Uniform Convergence, Integration, and Differentiation
Theorem 8.4.1
Suppose $f_n\in \mathscr{R}[a,b]$ for all $n\in\mathbb{N}$ ans suppose that the sequence $\{f_n\}$ converges uniformly to $f$ on $[a,b]$. Then $f\in\mathscr{R}[a,b]$ and
\[\begin{align*} \int_a^bf(x) dx=\lim_{n\to\infty}\int_a^bf_n(x)dx. \end{align*}\]<Proof>
Define
\[\begin{align*} \epsilon_n :=\sup_{x\in[a,b]}\lvert f_n(x) -f(x)\rvert \end{align*}\]for all $n\in\mathbb{N}$.
Since $f_n \rightrightarrows f$, we get $\lim_{n\to\infty}\epsilon_n=0$. For each $x\in [a,b]$,
\[\begin{align*} f_n(x)- \epsilon_n \leq f(x) \leq f_n(x) + \epsilon_n. \end{align*}\]It implies that
\[\begin{align*} \int_a^b (f_n-\epsilon_n) \leq \underline{\int}_a^b f \leq \overline{\int}_a^bf \leq \int_a^b (f_n+\epsilon_n). \end{align*}\]Therefore,
\[\begin{align*} 0\leq \overline{\int}_a^bf - \underline{\int}_a^b f \leq 2\epsilon_n(b-a) \end{align*}\]Since $2\epsilon_n (b-a) \to 0$ as $n\to \infty$, $f\in\mathscr{R}[a,b]$. Since $f\in\mathscr{R}[a,b]$,
\[\begin{align*} \int_a^b (f_n-\epsilon_n) \leq \int_a^b f \leq \int_a^b (f_n+\epsilon_n). \end{align*}\]Then,
\[\begin{align*} \left\lvert \int_a^b f(x)dx - \int_a^bf_n(x) \right\rvert \leq 2\epsilon_n(b-a). \end{align*}\]Since $2\epsilon_n(b-a)\to 0$ as $n\to\infty$,
\[\begin{align*} \int_a^b f_n(x) dx = \lim_{n\to\infty}\int_a^b f_n(x)dx. \end{align*}\] \[\tag*{$\square$}\]Corollary 8.4.2
If $f_k\in\mathscr{R}[a,b]$ for all $k\in\mathbb{N}$, and if
\[\begin{align*} f(x) = \sum_{k=1}^\infty f_k(x), \quad x\in [a,b], \end{align*}\]where the series converges uniformly on $[a,b]$, then $f\in\mathscr{R}[a,b]$ and
\[\begin{align*} \int_a^b f(x) dx = \sum_{k=1}^\infty f_k(x)dx. \end{align*}\]Apply the previous theorem to $S_n(x) = \sum_{k=1}^n f_k(x)$, which by Theorem 6.2.1 is integrable for each $n\in\mathbb{N}$.
\[\tag*{$\square$}\]Theorem 8.4.3 (Bounded Convergence Theorem)
Suppose $f$ and $f_n$ are Reimann integrable functions on $[a,b]$ for all $n\in\mathbb{N}$ with $\lim_{n\to\infty}f_n(x)=f(x)$ for all $x\in [a,b]$. Suppose also that there exists a positive constant $M$ such that $\lvert f_n(x) \rvert\leq M$ for all $x\in [a,b]$ and all $n\in\mathbb{N}$. Then
\[\begin{align*} \lim_{n\to\infty}\int_a^b f_n(x) dx = \int_a^b f(x)dx \end{align*}\]Theorem 8.5.1
Suppose $\{f_n\}$ is a sequence of differentiable functions on $[a,b]$. If
(a) $\{f^\prime_n\}$ converges uniformly on $[a,b]$, and
(b) $\{f_n(x_0)\}$ converges for some $x_0\in [a,b]$,
then $\{f_n\}$ converges uniformly to a function $f$ on $[a,b]$, with
\[\begin{align*} f^\prime(x) = \lim_{n\to\infty}f^\prime_n(x). \end{align*}\]<Proof>
Let $\epsilon>0$ be given. By our assumptions, there exists $n_0\in \mathbb{N}$ such that
\[\begin{align*} \lvert f_n(x_0) - f_m(x_0) \rvert <\frac{\epsilon}{2} \end{align*}\]for all $m,n \geq n_0$ and
\[\begin{align*} \lvert f^\prime_n (t) - f^\prime_m(t)\rvert < \frac{\epsilon}{2(b-a)} \end{align*}\]for all $m,n \geq n_0$ and for all $t\in[a,b]$.
