Transpose is weak star to weak star continuous

For bounded $T$, $\mathrm{\mid }(T^{\ast }f)(x)\mathrm{\mid }=\mathrm{\mid }f(Tx)\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$. Thus $T^{\ast }f\in X^{\ast }$ and $T^{\ast }$ is bounded and scalar-linear. For every $x\in X$, the composite of $T^{\ast }$ with evaluation at $x$ is evaluation at $Tx$. Its inverse scalar-open sets are weak-star open by the defining evaluation topology in F1. Thus $T^{\ast }$ is weak-star continuous.

For the converse, fix $x\in X$. The map $f\mapsto (Sf)(x)$ is weak-star continuous and scalar-linear, since $S$ is and evaluation at $x$ is. F1 gives a unique point $Tx\in Y$ with $f(Tx)=(Sf)(x)$ for all $f\in Y^{\ast }$. Unique specification defines $T$ on all $X$ without choosing from a family of non-singleton sets. For scalars $a,b$, evaluation gives $f(T(ax+bz))=(Sf)(ax+bz)=af(Tx)+bf(Tz)$ for every $f$. Point separation makes $T(ax+bz)=aTx+bTz$.

By F2 and boundedness of $S$, $\mathrm{\parallel }Tx\mathrm{\parallel }={\sup }_{\mathrm{\parallel }f\mathrm{\parallel }\le 1}\mathrm{\mid }(Sf)(x)\mathrm{\mid }\le \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$. Hence $T$ is bounded, and its defining identity says $S=T^{\ast }$. Any other preadjoint has the same evaluations at each $x$ and therefore equals $T$ by point separation. Zero spaces and $S=0$ satisfy the same formulas, with $T=0$. $\mathrm{\square }$