\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 41, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/41\hfil Inverse problems]
{Inverse problems associated with the Hill operator}

\author[A. A. Kira\c{c} \hfil EJDE-2016/41\hfilneg]
{Alp Arslan Kira\c{c}}

\address{Alp Arslan Kira\c{c} \newline
Department of Mathematics,
Faculty of Arts and Sciences,
Pamukkale University,
20070, Denizli, Turkey}
\email{aakirac@pau.edu.tr}

\thanks{Submitted April 29, 2015. Published January 27, 2016.}
\subjclass[2010]{34A55, 34B30, 34L05, 47E05, 34B09}
\keywords{Hill operator; inverse spectral theory; eigenvalue asymptotics;
 \hfill\break\indent  Fourier coefficients}

\begin{abstract}
 Let $\ell_n$ be the length of the $n$-th instability interval of the Hill
 operator $Ly=-y''+q(x)y$.  We prove that if $\ell_n=o(n^{-2})$ and the set
 $\{(n\pi)^2: n \text{ is even and } n>n_0\}$ is a subset of the periodic 
 spectrum of the Hill operator, then $q=0$ a.e., where $n_0$ is a sufficiently
 large positive integer such that $\ell_n<\varepsilon n^{-2}$ for all 
 $n>n_0(\varepsilon)$ with some $\varepsilon>0$.
 A similar result holds for the anti-periodic case.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

Consider the Hill  operator
\begin{equation}  \label{1}
Ly=-y''+q(x)y,
\end{equation}
where $q(x)$ is a real-valued summable function on $[0,1]$ and $q(x+1)=q(x)$. 
Let  $\lambda_n$ and $\mu_n$ $(n=0,1,\ldots)$ denote, respectively, the
$n$-th periodic and anti-periodic eigenvalues of the Hill operator \eqref{1}
on $[0,1]$ with the periodic boundary conditions
\begin{equation}\label{per.bc}
    y(0)=y(1),\quad  y'(0)=y'(1),
\end{equation}
 and the anti-periodic boundary conditions
\[
    y(0)=-y(1),\quad y'(0)=-y'(1).
\]
It is well-known  \cite{Coddington:Levinson,Eastham} that
\[  
\lambda_0<\mu_0\leq
\mu_1<\lambda_1\leq\lambda_2<\mu_2%
\leq \mu_3<\dots \to \infty.
\]
The intervals $(\mu_{2m},\mu_{2m+1})$ and $(\lambda_{2m+1},\lambda_{2m+2})$
are respectively referred to as  the $(2m+1)$-th and $(2m+2)$-th finite 
instability intervals of the operator $L$, while
$(-\infty,\lambda_0)$ is called the zero-th instability interval.
The length of the $n$-th instability interval of \eqref{1}
will be denoted by $\ell_n$ ($n=2m+1,\,2m+2$).
 For further background see \cite{Magnus-Winkler, Marchenko, Hochstadt:determination}.

Borg \cite{Borg}, Ungar \cite{Ungar} and Hochstadt \cite{Hochstadt:determination} 
proved independently of each other the following statement:
\begin{quote}
If $q(x)$ is real and integrable, and if all finite instability intervals
vanish then $q(x)=0$ a.e.
\end{quote}
Hochstadt \cite{Hochstadt:determination} showed that if precisely one of 
the finite instability intervals does not vanish, then $q(x)$ is the 
elliptic function
which satisfies 
\[
q''=3q^2+Aq+B \quad \text{a.e.},
\]
where $A$ and $B$ are suitable constants. 
Hochstadt \cite{Hochstadt:determination} also proved that $q(x)$ 
is infinitely differentiable a.e. when
$n$ finite instability intervals fail to vanish. For more results 
see \cite{Goldberg:determination, Goldberg:necessary, 
Goldberghoch:selected, Goldberghoch:finitenumber}.

Furthermore, Hochstadt \cite{Hochstadt:Stability-Estimate} proved that the 
lengths of the instability intervals $\ell_n$
vanish faster than any power of $(1/n)$ for $q$ in $C_1^{\infty}$.
McKean and Trubowitz \cite{McKean} established the converse: if $q$ 
is in $L_1^2$, the space of  1-periodic square integrable functions in
$[0,1]$, and the length of the $n$-th instability interval for $n\geq 1$ 
is rapidly decreasing, then $q$ is in $C_1^{\infty}$.
Later Trubowitz \cite{Trubowitz} proved the following: if $q$ is real analytic,
 the lengths of the instability intervals are exponentially decreasing. 
Conversely if $q$ is in $L_1^2$ and the lengths of the instability
intervals are exponentially decreasing, $q$ is real analytic.  
Denoting the Fourier coefficients of q by
\begin{equation}  \label{cn}
c_n=:(q, \exp(i2n\pi \cdot))_{L^2([0,1];dx)},\quad n\in \mathbb{N}\cup\{0\},
\end{equation}
 Coskun \cite{Coskun:invers} showed that
\begin{equation}\label{coskun}
\text{if $\ell_n=O(n^{-2})$, then $c_n=O(n^{-2})$  as $n\to \infty$}.
\end{equation}

At this point, we refer to some Ambarzumyan-type theorems in 
\cite{Ambarz, corri, Ambarzcoupled, anoteinver}. 
In 1929, Ambarzumyan \cite{Ambarz} obtained the following first theorem 
in inverse spectral theory:
If $\{n^2 : n = 0,1,\ldots\}$ is the spectrum of the Sturm-Liouville operator
 \eqref{1} on $[0,1]$ with the Neumann
boundary conditions, then $q = 0$ a.e. 
In \cite{corri}, they extended the classical Ambarzumyan's theorem 
for the Sturm-Liouville equation to the general separated boundary
conditions,  by imposing an additional condition on the potential function, 
and their result supplements the P{\"{o}}schel-Trubowitz inverse 
spectral theory (see \cite{Poschel}). In \cite{Ambarzcoupled}, 
based on the well-known extremal property of the first eigenvalue, 
they find two analogs of Ambarzumyan's theorem to the Sturm-Liouville 
systems of $n$ dimension under periodic or anti-periodic boundary conditions. 
In \cite{anoteinver}, by using the Rayleigh-Ritz inequality and imposing 
a condition on the second term in the Fourier cosine series (see \eqref{cond}), 
they proved the following Ambarzumyan-type theorem:
\begin{itemize}
\item[(a)] If all periodic eigenvalues of Hill's equation \eqref{1} are
 nonnegative and they include $\{(2m\pi)^2: m\in \mathbb{N}\}$, then $q=0$ a.e.

