Discrete spectrum for product of operators

Question

Let $A$ and $B$ be two positive-definite self-ajoint operators on a Hilbert space $H$. If the spectrum of both $A$ and $B$ is discrete, can we affirm that the spectrum of $AB$ will be also discrete ? In finite dimension, it is true, but what happen in infinite dimension?

Answer 1

There is very little control on what the spectrum could be.

Consider first $$ A_0=\begin{bmatrix} 1&0\\0&2\end{bmatrix},\qquad B_t=\begin{bmatrix} \cos t&\sin t\\ \sin t&-\cos t\end{bmatrix}\begin{bmatrix} 1&0\\0&2\end{bmatrix}\begin{bmatrix} \cos t&\sin t\\ \sin t&-\cos t\end{bmatrix}. $$ Since $B_t$ is unitarily conjugate to $A$, the eigenvalues of $B_t$ are also $1$ and $2$. Both are positive definite.

Meanwhile, $A_0B_t$ has eigenvalues $$\tag1 \lambda_{\pm}(t)=\frac14\,\Big(9+\cos2t\pm\sqrt{17+18\cos2t+\cos^22t\vphantom{\vspace{1cm}}}\Big). $$ We have $$\lambda_+([0,\pi/2])=[2,4],\qquad \lambda_-([0,\pi/2])=[1,2].$$ Now let $\{q_n\}$ be an enumeration of $\mathbb Q\cap[0,2\pi]$, and put $$ A=\bigoplus_n A_0,\qquad B=\bigoplus_n B_{q_n}. $$ Both $A$ and $B$ are positive, and both have spectrum $\{1,2\}$. Meanwhile, the spectrum of $$ AB=\bigoplus_n A_0B_{q_n} $$ has all the numbers $\lambda_+(q_n)$ and $\lambda_-(q_n)$, which are dense in $[1,4]$. As the spectrum is compact (and these operators are block diagonal), $$ \sigma(AB)=[1,4]. $$