# On the Bands of the Schrödinger Operator with a Matrix Potential

O.A.Veliev
Dogus University, Istanbul, Turkey.
e-mail<EMAIL_ADDRESS>
###### Abstract

In this article we consider the one-dimensional Schrödinger operator $L(Q)$
with a Hermitian periodic $m\times m$ matrix potential $Q$. We investigate the
bands and gaps of the spectrum and prove that most of the positive real axis
is overlapped by $m$ bands. Moreover, we find a condition on the potential $Q$
for which the number of gaps in the spectrum of $L(Q)$ is finite.

Key Words: Self-adjoint differential operator, Spectral bands, Periodic matrix
potential.

AMS Mathematics Subject Classification: 34L05, 34L20.

##  1 Introduction and Preliminary Facts

Let $L(Q)$ be the differential operator generated in the space
$L_{2}^{m}(-\infty,\infty)$ of the vector functions
$y=\left(y_{1},y_{2},...,y_{m}\right)$ by the differential expression

$-y^{{}^{\prime\prime}}+Qy,$ (1)

where $y_{k}\in L_{2}(-\infty,\infty)$ for $k=1,2,...,m,$
$Q(x)=\left(q_{s,j}(x)\right)$ is a $m\times m$ Hermitian matrix for all
$x\in(-\infty,\infty)$, $q_{s,j}$ is the complex-valued locally square
summable function and $Q\left(x+1\right)=Q\left(x\right)$. It is well-known
that [2, Chap.XIII], [4, 10, 13] the spectrum $\sigma(L(Q))$ of the operator
$L(Q)$ is the union of the spectra of the operators $L_{t}(Q)$ for
$t\in(-\pi,\pi],$ where $L_{t}(Q)$ is the operator generated in
$L_{2}^{m}[0,1]$ by the differential expression (1) and the quasiperiodic
conditions $y\left(1\right)=e^{it}y\left(0\right),$
$y^{{}^{\prime}}\left(1\right)=e^{it}y^{{}^{\prime}}\left(0\right).$ For
$t\in(-\pi,\pi]$ the spectra $\sigma(L_{t}(Q))$ of the operators $L_{t}(Q)$
consist of the eigenvalues

$\lambda_{1}(t)\leq\lambda_{2}(t)\leq\cdot\cdot\cdot$ (2)

called the Bloch eigenvalues of $L(Q)$. The $n$-th band function $\lambda_{n}$
continuously depends on $t$ and its range

$I_{n}(Q)=\left\\{\lambda_{n}(t):t\in(-\pi,\pi]\right\\}$ (3)

is called the $n$-th band of the spectrum of $L$:

$\sigma(L(Q))={\textstyle\bigcup\limits_{n=1}^{\infty}}I_{n}(Q).$

The continuity of $\lambda_{n}$ in the case $m=1$ was proved in [12]. The
general case follows from the arguments of the perturbation theory described
in [6] and [12]. In the Remark 1 of the next section, for the independence of
this paper, we give a proof of this statement within the framework of this
paper. The bands $I_{n}$ approach infinity as $n\rightarrow\infty.$ The spaces
between the bands $I_{k}$ and $I_{k+1}$ (if exist) for $k=1,2,...,$ are called
the gaps in the spectrum of $L(Q).$

In this paper we investigate the set of the Bloch eigenvalues, bands and gaps
of $L(Q)$. For this first we consider the set of the Bloch eigenvalues of the
operators $L(O)$ and $L(C),$ where $O$ is the $m\times m$ zero matrix and

$C=\int\limits_{[0,1]}Q\left(x\right)dx.$ (4)

It is clear that

$\varphi_{k,1,t}=\left(\begin{array}[c]{c}e_{k,t}\\\ 0\\\ \vdots\\\
0\end{array}\right),\text{ }\varphi_{k,2,t}=\left(\begin{array}[c]{c}0\\\
e_{k,t}\\\ \vdots\\\ 0\end{array}\right),...,\text{
}\varphi_{k,m,t}=\left(\begin{array}[c]{c}0\\\ \vdots\\\ 0\\\
e_{k,t}\end{array}\right)$

are the eigenfunctions of the operator $L_{t}(O)$ corresponding to the
eigenvalue $\left(2\pi k+t\right)^{2}$, where $e_{k,t}(x)=e^{i\left(2\pi
k+t\right)x}.$ If $t\neq 0,\pi$ then the multiplicity of the eigenvalue
$\left(2\pi k+t\right)^{2}$ is $m$ and the corresponding eigenspace is

$E_{k}(t)=Span\left\\{\varphi_{k,1,t},\varphi_{k,2,t},...,\varphi_{k,m,t}\right\\}.$

In the cases $t=0$ and $t=\pi$ the multiplicities of the eigenvalues
$\left(2\pi k\right)^{2}$ for
$k\in\left(\mathbb{Z}\backslash\left\\{0\right\\}\right)$ and $\left(2\pi
k+\pi\right)^{2}$ for $k\in\mathbb{Z}$ are $2m$ and the corresponding
eigenspaces are

$E_{k}(0)=Span\left\\{\varphi_{n,j,0}:n=k,-k;\text{ }j=1,2,...m\right\\}$

and

$E_{k}(\pi)=Span\left\\{\varphi_{n,j,\pi}:n=k,-(k+1);\text{
}j=1,2,...m\right\\}$

respectively. Thus the points $\left(2\pi k\right)^{2}$ for
$k\in\left(\mathbb{Z}\backslash\left\\{0\right\\}\right)$ and $\left(2\pi
k+\pi\right)^{2}$ for $k\in\mathbb{Z}$ are the exceptional Bloch eigenvalues
of $L(O),$ in the sense that at these points the multiplicities of the
eigenvalues are changed. It is clear that $\left(2\pi k+t\right)^{2}$ is an
exceptional Bloch eigenvalue of $L(O)$ if and only if

$\left(2\pi k+t\right)^{2}=\left(2\pi n+t\right)^{2}$ (5)

for some $n\neq k.$ Since (5) holds only in the cases $t=0,n=-k,$ $k\neq 0$
and $t=\pi,$ $n=-k-1,$ only the points $\left(2\pi k\right)^{2}$ for
$k\in\left(\mathbb{Z}\backslash\left\\{0\right\\}\right)$ and $\left(2\pi
k+\pi\right)^{2}$ for $k\in\mathbb{Z}$ are the exceptional Bloch eigenvalues
of $L(O).$

To analyze the set of the Bloch eigenvalues and the spectrum of the operator
$L(C)$ we introduce the following notations, where $C$ is defined in (4).
Denote by $\mu_{1}<\mu_{2}<...<\mu_{p}$ the distinct eigenvalues of the
Hermitian matrix $C$. If the multiplicity of $\mu_{j}$ is $m_{j},$ then
$m_{1}+m_{2}+...+m_{p}=m$. Let $u_{j,1},$ $u_{j,2},...,u_{j,m_{j}}$ be the
eigenvectors of the matrix $C$ corresponding to the eigenvalue $\mu_{j}.$ It
is not hard to see that

$\Phi_{k,j,s,t}(x)=u_{j,s}e^{i\left(2\pi k+t\right)x}$

for $s=1,2,...,m_{j}$ are the eigenfunctions of $L_{t}(C)$ corresponding to
the eigenvalue

$\mu_{k,j}(t)=\left(2\pi k+t\right)^{2}+\mu_{j}.$ (6)

To consider the spectrum of $L(Q)$ we use the following result of [16].

