Coupling constant limits of Schrödinger operators with critical potentials

If (2) holds, then $P_0\ge 0$ in $L^2({\mathbb{R}}^d)$ (see [1]).

Under the assumption on V, we know that $P(\lambda )$ has discrete eigenvalues when λ is large enough, and each discrete eigenvalue tends to zero when λ tends to some ${\lambda }_0$ (see Section 2). We study the asymptotic behaviors of the discrete eigenvalues of $P(\lambda )$ in this paper. The asymptotic behaviors for Schrödinger operators with fast decaying potentials were studied by Klaus and Simon [2]. In [2], they studied the convergence rate of discrete eigenvalues of $H(\lambda )=-\mathrm{\Delta }+\lambda V$ when $\lambda \to {\lambda }_0$. ${\lambda }_0$ is the value at which some discrete eigenvalue $e_i(\lambda )$ tends to zero. The main method they used in their paper is the Birman-Schwinger technique.

In order to use the Birman-Schwinger technique to $P(\lambda )$, we need to get the asymptotic expansion of ${(P_0-\alpha )}^{-1}$ for α near zero, $\alpha <0$, which was studied by Wang [1]. In this paper, we first show that there exists some ${\lambda }_0$ such that when $\lambda >{\lambda }_0$, $P(\lambda )$ has discrete eigenvalues. Then, we define the Birman-Schwinger kernel $K(\alpha )$ for $P(\lambda )$ and find that there is one-to-one correspondence between the discrete eigenvalues of $K(\alpha )$ and the discrete eigenvalues of $P(\lambda )$. Hence, the asymptotic expansion of the discrete eigenvalue of $P(\lambda )$ can be got through the asymptotic expansion of the discrete eigenvalue of $K(\alpha )$. In our main results, we need to use that $K(\alpha )$ is a bounded operator from $L^2$ to $L^2$. To get that, we add a strong condition on V (i.e., ${\rho }_0>6$ in (1)). We show that $K(\alpha )$ is a family of compact operators converging to $K(0)$ and obtain the asymptotic expansions of the discrete eigenvalues of $K(\alpha )$ by functional calculus. After that, the convergence rate of the smallest discrete eigenvalue of $P(\lambda )$ is obtained.

Here is the plan of our work. In Section 2, we recall some results of $P_0$ and define the Birman-Schwinger kernel $K(\alpha )$ for $P(\lambda )$. The relationship between the eigenvalues of these two kinds of operators is studied. In Section 3, we first study the asymptotic behavior of the discrete eigenvalues of $K(\alpha )$. Then the convergence rate of the smallest discrete eigenvalue of $P(\lambda )$ is obtained. We get the leading term and the estimate of the remainder term of the smallest discrete eigenvalue.

Let us introduce some notations first.

Notation The scalar product on $L^2({\mathbb{R}}^{+};r^{d-1}dr)$ and $L^2({\mathbb{R}}^d)$ is denoted by $〈\cdot ,\cdot 〉$ and that on $L^2({\mathbb{S}}^{d-1})$ by $(\cdot ,\cdot )$. $H^{r,s}({\mathbb{R}}^d)$, $r\in \mathbb{Z}$, $s\in \mathbb{R}$, denotes the weighted Sobolev space of order r with volume element ${〈x〉}^{2s}dx$. The duality between $H^{1,s}$ and $H^{-1,-s}$ is identified with the $L^2$ product. Denote $H^{0,s}=L^{2,s}$. Notation $\mathcal{L}(H^{r,s},H^{r^{\mathrm{\prime }},s^{\mathrm{\prime }}})$ stands for the space of continuous linear operators from $H^{r,s}$ to $H^{r^{\mathrm{\prime }},s^{\mathrm{\prime }}}$. The complex plane ℂ is slit along positive real axis so that $z^{\nu }=e^{\nu lnz}$ and $lnz=ln|z|+iargz$ with $0<argz<2\pi$ are holomorphic there.

in $L^2({\mathbb{R}}^d)$ (see [1]).

