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The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology making $f\mapsto f(x)$ continuous for every $x\in X$. Its basic neighbourhoods of zero impose finitely many inequalities $|f(x_j)|<ϵ$.

Suppose first that zero had a countable neighbourhood base $(U_n)$ in the entire dual. Choose a basic neighbourhood $V_n\subset U_n$ controlled by a finite set $F_n\subset X$, and let $E_n=\mathrm{span}(F_1\cup \cdots \cup F_n)$. Given $x\in X$, the neighbourhood $W_x=\{f:|f(x)|<1\}$ contains some $U_n$, hence $V_n$. If $x\notin \mathrm{span}{F}_n$, the Hahn-Banach theorem supplies a bounded linear functional $h$ which vanishes on $F_n$ but has $h(x)\ne 0$. Indeed, the finite-dimensional span is closed, and the functional taking $e+tx$ to $t$ on its sum with $\mathbb{F}x$ is bounded because $\mathrm{dist}(x,\mathrm{span}{F}_n)>0$. Every scalar multiple of $h$ lies in $V_n$, contradicting $V_n\subset W_x$. Thus $x\in E_n$, and $X={\bigcup }_n{E}_n$ has countable Hamel dimension.

If $X$ is infinite-dimensional, every finite-dimensional $E_n$ is closed and has empty interior. Completeness rules out their union. For clarity, the Baire category theorem needed here has a direct nested-ball proof: inside a starting open ball choose a closed ball missing $E_1$, then inside its interior choose a closed ball missing $E_2$, and continue with positive radii tending to zero. The centres are Cauchy; their limit belongs to every ball and to none of the $E_n$, a contradiction. A metric topology would have a countable ball base at zero. HenceThis is the weak-star topology on an entire infinite-dimensional Banach dual is not metrizable result. It does not contradict weak-star metrizability of the dual ball for a separable predual: that assertion concerns a bounded subset.

The Banach-Alaoglu theorem states that $B_{X^{\ast }}$ is compact for $\sigma (X^{\ast },X)$ for every normed vector space $X$, whether complete or not. Embed it in the product ${\prod }_{x\in X}\{a\in \mathbb{F}:|a|\le \Vert x\Vert \}$ by its evaluations. Each factor is compact, so Tychonoff theorem makes the product compact Hausdorff. The equations expressing additivity and scalar homogeneity define a closed subset of the product. Its elements are exactly the bounded linear functionals of norm at most one, since the coordinate bounds give $|f(x)|\le \Vert x\Vert$. The induced product topology is precisely weak-star, proving the theorem.

The theorem printed as “Goldstein” is the standard Goldstine theorem: $J(B_X)$ is weak-star dense in $B_{X^{\ast \ast }}$ for the canonical embedding into the bidual. To prove it, fix $\Phi \in B_{X^{\ast \ast }}$ and $f_1,\dots ,f_m\in X^{\ast }$. The set $S=\{(f_1(x),\dots ,f_m(x)):x\in B_X\}$ is convex. If $b=(\Phi (f_1),\dots ,\Phi (f_m))$ were outside its closure, finite-dimensional Hahn-Banach separation theorem would supply coefficients $a_j$ such that, for $f={\sum }_j{a}_j{f}_j$,The real case omits real parts; in the complex case every real-linear separating functional on ${\mathbb{C}}^m$ has the displayed form. The supremum equals the dual norm by multiplying vectors by signs or unimodular scalars. The strict inequality contradicts $\Vert \Phi \Vert \le 1$. Thus $b\in \overline{S}$, which is exactly approximation on every prescribed finite family of weak-star coordinates.

Now take $K=B_{X^{\ast }}$ with its compact Hausdorff weak-star topology. For each $x\in X$, the evaluation $\hat{x}:f\mapsto f(x)$ is continuous on $K$, and Hahn-Banach theorem gives $\Vert \hat{x}{\Vert }_{\infty }=\Vert x\Vert$. Thereforeis the isometric evaluation embedding into continuous functions on a compact dual ball, over the same scalar field as $X$.

For the final sequential assertions, choose a norm-dense sequence $(f_j)$ in the separable dual $X^{\ast }$ and suppose $\Vert x_n\Vert \le M$. Each bounded scalar sequence $f_j(x_n)$ has a convergent subsequence: divide a bounding interval or square into finitely many smaller closed pieces, repeatedly retain one containing infinitely many terms, and let the diameters tend to zero. Successive extraction for $j=1,2,\dots$, followed by a diagonal subsequence $(y_n)$, makes every $f_j(y_n)$ converge.

