We shall concentrate solely on Fourier analysis in the one-dimensional setting $d=1$ for simplicity of notation, but each of the results here have natural extensions to higher dimensions. Depending upon the physical context, one can view the physical domain $R$ as representing either time or space; we will mostly think in terms of the prior interpretation, even though the standard language of”time-frequency evaluation”, which we will make more notable use of in later notes, obviously originates in the latter. We have often performed various localisations in either physical space or Fourier area $R$, for instance so as to take advantage of the uncertainty principle. An Individual can formalise these operations Concerning the operational calculus of two basic operations on Schwartz functions

$\mathcal{S}(\mathbf{R})$, the position operator $X: \mathcal{S}(\mathbf{R}) \rightarrow \mathcal{S}(\mathbf{R})$ defined by
$$(X f)(x):=x f(x)$$
and the momentum operator $D: \mathcal{S}(\mathbf{R}) \rightarrow \mathcal{S}(\mathbf{R})$, defined by
$$(D f)(x):=\frac{1}{2 \pi i} \frac{d}{d x} f(x)$$
(The terminology comes from quantum mechanics, where it is customary to also insert a small constant $h$ on the right-hand side of (1) in accordance with de Broglie’s law. Such a normalisation is also used in several branches of mathematics, most notably semiclassical analysis and microlocal analysis, where it becomes profitable to consider the semiclassical limit $h \rightarrow 0$, but we will not emphasise this perspective here.) The momentum operator can be viewed as the counterpart to the position operator, but in frequency space instead of physical space, since we have the standard identity
$$\widehat{D f}(\xi)=\xi \hat{f}(\xi)$$
for any $\xi \in \mathbf{R}$ and $f \in \mathcal{S}(\mathbf{R}) .$ We observe that both operators $X, D$ are formally self-adjoint in the sense that
$$\langle X f, g\rangle=\langle f, X g\rangle ; \quad\langle D f, g\rangle=\langle f, D g\rangle$$
for all $f, g \in \mathcal{S}(\mathbf{R})$, where we use the $L^{2}(\mathbf{R})$ Hermitian inner product
$$\langle f, g\rangle:=\int_{\mathbf{R}} f(x) \overline{g(x)} d x$$
Clearly, for any polynomial $P(x)$ of one real variable $x$ (with complex coefficients), the operator $P(X): \mathcal{S}(\mathbf{R}) \rightarrow \mathcal{S}(\mathbf{R})$ is given by the spatial multiplier operator
$$(P(X) f)(x)=P(x) f(x)$$
and similarly the operator $P(D): \mathcal{S}(\mathbf{R}) \rightarrow \mathcal{S}(\mathbf{R})$ is given by the Fourier multiplier operator
$$\widehat{P(D) f(\xi)}=P(\xi) \hat{f}(\xi)$$
Inspired by this, if $m: \mathbf{R} \rightarrow \mathbf{C}$ is any smooth function that obeys the derivative bounds
$$\frac{d^{j}}{d x^{j}} m(x) \lesssim_{m, j}\langle x\rangle^{O_{m, j}(1)}$$
for all $j \geq 0$ and $x \in \mathbf{R}$ (that is to say, all derivatives of $m$ grow at most polynomially), then we can define the spatial multiplier operator $m(X): \mathcal{S}(\mathbf{R}) \rightarrow \mathcal{S}(\mathbf{R})$ by the formula
$$(m(X) f)(x):=m(x) f(x)$$

Problem 1.

(i) Show that for $a$ obeying (6), that $a(X, D)$ does indeed map $\mathcal{S}(\mathbf{R})$ to $\mathcal{S}(\mathbf{R})$
(ii) Show that the symbol $a$ is uniquely determined by the operator $a(X, D)$. That is to say, if $a, b$ are two functions obeying $(6)$ with $a(X, D) f=b(X, D) f$ for all $f \in \mathcal{S}(\mathbf{R})$, then $a=b$. (Hint:
apply $a(X, D)-b(X, D)$ to a suitable truncation of a plane wave $x \mapsto e^{2 \pi i x \xi}$ and then take limits.)

Proof .

Proof. $\quad$ (a) Suppose $f \in \mathcal{S}(\mathbb{R})$, then for $i, j \geqslant 0$ we get

\begin{aligned}
x^{i} \frac{d^{j}}{d x^{j}} a(X, D) f(x) &=\int x^{i} \frac{\partial^{j}}{\partial x^{j}}\left[a(x, \xi) e^{2 \pi i \xi \cdot \xi}\right] \hat{f}(\xi) d \xi \
&=\int\left(\sum_{k=0}^{j}\left(\begin{array}{l}
j \
k
\end{array}\right) \frac{\partial^{k}}{\partial x^{k}} a(x, \xi) x^{i}(2 \pi i \xi)^{j-k} e^{2 \pi i x \xi}\right) \hat{f}(\xi) d \xi \
&=\sum_{k=0}^{j}\left(\begin{array}{l}
j \
k
\end{array}\right) \int \frac{\partial^{k}}{\partial x^{k}} a(x, \xi) x^{i}(2 \pi i \xi)^{j-k} e^{2 \pi i x \xi} \widehat{f}(\xi) d \xi
\end{aligned}

