# Power of an infinitesimal generator of a $C_0$-semigroup in a Banach space

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 $${\cal{D}}(A^n):=\{f: f\in {\cal{D}}(A^{n-1})\; \text{and} \; A^{n-1}f \in {\cal{D}}(A) \},$$ $$A^{n}f:=A (A^{n-1} f).$$

• Cross-posted from math.stackexchange.com. Mar 7, 2019 at 7:12
• 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. Mar 7, 2019 at 8:39
• @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. Mar 7, 2019 at 10:47

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 $$p$$.