If $M=0$ the conclusion is immediate. Otherwise, for any $f\in X^{\ast }$ and $ϵ>0$, choose $j$ with $\Vert f-f_j\Vert <ϵ/(4M)$. For sufficiently large $n,m$,Thus $\phi (f)={\lim }_{n}f(y_n)$ exists for every $f$. Taking limits proves linearity, and $|\phi (f)|\le M\Vert f\Vert$ provesThis diagonal argument includes the continuity estimate rather than appealing to a sequential compactness theorem.

If $X$ is not reflexive, choose $\Phi \in B_{X^{\ast \ast }}\setminus JX$, scaling a bidual element outside $JX$ if necessary. The Goldstine theorem proved above supplies $x_n\in B_X$ with $|f_j(x_n)-\Phi (f_j)|<1/n$ for $j\le n$. Norm density and the bound $\Vert J{x}_n-\Phi \Vert \le 2$ extend this convergence to every $f\in X^{\ast }$. Hence $J{x}_n\to \Phi$ weak-star. Every subsequence has the same coordinate limits, so a weakly convergent subsequence with limit $x\in X$ would give $Jx=\Phi$, a contradiction. This sequential Goldstine approximation for a separable dual provesThe construction uses the Goldstine proof already given, not an unproved weak sequential compactness result.

In an infinite-dimensional Hilbert space, every bounded sequence has a weakly convergent subsequence, but it need not have a norm-convergent one. For example an orthonormal sequence $e_n$ has pairwise distance $\sqrt{2}$ and therefore no strongly convergent subsequence, although $e_n\rightharpoonup 0$ by Bessel inequality.

More generally, weak sequential compactness of bounded sequences in a reflexive Banach space ensures a weakly convergent subsequence with its limit in that space. Conversely, this property for all bounded sequences characterizes reflexivity, using the Eberlein-Smulian theorem and weak compactness of the closed unit ball. Thus reflexivity replaces finite-dimensional norm compactness by weak sequential compactness.

In a nonreflexive Banach space there is no such general weak-subsequence guarantee. The coordinate unit vectors in ${\ell }^1$ give an example: by the Schur property, weak convergence would imply norm convergence, impossible for their pairwise distance two. Banach-Alaoglu theorem supplies weak-star compactness of a dual unit ball, which gives convergent subnets, not automatically convergent subsequences. For a separable predual, the weak-star compact dual ball is metrizable and subsequences do suffice. For an arbitrary Banach space one can also regard a bounded sequence in its bidual and extract a weak-star convergent subnet there; the limit need not belong to the original space. These qualifications distinguish weak from weak-star compactness.

The weak-star topology $\sigma (X^{\ast },X)$ is generated by the seminorms $p_x(g)=|g(x)|$, $x\in X$. Thus every evaluation is continuous, so $f_n\stackrel{w^{\ast }}{\to }f$ implies $f_n(x)\to f(x)$ for each $x$. Conversely, a basic neighborhood of $f$ imposes finitely many conditions $|(g-f)(x_j)|<\epsilon$. Pointwise convergence eventually satisfies all of them simultaneously. HenceThe Uniform boundedness principle, applied on the Banach space $X$, gives $M={\sup }_n\Vert f_n\Vert <\infty$. If $x_n\to x$ in the norm topology, thenTherefore $\boxed{f_n(x_n)\to f(x)}$. Completeness of $X$ is used in this uniform boundedness step.

If $C\subseteq X^{\ast }$ is norm compact, the identity map from $(C,\Vert \cdot \Vert )$ to $(C,w^{\ast })$ is a continuous bijection. The target is Hausdorff, since distinct functionals differ at some $x\in X$. The compact-to-Hausdorff continuous bijection theorem gives

The Banach-Alaoglu theorem states that the closed dual unit ball $B_{X^{\ast }}$ is compact in the weak-star topology; in fact $X$ need only be a normed vector space. To prove it, work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and formEach $D_x$ is compact, so $Q$ is compact by the Tychonoff theorem. Embed $B_{X^{\ast }}$ by $g\mapsto (g(x){)}_{x\in X}$. Its image consists exactly of those families $(z_x)$ satisfyingThese are closed conditions in the product topology. Conversely, any family satisfying them defines a linear functional bounded by $|z_x|\le \Vert x\Vert$, hence an element of $B_{X^{\ast }}$. The image is therefore closed and compact. The induced product topology is precisely the weak-star topology, proving the theorem. This is a compactness proof for arbitrary nets, with no separability assumption.

