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The bidual of a normed vector space $X$ is the continuous dual $X^{\ast \ast }=(X^{\ast }{)}^{\ast }$. Evaluation defines the canonical embedding into the bidual $J_X(x)(f)=f(x)$. The norm inequality gives $\Vert J_{X}x\Vert \le \Vert x\Vert$, and a norming functional from the Hahn-Banach theorem gives equality. Thus $J_X$ is isometric, and its image is dense in its closure, a completion of a normed space.

A linear functional on $X^{\ast }$ is continuous for the weak-star topology exactly when it is evaluation at an element of $X$. Continuity makes it vanish on the common kernel of finitely many evaluations, so it is a linear combination of those evaluations. The canonical embedding into the bidual identifies this dual with $X$.

If a bounded sequence $(y_n)$ in $X$ converges weak-star to $\varphi \in X^{\ast \ast }$ under the canonical embedding into the bidual, then any convex blocks $(u_n)$ have the same scalar limit on every $f\in X^{\ast }$. Thus $y_n-u_n$ is a weakly null sequence, even when $\varphi$ does not belong to the embedded space $X$.

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.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

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

The Hahn-Banach theorem for bounded linear functionals says that a bounded linear functional on any linear subspace of a normed vector space extends to the whole space with its norm unchanged. No closedness or completeness of the subspace is required. Question 1 uses real-valued ${\ell }_{\infty }(\Gamma )$, so its operator assertions are read over the real scalar field. The analogous complex statements use complex-valued indexed functions.

For $x\ne 0$, define $g(tx)=t\Vert x\Vert$ on its one-dimensional span. This is a norm-one functional, so Hahn-Banach theorem supplies an extension $f\in X^{\ast }$ withThe canonical embedding into the bidual is $Jx(f)=f(x)$. It is linear and $\Vert Jx\Vert \le \Vert x\Vert$. The bounded linear functional just constructed gives the reverse inequality for nonzero $x$, and the zero case is immediate. Thus $\boxed{\Vert Jx\Vert =\Vert x\Vert }$ and $J$ is injective.

For the coordinate functional representation of an operator into bounded indexed functions, let $e_{\gamma }$ evaluate a coordinate and put $f_{\gamma }=e_{\gamma }T$. If $T$ is bounded and linear, then $f_{\gamma }\in X^{\ast }$ and $\Vert f_{\gamma }\Vert \le \Vert T\Vert$. Conversely, if $M={\sup }_{\gamma }\Vert f_{\gamma }\Vert <\infty$, the formula $(Tx)(\gamma )=f_{\gamma }(x)$ defines a bounded scalar function for each $x$, is linear, and satisfies $\Vert Tx{\Vert }_{\infty }\le M\Vert x\Vert$. Combining the two estimates givesFor an empty index set both spaces/families have norm bound zero; the supremum of the empty nonnegative family is taken as zero.

Choose $\Gamma =B_{X^{\ast }}$ and $Tx=(f(x){)}_{f\in B_{X^{\ast }}}$. The Hahn-Banach theorem makes $\Vert Tx{\Vert }_{\infty }=\Vert x\Vert$, so this is a linear isometric embedding into bounded scalar functions on an index set.

If $X$ is nonzero and separable, choose a dense sequence $u_n$ in its unit sphere and supporting functionals $f_n$ with $\Vert f_n\Vert =f_n(u_n)=1$. For a unit vector $u$ and $ϵ>0$, some $u_n$ satisfies $\Vert u-u_n\Vert <ϵ$, whence $|f_n(u)|\ge 1-ϵ$. Thus $(f_n)$ is a countable norming family and $x\mapsto (f_n(x))$ is an isometry into the l-infinity sequence space. Completeness of $X$ is not needed.

If instead $X=E^{\ast }$ for a separable normed vector space $E$, choose a dense sequence $v_n$ in $B_E$. The evaluations $f_n(x)=x(v_n)$ lie in $B_{X^{\ast }}$ and continuity of $x\in E^{\ast }$ gives ${\sup }_n|x(v_n)|=\Vert x\Vert$. This again gives $\Gamma =\mathbb{N}$, even when $E^{\ast }$ is not separable. If $X=\{0\}$, use zero coordinates throughout.

