# Lipschitz functions sequence which converges pointwise, also converges uniformly

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.

• I think you should consider Ascoli's theorem which would answer your question. May 21, 2020 at 15:07
• 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.
– cmk
May 21, 2020 at 15:08
• @Dldier_ I'll be sure to read about that, but we did not study it so I cannot use it. May 21, 2020 at 15:08
• @cmk which site? We didn't study these terms either. (equicontinuous, compactness) though I somewhat understand what compactness is May 21, 2020 at 15:10
• Just read about the proof : in your case it is an easy adaptation of Ascoli's theorem. May 21, 2020 at 15:11

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.

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.)