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\begin{document}
\title{On the Riesz Basisness of the Root \ Functions of the Nonselfadjoint
Sturm-Liouville Operator }
\author{Ne\c{s}e Dernek, O.A. Veliev \\
%EndAName
{\small \ Depart. of Math, Fen-Ed. Fak, Marmara Univ., }\\
{\small \ G\"{o}ztepe Kamp\"{u}s\"{u}, 81040, Kadik\"{o}y, }\\
{\small \ Istanbul, Turkey.}\\
{\small \ e-mail: ndernek@marmara.edu.tr}\\
\ {\small oveliev@marmara.edu.tr}}
\date{}
\maketitle
\begin{abstract}
In this article we obtain the asymptotic formulas of arbitrary order for
eigenfunctions and eigenvalues of the nonselfadjoint Sturm-Liouville
operators with \ periodic and antiperiodic boundary conditions, when the
potential is a summable function. Then using these asymptotic formulas, we
find the conditions on Fourier coefficients of the potential for which the
eigenfunctions and associated functions of these operators form a Riesz
basis in $L_{2}(0,1)$.
\end{abstract}
Let $L_{t}(q)$ be the operator generated in $L_{2}[0,1]$ by the expression
\begin{equation}
-y^{^{\prime \prime }}+q(x)y,
\end{equation}
and the boundary conditions
\begin{equation}
y(1)=e^{it}y(0),\text{ }y^{^{\prime }}(1)=e^{it}y^{^{\prime }}(0),
\end{equation}
where $t=0,\pi $ and $q(x)$ is a complex-valued summable function.
In this article we obtain the asymptotic formulas \ of order $O(n^{-l})$ (
for all $l>0$ ) for $n$-th\ eigenvalue and corresponding eigenfunction of
the operator $L_{t}(q)$ \ with $q(x)\in L_{1}(0,1),$ for $t=0,\pi $ , that
is, for periodic and antiperiodic boundary conditions. Then using these
asymptotic formulae, we find the conditions on Fourier coofficients $%
q_{n}=(q(x),e^{i2\pi nx})$, where $(.,.)$ denotes inner product in $%
L_{2}(0,1),$ of $q(x)$ for which the root functions ( the eigenfunctions and
associated functions) of $L_{t}(q),$ for $t=0,\pi ,$ form a Riesz basis in
$%
L_{2}(0,1)$. Note that the periodic and antiperiodic boundary conditions are
regular boundary conditions, but are not strongly regular boundary
conditions. Therefore, in general, the eigenfunctions and associated
functions of $L_{0}(q)$ and $L_{\pi }(q)$ \ do not form a Riesz basis, they
form a basis with brachet \ ( see [4],[5]). In this paper we prove that if
\begin{equation}
\lim_{n\rightarrow \infty }\frac{ln\left| n\right| }{nq_{2n}}=0,\text{ \ }%
q_{2n}\sim q_{-2n},
\end{equation}
where \ $a_{n}\sim b_{n}$ means that $c_{1}\mid b_{n}\mid <\mid a_{n}\mid
&\left| 2(n-k)\pi \right| \left|
2(n+k)\pi \right| -c_{3}n^{\frac{1}{2}}>c_{4}n, \nonumber \\
\forall \text{ }j &=&1,2,\text{\ }\forall k\neq n,\text{ }k=0,1,...,
\end{eqnarray}
\begin{eqnarray}
\left| \mu _{n,j}-(2\pi k+\pi )^{2}\right| &>&\left| 2(n-k)\pi \right|
\left| 2(n+k+1)\pi \right| -c_{5}n^{\frac{1}{2}}>c_{6}n\text{ } \nonumber
\\
\forall \text{\ }j &=&1,2,\text{\ }\forall k\neq n,,\forall k=0,1,....
\end{eqnarray}
for $n\geq N.$ \ To obtain the asymptotic formula for eigenvalues $\lambda
_{n,j}$ and corresponding normalized eigenfunctions $\Psi _{n,j}(x)$ of $%
L_{0}(q)$ we use (7) and the following well-known relation
\begin{equation}
(\lambda _{N,j}-(2\pi n)^{2})(\Psi _{N,j}(x),e^{i2\pi nx})=(q(x)\Psi
_{N,j}(x),e^{i2\pi nx}).
