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AMS Mathematics Subject Classifications (2000): 34A55, 34B05,34L40, 
 47E05, 47A10, 47A75
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AKNS systems, determination of coefficients, inverse problems, 
selfadjoint operator, spectral theory
---------------0102231836954
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\documentstyle[leqno, 12pt , amssymb]{article}

\setlength{\textwidth}{16cm}
\setlength{\textheight}{21cm}
\setlength{\oddsidemargin}{0cm}
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\newfont{\BBFONT}{msbm10 scaled 1200}
\newcommand{\ERE}{\hbox{\BBFONT R}}
\newcommand{\CE}{\hbox{\BBFONT C}}
\newcommand{\ZE}{\hbox{\BBFONT Z}}
\newcommand{\ENE}{\hbox{\BBFONT N}}

\newfont{\bbfont}{msbm10}
\newcommand{\pequere}{\hbox{\bbfont R}}
\newcommand{\zeti}{\hbox{\bbfont Z}}

\newcommand{\UNO}{\mathbin{\upharpoonleft\mkern-5mu\rfloor}}

\def\bull{\begin{flushright} \vrule height 6pt width 6pt depth -.pt
\end{flushright}}

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\begin{document}
\title{Inverse spectral results for AKNS systems with partial information on the 
potentials}

\author{R. del Rio \thanks{
IIMAS-UNAM,
Circuito Escolar, Ciudad Universitaria,
04510, M\'exico, D.F., M\'exico. }
\and B. Gr\'ebert \thanks{ UMR 6629, D\'epartement de Math\'ematiques,
Universit\'e de Nantes, 2 rue de la Houssini\`ere, 44322 Nantes Cedex 03, France.}
}

\date{}
\maketitle{ }

\begin{abstract}
For the AKNS operator on $L^2 ([0, 1], \CE^2) $ it is well known that the data of two spectra
uniquely determine the corresponding potential $\varphi $ a.e. on $[0, 1]$
(Borg's type Theorem).
We prove that, in the case where $\varphi$ is a priori known on $[a,1]$, then only a part
(depending on $a$) of two spectra determine $\varphi$ on $[0, 1]$.Our results include generalizations
for Dirac Systems of classical results obtained by Hochstadt and Lieberman for the Sturm Liouville case,
where they showed that half of the potential and one spectrum determine all the potential function. An 
important ingredient in our strategy is the link between the rate of growth of an entire function
and the distribution of is zeros. 
\end{abstract}

\section{Introduction}

We study problems related to classical results by Borg \cite{B} and Hochstadt-Lieberman \cite{HL}.
 For Sturm Liouville operator there exists a vast literature on this type of inverse problems,
  cf. \cite{dRGS} and references quoted therein. In this note we want to address similar results 
  for AKNS systems. Actually we use a different approach than in \cite{dRGS},
in particular we do not use the Titchmarsh-Weyl function and Marchenko's Theorem.

%We think that our method is more elementary

For $\varphi \in L^2 ([0,1], \CE)$ we define the AKNS operator on $L^2 ([0, 1], \CE^2) $ by

\begin{equation}
H(\varphi ):= \left( \matrix{0 & -1 \cr 1 & 0 \cr } \right) \frac{d}{d x} + 
\left( \matrix{-q (x) & p(x) \cr \quad p(x) & q(x) \cr } \right) \label{akns}
\end{equation}

\noindent where $\varphi = q - i p $ and $q$ and $p$ are real valued.

Notice that $H (\varphi )$ is unitarily equivalent to the Zakharov-Shabat operator,

\begin{equation}
L (\varphi ):= i \left( \matrix{1 & 0 \cr 0 & -1 \cr } \right) \frac{d}{d x} + 
\left( \matrix{0 & \varphi \cr \bar \varphi  & 0 \cr } \right) \label{zasha}
\end{equation}

\noindent where $\bar \varphi $ is the complex conjugate of $\varphi $.

For each $\alpha  \in  [0, \pi )$ we consider $\sigma  (\varphi , \alpha )$ the spectrum of 
the selfadjoint operator $H (\varphi )$ with domain, 
$F= {Y \choose Z} \in  H^1 ([0,1], \CE^2)$ such that

\begin{equation} \begin{array}{l}
Z (0) = 0 \\
\cos \alpha \ Y (1) - \sin \alpha \ Z (1) = 0.  \label{zeta}
\end{array} \end{equation}

Following \cite{GG} (cf. also \cite{GK}), $\sigma  (\varphi , \alpha )$ is a sequence of
 real numbers
$\Big(\mu _n (\varphi , \alpha )\Big)_{n \in  \zeti}$ satisfying 
$\mu _n < \mu _{n + 1} (n \in \ZE) $ and $\mu _n = \alpha  + n \pi  - \frac{\pi }{2} + 
l^2 (n)$ \footnote{ $a_n = b_n + l^2 (n)$ means that  
$\sum\limits_{n \in \zeti} \mid a_n - b_n \mid ^2 < \infty $.}.

By $\varphi \mid _{[a, b]}$ we shall denote the restriction of $\varphi $ to the interval 
$[a, b]$. That is $\varphi \mid _{[a, b]} (x) = \varphi  (x)$ if $x \in  [a, b]$.

Our main result is

\begin{theorem}
Let $\varphi \in  L^2 ([0,1], \CE),$ $\alpha , \beta  \in [0, \pi )$
 with  $\alpha \not= \beta $, $0  \leq a \leq  1,$ $l, k \in \ENE \cup  \{\infty \}$ with $\frac{1}{l} + \frac{1}{k} \geq 2 a$. Then $\{ \mu_{ln} (\alpha ), \mu_{kn} (\beta ) \mid  n \in  \ZE \}$
\footnote{ if $l$ (resp. $k$) equals $\infty $ we shall understand that $\{ \mu _{l n} (\alpha ) \mid  n \in \ZE \}$ (resp. $\{ \mu _{k n} (\beta ) \mid n \in  \ZE \}$) is empty.}
and $\varphi \mid_{[a, 1]}$ uniquely determine $\varphi $ a.e. on $[0, 1]$ and $\alpha , \beta $.
\end{theorem}

