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A contraction semigroup on a Banach space $X$ consists of bounded linear operators $S(t)$ with $S(0)=I$, $S(t+s)=S(t)S(s)$, $\Vert S(t)\Vert \le 1$, and $S(t)x\to x$ as $t\downarrow 0$ for each $x\in X$. Its infinitesimal generator of a semigroup isFor density, Yosida averaging of a semigroup gives $x_h=h^{-1}{\int }_0^{h}S(s)xds$. Taking the difference quotient of this Bochner integral shows that $x_h\in D(A)$ and $A{x}_h=(S(h)x-x)/h$. Strong continuity gives $x_h\to x$, so $D(A)$ is dense.

For closedness, the semigroup restricted to its generator domain satisfiesIf $x_n\to x$ and $A{x}_n\to y$, pass to the limit in this identity. After dividing by $t$, strong continuity gives $(S(t)x-x)/t\to y$. Hence $x\in D(A)$ and $Ax=y$. This proves generator of a strongly continuous semigroup is closed and densely defined.

The contraction form of the Hille-Yosida theorem states that a linear operator $A$ generates a contraction semigroup if and only if it is densely defined and closed, every real $\lambda >0$ lies in its resolvent set, andHere the $n=1$ bound implies the others by taking powers of the same bounded resolvent.

The H1 space is $H^1(\mathbb{R})=\{u\in L^2:u^{\prime }\in L^2\}$ with weak derivative $u^{\prime }$ and squared norm $\Vert u{\Vert }_2^2+\Vert u^{\prime }{\Vert }_2^2$. The displayed weighted space is the one-dimensional harmonic oscillator form domain, with inner productIf $u_n$ is Cauchy in $X$, it converges in $H^1$ to $u$, and $x{u}_n$ converges in L2 space to some $w$. On each bounded interval, multiplication by $x$ is bounded, so $w=xu$ there. Thus $xu\in L^2$ and convergence holds in $X$, proving completeness and the Hilbert space property. Equivalently this is closedness of the multiplication operator by $x$.

For the Sobolev characterization by bounded difference quotients, if $f\in H^1$ thenThe last assertion uses continuity of translation of a function in L2 space. Conversely, if $f\in L^2$ and the quotients are uniformly bounded for $0<|h|<1$, take a weakly convergent subsequence as $h\to 0$. For every test function $v$,The weak limit is therefore the weak derivative $f^{\prime }\in L^2$. This proves the characterization.

The definitions in a unital C-star algebra areThus a Hermitian element of a C-star algebra is self-adjoint, a Unitary element of a C-star algebra has inverse $x^{\ast }$, and a Normal element of a C-star algebra commutes with its adjoint. We derive the needed C-star algebra facts directly from $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$, as required.

First $\Vert 1\Vert =\Vert 1{\Vert }^2$ gives $\Vert 1\Vert =1$, andgives $\Vert a\Vert \le \Vert a^{\ast }\Vert$; applying the same argument to $a^{\ast }$ gives equality. Thus the involution is an isometry. For a Unitary element of a C-star algebra $u$,The general Banach algebra spectrum bound gives $|\lambda |\le 1$ for $\lambda \in \sigma (u)$. Such a $\lambda$ is nonzero, and the inverse spectral mapping theorem gives ${\lambda }^{-1}\in \sigma (u^{-1})$, so also $|\lambda {|}^{-1}\le 1$. Hence

If $h=h^{\ast }$, the convergent exponential series and the isometric involution give $(e^{ith}{)}^{\ast }=e^{-ith}$. These commuting exponentials multiply to $1$, so $e^{ith}$ is unitary for real $t$. By the exponential spectral mapping theorem, if $\lambda \in \sigma (h)$ then $e^{it\lambda }\in \sigma (e^{ith})\subseteq \mathbb{T}$. Taking $t=1$ gives $e^{-\mathrm{Im}\lambda }=1$, soOnly general Banach algebra spectral mapping theorem and the defining C-star identity were used.

