# Uniformly Continuous and Piecewise Continuously Differentiable implies Lipschitz Continuous? [closed]

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?

## closed as unclear what you're asking by Surb, Ethan Bolker, Shailesh, JonMark Perry, steven gregoryJun 8 '18 at 3:08

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• 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. – Martin R Jan 2 '16 at 14:32
• Right, so let us only consider compact domains $\Omega$. I edit the question accordingly. – Antonio Jan 2 '16 at 15:16
• The definition of a domain that I know implies that it is an open set, so it cannot be compact. – Martin R Jan 2 '16 at 15:46
• Ouch! Then I think I should just say "set". I mean something like $\Omega=\{(x,y)\in\mathbb{R}^2: \sqrt{x^2+y^2}\le 1\}$. – Antonio Jan 2 '16 at 15:49

No we can't. $\sqrt{|x|}$ is a counterexample.
• Oh, right! But what if the gradient of $f$ is uniformly bounded by a constant wherever it is defined? – Antonio Jan 2 '16 at 15:21
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.
• 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? – Antonio Jan 2 '16 at 15:58