I am not sure what this question is asking.
A linear operator $T$ between normed spaces X and Y is bounded if and only if it is a continuous linear operator. But weak topology is not metrizable. I guess I should use Closed Graph Theorem, is this a correct direction?
Any help will be appreciated!
We need to show that if $V$ is open in the weak topology of $Y$ then $T^{-1}[V]$ is open in the weak topology of $X$.
It clearly suffices to show that if $\mathscr S$ is a sub-basis of neighborhoods of zero in the weak topology of $Y$ and $W\in\mathscr S$, then $T^{-1}[W]$ is weak open in $X$.
A sub-basis of neighborhoods of zero in in the weak topology of $Y$ is obtained as $$ \mathscr S=\{W_{\varphi,\varepsilon}:\varphi\in Y^*\,\&\,\varepsilon>0 \}, $$ where $$ W_{\varphi,\varepsilon}=\{y\in Y: \lvert\varphi(x)\rvert<\varepsilon\} =\varphi^{-1}[(-\varepsilon,\varepsilon)]. $$ Thus $$ T^{-1}[W_{\varphi,\varepsilon}]=T^{-1}\big[\varphi^{-1}[(-\varepsilon,\varepsilon)]\big] =(\varphi\circ T)^{-1}[(-\varepsilon,\varepsilon)]. $$ But $(\varphi\circ T)^{-1}[(-\varepsilon,\varepsilon)]$ is open in the weak topology of $X$ since $\varphi\circ T\in X^*$.
Thus, indeed $T$ is continuous with respect to the weak topologies.