y}\rangle^{2(n+2N+3)-n-1-|\alpha|}}$
$\displaystyle\leq\lambda^{-n/2-N-1}\sum_{|\alpha|\leq
n+2N+3}\sup_{{\mathbb{R}^{n}}}\frac{|\partial^{\alpha}a(y;\lambda)|}{\langle{y}\rangle^{n+4N+5-|\alpha|}}.$

Also, for the first remainder term in (4.8) we have

$\displaystyle\sum_{|\alpha|\leq
n+2N+3}\sup_{B(x_{0},1)}|\partial^{\alpha}a(\cdot;\lambda)|$
$\displaystyle\leq C_{x_{0},n,N}\sum_{|\alpha|\leq n+2N+3}\sup_{y\in
B(x_{0},1)}\frac{|\partial^{\alpha}a(y;\lambda)|}{\langle{y}\rangle^{n+4N+5-|\alpha|}}$
$\displaystyle\leq C_{x_{0},n,N}\sum_{|\alpha|\leq
n+2N+3}\sup_{{\mathbb{R}^{n}}}\frac{|\partial^{\alpha}a(y;\lambda)|}{\langle{y}\rangle^{n+4N+5-|\alpha|}}.$

Combining these with (4.8), we arrive at (4.10). The proof is complete. ∎

Based on Theorem 4.2, now we prove the more general case.

###### Proof of Theorem 4.9.

Without loss of generality we assume $x_{0}=0$, $\varphi(0)=0$ and $a(0)=1$.
Hence by Taylor’s expansion (4.2) we have

$\varphi(x)=\sum_{j,k\leq
n}x_{j}x_{k}\int_{0}^{1}(1-t)\,\partial_{jk}\varphi(tx)\,\mathrm{d}{t}=x^{T}\cdot\int_{0}^{1}(1-t)\varphi_{xx}(tx)\,\mathrm{d}{t}\cdot
x.$

Note that $|\varphi_{xx}(0)|\neq 0$ and $|\varphi_{xx}(x)|$ is continuous on
$x$ ($\varphi\in C^{2}$), thus there exists a positive constant $None$ such
that $|\varphi_{xx}(x)|>|\varphi_{xx}(0)|/2>0$ for all $x\in B(0,r)$. Fix a
cutoff function $\chi\in C_{c}^{\infty}({\mathbb{R}^{n}})$ such that
$\mathop{\rm supp}\chi\subset B(0,r)$ and $\chi\equiv 1$ in $B(0,r/2)$. Hence:

  * •

on $B(0,r)$, matrix $\varphi_{xx}$ is non-degenerate;

  * •

on $\mathop{\rm supp}a\backslash B(0,r)$, $|\nabla\varphi(x)|$ is uniformly
bounded away from $0$.

Step 1: cutoff singularity of the phase function. We divide $I$ into two parts

$I(\lambda)=\int_{{\mathbb{R}^{n}}}(1-\chi(x))a(x)e^{i\lambda\varphi(x)}\,\mathrm{d}{x}+\int_{{\mathbb{R}^{n}}}\chi(x)a(x)e^{i\lambda\varphi(x)}\,\mathrm{d}{x}:=I_{1}+I_{2},$
(4.22)

and we will show that $I_{1}$ is rapidly decreasing w.r.t. $\lambda$ while
$I_{2}$ can be analyzed by using Theorem 4.2.

Step 2: $I_{1}$ is rapidly decaying. For $I_{1}$, denote
$L=\sum_{j=1}^{n}\frac{\varphi_{x_{j}}}{|\nabla\varphi|^{2}}\partial x_{j},$
where $\varphi_{x_{j}}$ is short for $\partial_{x_{j}}\varphi$. Then
$\frac{1}{i\lambda}Le^{i\lambda\varphi}=e^{i\lambda\varphi}\text{ and
}{}^{t}Lf=\sum_{j=1}^{n}\partial_{x_{j}}\big{(}\frac{\varphi_{x_{j}}f}{|\nabla\varphi|^{2}}\big{)}$.
For any $N\in\mathbb{N}^{+}$, $I_{1}$ can be easily estimated as follows
(which requires $a\in C^{n+N+1}({\mathbb{R}^{n}})$ and $\varphi\in
C^{n+N+2}({\mathbb{R}^{n}})$)

