# Quasinilpotent , non-compact operators

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.

• 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? Mar 22, 2019 at 16:59
• @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.) Mar 22, 2019 at 17:02
• @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 $(\epsilon_n)$. Mar 22, 2019 at 17:05