Theorem 4$(a)$ of [16]. All large eigenvalues of $L_{t}\left(Q\right)$ lie in
$\varepsilon_{k}$ neighborhood

$U_{\varepsilon_{k}}(\mu_{k,j}(t)):=(\mu_{k,j}(t)-\varepsilon_{k},\mu_{k,j}(t)+\varepsilon_{k})$

of the eigenvalues $\mu_{k,j}(t)$ of $L_{t}(C),$ where
$\varepsilon_{k}=c_{1}(\mid\frac{\ln|k|}{k}\mid+q_{k}),$

$q_{k}=\max\left\\{\left|\int\limits_{[0,1]}q_{s,r}\left(x\right)e^{-2\pi
inx}dx\right|:s,r=1,2,...,m;\text{ }n=\pm 2k,\pm(2k+1)\right\\},$

$c_{1}$ is a constant and does not depend on $t\in(-\pi,\pi].$ Moreover, for
each large eigenvalue $\mu_{k,j}(t)$ of $L_{t}(C)$ there exists an eigenvalue
of $L_{t}(Q)$ lying in $\varepsilon_{k}$ neighborhood of $\mu_{k,j}(t)$.

Now let’s explain a brief outline of this paper. Using Theorem 4$(a)$ of [16],
we first prove that most of the positive real axis is overlapped by $m$ bands
and estimate the length of the gap between the bands (see Theorems 1 and 2).
Then, in order to investigate the spectrum of $L(Q)$ in detail by using the
asymptotic formulas and perturbation theory we consider the multiplicities of
the eigenvalues of $L_{t}(C)$ and the large exceptional points of the spectrum
of $L(C).$ We consider the operator $L(Q)$ as perturbation of $L(C)$ by $Q-C$
and prove that the perturbation $Q-C$ may generate the gaps in
$\sigma\left(L(Q)\right)$ only at the neighborhoods of the exceptional Bloch
eigenvalues (see Theorem 3 and Corollary 2) of $L(C)$. In Theorem 4 we find a
condition (see Condition 1) on the eigenvalues of the matrix $C$ for which the
number of gaps in the spectrum of $L(Q)$ is finite. Note that in [16] we
proved Theorem 4 under the assumption that the matrix $C$ has three simple
eigenvalues $\mu_{j_{1}},$ $\mu_{j_{2}}$ and $\mu_{j_{3}}$ satisfying
Condition 1. These assumption simplifies the proof of Theorem 4. In this paper
we prove Theorem 4 without any conditions on the multiplicity of these
eigenvalues. Finally, note that in [8, 14, 15] we studied the non-self-adjoint
operators with a periodic matrix potential. This paper can be considered as
continuation of the paper [16], in which the self-adjoint case was
investigated. The self-adjoint case, was considered also in [1], where the
main goal was to reformulate some spectral problems for the differential
operator with periodic matrix coefficients as problems of conformal mapping
theory.

Finally note that a great number of paper is devoted to the scalar case
($m=1)$ (see for example the monographs [3] and [7] and the paper [9]). In
this paper we consider the finite-zone potentials in the vectorial case.
Therefore let us only stress the significant difference between the scalar and
vectorial cases in the investigations of the finite-zone potentials. In case
$m=1$ the finite zone potentials are infinitely differentiable functions and
have a special form expressed by Riemann $\theta$ function (see [7, Chapters 8
and 9] and [5]), while in the vectorial case we guarantee finite number of
gaps under simple algebraic condition on the eigenvalue of the matrix $C$.
Moreover, the method used in this paper for the investigation of the vectorial
case is absolutely different from the methods used in the scalar case.

##  2 Main Results

One can easily verify that the set

$\left\\{\left(2\pi k+t\right)^{2}:t\in(-\pi,\pi],\text{
}k\in\mathbb{Z}\right\\}$

of Bloch eigenvalues $\left(2\pi k+t\right)^{2}$ of $L(O)$ overlap $2m$ times
(counting the multiplicity) the half line $(0,\infty)$, since $\left(2\pi
k+t\right)^{2}=\left(-2\pi k-t\right)^{2}$ for $t\in(0,\pi)$ and the
multiplicity of the eigenvalues $\left(2\pi k\right)^{2}$ for
$k\in\left(\mathbb{Z}\backslash\left\\{0\right\\}\right)$ and $\left(2\pi
k+\pi\right)^{2}$ for $k\in\mathbb{Z}$ is $2m.$ Since the both Bloch
eigenvalues $\left(2\pi k+t\right)^{2}$ and $\left(-2\pi k-t\right)^{2}$
belong to the same band of $L(O),$ any element of the half line $(0,\infty)$
is overlapped by $m$ bands of $L(O)$. To investigate this overlapping problem
for the operator $L(Q)$ let us note that the eigenvalues of $L_{t}(Q)$
numbered in non-decreasing order (see (2)) continuously depend on
$t\in(-\pi,\pi].$

###### Remark 1

Here we prove (within the framework of this paper) that the eigenvalues
$\lambda_{n}(t)$ defined in (2) continuously depend on $t$. The eigenvalues of
the operator $L_{t}(Q)$ are the roots of the characteristic determinant

$\Delta(\lambda,t)=\det(U_{v}(Y_{j}))_{j,\nu=1}^{2}=$
$e^{i2mt}+f_{1}(\lambda)e^{i(2m-1)t}+f_{2}(\lambda)e^{i(2m-2)t}+...+f_{2m-1}(\lambda)e^{it}+1$

which is a polynomial of $e^{it}$ with entire coefficients
$f_{1}(\lambda),f_{2}(\lambda),...$, where

$U_{v}(Y_{j})=Y_{j}^{(\nu-1)}(1,\lambda)-e^{it}Y_{j}^{(\nu-1)}(0,\lambda),$

$Y_{1}(x,\lambda)$ and $Y_{2}(x,\lambda)$ are the solutions of the matrix
equation

$-Y^{{}^{\prime\prime}}(x)+Q\left(x\right)Y(x)=\lambda Y(x)$

satisfying $Y_{1}(0,\lambda)=O$, $Y_{1}^{{}^{\prime}}(0,\lambda)=I$ and
$Y_{2}(0,\lambda)=I$, $Y_{2}^{{}^{\prime}}(0,\lambda)=O$ (see [11] Chapter 3).

