Spectral theorem for unbounded self-adjoint operators (PVM form)

By [A3] the operator $C$ is unitary with $\ker (I-C)=\{0\}$, and by [A4] its spectrum lies in $S^1$. Let $F_0$ be the regular spectral PVM of $C$ on $\sigma (C)$ supplied by [A2], and extend it to $S^1$ by $F(B):=F_0(B\cap \sigma (C))$ for Borel $B\subseteq S^1$. Then $F$ is a regular PVM on $S^1$, is carried by $\sigma (C)$, and $C={\int }_{S^1}zdF(z)$.

$F(\{1\})=0$: for a bounded normal operator the spectral projection at a point $\lambda$ is the orthogonal projection onto $\ker (C-\lambda )$. Indeed, if $F(\{\lambda \})x=x$, then $F_x$ is carried by $\{\lambda \}$, so the bounded calculus gives $\mathrm{\parallel }(C-\lambda )x{\mathrm{\parallel }}^2={\int }_{S^1}\mathrm{\mid }z-\lambda {\mathrm{\mid }}^{2}d{F}_x(z)=0.$ Conversely, if $Cx=\lambda x$, the same identity shows that $F_x$ is carried by $\{\lambda \}$; hence $\mathrm{\parallel }(I-F(\{\lambda \}))x{\mathrm{\parallel }}^2=F_x(S^1\setminus \{\lambda \})=0$ and $F(\{\lambda \})x=x$. Thus $\mathrm{ran}F(\{\lambda \})=\ker (C-\lambda )$; with $\lambda =1$ this kernel is $\{0\}$ by [A3], so the projection $F(\{1\})$ is zero.

Let $\psi (\lambda )=(\lambda -i)(\lambda +i{)}^{-1}$. Then $\psi$ is a homeomorphism of $\mathbb{R}$ onto $S^1\setminus \{1\}$, and $E(B):=F(\psi (B))$ is a PVM on the Borel sets of $\mathbb{R}$ with $E(\mathbb{R})=F(S^1\setminus \{1\})=I-F(\{1\})=I$; it is regular because $F$ is and $\psi$ is a homeomorphism, and $C=\int zdF=\int \psi (\lambda )dE(\lambda )$.

Let $A$ be the self-adjoint operator $\int \lambda dE$ with domain $D(A)=\{x:\int {\lambda }^{2}d{E}_x<\mathrm{\infty }\}$, given by [A1]; write $h(\lambda )=(\lambda +i{)}^{-1}$, a bounded Borel function. Then $1-C=\int (1-\psi )dE=\int 2ihdE=2ih(E)$, and $(A+i)h(E)=I$ because $(\lambda +i)h(\lambda )=1$ and $Ah(E)=(\lambda h)(E)$ on the natural domain; hence $h(E)=(A+i{)}^{-1}$ and $(A+i)(1-C)=2iI$, so $A+i=2i(1-C{)}^{-1}$ on $\mathrm{ran}(1-C)=D(A)$.

Using step 3.1 in the formula $T=i(I+C)(I-C{)}^{-1}$ of [A3] gives $T=(1\mathrm{/}2)(I+C)(A+i)$; now $I+C=\int (1+\psi )dE=\int 2\lambda h(\lambda )dE=2\lambda h(\lambda )(E)$, so $T=(\lambda h)(E)(A+i)=({\lambda }^{2}h)(E)+i(\lambda h)(E)=(\lambda )(E)=A$ on $D(A)$, because ${\lambda }^{2}h+i\lambda h=\lambda h(\lambda +i)=\lambda$.

Uniqueness: let $E^{\mathrm{\prime }}$ be a regular PVM on $\mathbb{R}$ with $D(T)=\{x:\int {\lambda }^{2}d{E}_x^{\mathrm{\prime }}<\mathrm{\infty }\}$ and $Tx=\int \lambda d{E}^{\mathrm{\prime }}x$ for $x\in D(T)$, and put $F^{\mathrm{\prime }}(B):=E^{\mathrm{\prime }}({\psi }^{-1}(B\setminus \{1\}))$ for Borel $B\subseteq S^1$. Then $F^{\mathrm{\prime }}$ is a regular PVM on $S^1$, $F^{\mathrm{\prime }}(\{1\})=0$, and $\int zd{F}^{\mathrm{\prime }}=\int \psi d{E}^{\mathrm{\prime }}=I-2ih(E^{\mathrm{\prime }})=I-2i(T+i{)}^{-1}=C$, where $(T+i{)}^{-1}=h(E^{\mathrm{\prime }})$ is proved as in step 3.1 with $E^{\mathrm{\prime }}$ in place of $E$. By the uniqueness clause of [A2] applied to the bounded normal operator $C$, $F^{\mathrm{\prime }}$ is carried by $\sigma (C)$ and agrees there with $F_0$, hence $F^{\mathrm{\prime }}=F$ on $S^1$. Therefore $E^{\mathrm{\prime }}(B)=F^{\mathrm{\prime }}(\psi (B))=F(\psi (B))=E(B)$.

Conversely, if $E$ is a regular PVM on $\mathbb{R}$, then $A=\int \lambda dE$ is self-adjoint by [A1] and $E$ represents $A$; by step 4.2 the representing PVM is unique, so $E$ is the spectral PVM of $A$.