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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, the continuity of $F_n$ should be proved and analogously of $F$. But I don't know about the uniform convergence and how to use the weak convergence hypothesis..

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    A more bare-hands proof can be found [here](http://math.stackexchange.com/questions/95260/is-this-functional-weakly-lower-semicontinuous/176028#176028)2012-08-13

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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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    Yes, I get it now. I didn't remember that fact, thanks for your kind reply and sorry for the stupid question.2013-10-29