Spectral Measures and Borel Functional Calculus — Examples

The space ${\mathrm{\ell }}^2(I;\mathbb{C})$ is complete. If $(x^{(n)})$ is Cauchy in its norm, each coordinate is Cauchy since $\mathrm{\mid }{x}_i^{(n)}-x_i^{(m)}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(n)}-x^{(m)}{\mathrm{\parallel }}_2$, so let $x_i$ be its unique complex limit. A Cauchy sequence is norm bounded, say by $C$. For finite $F$, passage to the limit in the finite sum gives ${\sum }_{i\in F}\mathrm{\mid }{x}_i{\mathrm{\mid }}^2\le C^2$; taking suprema shows $x\in {\mathrm{\ell }}^2(I)$. Given $\epsilon >0$, choose $N$ such that $\mathrm{\parallel }{x}^{(n)}-x^{(m)}{\mathrm{\parallel }}_2<\epsilon$ for $m,n\ge N$. For fixed $n\ge N$, passage to coordinate limits on every finite $F$ gives ${\sum }_{i\in F}\mathrm{\mid }{x}_i^{(n)}-x_i{\mathrm{\mid }}^2\le {\epsilon }^2$; taking suprema gives $\mathrm{\parallel }{x}^{(n)}-x{\mathrm{\parallel }}_2\le \epsilon$. Thus the sequence converges (use half a prescribed tolerance). Unique coordinate limits require no choice. Since $I$ is nonempty and $e_i$ has norm one, this Hilbert space is nonzero.

The formula $Tx=({\lambda }_i{x}_i)$ defines a bounded linear operator with $\mathrm{\parallel }Tx{\mathrm{\parallel }}^2={\sum }_i\mathrm{\mid }{\lambda }_i{\mathrm{\mid }}^2\mathrm{\mid }{x}_i{\mathrm{\mid }}^2\le M^2\mathrm{\parallel }x{\mathrm{\parallel }}^2$, so $\mathrm{\parallel }T\mathrm{\parallel }\le M$, and $\mathrm{\parallel }T\mathrm{\parallel }\ge {\sup }_i\mathrm{\mid }{\lambda }_i\mathrm{\mid }=M$ by testing on the basis vectors, so $\mathrm{\parallel }T\mathrm{\parallel }=M$; the adjoint is $T^{\ast }y=(\overline{{\lambda }_i}{y}_i)$ because $⟨Tx,y⟩={\sum }_i{\lambda }_i{x}_i\overline{y_i}={\sum }_i{x}_i\overline{\stackrel{‾}{{\lambda }_i}{y}_i}$, so $T^{\ast }T=T{T}^{\ast }$ is diagonal with entries $\mathrm{\mid }{\lambda }_i{\mathrm{\mid }}^2$ and $T$ is normal.

The map $E$ takes values in orthogonal projections: for square-summable $x$ the family $({\mathbf{1}}_B({\lambda }_i)x_i)$ is square-summable with $\mathrm{\parallel }E(B)x{\mathrm{\parallel }}^2={\sum }_i{\mathbf{1}}_B({\lambda }_i)\mathrm{\mid }{x}_i{\mathrm{\mid }}^2$, so $\mathrm{\parallel }E(B)\mathrm{\parallel }\le 1$ and $E(B{)}^2=E(B)=E(B{)}^{\ast }$ because the identity holds coordinatewise and the formula is symmetric.

$\sigma (T)=\overline{\{{\lambda }_i\}}$: if $z$ is outside the closure then $\delta :={\inf }_i\mathrm{\mid }z-{\lambda }_i\mathrm{\mid }>0$, the diagonal operator with entries $(z-{\lambda }_i{)}^{-1}$ is bounded with norm at most ${\delta }^{-1}$ and is a two-sided inverse of $zI-T$, so $z\in \rho (T)$; if $z\in \overline{\{{\lambda }_i\}}$ pick a sequence ${\lambda }_{i_k}\to z$, so $\mathrm{\parallel }(T-zI)e_{i_k}\mathrm{\parallel }=\mathrm{\mid }{\lambda }_{i_k}-z\mathrm{\mid }\to 0$ and $T-zI$ is not bounded below, whence $z\in \sigma (T)$.