Notice that
$$\left(I-\Delta_{\xi}\right) e^{2 \pi i x \xi}=\left(1+4 \pi^{2}|x|^{2}\right) e^{2 \pi i x \xi}$$
and if we define
$$L_{\xi}=\left(1+4 \pi^{2}|x|^{2}\right)^{-1}\left(I-\Delta_{\xi}\right)$$
using repeated integration by parts we get for every $N$ that
$$(1)=\int\left(L_{\xi}\right)^{N} \frac{\partial^{k}}{\partial x^{k}} a(x, \xi) x^{i}(2 \pi i \xi)^{j-k} e^{2 \pi i x \xi} \widehat{f}(\xi) d \xi$$
By derivative bounds of $a(x, \xi)$, we get
$$\begin{array}{l} \int\left(L_{\xi}\right)^{N} \frac{\partial^{k}}{\partial x^{k}} a(x, \xi) x^{i}(2 \pi i \xi)^{j-k} e^{2 \pi i x \xi} \hat{f}(\xi) d \xi \mid \ \lesssim_{j, k, N}\left(1+4 \pi^{2}|x|^{2}\right)^{-N} \int\langle x\rangle^{O_{i, k}(1)}\langle\xi\rangle^{O_{i, j, k, N}(1)}|\widehat{f}(\xi)| d \xi \ =\frac{\langle x\rangle^{O_{i, k}(1)}}{\left(1+4 \pi^{2}|x|^{2}\right)^{N}} \int\langle\xi\rangle^{O_{i, j, k, N}(1)}|\hat{f}(\xi)| d \xi \end{array}$$
Therefore, if we choose $N$ sufficiently large depending on $i, k$ and observe that $f \in \mathcal{S}(\mathbb{R})$, the above equation is further bounded by
$$\lesssim_{i, j, k, f} 1$$
So we get
$$|(1)| \lesssim_{i, j, k, f} 1$$
showing that $x^{i} \frac{d^{j}}{d x^{j}} a(X, D) f(x) \in L_{x}^{\infty}(\mathbb{R})$. Since $i, j$ are arbitrary, we get $a(X, D) f \in \mathcal{S}(\mathbb{R})$,
as desired.

(b) Let $\phi \in C_{c}^{\infty}(\mathbb{R})$ be the standard bump function such that $\int \phi=1$. Then for $\varepsilon>0$ we define
$$\phi_{\varepsilon}(x):=\frac{1}{\varepsilon} \phi\left(\frac{x}{\varepsilon}\right)$$
Thus $\phi_{\varepsilon}$ is again in $C_{c}^{\infty}(\mathbb{R}) .$ Given $\eta \in \mathbb{R}$, we can then define $f_{\varepsilon} \in \mathcal{S}(\mathbb{R})$ such that
$$\hat{f}{\varepsilon}(\eta-\xi)=\phi{\varepsilon}(\xi)$$

Now, by assumption we have
\begin{aligned} 0=[a(X, D)-b(X, D)] f_{\varepsilon}(x) &=\int_{\mathbb{R}}(a(x, \xi)-b(x, \xi)) \hat{f}{\varepsilon}(\eta-\xi) e^{2 \pi i x \xi} d \xi \ &=:\left(g{x} * \phi_{\varepsilon}\right)(\eta) \end{aligned}
where $g_{x}(\xi):=(a(x, \xi)-b(x, \xi)) e^{2 \pi i x \xi}$. By the derivative bounds (6) in the notes, $g_{x}(\xi) \in$ $L_{l o c}^{1}(\mathbb{R})$, so the approximate identity $\phi_{\varepsilon}$ gives us $g_{x} * \phi_{\varepsilon} \rightarrow g_{x}$ a.e. as $\varepsilon \rightarrow 0 .$ Note that $g_{x}$
is smooth, so we actually get $g_{x} * \phi_{\varepsilon} \rightarrow g_{x}$ as $\varepsilon \rightarrow 0$. Therefore, sending $\varepsilon \rightarrow 0$ we see that
$$(a(x, \eta)-b(x, \eta)) e^{2 \pi i x \eta}=0$$
Since $x, \eta$ are arbitrary, we get $a-b \equiv 0$ as desired.

Problem 2.

(Composition of pseudodifferential operators) Let $a \in \mathcal{S}^{\alpha}$ and $b \in \mathcal{S}^{\beta}$ for some $\alpha, \beta \in \mathbf{R}$.
(i) Show that $a * b-a b \in \mathcal{S}^{\alpha+\beta-1}$. (Hint: reduce as before to the case where $a, b$ are compactly supported, and use the fundamental theorem of calculus to write $a(x, \eta)=a(x, \xi)+(\eta-\xi) \int_{0}^{1} a_{\xi}(x, \xi+t(\eta-\xi)) d t$, where
$a_{\xi}(x, \xi):=\frac{\partial a}{\partial \xi}(x, \xi)$. Then use the Fourier inversion formula, integration by parts, and arguments similar to those used to prove Theorem 8 .
(ii) Show that $a * b-a b-\frac{1}{2 \pi i} a_{\xi} b_{x} \in \mathcal{S}^{\alpha+\beta-2}$, where
$a_{\xi}(x, \xi):=\frac{\partial a}{\partial \xi}(x, \xi)$ and $b_{x}(x, \xi):=\frac{\partial b}{\partial x}(x, \xi) .$ (Hint: now apply
the fundamental theorem of calculus once more to expand $\left.a_{\xi}(x, \xi+t(\eta-\xi)) .\right)$
(iii) Check (i) and (ii) directly in the classical case when $a(x, \xi)=A(x) \xi^{l}$ and $b(x, \xi)=B(x) \xi^{m}$ for some smooth $A, B$ obeying the bounds (9) and for $l, m \geq 0$. Based on this, for any integer $r \geq 0$, make a prediction for an approximation to $a * b$ as a polynomial combination of the symbols arbitrary $a \in \mathcal{S}^{\alpha}, b \in \mathcal{S}^{\beta}$ and finitely many of their derivatives which is accurate up to an error in $\mathcal{S}^{\alpha+\beta-r} .$ Then verify this prediction.

Proof .

