1. Introduction

In this paper we consider

$H : = -\frac{d^2}{dx^2}-x^2+V(x)$

(1)

in $L^p(\mathbb{R})$, where $V\in C(\mathbb{R})$ is a real-valued and satisfies $V(x)\geq -a(1+x^{2})$ for some constant $a\geq 0$ and

$\int_{\mathbb{R}}\frac{|V(x)|}{\sqrt{1+x^2}}\, dx<\infty.$

(2)

The operator (1) describes the quantum particle affected by a strong repulsive force from the origin. In fact, in the classical sense the corresponding Hamiltonian (functional) is given by $\hat{H}(x, p) = p^2-x^2$ and then the particle satisfying $\dot{x} = \partial _p\hat{H}$ and $\dot{p} = -\partial _x\hat{H}$ goes away much faster than that for the free Hamiltonian $\hat{H}_0(x, p) = p^2$.

In the case where $p = 2$, the essential selfadjointness of $H$, endowed with the domain $C_0^\infty(\Omega)$, has been discussed by Ikebe and Kato ^[7]. After that several properties of $H$ is found out in a mount of subsequent papers (for studies of scattering theory e.g., Bony et al. ^[2], Nicoleau ^[10] and also Ishida ^[8]).

In contrast, if $p$ is different from $2$, then the situation becomes complicated. Actually, papers which deals with the properties of $H$ is quite few because of absence of good properties like symmetricity. In the $L^p$-framework, it is quite useful to consider the accretivity and sectoriality of the second-order differential operators. In fact, the case $-\frac{d^2}{dx^2}+V(x)$ with a nonnegative potential $V$ is formally sectorial in $L^p$, and therefore one can find many literature even $N$-dimensional case (e.g., Kato ^[9], Goldstein ^[6], Tanabe ^[14], Engel-Nagel ^[5]). However, it seems quite difficult to describe such a kind of non-accretive operators in a certain unified theory in the literature.

The present paper is in a primary position to make a contribution for theory of non-accretive operators in $L^p$ as mentioned above. The aim of this paper is to give a spectral properties of $H = -\frac{d^2}{dx^2}-x^2+V(x)$ for the case where $V(x)$ can be regarded as a perturbation of the leading part $-\frac{d^2}{dx^2}-x^2$; note that if $V(x) = [\log (e+|x|)]^{-\alpha}$ $(\alpha\in\mathbb{R})$, then $\alpha < 1$ is admissible, which is same threshold as in the short range potential for $-\frac{d^2}{dx^2}-x^2$ stated in Bony ^[2] and also Ishida ^[8].

Here we define the minimal realization $H_{p, \min}$ of $H$ in $L^p = L^p(\mathbb{R})$ as

$\begin{cases} D(H_{p, \min}) : = C_0^\infty(\mathbb{R}), \\[3pt] H_{p, \min}u(x) : = -u''(x)-x^2u(x)+V(x)u(x). \end{cases}$

(3)

Theorem 1.1. For every $1 < p < \infty$, $H_{p, \min}$ is closable and the spectrum of the closure $H_p$ is explicitly given as

$\sigma(H_p) = \left\{ \lambda\in\mathbb{C} \;;\; |{\rm Im}\, \lambda|\leq \left|1-\frac{2}{p}\right| \right\}.$

Moreover, for every $1 < p < q < \infty$, one has consistence of the resolvent operators:

$(\lambda+H_p)^{-1}f = (\lambda+H_q)^{-1}f\ a.e.\ on\ \mathbb{R} \quad \forall\, \lambda \in \rho(H_p)\cap \rho(H_q), \quad \forall\, f\in L^p\cap L^q.$

Remark 1.1. If $p = 2$, then our assertion is nothing new. The crucial part is the case $p\neq 2$ which is the case where the symmetricity of $H$ breaks down. The similar consideration for $-\frac{d^2}{dx^2}+V$ (but in $L^2$-setting) can be found in Dollard-Friedman ^[4].

This paper is organized follows: In Section 2, we prepare two preliminary results. In Section 3, we consider the fundamental systems of $\lambda u+Hu = 0$, and estimate the behavior of their solutions. By virtue of that estimates, we will describe the resolvent set of $H_p$ in Section 4. In section 5, we prove never to be generated $C_0$-semigroups by $\pm iH_p$ under the condition $V = 0$.

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2. Preliminary results

First we state well-known results for the essentially selfadjointness of Schrödinger operators in $L^2$ which is firstly described in ^[7]. We would like to refer also Okazawa ^[12].

Theorem 2.1 (Okazawa [12, Corollary 6.11]). Let $V(x)$ be locally in $L^{2}(\mathbb{R})$ and assume that $V(x)\geq -c_1-c_2|x|^{2}$, where $c_1, c_2\geq 0$ are constants. Then $H_{2, \min}$ is essentially selfadjoint.

Next we note the asymptotic behavior of solutions to second-order linear ordinary differential equations of the form

$y''(x) = (\Phi(x)+\Psi(x))y(x)$

in which the term $\Psi(x)y(x)$ can be treated as a perturbation of the leading part $\Phi(x)y(x)$.

Theorem 2.2 (Olver [13, Theorem 6.2.2 (p.196)]). In a given finite or infinite interval $(a_1, a_2)$, let $a\in (a_1, a_2)$, $\Psi(x)$ a positive, real, twice continuously differentiable function, $\Psi(x)$ a continuous real or complex function, and

$F(x) = \int \left\{\frac{1}{\Phi(x)^{1/4}}\frac{d^2}{dx^2} \left(\frac{1}{\Phi(x)^{1/4}}\right)-\frac{\Psi(x)}{\Phi(x)^{1/2}}\right\}\, dx.$

Then in this interval the differential equation

$\frac{d^2w}{dx^2} = \{\Phi(x)+\Psi(x)\}w$

has twice continuously differential solutions

$\begin{align*} w_1(x) = &\frac{1}{\Phi(x)^{1/4}} \exp\left\{ i\int \Phi(x)^{1/2}\, dx \right\}(1+\varepsilon _1(x)), \\ w_2(x) = &\frac{1}{\Phi(x)^{1/4}} \exp\left\{-i\int \Phi(x)^{1/2}\, dx\right\}(1+\varepsilon _2(x)), \end{align*}$

such that

$|\varepsilon _{j}(x)|, \ \frac{1}{\Phi(x)^{1/2}}|\varepsilon _{j}(x)| \leq \exp\left\{\frac{1}{2}\mathcal{V}_{a_j, x}(F)\right\}-1\quad(j = 1, \, 2)$

provided that $\mathcal{V}_{a_j, x}(F) < \infty$ (where $\mathcal{V}_{a_j, x}(F) = \int |F'(t)|\, dt$ is the total variation of $F$). If $\Psi(x)$ is real, then the solutions $w_1(x)$ and $w_2(x)$ are complex conjugates.