The projection identities, $E(\mathrm{\varnothing })=0$, $E(\sigma (T))=I$, and $E(B\cap C)=E(B)E(C)$ hold coordinatewise. For disjoint $(B_n{)}_{n\ge 0}$ with union $B$, fix $x$ and $\epsilon >0$ and choose a finite coordinate set $F$ with ${\sum }_{i\mathrm{\notin }F}\mathrm{\mid }{x}_i{\mathrm{\mid }}^2<{\epsilon }^2$, using the small-tail property in [A1]. Choose $N$ so every ${\lambda }_i\in B$ with $i\in F$ belongs to some $B_n$ with $n\le N$ (a finite maximum suffices). Then the difference $E(B)x-{\sum }_{n\le N}E(B_n)x$ vanishes on $F$ and has other coordinates of modulus at most $\mathrm{\mid }{x}_i\mathrm{\mid }$, so its squared norm is less than ${\epsilon }^2$. This proves strong countable additivity. For regularity of $E_x$, choose finite $F$ with squared tail less than $\epsilon$. Given Borel $B$, set $K=\{{\lambda }_i:i\in F,{\lambda }_i\in B\}$ and $U=\sigma (T)\setminus \{{\lambda }_i:i\in F,{\lambda }_i\mathrm{\notin }B\}$. Then $K$ is compact, $U$ is open, $K\subseteq B\subseteq U$, and both $E_x(B\setminus K)$ and $E_x(U\setminus B)$ are at most the tail. Thus every finite positive scalar measure is inner and outer regular; it is locally finite since its total mass is $\mathrm{\parallel }x{\mathrm{\parallel }}^2$. The spectrum is compact Hausdorff by [A3], so this is exactly a regular PVM.

For every $x$, every $i\in I$ and Borel $B$, the scalar measure $E_{x,e_i}$ equals $x_i{\delta }_{{\lambda }_i}$, since $⟨E(B)x,e_i⟩={\mathbf{1}}_B({\lambda }_i)x_i$. Integration against this measure gives $\int fd{E}_{x,e_i}=f({\lambda }_i)x_i$: first for indicators and simple functions, then for bounded Borel functions by uniform simple approximation and finite variation. By the bounded PVM integral's pairing formula, $({\mathrm{\Phi }}_E(f)x{)}_i=f({\lambda }_i)x_i$. This family is square summable since $f$ is bounded. In particular ${\mathrm{\Phi }}_E(z)x=Tx$ coordinatewise.

The regular PVM on $\sigma (T)$ just constructed has ${\mathrm{\Phi }}_E(z)=T$, so uniqueness in the spectral theorem identifies it as the spectral PVM of $T$. Its bounded Borel calculus therefore has $(f(T)x{)}_i=f({\lambda }_i)x_i$ by step 4.1.

The diagonal operator is therefore bounded normal with $\sigma (T)=\overline{\{{\lambda }_i\}}$, its spectral projections act by ${\mathbf{1}}_B$ on the coordinates, and its bounded Borel calculus acts by the scalar values $f({\lambda }_i)$.

$M_m$ is a bounded linear operator with $\mathrm{\parallel }{M}_m\mathrm{\parallel }=\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}$: $\mathrm{\parallel }{M}_{m}h{\mathrm{\parallel }}_2^2=\int \mathrm{\mid }m{\mathrm{\mid }}^2\mathrm{\mid }h{\mathrm{\mid }}^{2}d\mu \le \mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}^2\mathrm{\parallel }h{\mathrm{\parallel }}_2^2$, and if $0\le c<\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}$, the set $A=\{\mathrm{\mid }m\mathrm{\mid }>c\}$ has positive measure, and choosing $B\subseteq A$ of positive finite measure gives $\mathrm{\parallel }{M}_m{\mathbf{1}}_B{\mathrm{\parallel }}_2\ge c\mathrm{\parallel }{\mathbf{1}}_B{\mathrm{\parallel }}_2$, so $\mathrm{\parallel }{M}_m\mathrm{\parallel }\ge c$ and hence $\mathrm{\parallel }{M}_m\mathrm{\parallel }=\mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}$ (when the essential norm is zero, the upper bound already gives equality). Since $\mu (X)>0$, [A4] also supplies a nonzero finite-measure indicator, so this Hilbert space is nonzero.

$M_m$ is normal: the adjoint is $M_{\bar{m}}$ because $⟨M_{m}h,g⟩=\int mh\bar{g}d\mu =\int h\overline{\stackrel{‾}{m}g}d\mu =⟨h,M_{\bar{m}}g⟩$, and $M_m{M}_{\bar{m}}=M_{\mathrm{\mid }m{\mathrm{\mid }}^2}=M_{\bar{m}}{M}_m$.

