Suppose $f_j \to f$ strongly in $L^2(E)$ and $g_j \to g \ $ weakly in $L^2(E)$, where $E \subset \mathbb{R}^d$ is measurable. Show that $ \ \int_Ef_jg_jdx \to \int_E fgdx$.
Answer:
I am quoting the following theorem:
$\text{Product of weak-strong converging sequence}:$
Let $1 < p < \infty$, $ \ u_n: \Omega \to \mathbb{R}$ be a sequence in $L^p(\Omega)$, and $u \in L^p(\Omega)$. Let $v_n: \Omega \to \mathbb{R}$ be a sequence in $L^{p'}(\Omega)$, $ \ \frac{1}{p}+\frac{1}{p'}=1$. Suppose,
$u_n$ converges to $u$ weakly in $L^p(\Omega)$,
$v_n$ converges to $v$ strongly in $L^{p'}(\Omega)$,
Then $u_n v_n \to uv$ weakly in $L^{1}(\Omega)$
This is the theorem.
By applying the above theorem, we get
$f_jg_j$ converges to $fg$ weakly in $L^{1}(\Omega)$.
Does this conclude that $ \int_E f_jg_j dx \to \int_E fgdx$?
Help me
Yes: weak convergence of a sequence $ \left(h_n\right)_{n\geqslant 1}$ in $\mathbb L^1$ to a function $h$ means that for each linear continuous functional $\Phi\colon\mathbb L^1\to\mathbb R$, the sequence $\left(\Phi\left(h_n\right)\right)_{n\geqslant 1}$ converges to $\Phi\left(h\right)$. This definition is applied to $\Phi\colon h\mapsto \int_E h(x)\mathrm dx$, which is linear and continuous.