$f \in C^1$ defined on a compact set $K$ is Lipschitz?

Question

Let $f: \Omega \subseteq \mathbb{R}^N \to \mathbb{R}^M$ be $C^1$, and $K \subseteq \Omega$. Prove that $f \mid_K$ is Lipschitz.

Letting $x,y \in K$, I know that $f$ is locally Lipschitz, I thought about taking a finite subcover of $K$ by the open cover of the balls which makes $f$ locally Lipschitz, and then take a sequence of $x_k, k=0,\dots, n$ in such that $x_0 =x$ and $x_n=y$, $x_k =y$ and $(x_i)$ are aligned, then try to use $|f(x)- f(y)| \le \sum_{i=1}^{n}|f(x_{i=1})- f(x_i)|$. But the very first problem is to take this sequence as $K$ is not necessary convex.

Again, if $K$ was convex I would make use of the Mean Value Inequality and that $K$ is compact to limit the value of $||Df_{p}||$, but as it is not the case I don't know how to proceed.

So how can I avoid that? Does it hold for a general $K$ compact?

Finally, I am aware of this question, but his question, at the end, was only concerning the continuity, which I already know how to deal with there was no useful answer, nor comment for me, I apologise if I acted wrong by opening a new one.

Answer 1

I assume that $\Omega$ shall be open.

In my opinion, it's more convenient to not bother with open coverings of $K$, though that works too.

For every $\eta > 0$, we can look at

$$L^{(1)}_{\eta} := \sup \biggl\{ \frac{\lVert f(x) - f(y)\rVert}{\lVert x-y\rVert} : (x,y) \in K\times K \setminus \Delta_{\eta}\biggr\},\tag{$\ast$}$$

where $\Delta_{\eta} = \{ (x,y) \in K\times K : \lVert x-y\rVert < \eta\}$. Since $K\times K \setminus \Delta_{\eta}$ is compact, $L^{(1)}_{\eta} < +\infty$.

To establish Lipschitz continuity, we now need only look at $\Delta_{\eta}$. Now, if we choose $\eta$ small enough, we can happily ignore all problems arising from non-convexity, since then the straight line connecting $x,y$ with $(x,y) \in \Delta_{\eta}$ stays inside a compact subset of $\Omega$, where the derivative of $f$ is uniformly bounded and hence we have Lipschitz continuity there.

Since $\Omega$ is open and $K\subset \Omega$ compact, there is an $\varepsilon > 0$ such that

$$B_{\varepsilon}(K) = \{ x\in \mathbb{R}^N : \operatorname{dist}(x,K) < \varepsilon\} \subset \Omega.$$

Then $\overline{B_{\eta}(K)} \subset \Omega$ for all $\eta < \varepsilon$, and $\overline{B_{\eta}(K)}$ is compact. With a bound $L^{(2)}_{\eta}$ of $\lVert Df\rVert$ on $\overline{B_{\eta}(K)}$, the mean value theorem yields

$$\lVert f(x) - f(y)\rVert \leqslant L^{(2)}_{\eta}\cdot \lVert x-y\rVert$$

for $x,y\in K$ with $\lVert x-y\rVert \leqslant \eta$, and together with $(\ast)$ we deduce that $f\lvert_K$ is Lipschitz continuous with Lipschitz constant $L \leqslant \max \{ L^{(1)}_{\eta}, L^{(2)}_{\eta}\}$.