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\section{Uniform Convergence}

\subsection{Limits of Continuous Functions}

In this section we consider a sequence of functions, $f_1(x)$, $f_2(x)$,
\dots, that we assume share a common domain $D$. In order to
understand the function obtained as the limit of this sequence, $f(x) =
\lim_{n\to\infty}f_n(x)$ it would be useful to have criteria that help
us decide such questions as
\begin{enumerate}
\item Is $f$ continuous/integrable/differentiable if the $f_n$ are?
\item Does $\lim_{n\to\infty}\int_a^b f_n(x)dx = \int_a^b f(x)dx$?
\item Does $f_n'(x)\to f'(x)$?
\end{enumerate}
The concept of of \emph{uniform convergence} plays a central role in such
questions.

\begin{definition}
A sequence of functions $f_n$ \emph{converges (pointwise)} to $f$, $f_n\to
f$, if, for each $x\in D$, and $\forall\, 1/m$, $\exists\, N$ (that may depend
on $x$ and $1/m$) such that $|f_n(x) - f(x)| < 1/m$, $\forall\, n\ge N$
\end{definition}

\emph{Example:} Let
\[ f_n(x) = \left\{ \begin{array}{ll}
0 & 0 \le x < 1 - \frac{1}{n}\\
\frac{n}{2}(x-1) + \frac{1}{2} & 1 - \frac{1}{n} \le x < 1 + \frac{1}{n}\\
1 & 1 + \frac{1}{n} \le x \le 2
\end{array} \right. \]
\[ \resizebox{2in}{!}{\includegraphics{Pictures/02.fig1.eps}}\qquad
\resizebox{2in}{!}{\includegraphics{Pictures/02.fig2.eps}} \]
If $0 \le x < 1/2$, then $\exists\, N$ such that $1-1/N < x$, so $f_n(x) = 0$,
$\forall\, n \ge N$;
if $1/2 < x \le 1$, then $\exists\, N$ such that $1+1/N < x$, so $f_n(x) = 1$,
$\forall\, n \ge N$; finally, if $x = 1/2$, then $f_n(x) = 1/2$, $\forall\, n$.
Therefore, $f_n \to f$ where
\[ f(x) = \left\{ \begin{array}{ll}
0 & 0 \le x < \frac{1}{2} \\
\frac{1}{2} & x = 1\\
1 &\frac{1}{2}  < x \le 2
\end{array} \right. \]
Note that the limit function $f$ is discontinuous, even though the
functions $f_n$ are continuous. A stronger version of convergence is needed
to preserve continuity in the limit.

\begin{definition}
A sequence of functions $f_n$ \emph{converges uniformly} to $f$, $f_n\to
f$, if $\forall\, 1/m$, $\exists\, N$ (that may depend only on $1/m$) such that
$|f_n(x) - f(x)| < 1/m$, $\forall\, n\ge N$ and $\forall\, x\in D$.
\end{definition}

\emph{Example:} $f_n(x) = 1 + x + \dots + x^n$ converges \emph{uniformly}
to $f(x) = \frac{1}{1-x}$ on $[-r,r]$ for any fixed $0<r<1$, not \emph{not}
on $(-1,1)$.

\proof $f_n(x) = (1-x^{n+1})/(1-x) = f(x) - x^{n+1}/(1-x)$.
If $|x| \le r < 1$, then $1/|1-x| \le 1/(1-r)$, so
$|f_n(x) - f(x)| \le |x|^{n+1}/(1-r) \le r^{n+1}/(1-r)$.
Now, given $1/m$, choose $N$ such that $r^{N+1} < (1-r)/m$. Then,
$\forall\, |x|\le r$ and $\forall\, n \ge N$,
\[ |f_n(x) - f(x)| \le r^{n+1}/(1-r) \le  r^{N+1}/(1-r) < 1/m \]

To show the convergence is \emph{not} uniform on $(-1,1)$, we must show
that $\exists\, 1/m$ such that $\forall\, N$, $|f_n(x) - f(x)| \ge 1/m$ for
some $x\in(-1,1)$ and some $n \ge N$. Since $|f_n(x)-f(x)| =
|x|^{n+1}/|1-x| \to \infty$ as $x\to 1$, this is easy to arrange. \qed

\begin{theorem}\label{02:UNIFORMLIMITS}
Assume $f_n$ are continuous and $f_n\to f$ uniformly on~$D$. Then $f$ is
continuous on $D$.
\end{theorem}

\proof Given $1/m$, choose $N$ such that $|f_n(x)-f(x)| <
1/(3m)$, $\forall\, x\in D$ and $\forall\, n \ge N$. Then
\begin{eqnarray*}
|f(x) - f(y)| &\le& |f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y) - f(y)|\\
 &<& \frac{1}{3m} + |f_n(x) - f_n(y)| + \frac{1}{3m}
\quad \forall\, x,y\in D
\end{eqnarray*}

Now fix $x\in D$.  Since $f_n$ is continuous at $x$, $\exists\, 1/k$ such
that if $|x-y|<1/k$ then $|f_n(x)-f_n(y)|<1/(3m)$. Note that $1/k$ may
depend on $x$.  Then \[ |f(x) - f(y)| < \frac{1}{3m} + \frac{1}{3m} +
\frac{1}{3m} = \frac{1}{m} \] whenever $|x - y| < 1/k$ proving that $f$
is continuous at $x$.  \qed

\emph{Note:} The above proof also shows that if $f_n$ are uniformly
continuous and $f_n\to f$ uniformly, then $f$ is uniformly continuous. The
condition $|x-y|<1/k$ in the proof can be removed and all inequalities hold
$\forall\, x,y\in D$.

