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We all know that if a function is continuously differentiable on a compact set $\Omega\subseteq \mathbb{R}^2$ then it is also Lipschitz continuous on that domain.

But now consider a function $f:\Omega \to \mathbb{R}$ that is uniformly continuous on but only piecewise continuously differentiable on the compact set $\Omega\subseteq \mathbb{R}^2$.

Can we say that such a function is Lipschitz continuous on $\Omega$? Why?

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    $\begingroup$ There is already a problem in your first statement. $f(x, y) = x^2$ is continously differentiable, but not Lipschitz continuous on $\Omega = \Bbb R^2$. You need the boundedness of the derivative, or you have to consider Lipschitz continuity on compact subsets. $\endgroup$ – Martin R Jan 2 '16 at 14:32
  • $\begingroup$ Right, so let us only consider compact domains $\Omega$. I edit the question accordingly. $\endgroup$ – Antonio Jan 2 '16 at 15:16
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    $\begingroup$ The definition of a domain that I know implies that it is an open set, so it cannot be compact. $\endgroup$ – Martin R Jan 2 '16 at 15:46
  • $\begingroup$ Ouch! Then I think I should just say "set". I mean something like $\Omega=\{(x,y)\in\mathbb{R}^2: \sqrt{x^2+y^2}\le 1\}$. $\endgroup$ – Antonio Jan 2 '16 at 15:49
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No we can't. $\sqrt{|x|}$ is a counterexample.

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  • $\begingroup$ Oh, right! But what if the gradient of $f$ is uniformly bounded by a constant wherever it is defined? $\endgroup$ – Antonio Jan 2 '16 at 15:21
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    $\begingroup$ @Antonio Yes, that will do. The mean value theorem or the fundamental theorem of calculus will then imply a Lipshitz bound. $\endgroup$ – Thomas Jan 2 '16 at 16:09
  • $\begingroup$ Great! Do you know if there is a book stating this whole result (i.e. continuous + bounded gradient implies Lipschitz) that I could cite? $\endgroup$ – Antonio Jan 2 '16 at 16:15
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    $\begingroup$ @Antonio Not really, but it should be a matter of some lines to write this down $\endgroup$ – Thomas Jan 2 '16 at 16:32
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On the other hand, if $f'(x)=g(x)$ is piecewise continuous and uniformly bounded by some constant $M$ across each discontinuity then the theory of Riemann integration provides enough information to prove that $f(x)$ is locally Lipschitz. You will need to use the Fundamental Theorem of Calculus for piecewise continuous integrands to fill in the details.

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  • $\begingroup$ I think this is what I was looking for. Let me just check if we mean the same thing. The function $f$ defined on the plane, and it is continuous and bounded. However, the gradient of $f$ is only piecewise continuous, and within each continuity region, it is uniformly bounded by a constant $M>0$. On the boundary between two continuity regions, the gradient is not defined. Is this the scenario that you had in mind? $\endgroup$ – Antonio Jan 2 '16 at 15:58

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