Pvm of a multiplication operator

$M_m$ 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 $m\mathrm{\notin }R(m)$.