Now using these statements we prove that for each $n$ the function
$\lambda_{n}$ defined in (3) is continuous at each point $t_{0}\in(-\pi,\pi].$
Since $\lambda_{n}(t_{0})\rightarrow\infty$ as $n\rightarrow\infty,$ there
exist $k\leq n$ and $p\geq n$ such that
$\lambda_{k-1}(t_{0})<\lambda_{k}(t_{0})=\lambda_{k+1}(t_{0})=...=\lambda_{p}(t_{0})<\lambda_{p+1}(t_{0})$
if $\lambda_{n}(t_{0})>\lambda_{1}(t_{0}).$ Then the boundaries of the
rectangles

$R_{1}=\left\\{c<x<d_{1},\text{ }\left|y\right|<1\right\\}\text{ and
}R_{2}=\left\\{c<x<d_{2},\text{ }\left|y\right|<1\right\\}$

belong to the resolvent set of the operator $L_{t_{0}}(Q),$ where
$\lambda_{k-1}(t_{0})<d_{1}<\lambda_{k}(t_{0}),$
$\lambda_{p}(t_{0})<d_{2}<\lambda_{p+1}(t_{0})$ and $c$ is a number for which
$\sigma(L)\subset(c,\infty).$ It implies that $\Delta(\lambda,t_{0})\neq 0$
for each $\lambda\in\partial(R_{1}).$ Since $\Delta(\lambda,t_{0})$ is a
continuous function on the compact $\partial(R_{1}),$ there exists $a>0$ such
that $\left|\Delta(\lambda,t_{0})\right|>a$ for all
$\lambda\in\partial(R_{1}).$ Moreover, $\Delta(\lambda,t)$ is a polynomial of
$e^{it}$ with entire coefficients. Therefore, there exists $\delta_{1}>0$ such
that $\left|\Delta(\lambda,t)\right|>a/2$ for all
$t\in(t_{0}-\delta_{1},t_{0}+\delta_{1})$ and $\lambda\in\partial(R_{1}).$ It
means that $\partial(R_{1})$ belong to the resolvent set of $L_{t}(Q)$ for all
$t\in(t_{0}-\delta_{1},t_{0}+\delta_{1}).$ Moreover

$\left(L_{t}-\lambda
I\right)^{-1}f(x)=\int_{0}^{1}G(x,\xi,\lambda,t)f(\xi)d\xi,$

where $G(x,\xi,\lambda,t)$ is the Green’s function of $L_{t}-\lambda I$
defined by formula

$G(x,\xi,\lambda,t)=g(x,\xi,\lambda)-\frac{1}{\Delta(\lambda,t)}\sum\limits_{j,v=1}^{2}Y_{j}(x,\lambda)V_{jv}(x,\lambda)U_{v}(g),$

(see formula (8) of [7, p.117]). Here $g$ does not depend on $t$ and $V_{jv}$
is the transpose of that mth-order matrix consisting of the cofactor of the
element $U_{v}(Y_{j})$ in the determinant $\det(U_{v}(Y_{j}))_{j,\nu=1}^{2}.$
Hence the entries of the matrices $V_{jv}(x,\lambda)$ and $U_{v}(g)$ either do
not depend on $t$ or have the form $u(1,\lambda)-e^{it}u(0,\lambda)$ and
$h(1,\xi,\lambda)-e^{it}h(0,\xi,\lambda)$ respectively, where the functions
$u$ and $h$ do not depend on $t.$ Therefore using these formulas and the last
inequality for $\left|\Delta(\lambda,t)\right|$ one can easily verify that
$\left(L_{t}-\lambda I\right)^{-1}$ continuously depend on
$t\in(t_{0}-\delta_{1},t_{0}+\delta_{1})$ for $\lambda\in\partial(R_{1}).$
This implies that the operators $L_{t}$ for each
$t\in(t_{0}-\delta_{1},t_{0}+\delta_{1})$ have $k-1$ eigenvalues in $R_{1},$
since $L_{t_{0}}$ have $k-1$ eigenvalues in $R_{1}$. It is clear that these
eigenvalues are $\lambda_{1}(t),\lambda_{2}(t),...,\lambda_{k-1}(t).$ In the
same way we prove that there exists $\delta_{2}>0$ such that the operators
$L_{t}$ for $t\in(t_{0}-\delta_{2},t_{0}+\delta_{2})$ have $p$ eigenvalues in
$R_{2}$ and they are $\lambda_{1}(t),\lambda_{2}(t),...,\lambda_{p}(t).$ Thus
the closed rectangle $R=\left\\{d_{1}\leq x\leq d_{2},\text{
}\left|y\right|\leq 1\right\\}$ contains $p-k+1$ eigenvalues of $L_{t}$ for
$t\in(t_{0}-\delta,t_{0}+\delta)$ and they are
$\lambda_{k}(t),\lambda_{k+1}(t),...,\lambda_{p}(t),$ where
$\delta=\min\left\\{\delta_{1},\delta_{2}\right\\}$ and $n\in[k,p].$

Now we are ready to prove that $\lambda_{n}$ is continuous at the point
$t_{0}$ if $\lambda_{n}(t_{0})>\lambda_{1}(t_{0}).$ Consider any sequence
$\left\\{\left(\lambda_{n}(t_{k}),t_{k}\right):k\in\mathbb{N}\right\\}$ such
that $t_{k}\in(t_{0}-\delta,t_{0}+\delta)$ for all $k\in\mathbb{N}$ and
$t_{k}\rightarrow t_{0}$ as $k\rightarrow\infty.$ Let
$\left(\lambda,t_{0}\right)$ be any limit point of the sequence
$\left\\{\left(\lambda_{n}(t_{k}),t_{k}\right):k\in\mathbb{N}\right\\}.$ Since
$\Delta$ is a continuous function with respect to the pair $(\lambda,t)$ and
$\Delta\left(\lambda_{n}(t_{k}),t_{k}\right)=0$ for all $k$ we have
$\Delta(\lambda,t_{0})=0.$ It means that $\lambda$ is an eigenvalue of
$L_{t_{0}}(Q)$ lying in the rectangle $R$, that is,
$\lambda=\lambda_{k}(t_{0})=$ $\lambda_{k+1}(t_{0})=...=\lambda_{p}(t_{0}),$
where $n\in[k,p].$ Thus $\lambda_{n}(t_{k})\rightarrow\lambda_{n}(t_{0})$ as
$k\rightarrow\infty$ for any sequence $\left\\{t_{k}:k\in\mathbb{N}\right\\}$
converging to $t_{0}$ and $\lambda_{n}$ is continuous at the point $t_{0}$ if
$\lambda_{n}(t_{0})>\lambda_{1}(t_{0}).$ To prove the case
$\lambda_{n}(t_{0})=\lambda_{1}(t_{0})$ it is enough to consider only the
rectangle $R_{2}=\left\\{c<x<d_{2},\text{ }\left|y\right|<1\right\\},$ where
$\lambda_{1}(t_{0})=\lambda_{2}(t_{0})=...=\lambda_{p}(t_{0})<d_{2}<\lambda_{p+1}(t_{0}).$

First using this Remark and Theorem 4$(a)$ of [16] we consider the overlapping
problem for $L(Q).$ For this in the following remark we explain Theorem 4$(a)$
of [16] for the family of the operators $L(C+\varepsilon(Q-C))$ for
$\varepsilon\in[0,1].$

###### Remark 2

To prove Theorem 4$(a)$ we used the formula

$(\lambda_{k,j}(t)-\mu_{n,i}(t))(\Psi_{k,j,t},\Phi_{n,i,t})=((Q(x)-C)\Psi_{k,j,t},\Phi_{n,i,t})$
(7)

and proved that

$\left(\Psi_{k,j,t}(x),(Q\left(x\right)-C)\Phi_{n,i,t}(x)\right)=O(\frac{\ln|k|}{k})+O(b_{k}).$
(8)

Moreover, the last estimation does not depend on $t$. If the potential $Q$ is
replaced by $C+\varepsilon(Q-C)$ then the formula (7) has the form

$(\lambda_{k,j}(t)-\mu_{n,i}(t))(\Psi_{k,j,t},\Phi_{n,i,t})=(\varepsilon(Q(x)-C)\Psi_{k,j,t},\Phi_{n,i,t})$
(9)

Instead of (7) using (9) and repeating the proof of (8) we get

$\left(\Psi_{k,j,t}(x),\varepsilon(Q\left(x\right)-C)\Phi_{n,i,t}(x)\right)=O(\frac{\ln|k|}{k})+O(\varepsilon
q_{k}).$

Therefore, repeating the proof of Theorem 4$(a)$ of [16] we obtain that all
large eigenvalues of $L_{t}(C+\varepsilon(Q-C))$ for all $\varepsilon\in[0,1]$
lie in $\varepsilon_{k}=:c_{1}(\mid\frac{\ln|k|}{k}\mid+q_{k})$ neighborhood
of the eigenvalues $\mu_{k,j}(t)$ for $\left|k\right|\geq N$ and $j=1,2,...p,$
where the constants $N$ and $c_{1}$ do not depend on $t$ and $\varepsilon.$

Now we are ready to prove the following theorem about the overlapping problem.

