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For a probability measure whose identity map is strongly measurable with integrable norm, its barycenter is the Bochner integral of that map. It is characterized by $\phi (b(\mu ))=\int \phi (x)d\mu (x)$ for every continuous linear functional. For a measure on a weakly compact set in a separable Banach space, its barycenter belongs to the norm-closed convex hull of that set.

If a countable family in $X^{\ast }$ separates points of $X$, the displayed metric induces the weak topology on every weakly compact set $K$. The coordinate map into the countable product of scalar lines is continuous and injective; a continuous injection from a compact space into a Hausdorff space is a homeomorphism onto its image. Equivalently, uniform convergence of the metric series makes the identity from weak $K$ to metric $K$ continuous. Compactness is essential: point separation alone need not generate the weak topology on an arbitrary bounded set.

The weak topology on a normed vector space $X$ is $\sigma (X,X^{\ast })$, the coarsest topology making every bounded linear functional continuous. A neighbourhood base at $x$ is given by finitely many inequalities $|f_j(y-x)|<\epsilon$, with $f_j\in X^{\ast }$. In the complex case, separation uses real parts of bounded linear functionals.

Mazur theorem states that the weak closure of a convex set equals its closure in the norm topology. The weak topology is coarser than the norm topology, so the norm closure is contained in the weak closure. Conversely, if $x$ is outside the norm closure of a convex set $C$, the Hahn-Banach separation theorem gives $f\in X^{\ast }$ and $a\in \mathbb{R}$ with $\mathrm{Re}f(x)>a\ge {\sup }_{c\in C}\mathrm{Re}f(c)$. The corresponding weak topology neighbourhood of $x$ misses $C$, so $x$ is outside its weak closure. This proves Mazur theorem, including the empty-set case. In particular, if $x_n$ converges weakly to $x$, then $x$ is in the weak closure of every tail and therefore in the norm closure of its convex hull. Choosing a finite convex combination of the $n$th tail within $1/n$ of $x$ proves the usual Mazur lemma formulation as well.

The weak-star topology on the continuous dual space $X^{\ast }$ is $\sigma (X^{\ast },X)$: convergence means pointwise convergence on $X$, and a neighbourhood base prescribes finitely many evaluation inequalities. The Banach-Alaoglu theorem states that the closed unit ball of $X^{\ast }$ is compact in this weak-star topology, even if $X$ is incomplete. Embed this closed unit ball intowhere $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Each factor is compact, so $P$ is compact by the Tychonoff theorem. Inside $P$, the equations $a_{x+y}=a_x+a_y$ and $a_{\lambda x}=\lambda a_x$ define a closed set. Every such point defines a linear functional satisfying $|a_x|\le \Vert x\Vert$, hence belongs to the closed unit ball of $X^{\ast }$. Thus this image is closed in $P$. The product topology on it is exactly the weak-star topology, proving Banach-Alaoglu theorem. Evaluations also separate its points, so the weak-star topology is Hausdorff.

For the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, we have $(J_{X}x)(f)=f(x)$. Restricting all evaluations at $f\in X^{\ast }$ therefore gives exactly $\sigma (X,X^{\ast })$. Thus the induced subspace topology is the weak topology on $X$.

Now identify $X$ with $J_{X}X$. Let $C$ be a bounded convex set, write $D$ for its norm closure in $X$, and let $K$ be its weak-star topology closure in $X^{\ast \ast }$. Boundedness places $K$ in a multiple of the closed unit ball of $X^{\ast \ast }$, so $K$ is compact by Banach-Alaoglu theorem. The preceding subspace topology identification and Mazur theorem giveIf $D$ is a weakly compact set, its image in the Hausdorff weak-star topology is compact and therefore closed. It contains $C$, so $K\subseteq D\subseteq X$. Conversely, if $K\subseteq X$, the displayed identity gives $K=D$, and its compactness is precisely weak compactness in $X$. Hence $\boxed{D\text{ is weakly compact}⟺K\subseteq X}$. This argument also covers $C=\varnothing$.

