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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 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.

The weak-star topology $\sigma (X^{\ast },X)$ is generated by the seminorms $p_x(g)=|g(x)|$, $x\in X$. Thus every evaluation is continuous, so $f_n\stackrel{w^{\ast }}{\to }f$ implies $f_n(x)\to f(x)$ for each $x$. Conversely, a basic neighborhood of $f$ imposes finitely many conditions $|(g-f)(x_j)|<\epsilon$. Pointwise convergence eventually satisfies all of them simultaneously. HenceThe Uniform boundedness principle, applied on the Banach space $X$, gives $M={\sup }_n\Vert f_n\Vert <\infty$. If $x_n\to x$ in the norm topology, thenTherefore $\boxed{f_n(x_n)\to f(x)}$. Completeness of $X$ is used in this uniform boundedness step.

If $C\subseteq X^{\ast }$ is norm compact, the identity map from $(C,\Vert \cdot \Vert )$ to $(C,w^{\ast })$ is a continuous bijection. The target is Hausdorff, since distinct functionals differ at some $x\in X$. The compact-to-Hausdorff continuous bijection theorem gives

The Banach-Alaoglu theorem states that the closed dual unit ball $B_{X^{\ast }}$ is compact in the weak-star topology; in fact $X$ need only be a normed vector space. To prove it, work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and formEach $D_x$ is compact, so $Q$ is compact by the Tychonoff theorem. Embed $B_{X^{\ast }}$ by $g\mapsto (g(x){)}_{x\in X}$. Its image consists exactly of those families $(z_x)$ satisfyingThese are closed conditions in the product topology. Conversely, any family satisfying them defines a linear functional bounded by $|z_x|\le \Vert x\Vert$, hence an element of $B_{X^{\ast }}$. The image is therefore closed and compact. The induced product topology is precisely the weak-star topology, proving the theorem. This is a compactness proof for arbitrary nets, with no separability assumption.

For the dual map of bounded linear operator $T^{\ast }:Y^{\ast }\to X^{\ast }$, each $x\in X$ satisfiesEvaluation at the fixed vector $Tx$ is weak-star continuous, so $\boxed{T^{\ast }\text{ is weak-star-to-weak-star continuous}}$.

Suppose $T$ is a compact operator. By the allowed Schauder theorem for compact operators, $T^{\ast }$ is compact. If $S\subseteq Y^{\ast }$ is bounded, the norm closure $C=\overline{T^{\ast }(S)}$ is norm compact. We already proved that its norm and weak-star topologies coincide. The weak-star continuous map $T^{\ast }{|}_S$ into $C$ is consequently norm continuous. Conversely, if $T^{\ast }$ is weak-star-to-norm continuous on bounded subsets, its restriction to $B_{Y^{\ast }}$ maps a weak-star compact set, by Banach-Alaoglu theorem, to a norm compact set. Thus $T^{\ast }$ is a compact operator, and the Schauder theorem for compact operators gives

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

Choose a countable dense subset $\{x_1,x_2,\dots \}$ of the normed vector space $X$. For each fixed $j$, the scalar sequence $(f_n(x_j))$ is bounded becauseRepeated use of the Bolzano-Weierstrass theorem followed by a diagonal subsequence argument produces a subsequence $(f_n{)}_{n\in \Lambda }$ for which $f_n(x_j)$ converges for every $j$.

Let $D$ be the linear span of the $x_j$. On every $x\in D$, defineFinite linear combinations show that $f:D\to \mathbb{R}$ is linear, and passage to the limit in $|f_n(x)|\le \Vert x\Vert$ gives $|f(x)|\le \Vert x\Vert$. Thus $f$ is a bounded linear functional on the dense linear subspace $D$ and extends uniquely to an element of the dual space $X^{\ast }$ with $\Vert f\Vert \le 1$.

For arbitrary $x\in X$ and $y\in D$,First choose $y$ close to $x$, then let $n\in \Lambda$ tend to infinity. This provesThis is the Sequential Banach-Alaoglu theorem for a separable predual.

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.

The continuous dual space of a real normed vector space $X$ iswith the operator normTwo normed spaces are isometrically isomorphic when there is a bijective linear map between them that preserves norms.