###### Theorem 1

There exists a positive integer $N_{1}$ such that if $s\geq N_{1}$ then the
intervals

$I(s):=\left[(s\pi)^{2}+\mu_{p}+\varepsilon(s),\left(s\pi+\pi\right)^{2}+\mu_{1}-\varepsilon(s)\right]$

are contained in each of the bands

$I_{sm+1}(Q),I_{sm+2}(Q),...,I_{sm+m}(Q),$

where $\varepsilon(s)=\varepsilon_{k}$ if $s\in\left\\{2k,2k+1\right\\}$ and
$\varepsilon_{k}$ is defined in Remark 2.

Proof. First consider the case $s=2k.$ One can easily verify that the number
of the periodic Bloch eigenvalues of $L(C)$ (the eigenvalues of $L_{0}(C)$
counting the multiplicity) lying in the interval

$\left[c,\left(s\pi\right)^{2}+\mu_{p}+\varepsilon(s)\right]$

is $sm+m,$ where $c$ is a constant such that the spectra of the operators
$L(C+\varepsilon(Q-C))$ are contained in $(c,\infty)$ for all
$\varepsilon\in[0,1].$ It follows from Remark 2 that there exist constants
$N_{1}$ and $c$ such that if $s\geq N_{1},$ then the boundary of the rectangle

$R_{1}=\left\\{c<x<\left(s\pi\right)^{2}+\mu_{p}+\varepsilon(s),\text{
}\left|y\right|<1\right\\}$

belong to the resolvent set of the operators $L_{0}(C+\varepsilon(Q-C))$ for
all $\varepsilon\in[0,1]$. Hence, the projection of
$L_{0}(C+\varepsilon(Q-C))$ defined by contour integration over the boundary
of $R_{1}$ depends continuously on $\varepsilon.$ It implies that the number
of eigenvalues (counting the multiplicity) of $L_{t}(C+\varepsilon(Q-C))$
lying in $R_{1}$ are the same for all $\varepsilon\in[0,1].$ Since $L_{0}(C)$
has $sm+m$ eigenvalues (counting the multiplicity) in $R_{1},$ the operator
$L_{0}(Q)$ has also $sm+m$ eigenvalues.

In the same way we prove that the rectangle

$R_{2}=\left\\{c<x<\left(s\pi\right)^{2}+\mu_{1}-\varepsilon(s),\text{
}\left|y\right|<1\right\\}$

contains $sm-m$ eigenvalues of $L_{0}(Q).$ Therefore the interval

$\left(\left(s\pi\right)^{2}+\mu_{1}-\varepsilon(s),\left(s\pi\right)^{2}+\mu_{p}+\varepsilon(s)\right)$
(10)

contains $2m$ periodic eigenvalues and they are

$\lambda_{sm-m+1}(0)\leq\lambda_{sm-m+2}(0)\leq\cdot\cdot\cdot\leq\lambda_{sm+m}(0).$

In the similar way we prove that the interval

$\left(\left(s\pi+\pi\right)^{2}+\mu_{1}-\varepsilon(s),\left(s\pi+\pi\right)^{2}+\mu_{p}+\varepsilon(s)\right)$
(11)

contains $2m$ antiperiodic eigenvalues and they are

$\lambda_{sm+1}(\pi)\leq\lambda_{sm+2}(\pi)\leq\cdot\cdot\cdot\leq\lambda_{sm+2m}(\pi)$

Thus the bands $I_{r}$ for $r=sm+1,sm+2,...,sm+m$ contain the point
$\lambda_{r}(0)$ from interval (10) and the point $\lambda_{r}(\pi)$ from
interval (11). Therefore the bands $I_{sm+1},I_{sm+2},...,I_{sm+m}$ contains
the interval $I(s).$ In the same way we prove the case $s=2k+1.$

Theorem 1 immediately imply the following.

###### Corollary 1

Any spectral gap $(\alpha,\beta)$ with $\alpha>\left(\pi N_{1}\right)^{2}$ (if
exists) is contained in the intervals

$U(s):=(\left(\pi s\right)^{2}+\mu_{1}-\varepsilon(s-1),\left(\pi
s\right)^{2}+\mu_{p}+\varepsilon(s)),$

for $s>N_{1}$. Moreover, the spectral gap $(\alpha,\beta)\subset U(s)$ lies
between the bands $I_{sm}(Q)$ and $I_{sm+1}(Q).$

Proof. By Theorem 1 the intervals $I(s)$ for $s\geq N_{1}$ are the subsets of
the spectrum. Therefore the spectral gap $(\alpha,\beta)$ with
$\alpha>\left(\pi N_{1}\right)^{2}$ is contained between $I(s-1)$ and $I(s)$
for some $s>N_{1}.$ It means that $(\alpha,\beta)\subset U(s).$ The bands
$I_{sm+j}(Q)$ and $I_{sm+j+1}(Q)$ for $j=1,2,...,m-1$ have common intervals
$I(s)$ and hence there is not gaps between they. It means that the gap
$(\alpha,\beta)$ is located between the bands $I_{ms}$ and $I_{ms+1}.$

Now using Corollary 1 and Theorem 4$(a)$ of [16] we prove the following.

###### Theorem 2

The length of the spectral gaps $(\alpha,\beta)$ lying in $U(s)$ for $s>N_{1}$
is not greater than $2\max\left\\{\varepsilon(s-1),\varepsilon(s)\right\\}.$

Proof. By Corollary 1 we have

$(\left(s\pi\right)^{2}+\mu_{1}-\varepsilon(s-1)\leq\alpha<\beta\leq\left(\pi
s\right)^{2}+\mu_{p}+\varepsilon(s).$

Now suppose on the contrary that the length of the gap $(\alpha,\beta)$ is
greater than $2\max\left\\{\varepsilon(s-1),\varepsilon(s)\right\\}.$ Then it
follows from last equalities that

$\frac{\alpha+\beta}{2}\in(\left(s\pi\right)^{2}+\mu_{1},\left(\pi
s\right)^{2}+\mu_{p}),\text{
}\frac{\beta-\alpha}{2}>\max\left\\{\varepsilon(s-1),\varepsilon(s)\right\\}$

Using (6) one can easily conclude that there exist $t\in(-\pi,\pi]$ and
$j\in\left\\{1,2,...,p\right\\}$ such that the equality
$\mu_{s,j}(t)=\frac{\alpha+\beta}{2}$ holds. On the other hands, by Theorem
4$(a)$ of [16] there exists an eigenvalue $\lambda$ of $L_{t}(Q)$ lying in
$\max\left\\{\varepsilon(s-1),\varepsilon(s)\right\\}$ neighborhood of
$\frac{\alpha+\beta}{2}.$ Therefore

$\lambda\in\left(\frac{\alpha+\beta}{2}-\frac{\beta-\alpha}{2},\frac{\alpha+\beta}{2}+\frac{\beta-\alpha}{2}\right)=(\alpha,\beta).$

Hence $(\alpha,\beta)$ is not a gap in the spectrum of $L(Q).$ This
contradiction imply the proof of the theorem.

