\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 $0<c<1$ there is $\gamma\geq
0$ such that
    \[
\int_{\mathbb{R}} |u|^2\, d|\mu| \leq c \|u'\|_2^2 + \gamma
\|u\|_2^2 \quad(u\in W_2^1(\mathbb{R})).
\]
\end{lemma}

\begin{proof}
    For $\delta\in(0,1)$ and $n\in \mathbb{Z}$ we have
    \[
\|u\|_{\infty,[n\delta,(n+1)\delta]}^2 \leq 4 \delta
\|u'\|_{L_2(n\delta,(n+1)\delta)}^2 + \frac{4}{\delta}
\|u\|_{L_2(n\delta,(n+1)\delta)}^2\]
    by Sobolev's inequality.

    Now, we estimate
    \begin{align*}
        \int_{\mathbb{R}} |u|^2\, d|\mu|
& = \sum_{n\in \mathbb{Z}} \int_{n\delta}^{(n+1)\delta} |u|^2\, d|\mu| \\
& \leq \sum_{n\in \mathbb{Z}}
\|u\|_{\infty,[n\delta,(n+1)\delta]}^2
\|\mu\|_{\rm loc} \\
& \leq \|\mu\|_{\rm loc} \sum_{n\in \mathbb{Z}} \Big(4 \delta
\|u'\|_{L_2(n\delta,(n+1)\delta)}^2
+ \frac{4}{\delta} \|u\|_{L_2(n\delta,(n+1)\delta)}^2\Big)  \\
& = 4\delta\|\mu\|_{\rm loc} \|u'\|_2^2 + \frac{4\|\mu\|_{\rm
loc}}{\delta} \|u\|_2^2.
    \end{align*}
\end{proof}

Let $\mu$ be a Gordon measure and define
\[
    D(\tau)  := W_2^1(\mathbb{R}),\quad
    \tau(u,v)  := \int u'\overline{v}' + \int u\overline{v}\, d\mu.
\]
Then $\tau$ is a closed symmetric semibounded form. Let $H$ be the
associated self-adjoint operator.

In \cite{BenAmorRemling2005}, Ben Amor and Remling  introduced a
direct approach for defining the Schr\"odinger operator $H =
-\Delta + \mu$. Since we will use some of their results we sum up
the main ideas: For $u\in W_{1,{\rm loc}}^1(\mathbb{R})$ define
$Au\in L_{1,{\rm loc}}(\mathbb{R})$ by
\[
Au(x) := u'(x) - \int_{0}^x u(t)\, d\mu(t),
\]
where
\[
\int_{0}^x u(t)\, d\mu(t) := \begin{cases}
 \int_{[0,x]} u(t)\, d\mu(t) & \text{if } x\geq 0,\\
                -\int_{(x,0)} u(t)\, d\mu(t) & \text{if } x< 0.
 \end{cases}
\]
Clearly, $Au$ is only defined as an $L_{1,{\rm
loc}}(\mathbb{R})$-element. We define the operator $T$ in
$L_2(\mathbb{R})$ by
\[
 D(T)  := \big\{u\in L_2(\mathbb{R});\; u,Au\in W_{1,{\rm loc}}
^1(\mathbb{R}),\, (Au)'\in L_2(\mathbb{R})\big\},\quad
    Tu  := -(Au)'.
\]

\begin{lemma} \label{lem:Hsubseteq_T}
    $H\subseteq T$.
\end{lemma}

\begin{proof}
    Let $u\in D(H)$. Then $u\in W_2^1(\mathbb{R})\subseteq W_{1,{\rm loc}}^1(\mathbb{R})$ and
 $Au\in L_{1,{\rm loc}}(\mathbb{R})$. Let
$\varphi\in C_c^\infty(\mathbb{R})\subseteq D(\tau)$.
    Using Fubini's Theorem, we compute
    \begin{align*}
  \int_{\mathbb{R}} (Au)(x) \varphi'(x)\, dx
 & = \int_{\mathbb{R}} \Big(u'(x)
   - \int_{0}^x u(t)\, d\mu(t)\Big) \varphi'(x)\, dx \\
 & = \int_{\mathbb{R}} u'(x)\varphi'(x)\, dx
   - \int_{\mathbb{R}}\int_{0}^x u(t)\, d\mu(t) \varphi'(x)\, dx \\
 & = \int_{\mathbb{R}} u'(x)\varphi'(x)\, dx
    + \int_{-\infty}^0\int_{(-\infty,t)}\varphi'(x)\, dx  u(t)\, d\mu(t)\\
 &\quad  - \int_{0}^\infty\int_{[t,\infty)}\varphi'(x)\, dx
   u(t)\, d\mu(t)\\
 & = \int_{\mathbb{R}} u'(x)\varphi'(x)\, dx
   + \int_{-\infty}^0 u(t) \varphi(t)\, d\mu(t)
   + \int_{0}^\infty u(t)\varphi(t)\, d\mu(t)\\
& = \int_{\mathbb{R}} u'\varphi'
   + \int_{\mathbb{R}} u(x)\varphi(x)\, dx\\
 &= \tau(u,\overline{\varphi}) = (Hu \mid \overline{\varphi})
 = \int_\mathbb{R} Hu(x) \varphi(x)\, dx.
    \end{align*}
    Hence, $(Au)' = -Hu\in L_2(\mathbb{R})$. We conclude that
$Au\in W_{1,{\rm loc}}^1(\mathbb{R})$ and therefore $u\in D(T)$,
$Tu = -(Au)' = Hu$.
\end{proof}

