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Given a strongly continuous semigroup $T : \mathbb{R}_+ \to L(B)$ on a Banach space $B$, its infinitesimal generator $A$ of a strongly continuous semigroup $T$ is defined as a mapping $B \to B$ as $$ A\,x = \lim_{t\downarrow0} \frac1t\,(T(t)- I)\,x , \forall x \in B $$ whenever the limit exists wrt the norm on the Banach space $B$. Let $D(A)$ be the set of $x$'s in $B$, where the limit exists.

  1. Can the definition of $A$ be rewritten as the right-derivative of $T: \mathbb{R}_+ \to L(B)$ at $t=0^+$, wrt some norm $\|\cdot\|_{L(D(A))}$on $L(D(A))$, as $$ \lim_{t\downarrow0} \frac1t\,\|T(t)- I - tA\|_{L(D(A))} = 0? $$ What is the norm $\|\cdot\|_{L(D(A))}$on $L(D(A))$ then?

  2. Can the generator "generate" back the one-parameter semigroup of operators? I was wondering why $A$ is called a "generator"? What can the generator generate?

    Can the generator $A$ "generate" back the one-parameter semigroup of operators?

References are also appreciated!

Thanks and regards!

Tim
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1 Answers1

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Ad 1.: I don't think so. You would at least have to include $A$ into the norm, since otherwise the domain of the right-hand side is $\mathbb{R}$. But even so, that derivative with respect to the norm topology should be stronger than the one with respect to the strong operator topology, at least in the infinite-dimensional case.

Edit: Here is an example that the two are not equivalent: Consider $T_t: L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R}),\, T_t f(x) = f(x-t)$. Then $t \mapsto T$ defines a strongly continuous semigroup. The limit of $(T_t f - f)/t$ for $t \to 0$ exists for instance for $C^1$-functions with bounded support, which form a dense subset of $L^2(\mathbb{R})$. Let $Af$ be the limit where it exists. If you let $f_t \neq 0$ be a $C^1_c$-function with support in $[0,t]$ then $$\|T_t f_t - f_t\|_{L^2} = \sqrt{2} \|f_t\|_{L^2},$$ so $\|T_t - I\|_{L(D(A))} \geq \sqrt{2}$. In particular, $T_t$ cannot be differentiable with respect to the norm topology.

Ad 2.: Doesn't the next paragraph in that Wikipedia article answer that question? If and only if $A$ is an infinitesimal generator of a strongly continuous semigroup, the semigroup can be pointwisely recovered as the unique mild solution to the Cauchy problem.

Thomas
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  • Thanks! Regarding 1, (1) what I wrote was not proper. I just tried to write it properly. Please check it. (2)why " that derivative with respect to the norm topology should be stronger than the one with respect to the strong operator topology" – Tim Apr 08 '13 at 17:37
  • (3) what does a derivative stronger than another derivative mean? – Tim Apr 08 '13 at 17:54
  • Now your alternative definition coincides with the derivative with respect to the norm topology. By »stronger« I mean that if the family is differentiable with respect to the norm topology, then it is differentiable with respect to the strong operator topology and the derivatives coincide. I've added a counterexample for the converse. – Thomas Apr 09 '13 at 14:26
  • Thanks, Thomas! (1) In "$$|T_t f_t - f_t|{L^2} = \sqrt{2} |f_t|{L^2},$$ so $|T_t - I|{L(D(A))} \geq \sqrt{2}$", why "$|T_t - I|{L(D(A))} \geq \sqrt{2}$"? Do you assume the norm $|\cdot|{L(D(A))}$ is the operator norm? (2) Since the strong operator topology is the weakest locally convex topology on the set of bounded operators on a Banach space such that the evaluation map is continuous for each vector in the Banach space, should the norm $|\cdot|{L(D(A))}$ I am looking for induce a topology at least stronger than the strong operator topology? – Tim Apr 11 '13 at 15:24
  • Oh, now I see that you are looking for some norm such that the derivatives coincide. Yes, I was thinking of the operator norm. I wouldn't expect a norm you are looking for to exist. Inducing a stronger topology than the strong operator topology would be a first start, but it is not obvious that this is necessary for a norm you are looking for, given the special structure of the nets you would like to be convergent. Why do you want to have a norm anyway? – Thomas Apr 11 '13 at 15:48
  • The reason I want such a norm is because the definition of the generator may be easier to remember. I haven't figured out a "natural" way to understand and memorize it, "natural" in the sense that the statement can be rephrased in a more simpler way such as using some norm on $L(D(A))$ that I am looking for. – Tim Apr 11 '13 at 16:40