Domain of Schrödinger operators Let $S$ be a Schrödinger operator on $\mathbb{R}$, $Su=-u''+Vu$ with $V\geq1$ continuous and going to $+\infty$ at infinity (you can think of it as $x^2+1$). I wondering which assumptions do I have to make in order to have $\mathcal{D}(S)=\{u\in W^{2,2}(\mathbb{R}), uV\in L^2(\mathbb{R})\}$ ? ($W^{2,2}(\mathbb{R})$ being the set of functions $f$ having $2$ weak derivatives with finite $L^2$ norm).
I just found this article from Davies, which makes the hypothesis $|V'(x)|^2\leq\alpha V(x)^3$ with $\alpha\in(0,2)$.
Is there some more general conditions for this result to hold ?
Edit 1: I know that in general $D(S)=\{u\in W^{1,2}(\mathbb{R}), -u''+Vu\in L^2(\mathbb{R})\}$ but my question is precisely:
For which $V$ does $-u''+Vu\in L^2(\mathbb{R})$ is equivalent to $u''\in L^2(\mathbb{R})$ and $Vu\in L^2(\mathbb{R})$ for all $u\in W^{1,2}(\mathbb{R}) ?$
 A: This is more an extended comment rather an answer (which I don't know), to point out where the difficulties I believe come from. Let us consider the Schroedinger operator with inverse square potential $-D^2+bx^{-2}$ in 1d. The operator is bounded form below if and only if $b \geq -\frac 14$, by Hardy inequality, but the domain is $H^2$ intersected with the domain of the potential (that is $u \in H^2$ such that $Vu \in L^2$) if and only if $b>\frac 34$. Of course this is a singular potential, but one can approximate the singularity and construct a smooth potential closer and closer  to $b(x-x_k)^{-2}$ in neighborhoods of appropriately chosen $x_k$, for which the same phenomenon holds. Davies' condition or the $B_2$ condition I mentioned in a comment, clearly exclude this kind of potentials but then the difficulty is how to characterize them with a property of some function space. It is however possible that in 1d the explicit construction of the resolvent as a Sturm Liouville problem, allows to prove in more generality the boundedness of the operator $V(-D^2+V)^{-1}$. Surprisingly enough, the answer is always positive in $L^1$ and in any dimension: as soon as the potential is positive, the domain of $-\Delta +V$ is always $D(-\Delta) \cap D(V)$.
A: What you take as the domain is to some extent a matter of choice.  You do want to be able to define $Su$, for $u$ in the domain, as a member of $L^2(\mathbb R)$.  The real questions, I think, are whether with a given domain the operator is self-adjoint, or essentially self-adjoint.
