Let $P$ be a positive, self-adjoint (unbounded) operator in a Hilbert space $H$ with $0\notin \sigma(P)$. Consider its spectral decomposition
$$P = \int\_{\sigma(P)} t dE(t).$$
Since $0 \notin \sigma(P)$, we can define (the functional calculus)
$$P^{-}= \int\_{\sigma(P)} t^{-1}dE(t).$$
This is an unbounded operator with domain
$$\left\{\xi \in H: \int\_{\sigma(P)} t^{-2}dE\_{\xi, \xi}(t) < \infty\right\}.$$


Is there a relation between this operator and the inverse operator $P(T)\to H: P(\xi)\mapsto \xi$? Are they equal? For example, it is not even clear to me that they have the same domains.



Unbounded positive self-adjoint without $0$ in its spectrum: can we construct its inverse using functional calculus?
$\newcommand\si\sigma\newcommand\D{\mathscr D}$Yes, $P^-=P^{-1}$.


Indeed, the resolvent set of $P$ is open and $0$ is in this set. So, $\si(P)$ is bounded away from $0$, and hence the domain $\D(P^-)$ of $P^-$ is the entire space $H$ -- because, by Theorem 13.24 (a) in [1],
$$\D(P^-)=\D\_g:=\Big\{x\in H\colon \int\_{\si(P)}g^2\,dE\_{x,x}<\infty\Big\},$$
where $g(t):=1/t$. So, by Theorem 13.24 (b) in [1] with $g$ as above and $f(t):=t$, and in view of Definition 12.17 in [1] of a resolution of the identity,
$$PP^-=\Psi(f)\Psi(g)=\Psi(fg)=\Psi(1)=I\_H,$$
where $I\_H$ is the identity operator on $H$ and $\Psi(h):=\int\_{\si(P)}h\,dE$. (Note the obvious typo in Theorem 13.24 (b) in [1], which has $\D\_{fg}\in\D\_g$ instead of $\D\_{fg}\subseteq\D\_g$.)


Thus, $P^-=P^{-1}$, as claimed.


[1] Functional Analysis: Rudin, Second Edition


