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%==================== END PREAMBLE ==============
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\vspace*{-50pt}
{\bf  Math 361, Fall 2015\hfill Jerry L. Kazdan}
\medskip

\begin{center}
{\large\bf Problem Set 3}
\smallskip

%\fbox{{\sc \large Draft}}
\end{center}

%\comments{
{\sc Due:} Thurs. Sept.~17 in class. [Late papers will be accepted (without penalty) until 1:00 PM Friday.]
%}
\medskip

 Please carefully read Sections 6.1--6.5 in the Marsden-Hoffman text. 
\medskip

The following short answer problems from Marsden-Hoffman are not
assigned, but you should know how to do them.  Some short answer
problems will be on our exams.

\bigskip

p.~244 \#3, \quad p.~272 \#1, \quad p.~272 \#2, \quad p.~282 \#4, \quad p.~283 \#8,

p.~317 \#7, \quad p.~317 \#8, \quad p.~317 \#18, \quad p.~320 \#29(b,c, d), \quad p.~320 \#32

\bigskip

\begin{problems}

\item {}[Parts of p.~317\#10].  For each of the sequences a)-d) in \#2,
when can we integrate or differentiate term-by-term?  Justify your
response.

\item {} [Marsden-Hoffman, p.~320 \#34]

 


\item Let $X$ and $Y$ be linear spaces and $L:X \to Y$ be a linear map.

Say ${ x_1}$ and ${ x_2}$ are particular solutions of the
equations $L{ x} ={ y_1}$ and $L{ x} ={ y}_2$,
respectively, while ${ z}\ne 0$ is a solution of the homogeneous
equation $L{ z}=0$.  Answer the following in terms of ${ x_1}$, $x_2$,
and { z}.

\begin{parts}
\item Find some solution of $L{ x}= 3{ y}_1$.
\item Find some solution of $L{ x}= -5{ y}_2$.
\item Find some solution of $L{ x}= 3{ y}_1 - 5{ y}_2$.
\item Find another solution (other than ${ z}$ and $0$) of the
homogeneous equation $L{ x}=0$.
\item Find {\it two\/} solutions of $L{ x}={ y}_1$.
\item Find another solution of $L{ x}= 3{ y}_1 - 5{ y}_2$.
\end{parts}

\item In class, we showed that iff $f(x)$ and $K(x,y)$ are continuous
function of $x$ and $y$ for $x$ and $y$ in the interval $0\le x \le 1$,
AND if  $\abs{\lambda}$ is sufficiently small, then the integral
equation
$$
u(x) = f(x) + \lambda\int_0^1 K(x,y)u(y)\,dy
$$
has a unique solution $u\in C([0,1])$.
\smallskip

By looking at the special case where $K(x,y)\equiv 1$, and $f(x)\equiv
2$, show that the assumption that $\lambda$ is sufficiently small
cannot be eliminated completely.

\newpage

\item For each of the following give an example of a sequence of
continuous functions.  If you prefer, a clear sketch of a graph will
be adequate.
\begin{parts}
\item $f_n(x)\to 0$ for all $x\in[0,\,1]$ but 
$\displaystyle \int_0^1 f_n(x)\,dx \ge 1$ for all $n=1,2,\dots$.
\item $g_n(x)\to 0$ for all $x\in[0,\,1]$ and $\displaystyle\int_0^1
g_n(x)\,dx \to 0$ but the $g_n$ do not converge uniformly to zero on
$[0,\,1]$.
\end{parts}

\item Suppose that $G:\R^n\to \R^n$ is a continuous function with the
property that for some real $M$
\begin{equation}\label{Lip}
\norm{G(x)-G(y)}\le M\norm{x-y}\qquad\text{for all\ }\ x,y\in\R^n.
\end{equation}
Here $\norm{x}$ is the standard Euclidean distance in $\R^n$.

If $\lambda> 0$ is small enough, show that the function $F: \R^n \to
\R^n$ defined by 
$$
F(x) = x - \lambda G(x)
$$
is one-to-one and onto, so for every $z\in\R^n$ the equation $F(x)=z$
has one and only one solution $x\in\R^n$. Note that a solution $x$  is
a fixed point of some map.
\smallskip
 

%\newpage
\vspace*{-2.0cm} 
\item \begin{minipage}[t]{3.9in} Let $f(x):\R \to \R$ be continuous
for all $x$ with $f(x)=0$ for $\abs{x}\ge 1$ and let $g_n(x)$ be the
sequence of functions in the figure.  Let
$$
h_n(t)= \int_{-1}^1 f(t-x)g_n(x)\,dx 
$$
\begin{parts}
\item Show that $h_n$ is uniformly continuous.
%\vskip 9cm

\item Show that \ $\displaystyle\lim_{n\to\infty}h_n(t) = f(t)$ 
\ \ {uniformly}.
\end{parts}
\end{minipage}
\begin{minipage}{1.1in}
\vspace{1.4cm}\vbox{\includegraphics[scale=.65]{integral-sequence3.eps}}
\end{minipage}


\end{problems}

\comments{
\vskip 20pt

\begin{center}{\large \bf Bonus Problem}
			
[Please give this directly to Professor Kazdan]
\end{center}

\begin{problems}

\item [B-1] Let $f(x)$ be a continuous function for $0\le x \le 1$.
Evaluate \ $\ \displaystyle  \lim_{n\to \infty} \int_0^1
nf(x)x^n\,dx.$  (Justify your assertions.)


 \end{problems}
}


\medskip
[Last revised: \today]
\end{document}