A useful modification of the definition of uniform convergence is
\emph{uniform convergence on compact subsets}. For example,
$f_n(x) = 1 + x + \dots + x^n$ converges uniformly on compact subsets of
$(-1,1)$ as the previous
example shows (any $[a,b]\subset(-1,1)$ is contained in $[-r,r]$ for
some $r<1$).

\subsection{Limits of Integrals}

The next theorem shows that we can interchange integration and
\emph{uniform} limits.

\begin{theorem}\label{02:INTEGRALS}
Assume $f_n$ are integrable and $f_n\to f$ uniformly on $[a,b]$. Then $f$
is integrable and $\lim_{n\to\infty}\int_a^b f_n(x)dx = \int_a^b f(x)dx$.
\end{theorem}

\proof Recall
\begin{eqnarray*}
\overline{I}(f)&=&
\inf\Big\{\int_a^b s(x)dx\ \Big|\ s(x)\ge f(x), s = \mbox{step function}\Big\}\\
\underline{I}(f)&=&
\sup\Big\{\int_a^b t(x)dx\ \Big|\ t(x)\le f(x), t = \mbox{step function}\Big\}
\end{eqnarray*}

Given $1/m$, find $N$ such that $|f_n(x)-f(x)|<1/[3m(b-a)]$, $\forall\,
x\in[a,b]$ and $\forall\, n \ge N$.
\[ \begin{array}{ccccccc}
\Rightarrow& f_n(x)-\frac{1}{3m(b-a)} &<& f(x) &<& f_n(x)+\frac{1}{3m(b-a)}\\
\Rightarrow& \underline{I}(f_n)-\frac{1}{3m} &\le& \underline{I}(f) &\le&
\underline{I}(f_n)+\frac{1}{3m}\\
\end{array} \]
and similarly for $\overline{I}(f)$ (by the linearity of
$\underline{I}$ and $\overline{I}$). Therefore,
\begin{eqnarray*}
|\underline{I}(f) - \underline{I}(f_n)| &\le& \frac{1}{3m}\\
|\overline{I}(f) - \overline{I}(f_n)| &\le& \frac{1}{3m}
\end{eqnarray*}
Since $\overline{I}(f_n) = \underline{I}(f_n)$,
\begin{eqnarray*}
|\overline{I}(f) - \underline{I}(f)|
&=&
|\overline{I}(f) - \overline{I}(f_n) +
\underline{I}(f_n) - \underline{I}(f)| \\
&\le&
|\overline{I}(f) - \overline{I}(f_n)| +
|\underline{I}(f_n) - \underline{I}(f)| \\
&\le&
\frac{1}{3m} + \frac{1}{3m} = \frac{2}{3m} < \frac{1}{m}
\end{eqnarray*}
Since $1/m$ is arbitrary, $\overline{I}(f) = \underline{I}(f) = I(f)$ so
$f$ is integrable and $I(f) = \int_a^bf(x)dx$. We have already shown
$|I(f) - I(f_n)| \le 1/(3m)$, $\forall\, n\ge N$, and this proves $I(f_n)\to
I(f)$. \qed

\emph{Example:} Index the rational numbers in $[0,1]$ by $r_1, r_2, r_3,
\dots$ and define
\[ f_n(x) = \left\{\begin{array}{ll}
1 & \mbox{if\ } x = r_1, r_2, \dots, r_n\\
0 & \mbox{otherwise}
\end{array}\right.  \]
then $f_n(x) \to f(x)$ where
\[ f(x) = \left\{\begin{array}{ll}
1 & \mbox{if\ } x \in[0,1]\cap\Q\\
0 & \mbox{otherwise}
\end{array}\right.  \]
Note that $\int_0^1 f_n(x)dx = 0$ and $\lim_{n\to\infty}\int_0^1
f_n(x)dx = 0$, but $f(x)$ is not integrable!

\subsection{Limits of Derivatives}

The corresponding theorem for derivatives is \emph{false}.

\emph{Example:} Let $f_n(x) = \sqrt{x^2 + 1/n^2}$. Note that the functions
$f_n(x)$ are differentiable and converge uniformly on $[-1,1]$ to $f(x)
= \sqrt{x^2} = |x|$.
\[ \resizebox{!}{1.5in}{\includegraphics{Pictures/02.fig3.eps}} \]
To see that the convergence is uniform, note that the maximum difference
between $f_n$ and $f$ occurs at $x=0$,
\[ \max_{x\in[0,1]}|f_n(x) - f(x)| = \sqrt{0^2 + 1/n^2} - |0| = \frac{1}{n} \]
However, $f(x)$ is not differentiable at~0.  To preserve differentiability
in the limit we need to assume the uniform convergence of $f_n'$.

\begin{theorem}\label{02:DERIVATIVES}
Assume $f_n$ and $f_n'$ are continuous on $[a,b]$ (i.e., $f_n\in C^1[a,b]$).
If $f_n\to f$ pointwise and $f_n'\to g$ uniformly, then $f$ and $g$ are
continuous on $[a,b]$ and $f' = g$. (i.e., $f\in C^1[a,b]$).
\end{theorem}

\proof
Fix $x_0\in [a,b]$. Then by the Fundamental Theorem of Calculus,
$f_n(x) = f(x_0) + \int_{x_0}^x f_n'(t)dt$. By Theorem~\ref{02:INTEGRALS},
as $n\to\infty$, we get $f(x) = f(x_0) + \int_{x_0}^x g(t)dt$. Therefore
$f'(x) = g(x)$. \qed

 \end{document}
