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The continuous functional calculus of a bounded normal operator $T$ is a representation of $C(K)$, with $K=\sigma (T)$. The Riesz-Markov-Kakutani representation theorem represents $f\mapsto ⟨f(T)v,v⟩$ by a positive regular measure ${\mu }_v$. The assignment $f\mapsto f(T)v$ is isometric for the $L^2({\mu }_v)$ norm, and extends to a unitary map onto the closed cyclic subspace it generates. Multiplication by $z$ and $\overline{z}$ represent $T$ and $T^{\ast }$. Orthogonal decomposition into cyclic reducing subspaces, allowing an uncountable family, constructs the Borel functional calculus for a normal operator without requiring separability.

A bounded normal operator admits a normal square root even when no continuous scalar square root exists on its Banach algebra spectrum. Choose a bounded Borel branch $s(z)=|z{|}^{1/2}\exp (i\mathrm{Arg}(z)/2)$, with $s(0)=0$. The Borel functional calculus for a normal operator gives $S=s(T)$; multiplicativity gives $S^2=T$ and preservation of the adjoint gives $S^{\ast }S=S{S}^{\ast }=|s{|}^2(T)$. Unlike a positive square root, this square root need not be unique.

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.

The Stone theorem for one-parameter unitary groups states that every strongly continuous unitary representation $U_t$ of $\mathbb{R}$ is uniquely $U_t=e^{itA}$ for a possibly unbounded self-adjoint operator $A$. Its domain isConversely every self-adjoint $A$ gives such a group by the Borel functional calculus for a normal operator.

Prove the forward direction by putting $G=iA$, initially defined as the norm derivative at zero. Its domain is dense: for $f\in C_c^{\infty }(\mathbb{R})$, the vector ${\xi }_f=\int f(t)U_t\xi dt$ has derivative $G{\xi }_f=-\int f^{\prime }(t)U_t\xi dt$. Smooth approximate identities make ${\xi }_f\to \xi$. The group law gives $U_{t}D(G)=D(G)$ and $G{U}_t=U_{t}G$ on that domain. Differentiation of preserved inner products shows that $G$ is skew-symmetric, so $A=-iG$ is symmetric.

The derivative operator is closed. Indeed, if ${\xi }_n\to \xi$ and $G{\xi }_n\to \eta$, integrate the orbit derivative to obtainTaking limits gives the same identity with $\xi ,\eta$; dividing by $t$ at zero proves $\xi \in D(G)$ and $G\xi =\eta$.

Now define bounded Laplace integralsShifting the integration variable in $U_h{R}_{\pm }\xi$ and taking the derivative at zero shows that $R_{\pm }\xi \in D(G)$ andHence $I-G$ and $I+G$ are both surjective. Since $G=iA$, this means $A+iI$ and $A-iI$ are surjective.

Here is why these range conditions establish self-adjointness rather than just symmetry. If $\eta \in D(A^{\ast })$, solve $(A-iI)\xi =(A^{\ast }-iI)\eta$. Then $\eta -\xi \in \ker (A^{\ast }-iI)=\mathrm{Ran}(A+iI{)}^{\perp }=\{0\}$. Thus $D(A^{\ast })\subseteq D(A)$, and symmetry gives equality with $A^{\ast }=A$.

The permitted functional calculus now gives $V_t=e^{itA}$. It is unitary, strongly continuous by dominated convergence against spectral measures, and differentiable with derivative $iA{V}_t\xi$ on $D(A)$. Differentiate $V_{-t}{U}_t\xi$ for $\xi \in D(A)$; the two derivative terms cancel because both groups preserve the domain and commute with their generator. Its value is consequently the constant $\xi$. Density of the domain implies $U_t=V_t$ on all of $H$.

For completeness, the converse and the exact domain follow from the same calculus. If $\xi \in D(A)$, dominated convergence applied to $|(e^{it\lambda }-1)/t|\le |\lambda |$ gives the derivative $iA\xi$. If that norm derivative exists, Fatou's lemma applied to its squared norm gives $\int {\lambda }^{2}d⟨E_A(\lambda )\xi ,\xi ⟩<\infty$, so $\xi \in D(A)$. The derivative characterization also proves uniqueness. Thus

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 $T^{\ast }$.

If the spectrum has one point, the preceding part gives $T=\lambda I$, and the linear span of any nonzero vector is a closed proper invariant subspace because $\dim H>1$.