By mean value theorem for $m, n \geq n_0$,
\[\begin{align} \lvert (f_n(x) - f_m(x)) - (f_n(y) - f_m(y))\rvert &= \lvert (f^\prime_n(t) - f^\prime_m(t))(x-y)\rvert \label{eq:1} \end{align}\]for some $t\in (x,y)$. Take $y=x_0$ in equation $\ref{eq:1}$. Then
\[\begin{align*} \lvert f_n(x) - f_m(x) \rvert &\leq \lvert f_n(x) - f_m(x) - (f_n(x_0) -f_m(x_0)) \rvert + \lvert f_n(x_0) - f_m(x_0)\rvert \\ &= \lvert (f^\prime_n (t) - f^\prime_m(t))(x-x_0)\rvert + \lvert f_n(x_0) - f_m(x_0)\rvert \\ &< \frac{\epsilon}{2(b-a)} (b-a) + \frac{\epsilon}{2}\\ &=\epsilon \end{align*}\]By Cauchy criterion, $\{f_n\}$ converges uniformly on $[a,b]$. Let $f(x) = \lim_{n\to\infty}f_n(x)$ for each $x\in [a,b]$. We want to show that $f$ is differentiable on $[a,b]$ and $f^\prime(x) = \lim_{n\to\infty}f^\prime_n(x)$ for all $x\in[a,b]$.
Fix $p\in [a,b]$. Define
\[\begin{align*} g_n(t) := \frac{f_n(t) - f_n(p)}{t-p}, \quad g(t) = \frac{f(t)-f(p)}{t-p}. \end{align*}\]Then $g_n(t) \to g_(t)$ for each $t\in [a,b] \setminus \{ p\}$ and
\[\begin{align*} \lim_{t\to p} g_n(t) = \lim_{t\to p} \frac{f_n(t) - f_n(p)}{t-p} = f^\prime_n (p). \end{align*}\]Let $A:= [a,b] \setminus \{p\}$. Take $y=p$ in equation $\ref{eq:1}$,
\[\begin{align*} \lvert g_n(t) - g_m (t) \rvert &= \left\lvert \frac{f_n(t) - f_n(p)}{t-p} - \frac{f_m(t) - f_n(t)}{t-p}\right\rvert \\ &=\frac{1}{\lvert t-p \rvert} \left\lvert (f_n(t) - f_m(t)) - (f_n(p) - f_m(p))\right\rvert \\ &=\frac{1}{\lvert t-p \rvert} \left\lvert (f^\prime_n(c) - f^\prime_m(c)) (t-p) \right\rvert \\ &< \frac{\epsilon}{2(b-a)} \end{align*}\]for some $c \in [t,p]$ and for all $m,n \geq n_0$.
$\therefore \{g_n\}$ converges uniformly to $g$ on $A$.
\[\begin{align*} f^\prime(p)&=\lim_{t\to p} g(t)\\ &= \lim_{t\to p} \left(\lim_{n\to\infty}g_n(t) \right)\\ &=\lim_{n\to\infty}\left(\lim_{t\to p} g_n(t) \right) \\ &=\lim_{n\to\infty}f^\prime_n (p) \end{align*}\] \[\tag*{$\square$}\]Example 8.5.2
Consider the series
\[\begin{align*} \sum_{k=1}^\infty \frac{\sin kx}{2^k}. \end{align*}\]Since $\lvert 2^{-k}\sin kx \rvert \leq 2^{-k}$ for all $x\in \mathbb{R}$ by the Weierstrass M-test this series converges uniformly to a function $S$ on $\mathbb{R}$. For $n\in\mathbb{N}$, let
\[\begin{align*} S_n(x) = \sum_{k=1}^n \frac{\sin kx}{2^k}. \end{align*}\]Then
\[\begin{align*} S^\prime_n(x) = \sum_{k=1}^n \frac{k\cos kx}{2^k}. \end{align*}\]Since $\sum k 2^{-k}$ converges, by the Weierstrass M-test the sequence $\{S^\prime_n\}$ converges uniformly on $\mathbb{R}$. Thus by Theorem 8.5.1,
\[\begin{align*} S^\prime(x) = \lim_{n\to\infty}S^\prime_n(x) = \sum_{k=1}^\infty \frac{k \cos kx}{2^k}. \end{align*}\]Continuous but nowhere differentiable function
Let $\varphi(x) = \lvert x \rvert, -1\leq x \leq 1$. Extend it to a periodic function $\varphi(x+2) = \varphi(x)$. Observe that
\[\begin{align*} \lvert \varphi(s) - \varphi(t) \rvert &= \lvert \lvert s \rvert - \lvert t\rvert \rvert \\ &\leq \lvert s- t\rvert \end{align*}\]for all $s,t \in [-1,1]$, i.e., $\varphi$ is Lipschitz continuous function.
