If $X$ is a separable Banach space and $(\epsilon_n)\downarrow 0$, can we find a quasinilpotent, noncompact 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.

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 BátkaiMar 22, 2019 at 16:59

2$\begingroup$ @AndrásBátkai: What is probably a bit more interesting is to find such an operator which is, in addition, not powercompact, and thus in particular not nilpotent. (An example would be the resolvent of a nilpotent vectorvalued shift semigroup.) $\endgroup$– Jochen GlueckMar 22, 2019 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 GlueckMar 22, 2019 at 17:05
1 Answer
On the ArgyrosHaydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.

3$\begingroup$ Just curious: is the property "quasinilpotent is compact" equivalent to ArgyrosHaydon property "every operator is scalar plus compact"? $\endgroup$ Mar 22, 2019 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. $\endgroup$ Mar 22, 2019 at 22:50