Here ${\nu }^{\mathrm{\prime }}=\nu -[\nu ]$, and $[\nu ]$ is the largest integer which is not larger than ν. For $\nu >0$, let ${[\nu ]}_{-}$ be the largest integer strictly less than ν. When $\nu =0$, set ${[\nu ]}_{-}=0$. Define ${\delta }_{\nu }$ by ${\delta }_{\nu }=1$, if $\nu \in {\sigma }_{\mathrm{\infty }}\cap \mathbb{N}$, ${\delta }_{\nu }=0$, otherwise. One has $[\nu ]={[\nu ]}_{-}+{\delta }_{\nu }$.

The following is the asymptotic expansion for the resolvent $R_0(z)={(P_0-z)}^{-1}$.

Theorem 2.1 (Theorem 2.2 [1])

First, we show that $P(\lambda )$ has discrete eigenvalues when λ is large enough. In fact, we need only to show that there exists a function $\psi \in L^2({\mathbb{R}}^d)$ such that $〈\psi ,P(\lambda )\psi 〉<0$.

when λ is large enough, one has $〈\psi ,P(\lambda )\psi 〉<0$. This means that $P(\lambda )$ has discrete eigenvalues when λ is large enough.

$P(\lambda )$ has a continuous spectrum $[0,\mathrm{\infty })$ for $\lambda \ge 0$ because ${lim}_{|x|\to \mathrm{\infty }}V(x)$ exists and equals zero (see [3]). We know that $\sigma (P(0))=\sigma (P_0)=[0,\mathrm{\infty })$. Hence, from the continuity of a discrete spectrum of $P(\lambda )$, we know that there exists some ${\lambda }_0$ such that when $\lambda >{\lambda }_0$, $P(\lambda )$ has eigenvalues less than zero, and when $\lambda \le {\lambda }_0$, $\sigma (P(\lambda ))=[0,\mathrm{\infty })$. So, $P(\lambda )$ has an eigenvalue $e_1(\lambda )<0$ at the bottom of its spectrum for $\lambda >{\lambda }_0$. In Section 3 (Proposition 3.1), we prove that $e_1(\lambda )$ is simple and the corresponding eigenfunction can be chosen to be positive everywhere. (There are many results about the simplicity of the smallest eigenvalue of the Schrödinger operator without singularity, but there is no result which can be used directly, because the potential we use in this paper has singularity at zero. Theorem XIII.48 [4] can treat the Schrödinger operator with the potential which has singularity at zero, but the positivity of potential is needed. Hence, we give this result.) From the discussion above and the continuity of a discrete spectrum, one has that $e_1(\lambda )$ tends to zero at some λ. The asymptotic behavior of $e_1(\lambda )$ is studied in this paper.

Then we have the following result.

It follows that ${|V|}^{1/2}$ is injective from A to B.

It follows that ${(P_0-\alpha )}^{-1}{|V|}^{1/2}$ is injective from B to A.

(b) From (a), one has $dimA=dimB$. This means that the multiplicity of α as the eigenvalue of $P(\lambda )$ is exactly the multiplicity of ${\lambda }^{-1}$ as the eigenvalue of $K(\alpha )$. □

From the last proposition, we know that there exists one-to-one correspondence between the discrete eigenvalues of $P(\lambda )$ and the discrete eigenvalues of $K(\alpha )$. Hence, we can study the eigenvalues of $K(\alpha )$ first.

If $P_0$ and V are defined as above, we show that if $P_0+V$ has the eigenvalue less than zero, then the smallest eigenvalue of $P_0+V$ is simple. We use Theorems XIII.44, XIII.45 [4] to prove it.

Proposition 3.1 Suppose $P_0+V$ has an eigenvalue at the bottom of its spectrum. Then this eigenvalue is simple and the corresponding eigenfunction can be chosen to be a positive function.