For particular values of $a, k, l$ we obtain for the AKNS systems

\begin{itemize}
\item Borg type Theorem: two spectra uniquely determine 
$ \varphi$  on $[0, 1]$ $(a = 1,\ l = k = 1) $.
\item Hochstadt-Liebermann type Theorem: one spectrum and $\varphi $ 
on $[1/2, 1]$ uniquely determine $\varphi $ on $[0, 1]$ $(a = 1/2,\ l = 1, k = 
\infty ).$ (cf. \cite{A}
for an other proof of this result.) \\
\end{itemize}
Actually our Theorem includes much more general results as for
example:
\begin{itemize}
\item Half of one spectrum and $\varphi $ on $[1/4, 1]$ uniquely 
determine $\varphi $ on $[0, 1]$
$(a = 1/4,\ l = 2, k = \infty ).$
\item Half of two spectra and $\varphi $ on $[1/2, 1]$ uniquely 
determine $\varphi $ on $[0, 1]$
$(a = 1/2,\ l = k = 2).$
\end{itemize}

For the shake of simplicity we only consider the case of two different boundary 
conditions. However the same method of proof applies to more general situations (for instance considering three spectra, with obvious
notation the condition would be $\frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} \geq  2 a$).

\medskip
In the same way we prove

\begin{theorem}
Let $\varphi \in  L^2 ([0,1], \CE),$ $\alpha , \beta  \in [0, \pi )$ with 
$\alpha \not= \beta,$ $ 0 \leq a \leq 1,$ $l,\ k \in  \ENE \cup  \{ \infty \}$ 
with $\frac{1}{l} + \frac{1}{k} \geq  4 a$. Then $ \{ \mu_{l n} (\alpha ), 
\mu _{k n} (\beta ) \mid  n \geq  0 \}$ and $\varphi \mid_{[a, 1]}$ uniquely 
determine $\varphi $ a.e. on $[0, 1]$ and $\alpha, \beta .$
\end{theorem}

Roughly speaking, Theorem 2 means that the ``positive'' part of one spectrum 
$\Big( (\mu _n)_{n \geq  0} \Big)$ gives the same information than $ (\mu _{2 n})_{n \in \zeti}$.

Of course in Theorems 1 and 2 the data $\varphi \mid_{[a, 1]} $ can be replaced 
by $\varphi \mid_{ [0, 1 - a]}$. Nevertheless the interval where $\varphi $ is a 
priori known must contain 0 or 1. Actually even the data of $Re \varphi $ on 
$[0, 1]$, $Im \varphi $
on $[1/2 - \varepsilon , 1/2 + \varepsilon ] $ ( $\varepsilon  \in  (0, 1/2)$ arbitray) 
and one spectrum do not uniquely determine $\varphi $ on $[0, 1]$. 
Let us construct a counter example: From \cite[Propositions 1.3 and 1.4]{GK}
we learn that for any $\varphi  \in  L^2 ([0, 1], \CE)$ and 
$n \in \ZE$, $\mu _n (\tilde\varphi , \frac{\pi }
{2}) = \mu _n (\varphi , \frac{\pi }{2})$, where 
$ \tilde \varphi  (x) = \bar \varphi  (1 - x)$. Let $\varphi  = q - i p$ with $q$ even,
 $p (1 - x) = -p (x)$ for $x \in  [1/2 - \varepsilon, 1/2 + \varepsilon ]$ and 
 $p (1 - x) \not=  p (x)$ for $x \in  [0,1/2 - \varepsilon )$. Define $\psi $ by 
 $\psi (x):= q (x) + i p (1 - x)$.

Then, $\sigma (\varphi , \pi /2) = \sigma ( \psi , \pi /2)$, 
$Re \varphi  = Re \psi $ in $[0, 1]$, $Im \varphi = Im \psi $ on 
$[1/2 -\varepsilon , 1/2 + \varepsilon ]$ nevertheless $\varphi \not\equiv \psi $.

Similar construction shows that the data of $Re \varphi $ on $[0, 1]$, $Im \varphi 
$ on 
$[0,1] \setminus ( 1/2 - \varepsilon , 1/2 + \varepsilon )$  
($ \varepsilon \in  (0, 1/2)$ arbitrary) and one spectrum do not uniquely determine 
$\varphi $ on $[0,1]$.

Our fundamental strategy can be described as follows. Let $\varphi $ and $\psi $ 
satisfying the conditions of Theorem 1.

\begin{enumerate}
\item[(a)] We construct an entire function $ f (\lambda , \varphi , \psi )$ (cf. 
section 2.1) which vanishes at each $\mu _{l n} (\alpha )$ and $\mu_{k n} (\beta )$,
 $n \in  \ZE$.
\item[(b)] The partial information on the potential allows us to bound the 
exponential type of $f$ obtaining 
$\mid  f (\lambda ) \mid  = o (e^{2 a \mid Im \lambda \mid })$ as $\mid
\lambda \mid  \rightarrow \infty  $ (cf. Section 2.2).
\item[(c)] We use the general principle that the growth of an entire 
function is related to the distribution of its zeros \footnote{This is a 
generalization of the fact that the rate of growth of a polynomial is given by its 
degree or equivalently its number of roots.}
to prove that steps (a) and (b) and the condition 
$\frac{1}{l} + \frac{1}{k} \geq  2 a$ imply $f \equiv  0$. (cf. Section 2.3)
\item[(d)] $f \equiv  0$ imply $ \varphi \equiv \psi $. (cf. Section 2.4)
\end{enumerate}

\section{ Proofs}

\subsection{{Construction of the entire function $f$.} }

For $\varphi \in  L^2 ([0, 1], \CE)$ and $\lambda \in  \CE $ let 
$F (\cdot , \lambda , \varphi ) \equiv  \left( \matrix{ Y (\cdot , \lambda , \varphi) \cr 
Z (\cdot , \lambda , \varphi) \cr } \right) \in  H^1 ([0, 1], \CE^2)$ 
be the unique vector valued function satisfying 
$H (\varphi ) F = \lambda F \mbox{ and }F (0, \lambda , \varphi ) = {1 \choose 0}$.

For each $0 \leq  x \leq  1$ and $\varphi \in  L^2 ([0, 1]) \CE)$, 
$\lambda \mapsto F ( x, \lambda , \varphi )$ is entire
 (cf. \cite{GG} or \cite{GK} for the construction of $F$).
 