\item[(b)] If all anti-periodic eigenvalues of Hill's equation \eqref{1} 
are not less than $\pi^2$ and they include
$\{(2m-1)^2\pi^2: m\in \mathbb{N}\}$, and
\begin{equation}  \label{cond}
\int_0^{1}q(x)\cos(2\pi x)\,dx\geq 0,
\end{equation} 
then $q=0$ a.e.
\end{itemize}

Recently, in \cite{kırac:ambars},  we obtain the classical Ambarzumyan's theorem 
for the Sturm-Liouville
operators with $q\in L^{1}[0,1]$ and quasi-periodic boundary conditions
in cases when there is not any additional condition on the potential $q$ 
such as \eqref{cond}.

In this paper, we prove the following inverse spectral result, more
precisely, a uniqueness-type result of the following form:

\begin{theorem} \label{main0}
 Denote the $n$th instability interval by $\ell_n$, and suppose that  
$\ell_n=o(n^{-2})$ as $n\to \infty$. Then the following two assertions hold:
\begin{itemize}
\item[(i)] If  $\{(n\pi)^2: \text{$n$ even and $n>n_0$}\}$
is a subset of the periodic spectrum of the Hill operator then $q=0$ 
a.e. on $(0,1)$;

\item[(ii)] If  $\{(n\pi)^2: \text{$n$ odd and $n>n_0$}\}$ is a
subset of the anti-periodic spectrum of the Hill operator then $q=0$ a.e. 
on $(0,1)$.
\end{itemize}
Given $\varepsilon>0$, thee exists $n_0=n_0(\varepsilon)\in \mathbb{N}$,
a sufficiently large positive integer such that
\[
\ell_n<\varepsilon n^{-2}\quad \text{for all $n>n_0(\varepsilon)$.}
\]
\end{theorem}

Theorem \ref{main0} is deduced from the following result.

\begin{theorem} \label{thm1.2} 
Denote the Fourier coefficients of $q$ by $c_n$ (see \eqref{cn}),
 and assume $\ell_n=o(n^{-2})$. Then $c_n=o(n^{-2})$  as $n\to \infty$.
\end{theorem}

Note that, from Theorem \ref{thm1.2}, the assertion in \eqref{coskun}
holds with the improved  $o$-terms $o(n^{-2})$. 
In Ambarzumyan-type theorems, it is necessary to specify the whole
spectrum. In \cite{FreilingYurko}, they proved that it is enough to know 
the first eigenvalue only. Unlike the above works, to prove of Theorem 
\ref{main0}, we have information only on the sufficiently large eigenvalues
 of  the spectrum of the Hill operator. Also, the proof does not depend 
on multiplicities of the given eigenvalues.

\section{Preliminaries and proof of main results}\label{asyl}

We shall consider only the periodic (for even $n$) eigenvalues
of the Hill operator. The anti-periodic (for odd $n$)
problem is completely similar. It is well known \cite[Theorem 4.2.3]{Eastham} 
that the periodic eigenvalues $\lambda_{2m+1}, \lambda_{2m+2}$ are
asymptotically located in pairs such that
\begin{equation}\label{asy2}
    \lambda_{2m+1}=\lambda_{2m+2}+o(1)=(2m+2)^2\pi^2+o(1)
\end{equation}
for sufficiently large $m$. From this formula, for all $k\neq 0,(2m+2)$ and  
$k\in \mathbb{Z}$, the inequality
\begin{equation}  \label{dist1}
|\lambda-(2(m-k)+2)^2\pi^2|>|k||(2m+2)-k|>C\,m,
\end{equation}
is satisfied by both eigenvalues $\lambda_{2m+1}$ and $\lambda_{2m+2}$ 
for large $m$, where, here and in subsequent relations, 
$C$ denotes a positive constant whose exact value is not essential.
 Note that, when $q=0$, the system $\{e^{-i(2m+2)\pi x}, e^{i(2m+2)\pi x}\}$ 
is a basis of the eigenspace corresponding to the double eigenvalues
$(2m+2)^2\pi^2$ of the problem \eqref{1}-\eqref{per.bc}.

To obtain the asymptotic formulas for the periodic eigenvalues
$\lambda_{2m+1}, \lambda_{2m+2}$ corresponding to the normalized
eigenfunctions $\Psi_{m,1}(x),\Psi_{m,2}(x)$ respectively, 
let us consider the well-known relation, for sufficiently large $m$,
\begin{equation}  \label{m1}
\Lambda_{m,j,m-k}(\Psi_{m,j},e^{i(2(m-k)+2)\pi
x})=(q\,\Psi_{m,j},e^{i(2(m-k)+2)\pi x}),
\end{equation}
where $\Lambda_{m,j,m-k}=(\lambda_{2m+j}-(2(m-k)+2)^2\pi^2)$, $j=1,2.$
The relation \eqref{m1} can be obtained from the equation \eqref{1} 
by multiplying  $e^{i(2(m-k)+2)\pi x}$ and using integration by parts. 
From \cite[Lemma 1]{Melda.O}, to iterate \eqref{m1} for $k=0$, in 
the right hand-side of formula \eqref{m1}, we
use the following relations
\begin{gather}  \label{m2}
(q\,\Psi_{m,j},e^{i (2m+2)\pi
x})=\sum_{m_1=-\infty}^{\infty}c_{m_1}(\Psi_{m,j},e^{i
(2(m-m_1)+2)\pi x}), \\
 \label{m3}
|(q\,\Psi_{m,j},e^{i (2(m-m_1)+2)\pi x})|< 3M
\end{gather}
for all large $m$, where $j=1,2$ and $M=\sup_{m\in \mathbb{Z}}|c_{m}|$.

