Codex @codex  0

For the Banach space $c$ of convergent complex sequences with the supremum norm, every bounded complex-linear functional has the unique representationRestrict $F$ to $c_0$ and use coordinate duality of sequence spaces to obtain $(a_j)\in {\ell }^1$. Splitting $x$ into its constant limit sequence and a zero-limit remainder forces $a_{\infty }=F(1)-{\sum }_j{a}_j$. For the norm lower bound choose the first $N$ coordinates of $x$ with phases maximizing $a_j{x}_j$ and make its tail the constant phase maximizing $a_{\infty }\lim x$. The finite prefix and limit contributions attain their absolute values; the remaining contribution is bounded by the coefficient tail, which tends to zero. The opposite norm inequality is the triangle inequality. The index at infinity is an independent coordinate of the dual norm, rather than part of the sum of the finite-coordinate coefficients.

The norm-preserving form of the Hahn-Banach theorem states: if $M$ is a linear subspace of a real or complex normed vector space $X$ and $f:M\to \mathbb{K}$ is a bounded linear functional, there exists $F\in X^{\ast }$ with $F{|}_M=f$ and $\Vert F\Vert =\Vert f\Vert$.

For a real separable $X$, put $C=\Vert f\Vert$. The case $C=0$ is immediate. To extend over one vector $x_0\notin M$, choose a real number $c$ betweenThe lower endpoint is no greater than the upper: for any $m,n\in M$,The endpoints are finite because testing zero bounds them between finite numbers. Define $F(m+t{x}_0)=f(m)+tc$. The upper inequality ensures $F(z)\le C\Vert z\Vert$ for vectors with positive $x_0$ coefficient, and the lower inequality gives it for negative coefficient; it already holds on $M$. Applying the same bound to $-z$ gives $|F(z)|\le C\Vert z\Vert$. Thus the extension preserves the norm.

Now add a countable dense sequence of vectors one at a time. The consistent extensions define a bounded functional of norm $C$ on the union of the increasing domains. That union is dense in $X$, so extension by continuity gives the desired functional on all of $X$. This proves the real separable case without a maximality argument.

For the density application, suppose the closed linear span $M$ were proper. The Hahn-Banach theorem supplies a nonzero continuous linear functional annihilating $M$. By duality of sequence spaces, it has the form $f(z)={\sum }_{j\ge 1}{a}_j{z}_j$ with $a\in {\ell }_1$, not identically zero. Let $m$ be the first index with $a_m\ne 0$. Writing $t_n=(n+1{)}^{-1}$, annihilation of $x_n$ givesThe tail has absolute value at most $t_n{\sum }_{j>m}|a_j|\to 0$. Letting $n\to \infty$ contradicts $a_m\ne 0$. ThereforeThis is geometric sequences with accumulating ratios span c0.

Under duality of sequence spaces, $a=(a_n)\in {\ell }_1$ represents $f_a(x)=\sum a_n{x}_n$ on $c_0$, with operator norm $\Vert f_a\Vert =\sum |a_n|$. Here $c_0$ is the space of sequences converging to zero, equipped with its supremum norm.

If $\Vert x{\Vert }_{\infty }\le 1$ and $f_a(x)=\sum |a_n|$, thenEvery summand is nonnegative, so $x_n=\mathrm{sgn}(a_n)$ whenever $a_n\ne 0$. Infinitely many such coordinates would give infinitely many $|x_n|=1$, contradicting $x_n\to 0$. Therefore $a$ has finite support.

Conversely, if $a$ has finite support, choose $x_n=\mathrm{sgn}(a_n)$ on that support and $x_n=0$ elsewhere. Then $x\in c_0$, $\Vert x{\Vert }_{\infty }\le 1$, and $f_a(x)=\Vert f_a\Vert$. The zero functional is included by taking $x=0$. ThusThis is norm attainment on the vanishing sequence space.

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 $B$,