To investigate the spectrum of $L(Q)$ in detail by using the asymptotic
formulas and perturbation theory we need to consider the multiplicities of the
eigenvalues of $L_{t}(C)$ and the exceptional points of the spectrum of
$L(C).$ The multiplicity of $\mu_{k,j}(t)$ is $m_{j}$ if
$\mu_{k,j}(t)\neq\mu_{n,i}(t)$ for all $(n,i)\neq(k,j).$ The multiplicity of
$\mu_{k,j}(t)$ is changed, that is, $\mu_{k,j}(t)$ is an exceptional point of
the spectrum of $L(C)$ if

$\left(2\pi k+t\right)^{2}+\mu_{j}=\left(2\pi n+t\right)^{2}+\mu_{i}$ (12)

for some $(n,i)\neq(k,j).$ Since $\left(2\pi k+t\right)^{2}=\left(-2\pi
k-t\right)^{2}$, it is enough to study the equality (12) for $t\in[0,\pi]$ and
$k\in\mathbb{Z}.$ Moreover, we investigate the large exceptional Bloch
eigenvalues $\left(2\pi k+t\right)^{2}+\mu_{j}$ of $\sigma(L(C)),$ because we
are going to investigate the spectrum of the operator $L(Q)$ by using the
asymptotic formulas for the large eigenvalues. In other words, we need to
consider (12) in the case when $\left|k\right|$ is a large number. Then (12)
has a solution $t\in[0,\pi]$ only in the cases $n=-k$ and $n=-k-1.$ In these
cases (12) implies that the large eigenvalue $\left(2\pi
k+t\right)^{2}+\mu_{j}$ is an exceptional Bloch eigenvalue of $L(C)$ if either

$\mu_{k,j}(t)-\mu_{-k,i}(t)=8\pi kt+\mu_{j}-\mu_{i}=0$ (13)

or

$\mu_{k,j}(t)-\mu_{-k-1,i}(t)=4\pi(2k+1)\left(t-\pi\right)+\mu_{j}-\mu_{i}=0$
(14)

for some $i=1,2,...,p.$ We denote by

$t(2k,j,i)=\frac{\mu_{i}-\mu_{j}}{4\pi(2k)}$

and

$t(2k+1,j,i)=\pi+\frac{\mu_{i}-\mu_{j}}{4\pi(2k+1)}$

the solutions of equations (13) and (14) lying in $[0,\pi]$. Thus eliminating
the sets $\left\\{t(2k,j,i):i=1,2,...,p\right\\}$ and
$\left\\{t(2k+1,j,i):i=1,2,...,p\right\\}$ from $[0,\pi]$ we conclude that, if
$t$ belongs to the remaining part of $[0,\pi],$ then the multiplicity of the
eigenvalue $\mu_{k,j}(t)$ is $m_{j}$, where $k$ is a large number. In other
word, $\mu_{k,j}(t)$ is a non-exceptional Bloch eigenvalue of $L(C).$ However,
to investigate the perturbation of these non-exceptional Bloch eigenvalues by
using the asymptotic formulas obtained in [16], we eliminate
$\delta_{k}$-neighborhoods $U_{\delta_{k}}(t(2k,j,i))$ and
$U_{\delta_{k}}(t(2k+1,j,i))$ of $t(2k,j,i)$ and $t(2k+1,j,i)$ from $[0,\pi],$
where $\delta_{k}=o(k^{-1}).$ Moreover, $\delta_{k}$ can be chosen so that the
remaining part of $[0,\pi]$ consists of the pairwise disjoint intervals
$\left[a(k,j,s),b(k,j,s)\right]$ for $s=1,2,...,v:$

$[0,\pi]\backslash\left({\textstyle\bigcup\limits_{i=1,2,...,p}}\left(U_{\delta_{k}}(t(2k,j,i))\cup
U_{\delta_{k}}(t(2k+1,j,i))\right)\right)=$ (15)
${\textstyle\bigcup\limits_{s=1}^{v}}\left[a(k,j,s),b(k,j,s)\right],$

where $a(k,j,s)<b(k,j,s)<a(k,j,s+1)<b(k,j,s+1)$ for $s=1,2,...,v.$ These
intervals have the following property.

###### Lemma 1

There exists $N_{2}$ such that if $\mid$ $k\mid>N_{2}$,

$4\pi(2\left|k\right|-2)\delta_{k}=2\max\left\\{\varepsilon_{k},\varepsilon_{-k},\varepsilon_{-k-1}\right\\},$
(16)

and the quasimomentum $t$ belongs to the intervals
$\left[a(k,j,s),b(k,j,s)\right]$ defined in (15), then the following
statements hold.

$(a)$ The inequality

$\left|\mu_{k,j}(t)-\mu_{n,i}(t)\right|\geq 4\pi(2\left|k\right|-1)\delta_{k}$
(17)

holds for $n=-k,-k-1$ and for all $i=1,2,...,p.$

$(b)$ The closed interval

$\overline{U_{\varepsilon_{k}}(\mu_{k,j}(t))}=\left[\mu_{k,j}(t)-\varepsilon_{k},\mu_{k,j}(t)+\varepsilon_{k}\right]$

has no common points with $\overline{U_{\varepsilon_{n}}(\mu_{n,i}(t))}\ $for
$\mid$ $n\mid\geq N_{2}$ and $(n,i)\neq\left(k,j\right)$.

Proof. $(a)$ Introduce the notations $f(t)=\mu_{k,j}(t)-\mu_{-k,i}(t)$ and
$g(t)=\mu_{k,j}(t)-\mu_{-k-1,i}(t).$ By the definition of $t(2k,j,i)$ and
$t(2k+1,j,i)$ we have $f(t(2k,j,i))=0$ and $g(t(2k+1,j,i))=0$ (see (13) and
(14)). On the other hand, the derivatives of the functions $f$ and $g$ are
$8\pi k$ and $4\pi(2k+1)$ respectively. Therefore if $t$ does not belong to
the $\delta_{k}$-neighborhood of $t(2k,j,i)$ and $t(2k+1,j,i)$ for
$i=1,2,...,p$ that is, if $t$ belong to the intervals
$\left[a(k,j,s),b(k,j,s)\right]$, then $\left|f(t)\right|\geq
8\pi\left|k\right|\delta_{k}$ and $\left|g(t)\right|\geq
4\pi\left|2k+1\right|\delta_{k}.$ These inequalities imply (17).