Notice that the spectrum $\sigma (\varphi , \alpha )$ is the root's set of

\begin{equation}
\cos \alpha \ Y(1, \lambda , \varphi ) - \sin \alpha \ Z (1, \lambda , \varphi ) = 0. \label{cossin}
\end{equation}

Let $G \equiv \left(  \matrix{G_1 (x, \lambda , \varphi ) \cr G_2 (x, \lambda , \varphi )} \right) $ with

\begin{eqnarray}
G_1 (x, \lambda , \varphi ) & = & Z (x, \lambda , \varphi ) - i Y (x, \lambda , \varphi ) \label{letge}
\\
G_2 (x, \lambda , \varphi ) & = & Z (x, \lambda , \varphi ) + i Y (x, \lambda , \varphi ). \nonumber
\end{eqnarray}

One has

\begin{equation}
L (\varphi ) G (\cdot , \lambda , \varphi ) = \lambda  G (\cdot , \lambda , \varphi ) \label{gezy}
\end{equation}

\noindent and furthermore for $\lambda  \in  \ERE$,

\begin{equation}
\bar  G_1 (x, \lambda , \varphi ) = G_2 (x, \lambda , \varphi ) . \label{gege}
\end{equation}

Let $ \varphi , \psi  \in  L^2 ([0, 1], \CE)$, from (\ref{gezy}) the cross product,

$$A(\lambda ):= < G (\cdot , \lambda , \psi ), (L (\varphi ) - \lambda ) G (\cdot , \lambda , \varphi ) > - <G (\cdot , \lambda , \varphi ), (L (\psi ) - \lambda ) G (\cdot , \lambda , \psi ) >$$

\noindent vanishes for all $\lambda  \in  \CE$.

On the other hand by direct calculation using (\ref{gege}), one obtains for $\lambda  \in  \ERE$

\[
\begin{array}{lll}
A(x) & = & < G (\psi ), L (\varphi ) G (\varphi ) > - < G (\varphi ), L (\psi ) G (\psi ) > \\
& = & \displaystyle{ \int^1_0 G_2 (\psi ) G_2 (\varphi ) (\varphi  - \psi ) d x } 
 +  \displaystyle{ \int^1_0  G_1(\psi ) G_1 (\varphi ) (\bar \varphi  - \bar \psi ) d x } \\
& + & \displaystyle i { \int^1_0 \frac{d}{d x} \Big( -G_1 (\psi ) G_2 (\varphi ) + G_2 (\psi ) G_1 (\varphi ) \Big) d x.}
\end{array}
\]

Thus defining for $\lambda  \in  \CE$,

\begin{eqnarray}
f (\lambda , \varphi , \psi ) & := & \int^1_0  G_2 (x, \lambda , \psi ) G_2 (x, \lambda , \varphi ) (\varphi  (x) - \psi  (x) ) d x \label{laince} \\
& + & \int^1_0  G_1 (x, \lambda , \psi ) G_1 (x, \lambda , \varphi ) 
(\bar \varphi  (x) - \bar \psi  (x) ) d x\ , \nonumber
\end{eqnarray}

\noindent one gets for $\lambda  \in  \ERE$,

\begin{equation}
f (\lambda , \varphi , \psi ) = - i \Big[ W \Big( G (\varphi ), G (\psi ) \Big) \Big]^1_0.  \label{dobleu}
\end{equation}

\begin{lemma}
Let $\varphi, \psi  \in  L^2 ([0, 1], \CE)$ and $ \alpha \in [0, \pi )$.
Assume that $ \mu \in \sigma  (\varphi , \alpha ) \cap \sigma (\psi , \alpha )$.
Then
$$f (\mu , \varphi , \psi ) = 0.$$
\end{lemma}

{\it Proof} In view of formula (\ref{dobleu}) one has to prove that 
$\left[ W (G (\cdot , \mu , \varphi ), G (\cdot , \mu , \psi ) )\right]^1_0 = 0$.\\
 Since $G ( 0, \mu ) = { -i \choose +i}$, one has 
 $$ W (G (0, \mu , \varphi ), G (0, \mu , \psi )) = 0.$$

\medskip
On the other hand
$$W \Big(G (1, \mu , \varphi ), G (1, \mu , \psi ) \Big) = 
2 i \Big( Z (1, \mu , \varphi ) Y (1, \mu , \psi ) - 
Y (1, \mu , \varphi ) Z (1, \mu , \psi ) \Big) = 0$$
\noindent where we used that (cf. (\ref{cossin}))

$$\cos \alpha \ Y (1) - \sin \alpha \ Z(1) = 0.$$
\bull %%%q.e.d.

%\newpage

For simplicity we introduce for $i = 1, 2$

\begin{equation}
g_i (x, \lambda ) \equiv  g_i (x, \lambda , \varphi , \psi ):= G_i (x, \lambda , \varphi ) G_i (x, \lambda , \psi ).  \label{simple}
\end{equation}

Then formula (\ref{laince}) becomes, with $r  = \varphi  - \psi $,

\begin{equation}
f (\lambda , \varphi , \psi ) = \int^1_0 \Big( g_2 (x, \lambda ) r (x) + g_1 (x, \lambda ) \bar r  (x) \Big) d x. \label{funet}
\end{equation}

Notice that since $ \lambda  \mapsto  F (x, \lambda , \varphi )$ is entire, $ \lambda  \mapsto  f (\lambda , \varphi , \psi )$ is entire too.