Otherwise choose distinct points ${\lambda }_1,{\lambda }_2\in K$ and disjoint nonempty relatively open neighborhoods $U,V\subseteq K$. The open-set result gives $P(U)\ne 0$ and $P(V)\ne 0$, whileThus $P(U)\ne I$. Set $L=\mathrm{ran}P(U)$. As the range of an orthogonal projection, $L$ is closed, and the preceding observations give $\{0\}\ne L\ne H$. The Borel functional calculus for a normal operator gives $TP(U)=P(U)T$, soIn fact both $L$ and its orthogonal complement are invariant.

Let $M^{+}(K)$ be the positive linear functionals on $C(K)$: those $\phi$ for which $f\ge 0$ implies $\phi (f)\ge 0$. For real $f$,so positivity gives $|\phi (f)|\le \phi (1_K)\Vert f{\Vert }_{\infty }$; decomposition into real and imaginary parts, or the positive-functional Cauchy--Schwarz inequality, gives the same bound for complex $f$. Hence $\phi$ is continuous andThe Riesz-Markov-Kakutani representation theorem says that there is a unique finite regular positive Borel measure $\mu$ withMore generally, $C(K{)}^{\ast }=M(K)$ is the Banach space of finite regular complex Borel measures with the total-variation norm.

Now let $A\subseteq \mathcal{B}(H)$ be commutative, unital, and C-star, and put $K={\Phi }_A$. The Gelfand transform is an isometric star-isomorphism $A\cong C(K)$. For $\xi ,\eta \in H$, the functionalon $C(K)$ is represented by a regular complex measure ${\mu }_{\xi ,\eta }$. The diagonal measures are positive. Polarization and the Riesz theorem assemble them into a projection-valued measure $E$ characterized byfor Borel sets $S\subseteq K$. The multiplication identities first hold for continuous functions and extend to bounded Borel functions by a monotone-class argument. Thereforedefines a unital star-homomorphism with $\Vert \Psi (f)\Vert \le \Vert f{\Vert }_{\infty }$, andfor every $T\in A$.

For a normal $T\in \mathcal{B}(H)$, apply this construction to the commutative C-star algebra $C^{\ast }(1,T)$. Its character space identifies with $\sigma (T)$, and the Gelfand transform of $T$ is the coordinate function $z(\lambda )=\lambda$. We obtain the Borel functional calculus for a normal operatorIf $\sigma (T)\subseteq \mathbb{T}$, then $\overline{z},z=z\overline{z}=1$ on the spectrum. Since $\Psi$ preserves products and involution,Thus $T$ is a unitary operator.

A projection-valued measure on the Borel sets of $\mathbb{C}$ is a map $E$ into the orthogonal projections on a separable Hilbert space $\mathcal{H}$ such thatand for pairwise disjoint $B_j$,for every $v\in \mathcal{H}$, with convergence in norm.

The spectral theorem for normal operators on a separable Hilbert space states that a bounded normal operator $A$ has a unique projection-valued measure supported on $\mathrm{Sp}(A)$ for whichMore generally, the Borel functional calculus for a normal operator is

For $v,w\in \mathcal{H}$, the scalar spectral measures areIf $A$ is self-adjoint, its spectrum and hence the support of $E$ lie in $\mathbb{R}$. Moreover,so ${\mu }_v$ is a positive measure, andThe paper prints total mass $\Vert v\Vert$; with the standard definition it is $\Vert v{\Vert }^2$, so the unsquared norm is a typographical error.

A closed subspace of a Hilbert space reduces a bounded operator when both the operator and its adjoint leave the subspace invariant. Equivalently, the orthogonal projection onto it commutes with the operator: the adjoint invariance makes its orthogonal complement invariant too. For a star-closed algebra of operators, the closed span of the orbit of any vector is reducing for the whole algebra. This enables an orthogonal direct-sum decomposition into cyclic actions when constructing Borel functional calculus for a normal operator.

For a bounded normal operator $T$ with spectral measure of a normal operator $P$, any projection $P(E)$ commutes with $T$ and $T^{\ast }$ by the Borel functional calculus for a normal operator. Its range is a closed reducing subspace. If the spectrum has two points, choose disjoint nonempty relative open sets around them. Full support of a faithful spectral measure makes both projections nonzero and their product zero, so either range is nonzero and proper, in particular an invariant subspace.

Every strongly continuous unitary group has a unique possibly unbounded self-adjoint operator $A$ as its generator. Its domain consists of the vectors for which the norm limit of $(U_t\xi -\xi )/(it)$ exists as $t\to 0$. Conversely, the Borel functional calculus for a normal operator constructs a strongly continuous unitary group from every self-adjoint $A$. Smooth averaging supplies a dense generator domain; Laplace integrals of $U_t$ and $U_{-t}$ prove that $A\pm i$ are surjective, which establishes self-adjointness.