MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading The author appears to use the following fact:

Let $H$ be a Hilbert space. For every $\zeta \in \mathbb{C}\setminus\mathbb{R}$ we have a bounded operator $R(\zeta): H \to H$. We also know that

(1) $R(\zeta)$ has nullity $0$

(2) $R(\zeta)$ has dense range

(3) $R(\zeta)$ satisfies the first resolvent identity $R(\zeta) - R(\zeta') = (\zeta-\zeta')R(\zeta)R(\zeta')$.

Then we claim that there exists a densely defined operator $T: H \to H$ such that $\sigma(T) \subset \mathbb{R}$ and for $\zeta \in \mathbb{C}\setminus\mathbb{R}$ the resolvent at $\zeta$ is $R(\zeta)$.

How is this proved? Alternatively, does anyone know a reference where this is proved?

share|cite|improve this question
up vote 3 down vote accepted

Such operator families $R(\zeta)$ are called pseudoresolvents. The result you are looking for is, for example, proved in Chapter III, Proposition 4.6 of the book "One-parameter semigroups for linear evolution equations" by K. J. Engel and R. Nagel:

Google books link:

share|cite|improve this answer
Thanks, I'm not sure why it didn't occur to me to set $A=R(i)^{−1}+i$... – Yakov Shlapentokh-Rothman Aug 17 '11 at 20:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.