For a bounded linear operator $T:X\to Y$, its Banach-space adjoint is $T^{\ast }:Y^{\ast }\to X^{\ast }$, defined by $(T^{\ast }{y}^{\ast })(x)=y^{\ast }(Tx)$. Evaluation at any fixed $x$ is thus evaluation at $Tx$ after applying $T^{\ast }$. Each is continuous in the relevant weak-star topology, proving that $T^{\ast }$ is weak-star continuous. Applying the same result to $T^{\ast }$ shows that $T^{\ast \ast }:X^{\ast \ast }\to Y^{\ast \ast }$ is weak-star continuous. Direct evaluation gives

Use $B_X$ for the closed unit ball; using the open ball gives the same norm closure of $T(B_X)$. Goldstine theorem says that $J_X{B}_X$ is weak-star dense in $B_{X^{\ast \ast }}$. Put $K=T^{\ast \ast }(B_{X^{\ast \ast }})$. It is compact and closed in the weak-star topology by Banach-Alaoglu theorem and the established continuity. It contains $J_{Y}T(B_X)$. Conversely, Goldstine theorem gives, for every $x^{\ast \ast }\in B_{X^{\ast \ast }}$, a net $(J_X{x}_{\alpha })$ from $J_X{B}_X$ converging weak-star to $x^{\ast \ast }$; its image converges weak-star to $T^{\ast \ast }{x}^{\ast \ast }$. ConsequentlyApply the preceding bounded convex set criterion in $Y$, and then scale the closed unit ball. We obtain the bidual characterization of weakly compact operators:The canonical embedding into the bidual on the right specifies exactly which copy of $Y$ is intended.

Use the canonical embedding into the bidual $J_{X}x(f)=f(x)$. It is an isometry by the Hahn-Banach theorem. The weak-star topology $\sigma (X^{\ast \ast },X^{\ast })$, restricted to $J_{X}X$, is exactly the weak topology $\sigma (X,X^{\ast })$: both are determined by the same scalar evaluations $f(x)$. Therefore if $K\subseteq X$ is a weakly compact set, $J_{X}K$ is compact in the Hausdorff weak-star topology of $X^{\ast \ast }$, and hence is closed for the weak-star topology. This explains the first assertion without treating the whole embedded space $J_{X}X$ as weak-star closed.

For a bounded linear operator $T:X\to Y$, its Banach-space adjoint isFor each $x\in X$, the scalar evaluation $y^{\ast }\mapsto (T^{\ast }{y}^{\ast })(x)=y^{\ast }(Tx)$ is weak-star continuous on $Y^{\ast }$. These are exactly the defining evaluations of the weak-star topology on $X^{\ast }$, so $T^{\ast }$ is weak-star-to-weak-star continuous. Equivalently, if $y_{\alpha }^{\ast }\stackrel{w^{\ast }}{\to }{y}^{\ast }$, then $(T^{\ast }{y}_{\alpha }^{\ast })(x)\to (T^{\ast }{y}^{\ast })(x)$ for every $x$.

The Goldstine theorem states that $J_X(B_X)$ is dense in $B_{X^{\ast \ast }}$ for $\sigma (X^{\ast \ast },X^{\ast })$, where the balls are closed unit balls. The Banach-Alaoglu theorem states that for any normed vector space $E$, the closed dual unit ball $B_{E^{\ast }}$ is compact for $\sigma (E^{\ast },E)$. Neither assertion claims norm compactness or, in general, sequential compactness.

For the weakly compact operator equivalences below, identify $X$ and $Y$ with their canonical images only when stated. In particular, the range condition means $T^{\ast \ast }(X^{\ast \ast })\subseteq J_{Y}Y$. We shall prove $(i)⟺(ii)$, $(ii)⟹(iii)⟹(iv)⟹(ii)$. The key identities, obtained by evaluating on $y^{\ast }\in Y^{\ast }$, are

The bidual range condition implies both weak compactness and the stronger continuity of $T^{\ast }$. First assume $T^{\ast \ast }(X^{\ast \ast })\subseteq J_{Y}Y$. The Banach-Alaoglu theorem makes $B_{X^{\ast \ast }}$ weak-star compact, and $T^{\ast \ast }$ is weak-star continuous. Hence its image is compact for the weak-star topology in $Y^{\ast \ast }$. Because it lies in $J_{Y}Y$, it corresponds to a weakly compact subset $C\subseteq Y$ under $J_Y^{-1}$.

