# One-parameter semigroup of operators of a $C^*$-algebra applied to positive self-adjoint element

Let $$X$$ denote a complex $$C^*$$-algebra and $$\{Z(t)\}_{t\geq 0}$$ is a $$C_0$$-semigroup of operators on $$X$$.

Let $$x\in X$$ satisfy have $$x=x^*$$ (x is self-adjoint), such that its spectrum satisfies $$\sigma(x)\subset [0,\infty)$$.

Then under what conditions does it follow that $$[Z^*(t)]=[Z(t)]$$, and $$\sigma(Z(t)x)\subset [0,\infty)$$?

• What does it mean for an element of a Banach space to be self-adjoint? – Michael Renardy Dec 16 '13 at 8:49
• let me re-phrase it. $X$ is a complex $C^*$-algebra. – Shinning Star Dec 16 '13 at 9:02
• What is $Z^\ast(f)$? And what is $f$? – András Bátkai Dec 16 '13 at 9:31
• Sorry sir it was typing error. $Z^*(t)$ is adjoint operator of $Z(t)$. – Shinning Star Dec 16 '13 at 9:44
• How do you define $Z^\ast (t)x$? $x\in X$ and not from $X^\ast$... – András Bátkai Dec 16 '13 at 9:58

Apart from the fact that I do not understand some parts of your question, self-adjoint elements with positive spectrum define a positive cone in your $$C^\ast$$ algebra. Positivity-preserving semigroups in $$C^\ast$$ and von Neumann algebras were extensively studies, you should consult the chapter written by Ulrich Groh in the book
• yes it actually defines a positive cone in $C^*$-algebra. I want isotonic $Z(t)$. i.e. whenever $x\geq 0$, $Z(t)x\geq 0$. and the order relation is defined above. – Shinning Star Dec 16 '13 at 18:02