$\displaystyle I_{1}$
$\displaystyle=\int_{\mathbb{R}^{n}}(1-\chi)a\cdot((i\lambda)^{-n-N-1}L^{N+1}e^{i\lambda\varphi(x)})\,\mathrm{d}{x}$
$\displaystyle=(i\lambda)^{-n-N-1}\int_{\mathbb{R}^{n}}({}^{t}L)^{n+N+1}((1-\chi)a)\cdot
e^{i\lambda\varphi(x)}\,\mathrm{d}{x}$
$\displaystyle=\mathcal{O}(\lambda^{-n-N-1}\sum_{|\alpha|\leq
n+N+1}\|{\partial^{\alpha}a}\|_{L^{1}({\mathbb{R}^{n}})}),\quad\lambda\to\infty.$
(4.23)

As mentioned before, due to the presence of $1-\chi$, the denominator
$|\nabla\varphi|^{2}$ in $L$ keeps a positive distance away from 0,
guaranteeing that $({}^{t}L)^{N}((1-\chi)a)$ is bounded and compactly
supported.

Step 3: Turn $I_{2}$ into quadratic phase form (e.g. “$J_{2}$”). Now we turn
to $I_{2}$. Because $\varphi\in C^{2}({\mathbb{R}^{n}})$, $\varphi_{xx}(x)$ is
symmetric and thus there exist orthogonal matrix $P(x)$ and diagonal matrix
$\Lambda(x)=(\alpha_{j}(x))_{j=1,\cdots,n}$ such that

$2\int_{0}^{1}(1-t)\varphi_{xx}(tx)\,\mathrm{d}{t}=P(x)\Lambda(x)P^{T}(x).$

Especially we have $P(0)\Lambda(0)P^{T}(0)=\varphi_{xx}(0).$ Denote
$\alpha_{j}=\alpha_{j}(0)$ and $n\times n$ diagonal matrix
$\Lambda:=(\alpha_{j})_{j=1,\cdots,n}$ for short. Thus

$\Lambda(x)=(\sqrt{\frac{\alpha_{j}(x)}{\alpha_{j}}})_{j=1,\cdots,n}\cdot\Lambda\cdot(\sqrt{\frac{\alpha_{j}(x)}{\alpha_{j}}})_{j=1,\cdots,n}.$

Note that we can choose the support of $\chi$ to be small enough such that, on
$\mathop{\rm supp}\chi$, $\alpha_{j}(x)$ doesn’t change sign, so
$\alpha_{j}(x)/\alpha_{j}$ will always be positive on $\mathop{\rm supp}\chi$.
This grants the use of the square root operation.

Make the change of variable:

$y=\Phi(x):=\left(\sqrt{\frac{\alpha_{j}(x)}{\alpha_{j}}}\right)_{j=1,\cdots,n}\cdot
P^{T}(x)\cdot x.$ (4.24)

Note that

$\varphi\in C^{n+2N+6}\Rightarrow\Phi\in C^{n+2N+4}.$ (4.25)

We have

$\displaystyle\varphi(x)$
$\displaystyle=\frac{1}{2}x^{T}\cdot\big{[}2\int_{0}^{1}(1-t)\varphi_{xx}(tx)\,\mathrm{d}{t}\big{]}\cdot
x=\frac{1}{2}x^{T}\cdot\big{[}P(x)\Lambda(x)P^{T}(x)\big{]}\cdot x$
$\displaystyle=\frac{1}{2}[P^{T}(x)\cdot
x]^{T}\cdot(\sqrt{\frac{\alpha_{j}(x)}{\alpha_{j}}})_{j=1,\cdots,n}\cdot\Lambda\cdot(\sqrt{\frac{\alpha_{j}(x)}{\alpha_{j}}})_{j=1,\cdots,n}\cdot[P^{T}(x)\cdot
x]$
$\displaystyle=\frac{1}{2}\big{[}\big{(}\sqrt{\frac{\alpha_{j}(x)}{\alpha_{j}}}\big{)}_{j=1,\cdots,n}\cdot
P^{T}(x)\cdot
x\big{]}^{T}\cdot\Lambda\cdot\big{[}\big{(}\sqrt{\frac{\alpha_{j}(x)}{\alpha_{j}}}\big{)}_{j=1,\cdots,n}\cdot
P^{T}(x)\cdot x\big{]}$ $\displaystyle=\frac{1}{2}\langle{\Lambda
y,y}\rangle.$

