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Does uniform convergence on a closed and bounded interval preserve Lipschitz functions?
(Assume that the sequence of functions has a common Lipschitz constant $K$).

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I must be missing something obvious, but won't ordinary pointwise convergence preserve $K$-Lipschitz functions? – Jesse Madnick Jul 26 '12 at 4:10
@JesseMadnickHow are you proving it ? – Ester Jul 26 '12 at 4:18
up vote 5 down vote accepted

It's late at night where I am, so maybe I'm missing something obvious, but....

If $f_n\colon [a,b] \to \mathbb{R}$ each satisfy $|f_n(x) - f_n(y)| \leq K|x-y|$ for all $x, y \in [a,b]$, then just by taking the (pointwise) limit as $n \to \infty$, we obtain $|f(x) - f(y)| \leq K|x-y|$.

This reminds me of the following fact: If $\{f_n\}$ is a sequence of (uniformly) equicontinuous functions $[a,b]\to \mathbb{R}$, then $\{f_n\}$ converges pointwise if and only if $\{f_n\}$ converges uniformly.

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I have one question: does it hold also for Holder functions? I mean: is the pointwise limit of $\alpha$-Holder functions still $\alpha$-Holder? – Romeo Jul 26 '12 at 8:28
@Romen: If they have uniform Holder bounds, yes. The proof is basically the same as that Jesse gave above. – Willie Wong Jul 26 '12 at 9:41
@Jesse Thanks . I thought that it is false ,so I was looking for a counterexample . So stupid of me ! – Ester Jul 26 '12 at 13:22
what if the K for each $f_n$ are not the same? – user119459 Feb 2 at 20:29
@user119459: The OP says specifically: "Assume that the sequence of functions has a common Lipschitz constant $K$." – Jesse Madnick Feb 3 at 9:15

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