# Generator of a generated $C_0$ semigroup.

Consider $C_0$-semigroup $S_t:\mathscr{B(H)} \to \mathscr{B(H)}$ with generator $U$. Now define $P_t:\mathscr{B_1(H)} \to \mathscr{B_1(H)}$ where $P_t(\rho)=S_t\rho S_t^*$. How can I prove $P_t$ to be $C_0$-semigroup and the generator of $P_t$ is given by $A$, where $A(\rho)=U\rho+\rho U^*$?

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This doesn't look like a research question. Bratteli and Robinson is my favorite source for the basic theory of semigroups of operators. – Nik Weaver Jan 29 '13 at 11:31
What is ${\mathcal B}_1({\mathcal H})$? – Delio Mugnolo Jan 29 '13 at 16:30
If U is bounded, this is rather obvious. However, if U is unbounded, there is in general an issue of domains. It is easy to come up with example where $U\rho$ is an unbounded operator, but $A(\rho)$ is bounded. Hence the generator of $P_t$ will in general be an extension of $A$. – Michael Renardy Jan 29 '13 at 16:45

Andras, I do not understand. Would this semigroup be in general only weak*-continuous even if $\mathcal H$ is a Hilbert space? I thought this may happen only if the semigroup acts on a non-reflexive Banach space. – Delio Mugnolo Jan 30 '13 at 20:53
$\mathcal B(\mathcal H)$ is in general a non-reflexive Banach space... – András Bátkai Jan 31 '13 at 13:04
Both the semigroup law and strong continuity follow directly from the semigroup law satisfied by $S_t$ and ${S_t}^*$ and by their strong continuity (here it is fundamental that $\mathcal H$ is a Hilbert space, strong continuity of the adjoint semigroup being false in the case of general Banach spaces). Also determining the generator is just an easy exercise in differentiation of a vector-valued function of one real variable (applying the chain rule).