(a) Using an limiting argument as in Theorem 8 , we may assume $a, b$ are compactly supported symbols. Then by the fundamental theorem of calculus
$$a(x, \eta)=a(x, \xi)+(\eta-\xi) \int_{0}^{1} a_{\xi}(x, \xi+t(\eta-\xi)) d t$$

So we get
\begin{aligned} (a * b)(x, \xi) &=\int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \eta) b(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta \ &=\int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi) b(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta \ &+\int_{\mathbb{R}} \int_{\mathbb{R}}\left(\int_{0}^{1} a_{\xi}(x, \xi+t(\eta-\xi)) d t\right)(\eta-\xi) b(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta \end{aligned}
To evaluate the first term, we notice by Fourier inversion that
\begin{aligned} \int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi) b(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta &=a(x, \xi) \int_{\mathbb{R}} e^{2 \pi i x(\eta-\xi)}\left(\int_{\mathbb{R}} b(y, \xi) e^{-2 \pi i y(\eta-\xi)} d y\right) d \eta \ &=a(x, \xi) \int_{\mathbb{R}} e^{2 \pi i x(\eta-\xi)} \mathcal{F}{y} b(\eta-\xi, \xi) d \eta \ &=a(x, \xi) \int{\mathbb{R}} e^{2 \pi i x \eta} \mathcal{F}{y} b(\eta, \xi) d \eta \ &=a(x, \xi) b(x, \xi) \end{aligned} For the second term, we have $$\frac{\partial}{\partial y}\left(e^{2 \pi i(y-x)(\xi-\eta)}\right)=-2 \pi i(\eta-\xi) e^{2 \pi i(y-x)(\xi-\eta)}$$ so integrating by parts we get $$\int{\mathbb{R}}(\eta-\xi) b(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y=\frac{1}{2 \pi i} \int_{\mathbb{R}} b_{x}(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y$$
Then the second term evaluates to
$$\frac{1}{2 \pi i} \int_{\mathbb{R}} \int_{\mathbb{R}}\left(\int_{0}^{1} a_{\xi}(x, \xi+t(\eta-\xi)) d t\right) b_{x}(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta$$
so we may write
$$(a * b)(x, \xi)=a(x, \xi) b(x, \xi)$$
5 )
$$+\frac{1}{2 \pi i} \int_{\mathbb{R}} \int_{\mathbb{R}}\left(\int_{0}^{1} a_{\xi}(x, \xi+t(\eta-\xi)) d t\right) b_{x}(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta$$
Now, adopting the notation suggested by Haiyu Huang, for every integer $k \geqslant 1$ we define the binary operation $*{k}$ on all symbols by $\left(a *{k} b\right)(x, \xi)$
$$:=\int_{\mathbb{R}}\left(\int_{[0,1]^{k}} s_{k} s_{k-2}^{2} \cdots s_{1}^{k-1} a\left(x, \xi+s_{k-1} \ldots s_{1}(\eta-\xi)\right) d s_{1} \ldots d s_{k}\right) b(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta$$
From (2) we see that
\begin{aligned} \left(a *{k} b\right)(x, \xi)-a(x, \xi) b(x, \xi) &=\frac{1}{2 \pi i} \int{\mathbb{R}} \int_{\mathbb{R}}\left(\int_{0}^{1} a_{\xi}(x, \xi+t(\eta-\xi)) d t\right) b_{x}(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta \ &=\frac{1}{2 \pi i} a_{\xi} *{1} b{x} \end{aligned}
so to conclude the proof we want to show
$$\frac{\partial}{\partial x^{j}} \frac{\partial}{\partial \xi^{l}}\left(a_{\xi} *{1} b{x}\right)(x, \xi) \lesssim_{a, b, \alpha, \beta, j, l}\langle\xi\rangle^{\alpha+\beta-1-j}$$

Similar to the proof of Theorem 8 in Notes 3, from differentiation under the integral sign and integration by parts we obtain the Leibniz identities
$$\frac{\partial}{\partial x}\left(a *{k} b\right)=\left(\frac{\partial}{\partial x} a\right) *{k} b+a *{k}\left(\frac{\partial}{\partial x} b\right)$$ and $$\frac{\partial}{\partial \xi}\left(a *{k} b\right)=\left(\frac{\partial}{\partial \xi} a\right) *{k} b+a *{k}\left(\frac{\partial}{\partial \xi} b\right)$$
So from this induction on $j+l$ we see that it suffices for us to do the $j=l=0$ case
$$\left(a_{\xi} *{1} b{x}\right)(x, \xi) \lesssim_{a, b, \alpha, \beta}\langle\xi\rangle^{\alpha+\beta-1}$$
Applying a smooth partition of unity in the $\eta$ variable to $a(x, \eta)$, it suffices to establish the above estimate in the two regions ${(x, \eta)|\eta-\xi| \leqslant\langle\xi\rangle / 2}$ and ${(x, \eta):|\eta-\xi| \geqslant\langle\xi\rangle / 4}$.
In the first region, we write
$$\left(a_{\xi} *{1} b{x}\right)(x, \xi)=\int K_{x}(x-y) b_{x}(y, \xi) d y$$
where
$$K_{x}(z):=\int_{\mathbb{R}}\left(\int_{0}^{1} a_{\xi}(x, \xi+t(\eta-\xi)) d t\right) e^{2 \pi i z \eta} d \eta$$
Integrating by parts in $\eta$ twice and applying triangular inequality, we get
$$\left|K_{x}(z)\right| \lesssim_{a, \alpha}\langle\xi\rangle^{\alpha-2}\langle z\rangle^{-2}$$
where we have utilized the observation that $\langle\xi+t(\eta-\xi)\rangle \sim\langle\xi\rangle$ in this region. Thus we get
$$\left|K_{x}\right|_{L^{1}(\mathbb{R})} \lesssim_{a}\langle\xi\rangle^{\alpha-1}$$
By assumption of $b$ we know that
$$b_{x}(y, \xi) \lesssim_{b}\langle\xi\rangle^{\beta}$$
and thus we conclude
$$\left(a_{\xi^{*} 1} b_{x}\right)(x, \xi) \lesssim_{a, b, \alpha, \beta}\langle\xi\rangle^{\alpha+\beta-1}$$
In the second region, repeated integration by parts in $y$ gives
$$\left(a_{\xi} *{1} b{x}\right)(x, \xi)=\int K_{x, j}(x-y) \partial_{y}^{j} b(y, \xi) e^{2 \pi i(y-x) \xi} d y$$
where
$$K_{x, j}(z):=\int(2 \pi i(\eta-\xi))^{-j}\left(\int_{0}^{1} a_{\xi}(x, \xi+t(\eta-\xi)) d t\right) d \eta$$
Then taking $j$ sufficiently large we get
$$K_{x, j}(z) \lesssim_{a, \alpha}\langle\xi\rangle^{\alpha-2}\langle z\rangle^{-2}$$
We still have
$$b_{x}(y, \xi) \lesssim_{b}\langle\xi\rangle^{\beta}$$
and thus
$$\left(a_{\xi} *{1} b{x}\right)(x, \xi) \lesssim_{a, b, \alpha, \beta}\langle\xi\rangle^{\alpha+\beta-1}$$
as desired.