We have $\Phi(0)=0$. It is easy to check that $\frac{\partial\Phi}{\partial
x}(0)=P^{T}(0).$ From (4.24) it is clear that there exists inverse of $\Phi$,
i.e. $\phi=\Phi^{-1}$ Note that $x=\phi(\Phi(x))$ and

$\Phi\in C^{n+2N+4}\Rightarrow\phi\in C^{n+2N+4}.$ (4.26)

We have

$\displaystyle I_{2}$
$\displaystyle=\int_{\mathbb{R}^{n}}\chi(\phi(y))a(\phi(y))\cdot
e^{i\lambda\langle{\Lambda y,y}\rangle/2}\,\mathrm{d}{\phi(y)}$
$\displaystyle=\int_{\mathbb{R}^{n}}\chi(\phi(y))a(\phi(y))|\det\nabla_{y}\phi(y)|\cdot
e^{i\lambda\langle{\Lambda y,y}\rangle/2}\,\mathrm{d}{y}$
$\displaystyle=\int_{\mathbb{R}^{n}}f(y)e^{i\lambda\langle{\Lambda
y,y}\rangle/2}\,\mathrm{d}{y}$

where

$f(y)=\chi(\phi(y))\cdot a(\phi(y))\cdot|\det\nabla_{y}\phi(y)|,$

Note that

$\phi\in C^{n+2N+4},\,a\in C^{n+2N+2}\Rightarrow f\in C^{n+2N+2}.$ (4.27)

Now we can conclude from (4.27), (4.27) and (4.27) that

$\varphi\in C^{n+2N+6},\,a\in C^{n+2N+3}\Rightarrow f\in C^{n+2N+3}.$ (4.28)

Step 4: Apply Theorem 4.2. By using Theorem 4.2, we can obtain

$\displaystyle I_{1}(\lambda)$
$\displaystyle=\left(\frac{2\pi}{\lambda}\right)^{n/2}\frac{e^{i\frac{\pi}{4}\mathop{\rm
sgn}\Lambda}}{|\det\Lambda|^{1/2}}\sum_{0\leq j\leq
N}\frac{\lambda^{-j}}{j!}\left(\frac{\langle{\Lambda^{-1}D,D}\rangle}{2i}\right)^{j}f(0)$
$\displaystyle\qquad+\mathcal{O}(\lambda^{-\frac{n}{2}-N-1}\times\sum_{|\alpha|\leq
n+2N+3}\sup_{{\mathbb{R}^{n}}}|\partial^{\alpha}f|)$
$\displaystyle=\left(\frac{2\pi}{\lambda}\right)^{n/2}\frac{e^{i\frac{\pi}{4}\mathop{\rm
sgn}\Lambda}}{|\det\Lambda|^{1/2}}\sum_{0\leq j\leq
N}\frac{\lambda^{-j}}{j!}\left(\frac{\langle{\Lambda^{-1}D,D}\rangle}{2i}\right)^{j}f(0)$
$\displaystyle\qquad+\mathcal{O}(\lambda^{-\frac{n}{2}-N-1}\times\sum_{|\alpha|\leq
n+2N+3}\sup_{{\mathbb{R}^{n}}}|\partial^{\alpha}a|\times\sum_{|\alpha|\leq
n+2N+6}\sup_{{\mathbb{R}^{n}}}|\partial^{\alpha}\varphi|).$ (4.29)

It can be checked that $\mathop{\rm sgn}\Lambda=\mathop{\rm
sgn}\varphi_{xx}(0)$ and $\det\Lambda=\det\varphi_{xx}(0)$.

