If f:[a,b]→R is continuous on a closed interval and differentiable on (a,b), then there exists ξ∈[a,b] such that f(b)−f(a)=f′(ξ)(b−a).
More generally, if g:[a,b]→R is similarly continuous on [a,b] and differentiable on (a,b), then there exists a ξ with (f(b)−f(c))g′(ξ)=(g(b)−g(a))f′(ξ). What this means graphically:
If {fn} is a sequence of functions where
- each fn is differentiable,
- there is some G such that ‖, 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 {\mathbb{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.