\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 111, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/111\hfil Gordon type theorem] {Gordon type theorem for measure perturbation} \author[C. Seifert\hfil EJDE-2011/111\hfilneg] {Christian Seifert} \address{Christian Seifert \newline Fakult\"at Mathematik\\ Technische Universit\"at Chemnitz\\ 09107 Chemnitz, Germany} \email{christian.seifert@mathematik.tu-chemnitz.de} \thanks{Submitted May 31, 2011. Published August 29, 2011.} \subjclass[2000]{34L05, 34L40, 81Q10} \keywords{Schr\"odinger operators; eigenvalue problem; \hfill\break\indent quasiperiodic measure potentials} \begin{abstract} Generalizing the concept of Gordon potentials to measures we prove a version of Gordon's theorem for measures as potentials and show absence of eigenvalues for these one-dimensional Schr\"odinger operators. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \section{Introduction}\label{sec:introduction} According to \cite{DamanikStolz2000}, the one-dimensional Schr\"odinger operator $H=-\Delta+V$ has no eigenvalues if the potential $V\in L_{1,{\rm loc}}(\mathbb{R})$ can be approximated by periodic potentials (in a suitable sense). The aim of this paper is to generalize this result to measures $\mu$ instead of potential functions $V$; i.e., to more singular potentials. Although all statements remain valid for complex measures we only focus on real (but signed) measures $\mu$, since we are interested in self-adjoint operators. In the remaining part of this section we explain the situation and define the operator in question. We also describe the class of measures we are concerned with. Section \ref{sec:solution_estimates} provides all the tools we need to prove the main theorem: $H=-\Delta+\mu$ has no eigenvalues for suitable $\mu$. In section \ref{sec:examples} we show some examples for Schr\"odinger operators with measures as potentials. We consider a Schr\"odinger operator of the form $H = -\Delta + \mu$ on $L_2(\mathbb{R})$. Here, $\mu=\mu_+ - \mu_-$ is a signed Borel measure on $\mathbb{R}$ with locally finite total variation $|\mu|$. We define $H$ via form methods. To this end, we need to establish form boundedness of $\mu_-$. Therefore, we restrict the class of measures we want to consider. \begin{definition} \label{def1.1} \rm A signed Borel measure $\mu$ on $\mathbb{R}$ is called \emph{uniformly locally bounded}, if $\|\mu\|_{\rm loc} := \sup_{x\in \mathbb{R}} |\mu|([x,x+1]) < \infty.$ We call $\mu$ a \emph{Gordon measure} if $\mu$ is uniformly locally bounded and if there exists a sequence $(\mu^{m})_{m\in\mathbb{N}}$ of uniformly locally bounded periodic Borel measures with period sequence $(p_m)$ such that $p_m \to \infty$ and for all $C\in \mathbb{R}$ we have $\lim_{m\to\infty} e^{Cp_m}|\mu-\mu^{m}|([-p_m,2p_m]) = 0;$ i.e., $(\mu^{m})$ approximates $\mu$ on increasing intervals. Here, a Borel measure is $p$-periodic, if $\mu = \mu(\cdot + p)$. \end{definition} Clearly, every generalized Gordon potential $V\in L_{1,{\rm loc}}$ as defined in \cite{DamanikStolz2000} induces a Gordon measure $\mu=V\lambda$, where $\lambda$ is the Lebegue measure on $\mathbb{R}$. Therefore, also every Gordon potential (see the original work \cite{Gordon1976}) induces a Gordon measure. \begin{lemma} \label{lem1.1} Let $\mu$ be a uniformly locally bounded measure. Then $|\mu|$ is $-\Delta$-form bounded, and for all $00$ we have $|v(x) - v(x+r)| \to 0 \quad (|x|\to \infty).