weakly-* convergence and dominated convergence

Question

Let us assume that $g\in C^{\infty}_c(\mathbb{R})$. Assume furthermore that there is some $g_n\rightarrow g$ strongly in $L^2(\mathbb{R})$ and we have $h_n \rightarrow h$ weakly-* in $L^{\infty}([0,T],H^s(\mathbb{R}))$ and $s\in (0,1)$. How can one show that $$ \int_{0}^{T}{(g_n, h_n(t))_{L^2}dt} \rightarrow \int_{0}^{T}{(g,h)_{L^2}dt}, $$ where $(.,.)_{L^2}$ is the inner product on $L^2$.

I tried to prove it via dominated convergence but it didn't work out (but I guess it should). We can assume that $h_n(t)\rightarrow h(t)$ weakly in $H^s$ for almost every $t\in [0,T]$.

I tried the following: $$ | (g_n, h_n(t))_{L^2} - (g,h)_{L^2} | \leq |(g_n-g,h_n(t)-h(t))_{L^2}| +|(g_n-g,h(t))_{L^2}| +|(g,h_n(t)-h(t))_{L^2}|=: A_1+A_2 + A_3. $$ Now I wanted to show that all $A_i$ vanish as $n\rightarrow 0$.

Any ideas?

Answer 1

You have $$\int_0^T (g_n, h_n(t))_{L^2} \, \mathrm dt = ( g_n, \int_0^T h_n(t)\,\mathrm d t)_{L^2}$$ and $\int_0^T h_n(t)\,\mathrm d t$ converges weakly in $H^s(\mathbb R)$, hence in $L^2(\mathbb R)$. Together with the strong convergence of $g_n$, this is enough to infer the convergence of the scalar product.