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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$\begingroup$ No, thanks, but non-negativity seems to be important. $\endgroup$– user21629Commented Feb 23, 2012 at 12:25
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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.
answered Feb 23, 2012 at 18:28
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$\begingroup$ The unnecessary assumption that $f_n \ge 0$ makes the problem harder. $\endgroup$ Commented Feb 23, 2012 at 18:29
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$\begingroup$ I fixed the final equation in the first paragraph (from $f_n$ to $f_n^2$). $\endgroup$ Commented Feb 23, 2012 at 20:05
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$\begingroup$ In the first sentence, do you really mean "even for non negative?" $\endgroup$ Commented Feb 23, 2012 at 21:28
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$\begingroup$ Sorry for the unnecessary assumptions $\endgroup$ Commented Feb 24, 2012 at 19:30