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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

2 Answers 2

up vote 4 down vote accepted

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.

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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
    
Thank you, Matt. –  Bill Johnson Feb 23 '12 at 20:18
    
In the first sentence, do you really mean "even for non negative?" –  Yemon Choi Feb 23 '12 at 21:28
    
Thanks, Yemon. I corrected. –  Bill Johnson Feb 24 '12 at 1:06

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 $F=f^{2}$

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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 '12 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 –  user21629 Feb 23 '12 at 12:30

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