\end{equation}
Moreover, we use the following relations, which can be easily proved arguing
as in the proof of \ Lemma 1 of [6],
\begin{equation}
(q(x)\Psi _{N,j}(x),e^{i2\pi nx})=\sum_{m=-\infty }^{\infty }q_{-m}(\Psi
_{N,j}(x),e^{i2\pi (n+m)x})
\end{equation}
\begin{equation}
\left| (q(x)\Psi _{N,j}(x),e^{i2\pi nx})\right| <4M,\text{ }\forall n,\text{
}\forall N\gg 1
\end{equation}
where $M=\stackunder{n\in Z}{sup}\left| q_{n}\right| .$ Note that $%
q_{n}\rightarrow 0$ as $\left| n\right| \rightarrow \infty $ and, without
loss of generality, we assume that $q_{0}=0.$
Using (10) in (9) we get
\begin{equation}
(\lambda _{n,j}-(2\pi n)^{2})(\Psi _{n,j}(x),e^{i2\pi nx})=\sum\Sb %
n_{1}=-\infty , \\ n_{1}\neq 0 \endSb ^{\infty }q_{n_{1}}(\Psi
_{n,j}(x),e^{i2\pi (n-n_{1})x}).
\end{equation}
Now isolating the term in right-hand side of (12) containing the
multiplicant $(\Psi _{n,j}(x),e^{-i2\pi nx})$ (i.e., case $n_{1}=2n$ ),
replacing
\bigskip\ $(\Psi _{n,j}(x),e^{i2\pi (n-n_{1})x})$ by $\frac{(q(x)\Psi
_{n,j}(x),e^{i2\pi (n-n_{1})x}))}{\lambda _{n,j}-(2\pi (n-n_{1}))^{2}},$ for
$n_{1}\neq 2n$
( this replacement can be obtained from (9) by taking $n$ and $n-n_{1}$
instead of $N$ and $n$ ) and using (10) for the numerator of the last
fraction we obtain
\[
(\lambda _{n,j}-(2\pi n)^{2})(\Psi _{n,j}(x),e^{i2\pi nx})-q_{2n}(\Psi
_{n,j}(x),e^{-i2\pi nx})=
\]
\begin{equation}
\sum\Sb n_{1},n_{2}=-\infty \\ n_{1}\neq 2n \endSb ^{\infty }\frac{%
q_{n_{1}}q_{n_{2}}(\Psi _{n,j}(x),e^{i2\pi (n-n_{1}-n_{2})x})}{\lambda
_{n,j}-(2\pi (n-n_{1}))^{2}}.
\end{equation}
Note that, since $q_{0}=0,$ here and in the forthcoming relations the sums
are taken under conditions $n_{1},n_{2},...,\neq 0$. Now we isolate the
terms in right-hand side of (13) containing one of the multiplicants $(\Psi
_{n,j}(x),e^{i2\pi nx})$ , $(\Psi _{n,j}(x),e^{-i2\pi nx})$ (i.e., cases $%
n_{1}+n_{2}=0,2n$ ) and apply the above replacement to other terms (i.e.,
cases $n_{1}+n_{2}\neq 0,2n$ ) to get
\[
(\lambda _{n,j}-(2\pi n)^{2})(\Psi _{n,j}(x),e^{i2\pi nx})-q_{2n}(\Psi
_{n,j}(x),e^{-i2\pi nx})-
\]
\[
\sum\Sb n_{1}=-\infty \\ n_{1}\neq 2n \endSb ^{\infty }\frac{%
q_{n_{1}}q_{-n_{1}}(\Psi _{n,j}(x),e^{i2\pi nx})}{\lambda _{n,j}-(2\pi
(n-n_{1}))^{2}}-\sum\Sb n_{1}=-\infty \\ n_{1}\neq 2n \endSb ^{\infty }%
\frac{q_{n_{1}}q_{2n-n_{1}}(\Psi _{n,j}(x),e^{-i2\pi nx})}{\lambda
_{n,j}-(2\pi (n-n_{1}))^{2}}=
\]
\begin{equation}
\sum\Sb n_{1},n_{2}=-\infty \\ n_{1}\neq 2n,n_{1}+n_{2}\neq 0,2n \endSb %
^{\infty }\frac{q_{n_{1}}q_{n_{2}}(q(x)\Psi _{n,j}(x),e^{i(2\pi
(n-n_{1}-n_{2})x})}{\lambda _{n,j}-(2\pi (n-n_{1}))^{2}(\lambda _{n,j}-(2\pi
(n-n_{1}-n_{2}))^{2})}
\end{equation}
Repeating this process $m-$times (i.e. applying (10) to the numerator of the
