$f_n \rightharpoonup f$ in $L^2(\mathbb{R})$ and $f_n^2 \rightharpoonup g$ in $L^1(\mathbb{R})$, then $f^2\leq g$ a.e.

Could you guys help me check the proof please, thanks!

Proof: to show $f^2 \leq g$ a.e. it suffices to show that for any measurable set $A$ with $\mu(A) >0$ we have $$\int_A f^2 \leq \int_A g. $$ $$\bigg(\int_A f^2 \bigg)^2 = \bigg(\lim_{n\rightarrow \infty} \int_A f_n f \bigg)^2 \leq \lim_{n\rightarrow \infty} \bigg(\int_A f_n^2 \bigg)\bigg(\int_A f^2 \bigg) = \bigg(\int_A g \bigg)\bigg(\int_A f^2 \bigg),$$ and because $\int_A f^2 \geq 0$, we have the desired result.

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    $\begingroup$ This looks fine. $\endgroup$ – Etienne Aug 1 '14 at 10:28
  • $\begingroup$ @Etienne Thank you very much! $\endgroup$ – Xiao Aug 1 '14 at 10:32
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    $\begingroup$ convergence in $L^p$ implies there is a subsequence which converges a.e., so $f^2 = g$ a.e. $\endgroup$ – Petite Etincelle Aug 1 '14 at 10:51
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    $\begingroup$ @LiuGang: You only have weak convergence here! In this case, you don't necessarily have convergence a.e. for a subsequence. $\endgroup$ – PhoemueX Aug 1 '14 at 11:13
  • $\begingroup$ @PhoemueX It's said as convergence in $L^2$ and $L^1$, isn't it? $\endgroup$ – Petite Etincelle Aug 1 '14 at 11:15

Yes, it is correct. Maybe some steps could be justified: for example, write: "Using Cauchy-Schwarz inequality, we have for each integer $n$, $$\tag{*}\left(\int_A f_nf\right)^2\leqslant \left(\int_Af_n^2\right)\left(\int_Af^2\right).$$ On one hand, since $f_n\to f$ weakly in $L^2$ and $\chi_Af\in L^2$, the LHS of (*) converges to $\left(\int_A f^2\right)^2$. On the other hand, since $\chi_A\in L^\infty$, we get $\lim_n\int_Af_n^2=\int_Ag$, hence $\int_Af^2\leqslant \int_Ag$."


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