Step 5: The leading term. We are now almost arrive at (4.11) except for the
explicit computation of the leading term in (4.11) and (4.29). From equality
$x=\phi(\Phi(x))$ we know $I=\nabla_{y}\phi(\Phi(x))\cdot\nabla_{x}\Phi(x)$.
Formula (4.24) implies $\Phi(0)=0$ and $\nabla_{x}\Phi(0)=P^{T}(0)$, hence
$\det\nabla_{y}\phi(0)=\det\nabla_{y}\phi(\Phi(0))=\big{(}\det\nabla_{x}\Phi(0)\big{)}^{-1}=\big{(}\det
P^{T}(0)\big{)}^{-1}=1$. Therefore,

$f(0)=\chi(\phi(0))\cdot a(\phi(0))\cdot|\det\nabla_{y}\phi(0)|=\chi(0)\cdot
a(0)=a(0).$ (4.30)

Combining (4.22), (4.23), (4.29) and (4.30), we arrive at the conclusion. ∎

### Exercise

######  Exercise 4.1.

Use (4.5) to derive (4.4).

######  Exercise 4.2.

Show details about how to derive (4.7) from $\int
e^{-\frac{i\xi^{2}}{2}h}\hat{a}(\xi)\,\mathrm{d}{\xi}$.

######  Exercise 4.3.

(optional) In (4.17), if we instead set

$J(h,\chi
f):=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}e^{\frac{\xi_{j}^{2}h}{i2\alpha_{j}}}\cdot\widehat{\chi
f}(\xi;1/h)\,\mathrm{d}{\xi},$

and later on expand $J$ w.r.t. $h$ at $h=0$, will the computations following
(4.17) still give the desired result? Explain the reason briefly.

######  Exercise 4.4.

Assume $a\in C_{c}^{\infty}(\mathbb{R}^{2n})$ and denote a Lebesgue integral

$I(y,\eta;\lambda):=(2\pi)^{-n}\int_{\mathbb{R}^{2n}}e^{i\lambda
x\cdot\xi}a(x+y,\xi+\eta)\,\mathrm{d}{x}\,\mathrm{d}{\xi}.$

  1. (1)

fix $y$ and $\eta$, and use Proposition 4.8 to find the asymptotic expansion
of $I$ w.r.t. $\lambda$ as $\lambda\to+\infty$;

  2. (2)

write down the first $1+n$ terms (the leading term $+$ the first order terms)
of the asymptotic expansion.

Hint: $x\cdot\xi=\frac{1}{2}\langle{Q(x,\xi),(x,\xi)}\rangle$ with
$Q=\begin{pmatrix}0&I_{n\times n}\\\ I_{n\times n}&0\end{pmatrix}$, where
$(x,\xi)$ is treated as a vertical vector.

######  Exercise 4.5.

Assume symbol $a\in S^{m}({\mathbb{R}^{n}}\times{\mathbb{R}^{n}})$ and denote
an oscillatory integral

$I(y,\eta):=(2\pi)^{-n}\int_{\mathbb{R}^{2n}}e^{ix\cdot\xi}a(x+y,\xi+\eta)\,\mathrm{d}{x}\,\mathrm{d}{\xi}.$

  1. (1)

is $I$ well-defined? If it is, should the cutoff function $\chi$ (cf. (3.12))
be chosen to cutoff $\xi$ alone using $\chi(\epsilon\xi)$, or cutoff $x$ alone
using $\chi(\epsilon x)$, or cutoff both $x$ and $\xi$ together using
$\chi(\epsilon x,\epsilon\xi)$?

  2. (2)

use Proposition 4.8 to find the asymptotic expansion of $I$ w.r.t.
$\langle{\eta}\rangle$ as $|\eta|\to+\infty$;

  3. (3)

write down the first $1+n$ terms (the leading term $+$ the first order terms)
of the asymptotic expansion.

