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In an infinite-dimensional Hilbert space, every bounded sequence has a weakly convergent subsequence, but it need not have a norm-convergent one. For example an orthonormal sequence $e_n$ has pairwise distance $\sqrt{2}$ and therefore no strongly convergent subsequence, although $e_n\rightharpoonup 0$ by Bessel inequality.

More generally, weak sequential compactness of bounded sequences in a reflexive Banach space ensures a weakly convergent subsequence with its limit in that space. Conversely, this property for all bounded sequences characterizes reflexivity, using the Eberlein-Šmulian theorem and weak compactness of the closed unit ball. Thus reflexivity replaces finite-dimensional norm compactness by weak sequential compactness.

In a nonreflexive Banach space there is no such general weak-subsequence guarantee. The coordinate unit vectors in ${\ell }^1$ give an example: by the Schur property, weak convergence would imply norm convergence, impossible for their pairwise distance two. Banach-Alaoglu theorem supplies weak-star compactness of a dual unit ball, which gives convergent subnets, not automatically convergent subsequences. For a separable predual, the weak-star compact dual ball is metrizable and subsequences do suffice. For an arbitrary Banach space one can also regard a bounded sequence in its bidual and extract a weak-star convergent subnet there; the limit need not belong to the original space. These qualifications distinguish weak from weak-star compactness.

On a real Hilbert space, for a proper lower-semicontinuous convex function $E$, the proximal operator isThe quadratic makes this objective strongly convex and coercive; the affine lower bound available for a proper closed convex function and weak lower semicontinuity give existence. The minimizer is unique. The subdifferential optimality condition isFor a positive parameter $t$, ${\mathrm{prox}}_{tE}$ uses $tE(x)$ in the same formula. On a general Banach space, the analogous norm-squared minimization is Banach-space proximal minimization; it can be set-valued and the identity-valued Hilbert derivative must be replaced by the duality mapping.

First use the usual Hilbert space convention for the proximal operator, identifying the dual element $z$ with its representing vector. For $c\ne 0$, completing the square givesSet $v=cx-y$ and multiply the objective by $c^2$. Its variable part becomes $\Vert v-[c(q-z)-y]{\Vert }^2/2+\alpha c^{2}J(v)$. Thus the proximal operator under affine rescaling isNegative $c$ is allowed. If $c=0$ and $J(-y)<\infty$, the penalty is constant and the answer is $q-z$. If $c=0$ and $J(-y)=\infty$, $E$ is identically infinite, so there is no proper proximal problem.

The PDF actually assumes an arbitrary Banach space, so the Hilbert input shift cannot be asserted literally under its hypotheses. Define ${\mathrm{Prox}}_F(s)=\arg {\min }_v\{F(v)+\Vert v-s{\Vert }^2/2\}$ as a possibly empty or set-valued minimizer set. Absolute homogeneity of the norm still gives the valid change of variablesThis retains the linear term, rather than shifting by a dual vector. Equivalently its optimality condition uses the duality mapping $\mathcal{J}=\partial (\Vert \cdot {\Vert }^2/2)$:For a concrete failure of the Hilbert shortcut, take $({\mathbb{R}}^2,\Vert \cdot {\Vert }_1)$, $J=0$, $q=y=0$, $c=1$ and $z=(1,1)$. Every $x_1,x_2\le 0$ with $x_1+x_2=-1$ minimizes $\Vert x{\Vert }_1^2/2+x_1+x_2$ with value $-1/2$. The naive shifted answer $(-1,-1)$ has value zero and is not a minimizer. Hence the simple shifted formula requires Hilbert geometry; general Banach proximal minimizers need not be unique. For $c=0$ with finite constant penalty the Banach condition is $-z\in \mathcal{J}(x-q)$, not generally $x=q-z$.

For a convex functional $J:U\to \mathbb{R}\cup \{+\infty \}$ on a real Banach space, its subdifferential at $v$ is empty unless $J(v)$ is finite; otherwiseGiven $p_v\in \partial J(v)$, the Bregman distance isIt can be infinite if $J(u)$ is infinite. If $J$ is differentiable at $v$, the choice is its derivative; otherwise the base subgradient must be specified. Given also $p_u\in \partial J(u)$, add both directional distances to obtain the symmetric Bregman distanceThe symmetric version here is the sum, not half the sum. These nonnegative quantities are not generally metrics and can vanish at distinct points.

