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

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

• "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?
– user711689
Commented Feb 23, 2021 at 19:07
• @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? Commented Feb 23, 2021 at 19:08

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 $$x \to b-$$.

• Basically the same idea, but my proof circumvents proving that $f$ is continuous.
– RRL
Commented Feb 23, 2021 at 19:19
• $|f_n(x) - f(x)| < \epsilon$ for all $n \geqslant N(\epsilon)$ and $x \in (a,b)$. Now take limits as $x \to a+$ and $x \to b-$.
– RRL
Commented Feb 23, 2021 at 19:20
• @Daniil: You stated but did not prove that $f$ is uniformly continuous on $(a,b)$. How would you do that?
– RRL
Commented Feb 23, 2021 at 19:31
• 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.
– RRL
Commented Feb 23, 2021 at 19:34
• 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.
– RRL
Commented Feb 23, 2021 at 19:40