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A bounded linear operator $T:X\to Y$ between Banach spaces is a weakly compact operator exactly when its second adjoint takes values in the canonical copy of $Y$ inside $Y^{\ast \ast }$. The Goldstine theorem proves necessity; the Banach-Alaoglu theorem, continuity of the second adjoint, and the Mazur theorem prove sufficiency.

Let $Z\subseteq X^{\ast }$ separate points of a Banach space $X$. If the closed unit ball of $X$ is compact for $\sigma (X,Z)$ and $Z$ has the norm inherited from $X^{\ast }$, then the evaluation map $X\to Z^{\ast }$ is an isometric isomorphism. Goldstine theorem makes its unit-ball image weak-star dense, while the assumed compactness makes that image closed.

For a finite-dimensional vector subspace $E\subseteq X^{\ast }$ and $x^{\ast \ast }\in B_{X^{\ast \ast }}$, exact interpolation on $E$ is possible by an $x\in X$ of norm less than $1+\epsilon$. The restriction map $X\to E^{\ast }$ is onto and open. Separation of the image of the radius-$1+\epsilon$ open ball would contradict $|x^{\ast \ast }(e)|\le \Vert e\Vert$. Scaling the interpolant slightly back into the closed unit ball proves the Goldstine theorem.

Consider the restriction operatorIt is onto: otherwise its range, a vector subspace of the finite-dimensional vector space $E^{\ast }$, would have a nonzero annihilator in $E^{\ast \ast }=E$. That annihilator would be an $e\in E$ vanishing on all of $X$, so $e=0$, a contradiction. Moreover $T$ is open. To see this directly, choose preimages of a basis of $E^{\ast }$; they define a linear right inverse $S:E^{\ast }\to X$, continuous because its domain is finite-dimensional. Small changes of an image can then be lifted by small changes using $S$.

Put $r=1+\epsilon$ and $C=T(\{x:\Vert x\Vert <r\})$. It is open and convex in $E^{\ast }$. If $a\notin C$, the Hahn-Banach separation theorem gives a nonzero real linear functional on $E^{\ast }$, represented by some $e\in E$, such thatBut $a(e)=x^{\ast \ast }(e)\le \Vert e\Vert$, contradicting $r>1$. Therefore $a\in C$, and the finite-dimensional interpolation form of Goldstine's theorem yieldsIf $E=\{0\}$ one simply takes $x=0$.

To recover the Goldstine theorem, start with $x^{\ast \ast }\in B_{X^{\ast \ast }}$ and finitely many tests $e_1,\dots ,e_m\in X^{\ast }$. Apply the result to their span with a small parameter $\delta >0$. The resulting $x$ need not lie in $B_X$, but $v=x/(1+\delta )$ does. For every test,Choosing $\delta$ sufficiently small puts $Jv$ in any prescribed basic weak-star neighborhood. This proves the asserted weak-star density of the closed unit ball. The reverse inclusion follows because $B_{X^{\ast \ast }}$ is weak-star closed, being the intersection of the conditions $|x^{\ast \ast }(e)|\le \Vert e\Vert$ for $e\in X^{\ast }$.

The weak-star topology $\sigma (X^{\ast },X)$ is the topology of pointwise convergence on $X$. A basic neighborhood of $f_0$ has the formLet $J:X\to X^{\ast \ast }$ be the canonical embedding into the bidual, $Jx(f)=f(x)$. It is an isometry. The Goldstine theorem says thatHere both balls are closed unit balls. Thus approximation is by finitely many dual evaluations at a time; the theorem does not assert norm density in the bidual space.

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

Weak compactness forces the second adjoint to take values in $Y$. Suppose $C={\overline{T(B_X)}}^{\Vert \cdot \Vert }$ is weakly compact in $Y$. By the first observation, $J_{Y}C$ is closed for the weak-star topology in $Y^{\ast \ast }$. The second adjoint $T^{\ast \ast }:X^{\ast \ast }\to Y^{\ast \ast }$ is weak-star-to-weak-star continuous, by the adjoint-continuity argument applied to $T^{\ast }$.

