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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?2012-07-26
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    @JesseMadnickHow are you proving it ?2012-07-26

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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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    I have one question: does it hold also for Holder functions? I mean: is the pointwise limit of $\alpha$-Holder functions still $\alpha$-Holder?2012-07-26
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    @Romen: If they have uniform Holder bounds, yes. The proof is basically the same as that Jesse gave above.2012-07-26
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    @Jesse Thanks . I thought that it is false ,so I was looking for a counterexample . So stupid of me !2012-07-26
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    what if the K for each $f_n$ are not the same?2016-02-02
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    @user119459: The OP says specifically: "Assume that the sequence of functions has a _common_ Lipschitz constant $K$."2016-02-03
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    What if he didn't?2016-02-04
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    The problem, as stated, does not tell you that Lipschitz constants are same for all the functions in the sequence. The functions x^n on [0,1] show that the limit may not even be continuous if the Lipschitz constants are not bounded.2016-12-30