Square root of positive elements in Banach *-algebras

banach-algebrasc-star-algebrasfunctional-analysisreference-request

Let $A$ be a Banach $^*$-algebra, i.e., a complex Banach algebra together with a conjugate linear map $^*:A\to A$ that satisfies $(a^*)^*=a$, $(ab)^*=b^*a^*$ and $||a^*||=||a||$ for all $a,b\in A$. Furthermore, we assume that $A$ has an identity element with $||1||=1$ and we denote the spectrum of an element $a\in A$ by $\sigma(a)$.

I would like to know whether the following claim holds (and, if possible, a reference to a proof):

Claim: Consider a self-adjoint element $a\in A$ with $\sigma(a)\subset [0,\infty)$. Then there exists a unique self-adjoint element $b\in A$ with $\sigma(b)\subset[0,\infty)$ such that $b^2=a$.

Remark: If $A$ would be a $C^*$-algebra, then the statement follows from continuous functional calculus.

Best Answer

Let $A=C^1[0,1]$, with $\|f\|=\|f\|_\infty+\|f'\|_\infty$, and $f^*(t)=\overline{f(t)}$.

You have $\sigma(f)=f([0,1])$.

Take the identity function $f(t)=t$. If $g^2=f$, then $$\tag1 g(t)g(t)=t,\qquad t\in[0,1].$$ Differentiating, $$\tag2 2g(t)g'(t)=1,\qquad t\in[0,1]. $$ From $(1)$ we get that $g(0)=0$. Putting this into $(2)$, we get $0=1$, a contradiction. So such $g$ cannot exist, real valued or not.

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