Resolvent Operators

The resolvent replaces an unbounded operator by a family of bounded inverse operators. It solves the forced equation $(A-z)\psi=f$, measures proximity to the spectrum, and provides a controlled language for perturbations. This page uses the convention $R_A(z)=(A-zI)^{-1}$; using $(zI-A)^{-1}$ changes several signs.

Helpful background. Spectra of Operators defines the resolvent set. Bounded Operators develops the Neumann series.

A bounded inverse with an operator-domain range

For closed densely defined $A$ and $z\in\rho(A)$,

$$R_A(z):\mathcal H\to D(A)$$

is bounded when regarded as an operator into $\mathcal H$. Its range is exactly $D(A)$, and

$$AR_A(z)=I+zR_A(z).$$

Thus it is also bounded into the graph-norm space $D(A)$. The inverse does more than produce a Hilbert-space vector: it produces one with enough regularity for applying $A$.

This observation avoids a domain error in algebraic manipulations. The equation $(A-z)R_A(z)=I$ holds on all of $\mathcal H$, whereas $R_A(z)(A-z)=I$ holds on $D(A)$. The identity symbols have different input domains.

An unbounded $A$ can therefore be reconstructed from a single resolvent value: its domain is the resolvent range, and its action there is $A=z+R_A(z)^{-1}$.

Resolvent identity and analyticity

For $z,w\in\rho(A)$, subtraction of the inverse equations gives

$$R_A(z)-R_A(w) =(z-w)R_A(z)R_A(w).$$

The two resolvents commute. One can verify the formula first on vectors produced by a resolvent, where all unbounded actions exist, and then use the bounded resulting expressions on the whole Hilbert space.

Near a fixed $z\in\rho(A)$,

$$A-w=\bigl[I-(w-z)R_A(z)\bigr](A-z)$$

on $D(A)$. If $|w-z|\|R_A(z)\|<1$, the bracket is invertible by a Neumann series, giving

$$R_A(w)=R_A(z)\sum_{n=0}^{\infty} \bigl[(w-z)R_A(z)\bigr]^n.$$

This proves that the resolvent set is open and the resolvent is operator-norm analytic there. Differentiating yields $R_A'(z)=R_A(z)^2$ in the chosen sign convention.

Self-adjoint norm and adjoint relations

For self-adjoint $A$,

$$R_A(z)^*=R_A(\overline z), \qquad \|R_A(z)\|=\frac{1}{\operatorname{dist}(z,\sigma(A))}.$$

The spectral calculus proves the norm formula by treating the resolvent as multiplication by $(\lambda-z)^{-1}$. In particular,

$$\|R_A(z)\|\leq\frac{1}{|\operatorname{Im}z|}$$

off the real axis. The inequality follows already from the elementary shifted norm estimate and self-adjoint surjectivity; equality with inverse spectral distance uses the spectral support.

For position multiplication,

$$(R_Q(z)f)(x)=\frac{f(x)}{x-z}.$$

Its norm is $|\operatorname{Im}z|^{-1}$ because the position spectrum fills the real line. The factor $x/(x-z)$ is bounded, so the output belongs to $D(Q)$, explicitly confirming the range regularity.

A limit $z\to\lambda+i0$ with $\lambda$ inside the spectrum generally does not exist in the operator norm on $\mathcal H$. Scattering boundary values need weighted spaces, weak limits, or distributional interpretations, as developed in Resolvent Methods.

Bounded perturbations and a convergent expansion

Let $V$ be bounded and let $z\in\rho(A)$. On $D(A)$,

$$A+V-z=\bigl[I+VR_A(z)\bigr](A-z).$$

If $\|VR_A(z)\|<1$, then

$$R_{A+V}(z) =R_A(z)\sum_{n=0}^{\infty}(-VR_A(z))^n.$$

This is a genuine norm-convergent expansion under the stated bound. Outside that regime the same series may be only formal; the resolvent can still exist even though this sufficient convergence condition fails.

Whenever both resolvents exist, the second resolvent identity is

$$R_{A+V}(z)-R_A(z) =-R_{A+V}(z)V R_A(z).$$

For unbounded $V$, neither the product nor the common-domain factorization is automatic. Relative operator or form bounds supply suitable replacements, but must be specified.

Resolvent convergence is a precise approximation claim

For self-adjoint operators, norm-resolvent convergence means convergence of $R_{A_n}(z)$ to $R_A(z)$ in operator norm at one nonreal $z$; standard resolvent arguments propagate the conclusion to other nonreal points. Strong-resolvent convergence asks only for convergence after application to each fixed vector.

The distinction parallels the bounded projection example in Bounded Operators. A finite-basis computation of several eigenvalues is not itself proof of either mode of convergence. Resolvent convergence states how the entire forced equation behaves under approximation.

Exercises

Derive the second resolvent identity without multiplying two unbounded operators.

Solution

Write $R_{A+V}(z)[(A-z)-(A+V-z)]R_A(z)$. The rightmost resolvent maps into the common domain, so the middle actions are defined. The first term is $R_{A+V}(z)$ and the second is $R_A(z)$. The bracket equals $-V$, giving the identity with bounded products only.

For a self-adjoint $A$ with $\sigma(A)\subseteq[0,\infty)$ and bounded self-adjoint $V$, find a sufficient condition ensuring a negative real $z$ remains outside the perturbed spectrum.

Solution

For $z<0$, the spectral distance is at least $|z|$, so $\|R_A(z)\|\leq1/|z|$. If $|z|>\|V\|$, the Neumann criterion applies. Thus $z<-\|V\|$ lies in $\rho(A+V)$, consistent with the lower bound $A+V\geq-\|V\|I$.

References

• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, doi:10.1007/978-94-007-4753-1 — spectra and perturbations of closed operators.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014, author’s book page — resolvent identities, convergence, and scattering applications.