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Let $T:=-\frac{d^2}{dx^2}$ on $C_c^{\infty}(0,1) \subset L^2(0,1),$ then I know that $T^*$ is defined on $H^2(0,1).$ What happens now if I consider instead $T+V$ where $V \in L^2(0,1)$? Is this implication then still true? Obviously, we have $(T+V)^* \supset T^*+V^*$ where the right hand side is defined on $H^2(0,1).$ Does the converse implication hold in general as well?

I don't know if it helps, but $L^2$ potentials are relatively $0-$bounded w.r.t. the second derivative.

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  • $\begingroup$ Yes, this is true, and it is discussed in many introductory books on these topics. $\endgroup$ – Christian Remling Oct 25 '16 at 20:14

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