$ Then $|v(x)|\to 0$ as $|x|\to \infty$. \end{lemma} \begin{proof} Without restriction, we can assume that $v\geq 0$. We prove this lemma by contradiction. Assume that $v(x)\to 0$ does not hold for $x\to \infty$. Then we can find $\delta>0$ and $(q_k)$ in $\mathbb{R}$ with $q_k\to \infty$ such that $v(q_k)\geq \delta$ for all $k\in\mathbb{N}$. By square integrability of $v$ we have $\|v\mathbf{1}_{[q_k,q_k+1]}\|_2 \to 0$. Therefore, we can find a subsequence $(r_n)$ of $(q_k)$ satisfying $\|v\mathbf{1}_{[r_n,r_n+1]}\|_2 \leq 2^{-3n/2} \quad(n\in \mathbb{N}).$ Now, Chebyshev's inequality implies $\lambda(\big\{x\in[r_n,r_n+1];\; v(x)\ge 2^{-n}\big\}) \leq 2^{2n} \|v\mathbf{1}_{[r_n,r_n+1]}\|_2^2 \leq 2^{-n} \quad(n\in\mathbb{N}).$ Denote $A_n:= \big\{x\in[r_n,r_n+1];\; v(x)\ge 2^{-n}\big\} - r_n \subseteq [0,1]$. Then $\lambda(A_n) \leq 2^{-n}$ and $\lambda\big(\cup_{n\geq 3} A_n\big) \leq \sum_{n\geq 3} \lambda(A_n) \leq 2^{-2} <1.$ Hence, $G:=[0,1]\setminus (\cup_{n\geq 3} A_n)$ has positive measure. For $r\in G$, $r>0$ it follows $v(r_n+r)\le 2^{-n}$ ($n\geq 3$). Therefore, $\liminf_{n\to\infty} |v(r_n)-v(r_n+r)| \geq \delta>0,$ a contradiction. \end{proof} \begin{lemma} \label{lem:konvergenz} Let $\mu$ be a Gordon measure, $E\in \mathbb{R}$, $u\in D(H)$ a solution of $Hu=Eu$. Then $u(x)\to 0$ as $x\to \infty$ and $u'(x)\to 0$ as $x\to \infty$. \end{lemma} \begin{proof} Since $u\in D(H)\subseteq D(\tau)\subseteq W_2^1(\mathbb{R})$ we have $u(x)\to 0$ as $|x|\to \infty$. Lemma \ref{lem:Hsubseteq_T} yields $u\in D(T)$ and $-(Au)' = Hu = Eu$. Let $r>0$. Then, for almost all $x\in \mathbb{R}$, \begin{align*} u'(x+r) - u'(x) & = Au(x+r) - Au(x) + \int_{(x,x+r]} u(t)\, d\mu(t) \\ & = \int_{x}^{x+r} (Au)'(y)\, dy + \int_{(x,x+r]} u(t)\, d\mu(t). \end{align*} Hence, \begin{align*} |u'(x+r) - u'(x)| & \leq |E| \int_x^{x+r} |u(y)|\, dy + \int_{(x,x+r]} |u(t)|\, d|\mu|(t) \\ & \leq |E|r\|u\|_{\infty,[x,x+r]} + \|u\|_{\infty,[x,x+r]}|\mu|([x,x+r]) \\ & \leq \|u\|_{\infty,[x,x+r]} \left(|E|r +(r+1)\|\mu\|_{\rm loc}\right). \end{align*} By Sobolev's inequality, there is $C\in \mathbb{R}$ (depending on $r$, but $r$ is fixed anyway) such that $\|u\|_{\infty,[x,x+r]} \leq C \|u\|_{W_2^1(x,x+r)} \to 0 \quad (|x|\to \infty).$ Thus, $|u'(x+r) - u'(x)|\to 0 \quad (|x|\to \infty).$ An application of Lemma \ref{lem:u'_konvergiert} with $v:=u'$ yields $u'(x)\to 0$ as $|x|\to \infty$. \end{proof} Now, we can state the main result of this paper. \begin{theorem}\label{thm:Gordon} Let $\mu$ be a Gordon measure. Then $H$ has no eigenvalues. \end{theorem} \begin{proof} Let $(\mu^m)$ be the periodic approximants of $\mu$. Let $E\in \mathbb{R}$ and $u$ be a solution of $Hu = Eu$. Let $(u_m)$ be the solutions for the measures $(\mu^{m})$. By Lemma \ref{lem:expest} we find $m_0\in \mathbb{N}$ such that $\Big\|\begin{pmatrix} u(x) \\ u'(x) \end{pmatrix} - \begin{pmatrix} u_m(x) \\ u_m'(x) \end{pmatrix}\Big\| \leq \frac{1}{4}$ for $m\geq m_0$ and almost all $x\in [-p_m,2p_m]$. By Lemma \ref{lem:estimate} we have $\limsup_{|x|\to \infty} \left(|u(x)|^2 + |u'(x)|^2\right) \geq \frac{1}{4} > 0.