For the dual map of bounded linear operator $T^{\ast }:Y^{\ast }\to X^{\ast }$, each $x\in X$ satisfiesEvaluation at the fixed vector $Tx$ is weak-star continuous, so $\boxed{T^{\ast }\text{ is weak-star-to-weak-star continuous}}$.

Suppose $T$ is a compact operator. By the allowed Schauder theorem for compact operators, $T^{\ast }$ is compact. If $S\subseteq Y^{\ast }$ is bounded, the norm closure $C=\overline{T^{\ast }(S)}$ is norm compact. We already proved that its norm and weak-star topologies coincide. The weak-star continuous map $T^{\ast }{|}_S$ into $C$ is consequently norm continuous. Conversely, if $T^{\ast }$ is weak-star-to-norm continuous on bounded subsets, its restriction to $B_{Y^{\ast }}$ maps a weak-star compact set, by Banach-Alaoglu theorem, to a norm compact set. Thus $T^{\ast }$ is a compact operator, and the Schauder theorem for compact operators gives

The Mazur theorem states that the weak closure and norm closure of a convex set $C$ in a normed vector space coincide:Norm closure is contained in weak closure because the weak topology is coarser. Conversely, if $x\notin {\overline{C}}^{\Vert \cdot \Vert }$, the Hahn-Banach separation theorem provides a continuous real linear functional strictly separating $x$ from that closed convex set. In a complex space this is the real part of a continuous complex linear functional. A weak neighborhood of $x$ then misses $C$, so $x\notin {\overline{C}}^w$. This proves the equality. It also gives the usual Mazur lemma: if $x_n\rightharpoonup x$, then $x$ lies in the norm closure of the convex hull of each tail, so one can choose tail convex combinations $y_n$ with $\Vert y_n-x\Vert <1/n$.

Now let $K$ be a weakly compact set in a normed space $X$. Each $\phi \in X^{\ast }$ is bounded on $K$ because it is weakly continuous. The family $\{Jx:x\in K\}$ in $\mathcal{B}(X^{\ast },\mathbb{F})$, where $J$ is the canonical embedding into the bidual, is therefore pointwise bounded. The space $X^{\ast }$ is Banach even if $X$ is not. Apply the Uniform boundedness principle and use $\Vert Jx\Vert =\Vert x\Vert$ to obtainThis proves that a weakly compact set is norm bounded without assuming completeness of the original space.

For the real-valued dual and integral formulas that follow, take $X$ to be real, as in the PDF. In a complex space the norming formula uses real parts, and the integral identities use complex-valued functionals instead.

If the separable Banach space $X$ is nonzero, choose a norm-dense sequence $(u_n)$ in its unit sphere. By the Hahn-Banach theorem, choose ${\phi }_n\in X^{\ast }$ with $\Vert {\phi }_n\Vert =1$ and ${\phi }_n(u_n)=1$. For any unit vector $u$, arbitrarily close $u_n$ satisfyScaling gives the countable norming family identityFor $X=\{0\}$ use the constant sequence of zero functionals. If $f:\Omega \to X$ is norm-Borel measurable, every ${\phi }_n\circ f$ is measurable, so its countable supremum $\Vert f(\cdot )\Vert$ is measurable. Equivalently, this also follows directly from continuity of the norm.

For any $\phi \in X^{\ast }$, continuity makes $\phi \circ f$ measurable, andThus the assumed integrability of the norm implies scalar integrability, anddefines a bounded linear functional on $X^{\ast }$. Use the granted weak-star continuity of $T_f$. By the continuous dual of a weak-star topology, $T_f$ is evaluation at a vector of $X$. Indeed, continuity gives finitely many $x_1,\dots ,x_m$ and $\epsilon >0$ such that $|T_f(\phi )|<1$ whenever $|\phi (x_j)|<\epsilon$ for all $j$. Scaling shows that $T_f$ vanishes on the common kernel of these evaluations. It therefore factors through their finite-dimensional coordinate map, so $T_f(\phi )={\sum }_j{a}_j\phi (x_j)=\phi ({\sum }_j{a}_j{x}_j)$. The Hahn-Banach theorem makes this representing vector unique. HenceIn this separable setting the vector is the Bochner integral.

Return to a weakly compact set $K\subseteq X$ and its inclusion $f:K\to X$. For each fixed $a\in X$, the identity $\Vert x-a\Vert ={\sup }_n{\phi }_n(x-a)$ makes $x\mapsto \Vert x-a\Vert$ weakly Borel measurable. Norm balls are consequently weakly Borel measurable. Separability gives a countable base of such balls, so every norm-open set is weakly Borel measurable. This proves measurability of $f$, and establishes the equality of the weak and norm Borel sigma-algebras in a separable Banach space.

