Does uniform convergence on a closed and bounded interval preserve Lipschitz functions?
(Assume that the sequence of functions has a common Lipschitz constant $K$).
Lipschitz Functions
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real-analysis
holder-spaces
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1I must be missing something obvious, but won't ordinary pointwise convergence preserve $K$-Lipschitz functions? – 2012-07-26
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0@JesseMadnickHow are you proving it ? – 2012-07-26
1 Answers
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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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0I 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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0@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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1@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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0what if the K for each $f_n$ are not the same? – 2016-02-02
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0@user119459: The OP says specifically: "Assume that the sequence of functions has a _common_ Lipschitz constant $K$." – 2016-02-03
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0What if he didn't? – 2016-02-04
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0The 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