fractions in the rigth-hand side of (14) and isolating the terms containing
one of the multiplicants $(\Psi _{n,j}(x),e^{i2\pi nx})$ , $(\Psi
_{n,j}(x),e^{-i2\pi nx})$ etc.) we get
\begin{equation}
(\lambda _{n,j}-(2\pi n)^{2}-A_{m}(\lambda _{n,j}))(\Psi _{n,j}(x),e^{i2\pi
nx})-
\end{equation}
\[
(q_{2n}+B_{m}(\lambda _{n,j})(\Psi _{n,j}(x),e^{-i2\pi nx})=R_{m},
\]
where $A_{m}(\lambda _{n,j})=\sum_{k=1}^{m}a_{k}(\lambda _{n,j}),$ $%
B_{m}(\lambda _{n,j})=\sum_{k=1}^{m}b_{k}(\lambda _{n,j})$%
\[
a_{k}(\lambda _{n,j})=\sum_{n_{1},n_{2},...,n_{k}}\frac{%
q_{n_{1}}q_{n_{2}}...q_{n_{k}}q_{-n_{1}-n_{2}-...-n_{k}}}{[\lambda
_{n,j}-(2\pi (n-n_{1}))^{2}]...[\lambda _{n,j}-(2\pi
(n-n_{1}-...-n_{k}))^{2}]},
\]
\[
b_{k}(\lambda _{n,j})=\sum_{n_{1},n_{2},...,n_{k}}\frac{%
q_{n_{1}}q_{n_{2}}...q_{n_{k}}q_{2n-n_{1}-n_{2}-...-n_{k}}}{[\lambda
_{n,j}-(2\pi (n-n_{1}))^{2}]...[\lambda _{n,j}-(2\pi
(n-n_{1}-..-n_{k}))^{2}]%
},
\]
\[
R_{m}=\sum_{n_{1},n_{2},...,n_{m+1}}\frac{%
q_{n_{1}}q_{n_{2}}...q_{n_{m}}q_{n_{m+1}}(q(x)\Psi _{n,j}(x),e^{i2\pi
(n-n_{1}-...-n_{m+1})x})}{[\lambda _{n,j}-(2\pi (n-n_{1}))^{2}]...[\lambda
_{n,j}-(2\pi (n-n_{1}-...-n_{m+1}))^{2}]},
\]
Here the sums are taken under the conditions
\[
n_{j}\neq 0,\forall j\text{ ; }\sum\limits_{j=1}^{k}n_{j}\neq 0,2n,\text{ }%
\forall k=1,2,...,m+1.
\]
Using (7) and arguing as in the proof of equalities (27), (28) of [6] we get
the following relations
\begin{equation}
a_{k}=O\left( (\frac{ln\left| n\right| }{n})^{k}\right) ,b_{k}=O\left( (%
\frac{ln\left| n\right| }{n})^{k}\right) ,R_{m}=O\left( (\frac{ln\left|
n\right| }{n})^{m+1}\right) .
\end{equation}
In the same way one can obtain the relations
\begin{equation}
(\lambda _{n,j}-(2\pi n)^{2}-A_{m}^{^{\prime }}(\lambda _{n,j}))(\Psi
_{n,j}(x),e^{-i2\pi nx})-
\end{equation}
\[
(q_{-2n}+B_{m}^{^{\prime }}(\lambda _{n,j}))(\Psi _{n,j}(x),e^{i2\pi
nx})=R_{m}^{^{\prime }},
\]
where $A_{m}^{^{\prime }}(\lambda _{n,j})=\sum_{k=1}^{m}a_{k}^{\prime
}(\lambda _{n,j}),$ $B_{m}^{^{\prime }}(\lambda
_{n,j})=\sum_{k=1}^{m}b_{k}^{^{\prime }}(\lambda _{n,j}),$%
\[
a_{k}^{^{\prime }}(\lambda _{n,j})=\sum_{n_{1},n_{2},...,n_{k}}\frac{%
q_{n_{1}}q_{n_{2}}...q_{n_{k}}q_{-n_{1}-n_{2}-...-n_{k}}}{[\lambda
_{n,j}-(2\pi (n+n_{1}))^{2}]...[\lambda _{n,j}-(2\pi
(n+n_{1}+...+n_{k}))^{2}]},
\]
\[
b_{k}^{^{\prime }}(\lambda _{n,j})=\sum_{n_{1},n_{2},...,n_{k}}\frac{%
q_{n_{1}}q_{n_{2}}...q_{n_{k}}q_{-2n-n_{1}-n_{2}-...-n_{k}}}{[\lambda
_{n,j}-(2\pi (n+n_{1}))^{2}]...[\lambda _{n,j}-(2\pi
(n+n_{1}+...+n_{k}))^{2}]},
\]
\begin{equation}
a_{k}^{^{\prime }}=O\left( (\frac{ln\left| n\right| }{n})^{k}\right)
,b_{k}^{^{\prime }}=O\left( (\frac{ln\left| n\right| }{n})^{k}\right)
,R_{m}^{^{\prime }}=O\left( (\frac{ln\left| n\right| }{n})^{m+1}\right) .
