It is a more general form of the question here, only here $U$ is not a convex set but an open and connected subset of $\mathbb{R}^n$. I need to show that $f$ is $M$ Lipschitz on any compact $K \subset U$.
My attempt goes like this:
$U$ is an open connected set, and $K \subset U$ so for any two points $x,y \in K$ there is a finite set of points $\{x_i\}_{i=1}^{r}\in K$ and we'll denote $x_1:=x, x_r:=y$ so for every $i=2,3,…,r$ the straight segment $[x_{i-1},x_{i}]$ is contained by $K$.
By the mean value theorem for several variables, $\forall i=2,3,…,r:\ \lvert f(x_{i-1})-f(x_i) \rvert= \lVert f'(s_i) ( x_{i-1} – x_i ) \rVert$ when $s_i\in U$. Notice that since $s_i ,\ i=2,3,…,r$ is finite thus bounded, $f'$ exists and continuous – thus $\exists T := \max_{i=1,2,..,r} \{ \lVert f'(s_i) \rVert \}$, and because $K$ is compact, $\exists D:= \ diam \{ K \} \geq \max_{i=2,3,…,r} \{ \lVert x_{i-1} -x_i \rVert \}$ hence:
$$\lvert f(x)-f(y) \rvert=\lvert f(x_{1})-f(x_2) +f(x_2)-f(x_3)+…. +f(x_{r-1})-f(y)\rvert \leq TB\cdot r$$ and since the diameter of $K$ is finite, every $x \neq y \ \in K$ have a real positive number $t$ so that $\lVert x-y \rVert \cdot t = B$ and that gives $||f(x)-f(y)|| \leq T \cdot t \cdot||x-y||$.
Is it o.k?
Best Answer
Suppose the desired conclusion is false. Then for every $m\in \mathbb {N},$ there exist $x_m,y_m \in K$ with $|f(x_m)-f(y_m)|> m |x_m-y_m|.$ Now $K$ is compact, so there is a subsequence $m_k$ such that $x_{m_k}\to x,y_{m_k}\to y$ for some $x,y\in K.$ By continuity, $f(x_{m_k})\to f(x), f(y_{m_k})\to f(y).$ If $x\ne y,$ you get $|f(x)-f(y)| = \infty,$ contradiction. Thus $x=y.$ Choose $r > 0$ such that $\overline {B(x,r)}\subset U.$ Now you're in a nice compact convex set, where you know $f$ is Lipschitz, and yet $x_{n_k},y_{n_k} \in \overline {B(x,r)}$ for large $k$ and $|f(x_{m_k})-f(y_{m_k})|>m_k|x_{m_k}-y_{m_k}|,$ contradiction.