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For a unital star-subalgebra $A\subseteq B(H)$, its unit ball is dense in the unit ball of $A^{\prime \prime }$ in the strong operator topology. The same holds for self-adjoint unit balls and for positive unit balls, and the full unit balls may be approximated strongly together with their adjoints. Passing first to the norm closure allows continuous functional calculus; resolvent identities prove strong continuity of continuous functions vanishing at infinity, even for unbounded approximating nets of self-adjoint operators. A cutoff function supplies the norm bound, and a two-by-two self-adjoint matrix removes the self-adjoint restriction.

Let $K=\sigma (T)$ for a bounded normal operator on a nonzero complex Hilbert space. The zero Hilbert space has the trivial calculus. Take the inner product linear in its first argument. The continuous functional calculus is an isometric unital star-representation $\pi :C(K)\to \mathcal{B}(H)$ with $\pi (z)=T$. The Borel calculus will act on $B_b(K)$, the algebra of actual bounded Borel measurable functions. A quotient by null functions is introduced only after specifying the spectral measure; there is no preferred scalar measure such as plane area on $K$.

Here is a construction that also covers a nonseparable Hilbert space. For $v\in H$, the positive functional $f\mapsto ⟨\pi (f)v,v⟩$ has a finite regular Borel measure ${\mu }_v$ by the Riesz-Markov-Kakutani representation theorem. On continuous functions,Continuous functions are dense in $L^2(K,{\mu }_v)$. Hence $f\mapsto \pi (f)v$ extends to a unitary map from that space onto $H_v=\overline{\pi (C(K))v}$. This subspace reduces $T$ and $T^{\ast }$, which correspond to multiplication by $z$ and $\overline{z}$. Decompose $H$ as an orthogonal direct sum of such cyclic reducing subspaces, using maximality: a nonzero vector in the remaining orthogonal complement would generate another summand. The family need not be countable. This is the cyclic multiplication model for a normal operator.

On each summand define $f(T)$ to be multiplication by $f$, for $f\in B_b(K)$, and take their bounded direct sum. It defines a unital star-homomorphismThus $(fg)(T)=f(T)g(T)$, $\overline{f}(T)=f(T{)}^{\ast }$, and positivity of $f$ gives positivity of $f(T)$. The continuous functional calculus is recovered on $C(K)$.

Set $E(S)=\Phi (1_S)$ for a Borel subset $S$ of $K$. Then $E(S)$ is an orthogonal projection, $E(K)=I$, $E(S)E(R)=E(S\cap R)$, and disjoint countable unions add in the strong operator topology. Indeed on every multiplication summand the indicators add pointwise, and the tail squared norm is the measure of a decreasing sequence of sets tending to the empty set. Thus $E$ is a projection-valued measure, andFor $v,w\in H$, polarization gives scalar measures ${\mu }_{v,w}(S)=⟨E(S)v,w⟩$ satisfying $⟨f(T)v,w⟩=\int fd{\mu }_{v,w}$. Regularity of these finite scalar measures and uniqueness in the Riesz-Markov-Kakutani representation theorem make the projection-valued measure unique: its values on continuous functions determine all scalar measures and hence every projection. This also makes the calculus independent of the chosen cyclic decomposition.

The correct convergence assertion is bounded pointwise convergence in the strong operator topology. If $f_n\to f$ pointwise and ${\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$, thenby the dominated convergence theorem. This need not be convergence in operator norm. For multiplication by $t$ on $L^2[0,1]$, the functions $1_{[0,1/n]}$ converge pointwise to $1_{\{0\}}$, their operators converge strongly to zero, but their operator norms are all one.

A Borel measurable function can vanish spectrally without vanishing pointwise. Precisely, $f(T)=0$ if and only if $E(\{|f|>\epsilon \})=0$ for every $\epsilon >0$, andThe upper norm bound follows by integrating $|f{|}^2$ against vector measures; a unit vector in any nonzero projection $E(\{|f|>C\})$ gives $\Vert f(T)\Vert \ge C$. The induced map on the quotient by these spectral-null Borel measurable functions is isometric. Every nonempty relatively open subset of $K$ has nonzero spectral projection, since otherwise a nonzero continuous function supported there would contradict faithfulness of the continuous functional calculus. This full support gives the usual continuous spectral mapping theorem, but for a Borel measurable function the exact statement is insteadIf an indicated neighborhood has zero projection, define a bounded reciprocal of $f-\lambda$ off it and zero on it; the calculus supplies an inverse. Conversely unit vectors in the nonzero projections for arbitrarily small neighborhoods give $\Vert (f(T)-\lambda )v\Vert$ arbitrarily small, contradicting invertibility. This is spectral essential range in Borel functional calculus. For example $1_{\{0\}}(T)=0$ in the multiplication model on $[0,1]$, although the pointwise range of that function is $\{0,1\}$.