  4. (4)

compare with the result in Exercise 4.4, and revise Remark 4.7.

Hint: Perform the change of variable $\xi\to\langle{\eta}\rangle\xi$.

##  Chapter 5 Symbolic calculus of $\Psi$DOs

In this chapter we show certain symbolic calculus of $\Psi$DOs. We need some
preparations.

######  Lemma 5.1.

Assume $a$, $b\in\mathbb{R}$ such that $|a|\geq 1$ and $|b|\geq 1$, then for
every $m\in\mathbb{R}$ there exists a constant independent of $a$, $b$ such
that

$\boxed{\langle{ab}\rangle^{m}\leq
C_{m}\langle{a}\rangle^{m}\langle{b}\rangle^{m},\quad|a|\geq 1,\ |b|\geq 1.}$

###### Proof.

When $m\geq 0$, we have

$\langle{ab}\rangle^{m}\simeq(1+|ab|)^{m}\leq(1+|a|)^{m}(1+|b|)^{m}\leq\langle{a}\rangle^{m}\langle{b}\rangle^{m}.$

When $m<0$, because $|a|$, $|b|\geq 1$, we have

$\displaystyle\langle{ab}\rangle^{m}$
$\displaystyle\simeq\frac{1}{(1+|ab|)^{|m|}}<\frac{1}{|ab|^{|m|}}=\langle{a}\rangle^{m}\langle{b}\rangle^{m}(\frac{\langle{a}\rangle}{|a|}\frac{\langle{b}\rangle}{|b|})^{|m|}$
$\displaystyle\lesssim\langle{a}\rangle^{m}\langle{b}\rangle^{m}.$

We proved the result. ∎

######  Lemma 5.2 (Peetre’s inequality).

For $\forall a,b\in{\mathbb{R}^{n}}\text{~{}and~{}}\forall m\in\mathbb{R}$,
there exists a constant $C_{m}$ independent of $a$ and $b$ such that

$\boxed{\langle{a\pm b}\rangle^{m}\leq
C_{m}\langle{a}\rangle^{m}\langle{b}\rangle^{|m|}.}$

###### Proof.

For any $a$, $b\in{\mathbb{R}^{n}}$, we have

$1+|a-b|\leq 1+|a|+|b|\leq(1+|a|)\cdot(1+|b|).$

Note that $\langle{a}\rangle\simeq 1+|a|$, so we can conclude Lemma 5.2 for
the case where $m\geq 0$.

When $m<0$, we use the fact:

$\displaystyle 1+|a|\leq 1+|a-b|+|b|\leq(1+|a-b|)\cdot(1+|b|)$
$\displaystyle\Rightarrow\ (1+|a-b|)\geq(1+|a|)\cdot(1+|b|)^{-1}.$

Now assume $m<0$, we have

$(1+|a-b|)^{m}\leq(1+|a|)^{m}\cdot(1+|b|)^{-m}=(1+|a|)^{m}(1+|b|)^{|m|}.$

The proof is complete. ∎

###  5.1. Composition of $\Psi$DOs

Assume $a\in S^{m_{1}}$ and $b\in S^{m_{2}}$. For notational convenience we
denote $T=T_{a}\circ T_{b}$, thus for any $\varphi\in\mathscr{S}$, we have