For the above theorem, see also Beals-Wong [1,10.12,p.355].

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3. Fundamental systems of $\lambda u-u''-x^2u+Vu=0$

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3.1. The case $\lambda\in\mathbb{R}$

We consider the behavior of solutions to

$\lambda u(x) - u''(x)-x^{2}u(x)+V(x)u(x) = 0, \quad x \in \mathbb{R},$

(4)

where $\lambda\in \mathbb{R}$.

Proposition 3.1. There exist solutions $u_{\lambda, 1}, u_{\lambda, 2}$ of (4) such that $u_{\lambda, 1}$ and $u_{\lambda, 2}$ are linearly independent and satisfy

$\begin{align*} |u_{\lambda, 1}(x)| \leq C_{\lambda}(1+|x|)^{-\frac{1}{2}}, \quad |u_{\lambda, 2}(x)| \leq C_{\lambda}(1+|x|)^{-\frac{1}{2}} \quad \forall x\in\mathbb{R}, \\ |u_{\lambda, 1}(x)| \geq \frac{1}{2}(1+|x|)^{-\frac{1}{2}}, \quad |u_{\lambda, 2}(x)| \geq \frac{1}{2}(1+|x|)^{-\frac{1}{2}} \quad \forall x\geq R_\lambda \end{align*}$

for some constants $C_\lambda, R_\lambda>0$ independent of $x$. In particular, $u_{\lambda, 1}$, $u_{\lambda, 2}\in L^p(\mathbb{R})$ if and only if $2 < p < \infty$.

Proof. First we consider (4) for $x>0$. Using the Liouville transform

$v(y): = (2y)^{\frac{1}{4}}u\left((2y)^{\frac{1}{2}}\right), \qquad \text{or equivalently}, \qquad u(x) = x^{-\frac{1}{2}}v\left(\frac{x^2}{2}\right),$

we have

$(\lambda -x^{2})x^{-\frac{1}{2}}v\left(\frac{x^2}{2}\right) = u''(x)-V(x)u(x) = x^{\frac{3}{2}}v''\left(\frac{x^2}{2}\right) + \frac{3}{4}x^{-\frac{5}{2}}v\left(\frac{x^2}{2}\right) - x^{-\frac{1}{2}}V(x)v\left(\frac{x^2}{2}\right).$

Therefore noting that $y = x^2/2$, we see that

$v''(y) = \left[ -\left(1-\frac{\lambda}{4y}\right)^2+\frac{\lambda^2-3}{16y^2} +\frac{V((2y)^{\frac{1}{2}})}{2y} \right]v(y) = (\Phi(y)+\Psi(y))v(y).$

(5)

Here we have put for $y>0$,

$\Phi(y) : = -\left(1-\frac{\lambda}{4y}\right)^{2}, \quad \Psi(y) : = \frac{\lambda^2-3}{16y^2}+\frac{V((2y)^{\frac{1}{2}})}{2y}.$

Let

$\Pi(y) : = |\Phi(y)|^{-\frac{1}{4}} \left(-\frac{d^2}{dx^2}+\Psi(y)\right) |\Phi(y)|^{-\frac{1}{4}}, \quad y\geq \lambda_+: = \max \{\lambda, 0\}.$

Then we see that for every $y\geq \lambda_+$,

$\left|\Pi(y)\right| \leq \left(1-\frac{\lambda}{4y}\right)^{-3}\frac{3\lambda^2}{64y^2} +\left(1-\frac{\lambda}{4y}\right)^{-2}\frac{\lambda}{4y^3} + \left(1-\frac{\lambda}{4y}\right)^{-1}\frac{|\lambda^2-3|}{16y^2} + \frac{|V((2y)^{\frac{1}{2}})|}{2y} \leq \frac{M_{\lambda}}{y^2} +\frac{|V((2y)^{\frac{1}{2}})|}{2y},$

where $M_\lambda$ is a positive constant depending only on $\lambda$. Therefore

$\int_{\lambda_+}^{\infty} \left|\Pi(y)\right| \, dy \leq M_{\lambda}\int_{\lambda_+}^{\infty} \frac{1}{y^2} \, dy + \int_{\sqrt{2\lambda_+}}^{\infty} \frac{|V(x)|}{x} \, dx<\infty.$

Thus $\Pi\in L^1((\lambda_+, \infty))$. By Theorem 2.2, we obtain that there exists a fundamental system $(v_{\lambda, 1}, v_{\lambda, 2})$ of (5) such that

$v_{\lambda, 1}(y)y^{i\frac{\lambda}{4}}e^{-iy}\to 1, \quad v_{\lambda, 2}(y)y^{-i\frac{\lambda}{4}}e^{iy}\to 1 \quad \text{as }\ y\to \infty$

(see also ^[11]). Taking $u_{\lambda, j}(x) = x^{-\frac{1}{2}}v_{\lambda, j}(x^2/2)$ for $j = 1, 2$, we obtain that $(u_{\lambda, 1}, u_{\lambda, 2})$ is a fundamental system of (4) on $(\lambda_+, \infty)$ and

$u_{\lambda, 1}(y)x^{\frac{1}{2}+i\frac{\lambda}{2}}e^{-i\frac{x^2}{2}}\to 2^{-i\frac{\lambda}{4}}, \quad u_{\lambda, 2}(x)x^{\frac{1}{2}-i\frac{\lambda}{2}}e^{i\frac{x^2}{2}}\to 2^{i\frac{\lambda}{4}},$

as $x\to\infty$. The above fact implies that there exists a constant $R_{\lambda}>\lambda_+$ such that