$\sigma (M_m)=R$: if $z\mathrm{\notin }R$ then $\mu (m^{-1}(B(z,\epsilon )))=0$ for some $\epsilon >0$, so $\mathrm{\mid }m-z\mathrm{\mid }\ge \epsilon$ almost everywhere, $r(x):=1\mathrm{/}(m(x)-z)$ on $\{\mathrm{\mid }m-z\mathrm{\mid }\ge \epsilon \}$ and $r(x):=0$ elsewhere defines a bounded measurable multiplier $M_r$ that is a two-sided inverse on a.e. classes of $M_m-zI$, and $z\in \rho (M_m)$; if $z\in R$ then for each $n$ the set $A_n=\{\mathrm{\mid }m-z\mathrm{\mid }<1\mathrm{/}(n+1)\}$ has positive measure, sigma-finiteness gives $B_n\subseteq A_n$ with $0<\mu (B_n)<\mathrm{\infty }$, and the unit vectors $h_n={\mathbf{1}}_{B_n}\mathrm{/}\mathrm{\parallel }{\mathbf{1}}_{B_n}{\mathrm{\parallel }}_2$ satisfy $\mathrm{\parallel }(M_m-zI)h_n{\mathrm{\parallel }}_2\le \frac{1}{n+1}$, so $M_m-zI$ is not bounded below and $z\in \sigma (M_m)$.

The set $R=\sigma (M_m)$ is nonempty compact by [A6] and steps 1.1–2.1. It is a second-countable LCH space, being a compact subspace of the Euclidean plane (intersections with rational-centre, rational-radius balls give a countable base). Moreover $m\in R$ almost everywhere: for every $z\mathrm{\notin }R$ there is an open ball about $z$ with null preimage; a countable rational-ball base refines all these balls. The union of those base balls having null preimage is exactly $\mathbb{C}\setminus R$, since any point in one has a smaller ball with null preimage. Its preimage is a countable union of null sets.

Define $E(B)=M_{{\mathbf{1}}_{m^{-1}(B)}}$ for Borel $B\subseteq R$; these sets are also Borel in $\mathbb{C}$ since $R$ is closed. Each $E(B)$ is an orthogonal projection by multiplication and the adjoint pairing; $E(\mathrm{\varnothing })=0$, $E(R)=I$ by step 3.1, and $E(B\cap C)=E(B)E(C)$. For disjoint $(B_n{)}_{n\ge 0}$ with union $B$, the squared norm of the additive remainder is the integral of $\mathrm{\mid }{\mathbf{1}}_{m^{-1}(B)}-{\sum }_{n\le N}{\mathbf{1}}_{m^{-1}(B_n)}{\mathrm{\mid }}^2\mathrm{\mid }h{\mathrm{\mid }}^2$. The integrands tend to zero and are bounded by the integrable function $\mathrm{\mid }h{\mathrm{\mid }}^2$, so dominated convergence gives strong countable additivity. Thus $E$ is a PVM. Its positive scalar measures $E_h(B)={\int }_{m^{-1}(B)}\mathrm{\mid }h{\mathrm{\mid }}^{2}d\mu$ have mass $\mathrm{\parallel }h{\mathrm{\parallel }}^2$ and are finite Borel measures on the second-countable LCH space $R$, hence regular by [A5].

For every $h,g\in L^2(\mu )$ the product $h\bar{g}$ is integrable, since $2\mathrm{\mid }h\bar{g}\mathrm{\mid }\le \mathrm{\mid }h{\mathrm{\mid }}^2+\mathrm{\mid }g{\mathrm{\mid }}^2$. The scalar measure $E_{h,g}(B)={\int }_{m^{-1}(B)}h\bar{g}d\mu$ therefore satisfies $\int \phi d{E}_{h,g}=\int (\tilde{\phi }\circ m)h\bar{g}d\mu$ first for indicators and simple functions, using zero extensions. For bounded Borel $f$ on $R$, uniformly approximating by simple functions proves the same equality: the right-side error is bounded by the uniform error times $\int \mathrm{\mid }hg\mathrm{\mid }$, and the left-side error by the uniform error times $\mathrm{\mid }{E}_{h,g}\mathrm{\mid }(R)<\mathrm{\infty }$. No boundedness restriction on $h,g$ or density assertion is needed. In particular for $f(z)=z$, step 3.1 makes $\tilde{f}\circ m=m$ almost everywhere, so the bounded PVM integral satisfies ${\mathrm{\Phi }}_E(z)=M_m$ by equality of all pairings.

