$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\supp}{\operatorname{supp}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ ##[11.2. Distributions: more](id:sect-11.2) ____________________________ > 1. [Supports and some remarks](#sect-11.2.1) > 2. [Non-linear change of variables](#sect-11.2.2) > 3. [Fourier transform](#sect-11.2.3) > 4. [Convolution](#sect-11.2.4) > 5. [Fourier series](#sect-11.2.5) ###[Supports](id:sect-11.2.1) **[Definition 1.](id:definition-11.2.1)** Let us consider ordinary function $f$. Observe that if $f=0$ on open sets $\Omega\_\iota$ (where $\iota $ runs any set of indices––finite, infinite or even non-enumerable) then $f=0$ on $\bigcup\_\iota \Omega\_\iota$. Therefore there exists a largest open set $\Omega$ such that $f=0$ on $\Omega$. Complement to this set is called *support of $f$* and denoted as $\supp(f)$. **[Definition 2.](id:definition-11.2.2)** a. Let us consider distribution $f$. We say that $f=0$ on open set $Omega$ if $f(\varphi)=0$ for any test function $\varphi$ such that $\supp \varphi \subset \Omega$. b. Then the same observation as in (a) holds and therefore there exists a largest open set $\Omega$ such that $f=0$ on $\Omega$. Complement to this set is called *support of $f$* and denoted as $\supp(f)$. **[Definition 3.](id:definition-11.2.3)** Observe that $\supp(f)$ is always a closed set. If it is also bounded we say that $f$ has a compact support. **[Exercise 1.](id:exercise-11.2.1)** a. Prove that for two functions $f,g$ and for $f\in \mathcal{D}'$, $g\in \mathcal{E}$ \begin{gather} \supp (gf)\subset \supp(f) \cap \supp(g),\label{eq-11.2.1}\\\\ \supp (\partial f)\subset \supp(f)\label{eq-11.2.2} \end{gather} where $\partial$ is a differentiation; b. Prove that $\supp(f)=\emptyset$ iff $f=0$ identiacally; c. Prove that $\supp(\delta\_a)=\\{a\\}$. Prove that the same is true for any of its derivatives. **[Remark 1.](id:remark-11.2.1)** In fact, the converse to [Exercise 1(c)](#exercise-11.2.1) is also true: if $\supp(f)=\{a\}$ then $f$ is a linear combination of $\delta (x-a)$ and its derivatives (up to some order). **[Remark 2.](id:remark-11.2.2)** In the previous section we introduced spaces of test functions $\mathcal{D}$ and $\mathcal{E}$ and the corresponding spaces of distributions $\mathcal{D}'$ and $\mathcal{E}'$. However for domain $\Omega\subset \mathbb{R}^d$ one can introduce $\mathcal{D}(\Omega):= \\{ \varphi \in \mathcal{D}:\, \supp \varphi \subset \Omega\\}$ and $\mathcal{E}=C^\infty (\Omega)$. Therefore one can introduce corresponding spaces of distributions $\mathcal{D}'(\Omega)$ and $\mathcal{E}'(\Omega)=\\{ f\in \mathcal{E}:\, \supp f\subset \Omega\\}$. As $\Omega=\mathbb{R}^d$ we get our "old spaces". ###[Non-linear change of variables](id:sect-11.2.2) **[Definition 4.](id:definition-11.2.4)** Let $f$ be a distribution with $\supp f\subset \Omega\_1$ and let $\Phi:\Omega\_1\to \Omega\_2 $ be one-to-one correspondence, infinitely smooth and with non-vanishing Jacobian $\det \Phi'$. Then $\Phi\_\* f$ is a distribution: \begin{equation} (\Phi\_\* f)(\varphi) = f( |\det \Phi'| \cdot \Phi^\*\varphi ) \label{eq-11.2.3} \end{equation} where $(\Phi^\*\varphi)(x)=\varphi(\Phi(x))$. **[Remark 3.](id:remark-11.2.3)** a. This definition generalizes [Definition 11.1.6](./S11.1.html#definition-11.1.6) and [Definition 