\documentclass[11pt]{article} 
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{amsxtra}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{enumitem}
\usepackage{verbatim}
\usepackage{url}
\usepackage{color,hyperref}
\hypersetup{colorlinks,breaklinks,
             linkcolor=blue,urlcolor=blue,
           anchorcolor=blue,citecolor=blue}

\usepackage[margin=3cm]{geometry}
\newcommand{\eps}{\varepsilon}
\newcommand{\Z}{\mathbb{Z}}
\renewcommand{\L}{\mathcal{L}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\x}{\mathbf{x}}
\newcommand{\n}{\mathbf{n}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Q}{\mathbb{Q}}
\renewcommand{\phi}{\varphi}
\newcommand{\limn}{\lim_{n\to \infty}}
\newenvironment{bp}{\color{blue}\begin{proof}}{\end{proof}}
\newenvironment{bs}{\color{blue}\begin{proof}[Solution]}{\end{proof}}
\renewcommand{\emph}[1]{\textbf{#1}}
\renewcommand{\bar}[1]{\overline{#1}}
\renewcommand{\Im}{\text{Im}}
\renewcommand{\Re}{\text{Re}}
\newcommand{\dudn}{\frac{\partial u}{\partial n}}

\newtheorem*{thm*}{Theorem}

\begin{document}

{ \large \textbf{\textsc{Math 5587 -- Homework 11 (Due Thursday Dec 8)}}}
\thispagestyle{empty}
\vspace{0.1in}

The terms distribution and generalized function are used interchangeably in the homework. Recall a function $f$ is \emph{locally integrable} if
\[\int_a^b |f(x)| \,dx < \infty \ \ \text{for all } a<b.\]
\begin{enumerate}
\item \begin{enumerate}
\item Show that $x\delta(x) = 0$. 
\item Show that $\delta(2x) =\frac{1}{2}\delta(x)$.
\item Let $\delta^{(k)}$ denote the $k^{\rm th}$ distributional derivative of the Delta function. Show that
\[\langle \delta^{(k)},\phi\rangle = (-1)^k\phi^{(k)}(0)\]
for all test functions $\phi$.
\end{enumerate}
\item Let 
\[f_n(x) = \begin{cases}
\frac{n}{2},& \text{if } -\frac{1}{n} < x < \frac{1}{n}\\
0,& \text{otherwise.}
\end{cases}\]
Show that $f_n \rightharpoonup \delta$ weakly as $n\to \infty$. Sketch the graph of the functions $f_n$.
\item Let $f_n$ be the sequence of generalized functions given by
\[f_n(x) = \frac{n}{2}\left(\delta\left(x+\tfrac{1}{n}\right) - \delta\left(x-\tfrac{1}{n}\right)\right).\]
Show that
\[f_n \rightharpoonup \delta'(x) \ \text{ weakly as } n\to \infty.\]
\item Let $\phi \in C^\infty_c(\R)$ and let $f$ be a generalized function. Show that the product rule 
\[(\phi f)' = \phi'f + \phi f'\]
holds in the distributional sense.
\item In this question, you will construct a test function that is positive on the interval $(-1,1)$ and vanishes outside of this interval. Recall a test function must be infinitely differentiable and compactly supported. 

Define
\[\psi(x) = \begin{cases}
0,& x \leq 0\\
e^{-\frac{1}{x^2}},& x > 0.
\end{cases}.\]
\begin{enumerate}
\item Use induction to show that for $x>0$ \emph{all} derivatives of $\psi$ of all orders are of the form
\[p\left(\tfrac{1}{x}\right) e^{-\frac{1}{x^2}},\]
where $p$ is a polynomial. [The polynomial $p$ will be different for each derivative. You don't need to find $p$ explicitly.]
\item For any polynomial $p$ show that
\[\lim_{x \to 0^+}p\left(\tfrac{1}{x}\right) e^{-\frac{1}{x^2}} = 0.\]
[Hint: By setting $y=1/x$, you may show instead that
\[\lim_{y\to \infty} p(y) e^{-y^2} = 0.\]
To prove this, first show that for any positive integer $k$, $\lim_{y\to \infty} y^k e^{-y^2} = 0$. There are many ways to do this. One way is to use the Taylor series for $e^x$ to show that $e^x \geq x^k/k!$ for any $x>0$. From this you can deduce that $e^{-y^2} \leq k!/y^{2k}$.]
\item Conclude from (b) that $\psi$ is infinitely differentiable.
\item Define
\[\phi(x) = \psi(x+1)\psi(1-x).\]
Show that $\phi$ is infinitely differentiable, positive in $(-1,1)$ and vanishes for $x \not\in (-1,1)$. [Hint: The product and composition of smooth functions is smooth.]
\item Fix $x_0 \in \R$ and $\eps>0$. Use part (d) to find a test function $g$ that is positive on the interval $(x_0-\eps,x_0+\eps)$ and vanishes outside of this interval. [Hint: Scale and translate $\phi$.]
\end{enumerate}
\item 
\begin{enumerate}
\item Show that $f(x) = \tfrac{1}{2}\log(x^2)$ is locally integrable, and thus defines a generalized function.
\item Show that the ordinary derivative $f'(x) = \frac{1}{x}$ is not locally integrable, and is therefore not a generalized function.
\item Show that the distributional derivative of $f$ is the distribution
\[\langle f',\phi\rangle = -\langle f,\phi' \rangle  =  \text{P.V.} \int_{-\infty}^\infty \frac{\phi(x)}{x}\, dx.\]
[Here, P.V.~stands for the Cauchy principle value of the integral, and is defined as
\[\text{P.V.}\int_{-\infty}^\infty \frac{\phi(x)}{x}\, dx := \lim_{\eps\to 0} \int_{|x|>\eps} \frac{\phi(x)}{x} \, dx,\]
where
\[\int_{|x|>\eps} \frac{\phi(x)}{x} \, dx = \int_{-\infty}^{-\eps}  \frac{\phi(x)}{x} \, dx + \int_{\eps}^\infty  \frac{\phi(x)}{x} \, dx.\]]
\item Show that the second distributional derivative of $f$ is the distribution
\[\langle f'',\phi\rangle = \langle f,\phi''\rangle = \text{P.V.} \int_{-\infty}^\infty \frac{\phi(0) - \phi(x)}{x^2} \, dx.\]
\end{enumerate}
\end{enumerate}


\end{document}

