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Let $f_n$ sequence of continuous functions on $[a,b]$ and $\lim_{n\to \infty}f_n(x)=f(x)$ uniformly on $(a,b)$. Show that $f_n\mapsto f$ uniformly on $[a,b]$.

I would like to know how to continue my proof, please.

First of all, as $f_n$ are continuous on closed interval, this sequence is uniformly continuous on $[a,b]$. By definition we have: $\forall \epsilon>0 \ \exists \delta_1>0 \ \forall x,y \in [a,b]$: $|x-y|<\delta_1 \implies|f_n(x)-f_n(y)|<\epsilon$

As $f_n$ converges uniformly to $f$ on $(a,b)$, $f$ is uniformly continuous on $(a,b)$. Moreover, the uniform continuity of $f$ on $(a,b)$ implies that $\lim_{x\to a^+}f(x)$ and $\lim_{x\to b^-}f(x)$ exist. So by extension by continuity we have that $f$ is uniofmrly continuous on $[a,b]$: $\forall \epsilon>0 \ \exists \delta_2>0 \ \forall x,y \in [a,b]$: $|x-y|<\delta_2 \implies |f(x)-f(y)|<\epsilon$

We would like to show the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x \in [a,b]$: $|f_n(x)-f(x)|<\epsilon$.

How can i conclude with all the hypothesises that I've done until now?

Edit: Probably I could have tried to prove by absurd and work with sequence $x_n$ which is bounded so by Bolzano-Weierstrass there is a converging sub-sequence(So cauchy) then use the uniform continuity to get a contradiction

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  • $\begingroup$ "First of all, as $f_{n}$ are continuous on closed interval, this sequence is uniformly continuous on $[a,b]$." Considering the question, this seems like a logical jump to me. How confident are you in your proof? $\endgroup$
    – user711689
    Feb 23, 2021 at 19:07
  • $\begingroup$ @Peter Morfe if $f$ is continuous on $[a,b]$, then it is uniformly continuous on $[a,b]$, no? Probably, to be more rigorous, i should have taken each $\delta_i$ of every function, and take it's min? $\endgroup$
    – Daniil
    Feb 23, 2021 at 19:08

1 Answer 1

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By the uniform Cauchy criterion, there exists $N(\epsilon)$ such that for all $m > n \geqslant N(\epsilon)$ and all $x \in (a,b)$, we have $|f_m(x) - f_n(x) | < \epsilon$.

By continuity, this implies that $|f_m(a) - f_n(a)| = \lim_{x \to a+}|f_m(x) - f_n(x)| \leqslant \epsilon$, and similarly as $x \to b-$.

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    $\begingroup$ Basically the same idea, but my proof circumvents proving that $f$ is continuous. $\endgroup$
    – RRL
    Feb 23, 2021 at 19:19
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    $\begingroup$ $|f_n(x) - f(x)| < \epsilon $ for all $n \geqslant N(\epsilon)$ and $x \in (a,b)$. Now take limits as $x \to a+$ and $x \to b-$. $\endgroup$
    – RRL
    Feb 23, 2021 at 19:20
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    $\begingroup$ @Daniil: You stated but did not prove that $f$ is uniformly continuous on $(a,b)$. How would you do that? $\endgroup$
    – RRL
    Feb 23, 2021 at 19:31
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    $\begingroup$ You would use the triangle inequality: $|f(x) - f(y)| \leqslant |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)|$. The you apply the uniform convergence and uniform continuity of $f_n$. Next you use the fact that $f$ can be extended continuously. This is a lot of work not shown. So that leads me back to why would you not prove this in 2 lines? Apparently my answer does not appear to be helpful to you. $\endgroup$
    – RRL
    Feb 23, 2021 at 19:34
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    $\begingroup$ If a function is uniformly continuous on the open interval $(a,b)$ then it can be extended continuously to the closed interval. This has a straightforward but lengthy proof, which by the way uses Cauchy criterion as you have to produce the existence of a limit as $x \to a,b$ out of thin air. $\endgroup$
    – RRL
    Feb 23, 2021 at 19:40

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