Lipschitz continuity and sup of derivative norm On this wikipedia page, it is stated:

For a differentiable Lipschitz map $f : U \rightarrow R^m$ the inequality
  $\|Df\|_{\infty,U}\le K$ holds for the best Lipschitz constant of $f$, and
  it turns out to be an equality if the domain $U$ is convex.

I have two questions about this statement.
Question 1: does a best Lipschitz constant necessarily exists? Shouldn't be $K$ the $\inf$ of all Lipschitz constants?
Question 2: do you know a proof of the wikipedia statement, or a reference where this is proved? 
I tried to prove it by using Taylor expansion with integral remainder (which seems to make sense with the convex assumption for equality), but it did not work...
Thank you.
 A: Let $S$ be the set of all Lipschitz constats for $f$.

Take $\left(K_n\right)_{n\in \Bbb N}\in S^\Bbb N$ a sequence of Lipschitz constats for $f$ converging to $K_\infty$.
$\forall n \in \Bbb N, \forall x,y \in U, \left\|f(x)-f(y)\right\| \le K_n \left\| x-y \right\|$
$\forall x,y \in U, \forall n \in \Bbb N, \left\|f(x)-f(y)\right\| \le K_n \left\| x-y \right\|$
$\forall x,y \in U, \left\|f(x)-f(y)\right\| \le K_\infty \left\| x-y \right\|$
So $K_\infty \in S$

So we get that $S$ is closed so the $\inf$ is in fact a $\min$.
A: Question 1:  let $K$ be the inf of all Lipschitz constants. For every $\epsilon>0$, there is $K\leq L<K+\epsilon$ with $L$ Lipschitz constant. So
$$
\|f(x)-f(y)\|\leq L\|x-y\|\leq (K+\epsilon)\|x-y\|\qquad \forall x,y\in U\quad \forall \epsilon>0.
$$
Lettting $\epsilon$ tend to $0$ shows that $K$ is itself a Lipschitz constant, hence it is a min. In other terms, the set of all Lipschitz constants is of the form $[K,+\infty)$. Of course, we could have observed from the beginning that it had to be $(K,+\infty)$ or $[K,+\infty)$ since $M$ is a Lipschitz constant for every $M\geq L$ if $L$ is a Lipschitz constant.
Question 2.1:  fix $x\in U$ and take any $h$ in the ambiant normed space containing $U$ (I guess it is $\mathbb{R}^n$ in your case, but everything works the same in the general case). Then for $t\in\mathbb{R}$ close enough to $0$, $x+th$ belongs to $U$ and 
$$\|f(x+th)-f(x)\|\leq K\|x+th-x\|=K|t|\|h\|\quad\Rightarrow\quad \frac{\|f(x+th)-f(x)\|}{|t|}\leq K\|h\|.$$
Letting $t$ tend to $0$ yields
$$
\|Df_x(h)\|\leq K\|h\|\quad \forall h\quad\Rightarrow \quad \|Df_x\|\leq K \quad \forall x\in U\quad\Rightarrow\quad \|Df\|_{\infty,U}\leq K.
$$  
Question 2.2:  now assume $U$ is convex. And set $L:=\|Df\|_{\infty,U}$. Fix $x,y\in U$ and note that the segment $[x,y]:=\{x_t=(1-t)x+ty\,;\, t\in[0,1]\}$ is contained in $U$ by assumption. We will show that $\|f(x)-f(y)\|\leq L\|x-y\|$, which will prove that $L$ is a Lipschitz constant, whence $K\leq L$ and finally $K=\|Df\|_{\infty,U}$ in this case.
Recall that we denote $x_t=(1-t)x+ty$. Fix $\epsilon>0$ and consider 
$$
I_\epsilon:=\{t\in[0,1]\,;\,\|f(x)-f(x_s)\|\leq (L+\epsilon)\|x-x_s\|\;\forall 0\leq s\leq t\}.
$$
This is clearly a closed interval of the form $[0,t_0]$ in $[0,1]$. We will show that $t_0=1$. Since $f$ is differentiable at $x_{t_0}$ and $\|Df_{x_{t_0}}\|\leq L$, there exists $\delta>0$ such that
$$
\|f(y)-f(x_{t_0})\|\leq \|Df_{x_{t_0}}(y-x_{t_0})\|+\epsilon \|y-x_{t_0}\|\leq (L+\epsilon)\|y-x_{t_0}\|\quad\forall \|y-x_{t_0}\|<\delta.
$$
It follows that 
$$
\|f(x)-f(x_s)\|\leq \|f(x)-f(x_{t_0})\|+\|f(x_{t_0})-f(x_s)\|\leq (L+\epsilon)(\|x-x_{t_0}\|+\|x_{t_0}-x_s\|)
$$
$$
=(L+\epsilon)(\|t_0(y-x)\|+\|(s-t_0)(y-x)\|)=(L+\epsilon)s\|y-x\|=(L+\epsilon)\|x-x_s\|
$$
for all $t_0\leq s\leq t_0+\frac{\delta}{\|y-x\|}$. This means that $t_0$ can not be the sup of $I_\epsilon$, unless $t_0=1$. In particular
$$
\|f(x)-f(x_1)\|\leq (L+\epsilon)\|x-x_1\| \quad \mbox{i.e.}\quad \|f(x)-f(y)\|\leq (L+\epsilon)\|x-y\| .
$$
It only remains to let $\epsilon$ tend to $0$ to deduce that $L$ is indeed a Lipschitz constant as desired.
Remark: what makes the argument lengthy in 2.2 is that there is no mean value theorem for functions with range in a space of dimension greater than one. As we have seen, we still have a "mean inequality theorem", though, when the domain is convex. But 2.2 is much easier if $Df$ is assumed to be continuous. In this case, it suffices to write
$$
f(y)-f(x)=\int_0^1Df_{x_t}(y-x)dt \Rightarrow  \|f(y)-f(x)\|\leq \int_0^1\|Df_{x_t}(y-x)\|dt\leq \|Df\|_{\infty,U}\|y-x\|.
$$
