Let $f:U\rightarrow\mathbb{R}^n$ be a differentiable function over an open convex set $U\subset\mathbb{R}^m$. Show that

$\sup_{x\neq y}\dfrac{|f(x)-f(y)|}{|x-y|}=\sup_{z\in U}|(Df)_z|.$

I know I must use the Mean Value theorem to show this, and since the set $U$ is convex I will have the equality, but how can I reach this equation?

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The difference quotient gets arbitrarily close to the derivative as $x \to y$. Does this help? I think you can do it just based on the definition of the derivative if you like. – Patrick Apr 3 '12 at 2:57

I'm supposing $f \in C^1(U)$. As you noted, by using the Mean Value Theorem one can show that for every $x,y \in U$ we have $$|f(x) - f(y)| \le \sup_{z \in U} \|Df(z)\| |x-y|.$$ For the version of Mean Value Theorem used, see the section "Mean value theorem for vector-valued functions" at http://en.wikipedia.org/wiki/Mean_value_theorem. This of course implies the inequality $$\frac{|f(x) - f(y)|}{|x-y|} \le \sup_{z \in U} \|Df(z)\|.$$ To show the other direction, let $\varepsilon > 0$ be arbitrary and let $z_n \in U$, $n=1,2,\dots$, be a sequence of points so that $$\|Df(z_n)\| \to \sup_{z \in U} \|Df(z)\|.$$ Then by the definition of differentiability for every $n \in \mathbb{N}$ there exists a neighbourhood $U$ of $0$ in $\mathbb{R}^n$ so that for every $t_n \in U$ we have $$\frac{|f(z_n + t_n) - f(z_n) - Df(z_n)t_n|}{|t_n|} \le \varepsilon.$$ By the triangle inequality we hence have that $$\left|\left|\frac{f(z_n + t_n) - f(z_n)}{|t_n|}\right| - \left|\frac{Df(z_n)t_n}{|t_n|}\right|\right| \le \varepsilon.$$ By choosing $t_n$ so that $\frac{|Df(z_n)t_n|}{|t_n|} = \|Df(z_n)\|$ and $z_n + t_n \in U$, we have found $x_n = z_n + t_n$ and $y_n = z_n$ so that $$\left|\left|\frac{f(x_n) - f(y_n)}{|x_n - y_n|}\right| - \|Df(z_n)\|\right| \le \varepsilon.$$ Because $\varepsilon > 0$ is arbitrary and $\|Df(z_n)\| \to \sup_{z \in U} \|Df(z)\|$ we get the result.
It looks ok, thank you! But where did you use the fact that $f$ is $C^1$? – Marra Apr 4 '12 at 0:02
@J I have one question: why can I choose $t_n$ such that $\dfrac{|Df(z_n)t_n|}{|t_n|}=||Df(z_n)||$? – Marra Apr 5 '12 at 13:06
@Gustavo Marra: The norm of an operator $T \colon \mathbb{R}^n \to \mathbb{R}^n$ is defined by $\|T\| = \sup_{|x| = 1} |Tx|$. Because the sphere $S^{n-1}$ is compact, there exists such $x$ for which $|Tx| = \|T\|$. Notice that we also have $\frac{|T\lambda x|}{|\lambda x|} = \|T\|$ for any $\lambda > 0$. In particular if $T = Df(z_n)$ and $x$ is a unit vector with $|Df(z_n) x| = \|Df(z_n)\|$. Because $U$ is open, we can choose $\lambda x = t_n$ so that $\frac{|Df(z_n) t_n|}{|t_n|} = \|Df(z_n)\|$ and also $z_n + t_n \in U$ (by choosing $\lambda > 0$ to be small enough). – J. J. Apr 7 '12 at 13:58