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Suppose the compact operator $K$ acts on $H$, and $h_n\rightharpoonup h$. The Uniform boundedness principle makes $(h_n)$ bounded, so its images form a relatively compact set. Bounded linearity gives $K{h}_n\rightharpoonup Kh$: for every $g$, $⟨K{h}_n,g⟩=⟨h_n,K^{\ast }g⟩\to ⟨Kh,g⟩$. Every norm-convergent subsequence of $(K{h}_n)$ must therefore converge to $Kh$. If the whole sequence did not converge to $Kh$ in norm, a subsequence staying a fixed positive distance from $Kh$ would have a norm-convergent further subsequence, a contradiction. This proves that compact operators send weak convergence to norm convergence.

Conversely, assume the stated complete continuity property. Its strong limit necessarily equals $Kh$, because bounded linearity still supplies the weak limit. Every bounded sequence in a Hilbert space has a weakly convergent subsequence by weak sequential compactness in a Hilbert space. Applying the assumption to that subsequence gives a norm-convergent subsequence of the images. Thus the image of the unit ball is relatively sequentially compact and hence relatively compact. Complete continuity and compactness are equivalent on a Hilbert space.

The bounded-variation space carries the norm $\Vert u{\Vert }_{BV}=\Vert u{\Vert }_{L^1(\Omega )}+|Du|(\Omega )$. Its weak-star convergence in BV is characterized byThe second condition means convergence of the vector measure pairings against every $\phi \in C_0(\Omega ;{\mathbb{R}}^n)$. Equivalently, strong convergence in $L^1$ together with ${\sup }_k|D{u}_k|(\Omega )<\infty$ suffices: integration by parts identifies the limit on smooth compactly supported tests, and uniform approximation extends this to $C_0$ tests.

The bounded-variation compactness theorem says thatguarantees a subsequence convergent in weak-star convergence in BV on a bounded Lipschitz domain. This is the uniform criterion for relative sequential compactness. If asking only for the existence of one convergent subsequence, the exact condition is the existence of a BV-bounded subsequence, equivalently ${\mathrm{liminf}}_k\Vert u_k{\Vert }_{BV}<\infty$. The entire sequence need not be bounded: alternating zero functions and constants tending to infinity gives a simple example. Conversely, a convergent subsequence has bounded $L^1$ norm and bounded total variation seminorm by the Uniform boundedness principle for its derivative-measure pairings.

Suppose $K$ is a compact operator and $f_n\rightharpoonup f$. The Uniform boundedness principle makes $(f_n)$ bounded. If $K{f}_n$ did not converge in norm to $Kf$, some subsequence would stay a fixed positive distance away. Compactness provides a further norm-convergent subsequence, say to $g$. On the other hand, for every $h$, $⟨K{f}_n,h⟩=⟨f_n,K^{\ast }h⟩\to ⟨Kf,h⟩$, so its norm limit must be $Kf$, a contradiction.

Conversely, if $K$ sends every weakly convergent sequence to a norm-convergent sequence, take any sequence in the closed unit ball. Weak compactness of that ball supplies a weakly convergent subsequence, whose images converge in norm by hypothesis. Thus every sequence in the image has a convergent subsequence in $H$. Its closure is also sequentially compact: approximate its $n$th member by an image point within $1/n$. Since $H$ is a metric space, that closure is compact. We concludeThis is the principle that compact operators send weak convergence to norm convergence.

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 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

By the Riesz-Markov-Kakutani representation theorem, the continuous dual space of $C(K)$ is isometrically the space of finite regular Borel measures, signed for real scalars and complex for complex scalars:Here $|\mu |$ is the variation measure, whose total mass is the total variation norm of a measure.

If $f_n\rightharpoonup 0$ in $C(K)$, evaluation at each $t\in K$ gives $f_n(t)\to 0$. The Uniform boundedness principle also gives $M={\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$. For any $\mu \in M(K)$, the dominated convergence theorem with respect to the finite positive measure $|\mu |$ yieldssince $|f_n(t){|}^2\to 0$ pointwise and $|f_n(t){|}^2\le M^2$. ThusThe same proof works for complex squares, because the absolute-value domination remains valid.

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

For every finite scalar sequence $a=(a_n)$ with $|a_n|\le 1$, define a bounded linear functional on the continuous dual space $X^{\ast }$ byFor each fixed $f\in X^{\ast }$, the hypothesis giveswhere the supremum is over all finite choices and the phases $a_n$ may be chosen to align the complex numbers $f(x_n)$.

The dual $X^{\ast }$ is a Banach space even when $X$ is incomplete. The Uniform boundedness principle therefore gives a constant $C$ such thatFor each $N$, choose $a_n$ for $n\le N$ so that $a_{n}f(x_n)=|f(x_n)|$. ThenLetting $N\to \infty$ provesThis is the uniform bound from pointwise absolute summability of dual evaluations.

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.

Assume $x_n\rightharpoonup x_{\infty }$. For each $n$, define the bounded functionalWeak convergence makes $(T_n(h){)}_n$ convergent, hence bounded, for every $h\in H$. The Uniform boundedness principle givesBy the Riesz representation theorem, $\Vert T_n\Vert =\Vert x_n\Vert$, so $(x_n)$ is bounded.

Taking $h=e_i$ in the definition of weak convergence in a Hilbert space givesfor every $i$. Thus (i) implies (ii).

