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It is well known (e.g. Courant, Hilbert - Methods of Mathematical Physics) that solutions of the Sturm-Liouville problem on an interval $J=(a,b)$

\begin{equation} \tag{1} \left(p y' \right)' - qy \; = \; -\lambda w y \end{equation} with the natural boundary condition \begin{equation} \tag{2} p y' \; = \; 0 \quad \text{at} \; a, b \end{equation}

are critical points of the variational optimization problem in $y$

\begin{align} \tag{3} \text{minimize} & \quad \int_a^b p (y')^2 + q y^2 \; dx \\ \text{subject to} & \quad \int_a^b w y^2 \; dx \; = \; 1 \tag{4} \end{align}

Assume at least one end point of the domain is infinite, i.e. the problem is singular, but has a spectrum discrete and bounded below (BD). I think the boundary condition (2) grants that the problem is self-adjoint if applied at regular and LC end points (A. Zettl - Sturm-Liouville Theory, Chapter 10.4). Assume $p$ and $w$ are strictly positive and decay to zero at infinity, i.e. at singular boundary points. Assume that $\tfrac{1}{p}$, $q$, $w$ are in $L_\text{loc}(J, \mathbb{R})$, $y$ in $L^2(J, w)$, $y$ and $py'$ in $AC_\text{loc}$ (locally absolutely continuous). These are standard assumptions, e.g (A. Zettl - Sturm-Liouville Theory, Chapter 10).

Assume $\int_a^b w \; dx = 1$ and that e.g. $b$ is a singular end point with $b = \infty$:

My question: Is there a criterion in terms of the coefficient functions that allows to conclude (5) below?

\begin{equation} \tag{5} \lim_\limits{x \to b} \; w y^2 \; = \; 0 \end{equation}

Or is it always true under the given assumptions? I think it is true for Hermite polynomials, so it is no unreasonable hypothesis.

I am specifically interested in the special case $p = w$ and $q = 0$ and all eigenvalues are $\geq 0$ (c.f. Hermite polynomials). If there can something be said about this special case it would be sufficient for me.

(5) is much like assuming that Barbalat's Lemma can be applied to (4). However, that requires $w y^2$ to be uniformly continuous and I don't see how this follows from the assumptions. I read around in various books but did not find an answer to this question. I also excercised calculus back and forth. I think I can at least show $w y \to 0$ at singular end points, but not (5) so far. I can give details on request.

Note that I posted a related question some months ago here, which lacks any answer or hints as of this writing. I am posting this here with adjusted background in the hope that this community can maybe help.

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  • $\begingroup$ To check if I have understood the question. Do you ask whether (5) holds when $y$ is an eigenfunction subject to the boundary condition (2)? And you assume that $b=\infty$? $\endgroup$
    – Giorgio Metafune
    Commented Aug 21, 2020 at 18:29
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    $\begingroup$ Consider $q=0$, $p=w$ and $b=\infty$ and $u=y \sqrt{u}$. Then (1) is equivalent to $$u''-\frac{u}{2} \left (\frac{p''}{p}-\frac{p'^2}{p^2} \right )=-\lambda u.$$ Now everything depends on the potential $q=p''/p-p'^2/p^2$. If this is bounded from below, then $u$ is in the Sobolev space $H^1$ and then tends to 0 at $\infty$. $\endgroup$
    – Giorgio Metafune
    Commented Aug 22, 2020 at 9:29
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    $\begingroup$ If you refer to the behavior at $\infty$ of $H^1$ functions in $1d$, then you can find it in any book on Sobolev spaces. Concerning the fact that $L^2$ eigenfunctions are in $H^1$, under condition on the potential $q$, this is not so difficult but requires more space (the key word is essential self-adjointness of Schr\"odinger operators; if this is true, the domain of the operator defined through a form coincides with the maximal one, hence if $u, D^2u-qu \in L^2$, then $u \in H^1$. $\endgroup$
    – Giorgio Metafune
    Commented Aug 23, 2020 at 17:35
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    $\begingroup$ At this point I can explain. Assume that $q$ is bounded below, then you define the Schroedinger operator $A$ through a form and $D(A)$ will be contained in $H^1$ and $\lambda -A$ will be invertible from $D(A)$ to $L^2$ for large $\lambda$. You can also define the maximal domain $D_{max}$ as the set of all functions $u\in L^2$ such that $Au \in L^2$ (usually you need some local regularity) and this set is larger than $D(A)$ since you do not require global integrability on $\nabla u$. $\endgroup$
    – Giorgio Metafune
    Commented Aug 27, 2020 at 7:30
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    $\begingroup$ Essential selfadjoiness means that the closure of $A$, initially defined on smooth functions with comapct support, is self adjoint and can be restated by saying that $\lambda -A$ is injective on the maximal domain if $\lambda$ is sufficiently large. Now that $u$ in the maximal domain and let $f=\lambda u-Au$; then you find $v \in D(A)$ such that $\lambda v-Av=f$ and then $w=u-v$ is in the maxiamal domain and $\lambda w-Aw=0$ which gives $w=0$ and $u=v \in D(A)$. $\endgroup$
    – Giorgio Metafune
    Commented Aug 27, 2020 at 7:33

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