11.1.7](./S11.1.html#definition-11.1.7). b. Mathematicians call $\Phi^\*\varphi$ *pullback of $\varphi$* and $\Phi\_\*f$ *pushforward* of $f$. **[Exercise 2.](id:exercise-11.2.2)** Check that for ordinary function $f$ we get $(\Phi\_\*f)(x)=f (\Phi^{-1}(x))$. ###[Fourier transform](id:sect-11.2.3) **[Definition 5.](id:definition-11.2.5)** Let $f\in \mathcal{S}'$. Then Fourier transform $\hat{f}\in \mathcal{S}'$ is defined as \begin{equation} \hat{f}(\varphi) = f(\hat{\varphi}) \label{eq-11.2.4} \end{equation} for $\varphi \in \mathcal{S}$. Similarly, inverse Fourier transform $\check{f}\in \mathcal{S}'$ is defined as \begin{equation} \check{f}(\varphi) = f(\check{\varphi}) \label{eq-11.2.5} \end{equation} **[Exercise 3.](id:exercise-11.2.3)** a. Check that for ordinary function $f$ we get a standard definition of $\hat{f}$ and $\check{f}$. b. To justify [Definition 5](#definition-11.2.5) one need to prove that $f\in \mathcal{S}\iff \hat{f}\in \mathcal{S}$. Do it! c. Prove that for $f\in \mathcal{E}'$ both $\hat{f}$ and $\check{f}$ are ordinary smooth functions \begin{gather} \hat{f}(k) = (2\pi)^{-d} f(e^{-ix\cdot k}), \label{eq-11.2.6}\\\\ \check{f}(k) = f(e^{ix\cdot k}).\label{eq-11.2.7} \end{gather} d. Check that all properties of Fourier transform (excluding with norms and inner products which may not exist are preserved. **[Exercise 4.](id:exercise-11.2.4)** a. Prove that Fourier transforms of $\delta (x-a)$ is $(2\pi)^{-d}e^{-ix\cdot a}$. b. Prove that Fourier transforms of $e^{ix\cdot a}$ is $\delta (x-a)$. **[Exercise 5.](id:exercise-11.2.5)** In dimension $1$ a. Prove that Fourier transforms of $\theta(x-a)$ and $\theta(-x+a)$ are respectively $(2\pi i)^{-1} (k-a-i0)^{-1}$ and $-(2\pi i)^{-1} (k-a+i0)^{-1}$ which are understood as limits in the sense of distributions of $(2\pi i)^{-1}(k-a\mp i\varepsilon)^{-1}$ as $\varepsilon\to+0$. Recall that $\theta(x)$ is a Heaviside function. b. As a corollary conclude that Fourier transform of $\operatorname{sgn}(x):=\theta(x)-\theta(-x)=x/|x|$ is $(2\pi i)^{-1} \bigl((k-a-i0)^{-1}+ (k-a+i0)\bigr)^{-1}= \pi^{-1}(k-a)^{-1}$ with the latter understood in as principal value (see [Exercise 11.1.4(f)](./S11.4.html#exercise-11.1.4)). c. As another corollary conclude that Fourier transform of $\theta(x)+\theta(-x)=1$ is $(2\pi i)^{-1} \bigl((k-a-i0)^{-1}- (k-a+i0)\bigr)^{-1}$ and therefore \begin{equation} (2\pi i)^{-1} \bigl((k-a-i0)^{-1}- (k-a+i0)\bigr)^{-1}=\delta(x-a). \label{eq-11.2.8} \end{equation} ###[Convolution](id:sect-11.2.4) Recall convolution (see [Definition 5.2.1](../Chapter5/S5.2.html#definition-5.2.1)) and its connection to Fourier transform. **[Definition 6.](id:definition-11.2.6)** Let $f,g\in \mathcal{D}'$ (or other way around), $\varphi\in \mathcal{D}$ Then we can introduce $h(y) \in \mathcal{E}$ as \begin{equation\*} h(y) = g ( T\_y\varphi ),\qquad T\_y \varphi(x):= \varphi (x-y). \end{equation*} Observe that $h \in \mathcal{D}$ provided $g\in \mathcal{E}'$. In this case we can introduce $h \in \mathcal{E}$ for $\varphi \in \mathcal{E}$. Therefore if either $f\in \mathcal{E}'$ or $g\in \mathcal{E}'$ we introduce $f\*g$ as \begin{equation\*} (f\*g)(\varphi) = f ( h ). \end{equation*} **[Exercise 6.](id:exercise-11.2.6)** a. Check that for ordinary function $f$ we get a standard definition of the convolution; b. Prove that convolution convolution has the