Lecture Notes

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\lhead{\textsc{Math 425 --- Real Analysis}}
\rhead{\textsc{Lecture 12}}
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  {\LARGE\bfseries Lecture 12: Uniform Convergence of\\Function Sequences and Series}\\[10pt]
  \begin{tabular}{rl}
    \textbf{Course:} & Math 425 --- Introduction to Real Analysis \\
    \textbf{Lecturer:} & Prof.\ David Kowalski \\
    \textbf{Date:} & October 22, 2025 \\
    \textbf{Scribe:} & Mei-Lin Chang \\
  \end{tabular}
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\textbf{Topics covered:} Pointwise vs.\ uniform convergence, the Weierstrass $M$-test, interchange of limits, continuity of uniform limits, term-by-term integration and differentiation.

\textbf{Reading:} Rudin, \textit{Principles of Mathematical Analysis}, Chapter 7, Sections 7.1--7.17.

\section{Pointwise and Uniform Convergence}

Recall that we are studying sequences of functions $\{f_n\}$ where each $f_n: E \to \mathbb{R}$ is defined on a common domain $E \subseteq \mathbb{R}$.

\begin{definition}[Pointwise Convergence]
A sequence $(f_n)$ \emph{converges pointwise} to $f: E \to \mathbb{R}$ if for every $x \in E$,
\[
  \lim_{n \to \infty} f_n(x) = f(x).
\]
Equivalently, for every $x \in E$ and every $\varepsilon > 0$, there exists $N = N(\varepsilon, x) \in \mathbb{N}$ such that $|f_n(x) - f(x)| < \varepsilon$ for all $n \geq N$.
\end{definition}

The key issue with pointwise convergence is that $N$ may depend on $x$---this can lead to pathological behavior.

\begin{example}
Let $f_n: [0,1] \to \mathbb{R}$ be defined by $f_n(x) = x^n$. Then
\[
  f(x) = \lim_{n \to \infty} x^n = \begin{cases} 0 & \text{if } 0 \leq x < 1, \\ 1 & \text{if } x = 1. \end{cases}
\]
Each $f_n$ is continuous, but the pointwise limit $f$ is discontinuous at $x = 1$. This shows that pointwise limits of continuous functions need not be continuous.
\end{example}

This motivates a stronger notion of convergence.

\begin{definition}[Uniform Convergence]
A sequence $(f_n)$ \emph{converges uniformly} to $f: E \to \mathbb{R}$ if for every $\varepsilon > 0$, there exists $N = N(\varepsilon) \in \mathbb{N}$ (independent of $x$) such that
\[
  |f_n(x) - f(x)| < \varepsilon \quad \text{for all } n \geq N \text{ and all } x \in E.
\]
Equivalently, $\|f_n - f\|_\infty = \sup_{x \in E} |f_n(x) - f(x)| \to 0$ as $n \to \infty$.
\end{definition}

\begin{remark}
Uniform convergence implies pointwise convergence, but the converse is false (as the previous example shows).
\end{remark}

\section{Continuity of Uniform Limits}

The following theorem is one of the central results justifying the study of uniform convergence.

\begin{theorem}[Uniform Limit of Continuous Functions]
\label{thm:unif-cont}
Let $E \subseteq \mathbb{R}$ and let $(f_n)$ be a sequence of continuous functions $f_n: E \to \mathbb{R}$ that converges uniformly to $f: E \to \mathbb{R}$. Then $f$ is continuous on $E$.
\end{theorem}

\begin{proof}
Fix $x_0 \in E$ and let $\varepsilon > 0$. Since $f_n \to f$ uniformly, there exists $N$ such that $\|f_N - f\|_\infty < \varepsilon/3$. Since $f_N$ is continuous at $x_0$, there exists $\delta > 0$ such that $|x - x_0| < \delta$ implies $|f_N(x) - f_N(x_0)| < \varepsilon/3$. Then for $|x - x_0| < \delta$:
\begin{align*}
  |f(x) - f(x_0)| &\leq |f(x) - f_N(x)| + |f_N(x) - f_N(x_0)| + |f_N(x_0) - f(x_0)| \\
  &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon.
\end{align*}
Since $x_0$ was arbitrary, $f$ is continuous on $E$.
\end{proof}

\section{The Weierstrass $M$-Test}

For series of functions $\sum f_n$, the following criterion provides a practical tool for establishing uniform convergence.

\begin{theorem}[Weierstrass $M$-Test]
\label{thm:m-test}
Let $(f_n)$ be a sequence of functions on $E$, and suppose there exist constants $M_n \geq 0$ such that $|f_n(x)| \leq M_n$ for all $x \in E$ and all $n$. If $\sum_{n=1}^{\infty} M_n < \infty$, then $\sum_{n=1}^{\infty} f_n$ converges uniformly and absolutely on $E$.
\end{theorem}

\begin{proof}
For any $x \in E$, absolute convergence follows from comparison: $\sum |f_n(x)| \leq \sum M_n < \infty$. For uniform convergence, let $S_N(x) = \sum_{n=1}^{N} f_n(x)$ and $S(x) = \sum_{n=1}^{\infty} f_n(x)$. Then
\[
  |S(x) - S_N(x)| = \left|\sum_{n=N+1}^{\infty} f_n(x)\right| \leq \sum_{n=N+1}^{\infty} M_n.
\]
The right side is the tail of a convergent series and is independent of $x$, so it tends to $0$ as $N \to \infty$.
\end{proof}

\begin{example}
Consider the series $\displaystyle\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^2}$ for $x \in \mathbb{R}$. Since $|\sin(nx)/n^2| \leq 1/n^2$ and $\sum 1/n^2 = \pi^2/6 < \infty$, the Weierstrass $M$-test implies uniform convergence on all of $\mathbb{R}$. By Theorem~\ref{thm:unif-cont}, the sum is continuous.
\end{example}

\section{Term-by-Term Integration}

\begin{theorem}
\label{thm:term-int}
Let $(f_n)$ be a sequence of Riemann-integrable functions on $[a,b]$ converging uniformly to $f$ on $[a,b]$. Then $f$ is Riemann-integrable and
\[
  \int_a^b f(x)\, dx = \lim_{n \to \infty} \int_a^b f_n(x)\, dx = \int_a^b \lim_{n \to \infty} f_n(x)\, dx.
\]
\end{theorem}

\begin{corollary}
If $\sum f_n$ converges uniformly on $[a,b]$ and each $f_n$ is integrable, then
\[
  \int_a^b \sum_{n=1}^{\infty} f_n(x)\, dx = \sum_{n=1}^{\infty} \int_a^b f_n(x)\, dx.
\]
\end{corollary}

\begin{example}
Since $\sum \sin(nx)/n^2$ converges uniformly on $[0, \pi]$, we may integrate term by term:
\[
  \int_0^{\pi} \sum_{n=1}^{\infty} \frac{\sin(nx)}{n^2}\, dx = \sum_{n=1}^{\infty} \frac{1}{n^2}\int_0^{\pi} \sin(nx)\, dx = \sum_{n=1}^{\infty} \frac{1}{n^2} \cdot \frac{1 - \cos(n\pi)}{n} = 2\sum_{k=0}^{\infty} \frac{1}{(2k+1)^3}.
\]
\end{example}

\vspace{8pt}
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\textbf{Next lecture:} Equicontinuity and the Arzel\`{a}--Ascoli theorem.\\
\textbf{Homework:} Rudin Chapter 7, Exercises 1, 2, 4, 6, 8, 10.

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