(b) Again by fundamental theorem of calculus we write
$$a_{\xi}(x, \xi+t(\eta-\xi))=a_{\xi}(x, \xi)+t(\eta-\xi) \int_{0}^{1} a_{\xi \xi}(x, \xi+s t(\eta-\xi)) d s$$
Then
\begin{aligned} & \frac{1}{2 \pi i} \int_{\mathbb{R}} \int_{\mathbb{R}}\left(\int_{0}^{1} a_{\xi}(x, \xi+t(\eta-\xi)) d t\right) b_{x}(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta \ =& \frac{1}{2 \pi i} \int_{\mathbb{R}} \int_{\mathbb{R}} a_{\xi}(x, \xi) b_{x}(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta \ +& \frac{1}{2 \pi i} \iint\left(\int_{0}^{1} t(\eta-\xi) \int_{0}^{1} a_{\xi \xi}(x, \xi+s t(\eta-\xi)) d s d t\right) b_{x}(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta \end{aligned}
Using Fourier inversion as in (a) we evaluate the first term as
$$\frac{1}{2 \pi i} a_{\xi}(x, \xi) b_{x}(x, \xi)$$
For the second term, integrating by parts in $y$ as in (a) evaluates it to
\begin{aligned} &-\frac{1}{4 \pi^{2}} \int_{\mathbb{R}} \int_{\mathbb{R}}\left(\int_{0}^{1} \int_{0}^{1} t a_{\xi \xi}(x, \xi+s t(\eta-\xi)) d s d t\right) b_{x x}(y, \xi) e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta \ =&-\frac{1}{4 \pi^{2}}\left(a_{\xi \xi} *{2} b{x x}\right)(x, \xi) \end{aligned}
Therefore, by (5) we are left to show $a_{\xi \xi} *{2} b{x x}$ is a pseudodifferential operator in $(x, \xi)$ of order $\alpha+\beta-2$. Induct using Leibniz identities for $*{k}$ shown in (a), our task is showing $$\left(a{\xi \xi} *{2} b{x x}\right)(x, \xi) \lesssim_{a, b, \alpha, \beta, r}\langle\xi\rangle^{\alpha+\beta-2}$$
As in (a), we apply a smooth partition of unity in $\eta$ to $a(x, \eta)$ and establish the claim in regions ${(x, \eta):|\eta-\xi| \leqslant\langle\xi\rangle / 2}$ and ${(x, \eta):|\eta-\xi| \geqslant\langle\xi\rangle / 4}$. Using a similar argument as
in (a) we get the desired result.
(c) Let $a(x, \xi)=A(x) \xi^{l} \in S^{l}$ and $b(x, \xi)=B(x) \xi^{m} \in S^{m}$. A direct computation using FubiniTonelli gives
\begin{aligned} a * b(x, \xi) &=\int_{\mathbb{R}} \int_{\mathbb{R}} A(x) \eta^{l} B(y) \xi^{m} e^{2 \pi i(y-x)(\xi-\eta)} d y d \eta \ &=A(x) \xi^{m} \int_{\mathbb{R}} \eta^{l} e^{2 \pi i x(\eta-\xi)}\left(\int_{\mathbb{R}} B(y) e^{-2 \pi i(\eta-\xi)} d y\right) d \eta \ &=A(x) \xi^{m} \int_{\mathbb{R}} \eta^{l} e^{2 \pi i x(\eta-\xi)} \hat{B}(\eta-\xi) d \eta \ &=A(x) \xi^{m} e^{-2 \pi i x \xi} \int_{\mathbb{R}} \eta^{l} \hat{B}(\eta-\xi) e^{2 \pi i x \eta} d \eta \end{aligned}
Repeatedly integrating by parts, the above equation evaluates to
\begin{aligned} \frac{A(x) \xi^{m} e^{-2 \pi i x \xi}}{(2 \pi i)^{l}} \frac{\partial^{l}}{\partial x^{l}}\left(e^{2 \pi i x \cdot \zeta} B(x)\right) &=\frac{A(x) \xi^{m} e^{-2 \pi i x \xi}}{(2 \pi i)^{l}} \sum_{k=0}^{l}\left(\begin{array}{l} l \ k \end{array}\right) B^{(k)}(x)(2 \pi i \xi)^{l-k} e^{2 \pi i x \xi} \ &=\sum_{k=0}^{l}\left(\begin{array}{l} l \ k \end{array}\right) A(x) B^{(k)}(x) \xi^{m} \xi^{l-k}(2 \pi i)^{-k} \ &=\sum_{k=0}^{l} \frac{\partial_{\xi}^{k} a(x, \xi) \partial_{x}^{k} b(x, \xi)}{k !(2 \pi i)^{k}} \end{aligned}