\bigskip

\subsection{{Order and type of the entire function $f$.} }

In this section we shall prove that the entire function $f$ defined in (\ref{funet})
satisfies the following

\begin{lemma}
Let $ 0 \leq  a \leq 1$. 
Assume $\varphi  (x) = \psi  (x)$ for $x \in  [a, 1]$, then\\
$$f (\lambda , \varphi , \psi ) = o ( e^{2 \mid Im \lambda \mid  a})$$ 
when $\mid \lambda \mid \rightarrow \infty $.
\end{lemma}

{\it Proof} From \cite{GG} (see also \cite{GK}) we learn that uniformly for $x \in  [0, 1]$ one has when $\mid \lambda \mid \rightarrow \infty $,

\begin{equation}
\left\Vert F (x, \lambda )  - \left( \matrix{ \quad \cos \lambda  x \cr -\sin \lambda  x \cr} \right) \right\Vert  = o (e^{\mid Im \lambda \mid  x}). \label{roche}
\end{equation}

Therefore we get from (\ref{simple}) and (\ref{letge})


$$
g_2 (x, \lambda , \varphi , \psi )  =  \displaystyle{ \Big[i e^{i \lambda  x} 
+ o (e ^{\mid Im \lambda \mid x})\Big]^2 } 
 =   \displaystyle{ - e^{2 i \lambda  x} + o (e ^{2 \mid Im \lambda \mid x}) } $$
 and 
$$g_1 (x, \lambda , \varphi , \psi )  =   \displaystyle{ - e^{- 2 i \lambda  x} + o (e ^{2 \mid Im \lambda \mid x}). }
$$


Using (\ref{funet}) we obtain (with $r = \varphi - \psi )$

\[
\begin{array}{lll}
f (\lambda , \varphi , \psi ) & = & \displaystyle{  \int^1_0 \Big(- e^{2 i \lambda  x} + o (e^{2 \mid Im \lambda \mid x}) \Big) r (x) d x } \\
& + &  \displaystyle{  \int^1_0 \Big(- e^{- 2 i \lambda  x} + 
o (e^{2 \mid Im \lambda \mid x}) \Big) \bar r (x) d x . }
\end{array}
\]

Thus if $ \varphi  \equiv  \psi $ on $[a, 1]$,

\begin{equation}
f (\lambda ) = - \int^a_0 r (x) e^{2 i \lambda  x} d x 
- \int^a_0 \bar r (x) e^{-2 i \lambda  x} d x
+ o (e^{2 \mid Im \lambda \mid a}) \label{efere}
\end{equation}

\noindent where we used that the error
term $o (e^{2 \mid Im \lambda \mid x})$ is uniform in $ x \in [0,1]$ and that $ r \in  L^1([0,1])$.

For $\alpha \in  C^1 ([0, 1])$ one obtains integrating by parts,

$$ \left\vert \int^a_0 \alpha  (x)  e^{2 i \lambda  x} d x \right\vert  = o \left( \frac{e^{2 \mid Im \lambda \mid a}}{\mid \lambda \mid } \right).$$

Now for $\alpha \in  L^2 ([0, 1])$ and $ \varepsilon  > 0$ let $\alpha_{\varepsilon }
$ in $C^1 ([0,1])$ such that $\parallel \alpha_{\varepsilon } - \alpha
\parallel _{L^2} < \varepsilon /2$. There exists $A > 0$ such that for 
$|\lambda |>A$,
$$ \left\vert \int^a_0 \alpha_{\varepsilon } (x) 
e^{2 i \lambda  x} d x \right\vert  \leq  
\frac{\varepsilon}{2} \  e^{2 \mid Im \lambda a \mid}$$

\noindent and thus

$$
\begin{array}{lll}
 \left\vert \int^a_0 \alpha (x) e^{2 i \lambda  x} d x \right\vert & 
\leq & \left\vert \int^a_0 \alpha_{\varepsilon } (x) e^{2 i \lambda  x}
 d x \right\vert  
 +  \int^a_0 \mid \alpha - \alpha_{\varepsilon }\mid  \ 
  e^{2 \mid Im \lambda \mid x} d x 
 \leq \varepsilon e^{2 \mid Im \lambda a \mid}.
\end{array}
$$

Therefore one has proved (cf. \cite[Problem 3, p. 15]{PT}) that for 
$\alpha \in  L^2 ([0, 1])$
$$ \int^a_0 \alpha (x) e^{2 i \lambda  x} d x = o (e^{2 \mid Im \lambda \mid a} )\ .$$

Thus, using (\ref{efere}),

$$ f(\lambda ) = o (e^{2 a \mid Im \lambda \mid } ).$$
\bull %%%q.e.d.

\underbar{Remark:} We have proved that $f$ is an {\it entire} fuction of order not greater 
than 1 and type not greater than $2 a$. We are going to use that such a function cannot 
have ``many'' zeros. (cf. \cite{L}).

\bigskip

\subsection{ Infinite product representation }

We begin with three auxiliary Lemmas on infinite products.

Given a sequence of complex numbers $(a_k)_{k \in \zeti}$ we say that the 
product $\prod _{k \in \zeti} a_k$ 
is convergent if the limit $\lim _{N \rightarrow \infty } 
\prod_{|k| \leq N }  a_k$ exists.
In such a case we write
   \[ \prod _{k \in \zeti} a_k := \lim _{N \rightarrow \infty }
      \prod_{|k| \leq N }
	  a_k  .
   \]
\begin{lemma}
Let $(z_n)_{n \in  \zeti}$ a complex sequence satisfying $ z_n = n \pi  + l^2 (n)$. Then the formula

$$ h (\lambda ) := (\lambda  - z_0) \prod\limits_{n \in  \zeti \setminus \{ 0 \}} \frac{z_n - \lambda }
{n \pi } $$

\noindent defines and entire function satisfying
uniformly for $ (n + 1/6) \pi  \leq  \mid \lambda \mid  
\leq  (n + 5/6) \pi $,

\begin{equation}
h(\lambda ) = \sin \lambda  (1 + o (1)),\quad  n \rightarrow + \infty .  \label{tete} 
\end{equation}


\end{lemma}


Lemma 5 is proved in \cite[Lemma I - 16]{GK} (cf. \cite{GG} and \cite{PT}), but the uniformity 
of (\ref{tete})
is proved only for $ (n + 1/4) \pi  \leq  \mid \lambda \mid \leq  (n + 3/4) \pi $. 
The generalization to $ (n + 1/6) \pi \leq  \mid \lambda \mid \leq (n + 5/6) \pi $ is 
straightforward (actually the only important fact is that $ \mid \lambda \mid $ is far away 
from $ n \pi $ $  (n \geq  0)$ ).

