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I am reading http://www.springerlink.com/content/l76542r216362714/. 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?

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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: http://books.google.de/books?id=xcYVVSyAOkgC&pg=PA207&lpg=PA207&dq=pseudoresolvent+semigroup&source=bl&ots=qkEyL0hmVG&sig=YOIV5d9z4PzTHUgUMzds8vFKVL4&hl=de&ei=MBxMTq0Cj8myBqb1yMoB&sa=X&oi=book_result&ct=result&resnum=10&ved=0CGIQ6AEwCQ#v=onepage&q=pseudoresolvent%20semigroup&f=false

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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
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