Proof Let $0\le \chi (t)\le 1$ be a smooth nonincreasing function such that $\chi (t)=1$ if $|t|<1$ and $\chi (t)=0$ if $t>2$. Let ${\chi }_n(t)=\chi (t/n)$. Set $V_n={\chi }_n(r)\frac{q(\theta )}{r^2}+V-〈x〉$, $H_0=-\mathrm{\Delta }+〈x〉$, $H_n=H_0+V_n$, $H=P=P_0+V$. From the proof of Theorem XIII.47 [4], we know that $e^{-t{H}_0}$ is positivity preserving and $\{e^{-t{H}_0}\}\cup L^{\mathrm{\infty }}$ acts irreducibly on $L^2$. Hence, by Theorem XIII.45 [4], if $H_n$ converges to H and $H-V_n$ converges to $H_0$ in the strong resolvent sense, then $e^{-tH}$ is positivity preserving and $\{e^{-tH}\}\cup L^{\mathrm{\infty }}$ acts irreducibly on $L^2$. By Theorems XIII.43 and XIII.44 [4], we can get the result. Since $C_0^{\mathrm{\infty }}({\mathbb{R}}^d)$ is the core for all $P_n$ and P, and for any $\psi \in C_0^{\mathrm{\infty }}({\mathbb{R}}^d)$, $V_n\psi \to (\frac{q(\theta )}{r^2}+V)\psi$ in $L^2$, then we have the necessary strong resolvent convergence by Theorem VIII.25(a) [5]. This ends the proof. □

Proposition 3.2 Assume that $0\notin {\sigma }_{\mathrm{\infty }}$. $P_0{F}_{0}u=u$ in $H^{-1,s}$ for any $u\in H^{-1,s}$, $s>1$.

Proof If $u\in H^{-1,s}$, then $F_{0}u\in H^{1,-s}$. For any test function $\varphi \in C_0^{\mathrm{\infty }}({\mathbb{R}}^n)$, we have $〈P_0{F}_{0}u,\varphi 〉=〈u,F_0{P}_0\varphi 〉$. If $0\notin {\sigma }_{\mathrm{\infty }}$, then we have ${lim}_{z\to 0}{(P_0-z)}^{-1}=F_0$ in $H^{-1,s}$ for $\mathrm{\Im }z>0$. It follows $〈u,F_0{P}_0\varphi 〉={lim}_{z\to 0}〈u,{(P_0-z)}^{-1}{P}_0\varphi 〉={lim}_{z\to 0}〈u,\varphi -z{(P_0-z)}^{-1}\varphi 〉=〈u,\varphi 〉$ because ϕ and $P_0\varphi$ belong to $H^{-1,s}$. Hence, $P_0{F}_{0}u=u$ in $H^{-1,s}$. □

Proposition 3.3 Assume that $0\notin {\sigma }_{\mathrm{\infty }}$. $K(\alpha )$ is a compact operator for $\alpha \le 0$. And $K(\alpha )$ converges to $K(0)$ in operator norm sense.

and if ${\rho }_0>2$, then ${|V|}^{1/2}{R}_0^{(0)}{|V|}^{1/2}=o({|\alpha |}^{ϵ})$ in $L^2({\mathbb{R}}^d)$. Hence, $K(\alpha )\to K(0)$ in operator norm sense as $\alpha \to 0$. This means that $K(0)$ is a compact operator. □

then $|{\mu }_n(A_1)-{\mu }_n(A_2)|\le \parallel A_1-A_2\parallel$.

This ends the proof. □

It follows that ${\mu }_k(\alpha )$ converges to the eigenvalue of $T_0$.

It follows that $\parallel P_{\alpha }^{(k)}{\varphi }_k-{\varphi }_k\parallel =O({|\alpha |}^{ϵ})$. Let ${\varphi }_k(\alpha )=\frac{P_{\alpha }^{(k)}{\varphi }_k}{\parallel P_{\alpha }^{(k)}{\varphi }_k\parallel }$. Then $〈{\varphi }_k(\alpha ),{\varphi }_j(\alpha )〉=0$ for $k\ne j$ because $P_{\alpha }^{(k)}{P}_{\alpha }^{(j)}=0$ if $k\ne j$, and $\parallel {\varphi }_k(\alpha )-{\varphi }_k\parallel \le \parallel P_{\alpha }^{(k)}{\varphi }_k-{\varphi }_k\parallel +\parallel (1-\frac{1}{\parallel P_{\alpha }^{(k)}{\varphi }_k\parallel })P_{\alpha }^{(k)}{\varphi }_k\parallel =O({|\alpha |}^{ϵ})$. This ends the proof. □

Then we have the following.