\begin{remark} \label{rmk1.3} \rm
    For $u\in D(H)$ we obtain
\[
u'(x) = Au(x) + \int_{0}^x u(t)\, d\mu(t)
\]
for a.a. $x\in\mathbb{R}$. Since $Au\in W_{1,{\rm
loc}}^1(\mathbb{R})$ and $x\mapsto \int_{0}^x u(t)\, d\mu(t)$ is
continuous at all $x\in\mathbb{R}$ with $\mu(\{x\}) = 0$, $u'$ is
continuous at $x$ for all $x\in \mathbb{R}\setminus \mathrm{spt}
\mu_p$, where $\mu_p$ is the point measure part of $\mu$.
\end{remark}

\section{Absence of eigenvalues} \label{sec:solution_estimates}


We show that $H$ has no eigenvalues. The proof is based on two
observations. The first one is a stability result and will be
achieved in Lemma \ref{lem:expest}, the second one is an estimate
of the solution for periodic measure perturbations, see Lemma
\ref{lem:estimate}.

As in \cite{DamanikStolz2000} we start with a Gronwall Lemma, but
in a more general version for locally finite measures. For the
proof, see \cite{EthierKurtz2005}.

\begin{lemma}[Gronwall]\label{lem:gronwall}
 Let $\mu$ be a locally finite Borel measure on $[0,\infty)$,
$u\in \mathcal{L}_{1,{\rm loc}}([0,\infty),\mu)$ and
$\alpha:[0,\infty)\to [0,\infty)$ measurable. Suppose, that
\[
u(x) \leq \alpha(x) + \int_{[0,x]} u(s)\, d\mu(s) \quad(x\geq0).
\]
Then
\[
u(x) \leq \alpha(x) +
\int_{[0,x]}\alpha(s)\exp\bigl(\mu([s,x])\bigr)\,d\mu(s) \quad
(x\geq0).
\]
\end{lemma}

For $x\in\mathbb{R}$ we abbreviate
\[
I_x:= [x\wedge0,x\vee0], \quad I_x(t):= I_x\cap([t,x]\cup[x,t])
\quad(t\in\mathbb{R}).
\]
Let $\mu$ be uniformly locally bounded. Then
\[
|\mu|(I_x) \leq (|x|+1)\|\mu\|_{\rm loc} \quad(x\in \mathbb{R}).
\]
Furthermore, if $\mu$ is periodic and locally bounded, $\mu$ is
uniformly locally bounded.

Let $H := -\Delta + \mu$ and $E\in \mathbb{R}$. Then $u\in
W_{1,{\rm loc}}^1(\mathbb{R})$ ($=D(A)$) is a \emph{solution} of
$Hu=Eu$, if $-(Au)' = Eu$ in the sense of distributions (i.e., $u$
satisfies the eigenvalue equation but without being an
$L_2$-function).

\begin{lemma} \label{lem:cont}
    Let $\mu_1,\mu_2$ be two uniformly locally bounded measures,
$E\in \mathbb{R}$ and $u_1$ and $u_2$ solutions of
\[
H_1 u_1 = E u_1,\quad  H_2 u_2 = E u_2
\]
subject to
    \[
u_1(0) = u_2(0),\quad u_1'(0{\scriptstyle +}) =
u_2'(0{\scriptstyle +}),\quad
 |u_1(0)|^2 + |u_1'(0{\scriptstyle +})|^2 = 1.
\]
Then there are $C_0,C\geq 0$ such that for all $x\in\mathbb{R}$
    \begin{align*}
        & \Big\|\begin{pmatrix}
                u_1(x) \\
            u_1'(x)
            \end{pmatrix}
        -  \begin{pmatrix}
                u_2(x) \\
            u_2'(x)
            \end{pmatrix} \Big\|  \\
        & \leq C_0 + \int_{I_x} |u_2(t)|\, d
|\mu_1-\mu_2|(t) \\
        &\quad +  C  \int_{I_x}  \Big( C_0 + \int_{I_t}  |u_2| d |\mu_1-\mu_2|\Big)e^{C(\lambda + |\mu_1 - E\lambda|)(I_x(t))} \, d(\lambda + |\mu_1 - E\lambda|)(t).
    \end{align*}
\end{lemma}

