\documentclass[12pt]{article}
\usepackage{amsfonts,amsmath}

\def\Cx{\mathbb{C}}
\def\Nl{\mathbb{N}}
\def\Ra{\mathbb{Q}}
\def\Rl{\mathbb{R}}
\def\Ts{\mathbb{T}}
\def\Ir{\mathbb{Z}}

\begin{document}

\centerline{\textbf{Problem Set 5}}
\centerline{\textbf{Math 201A}: \textbf{Fall 2016}}

\bigskip
\bigskip
\noindent
\textbf{Problem 1.}  Define $f : [1,\infty) \to [1,\infty)$ by
$f(x) = x + {1}/{x}$.
Show that $|f(x) - f(y)| < |x-y|$ for all $x,y\in [1,\infty)$
but $f$ has no fixed point. Why doesn't this example contradict the contraction mapping theorem?



\bigskip\noindent
\textbf{Problem 2.} Suppose that $X$ is a compact metric space and $T : X\to X$
satisfies $d\left(T(x),T(y)\right) < d(x,y)$ for all $x,y\in X$ with $x\ne y$.
Prove that $T$ has a unique fixed point in $X$.
Why doesn't the example in Problem 1 contradict this theorem?


\bigskip\noindent
\textbf{Problem 3.} Consider the nonlinear integral equation
\[
u(x) - \frac{1}{4}\int_0^1 \left(x+y\right) u^2(y)\, dy = \frac{1}{2}.
\]
Show that there is a unique continuous solution $u : [0,1] \to \Rl$ of this equation with the property
that $0 \le u(x) \le 1$ for all  $0 \le x \le 1$.


\bigskip
\noindent
\textbf{Problem 4.} (a) Let $A$ be a dense subset of $C([0,1])$ with respect to the sup-norm.
Prove that $f=0$ if $f\in C([0,1])$ and
\[
\int_0^1 f(x) g(x)\, dx = 0\qquad \text{for every $g\in A$}.
\]
(b) Prove that $f=0$ if $f\in C([0,1])$ and
\[
\int_0^1 x^n f(x)\, dx = 0\qquad\text{for every $n\in \Nl$}.
\]

\bigskip
\noindent
\textbf{Problem 5.} Let $R= [0,1]\times [0,1]$ and suppose that $\mathcal{F}\subset C(R)$ consists
of all functions $f\in C(R)$ of the form
\[
f(x,y) = \sum_{n=1}^N g_n(x) h_n(y)
\]
for some $g_n,h_n\in C([0,1])$ and some $N\in \Nl$. Show that $\mathcal{F}$ is dense in $C(R)$.
(You can use any appropriate theorems.)



\bigskip
\noindent
\textbf{Problem 6.}
Let $C_0(\Rl)$ denote the Banach space of continuous functions
$f : \Rl\to\Rl$ such that $f(x) \to 0$ as $|x|\to \infty$, equipped with the sup-norm.

\smallskip\noindent
(a) For $n\in \Nl$, define $f_n \in C_0(\Rl)$ by
\[
f_n(x) =
\begin{cases}
1 & \text{if $|x| \le n$},
\\
n/|x|& \text{if $|x| > n$}.
\end{cases}
\]
Show that $\mathcal{F} = \{f_n : n\in \Nl\}$ is a bounded, equicontinuous subset of $C_0(\Rl)$,
but $(f_n)$ has no uniformly convergent subsequence. Why
doesn't this example contradict the Arzel\`a-Ascoli theorem?

\smallskip\noindent
(b) A family of functions $\mathcal{F} \subset C_0(\Rl)$ is said to be tight if for every $\epsilon > 0$
there exists $R > 0$ such that
$|f(x)| < \epsilon$ for all $x\in \Rl$ with $|x| \ge R$ and all $f \in \mathcal{F}$.
Prove that $\mathcal{F} \subset C_0(\Rl)$ is precompact in $C_0(\Rl)$ if it is bounded, equicontinuous, and tight.


\bigskip
\noindent
\textbf{Problem 7.}  Let $X$ be a compact metric space and $(f_n)$ an increasing sequence of continuous functions
$f_n : X \to \Rl$. If $f_n \to f$ pointwise and $f:X\to \Rl$ is continuous, prove that $f_n\to f$ uniformly.

\smallskip\noindent
\textsc{Hint.} Given $\epsilon >0$, for each $x\in X$ choose $N(x)\in \Nl$ such that
\[
0\le f(x) - f_{N(x)}(x) <\epsilon,
\]
then choose $\delta(x,\epsilon)>0$ such that $y \in B_{\delta(x,\epsilon)}(x)$ implies that
\[
|f_{N(x)}(x) - f_{N(x)}(y)| <\epsilon \quad\text{and}\quad |f(x) - f(y)| <\epsilon.
\]

\end{document}
