Pvm of a diagonal normal operator

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 $f({\lambda }_i)$.