Put $M={\sup }_{x\in K}\Vert x\Vert <\infty$. For every finite signed Borel measure $\mu$ on $K$,For positive measures this is the integral in the question. For signed measures, the correct integrability condition uses the variation measure; define the integral by taking the difference of the positive and negative integrals. The printed $\Omega$ in this clause should be $K$, the domain of the inclusion.

The Riesz-Markov-Kakutani representation theorem now defines the bounded linear mapFor each $\phi \in X^{\ast }$ the restriction $\phi {|}_K$ is in $C(K)$, andThe right side is weak-star continuous in $\mu$. The defining property of the weak topology therefore proves that $T$ is weak-star-to-weak continuous, for arbitrary nets. For a Dirac measure, $\boxed{T({\delta }_x)=x}$.

If $K\ne \varnothing$, let $\mathcal{P}(K)$ be its regular probability measures. This is a weak-star closed subset of $B_{C(K{)}^{\ast }}$: its conditions are $\mu (1)=1$ and $\mu (g)\ge 0$ for every nonnegative $g\in C(K)$. It is compact by Banach-Alaoglu theorem. Thus $T(\mathcal{P}(K))$ is weakly compact and convex, and contains $K$ because it contains all $T({\delta }_x)$. It is weakly closed, hence norm closed, so it contains $C={\overline{\mathrm{conv}K}}^{\Vert \cdot \Vert }$. By Mazur theorem, $C$ is weakly closed. Therefore it is a closed subset of the weakly compact set $T(\mathcal{P}(K))$, provingThe empty case is immediate. In fact $T(\mathcal{P}(K))=C$: a barycenter of a measure on a Banach space outside $C$ would be strictly separated by a functional $\phi$, contradicting $\phi (T(\mu ))={\int }_K\phi d\mu \le {\sup }_K\phi$.

For the first norm-one assertion, use the usual unital Banach algebra convention $\Vert 1\Vert =1$. A character of an algebra is a nonzero complex linear multiplicative functional, initially without any continuity assumption. It satisfies $\phi (1)=1$, since otherwise $\phi (a)=\phi (a1)=0$ for every $a$. Moreover, $a-\phi (a)1$ cannot be invertible: applying $\phi$ to an inverse identity would give $0=1$. ThusThis proves automatic continuity of characters and $\Vert \phi \Vert \le 1$. Evaluation at the normalized identity gives the reverse inequality, soIf a convention permits $\Vert 1\Vert >1$, only $\Vert \phi \Vert \le 1$ follows in general: on $\mathbb{C}$ with norm $2|z|$, the identity character has norm $1/2$. The later completion argument needs only the upper bound.

If $a$ is invertible, $\phi (a)\phi (a^{-1})=1$, so every character is nonzero on $a$. Conversely, if $a$ is not invertible, its principal ideal is proper and lies in a maximal ideal $M$ by Zorn lemma. The ideal $M$ is closed: if its closure were the whole algebra, it would contain $m$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series, which is impossible for a proper ideal. Maximality therefore gives $\overline{M}=M$. The quotient $A/M$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. The quotient map yields a character vanishing at $a$. Consequently

Now let $A=C(K)$. Its evaluation characters give the map $x\mapsto {\delta }_x\in {\Phi }_A$. Every character has this form. Indeed, its kernel is a proper ideal. If the functions in that kernel had no common zero, compactness would give finitely many $f_1,\dots ,f_m$ in the kernel with no common zero. Then $h={\sum }_j\overline{f_j}{f}_j$ belongs to the kernel and is everywhere positive, hence is invertible in $C(K)$, a contradiction. At a common zero $x$, the fact that $f-\phi (f)1$ belongs to the kernel gives $f(x)=\phi (f)$ for every $f$. Continuous functions distinguish points of the compact Hausdorff space $K$, so the map is bijective. It is continuous for the Gelfand topology, and the compact-to-Hausdorff continuous bijection theorem givesFor $K=\varnothing$ the norm comparison is trivial; assume henceforth that $K\ne \varnothing$.

Let $B$ be the completion for the other algebra norm $\Vert \cdot {\Vert }_1$. It remains commutative and unital, with $A$ embedded densely. The restriction of any $\psi \in {\Phi }_B$ is a nonzero character of $A$, since it takes $1$ to $1$, and is automatically continuous for the original supremum norm. Thus $R(\psi )=\psi {|}_A$ maps into ${\Phi }_A$. It is injective because two continuous characters agreeing on the dense subalgebra $A$ agree on $B$, and it is continuous because each coordinate $R(\psi )(f)=\psi (f)$ is weak-star continuous.

