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If $f:{\Bbb R}^m \to {\Bbb R}^n$ is Lipschitz on all compact sets, under what conditions is $f$ a $C^1$ map?

Background: I have proved the following result, and I am looking for a converse.

Proposition. Let $f: {\Bbb R}^m \to {\Bbb R}^n$ be a $C^1$ mapping and let $K \subset {\Bbb R}^m$ be compact. Then the restriction $f|_K$ of $f$ to $K$ is Lipschitz continuous.

The proof of this result, and some special cases, can be found here: Post 1, Post 2, Post 3, and Post 4.

In general, differentiability is a stronger condition than continuity, so there is no reason to expect the "Lipschitz on all compact sets" assumption above to imply differentiability. What other additional assumptions on $f$ would be required to ensure that it is a $C^1$ mapping?

Rademacher's Theorem is a related result; but I am clearly looking for something stronger.

The goal is to characterize $C^1$ maps in terms of Lipschitz continuity on compact sets, and other additional conditions if required.

Thank you, and I am excited to see where this goes!

stoic-santiago
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    Note $f(x) = |x|$ is Lipschitz on $\mathbb R$ (and therefore on all compact subsets) but not differentiable at $x=0$. – GEdgar Sep 15 '21 at 17:32
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    After thinking about this more carefully, I think I have identified a better example: Volterra's function. It is everywhere differentiable with bounded derivative, and is therefore Lipschitz, but its derivative is very discontinuous. – kahen Sep 15 '21 at 20:56
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    Unfortunately, being $C^1$ is really stronger than being Lipschitz ...in the same way that if a function is bounded, then it tells you nothing about its discontinuities. – LL 3.14 Sep 15 '21 at 21:12
  • @LL3.14 is spot on. In fact it is easier to turn things around. Given any bounded and (locally) integrable function $f$ on $R$, the function $F(x):=\int_0^x f(t) dt$ is Lipschitz and its derivative... is it continuous?! Of interest to the OP will be the fact that after removing a set of measure as small as we want, a Lipschitz function coincides with a C^1 function! This is probably the best one can get. – Behnam Esmayli Dec 27 '21 at 08:38

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