\begin{proof}
    Write
    \[
u_1(x) - u_2(x) = \int_{0}^x (u_1'(t) - u_2'(t))\, dt
\]
    and
\begin{align*}
u_1'(x) - u_2'(x) & = u_1'(0{\scriptstyle +}) -
u_2'(0{\scriptstyle +})
  - \left(u_1(0)\mu_1(\{0\}) - u_2(0)\mu_2(\{0\})\right) \\
&\quad + \int_0^x u_1(t)\, d\mu_1(t) - \int_0^x u_2(t)\, d\mu_2(t) - \int_0^x E(u_1(t) - u_2(t))\, dt \\
& = u_2(0)\left(\mu_2(\{0\}) - \mu_1(\{0\})\right) \\
& \quad + \int_0^x u_2(t)\, d(\mu_1-\mu_2)(t)
 + \int_0^x (u_1(t)-u_2(t))\, d(\mu_1-E\lambda)(t).
\end{align*}
    Hence,
    \begin{align*}
        \begin{pmatrix}
            u_1(x) - u_2(x)\\
            u_1'(x) - u_2'(x)
        \end{pmatrix}
& =  \begin{pmatrix} 0 \\ u_2(0)\left(\mu_2(\{0\})
- \mu_1(\{0\})\right)\end{pmatrix} + \int_0^x \begin{pmatrix} 0 \\ u_2(t) \end{pmatrix} d(\mu_1-\mu_2)(t) \\
&\quad  + \int_{0}^x \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} u_1(t) - u_2(t) \\ u_1'(t) - u_2'(t)\end{pmatrix}
\, d\begin{pmatrix} \lambda \\ \mu_1 - E\lambda \end{pmatrix}(t).
    \end{align*}
    We conclude, that
    \begin{align*}
         \Big\|\begin{pmatrix}
                u_1(x) \\
            u_1'(x)
            \end{pmatrix}
        - \begin{pmatrix}
                u_2(x) \\
            u_2'(x)
            \end{pmatrix}\Big\|
 & \leq C_0 + \int_{I_x} |u_2(t)|\, d |\mu_1-\mu_2|(t) \\
 & \quad + C\int_{I_x} \Big\|\begin{pmatrix}
                u_1(t) \\
            u_1'(t)
            \end{pmatrix}
        - \begin{pmatrix}
                u_2(t) \\
            u_2'(t)
            \end{pmatrix}\Big\|
\, d (\lambda + |\mu_1 - E\lambda|)(t).
    \end{align*}
    An application of Lemma \ref{lem:gronwall} with
$\alpha(x) = C_0 + \int_{I_x} |u_2(t)|\, d |\mu_1-\mu_2|(t)$ and
$\mu = C(\lambda + |\mu_1 - E\lambda|)$ yields the assertion.
\end{proof}

\begin{remark} \label{rem:further_est} \rm
    Regarding the proof of Lemma \ref{lem:cont} we can further
estimate $C_0 \leq |u_2(0)||\mu_1-\mu_2|(I_x)$ ($x\in\mathbb{R}$).
\end{remark}

\begin{lemma} \label{lem2.4}
    Let $E\in \mathbb{R}$ and $u_0$ be a solution of $-\Delta u_0 = E u_0$. Then
    there is $C\geq 0$ such that $|u_0(x)| \leq Ce^{C|x|}$ for all
    $x\in \mathbb{R}$.
\end{lemma}

In the following lemmas and proofs the constant $C$ may change
from line to line, but we will always state the dependence on the
important quantities.

