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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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    $\begingroup$ I must be missing something obvious, but won't ordinary pointwise convergence preserve $K$-Lipschitz functions? $\endgroup$ Jul 26 '12 at 4:10
  • $\begingroup$ @JesseMadnickHow are you proving it ? $\endgroup$
    – Ester
    Jul 26 '12 at 4:18
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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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  • $\begingroup$ I have one question: does it hold also for Holder functions? I mean: is the pointwise limit of $\alpha$-Holder functions still $\alpha$-Holder? $\endgroup$
    – Romeo
    Jul 26 '12 at 8:28
  • $\begingroup$ @Romen: If they have uniform Holder bounds, yes. The proof is basically the same as that Jesse gave above. $\endgroup$ Jul 26 '12 at 9:41
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    $\begingroup$ @Jesse Thanks . I thought that it is false ,so I was looking for a counterexample . So stupid of me ! $\endgroup$
    – Ester
    Jul 26 '12 at 13:22
  • $\begingroup$ what if the K for each $f_n$ are not the same? $\endgroup$
    – user119459
    Feb 2 '16 at 20:29
  • $\begingroup$ @user119459: The OP says specifically: "Assume that the sequence of functions has a common Lipschitz constant $K$." $\endgroup$ Feb 3 '16 at 9:15

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