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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.

For a convex subset of a Banach space, weak closure equals norm closure. The Mazur theorem states that for every convex set $C\subseteq X$,Since every bounded linear functional is norm-continuous, the weak topology is weaker than the norm topology, giving ${\overline{C}}^{\Vert \cdot \Vert }\subseteq {\overline{C}}^w$.

For the reverse inclusion, use this form of the Hahn-Banach separation theorem: if $D$ is a nonempty norm-closed convex set in a real normed vector space and $x\notin D$, there is a continuous real-linear functional $F$ such thatApply it to $D={\overline{C}}^{\Vert \cdot \Vert }$. In a complex Banach space, use the underlying real space; every continuous real-linear $F$ is the real part of the complex bounded linear functional $f(v)=F(v)-iF(iv)$. The separating strict inequality gives a neighbourhood open for the weak topology of $x$ disjoint from $D$, so $x\notin {\overline{C}}^w$. The empty set case is immediate. This proves the Mazur theorem.

A weakly null sequence has disjoint convex blocks converging to zero in norm. If $x_n\stackrel{w}{\to }0$, then for every starting index $m$,The Mazur theorem places $0$ in the norm closure of that tail convex hull. Thus a finite convex combination of vectors from any tail can have norm less than any prescribed $\epsilon >0$.

Choose the convex blocks recursively. After selecting the previous terminal index $q_{n-1}$, let $p_n>q_{n-1}$, and choose a finite convex combination from $\{x_i:i\ge p_n\}$ with norm less than $2^{-n}$. Choose $q_n$ beyond its largest used index, padding the intervening coefficients and the final coefficient with zero. This ensures the printed strict condition $p_n<q_n$ as well as $q_n<p_{n+1}$. Setting unused coefficients to zero yieldsTherefore $u_n\to 0$ in norm. The recursive tail selection is what makes these convex blocks, rather than merely unrelated convex combinations.

Next suppose $X^{\ast }$ is separable and $(x_n)$ is bounded. Use the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, where $J_{X}x(f)=f(x)$. If $\Vert x_n\Vert \le M$, then $J_X{x}_n\in M{B}_{X^{\ast \ast }}$. The Banach-Alaoglu theorem makes this ball compact in $\sigma (X^{\ast \ast },X^{\ast })$, and weak-star metrizability of the dual ball makes it metrizable because its predual $X^{\ast }$ is separable. A compact space that is a metric space is sequentially compact, so some subsequence $(y_n)$ satisfiesThe limit need not belong to $J_{X}X$; it is precisely the use of the bidual space that supplies compactness without assuming reflexivity.

For a convex block $u_n={\sum }_{i=p_n}^{q_n}{a}_i{y}_i$, every $f\in X^{\ast }$ satisfiesBoth $f(y_n)$ and $f(u_n)$ tend to $\varphi (f)$, soThis is the useful convex-block cancellation of a weak-star limit; it does not require the limit to be a vector in $X$.

The quotient sequence has approximate lifts in $3B_X$ that are weakly null after passing to a subsequence. Let $q:X\to Z=X/Y$ be the quotient map, with the quotient norm. Since $\Vert z_n\Vert \le 1$, choose $v_n\in X$ such thatThis uses the infimum defining the quotient norm and does not assume that it is attained. By the preceding compactness argument, pass to a subsequence $y_n=v_{k_n}$ with a common weak-star limit in $X^{\ast \ast }$. Its quotient images $q(y_n)=z_{k_n}$ form a weakly null sequence.

Apply the convex block construction to $(q(y_n))$ in the quotient Banach space. Using the same coefficients on $(y_n)$ gives convex blocksSet $x_n=y_n-u_n$. The convex-block cancellation of a weak-star limit proves $x_n\stackrel{w}{\to }0$, whileThus the approximate weakly null lifting through a quotient is achieved with the required bound:

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

Weak-star-to-weak continuity of the adjoint implies its weak compactness. By the Banach-Alaoglu theorem, $B_{Y^{\ast }}$ is compact for the weak-star topology. Under the assumed continuous map $T^{\ast }:(Y^{\ast },w^{\ast })\to (X^{\ast },w)$, its image $T^{\ast }(B_{Y^{\ast }})$ is weakly compact in $X^{\ast }$.

This image is closed for the weak topology, and hence for the norm topology, because the weak topology is Hausdorff and weaker than the norm topology. Thus its norm closure is the same weakly compact set. By the definition of a weakly compact operator, $T^{\ast }$ is weakly compact, proving $(iii)⟹(iv)$.

