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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?

Julian
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    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

2 Answers2

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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 $L = \max \{L_1, L_2\}$.

N.U.
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  • Follow the link above: you need locally bounded derivative (which may be implied in your answer, but deserves mention). – gnometorule May 19 '13 at 12:12
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    @gnometorule I didn't use that. For what part of my proof do I need that? Isn't that counterexample for global Lipschitz continuity? Here it is just one point. – N.U. May 19 '13 at 12:15
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It is not true as this example from Wikipedia (https://en.wikipedia.org/wiki/Lipschitz_continuity) proves:

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 \begin{equation} \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 \, , \end{equation} 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 $|f(x)-f(y)|\leq L|x-y|\, \forall\, x,\,y\in[0,\epsilon]$.

mortenmcfish
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