% In particular for any integer $k \geq 1$
%\[
%\begin{array}{lll}
%\sin \frac{\lambda }{k} & = & \frac{\lambda }{k} \prod\limits_{n \in  \zeti \setminus \{ 0 \}} \frac{n \pi - %\frac{x}{k}}{n \pi } \\
%& = &  \frac{\lambda }{k} \prod\limits_{n \in  \zeti \setminus \{ 0 \}} \frac{n k \pi - \lambda }{n k \pi }.
%\end{array}
%\]

From Lemma 5 follows

\begin{lemma}
Let $ (z_n)_{n \in \zeti}$ a sequence of complex numbers satisfying $z_n = n \pi  + l^2 (n)$ and $k \geq 1$. Then the formula

$$ h_k (\lambda ) := (\lambda  - z_0) \prod\limits_{n \in  \zeti \setminus \{ 0 \}} \frac{z_{nk} - \lambda }{n k \pi } $$

\noindent defines an entire function satisfying uniformly 
on $k (n + 1/6) \pi  \leq \mid \lambda \mid \leq k (n + 5/6) \pi $

$$ h_k (\lambda ) = k \sin \Big( \frac{\lambda }{k} \Big) (1 + o (1)),
\quad n \rightarrow + \infty . $$

\end{lemma}

{\it Proof} For $ n \in \ZE$ we define, $ \tilde z_n = \frac{z_{n k}}{k}$. One has

$$h_k (\lambda ) = (\lambda - k \tilde z_0) \prod\limits_{n \in  \zeti \setminus \{ 0 \}} \frac{k \tilde z_n -\lambda }{n k \pi }. $$

By Lemma 5, the function

$$ h (\lambda ) = (\lambda  - \tilde z_0) \prod\limits_{n \in  \zeti \setminus \{ 0 \}} \frac{\tilde z_n - \lambda}{n \pi } $$

\noindent satisfies

$$ h (\lambda ) = \sin \lambda  (1 + o (1)),\quad n \rightarrow + \infty $$

\noindent uniformly on $(n + 1/6) \pi  \leq \mid \lambda \mid \leq (n + 5/6) \pi $.

Notice that $h_k (\lambda ) = k h (\frac{\lambda }{k})$, hence

$$ h_k (\lambda ) = k \sin (\frac{\lambda}{k} ) (1 + o (1)),\quad n \rightarrow + \infty $$

uniformly on $ k (n + 1/6) \pi  \leq \mid \lambda \mid \leq k (n + 5/6) \pi $.
\bull %%%q.e.d.

% As an application of Lemma 2, one has \underbar{Lemma 3} Let $k_1, K-2 \geq 1$ then the %function defined by
%\[
%\begin{array}{ll}
%h(\lambda ) := (\lambda  - \mu _0) (\lambda  - \nu _0)  & \prod\limits_{n \in  \zeti \setminus \{ 0 \}} %\frac{\mu _{k_1 n} - \lambda }{k_1 n} \times \\
%& \prod\limits_{n \in  \zeti \setminus \{ 0 \}} \frac{\nu _{k_2 n} - \lambda }{k_2 n}
%\end{array}
%\]
%$ is entire. Furthermore there exist $n_0 > 1$ and $C >0$ such that uniformly for $\mid \lambda %\mid  = (n + 1/2) \pi $ and $n \geq n_0$
%$$ \mid h (\lambda ) \mid  \geq  C \exp \left( (\frac{1}{k_1} + \frac{1}{k_2} ) \mid Im \lambda \mid %\right). $$

As an application of Lemma 6, one has

\begin{lemma}
Let $ - \frac{\pi }{2} \leq a_ j < \frac{\pi }{2}$  $(j = 1, 2)$,
 $k_1 \geq 1,\ k_2 \geq 1$ and $(\mu _n)_{n \in \zeti}$,
 $(\nu _n)_{n \in \zeti}$ two sequences of real numbers satisfying 
 $\mu _n = n \pi  + l^2 (n)$,
$\nu _n = n \pi  + l^2 (n)$. Then the function defined by
$$
h (\lambda )  :=  \quad (\lambda  - \mu _0 - a_1) (\lambda  - \nu _0 - a_2)\ 
 \prod\limits_{n \in  \zeti \setminus \{ 0 \}} \frac{\mu _{k_1 n} + a_1 - 
 \lambda }{k_1 n \pi } \times \frac{\nu_{ k_2 n} + a_2 - \lambda }{k_2 n \pi } 
$$
is entire.\\
Furthermore there exists $ (\gamma _p)_{p \geq 1}$ a sequence of positive real 
numbers with $ \gamma _p  {\longrightarrow } \infty $, and $ C > 0$ such that 
uniformly on $\mid \lambda \mid = \gamma _p$ and $ p \geq 1,$  
$$\mid h (
\lambda ) \mid \geq  C \exp \Big( (\frac{1}{k_1} + \frac{1}{k_2} ) \mid  
Im \lambda \mid \Big). $$
\end{lemma}

{\it Proof} By Lemma 6, the functions

$$h_1 (\lambda ) := (\lambda  - \mu _0 - a_1) \prod\limits_{n \in  \zeti \setminus \{ 0 \}} 
\frac{\mu_{ k_1 n} + a_1 - \lambda }{k_1 n \pi } $$

\noindent and

$$h_2 (\lambda ) := (\lambda  - \nu _0 - a_2)
\prod\limits_{n \in  \zeti \setminus \{ 0 \}} \frac{\nu_{k_2 n} + a_2 - \lambda }{k_2 n \pi } $$

\noindent are entire and satisfy as $n \rightarrow \infty $

\begin{equation}
h_1 (\lambda ) = k_1 \sin \Big( \frac{\lambda - a_1}{k_1} \Big) (1 + o (1)) \label{yupi1}
\end{equation}

\noindent uniformly on $k_1 (n + 1/6) \pi \leq \mid \lambda  - a_1 \mid \leq k_1 (n + 5/6) \pi $ and

\begin{equation}
h_2 (\lambda ) = k_2 \sin \Big( \frac{\lambda - a_2}{k_2} \Big) (1 + o (1)) \label{yupi2}
\end{equation}

\noindent uniformly on $k_2 (n + 1/6) \pi \leq \mid \lambda  - a_2 \mid \leq k_2 (n + 5/6) \pi $.

