Positive scaling of a self-adjoint operator: the spectral projection-valued measure transforms by the preimage of the scaling map

[$F$ is a PVM] For Borel $B$ the set $c^{-1}B$ is Borel because $m_c$ is Borel measurable [F8], so $F(B)=E(c^{-1}B)$ is an orthogonal projection; $F(\mathrm{\varnothing })=E(\mathrm{\varnothing })=0$ and $F(\mathbb{R})=E(\mathbb{R})=I$; $c^{-1}(B\cap C)=c^{-1}B\cap c^{-1}C$ gives $F(B\cap C)=F(B)F(C)$; and for pairwise disjoint $B_n$ with union $B$ the sets $c^{-1}{B}_n$ are pairwise disjoint with union $c^{-1}B$, so strong countable additivity of $E$ [F2] gives ${\sum }_{n}F(B_n)x={\sum }_{n}E(c^{-1}{B}_n)x=E(c^{-1}B)x=F(B)x$ for every $x\in H$. Thus $F$ is a PVM on the Borel sets of $\mathbb{R}$.

[$F$ is regular] Fix $x\in H$ and write $\nu :=F_x$, so that $\nu (B)=E_x(c^{-1}B)$. By [F3], $E_x$ is a finite measure with $E_x(\mathbb{R})=\mathrm{\parallel }x{\mathrm{\parallel }}^2$, so $\nu (\mathbb{R})=\mathrm{\parallel }x{\mathrm{\parallel }}^2<+\mathrm{\infty }$. For inner regularity let $B$ be Borel and $\epsilon >0$; by regularity of $E_x$ [F7] there is compact $K\subseteq c^{-1}B$ with $E_x(c^{-1}B\setminus K)<\epsilon$. Then $K^{\mathrm{\prime }}:=m_c(K)$ is compact [F8], $K^{\mathrm{\prime }}\subseteq B$, and by $c^{-1}{K}^{\mathrm{\prime }}=K$ one has $\nu (K^{\mathrm{\prime }})=E_x(K)>E_x(c^{-1}B)-\epsilon =\nu (B)-\epsilon$. For outer regularity apply the same statement to the complement: choose open $V\supseteq c^{-1}B$ with $E_x(V\setminus c^{-1}B)<\epsilon$; then $V^{\mathrm{\prime }}:=m_c(V)$ is open [F8], $V^{\mathrm{\prime }}\supseteq B$, and $\nu (V^{\mathrm{\prime }})-\nu (B)=E_x(V\setminus c^{-1}B)<\epsilon$. Hence every $\nu =F_x$ is a Radon measure that is inner regular on Borel sets, i.e. a regular Borel measure [F7], and $F$ is a regular PVM. This proves (a).

[Bounded functional identity] Let $f$ be bounded Borel. The functional calculi of $E$ and of $F$ exist by [F4], since $F$ is a PVM by step 1.1. For $f={\mathbf{1}}_B$ and all $x,y\in H$, $⟨{\mathrm{\Phi }}_F({\mathbf{1}}_B)x,y⟩=\int {\mathbf{1}}_{B}d{F}_{x,y}=F_{x,y}(B)=⟨F(B)x,y⟩=⟨E(c^{-1}B)x,y⟩=E_{x,y}(c^{-1}B)=\int {\mathbf{1}}_{c^{-1}B}d{E}_{x,y}=\int ({\mathbf{1}}_B\circ m_c)d{E}_{x,y}=⟨{\mathrm{\Phi }}_E({\mathbf{1}}_B\circ m_c)x,y⟩$, so the uniqueness clause of [F4] gives ${\mathrm{\Phi }}_F({\mathbf{1}}_B)={\mathrm{\Phi }}_E({\mathbf{1}}_B\circ m_c)$. Both sides are linear in the bounded measurable function [F4], so the identity holds for every simple function. For bounded Borel $f$ choose simple $s_n$ with $\mathrm{\parallel }f-s_n{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$; then $\mathrm{\parallel }f\circ m_c-s_n\circ m_c{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }f-s_n{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$, and the norm-convergence clause of [F4] applied on both sides gives ${\mathrm{\Phi }}_F(f)=\lim {\mathrm{\Phi }}_F(s_n)=\lim {\mathrm{\Phi }}_E(s_n\circ m_c)={\mathrm{\Phi }}_E(f\circ m_c)$.