By the uniqueness clause of the spectral theorem the regular PVM $E$ on $R=\sigma (M_m)$ with $\int zdE=M_m$ is the spectral PVM of $M_m$; for every bounded Borel $f$ on $R$, let $\tilde{f}$ be its zero extension to $\mathbb{C}$. The same approximation argument gives $⟨f(M_m)h,g⟩=\int fd{E}_{h,g}=\int (\tilde{f}\circ m)h\bar{g}d\mu =⟨M_{\tilde{f}\circ m}h,g⟩$, hence $f(M_m)=M_{\tilde{f}\circ m}$ for all $h,g$; because $m\in R$ almost everywhere, this class is independent of the extension off $R$.

The multiplication operator has spectrum the essential range of $m$, spectral projections given by multiplication by the pulled-back indicators, and Borel calculus $f(M_m)=M_{\tilde{f}\circ m}$, customarily written $M_{f\circ m}$ modulo the null set where $m\mathrm{\notin }R(m)$.

First $C_b(K)$ is Banach in the supremum norm. If $(f_m)$ is Cauchy, then $(f_m(\zeta ))$ converges for every $\zeta$; call the limit $f(\zeta )$. Fixing one sufficiently late index bounds $f$ uniformly, and letting the other index tend pointwise to its limit in the Cauchy estimate gives $\mathrm{\parallel }{f}_m-f{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Thus $f$ is continuous by [A7], proving completeness. Now let $\delta =\mathrm{dist}(\gamma ,K)>0$, positive since the two compact sets are disjoint. For $z,w$ on the circle, $g_z\in C_b(K)$, $\mathrm{\parallel }{g}_z{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\delta }^{-1}$ and $\mathrm{\parallel }{g}_z-g_w{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }z-w\mathrm{\mid }{\delta }^{-2}$, by subtracting the reciprocals pointwise. Thus $t\mapsto g_{\gamma (t)}\gamma \mathrm{\prime }(t)$ is continuous, indeed uniformly continuous, into $C_b(K)$ on $[0,2\pi ]$. Step functions on successively finer equal subdivisions, with endpoint values as coefficients, approximate it uniformly, hence also in integral norm (error at most $2\pi$ times the uniform error); they show strong measurability and Bochner integrability by definition. Applying the bounded map ${\mathrm{\Phi }}_E:C_b(K)\to \mathcal{B}(H)$ also proves continuity and Bochner integrability of the resolvent contour integrand, since ${\mathrm{\Phi }}_E(g_z)=(zI-T{)}^{-1}$.

The scalar Cauchy kernel of the contour is the indicator of the enclosed disc: for every $\zeta \in \mathbb{C}\setminus \gamma ([0,2\pi ])$ one has $c(\zeta ):=\frac{1}{2\pi i}{\oint }_{\gamma }(z-\zeta {)}^{-1}dz=1$ if $\mathrm{\mid }\zeta -\lambda \mathrm{\mid }<r$ and $c(\zeta )=0$ if $\mathrm{\mid }\zeta -\lambda \mathrm{\mid }>r$, by the Cauchy integral formula applied to $f\equiv 1$ in the first case and Cauchy's theorem on the convex disc $D(\lambda ,\mathrm{\mid }\zeta -\lambda \mathrm{\mid })$ in the second.

By step 1.1 and Bochner commutation, $\frac{1}{2\pi i}{\oint }_{\gamma }(zI-T{)}^{-1}dz={\mathrm{\Phi }}_E(\frac{1}{2\pi i}{\int }_0^{2\pi }{g}_{\gamma (t)}\gamma \mathrm{\prime }(t)dt)$. For every $\zeta \in K$, evaluation $u\mapsto u(\zeta )$ is bounded linear on $C_b(K)$ with norm at most one; applying Bochner commutation once more identifies the function inside ${\mathrm{\Phi }}_E$ pointwise with $c{\mathrm{\mid }}_K$. All contour integrals here include the derivative of the parametrization.

Evaluation on the spectrum: the closed disc $\overline{D(\lambda ,r)}$ meets $\sigma (T)$ only at $\lambda$, so for $\zeta \in \sigma (T)$ the value $c(\zeta )$ is $1$ exactly at $\zeta =\lambda$ and $0$ otherwise; hence $c{\mathrm{\mid }}_{\sigma (T)}={\mathbf{1}}_{\{\lambda \}}$ and the contour integral equals ${\mathrm{\Phi }}_E({\mathbf{1}}_{\{\lambda \}})=E(\{\lambda \})$.

