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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

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

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  • $\begingroup$ Sorry to bother after such a long time, but from what characterisation of positivity does it follow that $a^{1/2}ba^{1/2} \geq 0$? $\endgroup$ – pabk Dec 17 '19 at 21:45
  • $\begingroup$ If $x\geq0$, you have $x=z^*z$ for some $z$. Then $y^*xy=y^*z^*zy=(zy)^*zy\geq0$. $\endgroup$ – Martin Argerami Dec 17 '19 at 22:16
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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.

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