# 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

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 $n$.