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


    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?

  • $\begingroup$ 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? $\endgroup$
    – LL 3.14
    Oct 10, 2023 at 14:07
  • $\begingroup$ @LL3.14 I changed my initial assumptions. $\endgroup$
    – B.Hueber
    Oct 10, 2023 at 15:13
  • 3
    $\begingroup$ 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. $\endgroup$ Oct 10, 2023 at 15:39
  • 1
    $\begingroup$ @SeverinSchraven The bound should rather be $\frac{\lVert\omega\rVert^2}{a^2}$. $\endgroup$
    – MaoWao
    Oct 11, 2023 at 19:56
  • $\begingroup$ @MaoWao As usual, you are correct. $\endgroup$ Oct 11, 2023 at 21:04

1 Answer 1


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.


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