\end{equation}
Here the sums are taken under the conditions
\[
n_{j}\neq 0,\forall j\text{ ; }\sum\limits_{j=1}^{k}n_{j}\neq 0,-2n,\forall
k=1,2,...,m+1.
\]
It follows from (7), (9) and (11) that
\[
\sum_{k\in Z,k\neq n,-n}\left| (\Psi _{n,j}(x),e^{i2\pi kx})\right|
^{2}=\sum_{k\in Z,k\neq n,-n}\frac{(4M)^{2}}{[\lambda _{n,j}-(2\pi
k)^{2}]^{2}}=O(\frac{1}{n^{2}}).
\]
Therefore $\Psi _{n,t}(x)$ has an expansion of the form
\begin{equation}
\Psi _{n,j}(x)=u_{n,j}e^{i2\pi nx}+v_{n,j}e^{-i2\pi nx}+h(x),
\end{equation}
where $u_{n,j}=(\Psi _{n,j}(x),e^{i2\pi nx}),$ $v_{n,j}=(\Psi
_{n,j}(x),e^{-i2\pi nx}),$ $\left\| h(x)\right\| =O(\frac{1}{n}),$%
\begin{equation}
\left| u_{n,j}\right| ^{2}+\left| v_{n,j}\right| ^{2}=1+O(\frac{1}{n^{2}})
\end{equation}
Using (16), (18), (20) from the system of equations (15), (17) we can find
asymptotic formulas \ of order $O(n^{-l})$ ( for all $l>0$ ) for
eigenfunctions $\Psi _{n,j}(x)$ and eigenvalues $\lambda _{n,j}$\ of the
operator $L_{0}(q)$ \ with $q(x)\in L_{1}(0,1).$ We do it under conditions
(3), since in these cases the asymptotic formulas give the Riesz basisness.
\bigskip
\begin{theorem}
Let the conditions (3) hold. Then:
$(a)$ The big eigenvalues of the operator $L_{0}$ are simple. They consists
of two sequences \ $\{\lambda _{n,1}:n>N\}$ and $\{\lambda _{n,2}:n>N\}$\
satisfying
\begin{equation}
\lambda _{n,j}=(2\pi n)^{2}+(-1)^{j}p_{n}+O\left( \frac{ln\left| n\right|
}{n%
}\right) ,\forall j=1,2,
\end{equation}
where $p_{n}=(q_{2n}q_{-2n})^{\frac{1}{2}}.$
The corresponding eigenfunction $\varphi _{n,j}(x)$ satisfies
\begin{equation}
\varphi _{n,j}(x)=e^{i2\pi nx}+\alpha _{n,j}e^{-i2\pi nx}+O(\frac{1}{n}%
),\forall j=1,2,
\end{equation}
where
\begin{eqnarray}
\alpha _{n,j} &\sim &1,\forall j=1,2, \\
\alpha _{n,j} &=&\frac{(-1)^{j}p_{n}}{q_{2n}}+O\left( \frac{ln\left|
n\right| }{nq_{2n}}\right) ,\forall j=1,2,
\end{eqnarray}
$(b)$\ The root functions of $L_{0}(q)$ form a Riesz basis in $L_{2}(0,1).$
\end{theorem}
%TCIMACRO{
%\TeXButton{Proof}{\proof%
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%BeginExpansion
\proof%
%
%EndExpansion
First we prove that if the conditions (3) hold then
\begin{equation}
u_{n,j}\sim v_{n,j}\sim 1,\forall j=1,2
\end{equation}
( for the definitions of $u_{n,j}$ and $v_{n,j}$ see (19)) and
\begin{equation}
\lambda _{n,j}-(2\pi n)^{2}=a_{j}p_{n}+O\left( \frac{ln\left| n\right| }{n}%
\right) .
\end{equation}
where $a_{j}$\ is either $1$ or $-1.$ From (15)- (18) we obtain the system
of equations
\begin{equation}
(\lambda _{n,j}-(2\pi n)^{2})u_{n,j}=q_{2n}v_{n,j}+O\left( \frac{ln\left|
n\right| }{n}\right) ,
\end{equation}
\begin{equation}
(\lambda _{n,j}-(2\pi n)^{2})v_{n,j}=q_{-2n}u_{n,j}+O\left( \frac{ln\left|
n\right| }{n}\right) .
\end{equation}
Without loss of generality we assume that \ $\left| v_{n,j}\right|
>\frac{1}{%
2}$ (see (20)) and hence $v_{n,j}\sim 1.$ Then from (28) and (3) we obtain
$%
\left| \lambda _{n,j}-(2\pi n)^{2}\right| \frac{1}{2},$ and (27) imply that $\left| \lambda
_{n,j}-(2\pi n)^{2}\right| >c_{8}\left| q_{2n}\right| .$\ Therefore using
(3) \ from these inequalities we get
\begin{equation}
\lambda _{n,j}-(2\pi n)^{2}\sim q_{2n}\sim q_{-2n}\sim p_{n},\text{ }%
\lim_{n\rightarrow \infty }\frac{ln\left| n\right| }{np_{n}}=0.
