 Solution

Write $h(t)=t/(1+t)$ for $t\ge 0$. Every summand in $d$ is nonnegative and at most $2^{-m}$, so the series converges. Symmetry is immediate. If $d(x,y)=0$, then ${\varphi }_m(x-y)=0$ for every $m$. For arbitrary $\varphi \in ({\ell }^p{)}^{\ast }$, choose a sequence of ${\varphi }_m$ converging to $\varphi$ in operator norm. Since $x-y$ is fixed, $\varphi (x-y)=0$. The continuous dual space separates points, so $x=y$.

The function $h$ is increasing and subadditive:Combining this with the triangle inequality for each $|{\varphi }_m(x-y)|$ and summing proves the triangle inequality for $d$. Hence $d$ is a metric on $B$.

Suppose first that $x^{(n)}$ converges weakly to $x$, meaning thatFor each fixed $m$, the $m$th summand of $d(x^{(n)},x)$ tends to zero. A finite-head and geometric-tail estimate, using that every summand is bounded by $2^{-m}$, then gives $d(x^{(n)},x)\to 0$.

Conversely, if $d(x^{(n)},x)\to 0$, each nonnegative summand tends to zero, and thereforefor every $m$. Given $\varphi \in ({\ell }^p{)}^{\ast }$ and $\epsilon >0$, choose $m$ with $\Vert \varphi -{\varphi }_m\Vert <\epsilon$. Since $x^{(n)},x\in B$,Taking the limit superior and then letting $\epsilon \downarrow 0$ proves weak convergence. ThusThe same approximation argument for neighbourhoods shows that $d$ induces the weak topology on $B$, as in the metrization of the weak topology on a bounded set.

If $q$ is conjugate to $p$, the given duality of sequence spaces applied twice givesso ${\ell }^p$ is a reflexive Banach space. Its closed unit ball is weakly compact by the weak compactness characterization of reflexivity. Since $d$ induces that topology on $B$,