If $X$ is a separable Banach space and $(\epsilon_n)\downarrow 0$, can we find a quasinilpotent, non-compact operator on $X$ such that $||T^n||^{1/n}<\epsilon_n$ for all $n$? I suspect the answer is positive, but cannot come up with an example.
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2$\begingroup$ What is wrong with a nilpotent shift on $L^p[0,1]$? For example $(Tf)(x):=f(x+\frac{1}{2})$ if $x<\frac{1}{2}$ and $(Tf)(x)=0$ elswhere? $\endgroup$– András BátkaiCommented Mar 22, 2019 at 16:59
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2$\begingroup$ @AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not power-compact, and thus in particular not nilpotent. (An example would be the resolvent of a nil-potent vector-valued shift semigroup.) $\endgroup$– Jochen GlueckCommented Mar 22, 2019 at 17:02
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$\begingroup$ @AndrásBátkai: Oh, I think we misunderstood the question. I guess, Markus asks whether such an operator exists for every separable space $X$ and every sequence $(\epsilon_n)$. $\endgroup$– Jochen GlueckCommented Mar 22, 2019 at 17:05
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1 Answer
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On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
answered Mar 22, 2019 at 17:03
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3$\begingroup$ Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"? $\endgroup$ Commented Mar 22, 2019 at 17:35
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$\begingroup$ Probably not, Fedor. Maybe there is a space with unconditional basis on which every operator is diagonal plus compact. Gowers built a space with unconditional basis on which every operator is diagonal plus strictly singular. $\endgroup$ Commented Mar 22, 2019 at 22:50