$ Hence, $u$ cannot be in $D(H)$ by Lemma \ref{lem:konvergenz}. \end{proof} \section{Examples} \label{sec:examples} \begin{remark}[periodic measures] \label{rmk} \rm Every locally bounded periodic measure on $\mathbb{R}$ is a Gordon measure. Thus, for $\mu:= \sum_{n\in\mathbb{Z}} \delta_{n+\frac{1}{2}}$ the operator $H:= -\Delta + \mu$ has no eigenvalues. \end{remark} Some examples of quasi-periodic $L_{1,{\rm loc}}$-potentials can be found in \cite{DamanikStolz2000}. For a measure $\mu$ and $x\in\mathbb{R}$ let $T_x\mu:= \mu(\cdot-x)$. If $\mu$ is periodic with period $p$, then $T_p\mu = \mu$. \begin{example} \label{examp3.2} \rm Let $\alpha\in(0,1)\setminus \mathbb{Q}$. There is a unique continued fraction expansion $\alpha = \frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{\ldots}}}}$ with $a_n\in\mathbb{N}$. For $m\in\mathbb{N}$ we set $\alpha_m = \frac{p_m}{q_m}$, where \begin{gather*} p_0 = 0, \quad p_1 = 1,\quad p_m = a_mp_{m-1}+p_{m-2}, \\ q_0 = 1, \quad q_1 = a_1,\quad q_m = a_mp_{m-1}+q_{m-2} . \end{gather*} The number $\alpha$ is called \emph{Liouville number}, if there is $B\geq 0$ such that $|\alpha-\alpha_m| \leq Bm^{-q_m}.$ The set of Liouville numbers is a dense $G_\delta$. Let $\nu,\tilde{\nu}$ be $1$-periodic measures and assume that there is $\gamma>0$ such that $|\nu(\cdot-x)-\nu|([0,1])\leq |x|^\gamma \quad(x\in\mathbb{R}).$ Define $\mu:=\tilde{\nu} + \nu\circ \alpha$ and $\mu^{m}:=\tilde{\nu} + \nu\circ\alpha_m$ for $m\in\mathbb{N}$. Then $\mu^m$ is $q_m$-periodic and \begin{align*} |\mu-\mu^m|([-q_m,2q_m]) & = |\nu\circ \alpha - \nu\circ\alpha_m|([-q_m,2q_m]) \\ & = \big\|\nu\circ\frac{\alpha}{\alpha_m}-\nu\big|([-p_m,2p_m]) \\ & \leq \sum_{n=-p_m}^{2p_m-1} \big|\nu\circ\frac{\alpha}{\alpha_m}-\nu\big|([n,n+1]). \end{align*} Now, we have \begin{align*} \big|\nu\circ\frac{\alpha}{\alpha_m}-\nu\big|([n,n+1]) & = T_{-n}\big|\nu\circ\frac{\alpha}{\alpha_m}-\nu\big|([0,1]) \\ & = \big|T_{-n}(\nu\circ\frac{\alpha}{\alpha_m})-T_{-n}\nu\big|([0,1]) \\ & = \big|T_{-n}(\nu\circ\frac{\alpha}{\alpha_m})-\nu\big|([0,1]). \end{align*} With $g_{m,n}(y):= y+(\frac{\alpha}{\alpha_m} - 1)(y+n)$ and using periodicity of $\nu$ we obtain $T_{-n}(\nu\circ\frac{\alpha}{\alpha_m}) = \nu\circ g_{m,n}.$ Hence, $|\nu\circ\frac{\alpha}{\alpha_m}-\nu|([n,n+1]) = |\nu\circ g_{m,n} - \nu|([0,1]).$ For $y\in[0,1]$ and $n\in\{-p_m,\ldots,2p_m-1\}$ we have $|\big(\frac{\alpha}{\alpha_m} - 1\big)(y+n)| \leq |\frac{\alpha}{\alpha_m} - 1|\leq 2q_mBm^{-q_m}.$ Thus, $|\nu\circ g_{m,n} - \nu|([0,1]) \leq \left(2q_mBm^{-q_m}\right)^\gamma = (2q_mB)^\gamma m^{-q_m\gamma}.$ We conclude that $|\mu-\mu^m|([-q_m,2q_m]) \leq 3p_m (2q_mB)^\gamma m^{-q_m\gamma}$ and therefore for arbitrary $C\geq 0$ $e^{Cq_m}|\mu-\mu^m|([-q_m,2q_m])\to 0 \quad(m\to \infty).$ Hence $\mu$ is a Gordon potential and $H:= -\Delta+\mu$ does not have any eigenvalues. \end{example} \subsection*{Acknowledgements} We want to thank Peter Stollmann for advices and hints and especially for providing a proof of Lemma \ref{lem:u'_konvergiert}. 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