Timeline for Reference request: The resolvent is analytic in the resolvent set
Current License: CC BY-SA 3.0
9 events
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| Apr 1, 2018 at 12:57 | comment | added | Jochen Glueck | @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. | |
| Apr 1, 2018 at 12:46 | comment | added | Jochen Glueck | @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? | |
| Mar 31, 2018 at 15:08 | comment | added | Mark | @JochenGlueck Isn't it true that the resolvent is analytic with respect to all three(weak, strong operator, uniform operator) topologies? | |
| Mar 21, 2018 at 0:39 | comment | added | Jochen Glueck | @DJS: The resolvent $\mathcal{R}(\lambda,A)$ of a closed linear operator $A$ is an injective operator for each $\lambda$ in the resolvent set of $A$ (since it is the inverse of $\lambda - A$). But clearly, the zero operator is not injective (unless the space is $\{0\}$). | |
| Mar 20, 2018 at 20:22 | vote | accept | user860374 | ||
| Mar 20, 2018 at 20:19 | comment | added | user860374 | @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 $T$? | |
| Feb 28, 2018 at 7:02 | comment | added | András Bátkai | Nice answer, Jochen. You make a good teacher. | |
| Feb 28, 2018 at 0:55 | comment | added | paul garrett | 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...) | |
| Feb 28, 2018 at 0:40 | history | answered | Jochen Glueck | CC BY-SA 3.0 |