Now prove spectral permanence for C-star algebras. Let $B\subseteq A$ be the norm-closed C-star subalgebra with the same identity. The algebraic inclusion always gives ${\sigma }_A(b)\subseteq {\sigma }_B(b)$. For $h=h^{\ast }\in B$, both spectra lie in $\mathbb{R}$. If $\lambda \notin {\sigma }_A(h)$, choose nonreal ${\lambda }_n\to \lambda$. Since ${\sigma }_B(h)\subseteq \mathbb{R}$, each $({\lambda }_{n}1-h{)}^{-1}$ belongs to $B$. By continuity of inversion in a Banach algebra, these inverses converge in $A$ to $(\lambda 1-h{)}^{-1}$, which belongs to $B$ because $B$ is closed. Thus $\lambda \notin {\sigma }_B(h)$, proving equality for hermitian elements.

For the requested normal $x\in B$, suppose $b=x-\lambda 1$ is invertible in $A$. The element $b^{\ast }b$ is hermitian and invertible in $A$, so the equality just proved places $(b^{\ast }b{)}^{-1}$ in $B$. ConsequentlySince $b$ already has a two-sided inverse in $A$, multiplying $yb=1$ by that inverse gives $y=b^{-1}$. Hence its inverse belongs to $B$, and

We need one more elementary consequence of the C-star identity: the spectral radius norm equality for normal elements. If $h=h^{\ast }$, then $\Vert h^2\Vert =\Vert h{\Vert }^2$. If $a$ is normal, commutativity of $a,a^{\ast }$ givesEvery power of $a$ is a Normal element of a C-star algebra. Induction gives $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The general spectral radius formula therefore implies

The continuous functional calculus is the unique unital $\ast$-homomorphism $C(K)\to \mathcal{B}(H)$ taking the coordinate function to $T$. For bounded linear operators on a Hilbert space, the adjoint identity gives $\Vert S\Vert ={\sup }_{\Vert v\Vert =\Vert w\Vert =1}|⟨Sv,w⟩|=\Vert S^{\ast }\Vert$. ThereforeThis proves the C-star identity for $\mathcal{B}(H)$ directly; completeness and submultiplicativity come from the operator norm. LetBecause $T$ is a normal operator, this is a commutative unital C-star algebra. Every element of $C$ is a Normal element of a C-star algebra, so the preceding norm identity says its Gelfand transform is isometric:Here $\Delta (C)$ is the compact character space from the general Gelfand representation theorem for commutative Banach algebras.

Each algebra character preserves the involution. Indeed, write $a=h+ik$ with $h,k$ hermitian; $\chi (h)\in {\sigma }_C(h)\subseteq \mathbb{R}$ and similarly for $k$, so $\chi (a^{\ast })=\overline{\chi (a)}$. Therefore the continuous mapis injective: its value determines $\chi (T^{\ast })$ and hence its value on all the dense polynomials. It is surjective because the general character description of the spectrum gives $\{\chi (T):\chi \in \Delta (C)\}={\sigma }_C(T)$, and spectral permanence for C-star algebras identifies this with ${\sigma }_{\mathcal{B}(H)}(T)=K$. A continuous function that is a bijection from a compact space to a Hausdorff space is a homeomorphism, so we identify $\Delta (C)$ with $K$.

Under this identification, the Gelfand transform sends $T$ to $u(z)=z$ and $T^{\ast }$ to $\overline{u}$. Its image is an isometric, and therefore closed, unital self-adjoint subalgebra of $C(K)$. It separates points because it contains $u$. The complex Stone-Weierstrass theorem makes the image dense, hence equal to all of $C(K)$. Inverting the Gelfand transform and including $C$ into $\mathcal{B}(H)$ givesan isometric unital C-star homomorphism with $u(T)=T$.

To prove uniqueness without assuming automatic continuity of C-star homomorphisms, let $\Psi$ be any other such map. Since every $f\in C(K)$ is normal, $\Psi (f)$ is normal. A unital algebra homomorphism preserves inverses, soUsing the normal-element norm identity yields $\Vert \Psi (f)\Vert =r(\Psi (f))\le \Vert f{\Vert }_{\infty }$. Thus $\Psi$ is contractive. It agrees with $\Phi$ on polynomials in $u,\overline{u}$, since both send them to the same polynomials in $T,T^{\ast }$. These polynomials are dense by the Stone-Weierstrass theorem, so continuity proves $\Psi =\Phi$. This establishes the continuous functional calculus with no unproved theorem specific to C-star algebras.