$\frac{1}{2}x^{-\frac{1}{2}}\leq |u_{\lambda, j}(x)| \leq \frac{3}{2}x^{-\frac{1}{2}}, \quad x\geq R_{\lambda}, \quad j = 1, 2.$

We can extend $(u_{\lambda, 1}, u_{\lambda, 2})$ as a fundamental system on $\mathbb{R}$. By applying the same argument as above to (4) for $x < 0$, we can construct a different fundamental system $(\tilde{u}_{\lambda, 1}, \tilde{u}_{\lambda, 2})$ on $\mathbb{R}$ satisfying

$\frac{1}{2}|x|^{-\frac{1}{2}}\leq |\tilde{u}_{\lambda, j}(x)| \leq \frac{3}{2}|x|^{-\frac{1}{2}}, \quad x\leq -\tilde{R}_{\lambda}, \quad j = 1, 2.$

By definition of fundamental system, $u_{\lambda, j}$ can be rewritten as

$u_{\lambda, 1}(x) = c_{11}\tilde{u}_{\lambda, 1}(x) +c_{12}\tilde{u}_{\lambda, 2}(x), \quad u_{\lambda, 2}(x) = c_{21}\tilde{u}_{\lambda, 1}(x) +c_{22}\tilde{u}_{\lambda, 2}(x).$

Hence we have the upper and lower estimates of $u_{\lambda, j}$ ($j = 1, 2$), respectively.

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3.2. The case $\lambda\in\mathbb{C}\setminus\mathbb{R}$

We consider the behavior of solutions to

$\lambda u(x) - u''(x)-x^{2}u(x)+V(x)u(x) = 0,$

(6)

where $\lambda\in \mathbb{C}\setminus\mathbb{R}$ with ${\rm Im}\, \lambda>0$. The case ${\rm Im}\, \lambda < 0$ can be reduced to the problem ${\rm Im}\, \lambda>0$ via complex conjugation.

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3.2.1. Properties of solutions to an auxiliary problem

We start with the following function $\varphi_{\lambda}$:

$\varphi_{\lambda}(x): = x^{-\frac{1+\lambda i}{2}}e^{i\frac{x^2}{2}}, \quad x>0.$

(7)

Then by a direct computation we have

Lemma 3.2. $\varphi_{\lambda}$ satisfies

$\lambda \varphi_{\lambda}-\varphi_{\lambda}''-x^{2}\varphi_\lambda +g_{\lambda}\varphi_{\lambda} = 0, \quad x\in (0, \infty),$

(8)

where $g_{\lambda}(x): = \frac{(1+\lambda i)(3+\lambda i)}{4x^2}$, $x>0$.

Remark 3.1. If $\lambda = i$ or $\lambda = 3i$, then $\varphi_{\lambda}$ is nothing but a solution of the original equation (6) with $V = 0$.

Next we construct another solution of (8) which is linearly independent of $\varphi_{\lambda}$. Before construction, we prepare the following lemma.

Lemma 3.3. Let $\lambda$ satisfy ${\rm Im}\, \lambda>0$ and let $\varphi_\lambda$ be given in (7). Then for every $a>0$, there exists $F_a^\lambda\in \mathbb{C}$ such that

$\int_{a}^x \varphi_{\lambda}(t)^{-2}\, dt \to F_a^\lambda\quad \text{as} x\to \infty$

and then $x\mapsto \int_{a}^x \varphi_{\lambda}(t)^{-2}\, dt-F_a^{\lambda}$ is independent of $a$. Moreover, for every $x>0$,

$\left |\int_{a}^x \varphi_{\lambda}(t)^{-2}\, dt -F_a^{\lambda} - \frac{i}{2}x^{\lambda i}e^{-ix^2} \right| \leq C_{\lambda}x^{-{\rm Im}\, \lambda-2},$

where $C_{\lambda} : = \frac{|\lambda|}{4} \left(1+ \sqrt{ 1+ \left(\frac{{\rm Re}\, \lambda} {{\rm Im}\, \lambda+2} \right)^2 } \right).$

Remark 3.2. If $a = 0$ and $\lambda = i$, then $F_0^i$ gives the Fresnel integral $\lim_{x\to\infty}\int_{0}^xe^{-it^2}\, dt$. Hence $F_0^i = \sqrt{\pi/8}(1-i)$.

Proof. By integration by part, we have

$\int_{a}^x t^{1+\lambda i}e^{-it^2}\, dt = \left( \frac{i}{2}x^{\lambda i}e^{-ix^2} - \frac{i}{2}a^{\lambda i}e^{-ia^2} \right) + \frac{\lambda i}{4} \left( x^{\lambda i-2}e^{-ix^2} - a^{\lambda i-2}e^{-ia^2} \right) -\frac{\lambda i(\lambda i-2)}{4} \int_{a}^x t^{\lambda i-3}e^{-it^2}\, dt.$

Noting that $t^{\lambda i-3}e^{-it^2}$ is integrable in $(a, \infty)$, we have

$\int_{a}^x t^{1+\lambda i}e^{-it^2}\, dt \to - \frac{i}{2}a^{\lambda i}e^{-ia^2} - \frac{\lambda i}{4} a^{\lambda i-2}e^{-ia^2} -\frac{\lambda i(\lambda i-2)}{4} \int_{a}^\infty t^{\lambda i-3}e^{-it^2}\, dt = :F_a^\lambda$

as $x\to \infty$. And therefore $\int_{a}^x t^{1+\lambda i}e^{-it^2}\, dt-F_a^\lambda$ is independent of $a$ and

$\left |\int_{a}^x t^{1+\lambda i}e^{-it^2}\, dt -F_a^{\lambda} - \frac{i}{2}x^{\lambda i}e^{-ix^2} \right| = \left|\frac{\lambda}{4} x^{-\lambda-2}e^{-ix^2} + \frac{\lambda i(\lambda i-2)}{4} \int_{x}^\infty t^{\lambda i-3}e^{-it^2}\, dt\right| \leq C_\lambda x^{-{\rm Im}\, \lambda-2}.$

This is nothing but the desired inequality.