For real $u\in L^1(\Omega )$, its total variation seminorm on a domain isThe vector fields have compact support inside $\Omega$, so no boundary term is charged. Taking both $\xi$ and $-\xi$ makes this equivalent to a supremum of absolute pairings. When finite, the distributional derivative $Du$ is a finite vector-valued Radon measure, and the displayed supremum is its total variation norm of a measure. For a smooth function it equals ${\int }_{\Omega }|\nabla u{|}_{2}dx$.

The function of bounded variation on a domain belongs to the Banach spaceThe L1 norm is needed because variation alone vanishes on constant functions and is only a seminorm. No smoothness of $u$ is part of this definition.

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.

One form of the Baire category theorem is: a nonempty complete metric space cannot be a countable union of closed nowhere dense subsets. Equivalently, a countable intersection of open dense subsets is dense. A Banach space is a complete metric space for its norm metric, so this form applies directly to the next part.

Suppose $(x_k{)}_{k\ge 1}$ were a countable Hamel basis. Put $E_n=\mathrm{span}(x_1,\dots ,x_n)$. Each $E_n$ is a finite-dimensional, hence closed, linear subspace of $X$. It is proper because $X$ is infinite dimensional, and therefore a nowhere dense set. A Hamel basis means that every vector is a finite linear combination, so $X={\bigcup }_n{E}_n$. This contradicts the Baire category theorem for the Banach space $X$. Hence an infinite-dimensional Banach space has no countable Hamel basis. The same proof excludes any countable algebraic spanning set.

If the bounded injective Fourier transform $L^1\to C_0$ were surjective, the open mapping theorem for Banach spaces would make its inverse bounded, yielding $\Vert h{\Vert }_1\le C\Vert \hat{h}{\Vert }_{\infty }$. But the preceding $g_n$ have uniform norm $2$, whereas $\Vert h_n{\Vert }_1$ diverges.

For a quantitative lower bound, on $0<x\le 1/4$ we have $\sin (2\pi x)\ge 4x$. On each interval $I_j=[(j+1/12)/n,(j+5/12)/n]$, $|\sin (2\pi nx)|\ge 1/2$. For $1\le j\le \lfloor n/8\rfloor$ these intervals lie below $1/4$ for large $n$, soThis contradicts boundedness of the inverse. Therefore the Fourier transform has proper dense image in $C_0(\mathbb{R})$.

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 Commutative Gelfand--Naimark theorem states that a complex commutative unital C-star algebra $A$ is isometrically star-isomorphic to $C({\Phi }_A)$ through its Gelfand transform, with ${\Phi }_A$ a compact Hausdorff space. The l-infinity sequence space is such an algebra under coordinatewise multiplication and conjugation, with the supremum norm. Thereforegives the required isometric isomorphism.

Uniqueness as a Banach space representation uses the Banach–Stone theorem, rather than just uniqueness as an algebra representation. To justify the relevant theorem, the Riesz-Markov-Kakutani representation theorem givesthe extreme points of the dual unit ball of C(K). A norm-one measure whose variation measure is not concentrated at one point splits into two normalized restrictions to disjoint sets of positive variation, and so is not extreme. Conversely, equality in the variation bound shows that any decomposition of $\alpha {\delta }_t$ into the average of two dual-unit-ball elements forces both to be $\alpha {\delta }_t$: after removing the phase, the two measures must be positive and their average is concentrated at $t$.

If $U:C(K)\to C(L)$ is a surjective linear isometry, its dual map preserves these extreme points and their scalar orbits. Hence $U^{\ast }{\delta }_s=u(s){\delta }_{h(s)}$ for a bijection $h:L\to K$ and $|u(s)|=1$. Evaluating at $1$ gives the continuous function $u=U1$, whileshows that $h$ is continuous, since continuous functions determine the topology of a compact Hausdorff space. It is consequently a homeomorphism. Thus any other compact space representing ${\ell }^{\infty }$ is homeomorphic to this $K$. The Banach–Stone theorem is also stated in Leonard Tomczak's notes on András Zsák's functional analysis lectures.

