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\date{26 March 2018}
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\begin{document}

\begin{frame}{Pointwise and uniform convergence}
\textbf{Definition} (``pointwise convergence"): Let $f_n \colon [a,b] \rightarrow \mathbb{R}$ be a sequence of functions. We say that $f_n$ converges pointwise to $f$ provided that $\forall x \in[a,b] \forall \epsilon>0 \exists N \in \mathbb{N}$ so that for all $n \geq N, |f_n(x)-f(x)|<\epsilon$. For this, we write $f_n \rightarrow f$. 

\textbf{Definition} (``uniform convergence"): Let $f_n \colon [a,b] \rightarrow \mathbb{R}$ be a sequence of functions. We say that $f_n$ converges uniformly to $f$ provided that $\forall \epsilon > 0 \exists N \in \mathbb{N}$ so that for all $n \geq N$ and $x \in [a,b]$, $|f_n(x)-f(x)| < \epsilon$. For this, we write $f_n \rightrightarrows f$.
\end{frame}

\begin{frame}{Cauchy sequence-like theorem}
\textbf{Theorem}: Suppose that $f_n \colon[a,b] \rightarrow \mathbb{R}$. Then $f_n \rightrightarrows f$ if and only if $\forall \epsilon > 0 \exists N \in \mathbb{N}$ such that for all $m, n \geq N$ and for all $x \in [a,b]$, $|f_n(x)-f_m(x)| < \epsilon$.  \\
\onslide<2->{\textit{Proof}: $(\rightarrow)$ Let $\epsilon > 0$. Suppose that $f_n \rightrightarrows f$. Then there is $N \in \mathbb{N}$ so that if $n, m \geq N$ and $x \in [a,b]$, then 
$$|f_n(x) - f(x)| < \dfrac{\epsilon}{2}$$
and
$$|f_m(x) - f(x)| < \dfrac{\epsilon}{2}.$$
\onslide<3->{Therefore 
$$|f_n(x)-f_m(x)| \leq |f_n(x)-f(x)| + |f_m(x)-f(x)| < \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} = \epsilon.$$}}
\end{frame}

\begin{frame}
$(\leftarrow)$ Let $\epsilon > 0$. By assumption, there is $N \in \mathbb{N}$ so that for all $m, n \geq N$ and $x \in [a,b]$, 
$$|f_n(x)-f_m(x)| < \epsilon.$$
\onslide<2->{Since $f_m(x)$ is a sequence in $\mathbb{R}$ that converges (why: we showed ``being a convergent sequence" is the same as ``being a Cauchy sequence" a long time ago) to some $f(x)$, we take the limit as $m \rightarrow \infty$ of the above expression which yields
$$|f_n(x)-f(x)| < \epsilon. \blacksquare$$}
\end{frame}

\begin{frame}{Properties of uniform convergence}
\textbf{Theorem}: Suppose that $f_n \colon [a,b] \rightarrow \mathbb{R}$, that $f_n \rightrightarrows f$, let $x \in [a,b]$, and suppose that $\displaystyle\lim_{t \rightarrow x} f_n(t)=A_n$. Then $A_n$ is a convergent sequence and 
$$\displaystyle\lim_{t \rightarrow x} f(t)=\displaystyle\lim_{n \rightarrow \infty} A_n.$$
\onslide<2->{\textit{Proof}: Let $\epsilon > 0$. Since $f_n \rightrightarrows f$, there exists $N \in \mathbb{N}$ so that for all $n,m \geq N$ and for all $t \in [a,b]$, 
$$|f_n(t)-f_m(t)| \leq \epsilon.$$
Taking the limit as $t \rightarrow x$, we get
$$|A_n - A_m| \leq \epsilon.$$
\onslide<3->{\textbf{Goal:} show that there is large enough $n$ and $t \in [a,b]$ close enough to $x$ (i.e. find a ``$\delta$") so that 
$$|f(t) - A_n| < \epsilon.$$}}
\end{frame}

\begin{frame}{Properties of uniform convergence}
So, since $f_n \rightrightarrows f$, we may choose $n$ large enough so that for all $t \in [a,b]$,
$$|f(t) - f_n(t)| < \dfrac{\epsilon}{3}$$
and
$$|A_n-A| < \dfrac{\epsilon}{3}.$$
Since $\displaystyle\lim_{t \rightarrow x} f_n(t)=A_n$, we may pick a $\delta > 0$ so that if $|t-x| < \delta$, then $|f_n(t)-A_n| < \dfrac{\epsilon}{3}$. 
\end{frame}

