Square root in Banach algebra

Question

Suppose we are given a unital Banach algebra $A$ and an element $a\in A$ such that the spectrum is a subset of the positive reals $\mathbb{R}_{>0}$. Then by a theorem (see for example W. Rudin 10.30) we know that there exists a "square root" of $a$, i.e. an element $b\in A$ such that $b^2 = a$.

Question: does there exist such a $b$ with a spectrum $\sigma(b)$ also contained in the positive reals $\mathbb{R}_{>0}$?

My thoughts:

• From the spectral mapping theorem we know that since the square root is holomorphic in a neighborhood of the spectrum, $\sigma(\sqrt a)=\sqrt{\sigma(a)}$. However, is it trivial that $\sqrt a= \pm b$ as in the complex case?
• Also, if the spectrum of $b$ had to be connected then it would be clear, but I don't think this is true in general.

Answer 1

Question: does there exist such a $b$ with a spectrum $\sigma(b)$ also contained in the positive reals $\mathbb{R}_{>0}$?

Yes, that's an important point that you have a positive square root of a positive element.

You take the principal branch of the square root, defined in $\mathbb{C}\setminus (-\infty,0]$, with positive real part to construct it.