Moreover, for $j = 1, 2$, there exist $C_j >0$ such that uniformly on

$$ I_j := \bigcup_{n \in \ZE}\big\{\lambda \ \mid \ k_j  (n + 1/6) \pi 
\leq \mid \lambda  - a_ j \mid 
\leq k_j (n + 5/6) \pi 
 \big\}, $$
\noindent one has (cf. \cite[Lemma 1 p. 27]{PT})
\begin{equation}
\left\vert \sin \left( \frac{\lambda - a_j}{k_j} \right) \right\vert > 
C_j \exp \left( \frac{\mid Im (\lambda  - a_j)\mid }{k_j} \right)\ . \label{potru}
\end{equation}

Noticing that $ h(\lambda ) = h_1 (\lambda ) h_2 (\lambda )$ and in view of 
(\ref{yupi1}) - (\ref{potru}), it remains to prove that there exists 
a sequence $(\gamma _p)_{ p \geq 1}$ with $ \gamma _p > 0 \; (p \geq 1)$ and
$ \gamma _p {\longrightarrow  \atop p \rightarrow \infty } \infty $ such that for each $p \geq 1$

$$ \gamma _p \in  I_1 \cap  I_2 \cap \ERE. $$

As $ I_j \cap \ERE$ is the union of segments whose wide is $2/3\ k_j \pi $ and 
the distance between two consecutive segments is $1/3\ k_j \pi $, the existence of such 
sequence $(\gamma _p) _{p \geq 1}$ is clear.

\bull %%%q.e.d.

We can now state the main result of this section. 
Recall that $ \sigma  (\varphi , \alpha ) \equiv
  (\mu _n (\varphi , \alpha ))_{n \in \zeti}$ is the spectrum of 
  $H (\varphi )$ with domain, $ F = {Y \choose Z} \in H^1 (0,1)$ 
  such that $ Z (0) = 0$ and $\cos \alpha \ Y (1) - \sin \alpha \ Z (1) = 0$.

\begin{proposition}
Let $\varphi ,\ \psi  \in  L^2 ([0,1], \CE)$, $0 \leq a \leq 1,\ 0 \leq \alpha ,\
 \beta < \pi $ with $ \alpha \not= \beta $ and $k_1,\ k_2 \geq 1$ with $\frac{1}{k_1} + 
 \frac{1}{k_2} \geq  2 a$. Assume that
\begin{enumerate}
\item[(i)] $\varphi \equiv \psi  \mbox{ on } [a, 1]$
\item[(ii)] $\mu _{k_1 n} (\varphi , \alpha ) = \mu _{k_1 n} (\psi , \alpha ), n \in \ZE$
\item[(iii)] $\mu _{k_2 n} (\varphi , \beta ) = \mu _{k_2 n} (\psi , \beta ), n \in \ZE$
\end{enumerate}
Then $f (\cdot , \varphi , \psi ) \equiv 0$.
\end{proposition}

{\it Proof} As mentioned in the introduction, following \cite{GK} (see also \cite{GG}), 
one deduces from Rouch\'e's Theorem and formula (\ref{roche})

\begin{equation}
\mu _n (\varphi , \alpha ) = n \pi + \alpha  - \frac{\pi}{2} + l^2 (n) \label{twin}
\end{equation}

\noindent and

\begin{equation}
\mu _n (\varphi , \beta ) = n \pi + \beta - \frac{\pi}{2} + l^2 (n). \label{twiny}
\end{equation}

Therefore by Lemma 7, the entire function

$$h (\lambda ) :=  (\lambda  - \mu _0 ) (\lambda  - \nu _0 )
 \prod\limits_{n \in  \zeti \setminus \{ 0 \}} \frac{\mu _{k_1 n} - \lambda }{k_1 n \pi } \; \frac{\nu_{ k_2 n} - \lambda }{k_2 n \pi }, $$

\noindent with $\mu _j := \mu _j (\varphi , \alpha )$, 
$\nu _j := \mu _j (\varphi , \beta )\ (j \in \ZE)$, satisfies for some constant $ C >0 $

\begin{equation}
\mid h (\lambda )\mid  \geq  C \exp \Big( (\frac{1}{k_1} + \frac{1}{k_2}) 
 \mid Im \lambda \mid \Big)\geq C \exp \Big(2a\mid Im \lambda \mid \Big)
\label{faro}
\end{equation}

\noindent uniformly on $ \mid \lambda \mid  = \gamma _p$ and $p \geq 1$ where $(\gamma _p)_{p \geq 1}$ is a sequence of positive real numbers with $\gamma _p {\longrightarrow  \atop p \rightarrow \infty } \infty $.

Furthermore, as $ \alpha  \not= \beta $, $\sigma (\varphi, \alpha ) \cap  \sigma (\varphi , \beta ) = \emptyset $. Thus $(\mu _{k_1 n})_{n \in \zeti}$ and $(\nu _{k_2 n})_{n \in \zeti}$ are simple roots of $h$.

On the other hand, by Lemma 3 and hypothesis (ii) (iii), $f (\mu _{k_1 n}) = f (\nu _{k_2 n}) = 0$ for all $n \in  \ZE$. Besides, by Lemma 4 and Hypothesis (i),

\begin{equation}
f (\lambda ) = o (e^{2 a \mid  Im \lambda \mid }) \label{coco}
\end{equation}

\noindent when $ \mid \lambda \mid  \rightarrow \infty $.

Therefore $ \lambda  \mapsto \frac{f (\lambda )}{h (\lambda )}$ is entire and 
combining
(\ref{faro}), (\ref{coco})  we get $\mid \frac{f (\lambda )}{h (\lambda )} \mid  = o (1)$ as $p \rightarrow \infty $ uniformly on $\mid \lambda \mid = \gamma _p$.

Hence by the maximum principle we conclude that $f \equiv 0$.
\bull %%%q.e.d.

\subsection{Integral representation }

The main result of this section is

\begin{proposition}
Let $\varphi , \psi  \in  L^2 ([0, 1], \CE)$. Assume that $f (\lambda, \varphi , \psi )= 0$ for $\lambda \in \ERE$, then $\varphi \equiv \psi $.
\end{proposition}

We follow the same strategy as in B.~Levin  \cite[Appendix 4]{L}.