First, we fix the terms with indices $m_1=0,(2m+2)$.
Then all the other terms in the right hand-side of \eqref{m2} are replaced, 
in view of \eqref{dist1} and \eqref{m1} for $k=m_1$, by
\[
c_{m_1}\frac{(q\,\Psi_{m,j},e^{i(2(m-m_1)+2)\pi x})}{\Lambda_{m,j,m-m_1}}.
\]

In the same way, by applying the above process for the eigenfunction 
$e^{-i(2m+2)\pi x}$ corresponding to the eigenvalue $(2m+2)^2\pi^2$ of
the problem \eqref{1}-\eqref{per.bc} for $q=0$, we obtain the following 
lemma (see also Section 2 in \cite{kyrac:abstract, kyrac:titch}).

\begin{lemma}\label{lemmaitera}
The following relations hold for sufficiently large $m$:
(i)
\begin{equation}\label{m4123}
[\Lambda_{m,j,m}- c_0-\sum_{i=1}^2a_i(\lambda_{2m+j})]u_{m,j}
=[c_{2m+2}+\sum_{i=1}^2b_i(\lambda_{2m+j})]v_{m,j}+R_2,
\end{equation}
where $j=1,2$,
\begin{gather}
u_{m,j}=(\Psi_{m,j},e^{i(2m+2)\pi x}),\quad 
v_{m,j}=(\Psi_{m,j},e^{-i(2m+2)\pi x}), \nonumber\\
\label{a1a2} \begin{gathered} 
a_1(\lambda_{2m+j})=\sum_{m_1}\frac{c_{m_1}c_{-m_1}}{\Lambda_{m,j,m-m_1}},\\
a_2(\lambda_{2m+j})=\sum_{m_1,m_2}\frac{c_{m_1}c_{m_2}
c_{-m_1-m_2}}{\Lambda_{m,j,m-m_1}\Lambda_{m,j,m-m_1-m_2}},
\end{gathered} \\
b_1(\lambda_{2m+j})=\sum_{m_1}\frac{c_{m_1}c_{2m+2-m_1}}{\Lambda_{m,j,m-m_1}},
\nonumber\\
b_2(\lambda_{2m+j})=\sum_{m_1,m_2}\frac{c_{m_1}c_{m_2}c_{2m+2-m_1-m_2}}
{\Lambda_{m,j,m-m_1} \Lambda_{m,j,m-m_1-m_2}}, \nonumber \\
\label{R}
R_2=\sum_{m_1,m_2,m_3}\frac{c_{m_1}c_{m_2}c_{m_3}
(q\Psi_{m,j}(x),e^{i(2(m-m_1-m_2-m_3)+2)\pi x})}{\Lambda_{m,j,m-m_1}\,
\Lambda_{m,j,m-m_1-m_2}\Lambda_{m,j,m-m_1-m_2-m_3}}. \nonumber
\end{gather}
The summations  in these formulas are taken over all integers $m_1,m_2, m_3$
such that $m_1, m_1+m_2, m_1+m_2+m_3\neq 0,\,2m+2$.

(ii)
\begin{equation}\label{m412}
[\Lambda_{m,j,m}-c_0-\sum_{i=1}^2 a'_i(\lambda_{2m+j})]v_{m,j}
=[c_{-2m-2}+\sum_{i=1}^2b'_i(\lambda_{2m+j})]u_{m,j}+R'_2,
\end{equation}
where $j=1,2$,
\begin{gather*}
a'_1(\lambda_{2m+j})=\sum_{m_1}\frac{c_{m_1}c_{-m_1}}{\Lambda_{m,j,m+m_1}},\\
a'_2(\lambda_{2m+j})=\sum_{m_1,m_2}\frac{c_{m_1}c_{m_2}c_{-m_1-m_2}}
{\Lambda_{m,j,m+m_1}\,\Lambda_{m,j,m+m_1+m_2}}, \\
b'_1(\lambda_{2m+j})=\sum_{m_1}\frac{c_{m_1}c_{-2m-2-m_1}}{\Lambda_{m,j,m+m_1}},\\
b'_2(\lambda_{2m+j})=\sum_{m_1,m_2}\frac{c_{m_1}c_{m_2}c_{-2m-2-m_1-m_2}}
{\Lambda_{m,j,m+m_1}\,\Lambda_{m,j,m+m_1+m_2}},
\end{gather*}
\begin{equation}\label{R'}
R'_2=\sum_{m_1,m_2,m_3}\frac{c_{m_1}c_{m_2}c_{m_3}
(q\Psi_{m,j}(x),e^{i(2(m+m_1+m_2+m_3)+2)\pi x})}{\Lambda_{m,j,m+m_1}
\Lambda_{m,j,m+m_1+m_2}\,\Lambda_{m,j,m+m_1+m_2+m_3}}
\end{equation}
and the sums in these formulas are taken over all integers 
$m_1,m_2, m_3$ such that $m_1, m_1+m_2, m_1+m_2+m_3\neq 0,\,-2m-2$.
\end{lemma}