Notice that the $k=0$ term is precisely $a b$, so it follows that
$$a * b-a b=\sum_{k=1}^{r} \frac{\partial_{\xi}^{k} a(x, \xi) \partial_{x}^{k} b(x, \xi)}{k !(2 \pi i)^{k}}$$
By definition of $a$ and $b$, the above equation is a polynomial in $\xi$ of degree at most $r+$ $m-1$, and the coefficients of the polynomial are functions of $x$ respecting a polynomial derivative bounds of any order. Therefore, $a * b-a b \in S^{r+m-1}$. Moreover, pulling out the first term we get
$$a * b-a b-\frac{1}{2 \pi i} a_{\xi} b_{x}=\sum_{k=2}^{r} \frac{\partial_{\xi}^{k} a(x, \xi) \partial_{x}^{k} b(x, \xi)}{k !(2 \pi i)^{k}}$$
and repeating the above reasoning we see that this equation lies in $S^{r+m-2}$. Now, let $r \in \mathbb{N}, \alpha^{\alpha}$, and $b \in \mathcal{S}^{\alpha}$. We claim that
$$a * b-\sum_{k=0}^{r-1} \frac{\partial_{\xi}^{k} a(x, \xi) \partial_{x}^{k} b(x, \xi)}{k !(2 \pi i)^{k}} \in \mathcal{S}^{\alpha+\beta-r}$$
To verify the claim, we use induction and fundamental theorem of calculus as in the previous two parts to get
$$a * b-\sum_{k=0}^{r-1} \frac{\partial_{\xi}^{k} a(x, \xi) \partial_{x}^{r} b(x, \xi)}{k !(2 \pi i)^{k}}=\frac{\left(\partial_{\xi}^{r} a {r} \partial{x}^{r} b\right)(x, \xi)}{r !(2 \pi i)^{r}}$$
So by induction and Leibniz identities for ${r}$, it suffices for us to show $$\left(\partial{\xi}^{r} a *{k} \partial{x}^{r} b\right)(x, \xi) \lesssim_{a, b, \alpha, \beta, k}\langle\xi\rangle^{\alpha+\beta-k}$$
As in (a), we apply a smooth partition of unity in $\eta$ to $a(x, \eta)$ and establish the claim in regions ${(x, \eta):|\eta-\xi| \leqslant\langle\xi\rangle / 2}$ and ${(x, \eta):|\eta-\xi| \geqslant\langle\xi\rangle / 4}$. Using a similar argument as
in (a) we get the desired result.

Problem 3.

(Gabor-type transforms and pseudodifferential operators) Given any function $\phi \in \mathcal{S}(\mathbf{R})$ with the $L^{2}$ normalisation $|\phi|_{L^{2}(\mathbf{R})}=1$, and any $f \in \mathcal{S}(\mathbf{R})$, define the Gabor-type transform $T_{\phi} f: \mathbf{R} \times \mathbf{R} \rightarrow \mathbf{C}$ by the formula
$$T_{\phi} f(x, \xi):=\int_{\mathbf{R}} f(y) \bar{\phi}(y-x) e^{-2 \pi i y \xi} d y$$
thus $T_{\phi} f(x, \xi)$ is the inner product of $f$ with the function $y \mapsto \phi(y-x) e^{2 \pi i y \xi}$, which is the “wave packet” formed from function $\phi$ by translating by $x$ and then modulating by $\xi$. (Intuitively, $T_{\phi} f(x, \xi)$ measures the extent to which $f$ lives at spatial location $x$ and frequency location $\xi$.) We also define the adjoint map $T_{\phi}^{} F: \mathbf{R} \rightarrow \mathbf{C}$ for $F \in \mathcal{S}(\mathbf{R} \times \mathbf{R})$ by the formula $$T_{\phi}^{} F(y):=\int_{\mathbf{R}} \int_{\mathbf{R}} F(x, \xi) \phi(y-x) e^{2 \pi i y \xi} d x d \xi$$
(i) Show that for any $f \in \mathcal{S}(\mathbf{R}), T_{\phi} f$ is a Schwartz function on $\mathbf{R} \times \mathbf{R}$, thus $T_{\phi}$ is a linear map from $\mathcal{S}(\mathbf{R})$ to $\mathcal{S}(\mathbf{R} \times \mathbf{R})$.
Similarly, show that for $F \in \mathcal{S}(\mathbf{R} \times \mathbf{R}), T_{\phi}^{} F$ is a Schwartz function on $\mathbf{R}$, thus $T_{\phi}^{}$ is a linear map from $\mathcal{S}(\mathbf{R} \times \mathbf{R})$ to $\mathcal{S}(\mathbf{R})$.
(ii) Establish the identity $T_{\phi}^{} T_{\phi} f=f$ for any $f \in \mathcal{S}(\mathbf{R})$, and conclude inparticular that $$\left|T_{\phi} f\right|_{L^{2}(\mathbf{R} \times \mathbf{R})}=|f|_{L^{2}(\mathbf{R})}$$ for any $f \in \mathcal{S}(\mathbf{R})$, thus $T_{\phi}$ extends to a linear isometry from $L^{2}(\mathbf{R})$ into $L^{2}(\mathbf{R} \times \mathbf{R})$ (iii) For any smooth compactly supported $a \in C_{c}^{\infty}(\mathbf{R})$ and $f \in \mathcal{S}(\mathbf{R})$, establish the identity $$T_{\phi}^{}\left(a T_{\phi} f\right)=\left(a * W_{\phi}\right)(X, D) f$$
where $W_{\phi}: \mathbf{R} \times \mathbf{R} \rightarrow \mathbf{C}$ is the (Kohn-Nirenberg) Wigner distribution of $\phi$, defined by the formula
$$W_{\phi}(x, \xi)=\phi(x) \overline{\hat{\phi}}(\xi) e^{-2 \pi i x \xi}$$
and $a * W_{\phi}$ is the phase space convolution
$$a * W_{\phi}(x, \xi):=\int_{\mathbf{R}} \int_{\mathbf{R}} a(y, \eta) W_{\phi}(x-y, \xi-\eta) d y d \eta$$

Proof .

