# Locally continuously differentiable implies locally Lipschitz

I am interesting in the following result:

Let $X$ be a normed space and $f : X \to \mathbb{R}$. If $f$ is continuously differentiable in a neighborhood $V$ of a point $x_0 \in X$, then $f$ is locally Lipschitz at $x_0$.

Using mean value theorem, for all $x,y$ in an open ball $B \subset V$, there exists $z \in [x,y]$ such that $f(x)-f(y)= \langle f'(z), x-y \rangle$ so $|f(x)-f(y)| \leq \max\limits_{z \in B} ||f'(z)||. ||x-y||$.

In finite dimension, we can suppose that the closure of $B$ is in $V$ so that $\max\limits_{z \in B} ||f'(z)|| < + \infty$ because $z \mapsto f'(z)$ is continuous and the closure of $B$ is compact.

But what happens in infinite dimension? Is the result still true?

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I think you need to assume $\max\limits_{z \in B} ||f'(z)|| < + \infty$ in infinite dimension to make the result true. – Makoto Kato Sep 22 '12 at 10:52
Would it be possible to choose $f$ so that $Df$ is one of the functions constructed in math.stackexchange.com/questions/170615/… ? – Siminore Sep 22 '12 at 10:55
«locally continuously differentialble» is exacty the same thing as «continuously differentiable»! – Mariano Suárez-Alvarez Sep 23 '12 at 3:10

Yes, it is still true. Given $x_0$, use the continuity of $f'$ to find a neighborhood $U$ of $x_0$ such that $\|f'(x)-f'(x_0)\|\le 1$ for all $x\in U$. It follows that $\|f'(x)\|\le \|f'(x_0)\|+1$ in $U$. Now the Mean Value Theorem implies that $f$ is Lipschitz on $U$.