Lemma 3.4. Let $\varphi_{\lambda}$ be as in (7) and define $\psi_\lambda$ as

$\psi_{\lambda}(x): = \varphi_{\lambda}(x) \int_{a}^x\frac{1}{\varphi_{\lambda}(t)^2}\, dt -F_a^\lambda \varphi_{\lambda}(x), \quad x>0.$

(9)

Then $\psi_\lambda$ is independent of $a$ and $(\varphi_\lambda, \psi_\lambda)$ is a fundamental system of (8). Moreover, there exists $a_0>0$ such that

$\frac{1}{3}x^{-\frac{{\rm Im}\, \lambda+1}{2}} \leq |\psi_{\lambda}(x)| \leq x^{-\frac{{\rm Im}\, \lambda+1}{2}}, \quad x\in [a_0, \infty).$

Proof. From Lemma 3.3 we have

$x^{\frac{{\rm Im}\, \lambda+1}{2}} \left| \psi_{\lambda}(x)-\frac{i}{2}x^{-\frac{1-\lambda i}{2}}e^{-i\frac{x^2}{2}} \right| = x^{\frac{{\rm Im}\, \lambda+1}{2}} |\varphi_\lambda(x)| \left| \int_{a}^x\frac{1}{\varphi_{\lambda}(t)^2}\, dt -F_a^\lambda-\frac{i}{2}x^{\lambda i}e^{-ix^2} \right| \leq C_{\lambda}x^{-2}.$

Putting $a_0 = (6C_\lambda)^{\frac{1}{2}}$, we deduce the desired assertion.

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3.2.2. Fundamental system of the original problem

Next we consider

$\lambda w-w''-x^{2}w+g_{\lambda}w = \tilde{g}_{\lambda}h, \quad x>0$

(10)

with a given function $h$, where $g_{\lambda}$ is given as in Lemma 3.2 and $\tilde{g}_\lambda: = g_{\lambda}-V$. To construct solutions of (6), we will define two types of solution maps $h\mapsto w$ and consider their fixed points.

First we construct a solution of (6) which behaves like $\psi_\lambda$ at infinity.

Definition 3.5. For $b>0$, define

$Uh(x): = \psi_\lambda(x) -\psi_\lambda(x)\int_{b}^x \varphi_\lambda(s)\tilde{g}_\lambda(s)h(s) \, ds -\varphi_\lambda(x)\int_{x}^{\infty} \psi_\lambda(s)\tilde{g}_\lambda(s)h(s) \, ds, \quad x\in [b, \infty)$

for $h$ belonging to a Banach space

$X_{\lambda}(b): = \left\{h\in C([b, \infty))\;;\;\sup\limits_{x\in [b, \infty)}\Big(x^{\frac{{\rm Im}\lambda+1}{2}}|h(x)|\Big)<\infty\right\}, \quad\|h\|_{X_{\lambda}(b)}: = \sup\limits_{x\in [b, \infty)}\Big(x^{\frac{{\rm Im}\lambda+1}{2}}|h(x)|\Big).$

Remark 3.3. For arbitrary fixed $b>0$, all solutions of (10) can be described as follows:

$w_{c_1, c_2}(x) = c_1\varphi_\lambda(x)+c_2\psi_\lambda(x)+ \int_{b}^x \Big(\varphi_\lambda(x)\psi_\lambda(s) -\varphi_\lambda(s)\psi_\lambda(x)\Big) \tilde{g}_\lambda(s)h(s) \, ds,$

where $c_1, c_2\in\mathbb{C}$. Suppose that $h\in C_0^\infty((b, \infty))$ with ${\rm supp}\, h\subset [b_1, b_2]$. Then $w_{c_1, c_2}\in C([b, \infty))$. In particular, for $x\geq b_2$,

$w_{c_1, c_2}(x) = \left( c_1 + \int_{b_1}^{b_2} \psi_\lambda(s)\tilde{g}_\lambda(s)h(s) \, ds \right)\varphi_\lambda(x) + \left( c_2- \int_{b_1}^{b_2} \varphi_\lambda(s) \tilde{g}_\lambda(s)h(s) \, ds \right)\psi_\lambda(x).$

Therefore $w_{c_1, c_2}$ behaves like $\psi_\lambda$ (that is, $w_{c_1, c_2}\in X_{\lambda}(b)$) only when

$c_1 = -\int_{b_1}^{b_2} \psi_\lambda(s) \tilde{g}_\lambda(s)h(s) \, ds = -\int_{b}^{\infty} \psi_\lambda(s) \tilde{g}_\lambda(s)h(s) \, ds.$

In Definition 3.5 we deal with such a solution with $c_2 = 1$.

Well-definedness of $U$ in Definition 3.5 and its contractivity are proved in next lemma.

Lemma 3.6. The following assertions hold :

(ⅰ) for every $b>0$, the map $U:X_{\lambda}(b)\to X_{\lambda}(b)$ is well-defined $;$

(ⅱ) there exists $b_{\lambda}>0$ such that $U$ is contractive in $X_{\lambda}(b_\lambda)$ with

$\|Uh_1-Uh_2\|_{X_{\lambda}(b)}\leq \frac{1}{5}\|h_1-h_2\|_{X_{\lambda}(b)}, \quad h_1, h_2\in X_{\lambda}(b_\lambda)$

and then $U$ has a unique fixed point $w_1\in X_{\lambda}(b_\lambda)$ $;$

(ⅲ) $w_1$ can be extended to a solution of (6) in $\mathbb{R}$ satisfying

$\frac{1}{12}x^{-\frac{{\rm Im}\lambda+1}{2}}\leq |w_1(x)| \leq 2x^{-\frac{{\rm Im}\lambda+1}{2}}, \quad x\in [b_{\lambda}, \infty).$

Proof. (ⅰ) By Lemma 3.4 we have $\psi_{\lambda}\in X_{\lambda}(b)$. Therefore to prove well-definedness of $U$, it suffices to show that the second term in the definition of $U$ belongs to $X_{\lambda}(b)$.

