# Weak L_1-convergence of squares

Let $f_k$ be a sequence of non-negative functions from $L_2(\Omega)$, where $\Omega$ is a bounded open set. Assume that $f_k\to f$ weakly in $L_2$ and strongly in $L_p$, $\forall p<2$. Assume also that $f_k^2\to F$ weakly in $L_1$. Does it imply that $F=f^2$?

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No, thanks, but non-negativity seems to be important. – user21629 Feb 23 '12 at 12:25

The answer is "yes", even for non non-negative $f_n$. First assume $f=0$. Since $f_n\to 0$ in $L_1$, WLOG by passing to a subsequence $f_n\to 0$ a.e. and hence $f_n^2\to 0$ a.e. But $f_n^2$ converges weakly in $L_1$ hence is uniformly integrable, whence $\|f_n^2\|_1 \to 0$.
The general case follows from the special case once you verify that $(f-f_n)^2$ is weakly convergent in $L_1$. Since $f_n^2$ converges weakly in $L_1$, it is enough to show that $f_n f$ converges weakly in $L_1$ to $f^2$, which in turn follows from the facts that $f_n$ converges weakly in $L_2$ to $f$ and $fg$ is in $L_2$ for every $g$ in $L_\infty$
The unnecessary assumption that $f_n \ge 0$ makes the problem harder. – Bill Johnson Feb 23 '12 at 18:29
I fixed the final equation in the first paragraph (from $f_n$ to $f_n^2$). – Matthew Daws Feb 23 '12 at 20:05