# Weak convergence in $L^2$ and uniform covergence

I have this problem: let $f_n$ converge weakly to $f$ in $L^2[0,1]$ and let $$F_n(x)=\int_0^xf_n(t) \, \textrm{d}t,$$ $$F(x)=\int_0^xf(t) \, \textrm{d}t.$$ Then $F_n,F$ are continuous and $F_n$ converges uniformly to $F$.

Writing $$F_n(x)=\int_0^1 f_n(t) \mathbb{1}_{[0,x]} \, \textrm{d}t$$ and applying the Lebesgue dominated convergence theorem it should be proved the continuity of $F_n$ and analougosly of $F$. But I don't know about the uniform convergence and how to use the weak convergence hypotesis..

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A more bare-hands proof can be found here –  user31373 Aug 13 '12 at 22:44

It's possible there is a simpler approach, but I'd proceed as follows.

1. Show that $F_n \to F$ pointwise.

2. It would suffice to show that $\{F_n\}$ is equicontinuous. (Remember the Arzela-Ascoli theorem and/or its proof.)

3. Show that the weak convergence implies that $\sup_n \|f_n\|_{L^2} < \infty$. (Use the uniform boundedness principle.)

4. Use the previous step together with Cauchy-Schwarz to estimate $|F_n(x) - F_n(y)|$ independently of $n$.

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May I ask you what you mean exactly in point 2? Why would it suffice to show $\{F_n\}$ is equicontinuous? Consider for example the case in which the sequence $(f_n)_n$ is bounded in $L^2$. In that case, the $F_n$ would be equi-$\alpha$-holder, with $\alpha=\frac{1}{2}$... How would you conclude? Thank you very much and sorry for disturbing. –  Romeo Oct 29 '13 at 18:57
@Romeo: It is a fact that a sequence of functions which is equicontinuous and converges pointwise on a compact set must converge uniformly. This fact is contained in the proof of Arzela-Ascoli; you can also get it from the conclusion of Arzela-Ascoli by noting that compactness means the sequence has at least one uniform limit point, but pointwise convergence means this limit point is unique. –  Nate Eldredge Oct 29 '13 at 19:44
@Romeo: And by the way, $\{f_n\}$ must be bounded in $L^2$ by my point 3. –  Nate Eldredge Oct 29 '13 at 19:45
Yes, I get it now. I didn't remember that fact, thanks for your kind reply and sorry for the stupid question. –  Romeo Oct 29 '13 at 22:17