TITLE: Proof on showing function $f \in C^1$ on an open & convex set $U \subset \mathbb R^n$ is Lipschitz on compact subsets of $U$
QUESTION [6 upvotes]: The question is as follows:

Given:
(1) function $f: U \subset \mathbb R^n ==> \mathbb R$
(2) $U$ is open and convex set
(3) $f \in  C^1$ in $U$
Goal: Show that $f$ is Lipschitz on any compact subset of $U$

By now, I have various ideas come to mind, but I can't connect the dots >_<
Here are my thoughts so far:
(1) Recall definitions:
(i) $f \in C^k$ means all partial derivatives up to (and including) order $k$ exist and continuous.  Here $k = 1$
(ii) a set $U$ is convex if for any 2 points $x, y$ in $U$, the segment joining $x$ and $y$ is totally inside $U$
(iii) function $f$ is Lipschitz if there is a bound $M$ such that $|f(x) - f(y)| \leq M |x-y|$
(2)By a theorem in my book:

if function $f \in C^1$ on an open & convex set $U$, then for any 2
  points $x$ and $y$ in $U$, there is a point $s$ lying on the segment
  joining $x$ and $y$ such that $f(x) - f(y) = Df(s) * (x - y)$

I firstly have a feeling that I may need this theorem in the proof, but I can't see the connection between my desired bound $M$ with $Df(s)$.  So I tried to think of other ideas and ...
(3) It turns out that by all the given information, I think if I can show function $f$ is convex, then I'm done because there is a theorem which said: a convex function on a open, convex set $U$ should be Lipschitz on $U$, thus I think it must also be Lipschitz on any subset of $U$, either that subset is compact or not.  
However, how can I prove function $f$ is convex, based on given information?  I have a feeling that I have to use the fact that set $U$ is convex, but then I'm stuck on how to proceed further >_>
*Would someone please help me on this question?
Thank you very much ^_^*

REPLY [7 votes]: You need to use the Mean Value theorem in several variables, as you already stated it:
For $x,y\in U$ there exists $s$ which lies on the line segment between $x$ and $y$ ( this line segment is contained in $U$, since it is convex), such that:
$$f(x)-f(y)=\nabla f(s)\cdot (x-y)\Rightarrow\text{( Cauchy-Schwarz )}$$
$$|f(x)-f(y)|\leq|\nabla f(s)|\cdot |x-y|$$
Now, take a compact subset $S$ of $U$, and $x,y\in S$. Then take a convex compact set $S'$ such that $S\subset S'\subset U$.This is because we want $s\in S'.$
Now, since $f\in C^1$ , the first partial derivatives are continuous, and therefore, bounded in the compact set $S'$. This implies that $|\nabla f(s)|\leq M$
Concluding, for each $x,y\in S$ we have that :
$$|f(x)-f(y)|\leq M\cdot |x-y|$$
which proves that $f$ is Lipschitz in $S$.