# Differentiability

If $$f: [a, b] \to {\mathbf{R}}$$ is continuous on a closed interval and differentiable on $$(a, b)$$, then there exists $$\xi \in [a, b]$$ such that \begin{align*} f(b) - f(a) = f'(\xi)(b-a) .\end{align*}

More generally, if $$g: [a,b]\to {\mathbf{R}}$$ is similarly continuous on $$[a, b]$$ and differentiable on $$(a, b)$$, then there exists a $$\xi$$ with \begin{align*} \qty{ f(b) - f(c) } g'(\xi) = \qty{g(b) - g(a)} f'(\xi) .\end{align*} What this means graphically:

If $$\left\{{f_n}\right\}$$ is a sequence of functions where

• each $$f_n$$ is differentiable,
• there is some $$G$$ such that $${\left\lVert { \sum_{n\leq N} f_n' - G} \right\rVert}_\infty \overset{N\to\infty}\longrightarrow 0$$, and
• there exists at least one point 1 $$x_0$$ such that $$\sum f_n(x)$$ converges (pointwise),

then there exists an $$F$$ such that 2 \begin{align*} {\left\lVert { \sum_{n\leq N} f_n - F} \right\rVert}_\infty \overset{N\to\infty}\longrightarrow 0 && F' = g .\end{align*}

A function $$f: (a, b) \to {\mathbf{R}}$$ is Lipschitz $$\iff f$$ is differentiable and $$f'$$ is bounded. In this case, $${\left\lvert {f'(x)} \right\rvert} \leq C$$, the Lipschitz constant.

\begin{align*} f(x) \coloneqq \begin{cases} x^2 \sin\qty{1\over x^2} & x\neq 0 \\ 0 & x=0. \end{cases} .\end{align*}

Note that $$f$$ is differentiable at $$x=0$$ since $${1\over h}{\left\lvert {f(h) - f(0)} \right\rvert} = {\left\lvert { h\sin\qty{h^{-2}}} \right\rvert}\leq {\left\lvert {h} \right\rvert}\to 0$$, and \begin{align*} f'(x) = 2x\sin\qty{1\over x^2 } - \qty{2\over x}\cos\qty{1\over x^2} \chi_{x\neq 0} .\end{align*} now take the sequence $$x_n \coloneqq 1/\sqrt{k\pi}$$ to get $$f'(x_n) = 2\sqrt{k\pi}(-1)^k \overset{n\to\infty}\longrightarrow\infty$$.

Footnotes
1.
So this implicitly holds if $$f$$ is the pointwise limit of $$f_n$$.
2.
See Abbott theorem 6.4.3, pp 168.