$(b)$ If (16) holds then it follows from (17) that the distance
$\left|\mu_{k,j}(t)-\mu_{n,i}(t)\right|$ between the centres of the intervals
$U_{\varepsilon_{k}}(\mu_{k,j}(t))$ and $U_{\varepsilon_{n}}(\mu_{n,i}(t))$ is
greater than the total sum $\varepsilon_{k}+\varepsilon_{n}$ of the radii of
these intervals for $n=-k,-k-1$. Therefore
$\overline{U_{\varepsilon_{k}}(\mu_{k,j}(t))}$ has no common points with the
intervals $\overline{U_{\varepsilon_{n}}(\mu_{n,i}(t))}\ $for $n=-k,-k-1$.
Similarly, if $n\neq k,-k,-k-1,$ $\mid$ $k\mid>N_{2},$ $\mid$ $n\mid\geq
N_{2}$ and $t\in[0,\pi]$ then $\left|\mu_{k,j}(t)-\mu_{n,i}(t)\right|$ is a
large number and hence is greater than $\varepsilon_{k}+\varepsilon_{n}.$ If
$n=k$ and $i\neq j,$ then
$\left|\mu_{k,j}(t)-\mu_{n,i}(t)\right|=\left|\mu_{j}-\mu_{i}\right|>\varepsilon_{k}+\varepsilon_{n},$
since $\varepsilon_{k}\rightarrow 0$ as $k\rightarrow\infty.$ The lemma is
proved.

Now using this lemma we consider the spectrum of $L(Q).$

###### Theorem 3

Let $j=1,2,...,p$ be fixed. For any interval
$\left[a,b\right]:=\left[a(k,j,s),b(k,j,s)\right]$ of (15) the following
statements hold.

$(a)$ If $t\in[a,b],$ then the operator $L_{t}(Q)$ has $m_{j}$ eigenvalues
(counting the multiplicity) lying in the interval
$U_{\varepsilon_{k}}(\mu_{k,j}(t))=\left(\mu_{k,j}(t)-\varepsilon_{k},\mu_{k,j}(t)+\varepsilon_{k}\right).$

$(b)$ There exists $l$ such that the eigenvalues of $L_{t}(Q)$ lying in
$U_{\varepsilon_{k}}(\mu_{k,j}(t))$ are
$\lambda_{l+1}(t),\lambda_{l+2}(t),...,\lambda_{l+m_{j}}(t)$ for all
$t\in[a,b].$

$(c)$ If $k>0$ ($k<0$), then the interval
$\left[\mu_{k,j}(a)+\varepsilon_{k},\mu_{k,j}(b)-\varepsilon_{k}\right]$

($\left[\mu_{k,j}(b)+\varepsilon_{k},\mu_{k,j}(a)-\varepsilon_{k}\right]$) is
a subset of the bands $\Gamma_{l+1},$ $\Gamma_{l+2},...,\Gamma_{l+m_{j}}$ of
the spectrum of $L(Q).$

Proof. $(a)$ Theorem 4$(a)$ of [16] implies that there exist $n$ and $N$ such
that the eigenvalues $\lambda_{s}(t)$ for $s>n$ lie in
$U_{\varepsilon_{k}}(\mu_{k,i}(t))$ for $\mid$ $k\mid>N$ and $i=1,2,...,p.$ On
the other hand, by Lemma 1 the integer $N$ can be chosen so that the closed
interval $\overline{U_{\varepsilon_{k}}(\mu_{k,j}(t))}$ for $t\in[a,b]$ and
$\mid$ $k\mid>N$ has no common points with the intervals
$\overline{U_{\varepsilon_{n}}(\mu_{n,i}(t))}\ $for $\mid$ $n\mid\geq N$ and
$(n,i)\neq\left(k,j\right).$ Therefore the circle

$\left\\{\lambda\in\mathbb{C}:\left|\lambda-\mu_{k,j}(t)\right|=\varepsilon_{k}\right\\}$
(18)

belong to the resolvent set of $L_{t}(Q).$ Repeating the proof of the case
$\varepsilon=1,$ one can easily verify that the circle (18) lies in the
resolvent sets of $L_{t}(C+\varepsilon(Q-C))$ for all $\varepsilon\in[0,1].$
Since $L_{t}(C)$ has $m_{j}$ eigenvalues (counting the multiplicity) in the
circle (18), the operator $L_{t}(Q)$ has also $m_{j}$ eigenvalues.

$(b)$ Let us denote the eigenvalues of $L_{t}(Q)$ lying in
$U_{\varepsilon_{k}}(\mu_{k,j}(t))$ by $\lambda_{l(t)+1}(t),$
$\lambda_{l(t)+2}(t),...,$ $\lambda_{l(t)+m_{j}}(t).$ We need to prove that
$l(t)$ does not depend on $t\in[a,b].$ Since we numerate the eigenvalues of
$L_{t}(Q)$ in nondecreasing order (see (2)) $\lambda_{l(t)}(t)$ and
$\lambda_{l(t)+m_{j}+1}(t)$ do not belong to the interval
$U_{\varepsilon_{k}}(\mu_{k,j}(t))$ and

$\lambda_{l(t)}(t)<\lambda_{l(t)+s}(t)<\lambda_{l(t)+m_{j}+1}(t)$ (19)

for all $s=1,2,...,m_{j}.$ It follows from the continuity of the band
functions and (19) that for each $t\in[a,b]$ there exists a neighborhood
$U(t)$ of $t$ such that

$\lambda_{l(t)}(y)<\lambda_{l(t)+s}(y)<\lambda_{l(t)+m_{j}+1}(y)$

for $y\in U(t).$ In the other words, $l(y)=l(t)$ for all $y\in U(t).$ Thus we
have

$\forall t\in[a,b],\exists U(t):l(y)=l(t),\forall y\in U(t).$ (20)

Let $U(t_{1}),U(t_{2}),...,U(t_{\omega})$ be a finite subcover of the open
cover $\\{U(t):t\in[a,b]\\}$ of the compact $[a,b],$ where $U(t)$ is the
neighborhood of $t$ satisfying (20). By (20), we have $l(y)=l(t_{i})$ for all
$y\in U(t_{i}).$ Clearly, if $U(t_{i})\cap U(t_{j})\neq\emptyset,$ then
$l(t_{i})=l(z)=l(t_{j}),$ where $z\in U(t_{i})\cap U(t_{j})$. Thus
$l(t_{1})=l(t_{2})=...=l(t_{\omega})$ and hence $l(t)$ does not depend on
$t\in[a,b].$

$(c)$ We consider the case $k>0.$ The case $k<0$ can be considered in the same
way. Since

$\lambda_{l+s}(a)\in\left(\mu_{k,j}(a)-\varepsilon_{k},\mu_{k,j}(a)+\varepsilon_{k}\right)$

and

$\lambda_{l+s}(b)\in\left(\mu_{k,j}(b)-\varepsilon_{k},\mu_{k,j}(b)+\varepsilon_{k}\right)$

for $s=1,2,...,m_{j}$ the interval
$\left[\mu_{k,j}(a)+\varepsilon_{k},\mu_{k,j}(b)-\varepsilon_{k}\right]$ is a
subset of $\Gamma_{l+s}$ for $s=1,2,...,m_{j}.$

Using this theorem and the construction of the intervals (15) we prove the
following consequence.