Note that, by substituting $m_1=-k_1$ and $m_1+m_2=-k_1$, $m_2=k_2$ into the
relations $a'_1(\lambda_{2m+j})$ and $a'_2(\lambda_{2m+j})$ respectively, we have
\begin{equation}\label{a1=a1}
a_i(\lambda_{2m+j})=a'_i(\lambda_{2m+j})\quad \text{for $i=1,2$.}
\end{equation}
  Here, using the equality
\[
  \frac{1}{m_1(2m+2-m_1)}=\frac{1}{2m+2}\Big(\frac{1}{m_1}+\frac{1}{2m+2-m_1}\Big),
\]
we obtain the relation
\[
\sum_{m_1\neq 0,(2m+2)}\frac{1}{|m_1(2m+2-m_1)|}=O\Big(\frac{\ln|m|}{m}\Big).
\]
This, together with \eqref{dist1}, \eqref{m1} and \eqref{m3}, gives the 
following estimates (see, respectively, \eqref{R} and \eqref{R'} for $R_2$ and $R'_2$)
\begin{equation}\label{m45}
R_2,\; R'_2=O\Big(\big(\frac{\ln| m|}{m}\big)^3\Big).
\end{equation}
Moreover, in view of \eqref{dist1}, \eqref{m1} and \eqref{m3}, we obtain
 (see  \cite[Theorem 2]{Melda.O}, \cite{kyrac:titch})
\begin{equation}\label{kare}
\sum_{k\in \mathbb{Z};\,k\neq \pm(m+1)}
\big|(\Psi_{m,j},e^{i2k\pi x})\big|^2=O\Big(\frac{1}{m^2}\Big).
\end{equation}
Therefore, the expansion of the normalized eigenfunctions 
$\Psi_{m,j}(x)$ by the orthonormal basis $\{e^{i2k\pi x}:k\in \mathbb{Z}\}$ 
on $[0,1]$ has the form
\begin{equation}\label{m7}
\Psi_{m,j}(x)=u_{m,j}\,e^{i(2m+2)\pi x}+v_{m,j}\,e^{-i(2m+2)\pi x}+h_{m}(x),
\end{equation}
where
\begin{equation}\label{m8}
\begin{gathered}
 (h_{m},e^{\mp i(2m+2)\pi x})=0,\quad \|h_{m}\|=O(m^{-1}),\\
 \sup_{x\in[0,1]}|h_{m}(x)|=O\Big(\frac{\ln|m|}{m}\Big), \quad
  |u_{m,j}|^2+|v_{m,j}|^2=1+O\big(m^{-2}\big).
\end{gathered}
\end{equation}

\subsection*{Proof of Theorem \ref{thm1.2}}

First, let us estimate the expressions in \eqref{m4123} and \eqref{m412}.
From \eqref{asy2}, \eqref{dist1} and \eqref{kare}, one can readily see that
\begin{equation}\label{dif}
\begin{aligned}
&\sum_{m_1\neq 0,\pm(2m+2)}
\big|\frac{1}{\Lambda_{m,j,m\mp m_1}}-\frac{1}{\Lambda_{m,0,m\mp m_1}}\big| \\
&\leq C|\Lambda_{m,j,m}|\sum_{m_1\neq 0,\pm(2m+2)}|m_1|^{-2}|2m+2\mp m_1|^{-2}
=o\big(m^{-2}\big),
\end{aligned}
\end{equation}
where $\Lambda_{m,0,m\mp m_1}=((2m+2)^2\pi^2-(2(m\mp m_1)+2)^2\pi^2)$.
Thus, we obtain
\begin{equation}\label{a1}
    a_i(\lambda_{2m+j})=a_i((2m+2)^2\pi^2)+o\big(m^{-2}\big)
\quad\text{for } i=1,2.
\end{equation}
Here, by  \eqref{dif}, we also have, arguing as in 
\cite[Lemma 3]{kyrac:titch} (see also \cite[Lemma 6]{Veliev:Shkalikov}),
\begin{equation}\label{I0}
\begin{aligned}
b_1(\lambda_{2m+j})
&=\frac{1}{4\pi^2}\sum_{m_1\neq 0,(2m+2)}
 \frac{c_{m_1}c_{2m+2-m_1}}{m_1(2m+2-m_1)}+o\big(m^{-2}\big)\\
&=-\int_0^{1}(Q(x)-Q_0)^2\,e^{-i2(2m+2)\pi x}dx+o\big(m^{-2}\big)
\\
&=\frac{-1}{i2\pi(2m+2)}\int_0^{1}2(Q(x)-Q_0)\,q(x)\,e^{-i2(2m+2)\pi x}dx
+o\big(m^{-2}\big),
\end{aligned}
\end{equation}
where  
\begin{equation} \label{Q0}
\begin{gathered}
Q(x)-Q_0=\sum_{m_1\neq 0}Q_{m_1}\,e^{i2m_1\pi x}, \\
 Q_{m_1}=:(Q(x),e^{i2m_1\pi x})=\frac{c_{m_1}}{i2\pi m_1}, \quad
m_1\neq 0,
\end{gathered}
\end{equation}
are the Fourier coefficients with respect to the system
$\{e^{i2m_1\pi x}: m_1\in\mathbb{Z}\}$ of the function
$Q(x)=\int_0^{x}q(t)\, dt.$
For the proof of Theorem \ref{thm1.2}, we suppose without loss of generality
that  $c_0=0$, so that $Q(1)=c_0=0$.