To prove 1-injectivity, take $T:Y\to {\ell }_{\infty }(\Gamma )$ on a subspace of $Z$. Extend every coordinate $f_{\gamma }\in Y^{\ast }$ to ${\tilde{f}}_{\gamma }\in Z^{\ast }$ by Hahn-Banach theorem, keeping its norm. Their uniform bound defines $\tilde{T}z=({\tilde{f}}_{\gamma }(z))$, and the coordinate norm identity yieldsSelection of these extensions uses the usual choice convention for arbitrary index sets. This verifies the lambda-injective normed space definition with $\lambda =1$.

For the final retraction characterization of lambda-injectivity, suppose first that $X$ is lambda-injective and $i:X\to Z$ is a linear isometry. The inverse $i(X)\to X$ has norm one when $X\ne 0$; extend it to $P:Z\to X$ with $\Vert P\Vert \le \lambda$. Then $Pi=I_X$, and $iP$ is a bounded projection onto $i(X)$. For the zero space take $P=0$.

Conversely, suppose every linear isometry out of $X$ has such a left inverse. Fix an isometry $j:X\to {\ell }_{\infty }(\Gamma )$ as above and a left inverse $P$ with $\Vert P\Vert \le \lambda$. For any $T:Y\to X$, extend $jT$ to $S:Z\to {\ell }_{\infty }(\Gamma )$ by 1-injectivity. Then $\tilde{T}=PS$ extends $T$, and $\Vert \tilde{T}\Vert \le \lambda \Vert S\Vert =\lambda \Vert T\Vert$. Therefore $X$ is lambda-injective exactly when every isometric embedding $i:X\to Z$ admits a bounded map $P:Z\to X$ satisfying

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.

Take $Y=X^{\ast \ast }$, which is a Banach space by the completeness of the dual just proved. The canonical embedding into the bidual isFor fixed $x$ this is a bounded linear functional on $X^{\ast }$, since $|f(x)|\le \Vert f\Vert \Vert x\Vert$, and $\Phi$ is linear in $x$. This gives $\Vert \Phi (x)\Vert \le \Vert x\Vert$. For $x\ne 0$, the norming functional from the Hahn-Banach theorem has $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$, giving the reverse inequality. At zero equality is immediate. ThusIn particular $\Phi$ is injective and isometric. If a dense isometric embedding is wanted, rather than merely the stated embedding, take the norm closure of $\Phi (X)$ in $X^{\ast \ast }$; a closed subspace of a Banach space is complete and supplies a completion of a normed space.

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 Uniform boundedness principle states that if $E$ is a Banach space and a family of bounded linear operators $S_{\alpha }:E\to Y$ is pointwise bounded, then ${\sup }_{\alpha }\Vert S_{\alpha }\Vert <\infty$.

Suppose $x_n\rightharpoonup x$ in the normed vector space $X$. For each $n$, define the evaluation functionalThe completeness of the dual space says that $X^{\ast }$ is Banach even when $X$ is not. For each $f\in X^{\ast }$, the scalar sequence $J_n(f)=f(x_n)$ converges to $f(x)$ and is therefore bounded. Uniform boundedness givesBy the canonical embedding into the bidual and the Hahn-Banach theorem, $\Vert J_n\Vert =\Vert x_n\Vert$. Hencewhich proves that a weakly convergent sequence is bounded.

For $x\in X$, define $e_x\in X^{\ast \ast }$ by $e_x(\varphi )=\varphi (x)$. The map $J:x\mapsto e_x$ is linear, andshows $\Vert Jx\Vert \le \Vert x\Vert$, so $J$ is continuous. The granted equality $\Vert Jx\Vert =\Vert x\Vert$ is the canonical embedding into the bidual.

Suppose, contrary to the claim, that exactly the same linear functionals are continuous for $\Vert \cdot \Vert$ and $\Vert \cdot {\Vert }^{\prime }$. Their common continuous dual has two operator norms, say $\Vert \cdot {\Vert }_{\ast }$ and $\Vert \cdot {\Vert }_{\ast }^{\prime }$. Both dual spaces are Banach, even if the second normed space $X$ is not complete. The identity maphas a closed graph: convergence in either operator norm implies pointwise convergence on $X$, so two limits must agree. The closed graph theorem makes the two dual norms equivalent.

Applying the isometric bidual formula for each primal norm then gives constants $c,C>0$ such thatThis says the two norms are equivalent norms, contrary to the hypothesis. Their continuous duals must therefore differ as sets, so a linear functional belonging to one and not the other is continuous for exactly one of the two norms.