\end{equation}
This with (27) implies that $u_{n,j}\sim v_{n,j}.$ \ Thus (25) is proved.
Now dividing both sides of (27) and (28) by $u_{n,j}$ $q_{2n}$\ and $%
(\lambda _{n,j}-(2\pi n)^{2})u_{n,j}$ respectively we get
\begin{equation}
\frac{v_{n,j}}{u_{n,j}}=\frac{\lambda _{n,j}-(2\pi n)^{2}}{q_{2n}}+O\left(
\frac{ln\left| n\right| }{nq_{2n}}\right) =\frac{q_{-2n}}{\lambda
_{n,j}-(2\pi n)^{2}}+O\left( \frac{ln\left| n\right| }{nq_{2n}}\right) ,
\end{equation}
from which using (29) by simple calculation we get (26).
The proof of $(a).$ Suppose $\lambda _{n,j},$ where $n>N,N\gg 1,$ is
multiple. There are two cases:
1. There is associated functions $\Psi _{n,j}^{1}(x)$ corresponding to the
eigenfunctions $\Psi _{n,j}(x),$ that is,
\begin{equation}
(L_{0}-\lambda _{n,j})\Psi _{n,j}^{1}(x)=\Psi _{n,j}(x).
\end{equation}
Since the boundary condition (2) for $t=0$ is selfadjoint $\overline{\lambda
_{n,j}}$ and $\overline{\Psi _{n,j}}$ are eigenvalue and eigenfunction of
adjoint operator $L_{0}^{\ast }.$ Therefore multiplying both sides of (31)
by $\overline{\Psi _{n,j}}$ we get $(\Psi _{n,j}(x),\overline{\Psi _{n,j}}%
)=0.$ This with (19) give the equality $u_{n,j}v_{n,j}=O(\frac{1}{n})$ which
contradicts (25).
2. There are two eigenfunctions corresponding $\lambda _{n,j}.$ Then all
solutions of the equation
\[
-y^{^{\prime \prime }}+q(x)y=\lambda _{n,j}y
\]
are eigenfunctions. In particular, the solution
\[
y(x,\lambda _{n,j})=e^{\rho _{n,j}x}+O(\frac{1}{n})=e^{2\pi
nx}+O(\frac{1}{n}%
),
\]
where $\rho _{n,j}=(\lambda _{n,j})^{\frac{1}{2}}=2\pi n+(\frac{1}{n}),$ (
see page 52 of [1]), and (26)) is eigenfunction. In (27) instead of using $%
\Psi _{n,j}$ taking this eigenfunction we get
\[
\lambda _{n,j}=(2\pi n)^{2}+O\left( \frac{ln\left| n\right| }{n}\right) ,
\]
which contradicts (29). Thus the eigenvalues $\lambda _{n,j}$ for
$n>N,j=1,2$
are simple and satisfy (26). Now, in order to obtain (21) from (26) it
remains to prove that the numbers $a_{1},a_{2}$ in (26) are different, say
$%
a_{1}=-1,a_{2}=1.$ Suppose both $a_{1}$ and $a_{2}$ are $-1$. Then (26),
(27) yield
\begin{equation}
-p_{n}u_{n,1}=q_{2n}v_{n,1}+O\left( \frac{ln\left| n\right| }{n}\right)
,-p_{n}u_{n,2}=q_{2n}v_{n,2}+O\left( \frac{ln\left| n\right| }{n}\right) ,
\end{equation}
Multiplying both sides of f\i rst and second equalities by $-v_{n,2}$ and $%
v_{n,1}$ respectively and then summing the obtained equalities we get
\begin{equation}
u_{n,1}v_{n,2}-u_{n,2}v_{n,1}=O\left( \frac{ln\left| n\right| }{np_{n}}%
\right) .
