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Let $K \subset \mathbb R^d$ be a compact. Let $\phi_{\varepsilon} \colon K \rightarrow \mathbb R$ be continuous and converge uniformly to $\phi$. Suppose further that $\phi$ is Lipschitz continuous. Can we deduce that $\phi_{\varepsilon}$ are Lipschitz continuous for $\varepsilon$ small?


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I don't get how you could say something about the terms of the sequence from the limit... If you change the first term, you still converge towards the same limit... – xavierm02 Nov 30 '12 at 15:39
Arzela-Ascoli says that the sequence is bounded and equicontinuous. My question is, since Lipschitz converge to Lipschitz, does the converse apply. – dcs24 Nov 30 '12 at 15:41
up vote 2 down vote accepted

For $d=1$ and $K=[0,1]$ take $\phi _\epsilon =\epsilon \sqrt x$

Then $\phi _ {\epsilon } $ converges uniformly to the constant function $f\equiv 0$ which is a lipschitz function

But no $\phi _ \epsilon $ is lipschitz.

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