Locally differentiable implies locally Lipschitz for multivariable functions

Question

Let $f:\mathbb{R}^{n}\to \mathbb{R}^{n}$ s.t. $f$ is differentiable at $\underline{x}$ with $D_{\underline{x}}f=I$ and $f(\underline{x})= \underline{0}$. I want to prove that there exists $r>0$ and $C>0$ s.t. $$ \|f(x_{1})-f(x_{2})\|\leq C \|x_{1}-x_{2}\| \quad \forall x_{1},x_{2}\in B(\underline{x},r) $$ With differentiable I mean the following $$ \begin{align} &\lim_{ \|x-\underline{x}\| \to 0 } \frac{\|f(x)-f(\underline{x})-D_{\underline{x}}f(x-\underline{x})\|}{\|x-\underline{x}\|} \\ =&\lim_{ \|x-\underline{x}\| \to 0 } \frac{\|f(x)-(x-\underline{x})\|}{\|x-\underline{x}\|}=0 \end{align} $$

I don't have the differentiability in other points, in that case I found a proof here

From the definition of differentiable follows that for any $\epsilon>0$ holds $$ \|f(x)-(x-\underline{x})\| \leq \epsilon\|x-\underline{x}\| $$ From there, using the triangular inequality, follows $$ \begin{align} \|f(x_{1})-f(x_{2})\|&=\|f(x_{1})-(x_{1}-\underline{x})+(x_{1}-x_{2})-f(x_{2})-(x_{2}-\underline{x})\| \\ &\geq \|f(x_{1})-(x_{1}-\underline{x})\|+\|x_{1}-x_{2}\|+\|f(x_{2})-(x_{2}-\underline{x})\| \\ &\geq\epsilon(\|x_{1}-\underline{x}\|+\|x_{2}-\underline{x}\|) + \|x_{1}-x_{2}\| \end{align} $$ which it doesn't help me so much.

So what have you tried? There are some standard things you should be writing down. Where do you get stuck? Edit your question to include these. — Ted Shifrin, Feb 01 '23 at 16:34
Thank you for the comment, I updated the question as you asked — Davide Modesto, Feb 01 '23 at 16:45
Try direct triangle inequality? Consider $|f(x_1)-f(x_2)-(x_1-x_2)|$, perhaps? — Ted Shifrin, Feb 01 '23 at 17:25
If I try to use your suggestion I get $|f(x_1)-f(x_2)| \leq \epsilon(|\underline{x}-x_1|+|\underline{x}-x_2|)$ which is similar to the result I wrote in the question. — Davide Modesto, Feb 01 '23 at 17:47
Did you try to find a counter example for functions of one variable? (A function differentiable at one point and discontinuous elsewhere.) — Moishe Kohan, Feb 01 '23 at 17:48
here they give an example of such a function. So it's not true because lipschitz in an open set implies continuity in the same set. How can I mark as solved? — Davide Modesto, Feb 01 '23 at 17:57
You can either delete or post your own answer and accept it. — Moishe Kohan, Feb 01 '23 at 19:40

score 1 · Answer 1 · answered Feb 02 '23 at 08:50

Here is given an example of a function differentiable in one point and discontinuous everywhere else. In particular, since Lipschitz implies continuity, it isn't true that a function differentiable in one point then it's continuous in its neighbourhood.

Locally differentiable implies locally Lipschitz for multivariable functions

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