###### Corollary 2

There exists $N_{3}>\max\left\\{N,N_{1},N_{2}\right\\}$ and a sequence
$\left\\{\gamma_{k}\right\\}\rightarrow 0$ such that $\gamma_{k}\rightarrow 0$
as $k\rightarrow\infty$ and the spectral gap $(\alpha,\beta)$ defined in
Corollary 1 and lying in $U(k)$ for $k>N_{3}$ is contained in the intersection
of the sets $S(1,k),S(2,k),...,S(p,k),$ where

$S(j,k)={\textstyle\bigcup\limits_{i=1,2,...,p}}\left(\left(\pi
k\right)^{2}+\frac{\mu_{i}+\mu_{j}}{2}-\gamma_{k},\left(\pi
k\right)^{2}+\frac{\mu_{i}+\mu_{j}}{2}+\gamma_{k}\right).$

Proof. We say that the intervals $[A+\varepsilon,B-\varepsilon]$ and
$(A-\varepsilon,B+\varepsilon)$ are respectively the $\varepsilon>0$
contraction and extension of the intervals $[A,B]$ and $(A,B).$ In Theorem 3
$(c)$ we proved that the $\varepsilon_{k}$ contraction of the images
$\mu_{k,j}([a,b])$ of the intervals $[a,b]$ of (15) is a subset of the
spectrum of $L(Q).$ On the other hand, the intervals of (15) are obtained from
$[0,\pi]$ by eliminating the open intervals

$(t(2k,j,i)-\delta_{k},t(2k,j,i)+\delta_{k}),\text{
}(t(2k+1,j,i)-\delta_{k},t(2k+1,j,i)+\delta_{k})$ (21)

for $i=1,2,...,p.$ Therefore, it follows from (6) that the gaps in the
spectrum are the subset of the $\varepsilon_{k}$ extension of the images
$\mu_{k,j}((c,d))$ of the intervals $(c,d)$ of (21). Since

$\mu_{k,j}(t(2k,j,i))=\left(2\pi
k\right)^{2}+\frac{\mu_{i}+\mu_{j}}{2}+\left(\frac{\mu_{i}-\mu_{j}}{8\pi
k}\right)^{2}$

and

$\mu_{k,j}(t(2k+1,j,i))=\left(2\pi
k+\pi\right)^{2}+\frac{\mu_{i}+\mu_{j}}{2}+\left(\frac{\mu_{i}-\mu_{j}}{4\pi(2k+1)}\right)^{2},$

using (6) and (16) we obtain that for any interval $(c,d)$ of (21) the
interval $\mu_{k,j}((c,d))$ and hence its $\varepsilon_{k}$ extension are
contained in $S(j,k)$ for each $j=1,2,...,p.$ The corollary is proved.

Now we find a condition on the eigenvalues of the matrix $C$ for which the
spectrum of $L(Q)$ contains the interval $(H,\infty)$ for some constant $H$.
If the matrix $C$ has only one eigenvalue $\mu$ with multiplicity $m,$ then it
is possible that the spectrum of $L(Q)$ has infinitely many gaps. For example,
if $Q=qI,$ where $q$ is not a finite zone scalar potential and $I$ is the
$m\times m$ unit matrix then the spectrum of $L(Q)$ has infinitely many gaps.
If the matrix $C$ has only two eigenvalues $\mu_{1}$ and $\mu_{2},$ then the
sets $S(1,k)$ and $S(2,k)$ have a common interval

$\left(\left(2\pi k\right)^{2}+\frac{\mu_{1}+\mu_{2}}{2}-\gamma_{k},\left(2\pi
k\right)^{2}+\frac{\mu_{1}+\mu_{2}}{2}+\gamma_{k}\right).$

Therefore, Corollary 2 does not imply that the number of the gaps in the
spectrum of $L(Q)$ is finite. However, we prove that if the number of
different eigenvalues of the matrix $C$ is greater than $2,$ then three sets
$S(j_{1},k),S(j_{2},k)$ and $S(j_{3},k)$ for the large values of $k$ have no
common intervals if the following condition holds.

###### Condition 1

Suppose that there exists a triple $(j_{1},j_{2},j_{3})$ such that

$\min_{i_{1},i_{2},i_{3}}\left(diam(\\{\mu_{j_{1}}+\mu_{i_{1}},\mu_{j_{2}}+\mu_{i_{2}},\mu_{j_{3}}+\mu_{i_{3}}\\})\right)=d\neq
0,$

where minimum is taken under condition $i_{s}\in\left\\{1,2,...,p\right\\}$
for $s=1,2,3$ and

$diam(E)=\sup_{x,y\in E}\mid x-y\mid.$

Let us first discuss why gaps in $\sigma(L(Q))$ do not appear in the interval
$(H,\infty)$ if $H$ is a large number and Condition 1 holds. Then in Theorem 4
we give the mathematical proof of this statement. For each
$j\in\left\\{1,2,3\right\\}$ the set

$\sigma_{j}\left(L(C)\right)=\left\\{\left(2\pi
k+t\right)^{2}+\mu_{j}:k\in\mathbb{Z},\text{ }t\in(-\pi,\pi]\right\\}$

(let us call it $j$ spectrum) cover the interval $(H,\infty).$ The
perturbation $Q-C$ may generate a gap in $\sigma_{j}\left(L(C)\right)$ only at
the neighborhood of the exceptional Bloch eigenvalues $\left(2\pi
k+t\right)^{2}+\mu_{j}$ (let us call it $j$ exceptional Bloch eigenvalue). On
the other hand, Condition 1 implies that the $j_{1},$ $j_{2}$ and $j_{3}$
exceptional Bloch eigenvalues have no common points. That is why, for each
$\lambda\in(H,\infty)$ there exists $s\in\left\\{1,2,3\right\\}$ such that
$\lambda$ does not belong to the neighborhood of $j_{s}$ exceptional Bloch
eigenvalues. Hence the perturbation $Q-C$ does not generate a gap in
$\sigma_{j_{s}}\left(L(C)\right)$ at the neighborhood of $\lambda.$

Now using Condition 1 and Corollary 2 we prove the following.

###### Theorem 4

If the matrix $C$ has three eigenvalues $\mu_{j_{1}},$ $\mu_{j_{2}}$ and
$\mu_{j_{3}}$ satisfying Condition 1, then there exists a number $H$ such that
$(H,\infty)\subset\sigma(L(Q)),$ that is, the number of the gaps in the
spectrum of $L(Q)$ is finite.

Proof. By Corollary 2 the gap $(\alpha,\beta)$ lying in $U(k)$ for $k>N_{3}$
belong to the set $S(j_{s},k,)$ for all $s\in\left\\{1,2,3\right\\}$ and for
some $k>N_{3}.$ Therefore it is enough to prove that

${\textstyle\bigcap\limits_{s=1,2,3}}S(j_{s},k)$ (22)

is an empty set for $k>N_{3}$. Since $\gamma_{k}\rightarrow 0$ the number
$N_{3}$ can be chosen so that $4\gamma_{k}<d$ for $k>N_{3}$. If the set (22)
contains an element $x,$ then using the definitions of $S(j_{s},k)$, we obtain
that there exist $k>N_{3}$ and $i_{s}\in\left\\{1,2,...,p\right\\}$ such that

$\mid x-(\pi k))^{2}-\frac{\mu_{j_{s}}+\mu_{i_{s}}}{2}\mid<\gamma_{k}$

for all $s=1,2,3$. This implies that

$\left|\left(\mu_{j_{u}}+\mu_{i_{u}}\right)-\left(\mu_{j_{v}}+\mu_{i_{v}}\right)\right|<4\gamma_{k}<d$

for all $u,v\in\left\\{1,2,3\right\\}$, where $v\neq u.$ This contradicts
Condition 1.