A disconnected spectrum gives a nontrivial closed invariant subspace. Write $K=K_1\bigsqcup K_2$ with $K_1,K_2$ nonempty and both open and closed in $K$. The indicator function $f=1_{K_1}$ is continuous on $K$, even though no such continuity is needed across the gap outside $K$. Let $P=f(T)$. The continuous functional calculus givesIts isometry gives $\Vert P\Vert =1$ and $\Vert I-P\Vert =1$, so $P\ne 0,I$. The linear projection has closed range $Y=PH=\ker (I-P)$. Its range is nonzero and proper, and $T(Px)=P(Tx)\in Y$ proves invariance. Since it also commutes with $T^{\ast }$, the same subspace is reducing. This proves disconnected spectrum gives a reducing subspace.

The figure can be realized without any eigenvalues: take $H=L^2(K,dA)$, where $dA$ is planar area on the two closed disks, and let $T$ multiply by $z$. Its adjoint multiplies by $\overline{z}$, so it is a normal operator. An eigenvector for $\lambda$ would be supported on the area-zero singleton $\{\lambda \}$, hence would be zero in $L^2$. Outside $K$, multiplication by $1/(z-\lambda )$ is a bounded inverse to $T-\lambda I$. For $\lambda \in K$, normalized indicator functions of $K\cap \{|z-\lambda |<\epsilon \}$ have $\Vert (T-\lambda I)h\Vert \le \epsilon$, excluding a bounded inverse. Thus its spectrum is exactly $K$. The linear projection in the figure multiplies by $1_{K_1}$, and its range consists of functions supported on the left disk.

Use $c>0$, the necessary parameter range for the bounds involving $1/c$. The construction comes from the singular value decomposition of matrix Fourier blocks. For each inequivalent unitary irreducible representation $\rho$ of dimension $d_{\rho }$, consider the Hilbert space ${\mathcal{H}}_{\rho }=M_{n\times d_{\rho }}(\mathbb{C})$ with the Hilbert-Schmidt inner product $⟨A,B⟩=\mathrm{tr}(A^{\ast }B)$ and the operatorIts operator norm is at most one, since $\rho (x)$ is a unitary matrix and $\Vert f(x){\Vert }_{\mathrm{op}}\le 1$. Thus all its singular values ${\lambda }_{\rho ,j}$ lie in $[0,1]$.

Two identities control their weighted moments:Here the expectation in $S_4$ is uniform on the $|G{|}^3$ solutions of the constraint, and the trace on ${\mathcal{H}}_{\rho }$ used to compute each moment is the ordinary operator trace. We justify the identities explicitly to fix their normalizations and the noncommutative order.

The adjoint operator is $T_{\rho }^{\ast }(A)={\mathbb{E}}_{y}f(y{)}^{\ast }A\rho (y)$, soFor rectangular matrices, the operator $A\mapsto LAR$ has trace $\mathrm{tr}(L)\mathrm{tr}(R)$, as is seen on the matrix-unit basis. Squaring the previous operator therefore givesThe required representation theory consists of unitarization of a finite-group representation, the Schur orthogonality relations, and the regular representation decomposition. The last gives the character expansion of the identity deltaSumming the two operator traces with weights $d_{\rho }$ selects $x=y$ in the first, and $w=x{y}^{-1}z$ in the second. Multiplication by $|G|$ converts each independent-variable expectation to its conditional expectation. This proves $S_2,S_4$. In particular $S_4$ is real and nonnegative, despite the apparently complex summands in its original formula. Since $S_4\le S_2\le n$, the nonvacuous hypothesis has $0<c\le 1$.

