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?
Best Answer
As OP argued, we have:
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):
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. $\square$