# Range of the generator of a one parameter semigroup of operators?

From Wikipedia:

If $X$ is a Banach space, a one-parameter semigroup of operators on $X$ is a family of operators indexed on the non-negative real numbers $\{T(t)\} t ∈ [0, ∞)$ such that $$T(0)= I \quad T(s+t)= T(s) \circ T(t), \quad \forall t,s \geq 0.$$ The infinitesimal generator of a one-parameter semigroup $T$ is an operator $A$ defined on a possibly proper subspace of $X$ as follows:

• The domain of $A$ is the set of $x ∈ X$ such that $$h^{-1}\bigg(T(h) x - x\bigg)$$ has a limit as $h$ approaches $0$ from the right.

• The value of $A x$ is the value of the above limit. In other words $A x$ is the right-derivative at 0 of the function $$t \mapsto T(t)x.$$

I was wondering if the generator $A$ of a one-parameter semigroup of operators on $X$ is an operator on its domain $D(A)$ in the sense that $A(D(A)) \subseteq D(A)$?

Thanks and regards!

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You should consider some examples to convince yourself that this very rarely happens. – Martin Mar 2 '13 at 21:03
@Martin: Generally, when one mapping is said to be an operator on a subspace of a vector space or a TVS, does it mean that the mapping's range is also in the subspace, or not necessarily? – Tim Mar 2 '13 at 21:05
No, not at all. In the context of semigroups $A$ is typically an unbounded operator $A \colon D(A) \subseteq X \to X$. No further requirement on the range than $A(D(A)) \subseteq X$. – Martin Mar 2 '13 at 21:19

No, $A(D(A)) \subseteq D(A)$ rarely happens.
One of the simplest examples of a semi-group is the translation semigroup on $C_0(\mathbb{R})$ given by $T_tf(x) = f(x+t)$ for which one has $$D(A) = \{f \in C_0(\mathbb R) \mid f \text{ is differentiable and } f' \in C_0(\mathbb R)\}$$ and $Af = f'$. Clearly, $A$ doesn't map $D(A)$ into itself.