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\author{Math 25}

\title{Homework 5 --- due Oct 24th}

\begin{document}

\maketitle
\setlength{\baselineskip}{18pt}

There are three sections: A,B and C 
(plus a fourth ``optional''
one which won't be graded). 
Please return the three parts
\emph{separately} to the proper CA.

\renewcommand{\thesection}{\Alph{section}}

\section{Problems graded by Jennifer}

\subsection{The AM-GM inequality} 
Let  $a_1,\cdots,a_n$ be
positive real numbers.  
The arithmetic mean of the $a_k$'s is $(a_1 + \cdots + a_n)/n$ and
their geometric mean is $\sqrt[n]{a_1 \cdots a_n}$.
The AM-GM inequality says that
\[ \sqrt[n]{a_1 \cdots a_n} 
\leq \frac{a_1 + \cdots + a_n}{n}. \] The goal of this problem is to
prove the above inequality:
\begin{enumerate}
\item If $a,b \geq 0$, compute $(\sqrt{a}+\sqrt{b})^2-(\sqrt{a}-\sqrt{b})^2$ 
and deduce that $\sqrt{ab} \leq (a+b)/2$;
\item Use this to prove that if $k \geq 1$ and $a_1,\cdots,a_{2^k}$
  are positive real numbers, then \[ \sqrt[2^k]{a_1 \cdots a_{2^k}} 
\leq \frac{a_1 + \cdots + a_{2^k}}{2^k}. \] 
\item prove the AM-GM inequality. 
Hint: given $a_1,\cdots,a_n$ choose $k$ such
  that $2^k \geq n$ and set $a_{n+1},\cdots,a_{2^k} = \sqrt[n]{a_1
  \cdots a_n}$. 
\end{enumerate}

\subsection{} Let $\{ f_n \}_{n \geq 0} : \RR_{\geq 0} \ra \RR$ 
be the sequence of functions
defined by $f_0(x)=0$ and \[  f_{n+1}(x) = f_n(x) + \frac{1}{2} 
(x-f_n^2(x)). \] 
\begin{enumerate}
\item
Prove that for every $x \in [0,1]$, \[ 0 \leq \sqrt{x} - f_n(x) \leq
\sqrt{x} \left( 1 - \frac{\sqrt{x}}{2} \right)^n. \]
\item Prove that $f_n$ converges uniformly to $\sqrt{x}$ on $[0,1]$.
\end{enumerate}

\section{Problems graded by Benjamin}

\subsection{Simple convergence and uniform convergence I}
Let $I=[a,b]$ be an interval and let $\{ f_n \}_{n \geq 1}$ and $f$ be
functions from $I$ to $\RR$. 
We say that the sequence $\{ f_n \}_{n
\geq 1}$

\begin{itemize}
\item converges simply to $f$ if for every $x \in I$, we have
$f_n(x) \ra f(x)$. 
\item converges uniformly to $f$ if $\sup_{x \in I} |f_n(x)-f(x)|
\ra 0$ as $n \ra +\infty$.
\end{itemize} 

\begin{enumerate}
\item Prove that if $\{ f_n \}_{n
\geq 1}$ converges uniformly to $f$ then it converges simply to $f$;
\item prove that the converse is not true as follows: let $I=[0,1]$ and 
define \[ f_n(x)=x^n \quad\text{and}\quad f(x) = \begin{cases} 
0 & \text{if $x \neq 1$} \\ 1 & \text{if $x = 1$.} \end{cases} \]
Show that  $\{ f_n \}_{n
\geq 1}$ converges simply to $f$ but that the convergence is not
uniform. 
\item prove that even if $f$ is continuous then simple convergence
  does not necessarily imply uniform convergence, by considering \[
  f_n(x) = \begin{cases}
nx &  \text{if $0 \leq x \leq 1/(2n)$} \\
1-nx & \text{if $1/(2n) \leq x \leq 1/n$} \\
0 & \text{if $1/n \leq x \leq 1$.} 
\end{cases} \]
\end{enumerate}

\subsection{Simple convergence and uniform convergence II}
Assume that $\{ f_n : I \ra \RR \}_{n \geq 1}$ is a sequence of 
continuous functions which converges simply to $f:I \ra \RR$, where
$f$ is also continuous, and that in addition, $f_{n+1}(x) \geq f_n(x)$
for every $n \geq 1$. 
\begin{enumerate}
\item show that the
sequence $\{\sup_{x \in I} |f_n(x)-f(x)|\}_{n \geq 1}$ is non-increasing; 
\item prove that $\{f_n\}_{n \geq 1}$ converges uniformly to $f$.
\end{enumerate}

\subsection{Simple convergence and uniform convergence III}
Assume that $\{ f_n : I \ra \RR \}_{n \geq 1}$ is a sequence of 
functions which converges simply to $f:I \ra \RR$, and that in
addition, for every $x \in I$ and every sequence $\{ x_n \}_{n \geq
1}$ which converges to $x$, we have $f_n(x_n) \ra f(x)$
\begin{enumerate}
\item prove that for every $\epsilon >0$, there exists
$\delta>0$ and $N$ such that $|x-y|<\delta$ implies that 
$|f_n(x)-f_n(y)|<\epsilon$ for all $n > N$;
\item prove that $f$ is continuous;
\item prove that $\{f_n\}_{n \geq 1}$ converges uniformly to $f$.
\end{enumerate}


\section{Problems graded by Inna}

\subsection{Homeomorphisms of $[0,1]$} Let $f : [0,1] \ra [0,1]$ be a continuous
bijection. Prove that either $f(0)=0$ and $f(1)=1$ and $f$ is strictly
increasing, or  $f(0)=1$ and $f(1)=0$ and $f$ is strictly
decreasing.

\subsection{} Let $f : [0,1] \ra [0,1]$ be a function such that for
every $x \in [0,1]$, there exists $n_x \geq 1$ with $f^{n_x} (x)=x$
(where $f^n$ is $f$ iterated $n$ times). 
\begin{enumerate}
\item
Show that $f$ is a bijection; 
\item 
assume that $f$ is continuous and that $f(0) < 1$. Find $f$.
\end{enumerate}

\subsection{Periods of functions} 
Let $f,g$ be continuous functions from $\RR$ to $\RR$ and let
$\alpha > 0$ be such that $f(x+1)=f(x)$ and $g(x+\alpha)=g(x)$. 
In (3), you can assume
that $1$ is the smallest period of $f$ and that $\alpha$ is the
smallest period of $g$.
\begin{enumerate}
\item prove that if $\alpha \in \QQ$ then $f+g$ is periodic;
\item prove that if $h$ is a continuous function which is periodic
with two periods $1$ and $\alpha \in \RR \setminus \QQ$, then $h$ is constant;
\item prove that $f+g$ is a periodic function if and only if $\alpha \in
\QQ$. Hint: if $f+g$ is periodic of period $\lambda$, what can you say
about the function $g(x+\lambda)-g(x)=-f(x+\lambda)+f(x)$?
\end{enumerate}


\section{Supplementary problems}

\subsection{Ascoli's theorem} Let $E$ be the space of all continuous
functions $f: [0,1] \ra \RR$ and let $d(f,g) = \sup_{x \in
  [0,1]}|f(x)-g(x)|$. Prove that the following are equivalent for a
  subset $K$ of $E$:
\begin{enumerate}
\item $K$ is closed, bounded, and equicontinuous (meaning:
that for every $\epsilon >0$, there exists
$\delta$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$
for all $f \in K$);
\item $K$ is compact.
\end{enumerate}

\end{document}