Let $A$ be the infinitesimal generator of a $C_0$ semigroup of linear operators in a Banach space. Let $n$ be a positive integer $n \geq 2$? Is the power operator $A^n$ 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 $$ {\cal{D}}(A^n):=\{f: f\in {\cal{D}}(A^{n-1})\; and \; A^{n-1}f \in {\cal{D}}(A) \}, $$ $$ A^{n}f:=A (A^{n-1} f). $$


1 Answer 1


It is. Choose any element $\lambda \in \rho(A)$ in the resolvent set of $A$ (depending on your sign convention, you've got an entire left or right halfplane in $\mathbb{C}$ at your disposal) and rewrite \begin{equation} A^n = (\lambda - A)^{n} \bigl(A(\lambda-A)^{-1}\bigr)^n. \end{equation} The first factor on the right-hand side is closed as inverse of a bounded operator and the second factor is in fact bounded. The composition, first bounded, afterwards closed, is therefore closed. One may wonder that the above claimed equality holds (pay attention to the domains). Well, the inclusion left side in right side is quickly established. The other way around follows from $x \in \mathcal{D}(A^n) \Leftrightarrow \bigl(A(\lambda-A)^{-1}\bigr)^n x \in \mathcal{D}(A^n)$ which in turn follows from \begin{equation} \bigl(A(\lambda-A)^{-1}\bigr)^n = \bigl(-1+\lambda(\lambda-A)^{-1}\bigr)^n \end{equation} and an application of the binomial theorem. Hope that helps.

Best, Jan

ps: the argument is not restricted to semigroup generators. It works for all all (closed) operators with non-empty resolvent set.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .