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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 $X$ be a Banach space and let $Z$ be a linear subspace of $X^{\ast }$ that separates points, with the inherited operator norm. If the closed unit ball $B_X$ is compact for $\sigma (X,Z)$, then the evaluation map $J:X\to Z^{\ast }$ is an isometric isomorphism of normed spaces.

For a unit-ball element $\Phi \in Z^{\ast }$, the Hahn-Banach theorem extends $\Phi$ to an element of $X^{\ast \ast }$ of norm at most one. Goldstine theorem approximates that extension weak-star by elements of $B_X$. Restricting the pairings to $Z$ shows that $J(B_X)$ is weak-star dense in $B_{Z^{\ast }}$. The assumed compactness also makes $J(B_X)$ weak-star compact and hence closed, so $J(B_X)=B_{Z^{\ast }}$.

Separation makes $J$ injective. Equality of the two unit balls makes it surjective and norm preserving: a smaller image norm could otherwise place a vector outside $B_X$ in the image of $B_X$. This proves the claimed dual representation without assuming in advance that $Z$ is norming.

If $Z$ is a Banach space and $X$ is a norm-closed reflexive linear subspace of $Z^{\ast }$, evaluation gives the displayed surjection. Restriction $R:Z^{\ast \ast }\to X^{\ast }$ is weak-star continuous and maps the closed unit ball onto the target ball by the Hahn-Banach theorem. The Goldstine theorem then makes $T(B_Z)$ weak-star dense in $B_{X^{\ast }}$, where $T=R{J}_Z$. Reflexivity of $X$ makes this target weak-star topology equal to its weak topology. Convexity and the Hahn-Banach separation theorem convert weak density of this ball image to norm density.

To pass from density to surjectivity, approximate a desired $f$ by $T{z}_1$ with $\Vert z_1\Vert \le \Vert f\Vert$ and residual at most $\Vert f\Vert /2$. Repeat on each residual. The chosen vectors have summable norms, with sum at most $2\Vert f\Vert$, and the residuals tend to zero geometrically. Completeness of $Z$ produces $z={\sum }_n{z}_n$ with $Tz=f$. Using any smaller residual ratio proves the infimum lifting norm equals $\Vert f\Vert$, although that infimum need not be attained. This argument proves onto-ness, rather than merely density of the evaluation functionals.

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.

An infinite-dimensional reflexive example is ${\ell }^2$, by the Riesz representation theorem.

The weak topology on $X$ is the coarsest topology making every $f\in X^{\ast }$ continuous. The weak-star topology on $X^{\ast }$ is the coarsest making every evaluation $f\mapsto f(x)$, $x\in X$, continuous. In each case basic neighborhoods impose finitely many evaluation inequalities. The Banach-Alaoglu theorem states that the closed unit ball of a dual space is compact for its weak-star topology. The separation theorem needed here says: if $K$ is a nonempty weak-star compact convex subset of a dual space $E^{\ast }$ and $z\notin K$, then some $e\in E$ strictly separates them,The real-part convention covers complex spaces; for real spaces omit it.

Let $K$ be the weak-star closure of $J_X(B_X)$ inside $B_{X^{\ast \ast }}$. This is compact by Banach-Alaoglu, convex, and nonempty. If $z\in B_{X^{\ast \ast }}\setminus K$, separation in the dual pair $(X^{\ast \ast },X^{\ast })$ gives $f\in X^{\ast }$ withBut $|z(f)|\le \Vert z\Vert \Vert f\Vert \le \Vert f\Vert$, a contradiction. Therefore $K=B_{X^{\ast \ast }}$, proving the required Goldstine theorem density.

If $X$ is reflexive, $J_X$ identifies its weak topology with the weak-star topology of $X^{\ast \ast }$ and carries $B_X$ onto $B_{X^{\ast \ast }}$, so Banach-Alaoglu gives weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact, hence closed in the Hausdorff bidual topology. Density then forces $J_X(B_X)=B_{X^{\ast \ast }}$. Scaling gives surjectivity on the entire bidual. Thus

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology making every $f\in X^{\ast }$ continuous. Its basic neighbourhoods impose finitely many inequalities $|f_j(x-x_0)|<\epsilon$. The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology making each evaluation $f\mapsto f(x)$ continuous, for $x\in X$.

The canonical embedding into the bidual isIt is linear and $\Vert J_{X}x\Vert \le \Vert x\Vert$. For $x\ne 0$, the Hahn-Banach theorem supplies a norming functional $f$ with $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$, proving equality. Thus $J_X$ is an isometric embedding. The coordinates defining the weak-star topology on $X^{\ast \ast }$, restricted to $J_X(X)$, are precisely $J_{X}x\mapsto f(x)$ for $f\in X^{\ast }$. These are exactly the coordinates defining the weak topology on $X$, so $J_X$ is also a topological embedding for these topologies.

A normed vector space is a reflexive space if $J_X$ is onto $X^{\ast \ast }$. Such a space is automatically a Banach space, since it is isometric to the complete continuous dual space $X^{\ast \ast }$. The Banach-Alaoglu theorem states that the closed unit ball of a continuous dual space is compact for its weak-star topology. The Goldstine theorem states that $J_X(B_X)$ is dense in the weak-star topology in $B_{X^{\ast \ast }}$. ConsequentlyIf $J_X$ is onto, its weak-to-weak-star homeomorphism carries $B_X$ onto the Banach-Alaoglu theorem's compact ball $B_{X^{\ast \ast }}$. Conversely, if $B_X$ is weakly compact, $J_X(B_X)$ is compact and hence closed in the Hausdorff weak-star topology. Its density from the Goldstine theorem makes it all of $B_{X^{\ast \ast }}$, and scaling gives surjectivity of $J_X$.

