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Let $X$ be Banach space, and $\{Z(t)\}_{t\geq 0}\subseteq B(X)$ be the $C_0$-semigroup of operators defined on $X$. Moreover, let $A$ be the infinitesimal generator of $\{Z(t)\}_{t\geq 0}$. A fractional power of any closed linear operator $F$ is defined, when $(-\infty,0)\subset \rho(F)$(the resolvent set.) and the set $\{\lambda(\lambda-F)^{-1}:0<\lambda<\infty\}$ is bounded. It is noted that any such conditions do not imply that $F$ generates a semigroup.

My problem is, that I have to consider the fractional powers of operators in $\{Z(t)\}_{t\geq 0}$. i.e. $[Z(t)]^r$ , for $r\in \mathbb{R}$ and $t>0$. Then when is it possible? If I consider, only those semigroups whose generator satisfies above conditions and its fractional powers are defined. As we know $Z(t)=e^{tA}$, then is it possible to take its fractional powers? because integer powers are very well defined through Banach algebra. The problem is just with its fraction powers.

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  • $\begingroup$ You could use spectral calculus to get the fractional power of the bounded operator $Z(t)$ and then use spectral mapping theorems (that are theorems that relate the spectrum of the generator with that of the semigroup operators) to obtain information on the generator. $\endgroup$
    – Uwe Stroinski
    Dec 3 '13 at 13:19
  • $\begingroup$ can you please give me references regarding this theory? $\endgroup$
    – Shinning Star
    Dec 3 '13 at 19:23
  • $\begingroup$ Why don't you ask your question here then? $\endgroup$
    – András Bátkai
    Dec 19 '13 at 19:07
  • $\begingroup$ I thought this is little different and if ask in a more clear way, it may help others like me too! as these questions also appear in search engines (e.g. google). so I asked it in a more precise manner to gain attention of other experts so that I get sure of it. If this is not correct, I'll take care of it next time. $\endgroup$
    – Shinning Star
    Dec 20 '13 at 9:52
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Observe that for $r>0$ the semigroup law implies $Z(t)^r = Z(r t)$. If $Z(t)$ is a $C_0$-group, then this is true for all $0\not=r\in\mathbb{R}$. The generator of $Z(t)^r$ is $(r A, D(A))$. You can find this construction sometimes called 'rescaled semigroup' e.g. in the book of Engel, Nagel, One-parameter semigroups for linear evolution equations, Springer 2000, p. 43 and p. 60. The book also contains a comprehensive treatment of fractional powers of generators pp. 137.

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  • $\begingroup$ Sir by semigroup law $Z(t)^r=Z(rt)$ when $r\in N$. but how is it possible when $r\in R$ ? Can you please guide me? I am little confused. $\endgroup$
    – Shinning Star
    Dec 4 '13 at 9:28
  • $\begingroup$ Since $Z(t+s)=Z(t)Z(s)$ for all $0\leq t,s\in\mathbb{R}$ you can e.g. take $Z(1/2 t)Z(1/2 t) = Z(t)$ and thus you find $Z(t)^{1/2}=Z(1/2 t)$. The inverse of $Z(t)$ is given as $Z(-t)$. You might really want to have a look into the book of Engel and Nagel. As I have learned from a comment of Andr'as B'atkai in another question, it can be found here fa.uni-tuebingen.de/research/publications/1999/… $\endgroup$
    – Uwe Stroinski
    Dec 4 '13 at 10:41
  • $\begingroup$ Alright Sir! but if I want such fraction powers only for a $C_0$semigroup, not a group. then does it makes sense for $r<0$ ?? $\endgroup$
    – Shinning Star
    Dec 4 '13 at 11:30
  • $\begingroup$ No. Take e.g. $r=-1$, then $Z(t)^{-1}$ must exist and thus your semigroup is indeed a group. $\endgroup$
    – Uwe Stroinski
    Dec 4 '13 at 12:46

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