Spectral Inclusion

Question

Let $H$ be a real Hilbert space, $B(H)$ be the set of all bounded linear operators on $H$. An operator $A$ in $B(H)$ is said to be positive if $\langle Ax, x\rangle > 0$ $\forall x \in H,$ where $\langle,\rangle$ is the inner product in $H$. Also $A,B \in B(H)$ is said to commute if $(AB)x = (BA)x$ $\forall x \in H$. Let $\sigma(A)$ denote the spectrum of $A$. The question is whether we have the following relation: Let $A,B \in B(H)$ such that they are positive and they commute. Then, $$\sigma(AB) \subset \sigma(A)\sigma(B).$$

Answer 1

More generally, if $\mathcal{A}$ is a unital and abelian Banach Algebra, then $\sigma(AB)\subset\sigma(A)\sigma(B)$ for all $A,B\in\mathcal{A}$.

To see this, note that in general $$\sigma(X)=\{\phi(X): \phi\in\Omega(\mathcal{A})\},$$ where $\Omega(\mathcal{A})$ denotes the character space of $\mathcal{A}$, i.e. $\phi\in\Omega(\mathcal{A})$ if and only if $\phi:\mathcal{A}\to\mathbb{C}$ is a non-zero $*$-homomorphism. Now $$\sigma(AB)=\{\phi(AB):\phi\in\Omega(\mathcal{A})\}=\{\phi(A)\phi(B):\phi\in\Omega(\mathcal{A})\}\subset\{\phi(A):\phi\}\cdot\{\phi(B):\phi\}=\sigma(A)\sigma(B)$$ Your problem follows by considering $\mathcal{A}$ to be the $C^*$-subalgebra of $B(H)$ generated by $1_H,A,B$. This is commutative since the generators commute and since the spectrum is invariant by passing to unital $C^*$-subalgebras, you get your result.