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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.

Given $x\in {\ell }^p$, choose a sequence $x^{(m)}\in F$ with $x^{(m)}\to x$, which is possible by part (a). The bound givesso $(T{x}^{(m)})$ is a Cauchy sequence. Since ${\ell }^p$ is a Banach space, it converges. DefineThe same estimate shows that this limit does not depend on the approximating sequence. Passing to the limit proves linearity andso $\Vert \tilde{T}\Vert \le C$. It agrees with $T$ on $F$. If two bounded extensions existed, their difference would vanish on the dense subspace $F$ and, by continuity, on its closure ${\ell }^p$. Thus the extension is unique. This is the extension of a bounded linear operator from a dense subspace.

Write $B$ for the open unit ball. Because $T$ is a linear map, the hypothesis scales toFix $y\in B$. Choose numbers $r$ and $\theta$ such thatthen $r/(1-\theta )<1$.

Set $z_0=y$. Inductively, if $\Vert z_{j-1}\Vert <r{\theta }^{j-1}$, the scaled density statement lets us chooseand we put $z_j=z_{j-1}-T{x}_j$. The geometric series givesBecause $X$ is a Banach space, $x={\sum }_{j\ge 1}{x}_j$ exists and lies in $B$. Since $T$ is a bounded linear operator, it is continuous, whileTherefore $Tx=y$. As $y\in B$ was arbitrary,This is the dense open-unit-ball image criterion.

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 Baire category theorem says that a complete metric space cannot be a countable union of nowhere dense sets, equivalently that every countable intersection of open dense sets is dense. To prove it, let $G_1,G_2,\dots$ be open dense subsets and let $V$ be nonempty and open. Choose nested closed ballsstarting inside $V\cap G_1$. The centres are a Cauchy sequence; completeness gives a limit lying in every ball, hence in $V\cap {\bigcap }_n{G}_n$. Thus the intersection is dense.

Choose a countable sequence $q_j\uparrow p$ with $1\le q_j<p$. Since ${\ell }^q\subset {\ell }^r$ for $q<r$, every ${\ell }^q$ with $q<p$ lies in some ${\ell }^{q_j}$. For integers $m\ge 1$, setPointwise convergence and Fatou lemma for series show that $E_{j,m}$ is closed in ${\ell }^p$. It has empty interior: inside any ${\ell }^p$ ball, add a sufficiently long finite block whose ${\ell }^p$ norm is tiny but whose ${\ell }^{q_j}$ norm is larger than $m$. Therefore every $E_{j,m}$ is nowhere dense. Since ${\ell }^p$ is a Banach space, Baire gives

An explicit element of the difference isIts $p$th powers form a convergent logarithmic series, whereas for every $q<p$ the powers contain $n^{-q/p}$ with exponent below one and their series diverges. Hence $x\in {\ell }^p$ but $x\notin {\ell }^q$ for every $q<p$.

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.

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.

Write $C(K)={\bigcup }_{n\ge 1}{\mathcal{S}}_n$ with every ${\mathcal{S}}_n$ equicontinuous. Its closure $E_n=\overline{{\mathcal{S}}_n}$ is still equicontinuous, because uniform limits preserve the same oscillation inequalities. The Baire category theorem applied to the Banach space $C(K)$ shows that some $E_n$ has nonempty interior. Hence there are $f_0\in C(K)$ and $r>0$ such that the open ball $B(f_0,r)$ lies in $E_n$.

Fix $x\in K$. Equicontinuity of $E_n$ gives a neighborhood $U$ of $x$ on which every member of $E_n$ changes by less than $r/4$. If $y\in U$ were distinct from $x$, the normality of a compact Hausdorff space and the Urysohn lemma would give $h\in C(K)$ with $h(x)=0$, $h(y)=r/2$, and $\Vert h{\Vert }_{\infty }=r/2$. Both $f_0$ and $f_0+h$ belong to $E_n$, yeta contradiction. Thus $U=\{x\}$, so every point is isolated and $K$ is a discrete space. A compact discrete space is finite, since its cover by singleton open sets has a finite subcover.

The preceding estimate gives constants $A,B>0$ such that, whenever $\Vert w{\Vert }_{H^2}\le R$,Choose $R>0$ so that $B{R}^2\le 1/2$, and then choose $r>0$ so that $Ar\le R/2$. If $\Vert f{\Vert }_2\le r$, the closed ball $B_R(0)$ is mapped into itself.

For $w,z\in B_R(0)$, factor the cubic gradient nonlinearity asThe same Sobolev embedding theorem and the Holder inequality implyThe elliptic estimate therefore yieldsShrinking $R$ further makes $C^{\prime }{R}^2<1$, so $\Phi$ is a contraction mapping of the closed ball. This ball is complete because $H^2(U)\cap H_0^1(U)$ is a Banach space. The contraction mapping theorem gives a fixed point $u=\Phi (u)$, and its defining equation isThus the nonlinear elliptic boundary value problem has a solution for sufficiently small $\Vert f{\Vert }_2$.

A Gaussian measure $\mu$ on a real separable Banach space $X$ is a Borel probability measure such that ${\ell }_{\ast }\mu$ is a one-dimensional normal distribution for every continuous linear functional $\ell \in X^{\ast }$. Its mean $m\in X$ and covariance operator of a Gaussian measure $C:X^{\ast }\to X$ are characterized byandfor all ${\ell }_1,{\ell }_2\in X^{\ast }$.

The infinitesimal generator of a semigroup is the linear operatorwith generator domainFor example, the Bochner integralbelongs to $D(A)$ for every $u\in H$ and $t>0$, since $A{u}_t=U(t)u-u$.

To prove that $A$ is a closed linear operator, suppose $u_n\in D(A)$, $u_n\to u$, and $A{u}_n\to v$. For vectors in the generator domain,Passing to the limit in the Banach space givesAfter division by $t$, strong continuity makes the right side converge to $v$ as $t\downarrow 0$. Hence $u\in D(A)$ and $Au=v$, so $A$ is closed.

For $\mathrm{Re}z>\omega$, define the Bochner integralIt converges absolutely becauseIntegrating the semigroup difference quotient shows that $R(z)u\in D(A)$ and $(zI-A)R(z)u=u$. The same computation for $u\in D(A)$ gives $R(z)(zI-A)u=u$. Thus the Laplace-transform formula for a semigroup resolvent proves