# Differentiability implies Lipschitz continuity

Let $f:[0,1]\to\mathbb{R}$ be a continuous function and suppose $f$ is differentiable at $x_0\in [0,1]$. Is it true that there exists $L>0$ such that $\lvert f(x)-f(x_0)\lvert\leq L\lvert x-x_0\lvert$?

I know that local continuously differentiable implies local Lipschitz continuity. Is this still true in the case given above?

• This is not a duplicate. This question asks about a continuous function on a compact set. The cited question does not specify either continuity or compactness. – robjohn May 19 '13 at 17:43

From differentiability at $x_0$, you will find an $L_1$ such that $| f(x)-f(x_0) |\leq L_1| x-x_0|$ for $|x-x_0| < \delta$. Since $f$ is continuous on a compact set, $|f(x)|_{\infty} < \infty$. This will give you an $L_2$ such that $| f(x)-f(x_0) |\leq L_2| x-x_0|$ for $|x-x_0| \geq \delta$. Take $L = \max \{L_1, L_2\}$.
Let $$f:[0,1]\to\mathbb{R},\, f(x)=x^{\frac{3}{2}}\sin(\frac{1}{x}),\,x>0,\, f(0)=0$$. Then it's derivate outside $$0$$ is $$f^`(x)=\frac{3}{2}x^{\frac{1}{2}}\sin(\frac{1}{x})-x^{-\frac{1}{2}}\cos(\frac{1}{x})$$ and note that $$\limsup_{x\downarrow 0}|f^´(x)|=\infty$$.
Furthermore $$f$$ is differentiable in $$0$$ since $$$$\lim_{h\downarrow 0}\left\lvert\frac{h^{\frac{3}{2}}\sin(\frac{1}{h})}{h}\right\rvert\leq\lim_{h\downarrow 0}\left\lvert h^{\frac{1}{2}}\right\rvert=0 \, ,$$$$ so $$f$$ actually meets your requirements.
But the unbounded derivative outside $$0$$ gives us that there is no $$\epsilon > 0$$ and $$L\geq0$$ such that $$|f(x)-f(y)|\leq L|x-y|\, \forall\, x,\,y\in[0,\epsilon]$$.