Select every singular value satisfying ${\lambda }_{\rho ,j}\ge \sqrt{c/2}$. Index the selected pairs by $p=1,\dots ,r$, set ${\rho }_p=\rho$, $n_p=d_{\rho }$, and choose unit right singular vectors $A_p$ and unit left singular vectors $B_p$ with $T_{{\rho }_p}{A}_p={\lambda }_p{B}_p$. DefineBoth are $n\times n_p$ matrices, proving (i). Repeated group representations in this list are literally the same chosen representative, rather than different equivalent realizations.

Let $V$ be the closure of the simple predictable processes in the Hilbert space $L^2(\mathcal{P},\mu )$. Common refinement of deterministic partitions makes these processes a vector space. To prove density of simple predictable processes for finite measures, suppose $g\in V^{\perp }$. Since $\mu$ is a finite measure, the Cauchy-Schwarz inequality gives $g\in L^1(\mu )$, sodefines a finite signed measure. The indicator of each predictable generating rectangle, including each time-zero rectangle, is a simple predictable process; hence $\nu$ vanishes on these rectangles.

Finite intersections of the rectangles are again rectangles or empty. Add the whole space to this pi-system. Its $\nu$-mass is zero because $1_{\Omega \times [0,n]}$ is simple and converges to $1$ in $L^2(\mu )$ by the dominated convergence theorem. The zero sets of $\nu$ form a lambda-system now that its total mass is zero. The Pi-lambda theorem therefore gives $\nu (E)=0$ for every $E\in \mathcal{P}$. Taking the sets where $g>0$ and $g<0$ gives $g=0$ $\mu$-almost everywhere.

Thus $V^{\perp }=\{0\}$ and the orthogonal complement criterion for a closed subspace of a Hilbert space yieldsOmitting the time-zero term would make this conclusion false when $\mu$ is concentrated at time zero.

Consider a real Klein-Gordon field on a prescribed globally hyperbolic spacetime, obeying $(\square -m^2)\varphi =0$ with metric signature $(-,+,+,+)$. A curvature coupling can be included as $m^2\mapsto m^2+\xi R$. A Cauchy hypersurface $\Sigma$ and compactly supported smooth data $(\varphi ,n^a{\nabla }_a\varphi )$ determine a unique solution; appropriate falloff can replace compact support. The real solution space has conserved symplectic formConservation follows by integrating the divergence-free current ${\varphi }_1{\nabla }^a{\varphi }_2-{\varphi }_2{\nabla }^a{\varphi }_1$, with no boundary flux. On complex solutions the conserved Klein-Gordon inner product isIt is indefinite on the full complex solution space.

Choose a complete positive-norm mode subspace, with modes $u_j$ satisfyingEquivalently choose a compatible complex structure on the Klein-Gordon solution space. The mode labels may be continuous, in which case sums and Kronecker symbols become integrals and Dirac delta functions. Construct the one-particle Hilbert space from these modes and its bosonic Fock space. Promote the field to the operator-valued distributionFor a foliation with spatial metric determinant $h$, the conjugate momentum density is $\hat{\pi }=\sqrt{h}{n}^a{\nabla }_a\hat{\varphi }$. Mode completeness gives the equal-time canonical commutation relationsThe Fock vacuum obeys $a_j|0⟩=0$, and $a_j^{†}{a}_j$ counts particles in the chosen mode. For local products and a renormalized stress-energy tensor, physically admissible states are further restricted by the Hadamard condition. The field algebra exists without a preferred Fock vacuum.

A different admissible mode splitting can mix positive and negative norms:Orthonormality imposes the canonical identities for a bosonic Bogoliubov transformationThe same field then hasWithout a preferred notion of positive frequency, a nonstationary spacetime supplies no distinguished mode splitting: particle number and vacuum depend on the choice of modes, although the field equation and field algebra do not. With infinitely many modes the Bogoliubov transformation need not be unitarily implementable; finite total mixing requires a Hilbert-Schmidt operator $\beta$.

For a stable strictly stationary spacetime, a chosen future globally timelike Killing vector field $K$ gives a preferred time translation. Choose modes withWhen the corresponding conserved Killing energy is positive and the spectral problem has suitable boundary conditions and no problematic zero modes, this gives the preferred vacuum state in a stationary spacetime and particles relative to $K$. Positive-frequency mode mixing within that same subspace leaves the Fock vacuum unchanged. Rescaling $K$ by a positive constant changes the frequency units but not their sign.