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.

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

The direct method in the calculus of variations gives the following existence theorem. Suppose bounded sequences in the Banach space $\mathcal{U}$ have $\tau$-convergent subsequences, and $E$ is a proper extended-real function with coercivity and is $\tau$-sequentially lower semicontinuous. Then $E$ has a finite-valued global minimizer.

To prove this, properness makes $m=\inf E<+\infty$. First rule out $m=-\infty$: a sequence with $E(u_n)\to -\infty$ eventually lies in a fixed sublevel set, which is bounded by coercivity. A $\tau$-convergent subsequence would then have a limit $u$ with $E(u)\le -\infty$, contradicting the codomain. Hence $m$ is finite.

Choose a minimizing sequence with $E(u_n)\to m$. It eventually belongs to the bounded sublevel set $\{E\le m+1\}$. Extract $u_{n_j}\stackrel{\tau }{\to }\bar{u}$. By sequential lower semicontinuity,Thus $E(\bar{u})=m$. In particular, a reflexive Banach space with the weak topology supplies the required subsequence property by weak sequential compactness of bounded sequences in a reflexive Banach space. A strictly convex function has at most one minimizer; this is an additional property, not part of the existence theorem.

Take the weak topology on the reflexive Banach space $\mathcal{U}$. A bounded linear operator is weak-to-weak continuous: $u_n\rightharpoonup u$ implies $⟨K{u}_n,v⟩\to ⟨Ku,v⟩$ for every $v\in \mathcal{V}$, by its adjoint operator. The weak lower semicontinuity of the Hilbert norm therefore makes $D(u)=\Vert Ku-f\Vert$ weakly sequentially lower semicontinuous.

This functional is finite everywhere and nonnegative. If it has coercivity, the direct method in the calculus of variations supplies a global minimizer $\bar{u}$. Squaring the residual does not change its global minimizers. Here the adjoint operator has codomain ${\mathcal{U}}^{\ast }$, since the source is a Banach space. Differentiating the squared residual along every real direction $h\in \mathcal{U}$ gives $\mathrm{Re}⟨K^{\ast }(K\bar{u}-f),h⟩=0$ in the dual pairing, hence the normal equation for a linear inverse problem $K^{\ast }(K\bar{u}-f)=0$. HenceThe converse fails. For $K:{\mathbb{R}}^2\to \mathbb{R}$, $K(x,y)=x$, and $f=1$, the datum is in the operator range, but $D(1,n)=0$ while $\Vert (1,n)\Vert \to \infty$. Thus admissible data do not imply coercivity of the residual; an unpenalized null space already prevents it.

Let $X$ have basis $e_1,\dots ,e_d$, and write $x={\sum }_j{a}_j{e}_j$. On the compact coordinate sphere ${\sum }_j|a_j|=1$, the continuous positive functionhas a positive minimum $m$. Homogeneity and the triangle inequality therefore give constants $m,M>0$ such thatIn particular, each coordinate functional ${\varphi }_j(x)=a_j$ belongs to the continuous dual space $X^{\ast }$.

Every member of $X^{\ast }$ is norm-continuous, so the weak topology is coarser than the norm topology. Conversely, given $x_0\in X$ and $\epsilon >0$, the basic weak neighbourhoodsatisfies $U\subseteq \{x:\Vert x-x_0\Vert <\epsilon \}$. Thus every norm-open set is weakly open, proving that the two topologies coincide. This is the finite-dimensional weak and norm topologies coincide theorem.

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.

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.

By definition of the weak topology, $x_n\rightharpoonup x$ exactly when $f(x_n)\to f(x)$ for every $f\in X^{\ast }$.

Let $(x_n)$ be bounded and choose a norm-dense sequence $(f_j)$ in $X^{\ast }$. Successive subsequences make $f_1(x_n),f_2(x_n),\dots$ converge, and the diagonal argument produces a single subsequence $(y_n)$ on which every $f_j$ converges. Uniform boundedness of $(y_n)$ and norm approximation of an arbitrary $f$ by the $f_j$ show that $f(y_n)$ is Cauchy for every $f\in X^{\ast }$. Thus $(y_n)$ is weakly Cauchy, andfor every $f$, so its difference sequence is weakly null.

