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Let $A$ and $B$ be positive commuting bounded operators on a Hilbert space. It can be shown by functional calculus that $AB=A^{1/2}BA^{1/2},$ so that $AB$ is again positive. If $A$ and $B$ are not bounded, it is known to be false. (Can be easily proved by showing the existence of a certain "bad" $*$-representation of $\mathbb{C}[x,y]$.) It would be good to have an explicit example of such operators (e.g. differential operators on the Schwarz space). My question is:

give an example of linear operators $A,B$ on a (infinite-dimensional, complex) unitary space $V$ such that: 1) $\langle A\varphi,\varphi\rangle\geq 0,\ \langle B\varphi,\varphi\rangle\geq 0, \forall\varphi\in V;$
2) $AB\varphi=BA\varphi,\ \forall\varphi\in V;$ 3) $\langle AB\psi,\psi\rangle< 0$ for some $\psi\in V.$

Unfortunately, I even have no link for the existence of such operators.

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I have to confess that I do not completely understand your question. An oprator $A:V\to V$, which is positive, has to be bounded by Hellinger-Toeplitz. If $A$ and $B$ are not everywhere defined, unbounded commuting selfadjoint operators, then the product is still positive, but not necessarily selfadjoint. Is this your question? –  András Bátkai Feb 25 '11 at 19:55
    
Ups, I made a mistake, of course the product will be selfadjoint, even in the unbounded case. Sorry, sorry. So now I am really lost... Maybe I am a bit slow. –  András Bátkai Feb 25 '11 at 20:08
    
Does "unitary space" you mean "inner-product space" (i.e. differs from a Hilbert space by being not complete)? –  Matthew Daws Feb 25 '11 at 20:31
    
yes, I mean inner-product space –  Yurii Savchuk Feb 26 '11 at 20:57
    
Of course, if $V$ is a Hilbert space, then Hellinger-Toeplitz applies –  Yurii Savchuk Feb 26 '11 at 20:58
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