To check the defining Riesz cycle conditions, choose $R>r$ with $K\setminus \{\lambda \}$ disjoint from $\overline{D(\lambda ,R)}$: compactness gives such an $R$ if this complement is nonempty, and any $R>r$ works otherwise. Choose $r<R_1<R_2<R$, set $U_1=D(\lambda ,R_1)$ and $U_0=\{z:\mathrm{\mid }z-\lambda \mathrm{\mid }>R_2\}$, and define $\chi =1$ on $U_1$, $\chi =0$ on $U_0$. These are disjoint open neighborhoods of the respective spectral parts. The circle lies in $U_1\setminus K$, has index one at $\lambda$, zero at the other spectral points and zero outside $U_1\cup U_0$, by step 1.2. Thus it is a permitted cycle in the Riesz definition and $\chi =1$ on it. Consequently $P_{\{\lambda \}}=\frac{1}{2\pi i}{\oint }_{\gamma }(zI-T{)}^{-1}dz=E(\{\lambda \})$.

The isolated spectral point is an eigenvalue. Indeed ${\mathbf{1}}_{\{\lambda \}}$ is a nonzero continuous function on $K$ because the singleton is clopen there, so isometry of the continuous calculus gives $\mathrm{\parallel }{\mathbf{1}}_{\{\lambda \}}(T)\mathrm{\parallel }=1$. Agreement of the calculi and step 3.1 identify this operator with $E(\{\lambda \})$, which is therefore nonzero; since its range equals $\ker (T-\lambda I)$, that eigenspace is nonzero.

The spectral projection of the isolated eigenvalue $\lambda$ is therefore exactly the Riesz projection computed from the resolvent $(zI-T{)}^{-1}$ along a positively oriented circle separating $\lambda$ from the rest of the spectrum.

The three continuity identities pass to the calculus: $T={\mathrm{\Phi }}_E(\lambda )={\mathrm{\Phi }}_E(\max (\lambda ,0))-{\mathrm{\Phi }}_E(\max (-\lambda ,0))=T_{+}-T_{-}$ and $\mathrm{\mid }T\mathrm{\mid }={\mathrm{\Phi }}_E(\mathrm{\mid }\lambda \mathrm{\mid })=T_{+}+T_{-}$, by linearity of the Borel calculus applied to the pointwise scalar identities, since the involved functions are continuous on the compact spectrum.

Orthogonality of the parts: $T_{+}{T}_{-}={\mathrm{\Phi }}_E(\max (\lambda ,0)\max (-\lambda ,0))={\mathrm{\Phi }}_E(0)=0$ by multiplicativity.

Sign: $s(\lambda {)}^2={\mathbf{1}}_{\sigma (T)\setminus \{0\}}(\lambda )$, hence $\mathrm{sgn}(T{)}^2={\mathrm{\Phi }}_E({\mathbf{1}}_{\sigma (T)\setminus \{0\}})=E(\sigma (T)\setminus \{0\})=I-E(\{0\})$, and $E(\{0\})$ is the orthogonal projection onto $\ker T$, so $I-E(\{0\})$ is the orthogonal projection onto $(\ker T{)}^{\perp }$.

The absolute value agrees with the earlier definition: $\mathrm{\mid }T\mathrm{\mid }={\mathrm{\Phi }}_E(\mathrm{\mid }\lambda \mathrm{\mid })$ is positive because $\mathrm{\mid }\lambda \mathrm{\mid }\ge 0$, and $\mathrm{\mid }T{\mathrm{\mid }}^2={\mathrm{\Phi }}_E(\mathrm{\mid }\lambda {\mathrm{\mid }}^2)={\mathrm{\Phi }}_E({\lambda }^2)=T^2=T^{\ast }T$; by the uniqueness of the positive square root of $T^{\ast }T$, $\mathrm{\mid }T\mathrm{\mid }=(T^{\ast }T{)}^{1\mathrm{/}2}$.

The operators $T_{\pm }$ are the half-sum and half-difference: from the two identities of step 1.1, $\mathrm{\mid }T\mathrm{\mid }+T=2T_{+}$ and $\mathrm{\mid }T\mathrm{\mid }-T=2T_{-}$, so $T_{\pm }=\frac{1}{2}(\mathrm{\mid }T\mathrm{\mid }\pm T)$.

Therefore $T=T_{+}-T_{-}$, $\mathrm{\mid }T\mathrm{\mid }=T_{+}+T_{-}$, $T_{+}{T}_{-}=0$, $T_{\pm }=\frac{1}{2}(\mathrm{\mid }T\mathrm{\mid }\pm T)$, $\mathrm{sgn}(T{)}^2=I-E(\{0\})$ with $I-E(\{0\})$ the projection onto $(\ker T{)}^{\perp }$, and $\mathrm{\mid }T\mathrm{\mid }$ coincides with $(T^{\ast }T{)}^{1\mathrm{/}2}$.