The character space ${\Phi }_B$ is weak-star compact even if its identity was not normalized: the spectral bound puts all its characters in $B_{B^{\ast }}$, and the conditions $\psi (1)=1$, $\psi (bc)=\psi (b)\psi (c)$ define a weak-star closed set. Apply Banach-Alaoglu theorem. Consequently $R$ is a homeomorphism onto a closed subset $L\subseteq K$.

Suppose $x\in U=K\setminus L$. The compact Hausdorff space $K$ is normal, so choose an open $V$ with $x\in V\subseteq \overline{V}\subseteq U$. By the Urysohn lemma, choose $g\in C(K)$ with $g(x)=1$ and $g=0$ on $K\setminus V$, and choose $f\in C(K)$ with $f=0$ on $\overline{V}$ and $f=1$ on $L$. Then $fg=0$. Every character $\psi$ of $B$ restricts to evaluation at a point of $L$, so $\psi (f)=1$. The invertibility criterion proved above makes $f$ invertible in $B$. Multiplying $fg=0$ by $f^{-1}$ in $B$ gives $g=0$, contradicting $g(x)=1$. Therefore $U=\varnothing$ and $R$ is onto $K$.

For each $x\in K$, choose a character ${\psi }_x\in {\Phi }_B$ extending ${\delta }_x$. Its spectral bound givesTaking the supremum over $x$ proves the minimality of the supremum norm on C(K):

True. If $(x_n)$ is dense in $B_X$, then on $B_{X^{\ast }}$ the weak-star topology is induced by a metric such asThe Banach-Alaoglu theorem makes this metric space compact, and every compact metric space is sequentially compact.

Goldstine theorem states that the canonical image $J(B_X)$ of the closed unit ball of a normed vector space $X$ is weak-star dense in $B_{X^{\ast \ast }}$.

The Banach-Alaoglu theorem states that $B_{X^{\ast }}$ is compact in the weak-star topology. To prove it, map each $f\in B_{X^{\ast }}$ to its values inEvery factor is compact, so Tychonoff theorem makes $P$ compact. The image of $B_{X^{\ast }}$ is cut out by the closed linearity conditionsand is therefore closed in $P$. The product topology restricted to this image is exactly pointwise convergence on $X$, namely the weak-star topology. Hence the ball is compact.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology on $X$ for which every bounded linear functional in $X^{\ast }$ is continuous. Thus every $f\in X^{\ast }$ is weakly continuous. Conversely, if a linear functional $f$ is weakly continuous at zero, some basic weak neighbourhood gives $f_1,\dots ,f_n\in X^{\ast }$ and $\epsilon >0$ such thatIt follows that ${\bigcap }_j\ker f_j\subseteq \ker f$, and elementary linear algebra then gives $f\in \mathrm{span}\{f_1,\dots ,f_n\}\subseteq X^{\ast }$.

To prove Mazur theorem, let $K$ be norm-closed and convex and let $x\notin K$. The Hahn-Banach separation theorem strictly separates $x$ from $K$ by some member of $X^{\ast }$. The corresponding open half-space is weakly open, contains $x$, and misses $K$. Thus $K$ is weakly closed.

If $X$ is reflexive, the Banach-Alaoglu theorem makes $B_{X^{\ast \ast }}$ weak-star compact, and the canonical identification transports this to weak compactness of $B_X$. Conversely, if $B_X$ is weakly compact, then $J(B_X)$ is weak-star compact and hence weak-star closed in $B_{X^{\ast \ast }}$. Goldstine theorem says it is weak-star dense there, so it equals $B_{X^{\ast \ast }}$ and $X$ is reflexive.

When $X$ is reflexive, weak and weak-star topologies coincide on $X^{\ast }$, so Banach–Alaoglu makes $B_{X^{\ast }}$ weakly compact and $X^{\ast }$ is reflexive. If $Y\subseteq X$ is closed, then $B_Y$ is a weakly closed subset of $B_X$, hence weakly compact. The quotient map sends a suitable weakly compact ball of $X$ onto the unit ball of $X/Y$, which is therefore weakly compact. Thus $Y$ and $X/Y$ are reflexive as well.

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.

The useful forms of the Hahn-Banach separation theorem in a real locally convex space are:

The Banach-Alaoglu theorem says that the closed unit ball of $E^{\ast }$ is compact in $\sigma (E^{\ast },E)$. Goldstine theorem says that the canonical image of the closed unit ball of a normed space $E$ is weak-star dense in the closed unit ball of $E^{\ast \ast }$.