For $y=(y_j)\in {\ell }^{\infty }$, defineThe series is absolutely convergent andso ${\Phi }_y\in ({\ell }^1{)}^{\ast }$ and $\Vert {\Phi }_y\Vert \le \Vert y{\Vert }_{\infty }$. For every $\epsilon >0$, choose $j$ with $|y_j|>\Vert y{\Vert }_{\infty }-\epsilon$ and test on $\mathrm{sgn}(y_j)e_j$. This gives the reverse inequality, hence

Conversely, given $\Phi \in ({\ell }^1{)}^{\ast }$, set $y_j=\Phi (e_j)$. Then $|y_j|\le \Vert \Phi \Vert$, so $y\in {\ell }^{\infty }$. For $x\in {\ell }^1$, its finite partial sums converge to $x$ in norm; continuity of $\Phi$ therefore givesThe representing sequence is unique, and the construction is linear and norm preserving. Thus the duality of l1 and l infinity proves

A family $\mathcal{S}\subseteq C(K)$ is equicontinuous if for every $x\in K$ and $\epsilon >0$ there is a neighborhood $U$ of $x$ such thatfor every $y\in U$ and every $f\in \mathcal{S}$. The Arzela-Ascoli theorem says that, for a compact Hausdorff space $K$, a subset of $C(K)$ has compact closure in the uniform norm if and only if it is equicontinuous and pointwise bounded. On compact $K$, equicontinuity and pointwise boundedness together imply uniform boundedness.

For necessity, a compact closure is totally bounded. Given $\epsilon >0$, choose a finite $\epsilon /3$-net $f_1,\dots ,f_m$. The finitely many continuous functions $f_j$ are simultaneously continuous near each point, and comparison with a nearby $f_j$ proves equicontinuity of the whole family. Evaluation at any fixed point is a continuous map $C(K)\to \mathbb{R}$ or $\mathbb{C}$, so compactness also gives pointwise boundedness.

Conversely, equicontinuity gives, around each $x\in K$, a neighborhood on which every $f\in \mathcal{S}$ oscillates by less than $\epsilon /3$. Compactness supplies a finite such cover with selected points $x_1,\dots ,x_m$. Pointwise boundedness makes the set of vectorsa bounded subset of a finite-dimensional normed vector space, so it has a finite $\epsilon /3$-net. Choose one function of $\mathcal{S}$ for every nonempty cell of this net. If two functions have evaluation vectors in the same cell, then comparison at a selected $x_j$ and the two oscillation bounds show that their uniform distance is less than $\epsilon$. Thus $\mathcal{S}$ is totally bounded. Since $C(K)$ is a Banach space, its closure is complete; a complete totally bounded metric space is compact, proving the theorem.

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.

A Rademacher sum is a linear combination of independent Rademacher random variables, with scalar or normed-space-valued coefficients. Its moments are controlled by the Khintchine inequality in the scalar case and the Kahane-Khintchine inequality in a general normed vector space.

A normed vector space has the Radon-Riesz property when weak convergence $u_k\rightharpoonup u$ together with $\Vert u_k\Vert \to \Vert u\Vert$ implies strong convergence $u_k\to u$. The Hilbert-space case is the Radon-Riesz theorem. The condition prevents norm from remaining in weakly invisible directions.

In a metric space, applying the triangle inequality in both directions gives $|d(x,z)-d(y,z)|\le d(x,y)$. In a normed vector space, its corresponding form is $|\Vert x\Vert -\Vert y\Vert |\le \Vert x-y\Vert$. In particular, distance to a fixed point is a Lipschitz continuous real function.

If a normed vector space $X$ is separable, every bounded sequence in $X^{\ast }$ has a subsequence that converges pointwise on $X$. Equivalently, the closed unit ball of $X^{\ast }$ is sequentially compact in the weak-star topology. A diagonal subsequence argument on a countable dense subset proves the result.

In a normed vector space, this set consists of all points at a fixed positive norm distance $r$ from $x_0$. It also makes sense in an infinite-dimensional Hilbert space, unlike a specifically Euclidean sphere. In a real Hilbert space, the sphere in a normed vector space $\Vert u{\Vert }^2=2E_0>0$ has tangent variations $v$ satisfying $(u,v)=0$; this follows by differentiating the fixed squared norm. This is the geometric constraint used in fixed-energy initial-condition optimality.

The unit ball of a normed vector space is $\{x:\Vert x\Vert \le 1\}$ under the closed-ball convention.

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