Let $A$ be a commutative Hermitian * Banach algebra, that is, a commutative Banach algebra with involution such that every self-adjoint element has real spectrum. It is known that if $x\in A$, then $y := x^\ast x \geq 0$, that is, $y^\ast = y$ and $\sigma(y)\subset[0,\infty)$, see "Bonsall and Duncan, Complete normed algebras, 41.Th 5.
My question is the following: let $y\in A$ be a positive element as before. Can we assume the existence of $x\in A$ such that $y = x^\ast x$? I'm particularly interested in the case $A = L^1 (\mathbb{R})$, the set of integrable functions over the real line, with convolution as multiplication. Note that in this case, $A$ is a $A^\star$-Banach algebra, that is, it is continuously embedded in a $C^\star$-algebra.
It is well known that the answer to this question is affirmative if $A$ is a $C^\star$-algebra, but I haven't been unable to find its answer to this more general case.
Edit: It seems that the answer to this question is positive if we add the condition that $0 \notin \sigma(y)$, i.e. $\sigma(y) \subset (0,\infty)$, see Theorem 11.20 in "W. Rudin, Functional analysis". However, I am still interested in the case $0 \in \sigma(y)$, so any help will be well received.
Best Answer
I think this is an interesting question! I am not sure about the case $A=L^1(\mathbb{R})$, but I think I can give an example with $A=C^1(S^2)$, the $*$-algebra of continuously-differentiable, complex-valued functions on the standard 2-sphere embedded in $\mathbb{R}^3$. The multiplication is pointwise multiplication and the adjoint is pointwise conjugation.
I couldn't find a reference, so you should confirm that $C^1(S^2)$ is a Banach algebra. I am pretty sure that, if $M$ is any compact Riemann manifold, then $C^1(M)$ is a Banach algebra for the norm $\|f\|_1 := \|f\|_\mathrm{sup} + \|df\|_\mathrm{sup}$. The metric is needed to assign a norm to each $\mathbb{R}$-linear functional $df(x) : T_xM \to \mathbb{C}$ so that $\|df\|_\mathrm{sup} := \sup_{x \in M} \|df(x)\|$ is defined.
Here is the main claim:
Proposition: Suppose that $f \in C^1(S^2)$ vanishes at a single point $p \in S^2$ and is strictly positive everywhere else. Further suppose that, in some local coordinate system $(x,y)$ around $p$, we have $f(x,y) = x^2+y^2$. Then, there does not exist any $g \in S^2$ such that $|g|^2=f$.
Why do I think this proposition holds? This basically comes down to the following two claims.
Claim 1: Suppose $g : \mathbb{R}^2 \to \mathbb{C}$ is an $\mathbb{R}$-linear map satisfying $|g(x,y)|^2 = x^2+y^2$ for all $(x,y) \in \mathbb{R}^2$. Then, either $g(x,y) = x+iy$, $g(x,y) = x-iy$ or else it is a multiple of one of those by a complex phase of modulus one.
Claim 2: Suppose $g: \mathbb{R}^2 \to \mathbb{C}$ is a $C^1$ function with $ |g(x,y)|^2 = x^2+y^2$ for all $(x,y) \in \mathbb{R}^2$. Then, if $\gamma:[0,1] \to \mathbb{R}^2 \setminus \{(0,0\}$ is a small loop with winding number $1$, then $g \circ \gamma: [0,1] \to \mathbb{C} \setminus \{0\}$ has winding number $\pm 1$.
Claim 1 is elementary and I guess that the second claim should follow from the first using a linear approximation near $(0,0)$.
Now, suppose $f$ is as in the proposition above and $g \in C^1(S^2)$ has $|g|^2=f$. In particular, $g:S^2 \setminus \{p\} \to \mathbb{C} \setminus \{0\}$. By the second claim, if $\gamma:[0,1] \to S^2 \setminus \{p\}$ is a small loop traveling around the point $p$, then $g \circ \gamma:[0,1] \to \mathbb{C} \setminus \{0\}$ has winding number $\pm 1$. However, this is not possible because $S^2 \setminus \{p\}$ is contractible, so it should be possible to deform $g \circ \gamma$ to a constant loop.