Now suppose $X$ is separable and nonzero. Choose a countable dense set $(u_n)$ in its unit sphere and, using the Hahn-Banach theorem, choose norming functionals $f_n$ with $\Vert f_n\Vert =1$ and $f_n(u_n)=1$. These continuous linear functionals separate points: for $x\ne 0$, choose $\Vert u_n-x/\Vert x\Vert \Vert <1/2$ to get $f_n(x)>\Vert x\Vert /2$. DefineThe series converges uniformly, separates distinct points, and obeys the triangle inequality because $\min (c,a+b)\le \min (c,a)+\min (c,b)$. Thus $d$ is a metric. Its topology is coarser than the weak topology: a finite set of coordinate controls makes the finite initial sum small, while the remaining sum is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$. On a weakly compact set $K$, the identity $(K,\text{weak})\to (K,d)$ is therefore a continuous bijection from a compact space into a Hausdorff space, hence a homeomorphism. This proves that $K$ is metrizable in its weak topology, an instance of countable separating family metrizes a weakly compact set.

If $X$ is also a reflexive space, its closed unit ball is weakly compact and therefore metrizable by this construction. Compactness in a metric space implies sequential compactness, so every sequence in $B_X$ has a weakly convergent subsequence. Scaling proves the claim for every bounded sequence.

For the converse, assume every bounded sequence in $X$ has a weakly convergent subsequence. Use the same metric $d$ on $B_X$. A weak limit of points of $B_X$ remains in $B_X$: $|f(x)|\le 1$ for every $\Vert f\Vert \le 1$, and the Hahn-Banach theorem recovers $\Vert x\Vert \le 1$. Thus every sequence in $B_X$ has a subsequence convergent in $d$ to a point of $B_X$, so $(B_X,d)$ is sequentially compact and hence compact.

We must still recover the entire weak topology, rather than just the selected coordinates. Suppose $x_n\to x$ in $d$ but not weakly. There is a neighbourhood in the weak topology $U$ of $x$ and a subsequence outside $U$. By hypothesis that subsequence has a further subsequence converging weakly to $y\in B_X$. It then converges in $d$ to $y$, so uniqueness of metric limits gives $y=x$, contradicting eventual membership in $U$. Hence convergence in $d$ implies weak convergence. Every $f\in X^{\ast }$ is therefore sequentially continuous on $(B_X,d)$ and hence continuous, since the domain is a metric space. The weak topology and $d$-topology on $B_X$ coincide. The ball is weakly compact, so the criterion above provesThe zero space satisfies all statements directly. The converse did not assume completeness, nor use the full Eberlein-Šmulian theorem.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology on $X$ making every continuous linear functional $f\in X^{\ast }$ continuous. A basic neighborhood of $x_0$ is specified by finitely many inequalities $|f_j(x-x_0)|<ϵ$.

The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology on $X^{\ast }$ making every evaluation $f\mapsto f(x)$, for $x\in X$, continuous. Its basic neighborhoods use finitely many such evaluations. Nets converge in these topologies exactly when all the corresponding scalar evaluations converge.

Let $J_X:X\to X^{\ast \ast }$ be the canonical isometric embedding, $J_{X}x(f)=f(x)$. Goldstine theorem statesHere the balls are closed unit balls. Thus any bidual unit vector can be approximated in finitely many dual evaluations by unit vectors from $X$; this does not claim norm density or sequential density in complete generality.

Put $T=R\circ J_Z:Z\to X^{\ast }$. Then $(Tz)(x)=x(z)$ and $\Vert Tz\Vert \le \Vert z\Vert$. We must prove $T$ onto.

Take $f\in B_{X^{\ast }}$. Part (iv) supplies $\theta \in B_{Z^{\ast \ast }}$ with $R\theta =f$. Goldstine's theorem gives a net $z_{\alpha }\in B_Z$ with $J_Z{z}_{\alpha }\to \theta$ weak-star. By continuity of $R$, $T{z}_{\alpha }\to f$ in $\sigma (X^{\ast },X)$. Reflexivity of $X$ makes this the weak topology on $X^{\ast }$ by part (iii). Thus $B_{X^{\ast }}$ is contained in the weak closure of $T(B_Z)$.

The reverse containment holds because $T(B_Z)\subseteq B_{X^{\ast }}$, and the latter ball is weakly closed: it is the intersection of the closed evaluation constraints $|g(x)|\le 1$ for $x\in B_X$. The set $T(B_Z)$ is convex, so part (ii), now on the Banach space $X^{\ast }$, givesDensity of the ball gives exact surjectivity by a convergent correction argument. For nonzero $f\in X^{\ast }$, choose $z_1$ with $\Vert z_1\Vert \le \Vert f\Vert$ and $\Vert f-T{z}_1\Vert \le \Vert f\Vert /2$. Inductively approximate the current residual $r_{n-1}$ by $T{z}_n$ with $\Vert z_n\Vert \le \Vert r_{n-1}\Vert$ and $\Vert r_{n-1}-T{z}_n\Vert \le \Vert r_{n-1}\Vert /2$. If a residual is zero, stop.

The residual norms are bounded by $2^{-n}\Vert f\Vert$, while ${\sum }_n\Vert z_n\Vert \le 2\Vert f\Vert$. Completeness of $Z$ gives $z={\sum }_n{z}_n\in Z$, and continuity of $T$ gives $Tz=f$. For $f=0$ use $z=0$. ThereforeThis is evaluation onto the dual of a reflexive dual subspace. The correction argument is needed: a dense image alone would not imply surjectivity.

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.