Let $A$ be a $C^*$-algebra If $a\in A$ is positive, is it true that for any $0<\alpha<\frac{1}{2}$ we have
$$\left(a+\frac{1}{n}1\right)^{\frac{-1}{2}}a^{\frac{1}{2}-\alpha}$$is self adjoint?
A more general question, is it true that if $a$ is positive then $a^\alpha$ and $\alpha>0$ is poistive? Thank you for your help.
By definition, $a$ is positive if it is self-adjoint with spectrum in $[0,+\infty)$. The function $$ f(t)=\left(t+\frac{1}{n}\right)^{-1/2}t^{1/2-\alpha} $$ is continuous on $[0,+\infty)$. By continuous functional calculus, $f(a)$ is well-defined. Since $f$ is real-valued, $f(a)^*=\bar{f}(a^*)=f(a^*)=f(a)$ is self-adjoint. Moreover, the spectrum of $f(a)$ is in $f([0,+\infty))\subseteq [0,+\infty)$. So $f(a)$ is positive.
Yes to your other question for similar reasons: the function $g(t)=t^\alpha$ is continuous on $[0,+\infty)$ and takes values in $[0,+\infty)$. So $g(a)$ is positive for every positive $a$.