Spectral projections provide reducing subspaces, and $E(\{\lambda \})H=\ker (T-\lambda I)$: the squared norm of $(T-\lambda )v$ is $\int |z-\lambda {|}^{2}d{\mu }_v$. In finite dimension the calculus is simply $f(T)={\sum }_{j}f({\lambda }_j)P_j$, the sum over eigenspaces. In general it also permits discontinuous cuts and Borel choices of square roots, such as $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ with $s(0)=0$, giving a normal operator $s(T)$ with $s(T{)}^2=T$ even when a continuous scalar branch is unavailable.

Any bounded operator commuting with both $T$ and $T^{\ast }$ commutes with their continuous functional calculus and with all $E(S)$. To justify the latter, regularity approximates $1_S$ by uniformly bounded continuous functions in the $L^2$ norms of any finite list of vector measures; the resulting net converges strongly to $E(S)$. Thus the Borel calculus lies in the Von Neumann algebra generated by $T$. On a separable Hilbert space, a countable dense list of vectors gives a controlling measure $\mu ={\sum }_j{2}^{-j}{\mu }_{v_j}/(1+\Vert v_j{\Vert }^2)$; spectral-null sets are exactly its null sets, and the range is the represented $L^{\infty }(K,\mu )$, equal to that Von Neumann algebra. Without separability this last equality is not automatic: for the diagonal operator with one eigenvector for each $t\in [0,1]$, the generated Von Neumann algebra has arbitrary bounded diagonal functions, while the original Borel calculus has Borel diagonal functions. The spectral-measure construction and the strong convergence theorem themselves require no separability. This distinguishes the Borel functional calculus for a normal operator from its continuous predecessor without confusing pointwise functions, spectral equivalence classes and topological closure.

Put $F=\ker (I-U)$. For a unitary operator, $\ker (I-U^{\ast })=F$, soThis follows from the general identity $\mathrm{Ran}(A{)}^{\perp }=\ker A^{\ast }$. The averaging operators have $\Vert T_n\Vert \le 1$ and equal the identity on $F$. On a vector $(I-U)\eta$, telescoping givesThe uniform norm bound extends this convergence to the closure of the range. Thus $T_n$ tends to identity on $F$ and to zero on its orthogonal complement. Consequentlywhich is exactly convergence in the strong operator topology. This proves the Von Neumann mean ergodic theorem in the requested unitary case.

The Von Neumann double commutant theorem states that for a unital star-subalgebra $A\subseteq B(H)$,The commutant of an operator algebra consists of operators commuting with every element. Each commutation equation is closed in the weak operator topology, so $A^{\prime \prime }$ is weakly closed and contains both indicated closures.

Let $T\in A^{\prime \prime }$ and fix ${\xi }_1,\dots ,{\xi }_m\in H$. On $H^m$ let $A$ act diagonally and put $\xi =({\xi }_1,\dots ,{\xi }_m)$. The subspace $K=\overline{\{(a{\xi }_1,\dots ,a{\xi }_m):a\in A\}}$ reduces the diagonal star-algebra. Its orthogonal projection $P$ therefore commutes with every diagonal $a$. Writing $P$ as an $m\times m$ operator matrix shows that each entry belongs to $A^{\prime }$. Since $T\in A^{\prime \prime }$, the diagonal $T$ commutes with $P$. Unitality gives $\xi \in K$, so $P\xi =\xi$ and $P(T\xi )=T\xi$. Thus $T\xi \in K$. By definition of $K$, for any $\epsilon >0$ a single $a\in A$ satisfies ${\sum }_j\Vert (T-a){\xi }_j{\Vert }^2<{\epsilon }^2$. These are precisely the finite-vector neighbourhoods of the strong operator topology. Hence $T$ belongs to the strong closure, completing the proof. Unitality, or the corresponding nondegeneracy hypothesis, is essential to the theorem's stated form.

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 $T^{\ast }$.

