$C^1$ function on compact set is Lipschitz Let $\emptyset\ne A\subset\mathbb{R}^n$ be open and
let $f \in C^1 (A, \mathbb{R})$ be a function. Let $\emptyset\ne K\subset A$ be compact and convex. I want to prove that $f$ is 
Lipschitz on $K$; that is, prove there exists a 
constant $c > 0$ such that 
$| f ( x  ) - f( y ) | \leq 
c  \, \|  x - y  \|,   \forall  x , y \in K$.
My approach:

Let $x,y\in K$ be two arbitrary points. Then, since $K$ is convex, the line segment between $x$ and $y$, i.e. $Co(x,y)$, is in $K$. Thus, by MVT, there exists a vector $b\in Co(x,y)$ such that
$$\text{(*) } |f(x)-f(y)|=|\langle x-y, (\nabla f)(b)\rangle|\le \|x-y\|\|(\nabla f)(b)\|$$
Now, since $K$ is compact, $f$ takes a maximum and a minimum values on $K$, so that $\exists b'\in K$ such that $\|(\nabla f)(b') \|\ge \|(\nabla f)(b) \|$, for all $b\in K$. Let $c\in \mathbb{R}$, $c:=(\nabla f)(b')$, then
$$|f(x)-f(y)|\le\|x-y\|\|(\nabla f)(b)\|\le\|x-y\|\|(\nabla f)(b')\|=c\|x-y\|$$
This implies that $f$ is Lipschitz on $K$ for all $x,y\in K$.

Please let me know if you think my proof is correct or not very much? I'm somewhat concerned about the part with the gradient - how exactly is the maximality of the norm of the gradient related to the EVT, that is to $f$ taking maximum and minimum values? As far as I can tell, the maximum norm of the gradient exists because that is the direction to the maximum (or minimum) point of $f$.
 A: Here's how I would do it:
for $x, y \in K$, let $\gamma(t):[0, 1] \to K$ be the line segment
$\gamma(t) = x + t(y - x); \tag{1}$
then 
$\gamma(0) = x, \tag{2}$
$\gamma(1) = x +(y - x) = y, \tag{3}$
and
$\gamma'(t) = y - x; \tag{4}$
furthermore, since $K$ is convex, $\gamma(t)$ lies entirely within $K$, hence also in $A$.  
Now, for $x, y \in K$, we have:
$\Vert f(y) - f(x) \Vert = \Vert \displaystyle \int_0^1 \dfrac{d(f(\gamma(t))}{dt}dt \Vert = \Vert \displaystyle \int_0^1 \nabla f(\gamma(t)) \cdot \gamma'(t) dt \Vert$
$\le \displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \cdot \gamma'(t) \Vert dt \le  \displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \Vert \Vert \gamma'(t) \Vert dt.\tag{5}$
Since $K$ is compact and $\nabla f \in C^0(A, \Bbb R)$, so hence $\nabla f \in C^0(K, \Bbb R)$, $\Vert \nabla f \Vert$ is bounded by some $B$ on $K$, hence
$\displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \Vert \Vert \gamma'(t) \Vert dt \le \displaystyle \int_0^1 B \Vert \gamma'(t) \Vert dt; \tag{6}$
using (4), 
$\displaystyle \int_0^1 B \Vert \gamma'(t) \Vert dt  = \displaystyle \int_0^1 B \Vert y - x \Vert dt = B \Vert y - x \Vert; \tag{7}$
bringing together (5), (6), and (7) yields
$\Vert f(y) - f(x) \Vert \le B \Vert y - x \Vert, \tag{8}$
that is, $f(x)$ is Lipschitz continuous on $K$.
A: The convexity of $K$ is not needed. Suppose the conclusion fails. Then for each $m\in \mathbb N,$ there exist $y_m,x_m \in K$ such that
$$\tag 1 |f(y_m)- f(x_m)| > m|y_m-x_m|.$$
Because $K$ is compact, we can find a subsequence $m_k$ such that the sequences $y_{m_k},x_{m_k}$ converge to points $y,x\in K$ respectively.
Suppose $y\ne x.$ Then $|y-x| > 0.$ For large $k$ we then have
$$|f(y_{m_k})- f(x_{m_k})| > m_k|y_{m_k}-x_{m_k}|> m_k(|y-x|/2)\to \infty$$
This implies $f$ is not bounded on $K.$ But $f$ is continuous on $A,$ hence is continuous on $K,$ hence $f$ is bounded on $K$ by compactness. This contradiction shows $y=x.$
Because $A$ is open we can choose $r>0$ such that $B(x,r)\subset A.$ For large $k$ we then have $y_{m_k},x_{m_k}$ in the compact convex set $\overline {B(x,r/2)} \subset A.$ Let $M=\sup_{\overline {B(x,r/2)}}|\nabla f|.$ Then for large $k$ the mean value inequality gives
$$|f(y_{m_k})- f(x_{m_k})| \le M |y_{m_k}-x_{m_k}|.$$
But the right side $\to 0,$ contradicting $(1).$ Therefore $(1)$ cannot hold, proving the result.
A: This fact is not true. Consider the function $f(x) = \sqrt{x}$, over [0, 1]. It's not Lipschitz, as its slope goes to infinity as $x \rightarrow 0$.
