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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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    $\begingroup$ 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 Bátkai Mar 22 at 16:59
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    $\begingroup$ @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 Glueck Mar 22 at 17:02
  • $\begingroup$ @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 Glueck Mar 22 at 17:05
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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.

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    $\begingroup$ Just curious: is the property "quasinilpotent is compact" equivalent to Argyros-Haydon property "every operator is scalar plus compact"? $\endgroup$ – Fedor Petrov Mar 22 at 17:35
  • $\begingroup$ 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. $\endgroup$ – Bill Johnson Mar 22 at 22:50

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