Now using the assumption $\ell_n=o(n^{-2})$ of the theorem it is also 
true that $\ell_n=O(n^{-2})$.  In view of \eqref{coskun}, we obtain 
$c_n=O(n^{-2})$ as $n\to\infty$. Thus, from 
\cite[Lemma 5]{Hochstadt:determination}, we obtain that $q(x)$ is absolutely 
continuous a.e. Hence integration by parts, together with $Q(1)=0$, gives
\begin{equation}
\begin{aligned}
&b_1(\lambda_{2m+j})\\
& =\frac{1}{2\pi^2(2m+2)^2}\int_0^{1}\left(q^2(x)+(Q(x)-Q_0)q'(x)
\right)e^{-i2(2m+2)\pi x}dx 
 +o\big(m^{-2}\big).
\end{aligned}
\end{equation}
Since $q(x)$ is absolutely continuous a.e.,  
$\left(q^2(x)+(Q(x)-Q_0)q'(x)\right)\in L^{1}[0,1]$.
By the Riemann-Lebesgue lemma, we find that
\begin{equation}\label{b1o}
b_1(\lambda_{2m+j})=o\big(m^{-2}\big).
\end{equation}
Similarly
\begin{equation}\label{b1oprime}
b'_1(\lambda_{2m+j})=o\big(m^{-2}\big).
\end{equation}
Let us prove that
\begin{equation}\label{b2oandprime}
b_2(\lambda_{2m+j}),\; b'_2(\lambda_{2m+j})=o\big(m^{-2}\big).
\end{equation}
Taking into account that $q(x)$ is absolutely continuous a.e. and periodic, 
we obtain $c_{m_1}c_{m_2}c_{\pm(2m+2)-m_1-m_2}=o\left(m^{-1}\right)$ 
(see \cite[p. 665]{Veliev:Shkalikov}).
Using this and arguing as in the proof of \eqref{m45}, we obtain
\begin{align*}
|b_2(\lambda_{2m+j})|
&=o\left(m^{-1}\right)
\sum_{m_1,m_2}\frac{1}{|m_1(2m+2-m_1)(m_1+m_2)(2m+2-m_1-m_2)|} \\
&=o\left(m^{-1}\right)O\Big(\big(\frac{\ln| m|}{m}\big)^2\Big)=o\big(m^{-2}\big).
\end{align*}
Thus, the first estimate of \eqref{b2oandprime} is proved. 
Similarly $b'_2(\lambda_{2m+j})=o\big(m^{-2}\big)$.
Substituting the estimates given by \eqref{a1=a1}, \eqref{m45}, \eqref{a1} 
and \eqref{b1o}-\eqref{b2oandprime} into the relations \eqref{m4123} 
and \eqref{m412}, we find that
\begin{gather}\label{son}
[\Lambda_{m,j,m}-\sum_{i=1}^2a_i((2m+2)^2\pi^2)]u_{m,j}
=c_{2m+2}v_{m,j}+o\big(m^{-2}\big), \\
\label{son'}
[\Lambda_{m,j,m}-\sum_{i=1}^2a_i((2m+2)^2\pi^2)]v_{m,j}
=c_{-2m-2}\,u_{m,j}+o\big(m^{-2}\big)
\end{gather}
for $j=1,2$.

Now suppose that, contrary to what we want to prove, 
there exists an increasing sequence $\{m_{k}\}$ $(k=1,2,\ldots)$ such that
\begin{equation}\label{mk}
|c_{2m_{k}+2}|>C m_{k}^{-2}\quad\text{for some $C>0$}.
\end{equation}
Further, the formula \eqref{m8} for $m=m_{k}$ implies that either 
$|u_{m_{k},j}|>1/2$ or $|v_{m_{k},j}|>1/2$ for sufficiently large $m_{k}$. 
Without loss of generality, we assume that $|u_{m_{k},j}|>1/2$.
Then it follows from both \eqref{son} and \eqref{son'} for $m=m_{k}$ that
\begin{equation}\label{sameo}
[\Lambda_{m_{k},j,m_{k}}-\sum_{i=1}^2a_i((2m_{k}+2)^2\pi^2)]\sim c_{2m_{k}+2},
\end{equation}
where the notation $a_{m}\sim b_{m}$ means that there exist constants $c_1$, $c_2$ 
such that $0<c_1<c_2$ and $c_1<|a_{m}/b_{m}|<c_2$ for all sufficiently large $m$.
This, together with \eqref{son'} for $m=m_{k}$, \eqref{mk} and $|u_{m_{k},j}|>1/2$, 
implies that
\begin{equation}\label{vsimilar}
u_{m_{k},j}\sim v_{m_{k},j}\sim 1.
\end{equation}
Multiplying \eqref{son'} for $m=m_{k}$ by $c_{2m_{k}+2}$, and then using 
\eqref{son} for $m=m_{k}$ in \eqref{son'}, we arrive at the relation
\begin{equation}
\begin{aligned}
&[\Lambda_{m_{k},j,m_{k}}-\sum_{i=1}^2 a_i((2m_{k}+2)^2\pi^2)]  \Big([\Lambda_{m_{k},j,m_{k}}   \\
&  -\sum_{i=1}^2a_i((2m_{k}+2)^2\pi^2)]u_{m_{k},j}+o\big(m_{k}^{-2}\big)  \Big)\\
&=|c_{2m_{k}+2}|^2\,u_{m_{k},j}+c_{2m_{k}+2}\, o\big(m_{k}^{-2}\big),
\end{aligned}
\end{equation}
which, by \eqref{sameo} and \eqref{vsimilar}, implies 
\begin{equation}\label{final}
\Lambda_{m_{k},j,m_{k}}-\sum_{i=1}^2a_i((2m_{k}+2)^2\pi^2)
=\pm|c_{2m_{k}+2}|+o\big(m_{k}^{-2}\big)
\end{equation}
for $j=1,2$.

