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It's well known that a differentiale continuous function is Lipschitz if and only if its gradient is bounded. (Is a function Lipschitz if and only if its derivative is bounded?)

Can this result be generalized to non-differential case? The tricky thing here is to replace the gradient with Clarke subgradient (https://en.wikipedia.org/wiki/Clarke_generalized_derivative). In other word, is the following statement correct?

Consider a continuous function $f(x)$ whose Clarke subgradient is $\partial f(x)$ on $x$. $f(x)$ is $L$-Lipschitz continous, i.e. $$\|f(x)-f(y)\|\le L\|x-y\|, \forall x, y\in{\rm dom}(f)$$ if and only if $\|g\|\le L$ for all $g\in\partial f(x)$ and $x\in{\rm int}({\rm dom}(f))$, where ${\rm int}({\rm dom}(f))$ represents the interior of ${\rm dom}(f)$.

Five Mr
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    Yes, if the subgradient is bounded (as a set), then the local lipschitz norm is bounded. Hence, if you have a global bound on the subgradient, then you have global lipschitz function. – Severin Schraven Oct 08 '23 at 05:36
  • Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Oct 08 '23 at 05:37
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    Seems specific enough to me. Maybe could add a domain and "for all $x$ in domain", though I think that's reasonably easy to infer. – M W Oct 08 '23 at 05:39
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    A convex proper function $f\colon X \to \mathbb R$ is Lip. On bounded subsets of the B space $X$ if and only if the subfifferential maps bounded subsets of $X$ to bounded, nonempty, subsets in $X^*$ – Evangelopoulos Foivos Oct 08 '23 at 17:20
  • @SeverinSchraven Thanks for your answer. But when I tried to google the concept "local lipschitz norm", I found nothing. Could you kindly provide some references where I can find them? – Five Mr Oct 10 '23 at 20:20

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