Embed $\mathbb{N}$ by the evaluation characters $\iota (n)(a)=a_n$. They are distinct because the coordinate indicator functions $e_n$ distinguish them. Moreover, $e_n^2=e_n$, so every character has $\chi (e_n)\in \{0,1\}$. If $\chi (e_n)=1$, thenand hence $\chi =\iota (n)$. It follows that $\{\iota (n)\}=\{\chi :|\chi (e_n)-1|<1/2\}$ is open in $K$. Thus $\iota$ is a homeomorphism from the discrete natural numbers onto its image.

To prove density, suppose $D=\overline{\iota (\mathbb{N})}\ne K$. The Urysohn lemma provides a nonzero $v\in C(K)$ vanishing on $D$. Write $v=\Gamma (a)$. Then $a_n=v(\iota (n))=0$ for all $n$, so $a=0$ and $v=0$, a contradiction. Every bounded function $b:\mathbb{N}\to \mathbb{C}$ is an element of ${\ell }^{\infty }$, and $\Gamma (b)$ is its continuous extension to $K$. Density makes this extension unique. ThereforeThis identifies $K$ with the Stone-Čech compactification of the natural numbers.

The direct method in the calculus of variations gives the following existence theorem. Suppose bounded sequences in the Banach space $\mathcal{U}$ have $\tau$-convergent subsequences, and $E$ is a proper extended-real function with coercivity and is $\tau$-sequentially lower semicontinuous. Then $E$ has a finite-valued global minimizer.

To prove this, properness makes $m=\inf E<+\infty$. First rule out $m=-\infty$: a sequence with $E(u_n)\to -\infty$ eventually lies in a fixed sublevel set, which is bounded by coercivity. A $\tau$-convergent subsequence would then have a limit $u$ with $E(u)\le -\infty$, contradicting the codomain. Hence $m$ is finite.

Choose a minimizing sequence with $E(u_n)\to m$. It eventually belongs to the bounded sublevel set $\{E\le m+1\}$. Extract $u_{n_j}\stackrel{\tau }{\to }\bar{u}$. By sequential lower semicontinuity,Thus $E(\bar{u})=m$. In particular, a reflexive Banach space with the weak topology supplies the required subsequence property by weak sequential compactness of bounded sequences in a reflexive Banach space. A strictly convex function has at most one minimizer; this is an additional property, not part of the existence theorem.

Take the weak topology on the reflexive Banach space $\mathcal{U}$. A bounded linear operator is weak-to-weak continuous: $u_n\rightharpoonup u$ implies $⟨K{u}_n,v⟩\to ⟨Ku,v⟩$ for every $v\in \mathcal{V}$, by its adjoint operator. The weak lower semicontinuity of the Hilbert norm therefore makes $D(u)=\Vert Ku-f\Vert$ weakly sequentially lower semicontinuous.

This functional is finite everywhere and nonnegative. If it has coercivity, the direct method in the calculus of variations supplies a global minimizer $\bar{u}$. Squaring the residual does not change its global minimizers. Here the adjoint operator has codomain ${\mathcal{U}}^{\ast }$, since the source is a Banach space. Differentiating the squared residual along every real direction $h\in \mathcal{U}$ gives $\mathrm{Re}⟨K^{\ast }(K\bar{u}-f),h⟩=0$ in the dual pairing, hence the normal equation for a linear inverse problem $K^{\ast }(K\bar{u}-f)=0$. HenceThe converse fails. For $K:{\mathbb{R}}^2\to \mathbb{R}$, $K(x,y)=x$, and $f=1$, the datum is in the operator range, but $D(1,n)=0$ while $\Vert (1,n)\Vert \to \infty$. Thus admissible data do not imply coercivity of the residual; an unpenalized null space already prevents it.

