Friedrichs extension and powers conjecture

About 6 years old · traced to

Let AA be a closed semi-bounded symmetric operator, let A{\bf A} denote the associated operator, and let AF{\bf A}_F be its Friedrichs self-adjoint extension. For r∈Nr\in\mathbb{N}, compare the 2r2r-th left-definite space of AF{\bf A}_F with the domain of the Friedrichs extension of the rr-th power of A{\bf A}. Friedrichs powers conjecture.

dom⁡((AF)r)=dom⁡((Ar)F).\operatorname{dom}(({\bf A}_F)^r)=\operatorname{dom}(({\bf A}^r)_F).

The equality holds in the Jacobi differential-operator case, and every computed case mentioned in the source supports it, but its status in the stated generality is unclear.

References

Primary source

Dale Frymark and Constanze Liaw, “Perspectives on General Left-Definite Theory”, arXiv:2012.01014 (2020).

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