\end{equation}
On the other hand, since $\overline{\lambda _{n,j}}$ and $\overline{\Psi
_{n,j}}$ are eigenvalue and eigenfunction of adjoint operator $L_{0}^{\ast
}$
and $\lambda _{n,1}\neq \lambda _{n,2}$ \ we have
\begin{equation}
0=(\Psi _{n,1},\overline{\Psi
_{n,2}})=u_{n,1}v_{n,2}+u_{n,2}v_{n,1}+O(\frac{%
1}{n})
\end{equation}
(see (19)). It follows from (33), (34), (29) that $u_{n,1}v_{n,2}=O\left(
\frac{ln\left| n\right| }{np_{n}}\right) =o(1)$ which contradicts (25). If
we suppose that both $a_{1}$ and $a_{2}$ are $1$ then, in the same way, we
obtain the same contradiction. Thus one of these numbers, say $a_{1}$ is
$-1$
and \ the other $a_{2}$ is $1$. The formulas (21) are proved. Since $\lambda
_{n,1}$ satisfies (21), \ for $j=1,$ first equality in (32) holds. Therefore
\begin{equation}
\frac{v_{n,1}}{u_{n,1}}=\frac{-p_{n}}{q_{2n}}+O\left( \frac{ln\left|
n\right| }{nq_{2n}}\right)
\end{equation}
Denoting $\alpha _{n,j}=\frac{v_{n,j}}{u_{n,j}},$ $\varphi _{n,j}(x)=\frac{%
\Psi _{n,j}(x)}{u_{n,j}}$ \ from (19), (25) and (35) we get (22), (23), (24)
for $j=1$. Taking into account that $\lambda _{n,2}$ satisfies (21) for
$j=2$
in the same way we obtain (22), (23), (24) for $j=2$.
The proof of $(b).$ Now we prove that the eigenfunctions $\varphi
_{n,j}\equiv \varphi _{n,j}^{0}$and associated functions $\varphi
_{n,j}^{k}, $ where $n=1,2,...,$ and$\ k=1,2,,,,s(n,j),$ of $L_{0}(q)$ form
a Riesz basis in $L_{2}(0,1)$. For every $f(x)\in L_{2}(0,1)$, the
asymptotic formulas (22), (23) yield
\begin{equation}
\sum_{n=1}^{N}(\sum_{k=0}^{s(n,j)}\sum_{j=1}^{2}\left| (f,\varphi
_{n,j}^{k})\right| ^{2})+\sum_{j=1}^{2}\sum\limits_{n=N+1}^{\infty }\left|
(f,\varphi _{n,j})\right| ^{2}<\infty .
\end{equation}
Let $\chi _{n,j}\equiv \chi _{n,j}^{0}$ and $\chi _{n,j}^{k},$ where $%
k=1,2,,,,t(n,j),$ be the biorthonormal system of eigenfunctions and
associated functions of $L_{0}^{\ast }.$ Clearly $\chi _{n,j}(x)=\frac{%
\overline{\varphi _{n,j}(x)}}{(\varphi _{n,j},\overline{\varphi _{n,j}})}$
for $n>N$ . Therefore using (22), (23) we get
\begin{equation}
\sum_{n=1}^{N}(\sum_{k=0}^{t(n,j)}\sum_{j=1}^{2}\left| (f,\chi
_{n,j}^{k})\right| ^{2})+\sum_{j=1}^{2}\sum\limits_{n=N+1}^{\infty }\left|
(f,\chi _{n,j})\right| ^{2}<\infty .
\end{equation}
Since $\{\varphi _{n,j}^{k}:k=0,1,2,,,,s(n,j);j=1,2;n=1,2,...\}$, and the
biorthonormal system is total (see chap.1, sec.3 of [2]) by the well-known
theorem of Bari (see [3], chap. 6) for the Riesz basis, (36) and (37) imply
that the system of the eigenfunctions and associated functions of $L_{0}(q)$
forms a Riesz basis in $L_{2}(0,1)$ .The theorem is proved$\blacksquare $
Instead of (3), (7), and (9) using (4), (8) and
\begin{equation}
(\mu _{N,j}-(2\pi n+\pi )^{2})(\Phi _{N,j}(x),e^{i\pi (2n+1)x})=(q(x)\Phi
_{N,j}(x),e^{i\pi (2n+1)x}),
\end{equation}
where $\Phi _{N,j}(x)$ is normalized eigenfunction of $L_{\pi }(q),$ and
arguing as in proof of Theorem 1 we obtain
\bigskip
\begin{theorem}
Let the conditions (4) hold. Then:
$(a)$ The big eigenvalues of the operator $L_{\pi }$ are simple. They
consist of two sequences \ $\{\mu _{n,1}:n>N\}$ and $\{\mu _{n,2}:n>N\}$\
satisfying
\begin{equation}
\mu _{n,j}=(2\pi n+\pi )^{2}+(-1)^{j}p_{n}^{^{\prime }}+O\left( \frac{%
ln\left| n\right| }{n}\right) ,\text{ }\forall j=1,2, \nonumber
\end{equation}
where $p_{n}^{^{\prime }}=(q_{2n+1}q_{-2n-1})^{\frac{1}{2}}.$ The
corresponding eigenfunction $\phi _{n,j}(x)$ satisfies
\begin{equation}
\phi _{n,j}(x)=e^{i(2\pi n+\pi )x}+\alpha _{n,j}^{^{\prime }}e^{-i(2\pi
n+\pi )x}+O(\frac{1}{n}),\text{ }\forall j=1,2,
\end{equation}
where $\alpha _{n,j}^{^{\prime }}\sim 1,$ $\alpha _{n,j}^{^{\prime
}}=\frac{%
(-1)^{j}p_{n}^{^{\prime }}}{q_{2n+1}}+O\left( \frac{ln\left| n\right| }{%
nq_{2n+1}}\right) ,\forall $ $j=1,2.$
$(b)$\ The root functions of $L_{\pi }(q)$ form a Riesz basis in
$L_{2}(0,1)$%
.
