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if $f$ is differentiable at a point $x$, is $f$ also necessary Lipshitz at $x$?

Since $f$ is differentiable at $x$, $f$ is also continuous at $x$. Then we have the $\varepsilon$-$\delta$ definition of $f$ continuous at $x$, and also we have $f'(x)$ exists. But i have no idea how to connect them together. Since $f$ is only differentiable at a point, i don't think mean value theorem is gonna work here either.

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marked as duplicate by Jonas Meyer, Stefan Hansen, Alexander Gruber, Ittay Weiss, Davide Giraudo Feb 15 '13 at 10:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

what do you mean by Lipschitz-continuous? – Ittay Weiss Feb 14 '13 at 3:32
well, basically if f is Lipschitz-continuous at Xo that means there's some non-negative constant C and some interval I=(Xo-a,Xo+a)around x such that |f(x)-f(Xo)|is smaller than or equal to C|x-Xo|. – Yunhui Shi Feb 14 '13 at 3:37
right. Locally lipschitz-continuous – Yunhui Shi Feb 14 '13 at 3:42
up vote 2 down vote accepted

If $f$ is differentiable at $x_0$, then for all $\epsilon>0$, there exists a $\delta>0$ such that if $\|x-x_0\|< \delta$, then $\|f(x)-f(x_0)-DF(x_0)(x-x_0)\| \leq \epsilon \|x-x_0\|$. S, choose $\epsilon=1$, then the estimate gives $\|f(x)-f(x_0)\|-\|DF(x_0)(x-x_0)\| \leq \|x-x_0\|$, which can be written as $\|f(x)-f(x_0)\| \leq (1+\|Df(x_0)\|) \|x-x_0\|$. Setting $L = 1+\|Df(x_0)\|$ shows that $\|f(x)-f(x_0)\| \leq L \|x-x_0\|$ locally.

However, being differentiable at a point does not mean that $f$ is locally Lipschitz. For example, let $f(x) = x^2 \sin \frac{1}{x^2}$. $f$ is differentiable at $x=0$, but is not locally Lipschitz around $x=0$ ($f'(x)$ is unbounded near $x=0$).

If the function is differentiable, then it is locally Lipschitz in a neighborhood iff the derivative is bounded on the neighborhood .

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Thank you, that was helpful. – Yunhui Shi Feb 14 '13 at 4:13

Lipschitz continuity is stronger than mere continuity. If the function $f$ is continuously differentiable at $x_0$, you can find a Lipschitz constant using a Taylor series or a mean value theorem for the derivative which will assume its finite maximum on a small compact interval around $x_0$ (which will be the local Lipschitz constant). The same holds for any function for which the derivative is locally bounded near $x_0$. But there are functions which are differentiable at a point but not Lipshitz continuous. I blanked on an example, and stole this from Wikipedia - a differentiable function on the compact set $[0, 1]$ that is not Lipshitz:

$$f(x) = x^{\frac{3}{2}} \sin(\frac{1}{x}), \, x ≠ 0; \quad f(0) = 0$$

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thank you very much – Yunhui Shi Feb 14 '13 at 4:14
You're welcome! – gnometorule Feb 14 '13 at 4:19
I don't understand the part about Taylor series for $C^1$ functions. – Jonas Meyer Feb 15 '13 at 7:42
@JonasMeyer: Neither do I. :) Thanks for catching the imprecision. Should be $C^{\infty}$ then. – gnometorule Feb 15 '13 at 8:09

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