## 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 $F=f^2$?

-
 No, thanks, but non-negativity seems to be important. – unknown (yahoo) Feb 23 2012 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 $L_\infty$

Unfamiliar background can be found in standard text books; in particular, the book of Albiac and Kalton.

-
 The unnecessary assumption that $f_n \ge 0$ makes the problem harder. – Bill Johnson Feb 23 2012 at 18:29 I fixed the final equation in the first paragraph (from $f_n$ to $f_n^2$). – Matthew Daws Feb 23 2012 at 20:05 Thank you, Matt. – Bill Johnson Feb 23 2012 at 20:18 In the first sentence, do you really mean "even for non negative?" – Yemon Choi Feb 23 2012 at 21:28 Thanks, Yemon. I corrected. – Bill Johnson Feb 24 2012 at 1:06
show 1 more comment

It is obvious true. $f_{k}\rightarrow f$ in$L^{1}$ strongly,(for $\forall p <2$) $f_{k}^{2}\rightarrow f_{2}$ strongly, of course weakly, then $F=f^{2}$

-
This is not true-- if $\Omega=(0,1)$ say, and $f_k = k^{2/3}\chi_{(0,1/k)}$ then $\|f_k\| = k^{-1/3}\rightarrow 0$ but $\|f_k^2\| = k^{1/3} \not\rightarrow 0$. Note that this is not a counter-example to the OP. – Matthew Daws Feb 22 2012 at 20:01
Yes, yaoxiao is not right, but this is not a counterexample to OP; however, I like positivity of the functions in this example – unknown (yahoo) Feb 23 2012 at 12:30