A solution operator for a nonautonomous evolution equation is an evolution family $U(t,s)$ satisfyingand, on a suitable common domain $\mathcal{D}$,

One applicable nonautonomous generation theorem is the following. Suppose $\mathcal{D}$ is a dense linear subspace of $H$, each $A(t)$ has domain $\mathcal{D}$, the family is a stable family of semigroup generators with constants $M,0$, and $t\mapsto A(t)$ is continuously differentiable as a map from $[0,T]$ to $\mathcal{B}(\mathcal{D},H)$, where $\mathcal{D}$ carries one of the uniformly equivalent graph norms. Then there is a unique evolution family such that:

For the uniform partition $t_j=s+j(t-s)/N$, the frozen-generator product approximation isAs $N\to \infty$, $U_N(t,s)u\to U(t,s)u$ in the norm of $H$ for every $u\in H$, uniformly for $(t,s)$ in the compact time triangle $0\le s\le t\le T$. This is convergence in the strong operator topology, rather than convergence in the operator norm.

It remains to verify the second differential equation. The evolution family law gives, for $h>0$,Divide by $h$. Since $u\in \mathcal{D}$,while strong continuity gives $U(t,s+h)A(s)u\to U(t,s)A(s)u$. ThereforeThe left derivative follows in the same way, so $s\mapsto U(t,s)u$ is differentiable.

Here functional calculus convergence means that for every $f\in C(\mathbb{T})$,

Embed ${\mathcal{H}}_n$ in $\mathcal{H}$ and write ${\tilde{A}}_n=P_n^{\ast }{A}_n{P}_n$. The hypothesis gives ${\tilde{A}}_n\rightharpoonup A$ in the weak operator topology. Since $A_n$ and $A$ are unitary operators,Thus ${\tilde{A}}_n\to A$ in the strong operator topology. Applying the same argument to the adjoints gives ${\tilde{A}}_n^{\ast }\to A^{\ast }$ strongly.

Products of uniformly bounded strongly convergent operators converge strongly, so for every integer $k$,strongly, with negative $k$ interpreted through adjoints. Therefore convergence holds for every Laurent polynomial. The Stone-Weierstrass theorem says that Laurent polynomials are uniformly dense in $C(\mathbb{T})$. Since the continuous functional calculus is contractive, uniform approximation finishes the proof for every $f\in C(\mathbb{T})$.

A projection-valued measure $E$ assigns an orthogonal projection $E(B)$ to each measurable set, with $E(\varnothing )=0$, $E(X)=I$, $E(B\cap C)=E(B)E(C)$, and countable additivity in the strong operator topology on pairwise disjoint sets.

For a unital star-representation $\pi :C(K)\to \mathcal{B}(H)$, the Riesz-Markov-Kakutani representation theorem gives regular measures ${\mu }_{x,y}$ representing $f\mapsto ⟨\pi (f)x,y⟩$. Define $P(E)$ by $⟨P(E)x,y⟩={\mu }_{x,y}(E)$. Positivity and ${\mu }_{x,x}(K)=\Vert x{\Vert }^2$ make these positive contractions. The identities ${\mu }_{\pi (f)x,y}=f{\mu }_{x,y}$ imply commutation with $\pi (f)$ and ${\mu }_{P(E)x,y}=1_E{\mu }_{x,y}$; hence $P(F)P(E)=P(F\cap E)$. Thus each $P(E)$ is an orthogonal projection. Scalar countable additivity and orthogonality give countable additivity in the strong operator topology, constructing the normalized regular projection-valued measure.

An operator family $U(t)$ is strongly continuous when $t\mapsto U(t)x$ is norm-continuous for every vector $x$. This is continuity in the strong operator topology.

This topology on bounded operators is generated by the seminorms $(\Vert T\xi {\Vert }^2+\Vert T^{\ast }\xi {\Vert }^2{)}^{1/2}$. A net converges exactly when both it and its adjoint net converge in the strong operator topology. It is the topology used in the full-unit-ball version of the Kaplansky density theorem.

A unital star-subalgebra of the bounded operators on a Hilbert space that is closed in the weak operator topology, equivalently in the strong operator topology. The Von Neumann double commutant theorem makes these conditions equivalent to equality with the double commutant of an operator algebra. It is norm closed and hence a C-star algebra.

For a unital star-subalgebra $A$ of the bounded operators on a complex Hilbert space, its strong operator topology and weak operator topology closures equal its double commutant of an operator algebra. To approximate $T\in A^{\prime \prime }$ on finitely many vectors, take their tuple in $H^m$. The projection onto its diagonal-$A$ cyclic subspace has matrix entries in $A^{\prime }$, so it commutes with the diagonal action of $T$. Therefore the image tuple under $T$ lies in that cyclic subspace and can be approximated by a single $a\in A$.