## References

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# Free uniformly complete vector lattices

###### Abstract

We define a free uniformly complete vector lattice over a set $A$ of
generators and give its concrete representation as the space
$\bigcup\limits_{B\in{\cal P}_{fin}(A)}H(\Delta_{B})$ of continuous positively
homogeneous functions.

Eduard Emelyanov and Svetlana Gorokhova

Keywords: free vector lattice, relative uniform convergence, free uniformly
complete vector lattice

Mathematics Subject Classification: 46A40 (primary), 46B42 (secondary)

The existence of a free vector lattice $FVL(A)$ over a nonempty set $A$ of
generators is the long established fact going back to Birkghoff [3]. A
concrete representation of $FVL(A)$ as a vector lattice of real-valued
functions with point-wise operations was proved by Weinberg [8] and Baker [2]
(cf. also [4]). Recently, a free Banach lattice $FBL(A)$ over $A$ was
introduced and investigated by de Pagter and Wickstead [6] (see also [7] for a
simple alternative approach to a free norm on $FVL(A)$). In the present note,
we define a free uniformly complete vector lattice over $A$ and give its
simple concrete construction. For unexplained notations and terminology we
refer to [1, 6, 5]. Although we have not found a suitable reference in
literature, we think the following definition belongs to folklore.

######  Definition 1.

Let $A$ be a subset of generators of a uniformly complete vector lattice $X$.
Then $X$ is said to be a free uniformly complete vector lattice over $A$ if,
for any uniformly complete vector lattice $Y$ containing $A$, there exists a
unique vector lattice homomorphism $T:X\to Y$ such that $Ta=a$ for each $a\in
A$.

So the free uniformly complete vector lattice over $A$, if exists, is an
initial object in the category of uniformly complete vector lattices and
lattice homomorphisms defined similarly to $FVL(A)$ in a smaller subcategory.
By [2], $FVL(A)$ exists, and it is a vector sublattice of
$\mathbb{R}^{\mathbb{R}^{A}}$ generated by $\delta_{a}$, where $\delta_{a}$
denotes a map $\xi\to\xi(a)$ on $\mathbb{R}^{A}$. Recall that (cf. [6,
Prop.3.5(2)]) if $B\subseteq A$, then there exists a unique lattice
homomorphism projection $P_{B}:FVL(A)\to FVL(A)$

$P_{B}(\delta_{a})=\begin{cases}\delta_{a}&\text{if}\ \ a\in B\\\ 0&\text{if}\
\ a\in A\setminus B\end{cases}$

onto $FBL(B)$. In particular, $FVL(A)=\bigcup\limits_{B\in{\cal
P}_{fin}(A)}FVL(B)$. Denote by $H(\mathbb{R}^{A})$ the space of all continuous
positively homogeneous real-valued functions on $\mathbb{R}^{A}$ and by
$H(\Delta_{A})$ the space of restrictions of elements of $H(\mathbb{R}^{A})$
to $H(\Delta_{A})=[-1,1]^{A}$. Clearly $H(\Delta_{A})$ is a closed vector
sublattice of $C(\Delta_{A})$.

######  Lemma 1.

Every $ru$-Cauchy net $f_{\alpha}$ in $FVL(A)$ lies in $H(\Delta_{B})$ for
some $B\in{\cal P}_{fin}(A)$.

###### Proof.

Let $u\in FVL(A)_{+}$ be such that, for every $\varepsilon>0$, there exists an
$\alpha_{\varepsilon}$ with

$|f_{\alpha_{1}}-f_{\alpha_{2}}|\leqslant\varepsilon u\ \ \ \
(\forall\alpha_{1},\alpha_{2}\geqslant\alpha_{\varepsilon}).$

Take $S:=S(f_{\alpha_{1}})\cup S(u)$, where $S(f)=\\{p\in
H(\Delta_{A}):f(p)\neq 0\\}$ denotes the open support of $f\in FVL(A)$. The
fact $S=S(|f_{\alpha_{1}}|+u)$ and $|f_{\alpha_{1}}|+u\leqslant
M\sum\limits_{k=1}^{n}|\delta_{a_{k}}|$ for some $M\in\mathbb{R}_{+}$ and
$a_{1},\ldots,a_{n}\in A$ imply $S\subseteq B:=\\{a_{k}:k=1,\ldots,n\\}$.
Hence, by [2, Thm.3.5], the net $f_{\alpha}$ lies in the principle ideal of
$FVL(B)$ determined by $w:=|f_{\alpha_{1}}|+u$. Taking the completion $FBL(B)$
of $FVL(B)$ with respect to the free lattice norm $\|\cdot\|_{F}$ from [6,
Def.4.5] and using [6, Cor.5.4] we obtain that the net $f_{\alpha}$ lies in
the principle ideal of $H(\mathbb{R}^{B})$ determined by $w$. The restriction
map $R:H(\mathbb{R}^{B})\to H(\Delta_{B})$ is a lattice isomorphism because
the set $B$ is finite. Consequently, the net $f_{\alpha}$ lies in the
principle ideal $I_{w}$ of $FBL(B)$ determined by $w$, and hence in
$H(\Delta_{B})$ by [6, Thm.8.2(2)], as desired. ∎

######  Theorem 1.

Let $A$ be a non-empty set. Then the vector lattice
$X=\bigcup\limits_{B\in{\cal P}_{fin}(A)}H(\Delta_{B})$ is the free uniformly
complete vector lattice over $A$.

###### Proof.

It follows from Lemma 1 and [6, Thm.8.2(4)] that $X$ is uniformly complete. We
identify as usual $a\in A$ with $\delta_{a}\in FVL(A)$. Let $Y$ be another
uniformly complete vector lattice containing $A$. Let $x\in X$. Then $x\in
H(\Delta_{B})=FBL(B)$ for some $B\in{\cal P}_{fin}(A)$ and hence there exists
a sequence $f_{n}$ in $FVL(B)$ that converges relatively uniformly to $x$.
Therefore the sequence $\hat{f}_{n}$ in $Y$ (obtained from elements of $B$ by
using the same lattice-linear expressions in $Y$ as those are used for $f_{n}$
in $FVL(B)$) is $ru$-Cauchy in $Y$. By $ru$-completeness of $Y$, the sequence
$\hat{f}_{n}$ $ru$-converges in $Y$ to some $y_{x}\in Y$. Now define $T:X\to
Y$ by $Tx:=y_{x}$. Clearly, $T:X\to Y$ is a unique vector lattice homomorphism
such that $Ta=a$ for each $a\in A$. ∎

It worth mentioned that by similar arguments as in the proof of Lemma 1 every
order Cauchy net in $FVL(A)$ lies eventually in $FVL(B)\subseteq
H(\mathbb{R}^{B})$ for some $B\in{\cal P}_{fin}(A)$. As $H(\mathbb{R}^{B})$ is
lattice isomorphic to $FBL(B)$ by [6, Cor.5.4] and $FBL(B)$ is regularly
embedded in $FBL(A)$ by [6, Prop.5.9] it follows from Theorem 1 that the free
uniformly complete vector lattice over $A$ is regularly embedded in $FBL(A)$.

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