$\sup_{x,y}\frac{|f(x)-f(y)|}{|x-y|} = \sup_x \|D_f(x)\|$

Question

Let $B \subset \mathbb R^n$ be an open ball, $f:B \to \mathbb R^n$ differentiable.

Show that $\sup_{x,y \in B, x\neq y}\frac{|f(x)-f(y)|}{|x-y|} = \sup_{x \in B}\|D_f(x)\|$

What I did:

$\sup_{x \in B}\|D_f(x)\| = \sup_{x \in B} \sup_{|v| = 1}|D_f(x)v| = \sup_{x \in B, |v| = 1}|D_f(x)v| = \sup_{x \in B, |v|=1} |\lim_{t \to 0} \frac{f(x+tv)-f(x)}{t}| = \sup_{x \in B, |v|=1} \lim_{t \to 0} \frac{|f(x+tv)-f(x)|}{|t|}$

And here I am stuck.

If we could somehow magically get rid of the limit, we would be left with something that looks a lot like what we want to show, but how do we do that?

Answer 1

The inequality $$ \sup\limits_{x, y \in B,\ x \ne y} \frac{\lvert f(x) - f(y) \lvert}{\lvert x - y\rvert} \le \sup\limits_{x \in B}\, \lVert D_f(x) \rVert $$ follows from (indeed, is almost the precise statement) of Theorem 11 on p. 277 of: C. C. Pugh, Real Mathematical Analysis, Springer, 2002 (the proof rests on an application of the one-dimensional MVT to the function $g(t) = \langle u, f(y + t (x - y) \rangle$, where $u \in \mathbb{R}^n$ is any unit vector).

The second inequality is straightforward.