Give $Z$ its norm inherited from $X^{\ast }$ and defineIt is linear and contractive, and it is injective because $Z$ separates points. Goldstine's theorem followed by restriction from $X^{\ast }$ to $Z$ shows that $J(B_X)$ is weak-star dense in $B_{Z^{\ast }}$: a functional on $Z$ first extends norm-preservingly to $X^{\ast }$, and elements of $B_X$ approximate that extension on every finite subset of $Z$.

The topology induced by $\sigma (Z^{\ast },Z)$ on $J(B_X)$ is exactly the given topology $\sigma (X,Z)$. By hypothesis $B_X$ is compact, so $J(B_X)$ is weak-star compact and therefore closed in the Hausdorff space $B_{Z^{\ast }}$. Density now givesThus $J$ is surjective and maps closed unit ball onto closed unit ball, so it is an isometric isomorphism. Hence $X$ is a dual space. This is the compact norming dual-pair criterion.

If $X$ is reflexive, then $X^{\ast }$ is reflexive. The weak topology $\sigma (X^{\ast },X^{\ast \ast })$ and weak-star topology $\sigma (X^{\ast },X)$ therefore coincide under the canonical identification $X^{\ast \ast }=X$. Thus every weak-star convergent sequence in $X^{\ast }$ is weakly convergent, so $X$ is a Grothendieck space.

Conversely, suppose that $X$ is separable and Grothendieck. The Banach-Alaoglu theorem and weak-star metrizability of the dual ball make $B_{X^{\ast }}$ weak-star compact and metrizable, hence weak-star sequentially compact. Every convergent subsequence is weakly convergent by the Grothendieck property. Thus $B_{X^{\ast }}$ is weakly sequentially compact. The stated converse to part (a), equivalently the other direction of the Eberlein-Smulian theorem, makes $B_{X^{\ast }}$ weakly compact. Hence $X^{\ast }$ is reflexive, and therefore so is $X$.

Finally let $T:X\to Y$ be bounded and onto, with $X$ Grothendieck, and suppose $y_n^{\ast }\to y^{\ast }$ weak-star in $Y^{\ast }$. Thenweak-star in $X^{\ast }$, hence weakly. The open mapping theorem implies that $T^{\ast }:Y^{\ast }\to X^{\ast }$ is an isomorphism onto its closed range. Given $y^{\ast \ast }\in Y^{\ast \ast }$, the functionalis bounded on $T^{\ast }(Y^{\ast })$ and extends by the Hahn-Banach theorem to some $x^{\ast \ast }\in X^{\ast \ast }$. ThereforeThis is weak convergence in $Y^{\ast }$, so $Y$ is Grothendieck.

In a unital C-star algebra $A$, an element is Hermitian when $x=x^{\ast }$, unitary when $x^{\ast }x=x{x}^{\ast }=1$, and normal when $x^{\ast }x=x{x}^{\ast }$. It is a Positive element of a C-star algebra when $x=y^{\ast }y$ for some $y$, equivalently when it is Hermitian and its spectrum lies in $[0,\infty )$.

Suppose first that $\Vert \tau \Vert =\tau (1)$, and normalize so that both equal one. If $x=x^{\ast }$, the C-star identity givesWriting $\tau (x)=a+ib$, we obtainfor every real $t$, which forces $b=0$. Decomposing an arbitrary $x$ into its Hermitian real and imaginary parts now gives

Let $x=x^{\ast }$. Every character of an algebra on the commutative unital Banach algebra generated by $x$ has norm and value at one equal to one, so the preceding argument makes its value on $x$ real. The character description of the spectrum of an element therefore gives $\sigma (x)\subset \mathbb{R}$. Iterating the C-star identity,so $\Vert x\Vert =r(x)$. If $c=\Vert x\Vert$, then $x+c1$ has nonnegative spectrum and is positive, as is $c1$; henceis a difference of positive elements.

If $0\le x$ and $\Vert x\Vert \le 1$, then spectral translation givesThus $1-x$ is positive and, by the norm--spectral-radius equality for Hermitian elements, $\Vert 1-x\Vert \le 1$.

Return to a norm-one $\tau$ with $\tau (1)=1$. For $0\le x\le 1$, the preceding paragraph and reality on Hermitian elements giveScaling proves that $\tau$ is a positive functional on a C-star algebra. Every character has norm and value at one equal to one, so every character is positive. On $C[0,1]$, the functionalis positive but is not multiplicative, and hence is not a character.