We first establish an integral representation of $g_1$ and $g_2$ (cf. formula (\ref{simple})).

\begin{lemma}
The exists a kernel $K \in  L^2 ([-1, 1]^2, \CE)$ such that for $x \in  [-1, 1]$ 
and $\lambda \in \ERE$,
\begin{equation}
g_1 (x, \lambda ) = - e^{- 2 i \lambda x} + 
\int^x_{- x} \bar K (x, u) e^{- 2 i \lambda u}  d u \label{deu1}
\end{equation}
\noindent and
\begin{equation}
g_2 (x, \lambda ) = - e^{2 i \lambda x} + 
\int^x_{- x} K (x, u) e^{2 i \lambda u}  d u . \label{deu2}
\end{equation}
\end{lemma}

{\it Proof of Lemma 10}

Let $M (\cdot , \lambda , \varphi )$ be the fundamental ($2 \times  2 $ matrix) solution of
 $H (\varphi ) M = \lambda M$ satisfying $ M(0, \lambda , \varphi ) = I d_{2 \times  2 }$ 
 and $R (x, \lambda )$ be given by
$$
 R (x, \lambda ) = \left( \matrix{ \quad \cos \lambda  x & \sin \lambda  x \cr - \sin \lambda x &
 \cos \lambda  x \cr} \right). 
$$
From \cite[p. 514]{MckV}\footnote{In \cite{MckV} the authors do not use the same spectral 
variable and we have to transform $\lambda $ in $\lambda /2$ in our formula (\ref{mere})}%
(cf. also \cite{L}) one learns that there exists a Kernel 
$A \in  L^2 ([0, 1]^2, M_{2 \times  2 }(\ERE))$ (where $M_{2 \times  2 }(\ERE)$ 
denotes the space of $2 \times  2 $ matrix with real entries) such that

\begin{equation}
M (x, \lambda ) = R (x, \lambda ) + \int^x_0  R (x - 2 y, \lambda ) A (x, y) d y \label{mere}
\end{equation}

% - TO BE CHECKED -

By definition $F (x, \lambda , \varphi )$ is the first column of $ M(x, \lambda ,  \varphi )$. 
Therefore by (\ref{letge})

$$G_2 (x, \lambda ) =(i, 1) M (x, \lambda ) {1 \choose 0}$$

\noindent and by a straightforward calculation one gets

\begin{equation}
G_2 (x, \lambda , \varphi ) = i e^{i \lambda x} + \int^x_0 e^{i \lambda (x - 2 y)} K_{\varphi}  (x, y) d y
\label{sape}
\end{equation}

\noindent where

$$K_\varphi  (x, y) = (i, 1) A(x, y) {1 \choose 0}. $$

Since for $ \lambda \in  \ERE $, $ G_1 (x, \lambda , \varphi)  = \bar G_2 (x, \lambda , \varphi )$, one also has

\begin{equation}
G_1 (x, \lambda , \varphi ) = - i e^{- i \lambda  x} + \int^x_0 e^{- i \lambda (x - 2 y)} \bar K_{\varphi}  (x, y) d y.  \label{nosape}
\end{equation}

Inserting (\ref{sape}) in (\ref{simple}) one gets

\begin{equation}
g_2 (x, \lambda ) = - e^{2 i \lambda  x} + h_1 (x, \lambda ) + h_2 (x, \lambda )  \label{union}
\end{equation}

\noindent where

\begin{equation}
h_1 (x, \lambda ) = i \int^x_0 (K_{\varphi } (x, t) + K_{\psi } (x, t) ) e^{2 i \lambda  (x - t)} d t  \label{dete}
\end{equation}

\noindent and

\begin{equation}
h_2 (x, \lambda ) =
\int^x_0 d t \int^x_0 d s K_{\varphi } (x, t) K_{\psi } (x, s) e^{2 i \lambda  (x - t - s)}.  
\label{conese}
\end{equation}

By the change of variable $ u = x - t - s$ and $v = t - s $ in (\ref{conese}) one obtains
$$
h_2 (x, \lambda ) = \frac{1}{2} \int_D K_{\varphi } (x, \frac{v - u + x}{2}) 
K_{\psi } (x, \frac{- v - u + x}{2}) e^{2 i \lambda  u} d u \; d v 
$$
\noindent where
$$
D = D_1 \cup  D_2
$$
\noindent and with
$$
D_1 \; := \; \{ (u, v) \mid - x \leq u \leq 0 ; u \leq  v \leq  - u \}
$$
\noindent and
$$
D_2 \; := \; \{ (u, v) \mid 0 \leq u \leq x ; - u \leq  v \leq  u \}.
$$
Thus
\begin{equation}
h_2 (x, \lambda ) = \int^x_{- x} e^{2 i \lambda  u} K_1 (x, u) d u \label{bono}
\end{equation}

\noindent with

$$K_1 (x, u) :=  \int^x_{- x} K_{\varphi  } (x, \frac{v - u + x}{2}) 
K_{\psi } (x, \frac{- v - u + x}{2}) \UNO_D (u, v) d v $$

\noindent and where$\UNO_D$
denotes the characteristic function of the set $D$.

Similarly by the change of variable $u = x - t$ in (\ref{dete}) one has

$$ h_1 (x, \lambda ) = i \int^x_0 \Big( K_{\varphi } (x, x - u) + 
K_{\psi } (x, x - u) \Big) e^{2 i \lambda  u} d u. $$

Thus

\begin{equation}
h_1 (x, \lambda ) = \int^x_{- x} K_2 (x, u) e^{2 i \lambda  u} d u. \label{pisa}
\end{equation}

\noindent  where

$$ K_2 (x, u) := i \UNO_{[0, x]} (u) \Big( K_{\varphi } (x, x - u) + K_{\psi } (x, x- u) \Big). $$

Combinig (\ref{union}), (\ref{bono}) and (\ref{pisa})
one gets (\ref{deu2}) with

$$K (x, u) = K_1 (x, u) + K_2 (x, u).$$

We deduce (\ref{deu1}) from  (\ref{deu2}) recalling that

$$g_1 (x, \lambda ) = \bar g_2 (x, \lambda ) \mbox{ for } \lambda  \in \ERE.$$
\bull %%%q.e.d.