Let $h\in X_{\lambda}(b)$. Then for $x\in [b, \infty)$,

$x^{\frac{{\rm Im}\, \lambda+1}{2}}\left|\varphi_\lambda(x) \int_{x}^\infty \psi_\lambda(s)\tilde{g}(s)h(s) \, ds \right| \leq x^{{\rm Im}\, \lambda} \|h\|_X \int_{x}^\infty s^{-{\rm Im}\, \lambda-1}|\tilde{g}_\lambda(s)| \, ds \leq \|h\|_X \|s^{-1}\tilde{g}_{\lambda}\|_{L^1(b, \infty)}$

and

$x^{\frac{{\rm Im}\, \lambda+1}{2}} \left| \psi_\lambda(x)\int_{b}^x \varphi_\lambda(s)\tilde{g}(s)h(s) \, ds \right| \leq \|h\|_X \int_{b}^x s^{-1}|\tilde{g}_\lambda(s)| \, ds \leq \|h\|_X \|s^{-1}\tilde{g}_{\lambda}\|_{L^1(b, \infty)}.$

Hence we have $Uh\in C([b, \infty))$ and therefore $Uh\in X_{\lambda}(b)$, that is, $U:X_{\lambda}(b)\to X_{\lambda}(b)$ is well-defined.

(ⅱ) Let $h_{1}, h_{2}\in X_{\lambda}(b)$. Then we have

$Uh_1(x)-Uh_2(x) = -\psi_\lambda(x)\int_{b}^x \varphi_\lambda(s)\tilde{g}_{\lambda}(s)(h_1(s)-h_2(s)) \, ds -\varphi_\lambda(x) \int_{x}^\infty \psi_\lambda(s)\tilde{g}_{\lambda}(s)(h_1(s)-h_2(s)) \, ds.$

Proceeding the same computation as above, we deduce

$\|Uh_1-Uh_2\|_{X_{\lambda}(b)} \leq 2\|s^{-1}\tilde{g}_{\lambda}\|_{L^1(b, \infty)}\|h_1-h_2\|_{X_{\lambda}(b)}.$

Choosing $b$ large enough, we obtain $\|Uh_1-Uh_2\|_{X_{\lambda}(b)} \leq 5^{-1}\|h_1-h_2\|_{X_{\lambda}(b)}$, that is $U$ is contractive in $X_{\lambda}(b)$. By contraction mapping principle, we obtain that $U$ has a unique fixed point $w_1\in X_{\lambda}(b)$.

(ⅲ) Since $w_1$ satisfies (10) with $h = w_1$, $w_1$ is a solution of the original equation (6) in $[b, \infty)$. As in the last part of the proof of Proposition 3.1, we can extend $w_1$ as a solution of (6) in $\mathbb{R}$. Since $Uw_1 = w_1$ and $U0 = \psi_\lambda$, it follows from the contractivity of $U$ that

$\|w_1-\psi_\lambda\|_X = \|Uw_1-U0\|_X \leq \frac{1}{5}\|w_1\|_X \leq \frac{1}{5}\|w_1-\psi_\lambda\|_X +\frac{1}{5}\|\psi_\lambda\|_X.$

Consequently, we have $\|w_1-\psi_\lambda\|_X\leq 4^{-1}\|\psi_\lambda\|_X\leq 4^{-1}$ and then for $x\geq b$,

$|w_1(x)| \geq |\psi_\lambda(x)|-|w_1(x)-\psi_{\lambda}(x)| \geq \left( \frac{1}{3} - \|w_1-\psi_\lambda\|_X \right) x^{-\frac{{\rm Im}\lambda+1}{2}} \geq \frac{1}{12}x^{-\frac{{\rm Im}\lambda+1}{2}}.$

Next we construct another solution of (6) which behaves like $\varphi_\lambda$ at infinity.

Definition 3.7. Let $b>0$ be large enough. Define

$\widetilde{U}h(x): = \varphi_\lambda(x)+ \int_{b}^x \Big(\varphi_\lambda(x)\psi_\lambda(s) -\varphi_\lambda(s)\psi_\lambda(x)\Big) \tilde{g}_{\lambda}(s)h(s) \, ds$

for $h$ belonging to a Banach space

$Y_{\lambda}(b): = \left\{h\in C([b, \infty))\;;\; \sup\limits_{x\in [b, \infty)} \Big(x^{-\frac{{\rm Im}\lambda-1}{2}}|h(x)|\Big)<\infty\right\}, \ \ \|h\|_{Y_{\lambda}(b)}: = \sup\limits_{x\in [b, \infty)} \Big(x^{-\frac{{\rm Im}\lambda-1}{2}}|h(x)|\Big).$

Lemma 3.8. The following assertions hold $:$

(ⅰ) for every $b>0$, the map $\widetilde{U}:Y_{\lambda}(b)\to Y_{\lambda}(b)$ is well-defined $;$

(ⅱ) there exists $b_{\lambda}>0$ such that $\widetilde{U}$ is contractive in $Y_{\lambda}(b_\lambda)$ with

$\|\widetilde{U}h_1-\widetilde{U}h_2\|_{Y_{\lambda}(b)}\leq \frac{1}{5}\|h_1-h_2\|_{Y_{\lambda}(b)}, \quad h_1, h_2\in Y_{\lambda}(b_\lambda)$

and then $\widetilde{U}$ has a unique fixed point $\tilde{w}_1\in Y_{\lambda}(b_\lambda)$ $;$

(ⅲ) $\tilde{w}_1$ can be extended to a solution of (6) in $\mathbb{R}$ satisfying

$\frac{1}{2}x^{\frac{{\rm Im}\, \lambda-1}{2}} \leq |\tilde{w}_1(x)| \leq 2x^{\frac{{\rm Im}\, \lambda-1}{2}}, \quad x\in [b_{\lambda}, \infty).$

Proof. The proof is similar to the one of Lemma 3.6.

Considering the equation (6) for $x < 0$, we also obtain the following lemma.

