I am considering a integral of product of three functions i.e. $$\int_{\Omega}f_{\epsilon}^{2}g_{\epsilon}dx$$ where $f_{\epsilon}\to f$ weakly in $L^{2}(\Omega)$ and $g_{\epsilon}\to g$ almost everywhere and $0\leq g_{\epsilon}\leq C$ for some $C>0$ and all $\epsilon$ and $\Omega$ is a bounded domain.
Moreover, I also know that this integral is uniformly bounded that is $$\int_{\Omega}f_{\epsilon}^{2}g_{\epsilon}dx\leq C$$ for some $C>0$.
What am I expecting is whether can I have some estimate for the limit integral i.e. if I can have $$\int_{\Omega}f^{2}g dx\leq \liminf_{\epsilon} \int_{\Omega}f_{\epsilon}^{2}g_{\epsilon}dx$$ Many thanks for any help!

My attempt is as following: since $g_{\epsilon}$ is uniformly postive, so $$\int_{\Omega}f_{\epsilon}^{2}g_{\epsilon}dx$$ can be viewed as the $L^{1}$ norm of $f_{\epsilon}^{2}g_{\epsilon}$, then by the weak lower semicontinuous of norm, I can get my estimate.
Does this make sense? My another try according to @daw's reply.
\begin{align*} \int_{\Omega}f_{\epsilon}^{2}g_{\epsilon}dx=\int_{\Omega}f_{\epsilon}^{2}gdx+\int_{\Omega}f_{\epsilon}^{2}(g_{\epsilon}-g)dx. \end{align*} First $$\int_{\Omega}f^{2}gdx\leq \liminf \int_{\Omega}f_{\epsilon}^{2}gdx$$ by weakly lower semiconitnuous.
Then by Egorov's theorem, for all $\eta>0$ there exists $|\Omega_{\eta}|<\eta$ such that $g_{\epsilon}\to g$ uniformly on $\Omega-\Omega_{\eta}$. Then \begin{align*} |\int_{\Omega}f_{\epsilon}^{2}(g_{\epsilon}-g)dx|&\leq|\int_{\Omega_{\eta}}f_{\epsilon}^{2}(g_{\epsilon}-g)dx|+|\int_{\Omega-\Omega_{\eta}}f_{\epsilon}^{2}(g_{\epsilon}-g)dx|. \end{align*} Moreover, the integrablity of $f_{\epsilon}$ and $g_{\epsilon}$ implies the first term goes to zero as $\eta$ goes to zero.
For the second term \begin{align*} |\int_{\Omega-\Omega_{\eta}}f_{\epsilon}^{2}(g_{\epsilon}-g)dx|&\leq \lVert f_{\epsilon}\rVert_{L^{2}(\Omega)}^{2}\lVert g_{\epsilon}-g\rVert_{L^{\infty}(\Omega-\Omega_{\eta})}\\ &\leq C\lVert g_{\epsilon}-g\rVert_{L^{\infty}(\Omega-\Omega_{\eta})}\to0. \end{align*} So we are done. Does this make more sense?

Okay, maybe this estimate is not something I can expect? But what about if I assume that $g_{\epsilon}$ is strictly positive i.e. $0<C_{1}\leq g_{\epsilon}\leq C_{2}$. From this, I think the bounded implies that $$\lVert \sqrt{g_{\varepsilon}}f_{\varepsilon}\rVert_{2}\leq C$$ Then by the weak lower semicontinous and a.e. convergence, I can get the inequality I want.
For this, i think we first take arbitrary $\phi\in L^{2}(\Omega)$. Then \begin{align*} &|\int_{\Omega}(\sqrt{g_{\epsilon}}f_{\epsilon}-\sqrt{g}f)\phi dx|\\ \leq&|\int_{\Omega}\sqrt{g}(f_{\epsilon}-f)\phi dx|+|\int_{\Omega}(\sqrt{g_{\epsilon}}-\sqrt{g})f_{\epsilon}\phi dx| \end{align*} The first term goes to zero by weak convergence and the second due to dominated convergence theorem. Does this make sense?

  • $\begingroup$ You can apply weak lower semicontinuity to the term $\int f_\epsilon^2 g$ but not the other one. Maybe you can use Egorov's theorem to do something for $\int f_\epsilon^2 (g-g_\epsilon)$. $\endgroup$
    – daw
    Commented Feb 1 at 21:21
  • $\begingroup$ @daw, may i ask why i can not apply the weak lower semicontinuity directly? because i think in the boundedness $\int_{\Omega}f_{\epsilon}^{2}g_{\epsilon}dx\leq C$, i can view $f_{\epsilon}^{2}g_{\epsilon}$ as a whole? $\endgroup$
    – Emiya
    Commented Feb 1 at 21:30
  • 1
    $\begingroup$ I don't really understand this argument: "Moreover, the integrablity of fϵ and gϵ implies the first term goes to zero as η goes to zero.". You need some uniformity in epsilon. $\endgroup$
    – PhoemueX
    Commented Feb 2 at 5:47

1 Answer 1


As OP argued, we have:

Lemma. Consider measurable functions $(f_n)$ and $(h_n)$ on a measure space $(\Omega, \Sigma, \mu)$ such that

  1. $f_n \rightharpoonup f$ in $L^2(\Omega)$ for some function $f$, and
  2. $ h_n $ is uniformly bounded and $h_n \to h$ almost surely for some function $h$.

Then $ f_n h_n \rightharpoonup f h$ in $L^2(\Omega)$.

Proof. Let $M = \sup_n \|f_n\|_2$. By the uniform boundedness principle, we know that $M < \infty$. Now consider an arbitrary test function $\phi \in L^2(\Omega)$. Then

\begin{align*} \left| \int_{\Omega} f_n h_n \phi \, \mathrm{d}\mu - \int_{\Omega} f h \phi \, \mathrm{d}\mu \right| &\leq \left| \int_{\Omega} (f_n - f) h \phi \, \mathrm{d}\mu \right| + \left| \int_{\Omega} f_n (h - h_n) \phi \, \mathrm{d}\mu \right| \\ &\leq \left| \int_{\Omega} (f_n - f) h \phi \, \mathrm{d}\mu \right| + M \| (h - h_n) \phi \|_2. \end{align*}

The first term converges to $0$ because $f_n \rightharpoonup f$, and the second term converges to $0$ because $h_n \phi \to h \phi$ in $L^2(\Omega)$ by the dominated convergence theorem. $\square$

Now consider OP's question. Under OP's setting, we know that $ f_{\epsilon} \sqrt{g_{\epsilon}} \rightharpoonup f \sqrt{g} $ by the above lemma. Then the desired inequality is the consequence of the lower semicontinuity of the norm (with respect to the weak topology):

Theorem. If $f_n \rightharpoonup f$ in $L^2(\Omega, \Sigma, \mu)$, then $$ \|f\|_2 \leq \liminf_{n\to\infty} \|f_n\|_2. $$

Proof. We already know that $f$ is in $L^2$. Then taking $\liminf$ as $n \to \infty$ to the inequality

$$ \int_{\Omega} f_n f \, \mathrm{d}\mu \leq \|f_n\|_2 \|f\|_2. $$

gives $\|f\|_2^2 \leq \liminf_{n\to\infty} \|f_n\|_2 \|f\|_2$, from which the desired conclusion follows. $\square$


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