Let $H$ be a $\mathbb R$-Hilbert space and $(H_\lambda)_{\lambda\ge0}$ be a spectral decomposition of $H$ (see below). Now, let $$\mathcal D\left(A_\varphi\right):=\left\{x\in H:\int_0^\infty\varphi(\lambda)\:{\rm d}\langle\pi_\lambda x,x\rangle_H<\infty\right\}$$ and $$\langle A_\varphi x,y\rangle_H:=\color{red}{\varphi(0)\langle\pi_0 x,y\rangle_H}+\int_0^\infty\varphi(\lambda)\:{\rm d}\langle\pi_\lambda x,y\rangle_H\;\;\;\text{for all }x\in\mathcal D\left(A_\varphi\right)\text{ and }y\in H\tag1$$ for Borel measurable $\varphi:[0,\infty)\to\mathbb R$.
Question 1: I can vaguely imagine that it's possible to define a operator by duality, but why exactly does $(1)$ yield a well-defined (symmetric?) linear operator $\left(\mathcal D\left(A_\varphi\right),A_\varphi\right)$? Moreover, sometimes I've seen $A_\varphi$ defined without the red term on the right-hand side of $(1)$. Can anybody explain this to me?
Question 2: Depending on the answer to the first question, I get that $\left(\mathcal D\left(A_\varphi\right),A_\varphi\right)$ is symmetric. Can we show that it is even self-adjoint (for any (possibly unbounded) Borel measurable $\varphi$)?
Now, if I got it correctly, if $(\mathcal D(A),A)$ is a nonnegative self-adjoint operator on $H$, $(H_\lambda)_{\lambda\ge0}$ can be chosen such that $$A_{\operatorname{id}_{[0,\:\infty)}}=A\tag2.$$
Question 3: Is this correct under the given assumptions? Sometimes I've seen that $H$ is assumed to be separable. Moreover, sometimes I've seen that $\bigcup_{\lambda\ge0}H_\lambda=H$ is assumed. Is this always possible?
Definitions:
$(H_\lambda)_{\lambda\ge0}$ is called spectral decomposition of $H$ if
- $H_\lambda$ is a closed subspace of $H$ for all $\lambda\ge0$
- $(H_\lambda)_{\lambda\ge0}$ is nondecreasing and right-continuous, i.e. $$\bigcap_{\mu>\lambda}=H_\lambda\;\;\;\text{for all }\lambda\ge0$$
- $\bigcup_{\lambda\ge0}H_\lambda$ is dense
Let $\pi_\lambda$ denote the orthogonal projection of $H$ onto $H_\lambda$ for $\lambda\ge0$. It can be shown that
- $[0,\infty)\ni\lambda\mapsto\pi_\lambda$ is nondecreasing, i.e. $$\langle\pi_\lambda x,x\rangle_H\le\langle\pi_\mu x,x\rangle_H\;\;\;\text{for all }x\in H,$$ and right-continuous (with respect to the strong operator topology)
So,
- $[0,\infty)\ni\lambda\mapsto\langle\pi_\lambda x,x\rangle_H=\left\|\pi_\lambda x\right\|_H^2$ is bounded (by $\left\|x\right\|_H^2)$, nondecreasing and right-continuous for all $x\in H$
- $[0,\infty)\ni\lambda\mapsto\langle\pi_\lambda x,y\rangle_H=2^{-1}\left(\langle\pi_\lambda(x+y),x+y\rangle_H-\langle\pi_\lambda x,x\rangle_H-\langle\pi_\lambda y,y\rangle_H\right)$ is right-continuous and of bounded variation for all $x,y\in H$