Stationarity alone, if it only means a Killing field timelike near infinity, is insufficient for a global unique particle interpretation. In the Kerr ergoregion, ${\partial }_t$ is spacelike, so positive frequency relative to $t$ does not automatically select a positive-norm subspace throughout the geometry; superradiance illustrates the difficulty. Additional vacuum and boundary choices remain necessary. The customary stationary answer therefore assumes a suitable timelike stationary flow and a stable positive-energy quantization; it does not assert that every stationary black-hole extension has one globally preferred vacuum.

For any normalized pure state on finite-dimensional Hilbert spaces ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$, the Schmidt decomposition givesThe $u_j$ and $v_j$ form orthonormal sets in their respective spaces. The positive numbers $\sqrt{p_j}$ are the Schmidt coefficients, and $r$ is the Schmidt rank. Taking a partial trace gives the reduced density matricesThus the nonzero eigenvalues of the two reduced density matrices agree.

One construction is a singular value decomposition of the coefficient matrix $M$ in $|\psi ⟩={\sum }_{a,b}{M}_{ab}|a⟩|b⟩$. If $M=U\Sigma V^{†}$, the columns of $U$ give $u_j$, the complex conjugates of the columns of $V$ give $v_j$, and the singular values give $\sqrt{p_j}$. This also explains why the Schmidt coefficients are real and nonnegative.

The Browder convergence theorem for averaged operators states: if $T:{\mathbb{R}}^d\to {\mathbb{R}}^d$ is an averaged operator and its fixed-point set $\mathrm{Fix}T=\{z:Tz=z\}$ is nonempty, then for every starting point, $z^{k+1}=T{z}^k$ converges in norm to a fixed point of $T$. The theorem assumes a fixed point exists; averaging by itself does not create one.

The finite-dimensional mechanism is worth making explicit. For any fixed point $p$, the averaged-operator inequality givesThis is Fejér monotonicity with respect to the fixed-point set. It makes the sequence bounded and, by summation, makes ${\sum }_k\Vert z^{k+1}-z^k{\Vert }_2^2$ finite. Hence its residual tends to zero. Choose a convergent subsequence $z^{k_j}\to \bar{z}$; continuity of the nonexpansive mapping $T$ implies $T\bar{z}=\bar{z}$. Apply Fejér monotonicity with $p=\bar{z}$. The nonincreasing distances $\Vert z^k-\bar{z}{\Vert }_2$ have a subsequence tending to zero, so the whole sequence converges to $\bar{z}$. This last compactness step uses finite dimension; a corresponding general Hilbert space statement usually gives weak convergence.

Choose a singular system of a compact operator with $K{v}_j={\sigma }_j{u}_j$, $K^{\ast }{u}_j={\sigma }_j{v}_j$ and ${\sigma }_j>0$. The Picard criterion, together with range orthogonality, givesFor necessity, expand a solution's component in $(\ker K{)}^{\perp }$ in the orthonormal vectors $v_j$; its coefficients must be $⟨f,u_j⟩/{\sigma }_j$. Their square sum is finite by Bessel's inequality. Conversely, the displayed square sum defines a convergent Hilbert space seriesApplying the bounded linear operator $K$ to the partial sums yields the expansion of $f$ in $\overline{\mathrm{ran}K}=(\ker K^{\ast }{)}^{\perp }$. Thus its limit solves $Ku=f$. The perpendicularity condition is essential here: the Picard criterion alone characterizes the domain of the generalized inverse, which also includes components in $\ker K^{\ast }$.