Lemma 3.9. For every $\lambda\in\mathbb{C}$ with ${\rm Im}\, \lambda>0$, there exist a fundamental system $(w_1, w_2)$ of (6) and positive constants $c_{\lambda}, C_\lambda, R_{\lambda}$ such that

$|w_1(x)| \leq C_\lambda(1+|x|)^{\frac{{\rm Im}\, \lambda-1}{2}}, \quad x\leq 0, \quad \quad |w_1(x)| \leq C_\lambda(1+|x|)^{-\frac{{\rm Im}\, \lambda+1}{2}}, \quad x\geq 0,$

(11)

$|w_2(x)| \leq C_\lambda(1+|x|)^{-\frac{{\rm Im}\, \lambda+1}{2}}, \quad x\leq 0, \quad \quad |w_2(x)| \leq C_\lambda(1+|x|)^{\frac{{\rm Im}\, \lambda-1}{2}}, \quad x\geq 0$

(12)

and

$|w_1(x)| \geq c_\lambda(1+|x|)^{-\frac{{\rm Im}\, \lambda+1}{2}}, \quad x\geq R_{\lambda}, \quad |w_2(x)| \geq c_\lambda(1+|x|)^{-\frac{{\rm Im}\, \lambda+1}{2}}, \quad x\leq-R_{\lambda}.$

(13)

Proof. In view of Lemma 3.6, it suffices to find $w_2$ satisfying the conditions above.

Let $w_*$ and $\tilde{w}_*$ be given as in Lemmas 3.6 and 3.8 with $V(x)$ replaced with $V(-x)$. Noting that $w_1$ can be rewritten as $w_1(x) = c_1w_*(-x) +c_2\tilde{w}_*(-x)$, we see from Lemma 3.6 and 3.8 that (11) and the first half of (13) are satisfied. Set $w_2(x) = w_*(-x)$ for $x\in \mathbb{R}$. As in the same way, we can verify (12).

Finally, we prove the last half of (13). Since $H_{2, \min}$ is essentially selfadjoint in $L^2(\mathbb{R})$, $\lambda$ belongs to the resolvent set of $H_2$, that is, $N(\lambda+H_2) = \{0\}$. This implies that $w_2\notin L^2(\mathbb{R})$. Noting that $w_2\in L^2((-\infty, 0))$, we have $w_2\notin L^2((0, \infty))$. Now using the representation

$w_2(x) = c_1w_1(x) +c_2\tilde{w}_1(x), \quad x\in \mathbb{R},$

we deduce that $c_2\neq 0$. Therefore using Lemma 3.6 (ⅲ) and Lemma 3.8 (ⅲ), we have

$|w_2(x)| \geq |c_2|\, |\tilde{w}_1(x)|-|c_1|\, |w_1(x)| \geq \frac{|c_2|}{2}x^{\frac{{\rm Im}\, \lambda-1}{2}}-2|c_1|\, x^{-\frac{{\rm Im}\, \lambda+1}{2}} \geq \frac{|c_2|}{4}x^{\frac{{\rm Im}\, \lambda-1}{2}}$

for $x$ large enough.

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4. Resolvent estimates in $L^p$

The following lemma, verified by the variation of parameters, gives a possibility of representation of the Green function for resolvent operator $H$ in $L^p$.

Lemma 4.1. Assume that $\lambda\in \rho(\widetilde{H})$ in $L^p$, where $\widetilde{H}$ is a realization of $H$ in $L^p$. Then for every $u\in C_0^\infty(\mathbb{R})$,

$u(x) = \frac{w_1(x)}{W_\lambda} \int_{-\infty}^x w_2(s)f(s)\, ds + \frac{w_2(x)}{W_\lambda} \int_x^{\infty} w_1(s)f(s)\, ds, \quad x\in \mathbb{R},$

where $f: = \lambda u-u''-x^2u+Vu\in C_0^\infty(\mathbb{R})$ and $W_{\lambda}\neq 0$ is the Wronskian of $(w_1, w_2)$.

Proposition 4.2. Let $1 < p < \infty$. If $|1-\frac{2}{p}| < {\rm Im}\, \lambda$, then the operator defined as

$R(\lambda)f(x): = \frac{w_1(x)}{W_\lambda} \int_{-\infty}^x w_2(s)f(s)\, ds + \frac{w_2(x)}{W_\lambda} \int_x^{\infty} w_1(s)f(s)\, ds, \quad f\in C_0^\infty(\mathbb{R})$

can be extended to a bounded operator on $L^p$. More precisely, there exists $M_{\lambda}>0$ such that

$\label{reso_p} \|R(\lambda)f\|_{L^p}\leq M_{\lambda}\left[ |{\rm Im}\lambda|^2-\left(1-\frac{2}{p}\right)^2\right]^{-1} \|f\|_{L^p}, \quad f\in L^p(\mathbb{R}).$

(14)

In particular, $H_{p, \min}$ is closable and its closure $H_p$ satisfies

$\left\{ \lambda\in\mathbb{C}\;;\; |{\rm Im}\, \lambda|>\left|1-\frac{2}{p}\right| \right\} \subset\rho(H_p).$

Proof. Let $f\in C_0^\infty(\mathbb{R})$. Set

$u_1(x) : = w_1(x)\int_{-\infty}^x w_2(s)f(s)\, ds, \quad u_2(x) : = w_1(x)\int_x^{\infty} w_1(s)f(s)\, ds.$

We divide the proof of $u_1\in L^p(\mathbb{R})$ into two cases $x\geq 0$ and $x < 0$; since the proof of $u_2\in L^p(\mathbb{R})$ is similar, this part is omitted.

The case $u_1$ for $x\geq 0$, it follows from Lemma 3.9 and Hölder inequality that

$\begin{align} \nonumber |u_1(x)| &\leq C_\lambda^2 (1+|x|)^{-\frac{{\rm Im}\, \lambda+1}{2}} \left[ \int_{-\infty}^0 (1+|s|)^{-\frac{{\rm Im}\, \lambda+1}{2}}|f(s)|\, ds + \int_{0}^x (1+|s|)^{\frac{{\rm Im}\, \lambda-1}{2}}|f(s)|\, ds\right] \\ \nonumber &\leq C_\lambda^2 \left(\frac{{\rm Im}\, \lambda+1}{2}p'-1\right)^{-\frac{1}{p'}}\|f\|_{L^p(\mathbb{R}_-)} (1+|x|)^{-\frac{{\rm Im}\, \lambda+1}{2}} \\ \nonumber &\quad+C_\lambda^2(1+|x|)^{-\frac{{\rm Im}\, \lambda+1}{2}} \left( \int_{0}^x (1+|s|)^{\frac{{\rm Im}\, \lambda-1}{2}p'-\alpha p'}\, ds \right)^{\frac{1}{p'}} \left( \int_{0}^x (1+|s|)^{\alpha p}|f(s)|^p\, ds \right)^{\frac{1}{p}} \\ \nonumber &\leq C_\lambda^2 \left(\frac{{\rm Im}\, \lambda+1}{2}p'-1\right)^{-\frac{1}{p'}}\|f\|_{L^p(\mathbb{R}_-)} (1+|x|)^{-\frac{{\rm Im}\, \lambda+1}{2}} \\ \label{est.u1:1} &\quad+C_\lambda^2 \left( \frac{{\rm Im}\, \lambda-1}{2}p'-\alpha p'+1 \right)^{-\frac{1}{p'}} (1+|x|)^{-\frac{1}{p}-\alpha}\left( \int_{0}^x (1+|s|)^{\alpha p}|f(s)|^p\, ds \right)^{\frac{1}{p}} \end{align}$