\begin{lemma} \label{lem:periodic_solution}
    Let $\mu_1$ be a locally bounded $p$-periodic measure,
$E\in\mathbb{R}$, $u_1$ a solution of $H_1u_1 = E u_1$. Then there
is    $C\geq0$ such that
\[
|u_1(x)| \leq Ce^{C|x|} \quad(x\in\mathbb{R}).
\]
\end{lemma}

\begin{proof}
    Let $u_0$ be a solution of $-\Delta u_0 = E u_0$ subject to
the same boundary conditions at $0$ as $u_1$.
    By Lemma \ref{lem:cont} we have
    \begin{align*}
        & |u_1(x)-u_0(x)| \\
        & \leq C + \int_{I_x} |u_0(t)|\, d |\mu_1|(t) \\
        &\quad + C\int_{I_x} \Big(C + \int_{I_t} |u_0(s)|\,
 d |\mu_1|(s)\Big)e^{C(\lambda
 + |\mu_1 - E\lambda|)(I_x(t))}\, d(\lambda + |\mu_1 - E\lambda|)(t) \\
& \leq  C + |\mu_1|({I_x})Ce^{C|x|} \\
&\quad + \int_{I_x} \left(C + C|\mu_1|({I_t})e^{C|t|}\right)
 e^{(\lambda + |\mu_1 - E\lambda|)(I_x(t))}\, d(\lambda
+ |\mu_1 - E\lambda|)(t) \\
        & \leq \left(C + C|\mu_1|({I_x})e^{C|x|}\right)
\left(1+e^{(\lambda + |\mu_1 - E\lambda|)({I_x})} (\lambda +
|\mu_1 - E\lambda|)({I_x})\right).
    \end{align*}
Since $\mu_1$ is periodic and locally bounded it is uniformly
locally bounded and we have
    $|\mu_1|({I_x}) \leq (|x|+1)\|\mu_1\|_{\rm loc}$.
Furthermore,     $\mu_1-E\lambda$ is periodic and uniformly
locally bounded, so $|\mu_1
    - E\lambda|)({I_x}) \leq (|x|+1)\|\mu_1-E\lambda\|_{\rm loc}$.
We    conclude that
    \begin{align*}
 |u_1(x)-u_0(x)|
    & \leq \Big(C + C(|x|+1)\|\mu_1\|_{\rm loc}e^{C|x|}\Big) \\
    & \quad\times \left(1+e^{(|x|+1)(1+\|\mu_1-E\lambda\|_{\rm loc})}(|x|+1)(1+\|\mu_1-E\lambda\|_{\rm loc})\right) \\
    & \leq C e^{C|x|},
    \end{align*}
    where $C$ is depending on $E$, $\|\mu_1\|_{\rm loc}$ and
$\|\mu_1-E\lambda\|_{\rm loc}$.
    Hence,
 \[
|u_1(x)| \leq |u_1(x)-u_0(x)| + |u_0(x)| \leq Ce^{C|x|}.
\]
\end{proof}

\begin{lemma} \label{lem:expest}
    Let $\mu$ be a Gordon measure and $(\mu^{m})$ the $p_m$-periodic
    approximants, $E\in\mathbb{R}$. Let $u$ be a solution of $Hu = E u$, $u_m$ a
    solution of $H_mu_m = E u_m$ for $m\in\mathbb{N}$ (obeying the same boundary
    conditions at $0$).
    Then there is $C\geq 0$ such that
    \[\Big\|\begin{pmatrix}
                u(x) \\
            u'(x)
            \end{pmatrix}
        -
        \begin{pmatrix}
                u_m(x) \\
            u_m'(x)
            \end{pmatrix}\Big\|
 \leq Ce^{C|x|} |\mu - \mu^{m}|(I_x) \quad(x\in\mathbb{R}).
\]
\end{lemma}

