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Let $A$ be the infinitesimal generator of a $C_0$-semigroup of linear operators in a Banach space. Let $n$ be a positive integer, $n\geq2$. Is $A^n$ closed?

Here (setting $A^1$ $:=$ $A$, and denoting the domain of $A$ by $\cal{D}(A)$), the operator $A^n$ has been defined inductively for $n=2,3...,$, by $$ {\cal{D}}(A^n):=\{f: f\in {\cal{D}}(A^{n-1})\; \text{and} \; A^{n-1}f \in {\cal{D}}(A) \}, $$ $$ A^{n}f:=A (A^{n-1} f). $$

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    $\begingroup$ Cross-posted from math.stackexchange.com. $\endgroup$
    – Jochen Glueck
    Commented Mar 7, 2019 at 7:12
  • $\begingroup$ This is more an exercise and there are many arguments. For example you could show that the resolvent set of $A^n$ is not empty. $\endgroup$
    – András Bátkai
    Commented Mar 7, 2019 at 8:39
  • $\begingroup$ @AndrásBatkai: This was my first thought, too; but after all, I think it's a bit more subtle: the resolvent set of $A^n$ can indeed by empty (for instance, if the spectrum of $A$ is the left halfplane). So with this argument you can only show that $(A-\lambda)^n$ is closed for some scalar $\lambda$ - but then it is not obvious why this should imply that $A^n$ is closed. $\endgroup$
    – Jochen Glueck
    Commented Mar 7, 2019 at 10:47

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Here is a more general result:

Theorem. Let $A$ be a closed linear operator with non-empty resolvent set on a complex Banach space. Then $p(A)$ is a closed operator for every complex polynomial $p$.

Reference: Markus Haase, The Functional Calculus for Sectorial Operators (2006), Proposition A.6.2.

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