Define
\[\begin{align*} f(x) = \sum_{n=0}^\infty \left( \frac{3}{4}\right)^n \varphi(4^nx). \end{align*}\]By Weierstrass M-test, the series converges uniformly. Since $f_k(x) = \sum_{n=0}^k\left(\frac{3}{4}\right)^n\varphi(4^nx)$ is continuous, $f$ is continuous.
We want to show that $f$ is non-differentiable.
<Proof>
Pick any $x\in\mathbb{R}$. For each $m\in\mathbb{N}$, define
\[\begin{align*} \delta_m = \delta_m(x) := \pm\frac{1}{2}4^{-m}, \end{align*}\]the sign is chosen such that no integer lies between $4^mx$ and $4^m(x+\delta_m)$. Note that $\lvert 4^m\delta_m\rvert = 1/2$. Consider
\[\begin{align*} \left\lvert \frac{f(x+\delta_m) -f(x)}{\delta_m} \right\rvert &=\sum_{n=0}^\infty \frac{\left(\frac{3}{4}\right)^n[\varphi(4^n(x+\delta_m))-\varphi(4^nx)]}{\delta_m}. \end{align*}\]Define
\[\begin{align*} \gamma_n = \frac{\varphi(4^n(x+\delta_m))-\varphi(4^nx))}{\delta_m}. \end{align*}\](1) If $n>m$, then $4^n\delta_m$ is an even integer. Since $\varphi(x+2)=\varphi(x), \gamma_n=0$.
(2) If $n<m$,
\[\begin{align*} \lvert \varphi(4^n(x+\delta_m))-\varphi(4^n\delta_m)\rvert \leq \lvert 4^n\delta_m\rvert \end{align*}\]by Lipschitz continuity of $\varphi$.
(3) If $n=m$,
\[\begin{align*} \frac{\left\lvert \varphi(4^mx\pm \frac{1}{2})-\varphi(4^mx)\right\rvert}{\left\lvert\pm\frac{1}{2}\right\rvert} =1. \end{align*}\]In other words, $\lvert \varphi(4^mx\pm \frac{1}{2})-\varphi(4^mx)\rvert = \frac{1}{2}$. Thus, $\lvert \gamma_n\rvert = 4^n$.
\[\begin{align*} \left\lvert \frac{f(x+\delta_m) -f(x)}{\delta_m}\right\rvert &=\left\lvert \sum_{n=0}^m \left(\frac{3}{4} \right)^n \gamma_n\right\rvert \quad (\because \gamma_n=0, \forall n >m) \\ &=\left\lvert \left(\frac{3}{4}\right)^m4^m + \sum_{n=0}^{m-1}\left(\frac{3}{4}\right)^n \gamma_n\right\rvert \\ &\geq 3^m -\sum_{n=0}^{m-1}\left(\frac{3}{4}\right)^n \lvert \gamma_n\rvert \\ &\geq 3^m - \sum_{n=0}^{m-1}\left(\frac{3}{4}\right)^n 4^n \quad (\because \vert\gamma \rvert \leq 4^n, \forall n <m) \\ &=3^m -\frac{3^m-1}{2} \\ &=\frac{1}{2}(3^m+1) \end{align*}\]$\therefore f$ is nowhere differentiable.
\[\tag*{$\square$}\]Reference
- Manfred Stoll, 『Introduction to Real Analysis』