Let $X$ and $Y$ be Banach spaces. Show that if a linear operator $T$ from $X$ into $Y$ is $w$-continuous, then $T$ $\in$ $\mathcal B$$(X,Y)$.
So far, this is what I've done:
Given $f\in Y^*$, consider $V=f^{-1}(-1,1)$. Then $V$ is a $w$-neighborhood of zero in $Y$. By our assumption, $ T^{-1}(V)$ is a $w$-neighborhood of zero in $ X$. Therefore, $\lambda B_x\subset T^{-1}(V)$ for some $\lambda $; that is $\vert f(T(B_x))\vert\le\frac 1\lambda$. Thus $T(B_x)$ is w-bounded and also bounded in $Y$. On the other hand, if $\lim_{w}x_\alpha$=$0$ in $X$ and $f\in Y^*$, then $f\circ T\in X^*$ and thus $\lim_{w}T(x_\alpha)=0 \in Y$.
What I ask of is if anyone could provide a proof, analogous to the upper one. Thank you.
Let $\{x_n\}$ be bounded. Define $ S_n:Y^{*} \to \mathbb R$ by $S_n(y^{*})=y^{*}(T(x_n))$. Apply Uniform Boundedness Principle to show that $\{T(x_n)\} $ is bounded.