The identity $T^{\ast \ast }{J}_X=J_{Y}T$ gives $T(B_X)\subseteq C$. A weakly compact set in a Hausdorff space is closed for the weak topology, so $C$ is also norm-closed and contains the norm closure of $T(B_X)$. That closure is a convex set closed in the norm topology, hence weakly closed by the Mazur theorem. It is therefore a weakly closed subset of the weakly compact $C$, and is a weakly compact set. This proves $(ii)⟹(i)$.

For $(ii)⟹(iii)$, fix $F\in X^{\ast \ast }$. By the range assumption there is $y_F\in Y$ with $T^{\ast \ast }F=J_Y{y}_F$. For every $y^{\ast }\in Y^{\ast }$,The right side is a defining weak-star continuous evaluation on $Y^{\ast }$. Since the weak topology on $X^{\ast }$ is determined by all $F\in X^{\ast \ast }$, every scalar coordinate of $T^{\ast }$ for that topology is weak-star continuous. Therefore

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.

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 $\phi (T(\mu ))={\int }_K\phi d\mu \le {\sup }_K\phi$.

First suppose that $X$ is separable. Choose a norm-dense sequence $(x_n)$ in $B_X$. On every norm-bounded subset of $X$, a countable norm-dense subset $(f_n)$ of $B_{X^{\ast }}$ in the weak-star topology separates points, andmetrizes the weak topology. Indeed, convergence for all $f_n$, boundedness, and weak-star density imply convergence for every $f\in X^{\ast }$. Thus a weakly compact set $K$ is a compact metric space and hence is sequentially compact.

For general $X$, take a sequence $(x_n)$ in $K$ and letThe space $Y$ is separable and norm closed. The Hahn-Banach theorem shows both that $Y$ is weakly closed in $X$ and that its own weak topology is the subspace topology inherited from $X$. Hence $K\cap Y$ is weakly compact and, by the separable case, contains a weakly convergent subsequence of $(x_n)$. Its limit lies in $K\cap Y$. Therefore every weakly compact subset of a Banach space is weakly sequentially compact, which is one direction of the Eberlein-Šmulian theorem.

A bounded sequence $(x_n)$ lies in a scalar multiple of the closed unit ball of the reflexive Banach space $X$, which is weakly compact. Part (a) gives a subsequence $x_{k_n}\rightharpoonup z$ for some $z\in X$.

The compact operator $T$ is weak-to-weak continuous, so $T{x}_{k_n}\rightharpoonup Tz$. We claim that the convergence is in norm. Otherwise some further subsequence would satisfy $\Vert T{x}_{k_n}-Tz\Vert \ge \epsilon$. Compactness supplies a norm-convergent subsubsequence; its norm limit must also be its weak limit $Tz$, a contradiction. Hence

A Banach space $X$ is reflexive when its canonical embedding $J_X:X\to X^{\ast \ast }$ is surjective. Equivalently, its closed unit ball is weakly compact.

A subset $D$ of a normed vector space is weakly bounded when ${\sup }_{x\in D}|f(x)|<\infty$ for every $f\in X^{\ast }$. It is norm bounded: its canonical embedding into the bidual is a pointwise bounded family of functionals on the complete continuous dual space, so the Uniform boundedness principle applies. The converse is immediate. A weakly compact set is weakly bounded because each scalar evaluation has compact image.

Every weakly compact set in a normed vector space is norm bounded, even if the space is incomplete. Its canonical embedding into the bidual is pointwise bounded on the continuous dual space; the Uniform boundedness principle on that Banach dual gives the norm bound.

Coordinate evaluations on the l-infinity sequence space form a countable separating family, so a weakly compact set $K$ is metrizable by countable separating family metrizes a weakly compact set. Choose a countable weakly dense subset $D$. Then $K$ lies in the weak closure of $\mathrm{conv}D$, which equals its norm closure by Mazur theorem. Rational convex combinations make that norm closure separable, and every subset of a separable metric space is separable. The weakly dense subset itself need not be norm dense in a nonconvex $K$.