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{\bf Solutions to Homework 10}
\end{center}
\vspace*{.3in}

\noindent{\bf Supplementary problem 1.} 
Prove that $\cos(2 x)$ is analytic at every point.
\vspace*{.1in}

\noindent{\bf Solution.}
Let $f(x) = \cos(2x)$ and fix a point $a\in\R$.  Let $T_n(x)$ be the
$n$th order Taylor polynomial of $f$ centered at $a$.  We will show
that $\lim_{n\to\infty} T_n(x) = f(x)$ for every $x\in\R$---i.e.
$$
\lim_{n\to\infty} |f(x) - T_n(x)| = 0.
$$
By Taylor's theorem, we have
$$
|f(x) - T_n(x)| = \frac{|f^{n+1}(c)|}{(n+1)!}|x-a|^{n+1}
$$
for some $c$ between $x$ and $a$.  Moreover, the $(n+1)$st derivative of
$f$ at $c$ is always $2^{n+1}$ times $\pm\cos 2c$ or $\pm\sin 2c$, so
$|f^{n+1}(c)| \leq 2^{n+1}$.  Therefore 
$$
|f(x) - T_n(x)| \leq \frac{(2|x-a|)^{n+1}}{(n+1)!}
$$
which converges to $0$ as $n\to\infty$, regardless of $x$ and $a$.
Hence for all $x\in\R$ (especially those in a neighborhood of $a$) 
$f(x)$ agrees with its Taylor series (centered at $a$) evaluated at $x$.
We conclude that $f$ is analytic at $a$.  Since $a$ was arbitrary, $f$
is analytic at all points in $\R$.
\qed

\vspace*{.3in}
\noindent{\bf Supplementary problem 2}
(Not entirely irrelevant sequence problem)
Let $\{a_{i,j}\}_{i,j\in\N}\subset\R^+$ be a double sequence of positive
numbers.  Suppose that 
\begin{itemize}
\item for each fixed $i\in \N$ the sequence $\{a_{i,j}\}_{j\in\N}$ is 
      increasing and converges to to a number $A_i\in\R$; and
\item $\sum_{i=0}^\infty A_i$ converges (call the value of the sum $S$).
\end{itemize}
Show that 
\begin{itemize}
\item $\sum_{i=0}^\infty a_{i,j}$ converges for each fixed $j\in\N$ (call
the value of the sum $S_j$); and
\item $\lim_{j\to\infty} S_j = S$.
\end{itemize} 
The main point here is the second item, because it involves switching
two limits.

\vspace*{.1in}
\noindent{\bf Solution.}
Since $|a_{i,j}| = a_{i,j} \leq A_i$ and since $\sum A_i$ converges, we
can apply Theorem 3.25 to conclude that $\sum_{i=0}^\infty a_{i,j}$ converges.
So the first conclusion holds.

Now to show that $\lim_{j\to\infty} S_j = S$, let $s_n = \sum_{i=0}^n A_i$
and $s_{n,j} = \sum_{i=0}^n a_{i,j}$ be the partial sums of the series 
concerned.  Given $\epsilon > 0$, choose $N\in\N$ such that that $n\geq N$
implies that 
$$
\sum_{i=n+1}^\infty A_i = |S - s_n| \leq \epsilon/3.
$$
It follows that
$$
|S_j - s_{n,j}| = \sum_{i=n+1}^\infty a_{i,j} \leq \sum_{i=n+1} A_i \leq 
	\epsilon/3,
$$
as well.

Moreover, we can exchange a limit with a \emph{finite} sum to obtain
$$
\lim_{j\to\infty} s_{N,j} = s_N.
$$
Hence, there exists $J\in\N$ such that $j\geq J$ implies that
$$
|s_{N,j} - s_N| \leq \epsilon/3.
$$
Putting these observations together gives us for $j\geq J$ 
that
$$
|S_j - S| = |S_j - s_{N,j} + s_{N,j} - S_N + S_N - S|
\leq 
|S_j - s_{N,j}| + |s_{N,j} - S_N| + |S_N - S| 
< \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon.
$$
So $S_j\to S$, as desired.
\qed



