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I am busy reading through Taylor's paper Spectral Theory of Closed Distributive Operators. On page 192, he defines the resolvent and spectrum of $T$:

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Later on in the paragraph, he then proceeds by saying

When $\rho(T)$ is not empty it is known that $R_\lambda(T)$ is analytic in $\rho(T)$ as a function with values in $[X].$

I am, however, having trouble finding a reference which proves that the resolvent is analytic in the resolvent set. Can anyone please point me to a reference in which this is proven?

I did previously ask this question here on Math.SE, but noticed that the answer provided there assumes that the resolvent is continuous in $\lambda$, which is also a result that I do not have.

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    $\begingroup$ When you prove that the resolvent set is open, you usually show this using Neumann series representation. This implies that the resolvent is locally a power series. Details you find on page 240 in Engel-Nagel: One Parameter Semigroups..., math.uni-tuebingen.de/arbeitsbereiche/agfa/members/rana/… $\endgroup$ – András Bátkai Feb 27 '18 at 23:04
  • $\begingroup$ @AndrásBatkai: Good comment, of course. I included a few more details in an answer since I found it worthwhile to point out that this is actually a property of pseudo-resolvents rather than resolvents. $\endgroup$ – Jochen Glueck Feb 28 '18 at 0:43
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Although András' comment already answers the question, I think it is worthwile to give a few more details explicitely here, in order to point out that the analyticity is in fact a consequence of the resolvent identity only and has nothing to do with the operator whose resolvent we consider:

Let $X$ denote a complex Banach space, let $[X]$ denote the space of all bounded linear operators on $X$ and let $U \subseteq \mathbb{C}$ be non-empty and open.

Definition. A mapping $\mathcal{R}: U \to [X]$ is called a pseudo-resolvent if it fulfils the so-called resolvent identity \begin{align*} \mathcal{R}(\lambda) - \mathcal{R}(\mu) = (\mu - \lambda) \mathcal{R}(\lambda)\mathcal{R}(\mu) \end{align*} for all $\lambda, \mu \in U$.

Note that the resolvent of every closed linear operator on $X$ is a pseudo-resolvent, but there are also pseudo-resolvents which are not the resolvent of a closed linear operator (for instance the constant mapping $\lambda \mapsto 0$).

Proposition. Let $\mathcal{R}: U \to [X]$ be a pseudo-resolvent, let $\lambda,\lambda_0 \in U$ and assume that $|\lambda-\lambda_0| < \|\mathcal{R}(\lambda_0)\|^{-1}$ (where we define $0^{-1} := \infty$). Then \begin{align*} \mathcal{R}(\lambda) = \sum_{n=0}^\infty (\lambda_0 - \lambda)^n \mathcal{R}(\lambda_0)^{n+1}, \end{align*} where the series converges absolutely in $[X]$ (which is endowed with the operator norm). In particular, the mapping $\mathcal{R}: U \to [X]$ is analytic.

Proof. It follows from the resolvent identity that \begin{align*} \mathcal{R}(\lambda) \big[\operatorname{id} - (\lambda_0 - \lambda) \mathcal{R}(\lambda_0)\big] = \mathcal{R}(\lambda_0). \end{align*} The operator norm of $(\lambda - \lambda_0) \mathcal{R}(\lambda_0)$ is strictly smaller than $1$ by assumption, so we conclude from the Neumann series that the operator in square brackets is invertible and that its inverse is given by $\sum_{n=0}^\infty (\lambda_0 - \lambda)^n \mathcal{R}(\lambda_0)^n$. This proves the assertion.

A few references:

For the case where the pseudo-resolvent is actually a resolvent of a closed operator:

  • "K.-J. Engel and R. Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)", Proposition IV.1.3 (as already pointed out by András in the comments)

  • "W. Arendt, C. Batty, M. Hieber, F. Neubrander: Vector-Valued Laplace Transforms and Cauchy Problems (2011)", Corollary B.3 in the appendix

For pseudo-resolvents:

  • "M. Haase: The Functional Calculus for Sectorial Operators (2006)": in the appendix of this book, pseudo-resolvents are treated as resolvents of so-called multi-valued operators.

Personal remark concerning the Math.SE post mentioned in the question: To derive the analyticity of the resolvent as an "immediate" consequence of the resolvent identity without noting that this "proof" requires continuity to be shown first is, at least in my experience, an easily made mistake. I recall falling into this trap myself some time ago.

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  • $\begingroup$ In addition to other possible cognitive-dissonance issues, there is also the pervasive question of whether "complex differentiable operator-valued" implies the comparable corollaries we know for scalar-valued complex-differentiable functions. The answer is "yes", for by-now-standard reasons, going back to A. Grothendieck's early work in the early 1950s. (I and many others also have on-line discussions of such things.) Thus, the potential ambiguities or risked presumptions are harmlessly resolved. (I remember being very uneasy years ago during discussions that presumed everything...) $\endgroup$ – paul garrett Feb 28 '18 at 0:55
  • $\begingroup$ Nice answer, Jochen. You make a good teacher. $\endgroup$ – András Bátkai Feb 28 '18 at 7:02
  • $\begingroup$ @JochenGlueck - Apologies for resurrecting this question, but can you please elaborate for me why $\lambda \mapsto 0$ cannot be the resolvent of a closed linear operator $T$? $\endgroup$ – user860374 Mar 20 '18 at 20:19
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    $\begingroup$ @Mark: Since every (pseudo-)resolvent is analytic with respect to the operator norm topology, it is certainly also analytic with respect to every weaker topology, right? $\endgroup$ – Jochen Glueck Apr 1 '18 at 12:46
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    $\begingroup$ @Mark: By the way, it might be worthwhile to point out in this context that a function $f: \mathbb{C} \supseteq \Omega \to [X]$ which is analytic with respect to the weak or strong operator topology is, under appropriate technical assumptions, automatically analytic with respect to the operator norm topology. This follows, for instance, from Theorem 1.3 in "Arendt, Nikolski: Vector-valued holomorphic functions revisited (2000)". See also [op. cit., Theorem 3.1] for a much deeper result. $\endgroup$ – Jochen Glueck Apr 1 '18 at 12:57

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