Weak topology of normed space Let $X,Y$ be two normed space and $T:X\rightarrow Y$ be bounded linear operator.Now consider $X,Y$ with weak topology. My question is that does $T$ maps weakly compact set of $X$ to weakly compact set of $Y$ and second question is that does $T$  remains a continuous map if we equip $X,Y$ with weak topology.
 A: If $V$ is a subbasis element of $\tau_w$ in $Y$ containing $0_Y$, then there is a functional $\phi:Y\to \mathbb F$ and $\epsilon>0$ such that $V=\{y:\phi(y)<\epsilon\}$. Then, $T^{-1}(V)=\{x:(\phi\circ T)(x)<\epsilon\}$. Now $\phi\circ T:X\to \mathbb F$ is a (norm-) continuous linear functional so $T^{-1}(V)$ is weakly open in $X$ and contains $0_X$. It follows that $T$ is weak-weak continuous. This gives an affirmative answer to the second question, which it turn gives an affirmative answer to the first.
A: This answer doesn't provide anything new, but I think an explanation in terms of sequences might be clearer.  The compactness question follows from weak-to-weak continuity (the implication holds for arbitrary topologies), so it suffices to show the latter.
Suppose $\{x_n\}_n\rightharpoonup y$.  Then, for all $f\in X^*$, $\{f(x_n)\}_n\to f(y)$.  In particular, any dual of the form $g\circ T$, where $g\in Y^*$, will satisfy $$\{g(Tx_n)\}_n\to g(Ty)$$  But this is just $\{Tx_n\}_n\rightharpoonup Ty$.
