In the case of an infinite-dimensional complex banach space $X$, under what conditions can a quasinilpotent operator $T\in B(X)$ be determined to be nilpotent?
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4$\begingroup$ This is far too vague: it is nilpotent when it is nilpotent. Please try to formulate a more well-defined question, perhaps limiting yourself to certain classes of operators that you are interested in. $\endgroup$– Yemon ChoiCommented Aug 7, 2022 at 22:25
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$\begingroup$ You're right the question lacks some precision: I need to know if there are some "well-known" results giving this implication. Thank you @YemonChoi $\endgroup$– Phd mCommented Aug 7, 2022 at 23:13
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2$\begingroup$ As I am telling you: you seem to be hoping for a magic bullet. Since quasinilpotent operators need not be nilpotent, what kind of conditions are you hoping for? What applications do you have in mind? $\endgroup$– Yemon ChoiCommented Aug 8, 2022 at 0:21
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1$\begingroup$ "In my research I have a quasinilpotent operator with some particular properties ... I have to verify if maybe, under the given properties, it is nilpotent." So if this operator is given to you, can you not try to check directly if it is nilpotent, rather than trying to find some general theorem? $\endgroup$– Yemon ChoiCommented Aug 8, 2022 at 20:48
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1$\begingroup$ OK, let me get the ball rolling. "If T is quasinilpotent and T is idempotent then it is zero, hence it is nilpotent." If this is not satisfactory, could you indicate what kind of answer would be better? $\endgroup$– Yemon ChoiCommented Aug 8, 2022 at 20:49
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2 Answers
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I quasi-nilpotent operator $T \in B(X)$ is nilpotent if and only if $0$ is a pole of its resolvent $(\cdot - T)^{-1}$, which means that there exists a number $M \ge 0$ and an integer $n \ge 1$ such that $$ \|(\lambda-T)^{-1}\| \le \frac{M}{\rvert\lambda\lvert^n} $$ for all $0 \not= \lambda \in \mathbb{C}$ close to $0$. This follows from the Laurent series expansion of the resolvent about the point $0$.
answered Dec 6, 2023 at 11:31
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1$\begingroup$ I am confused by this answer. If $T^2 = 0$ then $(\lambda - T)^{-1} = \lambda^{-1} + \lambda^{-2} T$; I don't think it is uniformly bounded above by $M / |\lambda|^n$. Did you may be want the inequality to go the other way? Or am I misunderstanding something? $\endgroup$ Commented Dec 6, 2023 at 15:06
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$\begingroup$ @WillieWong: Thanks, you're of course right. I wasn't sufficiently careful with the formulation. The estimate needs to be true in a pointed neighbourhood of $0$. $\endgroup$ Commented Dec 6, 2023 at 15:09
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$\begingroup$ Ah, that makes much more sense now. Thanks. $\endgroup$ Commented Dec 6, 2023 at 15:10
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$\begingroup$ thank you for the response @JochenGlueck $\endgroup$– Phd mCommented Dec 8, 2023 at 15:22
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This is almost tautological, but suppose that the quasi-nilpotent operator $T$ has the property that, for some $p\in\mathbb N_+$, the equality $\|T^{np}\|=\|T^p\|^n$ holds for infinitely many $n\in\mathbb N$. Then $0=r(T)=\lim_{n\to0}\|T^{np}\|^{1/np}=\|T^p\|^{1/p},$ so $T^p=0$.
answered Dec 6, 2023 at 10:45
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2$\begingroup$ I'd like to say "quasi-tautoligical" $\endgroup$ Commented Dec 6, 2023 at 18:49
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