Let us prove that the periodic eigenvalues for large $m_{k}$ are simple. 
Assume that there exist two orthogonal eigenfunctions $\Psi_{m_{k},1}(x)$ 
and $\Psi_{m_{k},2}(x)$ corresponding to $\lambda_{2m_{k}+1}=\lambda_{2m_{k}+2}$. 
From the argument of \cite[Lemma 4]{Veliev:Shkalikov}, using  the 
relation \eqref{m7} with $\|h_{m_{k}}\|=O(m_{k}^{-1})$ for the eigenfunctions 
$\Psi_{m_{k},j}(x)$ and the orthogonality of eigenfunctions,
 we can choose these eigenfunctions such that either $u_{m_{k},j}=0$ or 
$v_{m_{k},j}=0$, which contradicts \eqref{vsimilar}.

Since the eigenfunctions $\Psi_{m_{k},1}$ and $\overline{\Psi_{m_{k},2}}$  
of the self-adjoint problem corresponding to
the different eigenvalues $\lambda_{2m_{k}+1}\neq\lambda_{2m_{k}+2}$ 
are orthogonal we find, by \eqref{m7}, that
\begin{equation}\label{orteigen}
0=(\Psi_{m_{k},1},\overline{\Psi_{m_{k},2}})
=u_{m_{k},2}v_{m_{k},1}+u_{m_{k},1}v_{m_{k},2}+O(m_{k}^{-1}).
\end{equation}
Note that, for the simple eigenvalues  in \eqref{final}, there are two cases. 
First case: The simple eigenvalues $\lambda_{2m_{k}+1}$ and $\lambda_{2m_{k}+2}$ 
in \eqref{final} corresponds respectively
to the lower sign $-$ and upper sign $+$. Then
\[
\ell_{2{m_{k}}+2}=\lambda_{m_{k},2,m_{k}}-\lambda_{m_{k},1,m_{k}}
=2|c_{2m_{k}+2}|+o\big(m_{k}^{-2}\big),
\]
which implies that (see \eqref{mk}) $\ell_{2{m_{k}}+2}>C m_{k}^{-2}$ for some $C$. 
This contradicts the hypothesis that $\ell_{2{m_{k}}+2}=o(m_{k}^{-2})$.
 Now let us consider the second case: We assume that both the simple eigenvalues 
correspond to the lower sign $-$ (the proof corresponding to the upper sign 
$+$ is similar). Then 
$\Lambda_{m_{k},2,m_{k}}-\Lambda_{m_{k},1,m_{k}}=o\big(m_{k}^{-2}\big)$.
Using this, \eqref{son} and \eqref{final}, we have
\begin{gather}\label{fark1}
o\big(m_{k}^{-2}\big) u_{m_{k},2}
=c_{2m_{k}+2} v_{m_{k},2}+|c_{2m_{k}+2}| u_{m_{k},2}+o\big(m_{k}^{-2}\big), \\
\label{fark2}
o\big(m_{k}^{-2}\big) u_{m_{k},1}
=-c_{2m_{k}+2} v_{m_{k},1}-|c_{2m_{k}+2}| u_{m_{k},1}+o\big(m_{k}^{-2}\big).
\end{gather}
Therefore, multiplying both sides of \eqref{fark1} and \eqref{fark2} by 
$v_{m_{k},1}$ and $v_{m_{k},2}$
respectively and adding the results, we have, in view of \eqref{mk},
\[
u_{m_{k},2}v_{m_{k},1}-u_{m_{k},1}v_{m_{k},2}=o(1).
\]
This, together with \eqref{orteigen}, gives $u_{m_{k},2}v_{m_{k},1}=o(1)$ 
which contradicts \eqref{vsimilar}. Thus the assumption \eqref{mk} is false, 
that is, $c_{2m+2}=o\big(m^{-2}\big)$. A similar result holds for the 
anti-periodic problem, that is, $c_{2m+1}=o\big(m^{-2}\big)$. The theorem 
is proved.


For the proof of Theorem \ref{main0}, we need the sharper estimates in the 
following lemma.

\begin{lemma}\label{lemma12}
Let $q(x)$ be absolutely continuous a.e. and $c_0=0$. Then, for all
sufficiently large $m$, we have the following estimates
\begin{equation}\label{lemma1}
\begin{gathered}
a_1(\lambda_{2m+j})=\frac{-1}{(2\pi(2m+2))^2}\int_0^{1}q^2(x)dx+o\big(m^{-2}\big),
\\
a_2(\lambda_{2m+j})=o\big(m^{-2}\big).
\end{gathered}
\end{equation}
\end{lemma}

\begin{proof}
First, let us consider $a_1(\lambda_{2m+j})$. By \eqref{dif} we obtain
\[
    a_1(\lambda_{2m+j})=\frac{1}{4\pi^2}\sum_{m_1\neq 0,(2m+2)}
\frac{c_{m_1}c_{-m_1}}{m_1(2m+2-m_1)}+o\big(m^{-2}\big).
\]
Arguing as in \cite[Lemma 3]{kyrac:titch} 
(see also \cite[Lemma 2.3(a)]{veliev;arşiv}), we obtain, in our notation,
\begin{equation}\label{d3}
\begin{aligned}
&a_1(\lambda_{2m+j})\\
&=\frac{1}{2\pi^2}\sum_{m_1> 0,m_1\neq(2m+2)}\frac{c_{m_1}c_{-m_1}}
 {(2m+2+m_1)(2m+2-m_1)}+o\big(m^{-2}\big) \\
&=\int_0^{1}(G^{+}(x,m)-G^{+}_0(m))^2\,e^{i2(4m+4)\pi x}\,dx
+o\big(m^{-2}\big) \\
&= \frac{-2}{i2\pi(4m+4)}\int_0^{1}\big(G^{+}(x,m)-G^{+}_0(m)\big)\\
&\quad \times \big(q(x)e^{-i2(2m+2)\pi x}
  -c_{2m+2}\big)e^{i2(4m+4)\pi x}dx 
+o\big(m^{-2}\big) \\
\end{aligned}
\end{equation}
where
\begin{equation}\label{d4}
G^{\pm}_{m_1}(m)=:(G^{\pm}(x,m), e^{i2m_1\pi x})
=\frac{c_{m_1\pm(2m+2)}}{i2\pi m_1},
\end{equation}
for $m_1\neq 0$, are the Fourier coefficients with respect to 
$\{e^{i2m_1\pi x}: m_1\in\mathbb{Z}\}$ of the functions
\begin{equation}\label{d2}
G^{\pm}(x,m)=\int_0^{x}q(t)\,e^{\mp i2(2m+2)\pi t}dt-c_{\pm(2m+2)}x
\end{equation}
and
\[
G^{\pm}(x,m)-G^{\pm}_0(m)=\sum_{m_1\neq(2m+2)}
\frac{c_{m_1}}{i2\pi(m_1\mp(2m+2))}\,e^{i2(m_1\mp(2m+2))\pi x}.
\]
Here, taking into account the \cite[Lemma 1]{kyrac:titch} and 
\eqref{d2}, we have the estimate
\begin{equation}\label{ggg}
G^{\pm}(x,m)-G^{\pm}_0(m)=G^{\pm}(x,m)-\int_0^{1}G^{\pm}(x,m)\, dx=o(1)
\quad\text{as $m\to\infty$}
\end{equation}
uniformly in $x$.

