$\sup_{x \neq y}\frac{||f(x)-f(y)||}{||x-y||}= \sup_{z \in U} ||f'(z)||$ 
Let $f:U \to \mathbb{R}^{m}$ differentiable in the open convex $U \subset \mathbb{R}^{m}$. Prove that $$\sup_{x \neq y}\frac{\|f(x)-f(y)\|}{\|x-y\|}= \sup_{z \in U} \|f'(z)\|$$

by the mean value theorem we have $$\sup_{x \neq y}\frac{\|f(x)-f(y)\|}{\|x-y\|} \leq \sup_{z \in U} \|f'(z)\|$$
the other implication I can't do it, any ideas?

You want a bound for the derivative. Let $$A = \sup_{x \neq y} \frac{\|f(x)-f(y)\|}{\|x-y\|}.$$You know that $$\frac{\|f(z+th)-f(z)\|}{\|th\|} \leq A$$for every $h \neq 0$ and $t \neq 0$ (small enough so that $z+th \in U$). Take the limit $t \to 0$ (using the continuity of $\|\cdot\|$ to get $$\frac{\|f'(z)h\|}{\|h\|} \leq A$$for all $h \neq 0$. Then $\|f'(z)h\| \leq A\|h\|$ for all $h \neq 0$. Then by definition of operator norm, $\|f'(z)\| \leq A$. But $z \in U$ was arbitrary, so take $\sup$ to obtain $\sup_{z \in U}\|f'(z)\| \leq A$.
