product $ab$ of postive elements $a,b$ is again positive, if $ab=ba$.

Question

Let $A$ be a $C^*$-algebra, $a,b\in A$ positive elements (this means self-adjoint and the spectrum lies in $[0,\infty)$). In general, $ab$ isn't positive, for example consider the matrices $a=\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}, b=\begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix} $. The element $ab$ isn't self-adjoint. But the following seems to be true: "Let $A$ be a $C^*$-algebra, $a,b\in A$ positive elements such that $ab=ba$. Then $ab$ is again positive." Clearly, $ab$ is self-adjoint. But how to prove, that $\sigma(ab)\subseteq [0,\infty)$... Maybe with the continuous functional calculus of $ab$? Regards

Answer 1

If $ab=ba$, then $a^nb=ba^n$ for all $n$, and this implies that $p(a)b=bp(a)$ for all polynomials $p$ with $p(0)$, and thus $f(a)b=bf(a)$ for all continuous functions on $\sigma(a)$ that vanish at $0$.

In particular, $$ ab=a^{1/2}a^{1/2}b=a^{1/2}b\,a^{1/2}\geq0. $$

Answer 2

Another way to look at this: since $a$ and $b$ are self-adjoint and $ab=ba$, it follows that the C*-subalgebra of $A$ which they generate is commutative. So, by spectral theory, $$C^*(a,b) \cong C_0(X)$$ for a locally compact Hausdorff space $X$.

Now, positive elements of $C_0(X)$ are just nonnegative-valued functions, so the desired statement follows from the fact that the product of two nonnegative-valued functions is another nonnegative-valued function.

I prefer Martin Argerami's answer, however, which is more hands on.

Answer 3

Another way: Assume that $a$ and $b$ are two positive elements of a $C^*$-algebra $A$ such that $ab=ba$. Then $(ab)^*=b^*a^*=ba=ab$, and $ab$ is selfadjoint. We also have \begin{equation} (1)~~~\sigma(ab)\subseteq\sigma(a)\sigma(b)\subset[0,\infty), \end{equation} and $ab$ is too positive. Note that the right inclusion in (1) is trivial since both $\sigma(a)$ and $\sigma(b)$ are contained in $[0,\infty)$. We also note that the left inclusion in (1) is always true even if $a$ and $b$ are not assumed to be positive. Indeed, assume that $a$ and $b$ are commuting elements of $A$, and let $B$ be the unital $C^*$-subalgebra generated by $a$ and $b$. Then $B$ is commutative and $$\sigma_A(x)=\sigma_B(x)=\{\chi(x):\chi\in\Delta(B)\}$$ for all $x\in B$, where $\Delta(B)$ is the set of all characters of $B$. In particular, $$\sigma(ab)=\{\chi(ab):\chi\in\Delta(B)\}=\{\chi(a)\chi(b):\chi\in\Delta(B)\}\subseteq\sigma(a)\sigma(b);$$ as claimed.