Spectral projections and resolution of the identity

$E(B)$ commutes with $T$ and $T^{\ast }$: $E(B)T={\mathrm{\Phi }}_E({\mathbf{1}}_B){\mathrm{\Phi }}_E(z)={\mathrm{\Phi }}_E({\mathbf{1}}_{B}z)=TE(B)$ and $E(B)={\mathrm{\Phi }}_E({\mathbf{1}}_B)={\mathrm{\Phi }}_E(\overline{{\mathbf{1}}_B})={\mathrm{\Phi }}_E({\mathbf{1}}_B{)}^{\ast }$ is self-adjoint, so taking adjoints gives $E(B)T^{\ast }=T^{\ast }E(B)$; hence $\mathrm{ran}E(B)$ and $\ker E(B)$ are invariant under $T$ and $T^{\ast }$.

Eigenvectors of spectral projections: if $x=E(\{\lambda \})x$ then $Tx={\mathrm{\Phi }}_E(z)x={\mathrm{\Phi }}_E(z{\mathbf{1}}_{\{\lambda \}})x={\mathrm{\Phi }}_E(\lambda {\mathbf{1}}_{\{\lambda \}})x=\lambda E(\{\lambda \})x=\lambda x$, so $\mathrm{ran}E(\{\lambda \})\subseteq \ker (T-\lambda I)$.

For self-adjoint $T$ define $F(t):=E(\sigma (T)\cap (-\mathrm{\infty },t])$ for real $t$; for $s\le t$ the set $A_s:=\sigma (T)\cap (-\mathrm{\infty },s]$ is contained in $A_t:=\sigma (T)\cap (-\mathrm{\infty },t]$, so $F(s)=E(A_s)=E(A_s\cap A_t)=F(s)F(t)=F(t)F(s)$, and $F(t)-F(s)=E(A_t\setminus A_s)$ is the image of an indicator and therefore an orthogonal projection: $F$ is increasing in the projection order.

Conversely, suppose $Tx=\lambda x$. For $x=0$ the identity $x=E(\{\lambda \})x$ holds by linearity, with no point evaluation. For $x\mathrm{\ne }0$, the operator $\lambda I-T$ has nonzero kernel and is not invertible, so $\lambda \in \sigma (T)$. Thus evaluation at $\lambda$ and the Dirac measure ${\delta }_{\lambda }$ on $\sigma (T)$ are defined. This Dirac measure is regular: a set containing $\lambda$ contains the compact singleton, and a set omitting it has the open superset $\sigma (T)\setminus \{\lambda \}$ of zero mass. Now for every continuous $f$ one has $\int fd{E}_x=⟨f(T)x,x⟩=f(\lambda )\mathrm{\parallel }x{\mathrm{\parallel }}^2$, so the positive measure $E_x$ and the mass $\mathrm{\parallel }x{\mathrm{\parallel }}^2{\delta }_{\lambda }$ are finite regular measures with equal integrals against all continuous functions and hence are equal; therefore $E_x(\sigma (T)\setminus \{\lambda \})=0$, which gives $\mathrm{\parallel }E(\sigma (T)\setminus \{\lambda \})x{\mathrm{\parallel }}^2=0$ and, since $E(\{\lambda \})+E(\sigma (T)\setminus \{\lambda \})=E(\sigma (T))=I$, the identity $x=E(\{\lambda \})x$; with the preceding step this proves $E(\{\lambda \})H=\ker (T-\lambda I)$.

Strong right continuity: fix $t$ and put $t_n=t+1\mathrm{/}(n+1)$ for $n\ge 0$. The Borel indicators of $\sigma (T)\cap (t,t_n]$ are bounded by one and converge pointwise to zero. Hence $F(t_n)-F(t)$ converges strongly to zero. For $t<s\le t_n$, the norm-square formula for indicators gives $\mathrm{\parallel }(F(s)-F(t))x{\mathrm{\parallel }}^2=E_x(\sigma (T)\cap (t,s])\le E_x(\sigma (T)\cap (t,t_n])=\mathrm{\parallel }(F(t_n)-F(t))x{\mathrm{\parallel }}^2$. This proves the full right-hand strong limit as $s\downarrow t$, for every $x$.

Limits at infinity: since $\sigma (T)\subseteq [-\mathrm{\parallel }T\mathrm{\parallel },\mathrm{\parallel }T\mathrm{\parallel }]$, for $t<-\mathrm{\parallel }T\mathrm{\parallel }$ the set $\sigma (T)\cap (-\mathrm{\infty },t]$ is empty and $F(t)=0$, while for $t\ge \mathrm{\parallel }T\mathrm{\parallel }$ it is all of $\sigma (T)$ and $F(t)=E(\sigma (T))=I$; hence the strong limits at the two infinities are $0$ and $I$.

The spectral projections reduce $T$, the eigenspace at $\lambda$ is exactly $E(\{\lambda \})H$, and for self-adjoint $T$ the family $F(t)=E(\sigma (T)\cap (-\mathrm{\infty },t])$ is increasing, strongly right continuous, with strong limits $0$ and $I$ at the two infinities.