Let $B\in B(H)$ be self-adjoint and let $A$ be a densely defined symmetric (and closed if needed) operator such that $A^2$ is densely defined. If $BA^2\subset A^2B$ say, is there a result which gives $BA\subset AB$? We may add positivity to $A$ to avoid trivialities.
Notice that I already have a counterexample when $A^2$ is not densely defined.
Cheers,
Hichem
the iterations of densely define is not necessary to be densely define that is clear on derivative operator , and we need densely define in unbounded operator in order to get adjoint of operator,
Yes, true. But I assumed from the beginning that $A^2$ was densely defined...
So if I understand the question, you want to know if the condition B(A^2) is an extension of (A^2)B implies that BA is an extension of AB. To state a little differently assume A and B are bounded. The this implies if the Lie bracket [A^2,B]=0, then [A,B]=0. I don't believe that is even true in general for finite dimensional vector spaces, i.e., matrices.
Thanks a lot Pr. Truman Prevatt, but I also assumed that $A$ was positive.
We know that a positive symmetric operator has a self adjoint positive extension, the Fredricks extension.
Outline: Assume the space is Hilbert and A is strictly positive (bounded from below). We'll first consider bounded operators for unbounded the same outline should work with more technical details required. Here A^2 is self adjoint (for unbounded use the Friedrichs extension). The the spectrum of A^2 is contained in a subset of [e, +infinity] where e>0. Since the Lie bracket of A^2 and B is zero, then the eigenspaces of A are invariant under B. Hence from the spectral theorem A^2= Integral(x, dEx) where Ex is the resolution of the identity associated with A^2. That implies that BEx=ExB for all x. Using the function calculus A=Integral(sqrt(x)dEx). Hence AB=integral(sqrt(x)dEx)B=integral (sqrt(x)BdEx)=BA. If A has a compact resolvent (as many differential operators do) then the same type of argument would carry over if the domains work out. Of course details need to be filled in.
In the bounded case using the Gelfand transform it is a two line proof. I would expect it will carry over fine in most applications like differential equations. However, not knowing your application I can't say.
I do not have any particular applications involving differential operators. I just wanted to establish a general result to use it to prove another one.
Regards,
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