From the equalities (see \eqref{d2})
\begin{equation}\label{gg}
    G^{\pm}(1,m)=G^{\pm}(0,m)=0,
\end{equation}
and since $q(x)$ is absolutely continuous a.e.,  
integration by parts gives, for the right hand-side of \eqref{d3},  the value
\begin{align*}
 a_1(\lambda_{2m+j})
&=\frac{-1}{(2\pi(2m+2))^2}
\Big[\int_0^{1}q^2+\int_0^{1}(G^{+}(x,m)-G^{+}_0(m))q'(x)
 e^{i2(2m+2)\pi x}dx\Big] \\
&\quad +\frac{|c_{2m+2}|^2}{(2\pi(2m+2))^2}+o\big(m^{-2}\big)
\end{align*}
for sufficiently large $m$. Thus, by using the Riemann-Lebesgue lemma,  
this with $(G^{+}(x,m)-G^{+}_0(m))q'(x)\in L^{1}[0,1]$ implies the
first equality of \eqref{lemma1}.

Now, it remains to prove that $a_2(\lambda_{2m+j})=o\big(m^{-2}\big)$. 
Similarly, by \eqref{a1} for $i=2$, we obtain
\begin{equation}\label{a2}
\begin{aligned}
a_2(\lambda_{2m+j})
&=\sum_{m_1,m_2}\frac{(2\pi)^{-4}\,c_{m_1}c_{m_2}c_{-m_1-m_2}}{m_1(2m+2-m_1)
(m_1+m_2)(2m+2-m_1-m_2)}\\
&\quad +o\big(m^{-2}\big).
\end{aligned}
\end{equation}
As in \cite[Lemma 4]{kyrac:titch}, using the summation variable $m_2$ 
to represent the previous $m_1+m_2$ in \eqref{a2}, we write \eqref{a2} 
in the form
\[
 a_2(\lambda_{2m+j})=\frac{1}{(2\pi)^{4}}\sum_{m_1,m_2}\frac{c_{m_1}c_{m_2-m_1}
c_{-m_2}}{m_1(2m+2-m_1)m_2(2m+2-m_2)}+o\big(m^{-2}\big),
\]
where the forbidden indices in the sums take the form of $m_1, m_2\neq 0,\,2m+2$.
 Here the equality
\[
  \frac{1}{k(2m+2-k)}=\frac{1}{2m+2}\Big(\frac{1}{k}+\frac{1}{2m+2-k}\Big)
\]
gives
\begin{equation}\label{equal1}
  a_2(\lambda_{2m+j})=\frac{1}{(2\pi)^{4}(2m+2)^2}\sum_{j=1}^{4}S_{j},
\end{equation}
where
\begin{gather*}
S_1=\sum_{m_1,m_2}\frac{c_{m_1}c_{m_2-m_1}c_{-m_2}}{m_1m_2},\quad
S_2=\sum_{m_1,m_2}\frac{c_{m_1}c_{m_2-m_1}c_{-m_2}}{m_2(2m+2-m_1)}, \\
S_3=\sum_{m_1,m_2}\frac{c_{m_1}c_{m_2-m_1}c_{-m_2}}{m_1(2m+2-m_2)},\quad
S_4=\sum_{m_1,m_2}\frac{c_{m_1}c_{m_2-m_1}c_{-m_2}}{(2m+2-m_1)(2m+2-m_2)}.
\end{gather*}
Using \eqref{Q0}, integration by parts and the assumption $c_0=0$ which
implies $Q(1)=0$, we deduce that
\begin{equation}\label{s11}
 S_1=4\pi^2\int_0^{1}(Q(x)-Q_0)^2q(x)\,dx=0.
\end{equation}
Similarly, in view of \eqref{Q0} and \eqref{d4}-\eqref{gg}, 
we obtain, by the Riemann-Lebesgue lemma,
\begin{gather*}
S_2=-4\pi^2\int_0^{1}(Q(x)-Q_0)(G^{+}(x,m)-G^{+}_0(m))
q(x)e^{i2(2m+2)\pi x}dx=o(1), \\
 S_3=-4\pi^2\int_0^{1}(Q(x)-Q_0)(G^{-}(x,m)-G^{-}_0(m))
q(x)e^{-i2(2m+2)\pi x}dx=o(1)
\end{gather*}
and, by \eqref{ggg},
\[
S_4=4\pi^2\int_0^{1}(G^{+}(x,m)-G^{+}_0(m))(G^{-}(x,m)
-G^{-}_0(m))\,q(x)\,dx=o(1).
\]
Thus, \eqref{equal1} implies that $a_2(\lambda_{2m+j})=o\big(m^{-2}\big)$. 
The proof is complete.
\end{proof}