$\displaystyle T\varphi$ $\displaystyle=(2\pi)^{-n}\int
e^{i(x-y)\cdot\xi}a(x,\xi)T_{b}\varphi(y)\,\mathrm{d}{y}\,\mathrm{d}{\xi}$
$\displaystyle=(2\pi)^{-n}\int e^{i(x-z)\cdot\eta}\big{(}(2\pi)^{-n}\int
e^{i(x-y)\cdot(\xi-\eta)}a(x,\xi)b(y,\eta)\,\mathrm{d}{y}\,\mathrm{d}{\xi}\big{)}\varphi(z)\,\mathrm{d}{z}\,\mathrm{d}{\eta}$
$\displaystyle=(2\pi)^{-n}\int e^{i(x-z)\cdot\eta}\big{(}(2\pi)^{-n}\int
e^{-iy\cdot\xi}a(x,\eta+\xi)b(x+y,\eta)\,\mathrm{d}{y}\,\mathrm{d}{\xi}\big{)}\varphi(z)\,\mathrm{d}{z}\,\mathrm{d}{\eta}$
$\displaystyle=(2\pi)^{-n}\int
e^{i(x-z)\cdot\eta}c(x,\eta)\varphi(z)\,\mathrm{d}{z}\,\mathrm{d}{\eta},$
(5.1)

where $c$ is defined as the oscillatory integral

$c(x,\eta):=(2\pi)^{-n}\int
e^{-iy\cdot\xi}a(x,\eta+\xi)b(x+y,\eta)\,\mathrm{d}{y}\,\mathrm{d}{\xi}.$
(5.2)

If we could show $c\in S^{m}$ for certain $m$, then it implies the composition
of $\Psi$DOs is still a $\Psi$DO. We use the stationary phase lemma under
oscillatory integrals developed in §4 to show this expectation.

To show $c\in S^{m}$, the task boils down to show the asymptotics of $c$ and
its derivatives w.r.t. $|\eta|$, thus we set $\lambda:=\langle{\eta}\rangle$,
so

$c(x,\eta)=(2\pi)^{-n}\lambda^{n}\int e^{-i\lambda
y\cdot\xi}a(x,\lambda(\tilde{\eta}+\xi))b(x+y,\eta)\,\mathrm{d}{y}\,\mathrm{d}{\xi},\quad\text{where}\quad\tilde{\eta}:=\eta/\langle{\eta}\rangle.$

To make better correspondence with the notations in §4, we set

$c_{x,\eta}(y,\xi):=a(x,\lambda(\tilde{\eta}+\xi))b(x+y,\eta),$

thus

$c(x,\eta)=(2\pi)^{-n}\lambda^{n}\int_{\mathbb{R}^{2n}}e^{i\lambda\langle{Q(y,\xi),(y,\xi)}\rangle/2}c_{x,\eta}(y,\xi)\,\mathrm{d}{(y,\xi)},$
(5.3)

where $(y,\xi)$ is treated as a $2n$-dim vertical vector and

$Q=\begin{pmatrix}0&-I\\\ -I&0\end{pmatrix}\quad(\Rightarrow\ Q^{-1}=Q,\
\mathop{\rm sgn}Q=0,\text{~{}and~{}}\det Q=\pm 1).$

In $c_{x,\eta}(y,\xi)$, we regard $(x,\eta)$ as irrelevant parameters make the
following correspondence:

| function | variable | fixed point | in total | dimension
---|---|---|---|---|---
In Prop. 4.8 | $a$ | $x$ | $x_{0}$ | $a(x-x_{0})$ | $n$
at here | $c_{x,\eta}$ | $(y,\xi)$ | $(y_{0},\xi_{0})=0$ | $c_{x,\eta}(y,\xi)$ | $2n$

To use Proposition 4.8, the only thing left to check is (4.9), namely, to
check

$\forall\alpha,\beta:|\alpha|+|\beta|\leq
2n+2N+3,\quad|\partial_{y}^{\alpha}\partial_{\eta}^{\beta}\big{(}c_{x,\eta}(y,\xi)\big{)}|\lesssim
C_{N,n,\alpha,\beta}(\lambda)\langle{(y,\xi)}\rangle^{2N+2}.$ (5.4)

For $|\xi|\geq 2$, we have

$\displaystyle|\partial_{y}^{\alpha}\partial_{\xi}^{\beta}\big{(}c_{x,\eta}(y,\xi)\big{)}|$
$\displaystyle=|\partial_{y}^{\alpha}\partial_{\xi}^{\beta}\big{[}a(x,\lambda(\tilde{\eta}+\xi))b(x+y,\eta)\big{]}|$
$\displaystyle\leq