\end{theorem}
\bigskip Now we prove asymptotic formulas \ of arbitrary order for
eigenfunctions and eigenvalues of the operators $L_{0}(q),$ $L_{\pi }(q)$ \
with $q(x)\in L_{1}(0,1).$
\begin{theorem}
For the eigenvalue $\lambda _{n,j}$ satisfying (21), the following formulas
hold
\begin{eqnarray}
\lambda _{n,j} &=&(2\pi n)^{2}+F_{m,j}+(-1)^{j}G_{m,j}+O\left( (\frac{%
ln\left| n\right| }{n})^{m+1}\right) , \\
\forall m &=&0,1,2,...,\forall j=1,2, \nonumber
\end{eqnarray}
where $F_{0,j}=0,G_{0,j}=p_{n},$ $F_{k,j}=\frac{1}{2}(A_{k}((2\pi
n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})+$
$A_{k}^{^{\prime }}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})),$
$G_{k,j}=[\frac{1}{4}(A_{k}((2\pi
n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})-A_{k}^{^{\prime }}((2\pi
n)^{2}+F_{k-1,j}+$
$(-1)^{j}G_{k-1,j}))^{2}+(B_{k}((2\pi
n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})+q_{2n})(B_{k}^{^{\prime }}((2\pi
n)^{2}+$
$F_{k-1,j}+(-1)^{j}G_{k-1,j})+q_{-2n})]^{\frac{1}{2}},$ $\forall k=1,2,...,$
For the eigenfunction $\varphi _{n,j}(x)$ satisfying (23) the following
formulas hold
\begin{eqnarray}
\varphi _{n,j}(x) &=&e^{i2\pi nx}+\alpha _{n,j}e^{-i2\pi nx}+A_{m}^{\ast
}((2\pi n)^{2}+F_{m,j}+(-1)^{j}G_{m,j}) \\
+ &&(B_{m}^{\ast }((2\pi n)^{2}+F_{m,j}+(-1)^{j}G_{m,j}))\alpha _{n,j}+O((%
\frac{\ln n}{n})^{m+1}), \nonumber
\end{eqnarray}
where $A_{m}^{\ast }$ , $B_{m}^{\ast }$ are obtained from $A_{m},$ $B_{m}$
respectively by replacing $q_{n_{1}}$ with $e^{i2\pi (n-n_{1})x}$ and
\begin{eqnarray*}
\alpha _{n,j} &=&\frac{v_{n,j}}{u_{n,j}}=\frac{%
F_{m,j}+(-1)^{j}G_{m,j}-A_{m}((2\pi n)^{2}+F_{m,j}+(-1)^{j}G_{m,j})}{%
q_{2n}+B_{m}((2\pi n)^{2}+F_{m,j}+(-1)^{j}G_{m,j})}+ \\
&&O\left( \frac{1}{q_{2n}}(\frac{ln\left| n\right| }{n})^{m+1}\right) .
\end{eqnarray*}
For the eigenvalue $\mu _{n,j}$ and eigenfunction $\phi _{n,j}(x)$ \ of $%
L_{\pi }(q)$ satisfying (39), (40) similar formulas hold.
\end{theorem}
\bigskip
%TCIMACRO{
%\TeXButton{Proof}{\proof%
%} }%
%BeginExpansion
\proof%
%
%EndExpansion
Arguing as in the proof of (30), using (15), (17) in place of (27), (28),
and denoting $\lambda =\lambda _{n,j}-(2\pi n)^{2}$ we get
\begin{eqnarray}
\frac{v_{n,j}}{u_{n,j}} &=&\frac{\lambda -A_{m}(\lambda _{n,j})}{%
(q_{2n}+B_{m}(\lambda _{n,j}))}+O\left( \frac{1}{q_{2n}}(\frac{ln\left|
n\right| }{n})^{m+1}\right) \nonumber \\
&=&\frac{(q_{-2n}+B_{m}^{^{\prime }}(\lambda _{n,j}))}{\lambda
-A_{m}^{^{\prime }}(\lambda _{n,j})}+O\left( \frac{1}{q_{2n}}(\frac{ln\left|
n\right| }{n})^{m+1}\right) ,
\end{eqnarray}
Solving this square equation with respect to $\lambda $ we obtain
\[
\lambda _{n,j}-(2\pi n)^{2}=\frac{1}{2}(A_{m}(\lambda
_{n,j})+A_{m}^{^{\prime }}(\lambda _{n,j}))\mp \lbrack \frac{1}{4}%
(A_{m}(\lambda _{n,j})-A_{m}^{^{\prime }}(\lambda _{n,j}))^{2}+
\]
\begin{equation}
(B_{m}(\lambda _{n,j})+q_{2n})(B_{m}^{^{\prime }}(\lambda
_{n,j})+q_{-2n}))+O\left( q_{2n}(\frac{ln\left| n\right| }{n})^{m+1}\right)
]^{\frac{1}{2}}.