(a) Suppose $f \in \mathcal{S}(\mathbb{R})$, then for $\alpha, \beta, i, j \geqslant 0$, direct computation yields
$$\left|x^{\alpha} y^{\beta} \frac{\partial^{i}}{\partial x^{i}} \frac{\partial^{j}}{\partial \xi^{j}} T_{\phi} f(x, \xi)\right|=\left|\int_{\mathbb{R}} x^{\alpha} \xi^{\beta}(2 \pi y)^{j} f(y) \overline{\phi^{(i)}}(y-x) e^{-2 \pi i y \xi} d y\right|$$
As in Exercise 1, we note that
$$\left(1-\Delta_{y}\right) e^{-2 \pi i y \xi}=\left(1+4 \pi^{2}|\xi|^{2}\right) e^{-2 \pi i y \xi}$$
Therefore, we define $L_{y}=\left(1+4 \pi^{2}|\xi|^{2}\right)^{-1}\left(1-\Delta_{y}\right) .$ Using repeated integration by parts as above, for every $N \in \mathbb{N}$ we get
\begin{aligned} (6) &=\left|x^{\alpha} y^{\beta} \int_{\mathbb{R}}\left(L_{y}\right)^{N}(2 \pi y)^{j} f(y) \overline{\phi^{(i)}}(y-x) e^{-2 \pi i y \xi} d y\right| \ & \lesssim_{j} \int_{\mathbb{R}} \frac{\left|x^{\alpha} \xi^{\beta}\right|}{\left(1+4 \pi^{2}|\xi|^{2}\right)^{N}}\left|\left(1-\Delta_{y}\right)^{N} y^{j} f(y) \overline{\phi^{(i)}}(y-x)\right| d y \ & \lesssim \alpha \int_{\mathbb{R}} \frac{\langle x-y\rangle^{2 \alpha}\langle y\rangle^{2 \alpha}|\xi|^{\beta}}{\left(1+4 \pi^{2}|\xi|^{2}\right)^{N}}\left|\left(1-\Delta_{y}\right)^{N} y^{j} f(y) \overline{\phi^{(i)}}(y-x)\right| d y \end{aligned}
where in the last inequality we utilized the identity
$$|y| \lesssim\langle y-x\rangle^{2}\langle x\rangle^{2}$$

Actually, we notice by triangular inequality that
\begin{aligned} |y| & \leqslant 1+|y| \leqslant 1+|y-x|+|x| \ & \leqslant 1+|y-x|+|x|+|x||y-x| \ & \leqslant(1+|y-x|)(1+|x|) \ & \lesssim\left(1+|y-x|^{2}\right)\left(1+|x|^{2}\right) \ &=\langle y-x\rangle^{2}\langle x\rangle^{2} \end{aligned}
so the identity holds. Now, notice that $f, \phi \in \mathcal{S}(\mathbb{R})$, so if we choose $N>0$ sufficiently large depending on $i, j, \alpha, \beta$, we can bound $(7)$ by
$$\begin{array}{l} \lesssim{\alpha}{\alpha, \beta, i, j} \frac{1}{\left(1+4 \pi^{2}|\xi|^{2}\right)} \int{\mathbb{R}}\langle y\rangle^{-C_{i, j, \alpha, \beta}}\langle x-y\rangle^{-C_{i, j, \alpha, \beta}^{\prime}} d y \ \lesssim{\alpha}{\alpha, \beta, i, j} 1 \end{array}$$ where $C{i, j, \alpha, \beta}, C_{i, j, \alpha, \beta}^{\prime}>0$ are sufficiently large constants. Thus we have proven
$$\left|x^{\alpha} y^{\beta} \frac{\partial^{i}}{\partial x^{i}} \frac{\partial^{j}}{\partial \xi^{j}} T_{\phi} f(x, \xi)\right| \lesssim_{\alpha, \beta, i, j} 1$$
and hence $T_{\phi} f(x, \xi) \in \mathcal{S}\left(\mathbb{R}^{2}\right)$ as desired. Suppose $F \in \mathcal{S}\left(\mathbb{R}^{2}\right)$, then for $i, j \geqslant 0$,
\begin{aligned} \left|y^{i} \frac{d^{j}}{d y^{j}} T_{\phi}^{} F(y)\right| &=\left|\int_{\mathbb{R}} \int_{\mathbb{R}} y^{i} F(x, \xi) \frac{d^{j}}{d y^{j}}\left(\phi(y-x) e^{2 \pi i y \xi}\right) d x d \xi\right| \ &=\left|\int_{\mathbb{R}} \int_{\mathbb{R}} y^{i} F(x, \xi) \sum_{k=0}^{j}\left(\begin{array}{l} j \ k \end{array}\right) \phi^{(k)}(y-x)(2 \pi i \xi)^{j-k} e^{2 \pi i y \xi} d x d \xi\right| \ &=\left|\sum_{k=0}^{j}\left(\begin{array}{l} j \ k \end{array}\right) \int_{\mathbb{R}} \int_{\mathbb{R}} y^{i} F(x, \xi) \phi^{(k)}(y-x)(2 \pi i \xi)^{j-k} e^{2 \pi i y \xi} d x d \xi\right| \ & \lesssim_{j} \sum_{k=0}^{j}\left(\begin{array}{l} j \ k \end{array}\right) \int_{\mathbb{R}} \int_{\mathbb{R}}|y|^{i}\left|F(x, \xi)\left|\phi^{(k)}(y-x)\right| \xi\right|^{j-k} d x d \xi \end{aligned} Since $F \in \mathcal{S}\left(\mathbb{R}^{2}\right)$ and $\phi \in \mathcal{S}(\mathbb{R})$, we can use the rapid decay of Schwartz functions to bound the above equation by $$\begin{array}{l} \lesssim_{j} \int_{\mathbb{R}} \int_{\mathbb{R}}|y|^{i}\langle x\rangle^{-100 i j-1}\langle\xi\rangle^{-100 j-1}\langle y-x\rangle^{-100 i j-1} d x d \xi \ \lesssim_{i, j} \int_{\mathbb{R}} \int_{\mathbb{R}}\langle y-x\rangle^{2 i}\langle x\rangle^{2 i}\langle x\rangle^{-100 i j-1}\langle\xi\rangle^{-100 j-1}\langle y-x\rangle^{-100 i j-1} d x d \xi \end{array}$$ $\lesssim_{i, j} 1$ where the second last inequality utilizes the identity $$|y| \lesssim\langle y-x\rangle^{2}\langle x\rangle^{2}$$ as before. Therefore, $y^{i} \frac{d^{j}}{d y^{j}} T_{\phi}^{} F(y) \in L_{y}^{\infty}(\mathbb{R})$, and $T_{\phi}^{} F \in \mathcal{S}(\mathbb{R})$ as desired. It follows imme- diately that $T_{\phi}$ is a linear map $\mathcal{S}(\mathbb{R}) \rightarrow \mathcal{S}\left(\mathbb{R}^{2}\right)$ and $T_{\phi}^{}$ is a linear map $\mathcal{S}\left(\mathbb{R}^{2}\right) \rightarrow \mathcal{S}(R)$.