The Moore–Penrose inverse of an operator is the linear mapthat assigns each admissible datum its minimum-norm least-squares solution. If $f=y+z$ with $y\in \mathcal{R}(K)$ and $z\in \mathcal{R}(K{)}^{\perp }$, then $K^{†}f$ is the unique $u\in \mathcal{N}(K{)}^{\perp }$ satisfying $Ku=y$. In particular it vanishes on $\mathcal{R}(K{)}^{\perp }$, andThe continuity criterion, with the inherited data norm on its domain, isFor a closed operator range, the restriction of $K$ from $\mathcal{N}(K{)}^{\perp }$ to $\mathcal{R}(K)$ is a bounded bijection between Banach spaces, so its inverse is bounded by the bounded inverse theorem. Conversely, if $K^{†}$ is bounded and $K{u}_n\to y$, replace each $u_n$ by $P_{\mathcal{N}(K{)}^{\perp }}{u}_n=K^{†}K{u}_n$. This is a Cauchy sequence, hence converges to $u$, and $Ku=y$. Thus every limit of range elements remains in the operator range.

For a functional $E$ on a real Banach space, its convex conjugate is defined on the dual space byHere $⟨p,u⟩$ is the dual pairing. For complex spaces, use its real part, equivalently regard the space as real for convex analysis. The convex conjugate is a convex function, being the supremum of affine functions of $p$, even when $E$ itself is not convex.

A bounded linear operator $T:X\to Y$ is a compact operator when the image of the closed unit ball has compact closure, equivalently when every bounded sequence $(x_n)$ has a subsequence for which $(T{x}_n)$ converges.

Each $T_i$ is compact because its image lies in a finite-dimensional subspace. Let $B_X$ be the unit ball and fix $\epsilon >0$. Choose $i$ with $\Vert T-T_i\Vert <\epsilon /3$. Since $T_i(B_X)$ is relatively compact, it has a finite $\epsilon /3$-net $y_1,\dots ,y_N$. For each $x\in B_X$, choose $y_j$ with $\Vert T_{i}x-y_j\Vert <\epsilon /3$; thenThus $T(B_X)$ is totally bounded. Its closure is complete because $Y$ is a Banach space, so it is compact. This proves the norm-closedness of compact operators and hence

Let $(T_m)$ be Cauchy in the operator norm on $\mathcal{B}(X,Y)$. For each $x\in X$,so $(T_{m}x)$ is Cauchy in the Banach space $Y$. DefinePointwise passage to the limit shows that $T$ is linear. A norm-Cauchy sequence is bounded, say $\Vert T_m\Vert \le M$, and henceThus $T$ is bounded. Finally, for fixed $m$,The right-hand coefficient tends to zero with $m$, so $T_m\to T$ in operator norm. ThereforeThis is the Banach space of bounded linear operators.

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.

For a Banach space $X$, weak convergence $u_k\rightharpoonup u$ meansFor its continuous dual space $X^{\ast }$, convergence in the weak-star topology meansWeak convergence on $X^{\ast }$ tests against all of $X^{\ast \ast }$, whereas weak-star convergence tests only against the canonical image of $X$. Thus weak convergence always implies weak-star convergence. If $X$ is a reflexive Banach space, that image is all of $X^{\ast \ast }$, so the weak and weak-star topologies on $X^{\ast }$ coincide. Weak-star topology is defined relative to a chosen predual; it is not an extra intrinsic topology on an arbitrary Banach space without such an identification.

Choose the common dominating probability measure $\nu =(\mu +{\mu }^{\prime })/2$ and let $p=d\mu /d\nu$, $q=d{\mu }^{\prime }/d\nu$ be the Radon-Nikodym derivatives. The assumed second moments imply first-moment integrability by the Cauchy-Schwarz inequality. Since $Y$ is a separable Banach space, the measurable $f$ is a strongly measurable function, and both expected values exist as Bochner integrals.

The difference of these Bochner integrals satisfiesThe first inequality is the norm bound for a Bochner integral, and the second is the scalar Cauchy-Schwarz inequality. Since $(\sqrt{p}+\sqrt{q}{)}^2\le 2(p+q)$, the first factor is at mostWith the Hellinger distance convention of the original PDF, the second factor is $\sqrt{2}{d}_{\mathrm{Hell}}(\mu ,{\mu }^{\prime })$. Multiplication proves the Hellinger bound for differences of expectations:The proof uses only the two laws and their common dominating measure; no absolute continuity of measures assumption between $\mu$ and ${\mu }^{\prime }$ is needed.