\vspace*{.3in}
\noindent{\bf Solution to \#2 on Page 165.}
By hypothesis and problem \#1 (on the same page), there exists a number
$M\in\R$ such that $|f_n(x)|,|g_n(x)|\leq M$ for all $x$ and for all
$n\in\N$.  Hence
\begin{eqnarray*}
|f_n(x)g_n(x) - f(x)g(x)| & = & 
	|f_n(x)g_n(x) - f_n(x)g(x) + f_n(x)g(x) - f(x)g(x)|
	\leq |f_n(x)||g_n(x) - g(x)| + |g(x)||f_n(x) - f(x)|
	\leq M(|g_n(x) - g(x)| + |f_n(x) - f(x)|).
\end{eqnarray*}
So if we are given $\epsilon > 0$, we can choose $N_1,N_2 \in\N$ such that
$n\geq N_1$ implies that
$$
|f_n(x) - f(x)| < \epsilon/2M
$$
for all $x$, and similarly, $n\geq N_2$ implies that
$$
|g_n(x) - g(x)| < \epsilon/2M
$$
for all $x$.  Let $N=\max\{N_1,N_2\}$.  Then $n\geq N$ and the estimates above
imply that
$$
|f_n(x)g_n(x) - f(x)g(x)| < M(\epsilon/2M + \epsilon/2M) = \epsilon
$$
for all $x$.  We conclude that $\{f_ng_n\}$ converges uniformly to 
$fg$.
\qed

\vspace*{.3in}
\noindent{\bf Solution to \#3 on page 165.}

\vspace*{.1in}
Let $f_n,g_n:\R\to\R$ be given by $f_n(x) = g_n(x) = x + 1/n$.  Then
both $\{f_n\}$ and $\{g_n\}$ converge uniformly to the identity function
$f(x) = x$.  On the other hand $f_n(x)g_n(x) = x^2 + 2x/n + 1/n^2$
converges \emph{pointwise} to $f(x)^2$ but not uniformly.  To see this
note that for $x = n$, we have 
$$
|f_n(x)g_n(x) - f(x)^2| = |2x/n + 1/n^2| \geq 2
$$
for all $n\in\N$.  So if we take $\epsilon = 2$, there is no $N\in\N$
such that $n\geq N$ implies that
$$
|f_n(x)g_n(x) - f(x)^2| < \epsilon
$$
for \emph{all} $x\in\R$.


\vspace*{.3in}
\noindent{\bf Solution to \#9 on page 166.}
Given $\epsilon > 0$, we must find $N\in\N$ such that
$$
|f_n(x_n) - f(x)| < \epsilon
$$
when $n\geq N$, so this is what we do: by hypothesis there exists
$N_1\in\N$ such that $n\geq N_1$ implies that
$$
|f_n(x) - f(x)| < \epsilon/2
$$
for all $x\in E$.  Moreover, the limit function $f(x)$ is continuous by
Theorem 7.12, so there exists $\delta > 0$ such that $|x_n - x| < \delta$ 
implies that 
$$
|f(x_n) - f(x)| < \epsilon/2
$$.  Finally, since $x_n \to x$,
there is $N_2\in\N$ such that $n\geq N_2$ implies that 
$n\geq N_2$ implies that
$$
|x_n - x| < \delta.
$$
So if we take $N = \max\{N_1,N_2\}$, we obtain for $n\geq N$ that
$$
|f_n(x_n) - f(x)| \leq |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)|
	< \epsilon/2 + \epsilon/2 = \epsilon
$$
as desired.
\qed

As for the `converse' statement, it depends on what Rudin means by
'converse' here.  I interpret it as follows: \emph{suppose that for
any $x\in E$ and any sequence $\{x_n\}$ converging to $x$ we have
that
$$
\lim_{n\to\infty} f_n(x_n) = f(x).
$$
Then $f_n$ converges uniformly to $f$.}

This is false.  As a counterexample, take for instance $f_n(x) = x/n$ and 
$f(x) = 0$ for all $x\in\R$.  If $\{x_n\}$ is a convergent sequence of
points with limit $x$, then we have
$$
\lim_{n\to\infty} |f_n(x_n) - f(x)| = \lim_{n\to\infty} 
	\left|\frac{x_n}{n}\right| 
\leq \lim_{n\to\infty} \frac{M}{n} = 0
$$
simply because a convergent sequence of real numbers is bounded ($M$ here 
is any upper bound for $\{|x_n|\}$).  On the other hand taking $x = n$
for each $n$, we see that there is always \emph{some} point at which
$|f_n(x) - f(x)| \geq 1$.  Hence $f_n$ does not converge to $f$ uniformly.
\qed



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