Conversely, let $\tau$ be positive. Writing a Hermitian element as a difference of positive elements shows that $\tau$ is real on Hermitian elements. Positivity offor every $\lambda \in \mathbb{C}$ says that the associated quadratic polynomial is nonnegative. Minimizing it in $\lambda$ gives the Cauchy--Schwarz inequalityTaking $y=1$ yields $|\tau (x){|}^2\le \tau (x^{\ast }x)\tau (1)$. Since $0\le x^{\ast }x\le \Vert x{\Vert }^{2}1$, positivity givesTogether with $\tau (1)\le \Vert \tau \Vert$, this proves $\Vert \tau \Vert =\tau (1)$.

Convex combinations preserve positivity and value one at the identity, so the state space $S(A)$ is convex. If $x$ is normal, the unital C-star subalgebra $C^{\ast }(1,x)$ is commutative and its Gelfand transform identifies it with $C(\sigma (x))$. Choose $\lambda \in \sigma (x)$ with $|\lambda |=\Vert x\Vert$. Evaluation at $\lambda$ is a state taking $x$ to $\lambda$. Its norm-preserving Hahn--Banach extension to $A$ still takes $1$ to one, so the norm criterion makes the extension a state $\tau$ satisfying $|\tau (x)|=\Vert x\Vert$.

The state space is nonempty, convex, and weak-star compact by the Banach-Alaoglu theorem. The Krein-Milman theorem gives an extreme point, so a pure state on a C-star algebra exists. For positive $x$, the preceding norm-attainment result makesa nonempty weak-star compact convex set. It is a face: if a convex combination has the maximal possible value on $x$, each summand does. An extreme point of $F$ exists by Krein--Milman and, because $F$ is a face, is extreme in $S(A)$. It is the required pure state.

For a closed unital subalgebra $A\subseteq B$ containing $x$, invertibility in $A$ implies invertibility in $B$, so ${\sigma }_B(x)\subseteq {\sigma }_A(x)$. On a connected component of the resolvent set of $x$ in $B$, the set of $\lambda$ for which $(x-\lambda 1{)}^{-1}\in A$ is both open, by a local Neumann series, and closed, by closedness of $A$. It contains all sufficiently large $\lambda$, hence the entire unbounded component. Thus spectrum in a closed unital subalgebra says that ${\sigma }_A(x)$ is ${\sigma }_B(x)$ with some bounded complementary components filled in.

Now let $A$ be the Banach subalgebra generated by one element $x$ and put $K={\sigma }_A(x)$. If $D$ were a bounded component of $\mathbb{C}\setminus K$, choose $\lambda \in D$. Since $(x-\lambda 1{)}^{-1}\in A$, polynomials $p_n(x)$ converge to it. The polynomialssatisfy $q_n(\lambda )=0$ and $q_n(x)\to 1$. Applying the contractive Gelfand transform gives $q_n\to 1$ uniformly on $K$, hence on $\partial D$. But the maximum modulus principle applied to $1-q_n$ givesa contradiction. Therefore $\mathbb{C}\setminus K$ is connected.

The mapis continuous and surjective by part a. It is injective because characters agreeing on $x$ agree on every polynomial in $x$, hence by continuity on their norm closure $A$. The character space is compact by the Banach-Alaoglu theorem, while $K$ is Hausdorff, so this continuous bijection is a homeomorphism.

Under this identification, the Gelfand transform $\theta :A\to C(K)$ obeysby the calculation in part b. Since $f(x)\in A$, choose polynomials $p_n$ with $p_n(x)\to f(x)$. Contractivity of $\theta$ givesThus every function holomorphic near $K$ is uniformly approximable there by polynomials.

The set $K$ is weak-star compact by the Banach-Alaoglu theorem and weak-star metrizable by part a. Choose a decreasing countable neighbourhood base $(U_n)$ at $f$. Since $f\in K_{\epsilon }^{\prime }$, the set $U_n\cap K$ cannot lie inside the closed norm ball of radius $\epsilon /2$ about $f$, for then its diameter would be at most $\epsilon$. ChooseThe neighbourhood-base property gives $g_n\stackrel{w^{\ast }}{\to }f$.

For a unital Banach algebra, $\phi (a)$ belongs to ${\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. Thereforeso every character of an algebra is continuous and has norm one. The nonunital case follows by extending the character to the unitization of an algebra.

The Gelfand topology on ${\Phi }_A$ is the weak-star topology inherited from $A^{\ast }$: a net ${\phi }_i$ converges to $\phi$ exactly when ${\phi }_i(a)\to \phi (a)$ for every $a\in A$. If $A$ is unital, ${\Phi }_A$ lies in the weak-star compact dual unit ball by the Banach-Alaoglu theorem. The equationsdefine a weak-star closed subset, so ${\Phi }_A$ is compact.

