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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?

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Uria Mor
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    I don't think it is ok: the Lipschitz constant seems to be dependent on $x,y$, while it shouldn't be. – Crostul Apr 17 '15 at 16:59
  • Why do you think $K$ contains any line segments at all? $K$ could be a weird Cantor-like set. – zhw. Apr 17 '15 at 17:09
  • @zhw. If a set is open and connected (region) then it is polygonally connected http://math.stackexchange.com/questions/847337/polygonally-connected – Uria Mor Apr 18 '15 at 07:31
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    But $K$ is an abitrary compact subset of $U,$ and may contain no line segments at all. – zhw. Apr 18 '15 at 16:21
  • I can't see how. Do you have any concrete example in mind for a situation like this? – Uria Mor Apr 18 '15 at 16:29
  • What if $K$ is countably infinite? – zhw. Apr 18 '15 at 16:59
  • oh, I get you now.. It doesn't seem to get in the way - there will still be a finite set of polygonal segments in $U$ that connects $x$ and $y$. Now $K$ is compact so every open cover has a finite subcover. The finite subcover is contained by $U$ and that makes me want to claim that the segments can be contained by $K$. But I see what you are talking about... – Uria Mor Apr 18 '15 at 18:02

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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.

zhw.
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  • You seem to assume that $x\in U$ when you write that $\overline{B(x,r)} \subset U$, don't you? – Lierre May 23 '16 at 16:54
  • $x,y$ are both in $K$, which is a subset of $U.$ – zhw. May 23 '16 at 17:12
  • Of course... I have a extension problem in mind but that's not the question. – Lierre May 23 '16 at 17:17
  • By the way, what does "M Lipschitz" mean? – zhw. May 23 '16 at 17:18
  • You started with the assumption 'desired conclusion is false', but for the case $x=y$, how tcan you say 'Now you're in a nice compact convex set, where you know $f$ is Lipschitz' ? – Riaz Nov 27 '20 at 00:26
  • @RIYASUDHEENT.K The OP knows the result holds for compact convex sets, and $\overline{B(x,r)}$ is compact and convex. – zhw. Nov 28 '20 at 19:16