For the countable family $(y_{m,n}{)}_n$, use the weak metric from part a. Delete a finite initial segment from the $m$th sequence so that every remaining term has weak distance less than $1/m$ from zero, and relabel that tail. Enumerate all these tails while preserving the order within each one. For every weak neighbourhood of zero, all terms from sufficiently large $m$ lie inside it, and only finitely many terms from each of the finitely many remaining sequences lie outside it. The resulting enumeration $(z_n)$ is weakly null and contains the relabelled $m$th sequence as a subsequence for every $m$. Equivalently, without relabelling, it contains a tail-subsequence of every original sequence, which is the form used below.

LetChoose positive ${\epsilon }_n$ with ${\prod }_n(1+{\epsilon }_n)<\infty$. We construct $x_n\in K$ inductively. Once $E_n=\mathrm{span}\{x_1,\dots ,x_n\}$ has been chosen, compactness of its unit sphere gives finitely many members of $X^{\ast }$ that almost norm every vector of $E_n$. Because $0\in {\overline{K}}^w$, the next $x_{n+1}$ may be chosen so that all those functionals are as small on it as required. Choosing the error relative to $d$ givesfor all scalars $a_1,\dots ,a_{n+1}$. Indeed, if the last coefficient could threaten this estimate, $\Vert x_{n+1}\Vert \ge d$ first bounds that coefficient by a fixed multiple of the norm of the preceding sum; the selected norming functional then gives the displayed inequality. Iteration and the finite product bound satisfy the standard basis criterion, so $(x_n)$ is a basic sequence contained in $K$.

The same proof works for any Hausdorff locally convex vector topology $\tau$ weaker than the weak topology: on each finite-dimensional $E_n$, the $\tau$-continuous linear functionals still norm the space, and $0\in {\overline{K}}^{\tau }$ supplies the next point. A canonical strictly weaker example arises on $X^{\ast }$ when $X$ is not reflexive: the weak-star topology $\sigma (X^{\ast },X)$ is then strictly weaker than the weak topology $\sigma (X^{\ast },X^{\ast \ast })$.

We next prove the Eberlein-Šmulian theorem. If the weak closure of a bounded set $K$ is weakly compact, take any sequence in $K$ and let $Y$ be its closed linear span. The relevant weak closure lies in the separable space $Y$. A countable weak-star dense subset of the dual unit ball separates points of this compact set, so its weak topology is metrizable. Compact metrizability gives a weakly convergent subsequence.

For the converse, suppose $K$ is not relatively weakly compact. In the canonical embedding into $X^{\ast \ast }$, chooseThe Hahn-Banach theorem gives $x^{\ast \ast \ast }\in X^{\ast \ast \ast }$ that vanishes on $J_X(X)$ but satisfies $x^{\ast \ast \ast }(x^{\ast \ast })=1$. Alternating Goldstine approximation with the fact that $x^{\ast \ast }$ lies in the weak-star closure of $J_{X}K$ constructs bounded $x_n^{\ast }\in X^{\ast }$ and $x_n\in K$ such that, up to errors tending to zero,If a subsequence $x_{n_j}$ converged weakly to $x$, then for each fixed $i$ the second relation would give $x_i^{\ast }(x)=1$. A weak-star cluster point $x^{\ast }$ of the bounded sequence $(x_i^{\ast })$ satisfies $x^{\ast }(x_{n_j})=0$ by the first relation, and hence $x^{\ast }(x)=0$ by weak convergence; but passing to the same cluster point in $x_i^{\ast }(x)=1$ gives $x^{\ast }(x)=1$. This contradiction produces a sequence in $K$ with no weakly convergent subsequence. Relative weak compactness is therefore equivalent to the subsequence condition.

Finally suppose $K$ is weakly sequentially compact and $x\in {\overline{K}}^w$. If $x$ lies in the norm closure of $K$, a norm-convergent sequence suffices. Otherwise apply the first part to $K-x$ and obtain a basic sequence $(y_n)$. A subsequence converges weakly by hypothesis. Its weak limit lies in the closed span of the basic sequence, while every coordinate functional is eventually zero on that subsequence. The limit is consequently zero, so the corresponding sequence from $K$ converges weakly to $x$.

This also shows that a weakly sequentially compact $K$ is weakly closed: any point of its weak closure is the limit of a sequence in $K$, and a weakly convergent subsequence has its limit in $K$. The relative form of the Eberlein-Šmulian theorem then makes $K$ weakly compact. The reverse implication follows from the first direction of that theorem. Thus weak compactness and weak sequential compactness are equivalent.