Does $T$ bounded linear operator necessarily imply weak sequential continuity

Question

Let $X,Y$ be Banach Spaces, show that $x_{n} \xrightarrow{w} x$ and $T \in BL(X,Y)\Rightarrow Tx_{n} \xrightarrow{w} Tx$

Question: Does sequential continuity (which $T$ clearly has) necessarily imply that $T$ is weak-sequentially continuous? If so, then the above is trivial.

Otherwise:

$\vert\ell(Tx_{n})-\ell(Tx)\vert=\vert\ell(Tx_{n}-Tx)\vert=\vert T^{*}\ell(x_{n}-x)\vert\xrightarrow{n\to \infty} 0$ since $T^{*}\ell \in X^{*}$. I am somewhat unsure about this, since I have not used boundedness of $T$ anywhere.

Answer 1

If $y^{*} \in Y^{*}$ then $x^{*}(x)=y^{*}(Tx)$ defines a continuous linear functional on $X$. Hence $y^{*}(Tx_n)\to y^{*}(Tx)$. This implies that $Tx_n \to Tx$ weakly.

Answer 2

As in the previous answer you can prove sequential continuity. A stronger result is the following: bounded linear operators are continuous in the weak topology.

Remark that the weak topology is not metrizable (no sequential characterization of continuity).