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We have functions $f_n\in L^2$ such that $f_n(x)$ tends to some $f(x)$ for almost all $x$, suppose that $||f_n||<M$. Does this mean that $f_n$ tend to $f$ weakly? In my previous question about $L^1$ I realized that the weak convergence fails; can one describe spaces for which this is true?

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Every closed and bounded set is weakly relatively compact in $L^2$ hence every bounded sequence has a weakly convergent subsequence. If a sequence converges strongly it converges weakly as well (the whole sequence). –  Jonas Teuwen May 18 '11 at 14:09
    
For the strong convergence this fails as well as for $L^1$. –  peno May 18 '11 at 14:12
    
I want convergence for the whole sequence, not for a subsequence. –  peno May 18 '11 at 14:18
    
your accounts have been merged. Please register your account so that you don't have login problems in the future and don't need to make new accounts. –  Qiaochu Yuan May 19 '11 at 6:49

1 Answer 1

up vote 4 down vote accepted

Yes.

By Fatou's lemma we have $f \in L^2$, so by replacing $f_n$ by $f_n - f$ we can assume that $f=0$.

Fix $g \in L^2$ and $\epsilon > 0$. By a standard dominated convergence argument, there is a set $A$ of finite measure with $\int_{A^c} |g|^2 < \epsilon$. There is also a $K$ so large that $\int_{|g| > K} |g|^2 < \epsilon$. Let $B = A \cap \{|g| \le K\}$; then $\lvert\int_{B^c} f_n g \rvert \le M \sqrt{2 \epsilon}$ by Cauchy--Schwarz, for every $n$.

On the other hand, for each $n$ we have $\int_B |f_n g|^2 \le K^2 M$. So $\{f_n g\}$ is an $L^2$-bounded sequence of functions on the finite measure space $B$, hence it is uniformly integrable on $B$. We also have $f_n g \to 0$ almost everywhere, and hence they converge in measure on $B$; by the Vitali convergence theorem $f_n g \to 0$ in $L^1(B)$, so $\int_B f_n g \to 0$. Letting $\epsilon \to 0$ completes the proof.

Edit: With respect to your other question (when can you say that an $L^1$-bounded, a.e. convergent sequence must converge weakly in $L^1$), the answer is: only in trivial cases. Essentially, if the measure space is anything but a finite union of atoms, you can construct a sequence of nonnegative $f_n \in L^1$ with $f_n \to 0$ a.e. but $||f_n||_{L^1} = 1$, so that taking $g=1 \in L^\infty$, you have $\int f_n g = 1$ for all $n$. (Idea: let $f_n = \mu(A_n)^{-1} 1_{A_n}$ where either $\mu(A_n) \uparrow \infty$, or $\mu(A_n) \downarrow 0$ fast (in particular try $\sum \mu(A_n) < \infty$)).

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Thank you so much! –  peno May 18 '11 at 19:10
    
Nice result! Could you state what are you saying only occurs in trivial cases? –  user1736 May 18 '11 at 23:58
    
@user1736: done. –  Nate Eldredge May 19 '11 at 1:56

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