Is the range a of a generator of a strongly continuous semigroup contained in the domain of the generator?

Question

Let $X$ be a Banach space and $A:D(A)\rightarrow X$ be a infinitesimal generator of a a $C_0$ semi group $\{S(t)\}_{t\geq 0}$. In this case is it possible that $\operatorname{Range}(A)\subset D(A)$?

I was trying the following:

$$D(A)=\left\{x\in X: \lim_{h\to0^+}\frac{S(h)x-x}{h}\space \text{exists}\right\},$$ as it is a generator of a $C_0$ semigroup, $A$ is densely defined and closed operator. Let $v\in D(A)$ . Now $$\lim_{h\to0^+}\frac{S(h)A(v)-A(v)}{h}= \lim_{h\to0^+}A\left(\frac{S(h)v-v}{h}\right).$$ From here can I conclude that above limit exists?

If so then $A(v)\in D(A)$.

Answer 1

Start with an example of the translation semigroup: $(T(t)f)(x)=f(x+t)$ for $f \in L^{1}[0,\infty)$. The generator $A$ of this semigroup is differentiation. So you do not have $\mathcal{R}(A)\subseteq\mathcal{D}(A)$ because that would force every $f \in \mathcal{D}(A)$ to be infinitely differentiable.

If the generator $A$ is bounded, then you do have $\mathcal{R}(A)\subseteq\mathcal{D}(A)$ because $\mathcal{D}(A)=X$ in that case.

Suppose you know $\mathcal{R}(A)\subseteq\mathcal{D}(A)$. The graph $\mathcal{G}(A)$ is a closed subspace of $X\times X$. So $Y=\mathcal{D}(A)$ is a Banach space when endowed with the norm $\|x\|_{Y}=\|x\|_{X}+\|Ax\|_{X}$. And $A : Y\rightarrow Y$ is bounded because $A$ can be seen to be closed on $Y$ as well. Therefore $A \in \mathcal{L}(Y)$. So there is a sense in which $A$ is bounded in this case: there is a densely embedded Banach space $Y\subset X$ with a stronger norm $\|\cdot\|_{Y}$ such that $A : Y\rightarrow Y$ is bounded.

Answer 2

If an operator is the generator of a $C_0$ semi-group then we will have the following properties (referring to Pazy, "semi-group Theory"):

1) For $z_0\in D(A)$ then $ T(t)z_0 \in D(A), \forall t>0$

2) $\bigcup\limits_{n=1}^{\infty}D(A^n)$ is dense in X

Note that the first property does not guarantee that $Az_0\in D(A)$, However, from the property 2, it can be readily deduced that $T(t)Az_0 \in D(A)$, which is a completely deference case.

Regarding to above answer, it think the question is more general and has nothing to do with translation semi-group. As mentioned above, $R(A)\subset D(A)$ cannot be true for a differential semi-group.