# Fractional power of operators in $C_0$-semigroup

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.

• 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. Dec 3, 2013 at 13:19
• can you please give me references regarding this theory? Dec 3, 2013 at 19:23
• Why don't you ask your question here then? Dec 19, 2013 at 19:07
• 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. Dec 20, 2013 at 9:52

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.
• 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. Dec 4, 2013 at 9:28
• 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/… Dec 4, 2013 at 10:41
• Alright Sir! but if I want such fraction powers only for a $C_0$semigroup, not a group. then does it makes sense for $r<0$ ?? Dec 4, 2013 at 11:30
• No. Take e.g. $r=-1$, then $Z(t)^{-1}$ must exist and thus your semigroup is indeed a group. Dec 4, 2013 at 12:46