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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

1 Answer 1

up vote 2 down vote accepted

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.

There are lots of good examples of semigroups with explicit discussion of their infinitesimal generators in the first two chapters of the very readable and detailed Short course on Operator Semigroups by Engel and Nagel.

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