Continuous dual of a weak star topology

Continuity at zero gives points $x_1,\dots ,x_m$ and $\epsilon >0$ such that $\mathrm{\mid }L(f)\mathrm{\mid }<1$ when $\mathrm{\mid }f(x_j)\mathrm{\mid }<\epsilon$ for all $j$. If all evaluations vanish, the same bound holds for every $tf$, forcing $L(f)=0$. Thus $L$ vanishes on the kernel of $A(f)=(f(x_1),\dots ,f(x_m))$.

The rule $\mathrm{\ell }(Af)=L(f)$ is therefore well-defined and linear on $A(X^{\ast })\subseteq {\mathbb{K}}^m$. Take a finite basis of this image and extend it to a basis of ${\mathbb{K}}^m$, adding standard coordinate vectors successively. Extend $\mathrm{\ell }$ by zero on added basis vectors. Writing its coordinate coefficients as $c_j$ gives $L(f)={\sum }_j{c}_{j}f(x_j)=f({\sum }_j{c}_j{x}_j)$. Set $x={\sum }_j{c}_j{x}_j$. Empty coordinates give $L=0$ and $x=0$. This finite construction needs no choice axiom.

Every evaluation at a given $x$ is weak-star continuous. If $x,y$ give the same evaluation then $f(x-y)=0$ for every $f\in X^{\ast }$. Under HB point separation implies $x=y$. This includes the zero space and proves the asserted identification and its uniqueness. $\mathrm{\square }$