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That sounds a lot like a tongue-twister. I know that there exist sequences of Lipschitz functions whose uniform limit is not Lipschitz (for instance, just use Weierstrass theorem on $[a,b]$). Clearly if the sequence is uniformly Lipschitz, then the uniform limit has to be Lipschitz.

I think something similar should happen when we replace Lipschitz with uniform continuity. Weierstrass-Stone theorem though cannot be used as it requires a compact interval of definition which, in turn grants uniform continuity of continuous functions.

Is there an easy counterexample or is it actually true that

Given $\{f_n\}$ a sequence of uniform continuous functions, let $f$ be such that $$ f_n \to f $$ uniformly. Is then $f$ uniformly continuous itself?

Paolo Intuito
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  • See this: https://math.stackexchange.com/questions/1007361/uniform-limit-of-uniformly-continuous-functions – sranthrop Jan 12 '18 at 12:37
  • On a compact it is true (follows from continuity of f). Otherwise see sranthrop's link. – Ian Jan 12 '18 at 12:50
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    That link has a weaker condition - that the convergence is uniform on all compact intervals. With that, a counterexample exists. I am not sure a counterexample exists if we assume uniform convergence everywhere. –  Jan 12 '18 at 12:58
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    @user8734617 Global uniform convergence implies uniform continuity of the limit. For the case of the domain being $\mathbb{R}$, I state that explicitly in the last paragraph of my answer over there. – Daniel Fischer Jan 12 '18 at 13:05

1 Answers1

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$f$ must be uniformly continuous.

Proof: Let's choose any $\epsilon\gt 0$.

  • Because of uniform convergence, there is one of the functions (say $f_n$) such that $|f_n(x)-f(x)|\lt\frac{\epsilon}{3}$ for all $x$.

  • Now, $f_n$ is uniformly continuous, so let's find $\delta\gt 0$ such that $|x-y|\lt\delta$ implies $|f_n(x)-f_n(y)|\lt\frac{\epsilon}{3}$.

  • Finally, for any $x,y$ such that $|x-y|\lt\delta$, we have $|f(x)-f(y)|\le|f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|\lt\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$

As $\epsilon\gt 0$ was arbitrarily chosen to start with, it follows that $f$ is uniformly continuous.