Does $g_n \rightarrow 0$ converge weakly? This is where I am stuck while solving another problem.
Let $T:L^1 \rightarrow X$ be an operator such that $T|_{L^2(\mu)}$ is compact. 
Suppose $f_n$ be a sequence in $L^1$ such that $f_n \rightarrow 0$ weakly. Then set $g_n=f_n\mathbb{1}_{A_n}$ where $A_n=\{x : |f_n(x)|<M\}$ for some fixed $M$.
Then how can we say that $\|Tg_n\| \rightarrow 0$ in norm?
It would be very nice if we could say $g_n \rightarrow 0$ weakly.
Thanks for the help!
 A: I assume that $T|_{L^2(\mu)}$ is compact as an operator from the Hilbert space $L^2(\mu)$. 
As Topology has pointed out in a comment, it suffices to prove that $g_n\to 0$ weakly in $L^2(\mu)$, since compact operators maps weakly convergent sequences to norm convergent ones. Then sequence $(T g_n)_{n\in\mathbb{N}}$ is weakly convergent to $0$ and norm convergent, so it is norm convergent to $0$.
First, observe that for each $n\in\mathbb{N}$ we have $g_n\in L^2(\mu)$:
$\|g_n\|_2^2=\int |f_n|^2 \chi_{A_n} d\mu\le M^2<+\infty.$
Now, take any $g\in L^2(\mu)$. We want to prove $\langle g| g_n\rangle\xrightarrow[n\to\infty]{} 0$. Take any $\epsilon>0$. There exists $g'\in L^2(\mu)\cap L^{\infty}(\mu)$ such that $\|g-g'\|_2\le \epsilon$. Take $n_0\in\mathbb{N}$ such that $\int |g' f_n| d\mu\le\epsilon$ for $n>n_0$. Such $n_0$ exists due to weak convergence $f_n\to 0$ in $L^1(\mu)$. We have
$|\langle g|g_n \rangle|\le \int |(g-g') g_n| d \mu+\int |g' g_n|d \mu\le \|g-g'\|_2 \bigl(\int |f_n|^2 \chi_{A_n} d\mu\bigr)^{\frac{1}{2}} +\epsilon\le (M+1) \epsilon$
for every $n>n_0$ which finishes the proof.
