Basic weak star neighborhoods

Each displayed set is a finite intersection of inverse scalar disks. Conversely, every finite intersection of subbasic sets containing $f$ contains such a set by shrinking each scalar open set to a disk and taking the smallest of the finitely many positive radii. The empty intersection needs no shrinking.

For a finite list set $p(h)={\max }_j\mathrm{\mid }h(x_j)\mathrm{\mid }$, with $p=0$ for an empty list. Linearity gives $p(h+k)\le p(h)+p(k)$ and $p(ah)=\mathrm{\mid }a\mathrm{\mid }p(h)$. Hence its open balls are balanced and real-convex; the bounds $p(h-h_0),p(k-k_0)<\epsilon \mathrm{/}2$ imply $p(h+k-h_0-k_0)<\epsilon$, proving addition is continuous.

To control scalar multiplication at $(a_0,h_0)$, use $p(ah-a_0{h}_0)\le \mathrm{\mid }a\mathrm{\mid }p(h-h_0)+\mathrm{\mid }a-a_0\mathrm{\mid }p(h_0)$. The requirements $\mathrm{\mid }a-a_0\mathrm{\mid }<\min (1,\epsilon \mathrm{/}(2(p(h_0)+1)))$ and $p(h-h_0)<\epsilon \mathrm{/}(2(\mathrm{\mid }{a}_0\mathrm{\mid }+1))$ make this less than $\epsilon$. Thus the topology is a locally convex vector topology.

If $f\mathrm{\ne }g$ as functions, some $x\in X$ has $f(x)\mathrm{\ne }g(x)$. Put $d=\mathrm{\mid }f(x)-g(x)\mathrm{\mid }>0$. The evaluation disks of radius $d\mathrm{/}3$ about these two values have disjoint inverse images containing $f$ and $g$, since a common member would give $d<2d\mathrm{/}3$. This proves Hausdorffness; on a singleton dual it is vacuous. No norming principle is involved. $\mathrm{\square }$