Show a bounded linear operator is weakly sequentially continuous

Question

Let $X$ and $Y$ be normed linear spaces, $T \in B(X,Y)$, and $\{x_{n}\}_{n=1}^{\infty}\subset X$. If $x_n \rightharpoonup x$, I need to prove that $T x_{n} \rightharpoonup T x$ in $Y$, where $\rightharpoonup$ denotes weak convergence.

Answer 1

Take $f\in Y^*$. Define $g\in X^*$ by $g(x)=f(Ax)$. Since $x_n\rightharpoonup x$ in $X$, it follows $g(x_n)\to g(x)$, or equivalently $$ f(Ax_n) \to f(Ax). $$ Since this holds for all $f\in Y^*$, we have verified that $Ax_n\rightharpoonup Ax$ in $Y$.