Dual ball weak-star metrizable for a separable predual

The series defining $d$ converges because its terms lie between $0$ and $2^{-n}$. Symmetry and the triangle inequality follow from those of the absolute value and from $\min (1,a+b)\le \min (1,a)+\min (1,b)$. If $d(f,g)=0$, then $(f-g)(x_n)=0$ for all $n$; for any $x\in X$, take for each $k\ge 1$ the least index $n_k$ with $\parallel x-x_{n_k}\parallel <1\mathrm{/}k$. Then $x_{n_k}\to x$ and $\mathrm{\mid }(f-g)(x)\mathrm{\mid }\le 2M\parallel x-x_{n_k}\parallel \to 0$. Thus $f=g$. This also covers $M=0$, when $A$ has at most one point.

Fix $f\in A$ and a basic weak-star neighborhood $U=\{g\in A:\mathrm{\mid }(g-f)(y_j)\mathrm{\mid }<\epsilon ,1\le j\le m\}$. If $m=0$, take any metric ball. Otherwise, when $M>0$, choose $n_j$ with $\parallel y_j-x_{n_j}\parallel <\epsilon \mathrm{/}(4M)$; when $M=0$ the assertion is immediate. Put $\delta ={\min }_j{2}^{-n_j}\min (1,\epsilon \mathrm{/}2)>0$. If $d(f,g)<\delta$, then $\mathrm{\mid }(g-f)(x_{n_j})\mathrm{\mid }<\epsilon \mathrm{/}2$, and hence $\mathrm{\mid }(g-f)(y_j)\mathrm{\mid }<2M\epsilon \mathrm{/}(4M)+\epsilon \mathrm{/}2=\epsilon$. Thus a $d$-ball about $f$ lies in $U$.

Conversely, given $\eta >0$, choose $N$ so that ${\sum }_{n>N}{2}^{-n}<\eta \mathrm{/}2$ and put $\rho =\min (1,\eta \mathrm{/}2)$. The weak-star neighborhood $V=\{g\in A:\mathrm{\mid }(g-f)(x_n)\mathrm{\mid }<\rho ,1\le n\le N\}$ satisfies $d(f,g)<\rho {\sum }_{n\le N}{2}^{-n}+\eta \mathrm{/}2<\eta$.

Step 2.1 makes every weak-star neighborhood contain a metric neighborhood, while step 2.2 makes every metric neighborhood contain a weak-star neighborhood. Hence the two relative topologies on $A$ agree.