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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 $u\in W^{1,2}(\mathbb{R}) ?$

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    $\begingroup$ These issues can get tricky only in dimension $>1$. In one space dimension, there is a complete theory on how to give such operators domains on which they are self-adjoint. In your case, there is a unique such domain, and it's the fairly obvious attempt $D(S)=\{ u\in L^2: -u''+Vu\in L^2\}$ (with $u''$ interpreted as a distribution initially). $\endgroup$
    – Christian Remling
    Commented Feb 9, 2023 at 17:53
  • $\begingroup$ This differs from your suggestion in that I don't require $u''$, $Vu$ to lie in $L^2$ separately. It could be that the two domains agree though under the assumption you specified (and if they do, it can't be hard to show). $\endgroup$
    – Christian Remling
    Commented Feb 9, 2023 at 17:54
  • $\begingroup$ One more general comment perhaps: the domain I gave makes $S$ self-adjoint if and only if $V$ is in the limit point case at both $\pm\infty$ (look that up perhaps if you're not familiar with the terminology). This will hold for any $V$ bounded below. Continuity is not needed, $V\in L^1_{\textrm{loc}}$ is enough. $\endgroup$
    – Christian Remling
    Commented Feb 9, 2023 at 17:56
  • $\begingroup$ Yes thanks, I'm more or less familiar with the theory so I know that in general D(S) has the following expression that you gave and that we know it is included in $W^{1,2}(\mathbb{R})$. The result I'm trying to show is for general $V$ and for that I would like $D(S)$ to be exactly $\{u\in W^{2,2}(\mathbb{R}), uV\in L^2(\mathbb{R}\}$. I know that in certain cases this is not true but also that in somes cases it's true. What I'd like is some hypothesis to make on $V$ to make this possible. $\endgroup$
    – BlueCharlie
    Commented Feb 9, 2023 at 18:36
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    $\begingroup$ I am quite sure that there is no characterization of such potentials in dimension $d>1$ and I also doubt that it exists for $d=1$. There is another class of potentials for which the result is true, namely the reverse Holder class $B_2$. A reference for that is the paper "P. Auscher, B. Ben Ali: Maximal inequalities and Riesz transform estimates on Lp spaces for Schrodinger operators with nonnegative potentials Annales de L'Institut Fourier, 57 n.6 (2007)", 1975-2013. All these results hold, however, in any dimension $\endgroup$
    – Giorgio Metafune
    Commented Feb 9, 2023 at 19:14

2 Answers 2

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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 $D(-\Delta) \cap D(V)$.

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  • $\begingroup$ Can you please add some detail to your final comment. In particular, what is "the domain" of $-\Delta +V$? (On $L^2$, there are at least two natural ways to obtain the domain $D=\{ u\in H^1: -u''+Vu\in L^2\}$: it's the closure of the operator on $C_0^{\infty}$ and also the only self-adjoint extension of this operator.) $\endgroup$
    – Christian Remling
    Commented Feb 12, 2023 at 18:15
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    $\begingroup$ @ChristianRemling I consider the semigroup $e^{t(\Delta-V)}$ generated by $-\Delta+V$ in $L^2$ which extends to all $L^p$ (here $V \geq 0$) and consider the domain of the generator in $L^1$. In few words I want $u-\Delta u+Vu=f$ uniquely solvable for $u$ in the $L^1$ domain, for every $f \in L^1$. It turns out that both $\Delta u, Vu$ are in $L^1$: just multiply the equation by $u/|u|$, while this is more subtle in $L^2$. $\endgroup$
    – Giorgio Metafune
    Commented Feb 12, 2023 at 18:36
  • $\begingroup$ @GiorgioMetafune Thank you very much for your previous comment and this answer. I read carefully the article of Auscher and Ben Ali as well as Davies' papers, so I see in which case we can say things. The Sturm-Liouville approach seems very interesting though, would you have a reference about this explicit construction of the resolvent ? Thanks! $\endgroup$
    – BlueCharlie
    Commented Feb 14, 2023 at 18:03
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    $\begingroup$ I have in mind elementary arguments on Sturm-Liouville problems (for example in the book by Birkhoff-Rota on ODE). If we call $u_1, u_2$ the solutions of $-u"+Vu=0$ vanishing at $\mp \infty$, then the resolvent is given by the kernel $G(x,s)=u_1(x)u_2(s)$ for $x \leq s$ and $u_1(s)u_2(x)$ for $x \geq s$. Then your question is equivalent to the $L^2$ boundedness of the operator with kernel $V(x)G(x,s)$. The book of Olver on asymptotic methods has results on the behaviour of $u_i$ and this should help. $\endgroup$
    – Giorgio Metafune
    Commented Feb 14, 2023 at 19:49
  • $\begingroup$ Thank you very much! I'm gonna look at it $\endgroup$
    – BlueCharlie
    Commented Feb 16, 2023 at 12:05
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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.

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  • $\begingroup$ As I commented above, I'm trying to show a result for general $V$, though I restrict myself to the $V$'s such that $S$ is essentially self-adjoint. So I guess my question is: when do $-u''+Vu\in L^2(\mathbb{R})$ is equivalent to $u''\in L^2(\mathbb{R})$ and $Vu\in L^2(\mathbb{R})$ ?? What conditions does it require on $V$ ?? $\endgroup$
    – BlueCharlie
    Commented Feb 9, 2023 at 18:52

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