\begin{proof}
    By Lemma \ref{lem:cont} and Remark \ref{rem:further_est} we know that
    \begin{align*}
 \Big\|\begin{pmatrix}
                u(x) \\
            u'(x)
            \end{pmatrix}
        -   \begin{pmatrix}
                u_m(x) \\
            u_m'(x)
            \end{pmatrix}\Big\|
& \leq |u_m(0)||\mu - \mu^m|(I_x)
+ \int_{I_x} |u_m(t)|\, d |\mu-\mu^{m}|(t) \\
        &\quad + C \int_{I_x} \Big(|u_m(0)||\mu - \mu^m|(I_t)
+ \int_{I_t} |u_m|d |\mu-\mu^{m}|\Big) \\
        & \quad  \times e^{C(\lambda
+ |\mu - E\lambda|)(I_x(t))}\, d(\lambda + |\mu - E\lambda|)(t).
    \end{align*}
We have
    \[
M:= \sup_{m\in \mathbb{N}} \|\mu^{m}\|_{\rm loc} < \infty,
\]
    since $(\mu^{m})$ approximates $\mu$. Hence, also
    \[
\sup_{m\in \mathbb{N}} \|\mu^{m}-E\lambda\|_{\rm loc} < \infty
\]
    and Lemma \ref{lem:periodic_solution} yields
    \[
|u_m(x)| \leq Ce^{C|x|},
\]
    where $C$ can be chosen independently of $m$. Therefore
    \begin{align*}
 \Big\|\begin{pmatrix}
                u(x) \\
            u'(x)
            \end{pmatrix}
        -  \begin{pmatrix}
                u_m(x) \\
            u_m'(x)
            \end{pmatrix}\Big\|
        & \leq \Big(Ce^{C|x|}|\mu-\mu^{m}|({I_x})
+ Ce^{C|x|}|\mu-\mu^{m}|({I_x})\Big) \\
        & \quad  \times \Big(1+e^{C(\lambda
+ |\mu - E\lambda|)(I_x)}(\lambda + |\mu - E\lambda|)({I_x})
\Big).
    \end{align*}
Since
    \[
|\mu - E\lambda|({I_x}) \leq (|x|+1)\|\mu-E\lambda\|_{\rm loc},
\]
    we further estimate
\[
\Big\|\begin{pmatrix}
                u(x) \\
            u'(x)
            \end{pmatrix}
        -
        \begin{pmatrix}
                u_m(x) \\
            u_m'(x)
            \end{pmatrix}\Big\|
 \leq Ce^{C|x|} |\mu-\mu^{m}|({I_x})
\]
    where $C$ is depending on $\|\mu-E\lambda\|_{\rm loc}$
(and of course on $M$, $\|\mu\|_{\rm loc}$ and $E$).
\end{proof}

Lemma \ref{lem:expest} can be regarded as a stability (or
continuity)  result: if the measures converge in total variation,
the corresponding solutions converge as well.

Now, we focus on periodic measures and estimate the solutions.
This will then be applied to the periodic approximations of our
Gordon measure $\mu$.

\begin{remark} \label{rmk2.7} \rm
    \begin{itemize}
        \item[(a)]
            Let $f,g$ be two solutions of the equation $Hu = Eu$. Define their Wronskian by $W(f,g)(x):= f(x)g'(x{\scriptstyle +}) - f'(x{\scriptstyle +})g(x)$. By \cite{BenAmorRemling2005}, Proposition 2.5, $W(f,g)$ is constant.
       \item[(b)]
            Let $u$ be a solution of the equation $Hu = Eu$.
Define the transfer matrix $T_E(x)$ mapping
$(u(0),u'(0{\scriptstyle +}))^\top$
 to  $(u(x),u'(x{\scriptstyle +}))^\top$. Consider now the two solutions $u_N$, $u_D$
            subject to
\[
                \begin{pmatrix} u_N(0)\\u_N'(0{\scriptstyle +})\end{pmatrix}
= \begin{pmatrix}
                1 \\ 0  \end{pmatrix},\quad
\begin{pmatrix}
  u_D(0)\\u_D'(0{\scriptstyle +})\end{pmatrix}
= \begin{pmatrix}
   0 \\ 1  \end{pmatrix}.
\]
Then
            \[
T_E(x) = \begin{pmatrix}
                        u_N(x) & u_D(x) \\
                    u_N'(x{\scriptstyle +}) & u_D'(x{\scriptstyle +})
                       \end{pmatrix}.
\]
 We obtain $\det T_E(x) = W(u_N,u_D)(x)$ and $\det T_E$ is constant,
hence equals $1$ for all $x\in \mathbb{R}$.
    \end{itemize}
\end{remark}


\begin{lemma} \label{lem:estimate}
    Let $\mu$ be $p$-periodic and $E\in \mathbb{R}$. Let $u$ be a
solution of $Hu = E u$ subject to
    \[
|u(0)|^2 + |u'(0{\scriptstyle +})|^2 = 1.
\]
    Then
    \[
\max\Big\{\Big\|\begin{pmatrix} u(-p) \\u'(-p{\scriptstyle +})
 \end{pmatrix}
\Big\|,
    \Big\|\begin{pmatrix} u(p) \\u'(p{\scriptstyle +})\end{pmatrix}
 \Big\|,
 \Big\|\begin{pmatrix}
    u(2p) \\
 u'(2p{\scriptstyle +})\end{pmatrix}\Big\|\Big\}
\geq \frac{1}{2}.
\]
\end{lemma}

The proof of this lemma is completely analoguous to the proof
 of \cite[Lemma 2.2]{DamanikStolz2000}.

\begin{lemma} \label{lem:u'_konvergiert}
    Let $v\in L_2\cap BV_{\rm loc}(\mathbb{R})$ and assume that for all $r>0$ 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}. Many thanks to the anonymous referee for
the useful comments.


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\end{document}