[Domain and operator identity] Put $f(\lambda ):=\lambda$, so that $f(E)=A$ and $f(E_F)=\int \lambda dF(\lambda )$ by [F1] and [F5], and put $g:=f\circ m_c=m_c$, so that $g(E)=cA$. For $n\ge 1$ the truncations $f_n:=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le n\}}$ are bounded Borel and step 2.2 with $f_n$ gives ${\mathrm{\Phi }}_F(f_n)={\mathrm{\Phi }}_E(f_n\circ m_c)={\mathrm{\Phi }}_E(g_n)$, where $g_n=f_n\circ m_c=g{\mathbf{1}}_{\{\mathrm{\mid }g\mathrm{\mid }\le n\}}$. For every $x\in H$ the norm identities of [F4] and [F5] give $\int \mathrm{\mid }{f}_n{\mathrm{\mid }}^{2}d{F}_x=\mathrm{\parallel }{\mathrm{\Phi }}_F(f_n)x{\mathrm{\parallel }}^2=\mathrm{\parallel }{\mathrm{\Phi }}_E(g_n)x{\mathrm{\parallel }}^2=\int \mathrm{\mid }{g}_n{\mathrm{\mid }}^{2}d{E}_x$; since $\mathrm{\mid }{f}_n{\mathrm{\mid }}^2\uparrow {\lambda }^2$ and $\mathrm{\mid }{g}_n{\mathrm{\mid }}^2\uparrow c^2{\lambda }^2$ pointwise, monotone convergence [F6] gives $\int {\lambda }^{2}d{F}_x=c^2\int {\lambda }^{2}d{E}_x$ for every $x\in H$. Hence $D(f(E_F))=\{x:\int {\lambda }^{2}d{F}_x<\mathrm{\infty }\}=D(A)$ by [F1] and [F5]. For $x\in D(A)$ the convergence clause of [F5] applied to $g_n\to g$ with $\mathrm{\mid }{g}_n\mathrm{\mid }\le \mathrm{\mid }g\mathrm{\mid }$ gives $f(E_F)x=\lim {\mathrm{\Phi }}_F(f_n)x=\lim {\mathrm{\Phi }}_E(g_n)x=g(E)x$; since $g_n\mathrm{/}c\to f$ pointwise with $\mathrm{\mid }{g}_n\mathrm{/}c\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$, [F5] gives ${\mathrm{\Phi }}_E(g_n\mathrm{/}c)x\to f(E)x=Ax$; linearity [F4] then makes the same limit equal to $c\lim {\mathrm{\Phi }}_E(g_n\mathrm{/}c)x=cAx$. Therefore $f(E_F)=cA$ with domain $D(A)$.

[Identification of the spectral measure] By steps 1.1 and 2.1, $F$ is a regular PVM on the Borel sets of $\mathbb{R}$, and by step 3.1 the operator $\int \lambda dF(\lambda )=f(E_F)$ equals $cA$ with domain $D(A)$. The converse clause of the spectral theorem [F1] applied to the regular PVM $F$ shows that $\int \lambda dF(\lambda )$ is self-adjoint with spectral PVM $F$; by the uniqueness clause of [F1] applied to the self-adjoint operator $cA$, its spectral PVM is $E_{cA}=F$, that is, $E_{cA}(B)=E(c^{-1}B)$ for every Borel $B$. This is (c) for general $B$; at $B=(-\mathrm{\infty },0)$ one has $c^{-1}(-\mathrm{\infty },0)=(-\mathrm{\infty },0)$, so the negative spectral subspaces agree.