Let $A$ be the infinitesimal generator of a $C_0$ semigroup of linear operators in a Banach space. Let $n$ pe 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 $$ {\cal{D}}(A^n):=\{f: f\in {\cal{D}}(A^{n-1})\; and \; A^{n-1}f \in {\cal{D}}(A) \}, $$ $$ A^{n}f:=A (A^{n-1} f). $$

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

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

Let $A$ be the infinitesimal generator of a $C_0$ semigroup of linear operators in a Banach space. Let $n$ pe 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 $$ {\cal{D}}(A^n):=\{f: f\in {\cal{D}}(A^{n-1})\; and \; A^{n-1}f \in {\cal{D}}(A) \}, $$ $$ A^{n}f:=A (A^{n-1} f). $$