\begin{frame}{Properties of uniform convergence}
Therefore, for $t$ so that $|t-x| < \delta$,
$$\begin{array}{ll}
|f(t)-A| &\leq |f(t)-f_n(t)| + |f_n(t)-A| \\
&\leq |f(t)-f_n(t)| + |f_n(t)-A_n| + |A_n-A| \\
&\leq \dfrac{\epsilon}{3} + \dfrac{\epsilon}{3} + \dfrac{\epsilon}{3} \\
&< \epsilon,
\end{array}$$
completing the proof. $\blacksquare$ \\
\textbf{Corollary}: The uniform limit of continuous functions is continuous.
\end{frame}

\begin{frame}{Properties of uniform convergence}
\textbf{Theorem}: Suppose that $f_n \colon [a,b] \rightarrow \mathbb{R}$ are Riemann-integrable and suppose that $f_n \rightrightarrows f$. Then $f$ is Riemann-integrable and
$$\displaystyle\int_a^b f(x) \mathrm{d}x = \displaystyle\lim_{n \rightarrow \infty} \displaystyle\int f_n(x) \mathrm{d}x.$$
\textit{Proof}: Define $\varepsilon_n = \sup\limits_{x \in [a,b]} |f_n(x)-f(x)|$. Then for all $x \in [a,b]$,
$$f_n(x) - \varepsilon_n \leq f(x) \leq f_n(x) + \varepsilon_n.$$
(note 1: we know $\lim \varepsilon_n = 0$) \\
(note 2: we don't know that $f$ is integrable yet! So we need to argue using lower and upper sums)
$$\displaystyle\int_a^b f_n(x) -\varepsilon_n \mathrm{d}x \leq \underline{\displaystyle\int} f(x) \mathrm{d}x \leq \overline{\displaystyle\int} f(x) \mathrm{d}x \leq \displaystyle\int_a^b f_n(x) + \varepsilon_n \mathrm{d}x.$$
\end{frame}

\begin{frame}{Properties of uniform convergence}
$$\displaystyle\int_a^b f_n(x)- \varepsilon_n \mathrm{d}x \leq \underline{\displaystyle\int} f(x) \mathrm{d}x \leq \overline{\displaystyle\int} f(x) \mathrm{d}x \leq \displaystyle\int_a^b f_n(x) + \varepsilon_n \mathrm{d}x.$$
Thus since $\displaystyle\int_a^b f_n(x)- \varepsilon_n \mathrm{d}x \leq \underline{\displaystyle\int} f(x) \mathrm{d}x$ implies
$$-\underline{\displaystyle\int} f(x) \mathrm{d}x \leq -\displaystyle\int_a^b f_n(x)- \varepsilon_n \mathrm{d}x,$$
we see
\begin{multline*}
0 \leq \overline{\displaystyle\int} f(x) \mathrm{d}x - \underline{\displaystyle\int} f(x) \mathrm{d}x \leq \displaystyle\int_a^b f_n(x) + \varepsilon_n \mathrm{d}x- \underline{\displaystyle\int} f(x) \mathrm{d}x \\
\leq \displaystyle\int_a^b f_n(x) + \varepsilon_n \mathrm{d}x-\displaystyle\int_a^b f_n(x)- \varepsilon_n \mathrm{d}x = 2\varepsilon_n (b-a),
\end{multline*}
taking the limit as $n \rightarrow \infty$ completes the proof. $\blacksquare$
\end{frame}

\begin{frame}{Another example}
Let $f_n \colon \mathbb{R} \rightarrow \mathbb{R}$ with $f_n(x) = \dfrac{\sin(nx)}{\sqrt{n}}$. 
\begin{enumerate}[1.]
\item Compute $\displaystyle\lim_{n \rightarrow \infty} f_n(x)$ (call this function $\mathcal{F}(x)$).\onslide<2->{Ans: $\mathcal{F}(x)=0$}
\item Is that convergence uniform? \onslide<2->{Ans: Yes!}
\item Compute $\mathcal{F}'(x)$. \onslide<2->{Ans: $0$}
\item Compute $f_n'(x)$. \onslide<2->{$f_n'(x)=\sqrt{n}\cos(nx)$}
\item Compute $\displaystyle\lim_{n \rightarrow \infty} f_n'(x)$. \onslide<2->{Ans: $\infty$, $-\infty$, or undefined depending on the sign of $\cos(nx)$ For instance, if $x=0$ you get $\displaystyle\lim_{n \rightarrow \infty} \sqrt{n}=\infty$.} \\
\onslide<2->{\textbf{Consequence}: Even the uniform limit of differentiable functions can fail to be differentiable!}
\end{enumerate}
\end{frame}

\end{document}