(15)

with $0 < \alpha < \frac{{\rm Im}\lambda+1}{2}+1/p'$. By the triangle inequality we have

$\|u_1\|_{L^p(\mathbb{R}_+)} \leq C_\lambda^2 \left(\frac{{\rm Im}\, \lambda+1}{2}p'-1\right)^{-\frac{1}{p'}} \left(\frac{{\rm Im}\, \lambda+1}{2}p-1\right)^{-\frac{1}{p}} \|f\|_{L^p(\mathbb{R}_-)} + \mathcal{I}_1(\alpha)$

and

$\begin{align*} \big(\mathcal{I}_1(\alpha)\big)^p & = C_\lambda^{2p} \left( \frac{{\rm Im}\, \lambda-1}{2}p'-\alpha p'+1 \right)^{-\frac{p}{p'}} \int_{0}^\infty (1+|x|)^{-1-\alpha p} \left( \int_{0}^x (1+|s|)^{\alpha p}|f(s)|^p\, ds \right)\, dx \\ & = C_\lambda^{2p} \left( \frac{{\rm Im}\, \lambda-1}{2}p'-\alpha p'+1 \right)^{-\frac{p}{p'}} \left(\alpha p \right)^{-1} \int_{0}^\infty |f(s)|^p\, ds. \end{align*}$

Choosing $\alpha = \frac{1}{pp'}\left(\frac{{\rm Im}\, \lambda-1}{2}p'+1\right)$, we obtain

$\|u_1\|_{L^p(\mathbb{R}_+)} \leq C_\lambda^2 \left(\frac{{\rm Im}\, \lambda+1}{2}p'-1\right)^{-\frac{1}{p'}} \left(\frac{{\rm Im}\, \lambda+1}{2}p-1\right)^{-\frac{1}{p}}\|f\|_{L^p(\mathbb{R}_-)} + C_\lambda^2 \left( \frac{{\rm Im}\, \lambda-1}{2}+\frac{1}{p'} \right)^{-1}\|f\|_{L^p(\mathbb{R}_+)}.$

The case $u_1$ for $x < 0$, by the same way as the case $x>0$, we have

$|u_1(x)|^p \leq C_\lambda^{2p} \left(\frac{{\rm Im}\lambda+1}{2}p'-\beta p'-1\right)^{-\frac{p}{p'}} (1+|x|)^{-1+\beta p} \int_{-\infty}^x (1+|s|)^{-\beta p}|f(s)|^p\, ds,$

(16)

where $0 < \beta < \frac{{\rm Im}\lambda +1}{2}-\frac{1}{p'}$. Taking $\beta = \frac{1}{pp'}\left(\frac{{\rm Im}\lambda+1}{2}p'-1\right)$, we have

$\|u_1\|_{L^p(\mathbb{R}_-)}\leq C_\lambda^{2} \left(\frac{{\rm Im}\lambda+1}{2}-\frac{1}{p'}\right)^{-1}\|f\|_{L^p(\mathbb{R}_-)}.$

Proceeding the same argument for $u_2$ and combining the estimates for $u_1$ and $u_2$, we obtain (14).

Corollary 4.3. Let $\mathcal{R}(\lambda)$ be as in Proposition 4.2. Then for every $f\in L^p(\mathbb{R})$, $\mathcal{R}(\lambda)f\in C(\mathbb{R})$ and

$\sup\limits_{x\in \mathbb{R}}\Big( (1+|x|)^{\frac{1}{p}}|\mathcal{R}(\lambda)f(x)| \Big) \leq \tilde{C}_{\lambda} \|f\|_{L^p}.$

(17)

Proof. Let $f\in C_0^\infty(\mathbb{R})$ and set $u_1$ and $u_2$ as in the proof of Proposition 4.2. Since the proof for $u_1$ and $u_2$ are similar, we only show the estimate of $u_1$. From (15), we have for $x\geq 0$,

$\begin{align*} (1+|x|)^{\frac{1}{p}}|u_1(x)| &\leq C_\lambda^2 \left(\frac{{\rm Im}\, \lambda+1}{2}p'-1\right)^{-\frac{1}{p'}} \|f\|_{L^p(\mathbb{R}_-)} (1+|x|)^{-\frac{{\rm Im}\, \lambda}{2}+\frac{1}{p}-\frac{1}{2}} \\ &\quad+C_\lambda^2 \left( \frac{{\rm Im}\, \lambda-1}{2}p'-\alpha p'+1 \right)^{-\frac{1}{p'}} (1+|x|)^{-\alpha}\left( \int_{0}^x (1+|s|)^{\alpha p}|f(s)|^p\, ds \right)^{\frac{1}{p}} \\ &\leq C_\lambda^2 \left(\frac{{\rm Im}\, \lambda+1}{2}p'-1\right)^{-\frac{1}{p'}} \|f\|_{L^p(\mathbb{R}_-)} +C_\lambda^2 \left( \frac{{\rm Im}\, \lambda-1}{2}p'-\alpha p'+1 \right)^{-\frac{1}{p'}} \|f\|_{L^p(\mathbb{R}_+)}, \end{align*}$

where $0 < \alpha < \frac{{\rm Im}\lambda+1}{2}+\frac{1}{p'}$. This implies (17) for $x\geq 0$. If $x\leq 0$, then from (16) we can obtain

$(1+|x|)^{\frac{1}{p}}|u_1(x)| \leq C_\lambda^{2} \left(\frac{{\rm Im}\lambda+1}{2}p'-\beta p'-1\right)^{-\frac{1}{p'}} \|f\|_{L^p(\mathbb{R}_-)},$

where $0 < \beta < \frac{{\rm Im}\lambda +1}{2}-\frac{1}{p'}$. This yields (17) for $x\leq 0$. The proof is completed.