A linear map $T:X\to Y$ is a weakly compact operator when $T(B_X)$ is relatively weakly compact. Every weakly compact subset of a Banach space is norm bounded, soThus even without assuming continuity initially, a weakly compact linear map is bounded.

The key bidual criterion isFor the forward implication, approximate $x^{\ast \ast }\in B_{X^{\ast \ast }}$ weak-star by points $J_X{x}_{\alpha }$ using the Goldstine theorem. Weak compactness supplies a subnet for which $T{x}_{\alpha }$ converges weakly to some $y\in Y$, while weak-star continuity of $T^{\ast \ast }$ makes $J_{Y}T{x}_{\alpha }$ converge to $T^{\ast \ast }{x}^{\ast \ast }$. Hence $T^{\ast \ast }{x}^{\ast \ast }=J_{Y}y$. Conversely, if the displayed inclusion holds, the weak-star compact set $T^{\ast \ast }(B_{X^{\ast \ast }})$ lies in $J_Y(Y)$, where the inherited weak-star topology is the weak topology of $Y$. It is a weakly compact set containing $J_{Y}T(B_X)$.

If $T$ is weakly compact and $y^{\ast \ast \ast }\in Y^{\ast \ast \ast }$, restriction of $y^{\ast \ast \ast }$ to $J_Y(Y)$ defines $y^{\ast }\in Y^{\ast }$. For $x^{\ast \ast }\in X^{\ast \ast }$,so $T^{\ast \ast \ast }{y}^{\ast \ast \ast }=J_{X^{\ast }}(T^{\ast }{y}^{\ast })$. The bidual criterion makes $T^{\ast }$ weakly compact. Conversely, if $T^{\ast }$ is weakly compact, then $T^{\ast \ast \ast }(Y^{\ast \ast \ast })\subseteq J_{X^{\ast }}(X^{\ast })$. Every $y^{\ast \ast \ast }$ annihilating $J_Y(Y)$ is sent to zero, so every $T^{\ast \ast }{x}^{\ast \ast }$ lies inThe criterion makes $T$ weakly compact. Hence $T$ is weakly compact exactly when $T^{\ast }$ is.

The bidual criterion also proves the structure of $\mathcal{W}(X,Y)$. It is closed under linear combinations. If $T_n\to T$ in operator norm and every $T_n$ is weakly compact, then $T_n^{\ast \ast }{x}^{\ast \ast }\to T^{\ast \ast }{x}^{\ast \ast }$ in norm; the canonical copy $J_Y(Y)$ is norm closed, so $T$ is weakly compact. For bounded composable maps $A$ and $B$, the image condition for $(ATB{)}^{\ast \ast }=A^{\ast \ast }{T}^{\ast \ast }{B}^{\ast \ast }$ proves the ideal property. Thus the weakly compact operators form a norm-closed operator ideal.

The Krein-Smulian theorem states that the norm-closed convex hull $\overline{\mathrm{co}}K$ of a weakly compact set $K$ is weakly compact. To prove it, view $K$ as a compact Hausdorff space in its weak topology. The probability measures on $K$ form a weak-star compact subset of $C(K{)}^{\ast }$ by the Banach-Alaoglu theorem. The barycentre mapis weak-star-to-weak continuous; scalar integration and weak compactness ensure that $b_{\mu }\in X$. Finitely supported probability measures are weak-star dense and their barycentres are precisely $\mathrm{co}K$. The image of all probability measures is therefore the weak closure of $\mathrm{co}K$, and it is weakly compact. A convex set has the same weak and norm closures by the Hahn-Banach separation theorem, proving the claim.

If $X$ is reflexive, $B_X$ is weakly compact and every bounded $T:X\to Y$ is weak-to-weak continuous, so $T$ is weakly compact. If $Y$ is reflexive, every bounded subset of $Y$ is relatively weakly compact, with the same conclusion. Thuswhenever either space is reflexive.

If $X={\ell }^1$ and $Y$ is nonreflexive, the Eberlein-Smulian theorem supplies a bounded sequence $(y_n)$ in $Y$ with no weakly convergent subsequence. The formuladefines a bounded operator $T:{\ell }^1\to Y$. Since $T{e}_n=y_n$, it cannot be weakly compact. Therefore $\mathcal{B}({\ell }^1,Y)\ne \mathcal{W}({\ell }^1,Y)$.