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The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology making every bounded linear functional continuous. A neighbourhood basis at $x$ consists of sets $\{y:|f_j(y-x)|<ϵ,1\le j\le m\}$ for finite families in $X^{\ast }$.

Mazur theorem says that for any convex subset $C$ of a real or complex normed vector space,The norm closure is convex. If $x$ is outside it, the Hahn-Banach separation theorem gives a bounded linear functional $f$ and a real $a$ with $\mathrm{Re}f(x)>a\ge {\sup }_{y\in C}\mathrm{Re}f(y)$. In the real case omit the real part. Thus $x$ has a weak neighbourhood missing $C$, proving that the weak closure lies in the norm closure. The other inclusion follows because the weak topology is weaker than the norm topology. In the complex case real separation is converted to a complex functional by $f(z)=h(z)-ih(iz)$.

The sequential formulation, Mazur lemma, follows as well. If $x_n\rightharpoonup x$, then $x$ lies in the weak closure of each tail and hence in the norm closure of its convex hull. Select a finite convex combination of the $n$th tail at norm distance less than $1/n$ from $x$.

A weakly bounded set $D$ satisfies ${\sup }_{x\in D}|f(x)|<\infty$ for every $f\in X^{\ast }$. Regard $J(D)$ as a pointwise bounded family of functionals on $X^{\ast }$. The completeness of the dual space holds even if $X$ is incomplete: a norm-Cauchy sequence of functionals has a pointwise bounded linear limit and then converges uniformly on the unit ball. The Uniform boundedness principle therefore gives ${\sup }_{x\in D}\Vert Jx\Vert <\infty$, and $\Vert Jx\Vert =\Vert x\Vert$ proves norm boundedness.

One can see the precise Baire argument here. The closed sets $F_n=\{f\in X^{\ast }:{\sup }_{x\in D}|f(x)|\le n\}$ cover the Banach space $X^{\ast }$. The Baire category theorem makes some $F_n$ contain a ball $f_0+r{B}_{X^{\ast }}$, after shrinking the radius. For $\Vert g\Vert \le r$, both $f_0$ and $f_0+g$ lie in $F_n$, so ${\sup }_D|g(x)|\le 2n$. Scaling and the dual norm formula yield ${\sup }_D\Vert x\Vert \le 2n/r$. A weakly compact set is weakly bounded because each functional has compact, hence bounded, image. Consequently it is norm bounded.

For a Banach space, let $J:X\to X^{\ast \ast }$ be its canonical isometry. We use two explicitly stated weak-star facts: Banach-Alaoglu theorem makes a dual unit ball weak-star compact, and Goldstine theorem makes $J(B_X)$ weak-star dense in $B_{X^{\ast \ast }}$. The weak topology on $X$ is carried by $J$ to $\sigma (X^{\ast \ast },X^{\ast })$ on its image, since their coordinates are the same evaluations $f(x)$.

If $X$ is a reflexive Banach space, $J(B_X)=B_{X^{\ast \ast }}$ and Banach-Alaoglu proves weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact and hence closed in the Hausdorff space $X^{\ast \ast }$. Goldstine density then forces $J(B_X)=B_{X^{\ast \ast }}$, which implies $JX=X^{\ast \ast }$. This proves the weak compactness characterization of reflexivity:Finally let $K\subset X$ be weakly compact and let $(f_n)\subset B_{X^{\ast }}$ separate points. If $K$ is empty there is nothing to prove; otherwise defineThe summands are bounded by $2^{-n}$ and separation makes $d(x,y)=0$ only for $x=y$. The coordinate maps are weakly continuous, so the series, being uniformly convergent, makes the identity from weak $K$ to metric $K$ continuous. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Thus $d$ induces precisely the weak topology on $K$, the countable separating family metrizes a weakly compact set result. No norm-density of the separating family in $X^{\ast }$ is claimed or required.

Put $d=\inf \{\Vert x\Vert :x\in C\}$. The nonemptiness of $C$ gives $0\le d<\infty$. By Mazur theorem, a norm-closed convex set is weakly closed. This applies to both $C$ and every closed norm ball.

For $n\ge 1$, the sets $C_n=C\cap (d+1/n)B_X$ are nonempty by the definition of the infimum, weakly closed and nested. They all lie in the weakly compact set $(d+1)B_X$, by the weak compactness characterization of reflexivity. Hence they have the finite intersection property, and compactness gives $x_0\in {\bigcap }_n{C}_n$. Then $x_0\in C$ and $\Vert x_0\Vert \le d+1/n$ for every $n$, soThis proves the norm minimizer in a closed convex subset of a reflexive Banach space assertion without assuming sequential weak compactness. It includes $d=0$; uniqueness is not asserted without an additional condition such as strict convexity of the norm.

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 $B$,

Suppose first that $X$ is separable, and choose a norm-dense sequence $(x_n)$ in its unit ball. On the dual unit ball defineUniform boundedness on the unit ball and density of the $x_n$ show that this metric induces the weak-star topology. Conversely, if $B_{X^{\ast }}$ is weak-star metrizable, the Banach-Alaoglu theorem makes it a compact metric space. Hence $C(B_{X^{\ast }})$ is separable. The evaluation mapis an isometry by the Hahn-Banach theorem. A subspace of a separable metric space is separable, so $X$ is separable. This proves the weak-star metrizability criterion for a dual ball.

If $X$ has a countable weakly dense subset $D$, the rational linear span of $D$ is weakly dense. Its norm closure is a convex set, so Mazur theorem says that its weak and norm closures agree. Thus $X$ is norm separable. The weak-star compact metric ball $B_{X^{\ast }}$ consequently has a countable weak-star dense subset, and the union of its integer dilates is weak-star dense in $X^{\ast }$. Therefore $X^{\ast }$ is weak-star separable.

It need not be weakly separable. Take $X={\ell }^1$, whose dual is ${\ell }^{\infty }$. A weakly separable normed space is norm separable by the preceding convex-closure argument, whereas ${\ell }^{\infty }$ is not norm separable.

If the Banach space $X$ is reflexive, its closed unit ball identifies with the weak-star compact ball of $X^{\ast \ast }$, hence is weakly compact. Conversely, if $B_X$ is weakly compact, its canonical image $J(B_X)$ is weak-star compact and therefore weak-star closed in $X^{\ast \ast }$. Goldstine theorem says it is weak-star dense in $B_{X^{\ast \ast }}$, soScaling proves that $J$ is surjective and $X$ is reflexive. This is the weak compactness characterization of reflexivity.

The Krein-Milman theorem says that a nonempty compact convex subset of a locally convex space is the closed convex hull of its extreme points. For reflexive $X$, the ball $B_X$ is weakly compact, sowhere weak and norm closure agree for the convex hull by Mazur's theorem.

For the final claim, let $\mathcal{H}$ be the set of functions ${\mathbb{Z}}^2\to [0,1]$ with the mean-value property. It is a compact convex subset of the product $[0,1{]}^{{\mathbb{Z}}^2}$. If $f$ is extreme, its four unit translates also lie in $\mathcal{H}$, and the mean-value identity writes $f$ as their average. Extremality forces every translate to equal $f$, so $f$ is constant. Every extreme point is therefore constant. Krein--Milman now makes every member of $\mathcal{H}$ a limit of convex combinations of constant functions, and hence constant. This is the bounded harmonic function theorem on the integer lattice.