By interpolation inequality, we deduce the following assertion.

Proposition 4.4. Let $1 < p < \infty$ and $p\le q \le \infty$. Then

$D(H_{p}) \subset \left\{w\in C(\mathbb{R}) \;;\; \langle x\rangle ^{\frac{1}{p}-\frac{1}{q}}w \in L^q \right\}.$

More precisely, there exists a constant $C_{p, q}>0$ such that

$\left\|\langle x\rangle^{\frac{1}{p}-\frac{1}{q}}u\right\|_{L^q} \leq C_{p, q}\Big(\|H_p u\|_{L^p}+\|u\|_{L^p}\Big), \quad u\in D(H_p).$

Proof. The assertion follows from Proposition 4.2 and Corollary 4.3.

Proposition 4.5. (ⅰ) If $2 < p < \infty$ and $0 < |{\rm Im}\, \lambda| < 1-\frac{2}{p}$, then $N(\lambda+H_p)\neq\{0\}$, and then

$\left\{\lambda\in\mathbb{C}\;;\;|{\rm Im}\, \lambda|\leq 1-\frac{2}{p}\right\} \subset \sigma(H_p);$

(ⅱ) If $1 < p < 2$ and $0 < |{\rm Im}\, \lambda| < \frac{2}{p}-1$, then $\overline{N(\lambda+H_p)}\subsetneq L^p$, and then

$\left\{\lambda\in\mathbb{C}\;;\;|{\rm Im}\, \lambda|\leq \frac{2}{p}-1\right\} \subset \sigma(H_p).$

Proof. (ⅰ) ($2 < p\leq \infty$, ${\rm Im}\, \lambda < 1-\frac{2}{p}$) Noting that

$\frac{{\rm Im}\, \lambda+1}{2}>\frac{1}{p}, \quad -\frac{{\rm Im}\, \lambda-1}{2}>\frac{1}{p},$

we have by (11),

$\begin{align*} \int_{-\infty}^\infty |w_1(x)|^p \, dx &\leq C_{\lambda} \left( \int_{-\infty}^0(1+|s|)^{\frac{{\rm Im}\, \lambda-1}{2}p}\, ds + \int_{0}^\infty(1+|s|)^{-\frac{{\rm Im}\, \lambda+1}{2}p}\, ds \right) \\ &\leq C_{\lambda} \left[ \left(\frac{1-{\rm Im}\, \lambda}{2}p-1\right)^{-1} + \left(\frac{{\rm Im}\, \lambda+1}{2}p-1\right)^{-1} \right] <\infty. \end{align*}$

This means that $w_1, w_2\in N(\lambda+H_{p})$.

(ⅱ) $(1 < p < 2, {\rm Im}\, \lambda < \frac{2}{p}-1$) Note that $H_p$ is the adjoint operator of $H_{p'}$. Since $w_1\in D(H_{p'})$ for every $u\in C_0^\infty(\mathbb{R})$,

$\int_{-\infty}^\infty(\lambda u+H_pu)w_1\, dx = \int_{-\infty}^\infty u (\lambda w_1+H_{p'}w_1)\, dx = 0,$

the closure of $R(\lambda+H_p)$ does not coincide with $L^p$, that is, $\overline{R(\lambda+H_p)}\subsetneq L^p$.

Since $\sigma(H_p)$ is closed in $\mathbb{C}$ and we can argue the same assertion for ${\rm Im}\lambda < 0$ via complex conjugation, we obtain the assertion.

Combining the assertions above, we finally obtain Theorem 1.1.

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5. Absence of $C_0$-semigroups on $L^p$ ($p\neq 2$, $V=0$)

In Theorem 1.1, we do not prove any assertions related to generation of $C_0$-semigroups by $\pm iH_p$. In this subsection we prove

Theorem 5.1. Neither $iH_p$ nor $-iH_p$ generates $C_0$-semigroup on $L^p$.

Proof. We argue by a contradiction. Assume that $iH_p$ generates a $C_0$-semigroup $T(t)$ on $L^p$. Then it follows from Theorem 1.1 (the coincidence of resolvent operators) that we have $T(t)f = S(t)f$ for every $t>0$ and $f\in L^2\cap L^p$, where $S(t)$ is the $C_0$-group generated by the skew-adjoint operator $iH_2$.

Fix $f_0\in L^2\cap L^p$ such that $\mathcal{F}f_0\notin L^p$ ($\mathcal{F}$ is the Fourier transform). Then by the Mehler's formula (see e.g., Cazenave [3,Remark 9.2.5]), we see that

$\big[S(t)\big]f(x) = \left(\frac{1}{2\pi \sinh(2t)}\right)^{\frac{N}{2}} e^{-i\frac{1}{2\tanh(2t)}|x|^2} \int_{-\infty}^\infty e^{-\frac{i}{\sinh(2t)}x\cdot y} e^{-i\frac{1}{2\tanh(2t)}|y|^2}f(y)\, dy.$

In other words, using the operators

$M_\tau g(x): = e^{-i\frac{|x|^2}{2\tau}}g(x), \quad D_{\tau}g(x): = \tau^{-\frac{N}{2}}g(\tau^{-1}x),$

we can rewrite $S(t)$ as the following form $S(t)f = M_{\tanh(2t)}\mathcal{F}D_{\sinh(2t)}M_{\tanh(2t)}f$. Taking $f_{t_0} = M_{\tanh(2t_0)}^{-1}D_{\sinh(2t_0)}^{-1}f_0\in L^p$, we have

$S(t_0)f_{t_0} = M_{\tanh(2t)}\mathcal{F}f_0\notin L^p.$

This contradicts the fact $T(t_0)f_{t_0}\in L^p$. This completes the proof.

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Acknowledgments

This work is partially supported by Grant-in-Aid for Young Scientists Research (B), No. 15K17558, 16K17619.

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Conflict of Interest

All authors declare no conflicts of interest in this paper.

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