${x_n} \to x$ weakly, why does $T{x_n} \to Tx$ weakly?

Question

If $T \in B(X,Y)$ and ${x_n} \to x$ weakly, why does $T{x_n} \to Tx$ weakly?

Answer 1

Note that $x_n \to x$ weakly means that for every $f \in X'$, we have $f(x_n) \to f(x)$.

Now, note that for any $g \in Y'$, we have $g(Tx_n) = g \circ T(x_n)$, and that $g \circ T \in X'$. From there, we simply apply the definition.