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Let $(f_n):[a,b] \rightarrow \mathbb{R}$ be a sequence of K-lipschitz functions pointwise converging to $f:[a, b] \rightarrow \mathbb{R}$ on $[a, b]$. Prove $(f_n)$ converges uniformly to $f$.

Intuitively my idea is that if it wouldn't converge uniformly there would be at least one point the function would be very "steep" around for some large $n$, which contradicts lipschitz. I tried to prove by contradiction (assuming it doesn't converge uniformly) but was stuck. I also thought about trying to use Cantor's Lemma but didn't accomplish much.

I'm looking for a small hint, and also would like feedback about the idea of using Cantor's Lemma.

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  • $\begingroup$ I think you should consider Ascoli's theorem which would answer your question. $\endgroup$
    – Didier
    May 21, 2020 at 15:07
  • $\begingroup$ If they're Lipschitz with the same constant, then they're equicontinuous. This + the compactness of the domain tells you that the pointwise convergence is actually uniform. You can find this question on this site, for example. $\endgroup$
    – cmk
    May 21, 2020 at 15:08
  • $\begingroup$ @Dldier_ I'll be sure to read about that, but we did not study it so I cannot use it. $\endgroup$ May 21, 2020 at 15:08
  • $\begingroup$ @cmk which site? We didn't study these terms either. (equicontinuous, compactness) though I somewhat understand what compactness is $\endgroup$ May 21, 2020 at 15:10
  • $\begingroup$ Just read about the proof : in your case it is an easy adaptation of Ascoli's theorem. $\endgroup$
    – Didier
    May 21, 2020 at 15:11

2 Answers 2

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Let $f$ be the pointwise limit of $f_n$. We want to prove that $\|f-f_n\|_{\infty} \to 0$.

All $f_n$ are $K-$Lipschitz, that means : $\forall x,y \in [a,b], \forall n, |f_n(x)-f_n(y)|\leqslant K|x-y|$

Take the limit with $n$ : you have that $|f(x) - f(y)| \leqslant K|x-y|$ so that $f$ is $K-$Lipschitz, thus continuous, and for all $n$, $f-f_n$ is continuous.

As $[a,b]$ is a compact, $M_n=||f-f_n||_{\infty} <\infty$. We want to prove that $M_n \to 0$. By continuity, there exists $x_n \in [a,b]$ such that $M_n = |f(x_n)-f_n(x_n)|$.

Moreover, $M_n \leqslant ||f||_{\infty} + ||f_n||_{\infty}$ by triange inequality, and $||f_n||_{\infty} \leqslant |f_n(a)| + K|b-a|$. As $f_n(a)$ is a convergent sequence, $||f_n||_{\infty}$ is bounded, so is $M_n$.

Right now, we know that $M_n$ is a bounded sequence in $\mathbb{R}_+$. To show it converges to $0$, we only have to show that every converging subsequence has limit $0$. Take $M'_n = M_{\varphi(n)}$ a converging subsequence, $f'_n = f_{\varphi(n)}$ and $x'_n = x_{\varphi(n)}$. To show that, use the fact that $[a,b]$ is compact, thus, $x'_n$ has a converging subsequence $x'_{\psi(n)}$ with limit $x_{\infty}$, and use : \begin{align} M'_{\psi(n)} &= |f(x'_{\psi(n)}) - f'_{\psi(n)}(x_{\psi(n)})| \\&\leqslant |f(x'_{\psi(n)})-f(x_{\infty})| + |f(x_{\infty}) - f'_{\psi(n)}(x_{\infty})|+|f'_{\psi(n)}(x_{\infty}) - f'_{\psi(n)}(x'_{\psi(n)})| \\ & \leqslant K|x'_{\psi(n)}-x_{\infty}| + |f(x_{\infty}) - f(x'_{\psi(n)})| + K|x'_{\psi(n)} - x_{\infty}| \end{align}

And all three terms are going to zero because of the hypothesis. As $M'_n$ is convergent, its limit is $0$, and every converging subsequence of $M_n$ has limit $0$. $M_n$ is then going to $0$, which means that $f_n$ converges uniformly.

Sorry, there is many notations, but it's the detailed proof.

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The functions $g_n$ defined by $g_n(x):=\sup_{k\ge n}|f_k(x)-f(x)|$ are (i) lower semi-continuous, and (ii) decrease pointwise to $0$. By Dini's Theorem, the convergence is uniform on $[a,b]$. (The only role played by the Lipschitz assumption is to ensure that $f$ is continuous.)

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