Thus $(T^{\ast \ast }{)}^{-1}(J_{Y}C)$ is closed for the weak-star topology in $X^{\ast \ast }$. The identity $T^{\ast \ast }{J}_X=J_{Y}T$ makes it contain $J_X(B_X)$. The Goldstine theorem therefore givesScaling any $F\in X^{\ast \ast }$ into the closed unit ball provesThis is the forward direction of the bidual characterization of weakly compact operators.

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 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.

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.

For $p\ge 2$, integrate the stated scalar inequality to obtain the Clarkson inequalityIf $f,g$ belong to the unit ball and $\Vert f-g{\Vert }_p\ge \epsilon$, thenThus $L^p[0,1]$ is uniformly convex.

Now let $X$ be uniformly convex. It is enough to show that every $\Phi \in S_{X^{\ast \ast }}$ lies in the canonical image of $X$. Given $\epsilon >0$, choose the corresponding uniform-convexity constant $\delta$, and choose $f\in B_{X^{\ast }}$ with $\Phi (f)>1-\delta$. If $x,y\in B_X$ both satisfy $f(x),f(y)>1-\delta$, thenso $\Vert x-y\Vert <\epsilon$. By Goldstine theorem, every weak-star neighbourhood of $\Phi$ contains some $Jx$ with $x\in B_X$. Directing these neighbourhoods produces a norm-Cauchy net $(x_{\alpha })$; completeness gives $x_{\alpha }\to x\in B_X$, and weak-star convergence then gives $Jx=\Phi$. Scaling handles the whole bidual ball, so $X$ is reflexive.

For $p\ge 2$, uniform convexity therefore makes $L^p$ reflexive. If $1<p<2$, its conjugate exponent $q$ is greater than two, so $L^q$ is reflexive. Since $(L^p{)}^{\ast }=L^q$ and a Banach space whose dual is reflexive is itself reflexive, $L^p$ is reflexive for every $1<p<\infty$.

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.

Let $p_C(x)=\inf \{t>0:x\in tC\}$ be the Minkowski functional of $C$. Openness, convexity, and $0\in C$ make $p_C$ sublinear, with $p_C(x)<1$ for $x\in C$ and $p_C(x_0)\ge 1$. Define a linear functional on $\mathbb{R}{x}_0$ by $f(t{x}_0)=t{p}_C(x_0)$. The real Hahn-Banach theorem extends it to $X$ with $f\le p_C$. Hence

Apply this separation to the open ball of radius $\Vert x_0\Vert$ and rescale to obtain $f\in S_{X^{\ast }}$ with $f(x_0)=\Vert x_0\Vert$. For a closed subspace $Y$, apply it toThe separator must vanish on $Y$ because $C$ contains every translate along $Y$, and normalization gives

Let $F\subseteq X^{\ast }$ be finite-dimensional and $\Phi \in B_{X^{\ast \ast }}$. ConsiderIf $\Phi {|}_F$ did not belong to $S((1+\epsilon )B_X)$, finite-dimensional strict separation would produce some $f\in F$ withcontradicting $\Vert \Phi \Vert \le 1$. Thus there is $x\in X$ with $\Vert x\Vert <1+\epsilon$ and $f(x)=\Phi (f)$ for every $f\in F$.

Given a basic weak-star neighbourhood of $\Phi$ in $B_{X^{\ast \ast }}$, apply this result to the finite-dimensional span of its defining functionals and then replace $x$ by $x/(1+\epsilon )$. As $\epsilon \downarrow 0$, the resulting points of $J(B_X)$ enter that neighbourhood. Hence $J(B_X)$ is weak-star dense in $B_{X^{\ast \ast }}$, proving Goldstine theorem.

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.

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.

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-Šmulian 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-Šmulian 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 $\mathcal{B}({\ell }^1,Y)\ne \mathcal{W}({\ell }^1,Y)$.