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 $F=f^2$?



Weak L_1-convergence of squares 
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.


