# liminf estimate of a integral of weak convergence and almost everywhere convergence

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. $$0. From this, I think the bounded implies that $$\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?

• 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 $\int f_\epsilon^2 (g-g_\epsilon)$.
– daw
Commented Feb 1 at 21:21
• @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? Commented Feb 1 at 21:30
• 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. Commented Feb 2 at 5:47

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. $$\square$$