same properties as multiplication; c. Prove that [Theorem 5.2.4](../Chapter5/S5.2.html#thm-5.2.4) holds; d. Prove that $f\*\delta=\delta\*f=f$; e. Prove that $\partial (f\*g)=(\partial f)\*g = f\*(\partial g)$; f. Prove that $T\_a (f\*g)=(T\_a f)\*g = f\*(T\_a g)$ for operator of shift $T\_a$; g. Prove that $\supp (f\*g) \subset \supp(f)+\supp(g)$ where *arithmetic sum* of two sets is defined as $A+B:=\\{x+y:\, x\in A,\, y\in B\\}$. **[Remark 4.](id:remark-11.2.4)** a. One can prove that if a linear map $L:\mathcal{E}'\to \mathcal{D}'$ commutes with all shifts: $T\_a (L f )=L(T\_a f)$ for all $f\in \mathcal{E}'$ then there exists $g\in \mathcal{D}'$ such that $L$ is an operator of convolution: $Lf= g\*f$; b. One can extend convolution if none of $f,g$ has a compact support but some other assumption is fulfilled. For example, in one–dimensional case we can assume that either $\supp(f)\subset [a,\infty)$, $\supp(g)\subset [a,\infty)$ or that $\supp(f)\subset (-\infty,a]$, $\supp(g)\subset (-\infty,a]$. Similarly in multidimensional case we can assume that $\supp(f)\subset C$, $\supp(g)\subset C$ where $C$ is a cone with angle $<\pi $ at its vertex $a$. ###[Fourier series](id:sect-11.2.5) **[Definition 7.](id:definition-11.2.7)** a. We call one-dimensional distribution $f$ *periodic with period $L$* if $f(x-L)=f(x)$. b. More generally, let $\Gamma$ be a lattice of periods (see [Definition 4.B.1](../Chapter4/S4.B.html#definition-4.B.1)). We call distribution $f$ *$\Gamma$-periodic* if $f(x-n)=f(x)$ for all $n\in \Gamma$. Periodic distributions could be decomposed into Fourier series: in one-dimensional case we have \begin{equation} f= \sum\_{-\infty< m<\infty} c\_n e^{\frac{2\pi imx}{L} } \label{eq-11.2.9} \end{equation} and in multidimensional case \begin{equation} f= \sum\_{ m\in \Gamma^\*} c\_m e^{im\cdot } \label{eq-11.2.10} \end{equation} where $\Gamma^\*$ is a dual lattice (see [Definition 4.B.3](../Chapter4/S4.B.html#definition-4.B.3)). To define coefficients $c\_m$ we cannot use ordinary formulae since integral over period (or elementary cell, again see the same definition) is not defined properly. Instead we claim that there exists $\varphi\in \mathcal{D}$ such that \begin{equation} \sum\_{n\in \Gamma} \varphi(x-n) =1. \label{eq-11.2.11} \end{equation} Indeed, let $\psi \in \mathcal{D}$ be non-negative and equal $1$ in some elementary cell. Then $\varphi (x)= \psi (x)/ \bigl(\sum\_{n\in \Gamma} \psi(x-n)\bigr)$ is an appropriate function. Then \begin{equation} c\_m= |\Omega|^{-1} (\varphi f)( e^{-im\cdot x}) \label{eq-11.2.12} \end{equation} where $|\Omega|$ is a volume of the elementary cell. **[Exercise 7.](id:exercise-11.2.7)** a. Find decomposition in Fourier series of one-dimensional distribution $f=\sum\_{-\infty< n<\infty} \delta (x-n L)$; b. Find Fourier transform of $f$ defined in (a); c. Find the connection to Poisson summation formula (see [Theorem 5.2.5](../Chapter5/S5.2.html#thm-5.2.5)). d. Find decomposition in Fourier series of $d$-dimensional distribution $f=\sum\_{n\in \Gamma } \delta (x-n)$; e. Find Fourier transform of $f$ defined in (d); f. Find the connection to multidimensional Poisson summation formula (see [Remark 5.2A.3](../Chapter5/S5.2.A.html#remark-5.2A.3)). --------------- [$\Leftarrow$](./S11.1.html)  [$\Uparrow$](../contents.html)  [$\Rightarrow$](./S11.3.html)