# Spectral theorem and invertible operators

Let $$\mathcal{H}$$ be a (infinite-dimensional) Hilbert space and $$A:\mathcal{D}(A)\to\mathcal{H}$$ be a densely-defined self-adjoint linear operator. Furthermore, assume that $$A$$ is injective and that its spectrum is strictly-positive, i.e. $$\sigma(A)\subset [a,\infty)$$ for some $$a>0$$. Now, consider the following two constructions:

1. Since $$A$$ is injective, $$A:\mathcal{D}(A)\to\mathrm{ran}(A)$$ is invertible and we get an inverse $$A^{-1}:\mathrm{ran}(A)\to\mathcal{H}$$. For an injective self-adjoint operator, the range is dense, which shows that $$A^{-1}$$ is a densely-defined operator.

2. By the assumption on the spectrum, the function $$\sigma(A)\ni\lambda\mapsto\lambda^{-1}$$ is well-defined and measurable. Hence, I can apply the spectral theorem to define an operator $$A^{-1}$$ via

$$A^{-1}=\int_{\sigma(A)}\,\lambda^{-1}\,\mathrm{d}P(\lambda)$$

with domain $$\mathcal{D}(A^{-1})=\{\omega\in\mathcal{H}\mid \int_{\sigma(A)}\,\lambda^{-2}\,\mathrm{d}\langle P(\lambda)\omega,\omega\rangle_{\mathcal{H}}<\infty\}$$, where $$P$$ denotes the spectral measure corresponding to $$A$$.

Are the two operators $$A^{-1}$$ defined above the same?

Of course, by the properties of the spectral calculus, it is clear that $$A^{-1}$$ constructed as in (2) has similar properties as an inverse, i.e. $$AA^{-1}=A^{-1}A=\mathrm{id}$$ on the right domains. Hence the two operators defined in (1) and (2) agree on their common domain, I guess. But are the two domain the same in general?

• If 2) is true, then the inverse is bounded (by $a^{-1}$) so the domain is the whole space. But I am not sure this follows from your initial assumption? Oct 10, 2023 at 14:07
• @LL3.14 I changed my initial assumptions. Oct 10, 2023 at 15:13
• If $0$ is not in the spectrum, then $A: \mathcal{D}(A)\rightarrow \mathcal{H}$ is a bijection and the domain of the inverse of the first construction is the entire Hilbert space. Using that the spectrum is bounded from below by $a>0$, we get $$\vert\int_{\sigma(A)} \lambda^{-2} d\langle P(\lambda)\omega,\omega\rangle_\mathcal{H}\vert\leq \left(\int_a^\infty \lambda^{-2} d\lambda\right) \Vert \omega\Vert^2<\infty.$$ Thus, also the domain of the second construction is the entire Hilbert space. Oct 10, 2023 at 15:39
• @SeverinSchraven The bound should rather be $\frac{\lVert\omega\rVert^2}{a^2}$. Oct 11, 2023 at 19:56
• @MaoWao As usual, you are correct. Oct 11, 2023 at 21:04

As $$0\notin [a,\infty)$$ and $$\sigma(A) \subseteq [a,\infty)$$ we get $$0\notin \sigma(A)$$. Hence, we have (by definition) that $$A: \mathcal{D}(A) \rightarrow \mathcal{H}$$ is a bijection. Thus, the domain of the first construction is $$\mathcal{H}$$.
On the other hand, we have \begin{align*} \int_{\sigma(A)} \lambda^{-2} d \langle P(\lambda) \omega, \omega \rangle &\leq a^{-2} \int_{a/2}^\infty d \langle P(\lambda)\omega, \omega \rangle = a^{-2} \Vert \omega\Vert^2<\infty. \end{align*} Thus, the domain of your second construction is also $$\mathcal{H}$$ and the two operators coincide.