For a bounded linear operator $K:U\to V$ between Hilbert spaces, the Moore–Penrose inverse of an operator assigns to each admissible datum its unique minimum-norm least-squares solution. Its domain isWrite $f=Ku+h$, where $u\in (\ker K{)}^{\perp }$ and $h\in (\mathrm{ran}K{)}^{\perp }=\ker K^{\ast }$. Then $K^{†}f=u$. Restricting to $(\ker K{)}^{\perp }$ removes the ambiguity among solutions differing by an element of the kernel. The orthogonal decomposition also shows that $K{K}^{†}f$ is the orthogonal projection of $f$ onto $\overline{\mathrm{ran}K}$, and $K^{†}Ku$ is the orthogonal projection of $u$ onto $(\ker K{)}^{\perp }$. The inverse is defined on a dense domain and need not be a bounded linear operator when the range is not closed.

For a proper convex function $E$ on a real Hilbert space, its subdifferential at $y$ isIn a Banach space, interpret the pairing with an element of the dual space. For a chosen subgradient $p\in \partial E(y)$, the Bregman distance isIf also $q\in \partial E(x)$, its symmetric Bregman distance isThe subgradient inequality proves nonnegativity. These quantities depend on the chosen subgradients and generally are not metrics: they need not distinguish different points or obey the triangle inequality.

Let $s=Kx$ and assume $s>0$. The scalar Poisson data fidelity integrand has derivative $1-y/s$. Applying the permitted interchange of differentiation and integration in a direction $z$ givesThus the ambient-space Frechet derivative, or its Hilbert space gradient when the pairing is an inner product, isThe quotient is pointwise, and the adjoint operator transports it back from the data domain to the source domain. The chosen spaces must ensure the integrals and derivative pairing are finite; an output bounded away from zero is a useful sufficient condition.

The printed set of probability density functions and positive cone are not vector spaces, so the notation for a bounded linear map between them is understood as a positive bounded linear operator on ambient function spaces, restricted to admissible inputs. If unit mass is a constraint, admissible directions satisfy ${\int }_{\Omega }z=0$. The displayed gradient is then a representative modulo an additive constant; optimization with that constraint additionally includes its normal cone.

For a proper convex function $E$ that is lower semicontinuous on a Hilbert space, its proximal operator at scale $\gamma >0$ isThe minimizer is unique because the objective is strongly convex; existence follows from the closed proper convex functional and the coercive quadratic, using an affine lower bound. The subgradient optimality condition is $0\in x-z+\gamma \partial E(x)$, which is exactly the displayed resolvent of a monotone operator relation. Setting $\gamma =1$ gives the unscaled proximity or resolvent operator requested here. This is a nonlinear resolvent of the subdifferential, distinct from the spectral resolvent of a linear operator.

The Riesz-Markov-Kakutani representation theorem identifies the dual of the complex space of continuous functions on a compact space $C(K)$ with finite regular complex Borel measures on $K$: each bounded linear functional has a unique representationHere $|\mu |$ is the variation measure; positive functionals correspond exactly to positive regular measures. This is the measure representation theorem, rather than the Hilbert-space representation of a functional by a vector.

For the spectral construction use the convention that the Hilbert inner product is linear in its first argument. A unital subalgebra of $\mathcal{B}(H)$ is understood to contain $I_H$: an algebra whose abstract identity is a smaller projection could not give a resolution normalized by $P(K)=I_H$. On a compact Hausdorff space, a resolution of the identity is understood to be a normalized regular projection-valued measure: its scalar measures are regular, its values are orthogonal projections, and it is countably additive in the strong operator topology. Regularity is part of the usual convention needed for the uniqueness assertion; the statement is interpreted in this sense.

The Commutative Gelfand--Naimark theorem makes the Gelfand transform an isometric unital star-isomorphism $A\to C(K)$, where $K={\Phi }_A$ is compact Hausdorff. Denote its inverse by $\pi :C(K)\to A\subset \mathcal{B}(H)$. For $x,y\in H$, the bounded linear functional $f\mapsto ⟨\pi (f)x,y⟩$ has norm at most $\Vert x\Vert \Vert y\Vert$. Riesz-Markov-Kakutani gives a regular complex measure ${\mu }_{x,y}$ such thatUniqueness makes these measures sesquilinear in $x,y$. For $f\ge 0$, the identity $\pi (f)=\pi (\sqrt{f})\pi (\sqrt{f}{)}^{\ast }$ shows positivity, so ${\mu }_{x,x}$ is positive with total mass $\Vert x{\Vert }^2$.

