Transpose is weak to weak continuous

For $g\in Y^{\ast }$, $g\circ T\in X^{\ast }$ and $\mathrm{\mid }(g\circ T)(x)\mathrm{\mid }\le \mathrm{\parallel }g\mathrm{\parallel }\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$. The inverse under $T$ of a weak subbasic set $g^{-1}(V)$ is $(g\circ T{)}^{-1}(V)$, which is weakly open in $X$. Inverse images preserve unions and finite intersections, proving continuity of $T$. Equivalently, F2 gives $g(T{x}_i)\to g(Tx)$ for every weakly convergent net.

Taking the supremum over $\mathrm{\parallel }x\mathrm{\parallel }\le 1$ in the bound of step 1.1 gives $\mathrm{\parallel }{T}^{\ast }g\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }g\mathrm{\parallel }$, so $T^{\ast }$ is bounded. Apply the step 1.1 argument to this bounded map: for every $\mathrm{\Phi }\in X^{\ast \ast }$, $\mathrm{\Phi }\circ T^{\ast }\in Y^{\ast \ast }$ since $\mathrm{\mid }\mathrm{\Phi }(T^{\ast }g)\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }g\mathrm{\parallel }$. Thus inverse weak subbasic sets are weakly open, proving weak-to-weak continuity of $T^{\ast }$. The estimates remain valid for zero maps and zero spaces. $\mathrm{\square }$