(b) Let $f \in \mathcal{S}(\mathbb{R})$. Note that
\begin{aligned} T_{\phi}^{} T_{\phi} f(y) &=\int_{\mathbb{R}} \int_{\mathbb{R}} T_{\phi} f(x, \xi) \phi(y-x) e^{2 \pi i y \xi} d x d \xi \ &=\int_{\mathbb{R}} \int_{\mathbb{R}}\left(\int_{\mathbb{R}} f(z) \bar{\phi}(z-x) e^{-2 \pi i z \xi} d z\right) \phi(y-x) e^{2 \pi i y \xi} d x d \xi \ &=\int_{\mathbb{R}} \int_{\mathbb{R}} \mathcal{F}{z}(f(z) \bar{\phi}(z-x))(\xi) e^{2 \pi i y \xi} \phi(y-x) d x d \xi \ &=\int{\mathbb{R}}\left(\int_{\mathbb{R}} \mathcal{F}{z}(f(z) \bar{\phi}(z-x))(\xi) e^{2 \pi i y \xi} d \xi\right) \phi(y-x) d x \ &=\int{\mathbb{R}} f(y) \bar{\phi}(y-x) \phi(y-x) d x \ &=f(y)|\phi|_{L^{2}(\mathbb{R})}=f(y) \end{aligned}
where the fourth equality uses Fubini-Tonelli, which is applicable here since $f, \phi \in \mathcal{S}(\mathbb{R})$, and the last equality utilizes our normalization assumption $|\phi|_{L^{2}(\mathbb{R})}=1$. In particular, we have
\begin{aligned} |f|_{L^{2}(\mathbb{R})} &=\langle f, f\rangle_{L^{2}(\mathbb{R})}^{1 / 2}=\left\langle T_{\phi}^{} T_{\phi} f, f\right\rangle_{L^{2}(\mathbb{R})}^{1 / 2} \ &=\left\langle T_{\phi} f, T_{\phi} f\right\rangle_{L^{2}\left(\mathbb{R}^{2}\right)}^{1 / 2}=\left|T_{\phi} f\right|_{L^{2}(\mathbb{R})} \end{aligned}
Now, for $f \in L^{2}(\mathbb{R})$, we can find $\left{f_{n}\right}_{n} \subset \mathcal{S}(\mathbb{R})$ such that $f_{n} \rightarrow f$ in $L^{2}$. Then $\left{f_{n}\right}_{n}$ is Cauchy, and by (8) we have $\left{T_{\phi} f_{n}\right}_{n}$ is Cauchy. Since $L^{2}\left(\mathbb{R}^{2}\right)$ is complete, $T_{\phi} f_{n}$ converges in $L^{2}\left(\mathbb{R}^{2}\right)$. We define $T_{\phi} f$ to be the $L^{2}$ limit of $T_{\phi} f_{n}$, and we need to verify that it is well-defined. Suppose $\left{g_{n}\right}_{n} \subset \mathcal{S}(\mathbb{R})$ is another sequence that converges to $f$ in $L^{2}$, then $\left|T_{\phi} g_{n}-T_{\phi} f_{n}\right|_{L^{2}\left(\mathbb{R}^{2}\right)} \rightarrow 0 .$ It follows that $T_{\phi} g_{n}$ also converges to $T_{\phi} f$ in $L^{2}(\mathbb{R} \times \mathbb{R})$ and $T_{\phi} f$
is well-defined. By writing
$$\left|T_{\phi} f_{n}\right|_{L^{2}\left(\mathbb{R}^{2}\right)}=\left|f_{n}\right|_{L^{2}\left(\mathbb{R}^{2}\right)}$$
and sending $n \rightarrow \infty$, we see that $T_{\phi}$ extends to an isometry $L^{2}(\mathbb{R}) \rightarrow L^{2}\left(\mathbb{R}^{2}\right)$.
(c) Let $a \in C_{c}^{\infty}\left(\mathbb{R}^{2}\right)$ and $f \in \mathcal{S}(\mathbb{R})$, then
\begin{aligned} T_{\phi}^{*}\left(a T_{\phi} f\right)(z) &=\int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi) T_{\phi} f(x, \xi) \phi(z-x) e^{2 \pi i z \xi} d x d \xi \ &=\int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi)\left(\int_{\mathbb{R}} f(y) \bar{\phi}(y-x) e^{-2 \pi i y \xi} d y\right) \phi(z-x) e^{2 \pi i z \xi} d x d \xi \ &=\int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi)\left(\int_{\mathbb{R}} f(y+x) \bar{\phi}(y) e^{-2 \pi i y \xi} d y\right) \phi(z-x) e^{2 \pi i(z-x) \xi} d x d \xi \ &=\int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi) \widehat{\tau_{x} f \cdot \bar{\phi}}(\xi) \phi(z-x) e^{2 \pi i(z-x) \xi} d x d \xi \ &=\int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi) \widehat{\tau_{x} f} * \hat{\bar{\phi}}(\xi) \phi(z-x) e^{2 \pi i(z-x) \xi} d x d \xi \ &=\int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi) \widehat{\tau_{x} f} * \hat{\bar{\phi}}(\xi) \phi(z-x) e^{2 \pi i(z-x) \xi} d x d \xi \ &=\int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi)\left(\int_{\mathbb{R}{x} f}(\eta) \hat{\bar{\phi}}(\xi-\eta) d \eta\right) \phi(z-x) e^{2 \pi i(z-x) \xi} d x d \xi \ &=\int{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi)\left(\int_{\mathbb{R}} e^{2 \pi i x \eta} \hat{f}(\eta) \overline{\widehat{\phi}}(\eta-\xi) d \eta\right) \phi(z-x) e^{2 \pi i(z-x) \xi} d x d \xi \end{aligned}