For every Borel set $B$, the bounded sesquilinear form $(x,y)\mapsto {\mu }_{x,y}(B)$ determines an operator $P(B)$ by Hilbert-space Riesz representation theorem, withIn particular $0\le P(B)\le I$ and $P(B)$ is self-adjoint. We now verify the projection identity rather than assume it.

For continuous $f$, uniqueness of the representing measure applied to continuous test functions givesThese identities imply $P(B)\pi (f)=\pi (f)P(B)$. Testing once more against a continuous $g$ givesThe restriction $1_B{\mu }_{x,y}$ of a finite regular Borel measure is still regular, so uniqueness in Riesz-Markov-Kakutani yields ${\mu }_{P(B)x,y}=1_B{\mu }_{x,y}$. Consequently, for any Borel $C$,Taking $C=B$ proves that $P(B)$ is an orthogonal projection. Also $P(K)=I$ and $P(\varnothing )=0$. For disjoint $B_j$, these projections are orthogonal, and scalar countable additivity gives weak countable additivity. If $B={\bigcup }_j{B}_j$, then $R_N=P(B)-{\sum }_{j\le N}P(B_j)$ is the projection of the remaining union andThus countable additivity holds in the strong operator topology. The scalar measures are the regular measures already constructed. This completes the scalar-measure construction of a projection-valued measure.

By the integral theorem permitted in the question, continuous $f$ satisfies $⟨(\int fdP)x,y⟩=\int fd{\mu }_{x,y}=⟨\pi (f)x,y⟩$, and hence $\int fdP=\pi (f)$. For $T\in A$, this proves the spectral theorem for a commutative operator algebra:Any other regular resolution giving these integrals has the same scalar integrals on all of $C(K)$; Riesz-Markov-Kakutani uniqueness forces the same scalar measures and therefore the same projections on every Borel set.

For nonempty open $U\subset K$, compact Hausdorff normality supplies a nonzero continuous function $f$ supported in $U$. If $P(U)=0$, the stated squared-norm identity for spectral integrals gives $\pi (f)=\int fdP=0$, contradicting the isometry of $\pi$. This proves the full support of a faithful spectral measure propertyFaithfulness of the representation is essential here.

The spectral theorem for normal operators says that a bounded normal operator $T$ on a nonzero complex Hilbert space has a unique regular projection-valued measure $E$ on $\sigma (T)$ such that $T=\int zdE(z)$. Its support is all of $\sigma (T)$, and bounded Borel functions have the associated Borel functional calculus for a normal operator.

For the proof sketch, $A=C^{\ast }(I,T)$ is commutative because $T$ commutes with $T^{\ast }$; polynomials in these two operators commute, as do their norm limits. The map ${\Phi }_A\to \sigma (T)$, $\phi \mapsto \phi (T)$, is onto by the character of an algebra formula and spectral permanence for C-star algebras. It is one-to-one because a character of an algebra preserves the star operation and its values on $T,T^{\ast }$ determine it on their dense polynomial algebra. It is therefore a homeomorphism from compact ${\Phi }_A$ to the Hausdorff spectrum. Transport the resolution just constructed through this homeomorphism. It gives the formula for $T$ and full support; conversely a regular resolution for $T$ gives the same integrals for polynomials in $T,T^{\ast }$, hence by density the same continuous functional calculus and the same resolution. This argument works without separability of $H$.

Finally choose disjoint nonempty relatively open sets $U,V\subset \sigma (T)$ around two distinct spectral points, and put $Q=E(U)$. Full support gives $Q\ne 0$ and $E(V)\ne 0$, while $QE(V)=0$, so $Q\ne I$. The spectral integral, or multiplicativity of its Borel calculus with $1_U$, gives $QT=TQ$ and also $Q{T}^{\ast }=T^{\ast }Q$. Thus $Y=QH$ is closed, nonzero and proper, and $T(Qx)=Q(Tx)\in Y$. The spectral projection gives a reducing subspace conclusion isIn fact it is a reducing subspace, since it is also invariant under $T^{\ast }$.