\end{equation}
Here upper and lower sign is taken for $j=1,$ and $j=2$ respectively, since
the expression in the square bracket is
\[
q_{2n}q_{-2n}+O(q_{2n}\frac{ln\left| n\right| }{n})=q_{2n}q_{-2n}(1+o(1))
\]
(see (3), (16), (18)), and (21) holds. Thus in (44) the $\mp $ can be
replaced by $(-1)^{j}.$ Now we prove (41) by induction. It is proved for $%
m=1 $ ( see (21) ). Assume that it is true for $m=k-1$. Substituting the
value of $\lambda _{n,j},$ given by this formula for $m=k-1,$ in (44) for $%
m=k$ and taken into account that
\begin{eqnarray}
&&A_{k}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j}+O\left( (\frac{ln\left|
n\right| }{n})^{k}\right) ) \nonumber \\
&=&A_{k}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})+O\left( (\frac{ln\left|
n\right| }{n})^{k+1}\right) ,
\end{eqnarray}
and the same formulas hold for $A_{k}^{^{\prime }},B_{k},B_{k}^{^{\prime }}$
we get the proof of (41) for $m=k.$ Note that the validity of (45) can be
easily verified by using the obvious relation
\[
\sum\Sb n_{1}=-\infty , \\ n_{1}\neq 0,2n \endSb ^{\infty }\mid \frac{1}{%
(2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j}+O\left( (\frac{ln\left| n\right|
}{n%
})^{k}\right) -(2\pi (n-n_{1})^{2}}-
\]
\[
\frac{1}{(2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j}-(2\pi (n-n_{1})^{2}}\mid
=O\left( (\frac{ln\left| n\right| }{n})^{k+1}\right) .
\]
Now we prove (42). Writing the decomposition of $\Psi _{n,j}(x)$ by
$\{e^{i2\pi (n-n_{1})x}:n_{1}\in Z\}$ we obtain
\begin{equation}
\Psi _{n,j}(x)-u_{n,j},e^{i2\pi nx}=\sum\Sb n_{1}=-\infty , \\ n_{1}\neq 0
\endSb ^{\infty }(\Psi _{n,j}(x),e^{i2\pi (n-n_{1})x})e^{i2\pi (n-n_{1})x}.
\end{equation}
The right side of this formula can be obtained from the right side of (12)
by replacing $q_{n_{1}}$ with $e^{i2\pi (n-n_{1})x}.$ Therefore doing the
iteration which was dane in order to obtain (15) from (12) ( namely, isolate
the terms in right-hand side of (46) containing the multiplicant $(\Psi
_{n,j}(x),e^{-i2\pi nx})$ (i.e., case $n_{1}=2n$ ) and apply (9), (10) to
other terms (i.e., cases $n_{1}\neq 2n$ ), then isolate the terms containing
one of the multiplicants $(\Psi _{n,j}(x),e^{-i2\pi nx})$, $(\Psi
_{n,j}(x),e^{-i2\pi nx})$ and apply (9), (10) to other terms , ets.) and the
estimation as in (16) we get
\begin{eqnarray}
\Psi _{n,j}(x) &=&u_{n,j}e^{i2\pi nx}+v_{n,j}e^{-i2\pi nx}+ \\
&&u_{n,j}A_{m}^{\ast }(\lambda _{n,j})+v_{n,j}B_{m}^{\ast }(\lambda
_{n,j})+O\left( (\frac{ln\left| n\right| }{n})^{m+1}\right) \nonumber
\end{eqnarray}
Dividing both sides of (47) by $u_{n,j}$ and using (43), (41), (45) we get
(42).
Instead of (3), (7), (9) and Theorem 1 using (4), (8), (38) and Theorem 2 in
the same way we get the similar asymptotic formulas for eigenvalues and
eigenfunctions of $L_{\pi }(q)\blacksquare $
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\end{document}