We know that $a(x, \xi)$ is smooth compactly supported, and $f, \phi \in \mathcal{S}(\mathbb{R})$, so we can use Fubini-Tonelli to evaluate the above integral as
\begin{aligned} & \int_{\mathbb{R}} \int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi) e^{2 \pi i x \eta} \hat{f}(\eta) \overline{\widehat{\phi}}(\eta-\xi) \phi(z-x) e^{2 \pi i(z-x) \xi} d x d \xi d \eta \ =& \int_{\mathbb{R}} \int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi) \hat{f}(\eta) \overline{\hat{\phi}}(\eta-\xi) \phi(z-x) e^{-2 \pi i(z-x)(\eta-\xi)} e^{2 \pi i z \eta} d x d \xi d \eta \ =& \int_{\mathbb{R}} \int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi) \hat{f}(\eta) W_{\phi}(z-x, \eta-\xi) e^{2 \pi i z \eta} d x d \xi d \eta \ =& \int_{\mathbb{R}}\left(\int_{\mathbb{R}} \int_{\mathbb{R}} a(x, \xi) W_{\phi}(z-x, \eta-\xi) d x d \xi\right) \hat{f}(\eta) e^{2 \pi i z \eta} d \eta \ =& \int_{\mathbb{R}} a * W_{\phi}(z, \eta) \hat{f}(\eta) e^{2 \pi i z \eta} d \eta \ =&\left(a * W_{\phi}\right)(X, D) f \end{aligned}
so we get
$$T_{\phi}^{*}\left(a T_{\phi} f\right)(z)=\left(a * W_{\phi}\right)(X, D) f(z)$$
as desired. This finishes the proof.

## Math 3205 fouerier analysis

Introduces the basics of Fourier analysis as a prelude to applications. The course develops Fourier analysis from a general pure mathematical perspective starting with Lebesgue integration and elements of the theory of Hilbert spaces, leading to Fourier series, Fourier integrals and the fast Fourier transform, and then to applications such as partial differential equations and sampling. These subjects are of great importance to the electrical engineering and physics communities. The course concludes with more modern topics such as Gabor and wavelet transforms.

• Basics of Lebesgue integration
• Elements of Hilbert spaces and orthogonal expansions
• Fourier series and Fourier transforms of continuous data; applications to partial differential equations, sampling and uncertainty
• Fast Fourier transform of discrete data
• Time-frequency and time-scale analysis

## STÆ211G Applied Fourier Analysis

Course Name:Applied Fourier AnalysisCourse Description:

Fourier series in trigonometric and exponential form. Cosine and sine series on bounded intervals. Discussion on the point-wise and uniform convergence of the Fourier series. Square error and minimum square error, optimal properties of the Fourier series, Bessel’s inequality, Parseval’s equality, and spectrum of the series. Fourier series for primitive functions and derivatives.

Emphasis is laid on physical applications  and on problem solving.Learning Outcomes:

Knowledge and understanding: To complete this course the students should be able to

1. Define, explain, and give examples of the main concepts of the course, such as Fourier series, minimum square error and its physical meaning, optimal properties of the Fourier series, Fourier series for integrals and derivatives.
2. State and explain the main theoretical results of the course, such as the convergence of the Fourier series and the Parseval’s equality.

Skills: To complete this course the student should be able to

1. Calculate Fourier series for simple functions
2. Recognize some basic properties of the functions and indicate whether we can expect the convergence of the corresponding Fourier series.
3. Compute minimum square error and draw the amplitude spectrum for simple functions
4. Use Parseval’s equality and convergence of the series to obtain the sum of simple numerical series.
5. Apply the results from the course to other topics such as optics, electrical circuits.

Supervision:

TutorRagnar Sigurðsson[email protected]Professor MorePreceding Courses / Prerequisites: Required preparation  STÆ101G Mathematical Analysis IA or STÆ104G Mathematical Analysis I or STÆ105G Practical Mathematical Analysis
Books:No textbooks have been specifiedOther reading material:Programmes:Physics, BS (180 ECTS) (First year, Spring)
Chemistry, BS (180 ECTS) (Year of study unspecified, Spring)
Engineering Physics, BS (180 ECTS) (First year, Spring)
Applied Mathematics, BS (180 ECTS) (First year, Spring)

# Winter 2021 STAT 31460 Applied Fourier Analysis

Course: STAT 31460=CAAM 31460

Title: Applied Fourier Analysis

Instructor(s): Benjamin Palacios

Teaching Assistant(s): TBA

Class Schedule: Sec 1: MW 4:10 PM–5:30 PM (Remote)

Textbook(s): Mallatt, A Wavelet Tour of Signal Processing: The Sparse Way (3rd ed)

Description: Decompositions of functions into frequency components via the Fourier transform, and related sparse representations, are fundamental tools in applied mathematics. These ideas have been important in applications to signal processing, imaging, and the quantitative and qualitative analysis of a broad range of mathematical models of data (including modern approaches to machine learning) and physical systems. Topics to be covered in this course include an overview of classical ideas related to Fourier series and the Fourier transform, wavelet representations of functions and the framework of multiresolution analysis, and applications throughout computational and applied mathematics.

Prerequisite(s): Graduate student in the Physical Sciences Division or consent of instructor.