For each $n$, define $J{x}_n:X^{\ast }\to \mathbb{R}$ by $J{x}_n(f)=f(x_n)$. Weak convergence makes $(J{x}_n(f){)}_n$ bounded for every $f\in X^{\ast }$. The Uniform boundedness principle gives

Let $K$ be the closed convex hull of the $x_n$. Given a sequence in $K$, approximate its terms in norm by finite convex combinations of the $x_n$. A diagonal subsequence makes every coefficient converge. Any loss of total coefficient mass is assigned to zero, which belongs to $K$ by Mazur theorem because $x_n\rightharpoonup 0$. Since $f(x_n)\to 0$ for every $f\in X^{\ast }$, splitting each sum into a finite head and a uniformly small tail proves weak convergence of this subsequence to the corresponding convex combination. Hence $K$ is weakly sequentially compact and, by the stated theorem, weakly compact.

DefineIt is bounded because $(x_n)$ is norm bounded, and its values lie in $c_0$ because $x_n\rightharpoonup 0$. Its adjoint-on-preduals map isIf $K$ had nonempty norm interior, then $K-K$ would contain a ball about zero. The quantitative open-mapping argument applied to the convex combinations above would make $T_{\ast }$ surjective, and hence make $T$ bounded below. Its range $R$ would be a closed infinite-dimensional subspace of $c_0$ whose unit ball is compact for coordinatewise convergence, since it lies in the coordinatewise compact image of a weak-star compact ball of $X^{\ast }$.

By the stated structural theorem, $R$ contains a closed subspace isomorphic to $c_0$. The bounded partial sums of the image of the standard $c_0$ basis would then have a coordinatewise convergent subnet. Uniform boundedness turns coordinatewise convergence in $c_0$ into weak convergence, and norm-closed subspaces are weakly closed. Pulling the limit back would make the partial sums of the standard basis converge weakly in $c_0$, impossible because their coordinate values force the putative limit to be the constant-one sequence. This contradiction proves that $K$ has empty norm interior.

Mazur theorem says that the weak closure and norm closure of a convex subset of a real or complex normed space coincide. The norm closure is contained in the weak closure because every norm-continuous linear functional is norm-continuous. 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 real number $\alpha$ such thatThis weakly open separation shows that $x$ is outside the weak closure.

By definition,For each $n$, regard $x_n$ as the bounded functional $J{x}_n:X^{\ast }\to \mathbb{F}$ given by $J{x}_n(f)=f(x_n)$. Pointwise convergence makes the family $(J{x}_n)$ pointwise bounded on the Banach space $X^{\ast }$. The Uniform boundedness principle gives

Suppose first that $x_n\rightharpoonup 0$. Every subsequence indexed by an infinite set $M$ also converges weakly to zero. Thus zero belongs to the weak closure of the convex hull of $\{x_n:n\in M\}$, and Mazur's theorem puts it in its norm closure. This gives the required finite convex combination of norm below any prescribed $\epsilon$.

Conversely, if weak convergence fails, some $f\in X^{\ast }$ and an infinite subsequence satisfy either $\mathrm{Re}f(x_n)\ge \epsilon$ throughout or $\mathrm{Re}f(x_n)\le -\epsilon$ throughout. Every convex combination from that subsequence then has norm at least $\epsilon /\Vert f\Vert$, contradicting the stated property.

Now let the $x_n\in {\ell }^p$, $1<p<\infty$, have pairwise disjoint supports and satisfy $\Vert x_n{\Vert }_p\le C$. For $N$ distinct terms,The convex-combination criterion therefore proves $x_n\rightharpoonup 0$. This is weak convergence of bounded disjointly supported sequences in lp.

The final implication for a commutative unital C*-algebra is true. By the Commutative Gelfand--Naimark theorem, $A\cong C(K)$ and its characters are the point evaluations. The hypotheses say that the uniformly bounded functions $a_n$ converge pointwise to zero. Every functional on $C(K)$ is integration against a finite regular measure by the Riesz-Markov-Kakutani representation theorem; the dominated convergence theorem givesHence $a_n\rightharpoonup 0$.

For each $n$, define the bounded linear functionalThe hypothesis is precisely that $x_n$ converges weakly to $x$. In particular, for each fixed $y\in H$, the scalar sequence $T_n(y)$ converges and is therefore bounded. Thus the family $\{T_n\}$ is pointwise bounded.

The Uniform boundedness principle givesBy the Riesz representation theorem, $\Vert T_n\Vert =\Vert x_n{\Vert }_H$. Henceso the sequence is bounded.

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.

Every weakly compact set in a normed vector space is norm bounded, even if the space is incomplete. Its canonical embedding into the bidual is pointwise bounded on the continuous dual space; the Uniform boundedness principle on that Banach dual gives the norm bound.

Every weakly convergent sequence in a normed vector space is a bounded sequence. Apply the Uniform boundedness principle on the Banach space $X^{\ast }$ to the evaluation maps $J{x}_n:f\mapsto f(x_n)$. They are pointwise bounded, while the canonical embedding into the bidual gives $\Vert J{x}_n\Vert =\Vert x_n\Vert$.

Convergence in the displayed sense combines strong function convergence with weak-star convergence of derivative vector measures against $C_0$ tests. Strong $L^1$ convergence and a uniform variation bound imply the derivative convergence by integration by parts and uniform test-function approximation. Conversely the Uniform boundedness principle bounds the derivative measures of a convergent sequence.