Fix a L2-bounded continuous martingale $M$. Its quadratic-variation measure on the predictable sigma-algebra isIt is finite, since ${\nu }_M(\Omega \times [0,\infty ))=\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2$. The source Hilbert space is $L^2(M)=L^2(\Omega \times [0,\infty ),\mathcal{P},{\nu }_M)$: predictable processes with finite $\mathbb{E}\int H^{2}d[M]$, identified when equal ${\nu }_M$-almost everywhere. Its norm is $\Vert H{\Vert }_{L^2(M)}^2=\mathbb{E}\int H^{2}d[M]$.

The target ${\mathcal{M}}_0^2$ consists of L2-bounded continuous martingales starting at zero, identified up to indistinguishability of stochastic processes, with norm $\Vert K{\Vert }_{{\mathcal{M}}_0^2}=\Vert K_{\infty }{\Vert }_2$. It is a Hilbert space: terminal values belong to the closed linear subspace of $L^2({\mathcal{F}}_{\infty })$ whose conditional expectation given ${\mathcal{F}}_0$ is zero and whose associated martingales have continuous versions. Closure of that continuous-version subspace follows from the Doob L2 maximal inequality and an almost surely uniformly convergent subsequence.

The Itô isometry says that integration extends uniquely from simple predictable processes to a linear isometry $L^2(M)⟶{\mathcal{M}}_0^2$, withFor every finite $t$, the same identity holds with the integral restricted to $[0,t]$ and the left side $\mathbb{E}|(H\cdot M{)}_t{|}^2$. No claim of surjectivity onto all of ${\mathcal{M}}_0^2$ is needed: that would require an additional Martingale representation theorem.

A real positive-definite kernel is a symmetric function $k:\mathcal{X}\times \mathcal{X}\to \mathbb{R}$ such that, for every finite collection $x_1,\dots ,x_n$ and every $a\in {\mathbb{R}}^n$,Thus every kernel matrix is a positive semidefinite matrix; “positive definite” here does not require strict positivity. A Reproducing-kernel Hilbert space is a Hilbert space $\mathcal{H}$ of functions on $\mathcal{X}$ in which every point evaluation is a continuous linear functional. Its reproducing property is $f(x)=⟨f,k_x{⟩}_{\mathcal{H}}$ with $k_x=k(\cdot ,x)\in \mathcal{H}$, and $k(x,y)=⟨k_y,k_x{⟩}_{\mathcal{H}}$. For a sum of positive-definite kernels, the quadratic form of its kernel matrix is a sum of nonnegative quadratic forms. Symmetry also survives summation. Consequently $\boxed{k=\sum _j{k}_j\text{ is a positive-definite kernel}.}$

For the representer theorem, put $V=\mathrm{span}\{k_{x_1},\dots ,k_{x_n}\}$. This is a closed finite-dimensional vector space. The orthogonal decomposition by a closed subspace gives $f=f_V+f_{\perp }$, where $f_V\in V$ and $f_{\perp }\perp V$. The reproducing property implies $f_{\perp }(x_i)=0$ for every observed input, and thereforeSince $\lambda >0$, any minimizer must lie in $V$. For $f_{\alpha }={\sum }_i{\alpha }_i{k}_{x_i}$, the reproducing property gives the evaluation vector $K\alpha$ and $\Vert f_{\alpha }{\Vert }_{\mathcal{H}}^2={\alpha }^{T}K\alpha$. Hence $Q_1(f_{\alpha })=M(\alpha )$. If $\hat{f}$ minimizes $Q_1$, its representation yields a minimizing $\hat{\alpha }$. Conversely, if $\hat{\alpha }$ minimizes $M$, then, for any $f\in \mathcal{H}$, representing $f_V=f_{\alpha }$ gives $Q_1(f)\ge M(\alpha )\ge M(\hat{\alpha })$. ThusThis is an equivalence of minimizers, not an existence or uniqueness assertion: the loss function is arbitrary. A singular kernel matrix also allows nonunique coefficients, since differences in $\ker K$ represent the zero function. Strict convexity of the loss was not used.