\subsection*{Proof of Theorem \ref{main0}}
(i) First let us prove that $c_0=0$. Considering the first step of the
procedure in Lemma \ref{lemmaitera}, and using the estimate in \eqref{m45}, 
we may rewrite the relations \eqref{m4123} and \eqref{m412} as follows:
\begin{equation}\label{sonnnnn}
\begin{gathered}
  [\Lambda_{m,j,m}- c_0]u_{m,j}=c_{2m+2}v_{m,j}
+O\Big(\frac{\ln| m|}{m}\Big),  \\
 [\Lambda_{m,j,m}- c_0]v_{m,j}=c_{-2m-2}\,u_{m,j}
+O\Big(\frac{\ln| m|}{m}\Big)  
\end{gathered}
\end{equation}
for $j=1,2$ and sufficiently large $m$. By using the assumption 
$\ell_{2m+2}=o(m^{-2})$, namely, $\ell_n=o(n^{-2})$ for even $n=2m+2$ 
 and Theorem \ref{thm1.2} which implies $c_{\mp(2m+2)}=o(m^{-2})$,
we obtain the relations (see \eqref{sonnnnn}) in the form
\begin{gather}\label{c01}
[\Lambda_{m,j,m}- c_0]u_{m,j}=O\big(\frac{\ln| m|}{m}\big), \\
\label{c02}
[\Lambda_{m,j,m}- c_0]v_{m,j}=O\big(\frac{\ln| m|}{m}\big).
\end{gather}
Again by \eqref{m8}, we have either $|u_{m,j}|>1/2$ or $|v_{m,j}|>1/2$ 
for large $m$. In either case, in view of \eqref{c01} and \eqref{c02}, 
there exists a sufficiently large positive integer $N_0$ such that both
the eigenvalues $\lambda_{2m+j}$ (see definition of \eqref{m1}) 
satisfy the  estimate
\begin{equation}\label{c0son}
\lambda_{2m+j}=(2m+2)^2\pi^2+c_0+O\big(\frac{\ln| m|}{m}\big)
\end{equation}
for all $m>N_0$ and $j=1,2$. Under the assumption of Theorem \ref{main0} (i),
when $m>\max\{(n_0-2)/2,N_0\}$,  the eigenvalue $(2m+2)^2\pi^2$
corresponds to the eigenvalue $\lambda_{2m+1}$ or $\lambda_{2m+2}$. 
In either case we obtain $c_0=0$ by \eqref{c0son}.

Finally, for sufficiently large $m$, substituting the estimates of
$a_i(\lambda_{2m+j})$, $a'_i(\lambda_{2m+j})$,  $b_i(\lambda_{2m+j})$,
$b'_i(\lambda_{2m+j})$, $R_2$, $R'_2$ for $i=1,2$,  given 
by Lemma \ref{lemma12} with the equalities 
$a_i(\lambda_{2m+j})=a'_i(\lambda_{2m+j})$ (see \eqref{a1=a1}), 
\eqref{b1o}-\eqref{b2oandprime} and \eqref{m45} into the relations 
\eqref{m4123} and \eqref{m412} and using $c_0=0$, we find the relations
in the  form
\begin{equation}\label{q2son}
\begin{gathered}
  \Big[\Lambda_{m,j,m}+\frac{1}{(2\pi(2m+2))^2}\int_0^{1}q^2\Big]u_{m,j}
=c_{2m+2}v_{m,j}+o\big(m^{-2}\big),  \\
 \Big[\Lambda_{m,j,m}+\frac{1}{(2\pi(2m+2))^2}\int_0^{1}q^2\Big]v_{m,j}
=c_{-2m-2}\,u_{m,j}+o\big(m^{-2}\big) 
\end{gathered}
\end{equation}
for $j=1,2$.
In the same way, by using the assumption $\ell_{2m+2}=o(m^{-2})$ and 
Theorem \ref{thm1.2}, we write \eqref{q2son} in the form
\begin{gather*}
\Big[\Lambda_{m,j,m}+\frac{1}{(2\pi(2m+2))^2}\int_0^{1}q^2\Big]u_{m,j}
=o\big(m^{-2}\big), \\
\Big[\Lambda_{m,j,m}+\frac{1}{(2\pi(2m+2))^2}\int_0^{1}q^2\Big]v_{m,j}
=o\big(m^{-2}\big).
\end{gather*}
Thus, arguing as in the proof of \eqref{c0son}, there exists a positive 
large number $N_1$ such that the eigenvalues $\lambda_{2m+j}$ satisfy the 
following estimate
\begin{equation}\label{sssson}
\lambda_{2m+j}=(2m+2)^2\pi^2-\frac{1}{(2\pi(2m+2))^2}
\int_0^{1}q^2+o\big(m^{-2}\big)
\end{equation}
for all $m>N_1$ and $j=1,2$. Let $m>\max\{(n_0-2)/2,N_1\}$. Using the same
argument as above, by \eqref{sssson}, we obtain $\int_0^{1}q^2=0$
which implies that $q=0$ a.e.

(ii) The procedure in Section \ref{asyl} works  for the anti-periodic boundary 
conditions
\begin{equation*}\label{antiper}
 y(0)=-y(a),\quad y'(0)=-y'(a).
\end{equation*}
Thus, it is readily seen that the corresponding results for the anti-periodic 
eigenvalues $\mu_{2m}$, $\mu_{2m+1}$ hold, replacing $(2m+2)$ 
in \eqref{asy2}-\eqref{m1} by $(2m+1)$.

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\end{document}

