# 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 '13 at 13:19
• can you please give me references regarding this theory? Dec 3 '13 at 19:23
• Why don't you ask your question here then? Dec 19 '13 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 '13 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 '13 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 '13 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 '13 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 '13 at 12:46