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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}{Mean value theorem, Cauchy Criterion, and limit theorem}
\textbf{Theorem}: (Mean Value Theorem) Let $f \colon [a,b] \rightarrow \mathbb{R}$ be continuous and differentiable on $(a,b)$. Then there exists $\theta \in (a,b)$ so that 
$$\dfrac{f(b)-f(a)}{b-a} = f'(\theta).$$

\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 $x \in [a,b]$, $|f_n(x)-f_m(x)| < \epsilon$.  \\

\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 
$$\displaystyle\lim_{t \rightarrow x} f(t)=\displaystyle\lim_{n \rightarrow \infty} A_n.$$

\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 $\mathcal{F}(x)$). Ans: $\mathcal{F}(x)=0$
\item Is that convergence uniform? Ans: Yes!
\item Compute $\mathcal{F}'(x)$. Ans: $0$
\item Compute $f_n'(x)$. Ans: $f_n'(x)=\sqrt{n}\cos(nx)$
\item Compute $\displaystyle\lim_{n \rightarrow \infty} f_n'(x)$. Ans: $\infty$, $-\infty$, or undefined depending on the sign of $\cos(nx)$ For instance, if $x=0$ you get $\displaystyle\lim_{n \rightarrow \infty} \sqrt{n}=\infty$. \\
\textbf{Consequence}: Even the uniform limit of differentiable functions can fail to be differentiable!
\end{enumerate}
\end{frame}

\begin{frame}{Uniform convergence and differentiation}
\textbf{Theorem}: Suppose $f_n \colon [a,b] \rightarrow \mathbb{R}$ is a sequence of differentiable functions for which $f_n(x_0)$ converges. If $f_n'$ converges uniformly on $[a,b]$, then $f_n$ converges uniformly on $[a,b]$ to a function $f$ and 
$$f'(x) = \displaystyle\lim_{n \rightarrow \infty} f_n'(x).$$
\textbf{Proof}: Let $\epsilon > 0$. Since $f_n(x_0)$ converges, choose $N \in \mathbb{N}$ so that if $n, m \geq N$ implies 
$$\left|f_n(x_0) - f_m(x_0) \right| < \dfrac{\epsilon}{2}.$$
Since $f_n'$ converges uniformly, the $N$ earlier can be chosen large enough so that for \textbf{all} $t \in [a,b]$,
$$|f_n'(t)-f_m'(t)| < \dfrac{\epsilon}{2(b-a)}.$$
\end{frame}

\begin{frame}{Uniform convergence and differentiation}
Since $f_n$ and $f_m$ are differentiable, $f_n-f_m$ is differentiable. Let $x, t \in [a,b]$ be any two points (without loss of generality, assume $t<x$). By the mean value theorem applied to the function $g(x)=f_n(x)-f_m(x)$, there exists $\theta \in [a,b]$ so that
$$\dfrac{g(x)-g(t)}{x-t} = g'(\theta),$$
in other words and taking absolute value,
$$\dfrac{\left|(f_n(x)-f_m(x)) - (f_n(t)-f_m(t)) \right|}{|x-t|} = |f_n'(\theta)-f_m'(\theta)|.$$
\end{frame}

\begin{frame}{Uniform convergence and differentiation}
So we have: for \textbf{all} $t \in [a,b]$, $|f_n'(t) - f_m'(t)| < \dfrac{\epsilon}{2(b-a)}$ and
\begin{multline*}
\left|(f_n(x)-f_m(x)) - (f_n(t)-f_m(t)) \right| = |x-t||f_n'(\theta)-f_m'(\theta)| < \dfrac{|x-t| \epsilon}{2(b-a)} < \dfrac{\epsilon}{2}. 
\end{multline*}
Compute (how?)
$$|f_n(x)-f_m(x)| \leq |f_n(x)-f_m(x) - f_n(x_0)+f_m(x_0)| + |f_n(x_0)-f_m(x_0)|.$$
\end{frame}

\begin{frame}{Uniform convergence and differentiation}
We have 
$$|f_n(x_0)-f_m(x_0)| < \dfrac{\epsilon}{2},$$
and for \textbf{all} $x, t \in [a,b]$, $\left|(f_n(x)-f_m(x)) - (f_n(t)-f_m(t)) \right| < \dfrac{\epsilon}{2}$
and $|f_n(x)-f_m(x)| \leq \underbrace{|f_n(x)-f_m(x) - f_n(x_0)+f_m(x_0)|}_{<\frac{\epsilon}{2}} + \underbrace{|f_n(x_0)-f_m(x_0)|}_{<\frac{\epsilon}{2}}.$
Therefore we have for any $x \in [a,b]$ and any $m, n \geq N$,
$$|f_n(x)-f_m(x)| < \epsilon.$$
Therefore by the Cauchy criterion, we may conclude that $f_n$ converges uniformly on $[a,b]$.  
\end{frame}

\begin{frame}{Uniform convergence and differentiation}
Define the limit function $f \colon [a,b] \rightarrow \mathbb{R}$ by $f(x) = \displaystyle\lim_{n \rightarrow \infty} f_n(x)$.  Fix $x \in [a,b]$ and define two functions $\phi_n, \phi \colon [a,b] \setminus \{x\} \rightarrow \mathbb{R}$:
$$\phi_n(t) = \dfrac{f_n(t)-f_n(x)}{t-x}, \quad \phi(t) = \dfrac{f(t)-f(x)}{t-x}.$$
By definition of differentiation, for any $n=1,2,3,\ldots$ we get
$$\displaystyle\lim_{t \rightarrow x} \phi_n(t) = f_n'(x).$$
\end{frame}

\begin{frame}{Uniform convergence and differentiation}
In the new notation, we may rewrite the inequality (which held for \textbf{all} $x,t \in [a,b]$)
$$\left|(f_n(x)-f_m(x)) - (f_n(t)-f_m(t)) \right| < \dfrac{|x-t|\epsilon}{2(b-a)}$$
by dividing by $|x-t|$, we get
$$\left| \dfrac{(f_m(t)-f_m(x))-(f_n(t)-f_n(x))}{|x-t|} \right| = |\phi_m(t)-\phi_n(t)| < \dfrac{\epsilon}{2(b-a)}.$$
This establishes by the Cauchy criterion that $\phi_n$ converges uniformly whenever $t \in [a,b] \setminus \{x\}$. 
\end{frame}

\begin{frame}{Uniform convergence and differentiation}
Since  $\phi_n$ converges uniformly and we know that $f_n \rightrightarrows f$, we can say by the definitions of $\phi_n$ and $\phi$ that
$$\displaystyle\lim_{n \rightarrow \infty} \phi_n(t) = \phi(t).$$
Therefore since we also know
$$\displaystyle\lim_{t \rightarrow x} \phi_n(t) = f_n'(x),$$
by the limit theorem,
$$\displaystyle\lim_{t \rightarrow x} \phi(t) = \displaystyle\lim_{n \rightarrow \infty} f_n'(x).$$
So, by definition of $\phi$ we conclude
$$f'(x) = \displaystyle\lim_{n \rightarrow \infty} f_n'(x). \blacksquare$$
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