8
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – robjohn May 19 '13 at 17:43
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Follow the link above: you need locally bounded derivative (which may be implied in your answer, but deserves mention). $\endgroup$ – gnometorule May 19 '13 at 12:12
  • 1
    $\begingroup$ @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. $\endgroup$ – N.U. May 19 '13 at 12:15
0
$\begingroup$

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 $|f(x)-f(y)|\leq L|x-y|\, \forall\, x,\,y\in[0,\epsilon]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.