{\it Proof of Proposition 9} Recall that, with $r = \varphi  - \psi $ (cf. (\ref{funet}))

$$f (\lambda ) \equiv f (\lambda , \varphi , \psi ) = \int^1_0 \Big( g_2 (x, \lambda ) r (x) +
 g_1 (x, \lambda ) \bar r (x) \Big) d x. $$

By Lemma 10 one gets for $\lambda \in  \ERE$

\begin{eqnarray}
f (\lambda ) & = & \int^1_0 \Big( -  e^{2 i \lambda  x} + 
\int^x_{- x} K (x, u) e^{2 i \lambda  u} d u \Big) r (x) d x
 \label{primera} \\
& + & \int^1_0 \Big( -  e^{- 2 i \lambda  y} + 
\int^y_{- y} \bar K (y, v) e^{- 2 i \lambda  v} d v \Big) \bar r (y) d y.
\nonumber
\end{eqnarray}

By the change of variable $x = - y$ and $ u= - v$ in the second term of the right side of 
(\ref{primera}) one obtains

\begin{eqnarray}
f (\lambda ) & = & \int^1_0 \Big( -  e^{2 i \lambda  x} + 
\int^x_{- x} K (x, u) e^{2 i \lambda  u} d u \Big) r (x) d x \label{segunda}  \\
& + & \int^0_{-1} \Big( -  e^{2 i \lambda  x} - 
\int^x_{- x} \bar K (- x, - u) e^{2 i \lambda  u} d u \Big) \bar r (- x) d x.
\nonumber
\end{eqnarray}

Thus with $m (x) := r (x)$ for $x \in [0,1]$ and 
$m (x) := \bar r (- x)$ for $ x \in [-1, 0]$ and with

\begin{equation}
\begin{array}{lll}
B (x, u) & : = K (x, u) & \mbox{ for } x \in  [0, 1] \mbox{ and } \nonumber \\
B (x, u) & : =  \bar K (- x, - u) & \mbox{ for }  x \in [- 1, 0],        \label{bebe}
\end{array}
\end{equation}

\noindent (\ref{segunda}) leads to

\begin{eqnarray*}
f (\lambda ) & = & \int^1_{- 1} \Big( -  e^{2 i \lambda  x} + 
\int^{\mid x\mid}_{-\mid x\mid} B (x, u) e^{2 i \lambda  u} d u \Big) m (x) d x \\
& = & \int^1_{- 1} e^{2 i \lambda  x} \Big( - m(x) + 
\int^{- \mid x \mid} _{- 1} B (u, x) m (u) d u
+ \int^1_{\mid x \mid} B (u, x) m (u) d u \Big) d x.
\end{eqnarray*}

Since $f (\lambda ) = 0$ for all $\lambda  \in \ERE$ and 
$\{ e^{2 i \lambda  x} \mid \lambda \in \ERE \}$ spans $L^2 ([-1,  1], \CE)$, 
the last formula implies that, for $x \in [-1, 1]$,

$$m (x) =  \int^{- \mid x \mid} _{- 1} B (u, x) m (u) d u + 
\int^1_{\mid x \mid} B (u, x) m (u) d u\ . $$

In particular for $x \in  [0, 1]$ one gets by defnition of $B$ and $m$,

$$r (x) = \int^1_x \Big( \bar K (u, - x) \bar r (u) + K (u, x) r (u) \Big) d u. $$

Therefore defining $P \in  L^2 ([-1, 1])$ by

$$ P (u, v) := \mid K (u, v) \mid + \mid \bar K (u, -v) \mid, $$

\noindent one obtains for $x \in [0,1]$

\begin{equation}
\mid r (x) \mid  \leq  \int^1_x P (u, x) \mid r (u) \mid  d u. \label{yoyo}
\end{equation}

Iterating formula (\ref{yoyo}) leads to, for each $ n \geq  1$,

$$ \mid r (x) \mid  \leq \int^1_x d u_1 \int^1_{u_1} d u_2 \ldots \int^1_{u_{n - 1}} d u_n\
 P (u_1, x) \ldots P(u_n, u_{n - 1}) \mid r (u_n) \mid \ . $$

%Therefore using Cauchy Schwarz
%$$\int^1 \mid r (x) \mid  d x \leq C^n \big( \int^1 d x \int^1 d u_1 \ldots \int^1 d u_n\
% \mid  r (u_n) \mid ^2 %\big)^{1/2}$$

Interchanging the order of integration we obtain\footnote{the following argument is analogous to parts of \cite[Theorem 6, Ch. 2]{H} } 

\begin{equation}
\mid r (x) \mid  \leq  \int^1_x d u_n \int^{u_n}_x d u_{n - 1} \ldots \int^{u_2}_x d u_1\
 P (u_1, x) \ldots
P (u_n, u_{n-1}) \mid r (u_n) \mid \ . \label{olas}
\end{equation}

Now setting $K_1 (u, x) = P (u, x)$ and defining

$$K_j (u, x) := \int^u_x d v\ K_{j - 1} (v, x)  P (u, v)$$

\noindent the inequality (\ref{olas}) can be written as

\begin{equation}
\mid  r (x) \mid \leq  \int^1_x d u K_n (u, x) \mid r (u) \mid \ . \label{pera}
\end{equation}

Defining for $0\leq x\leq u\leq 1$

\begin{eqnarray*}
B (x) & = & \int^1_x \mid P (z, x) \mid ^2 d z\ , \\
A (u) & = & \int^u_0 \mid P (u, z) \mid ^2 d z\ , \\
\rho  (x) & = & \int^x_0 A (u) d u\ ,
\end{eqnarray*}

\noindent one obtains by a straigtforward recurrence and the Cauchy-Schwarz inequality
$$ \mid K_n (u, x) \mid ^2 \leq  A (u) B(x) \frac{\rho (u)^{n - 2}}{(n - 2)!}\ .$$

Therefore

\begin{eqnarray*}
\int^1_x \mid K_n (u, x) \mid ^2 d u & \leq & \frac{B(x)}{(n - 2)!} 
\int^1_0 d u A (u) \rho (u)^{n - 2} \\
& = & \frac{B (x)}{(n - 1)!} \; \rho (1)^{n - 1} \ .
\end{eqnarray*}

Hence

\begin{equation}
\int^1_x \mid K_n (u, x) \mid ^2 d u \leq \frac{1}{(n - 1)!} 
\int^1_0 \mid  P (z, x) \mid ^2 d z \cdot  \Big( \int^1_0 d u 
\int^1_0 d z \mid P (u, z) \mid ^2 \Big)^{n - 1}\ . \label{dado}
\end{equation}

Using Cauchy--Schwarz in (\ref{pera}) we get

\begin{equation}
\mid r (x)\mid ^2 \leq  \int^1_x d u \mid K_n (u, x) \mid ^2 \cdot  
\int^1_x \mid  r (u) \mid ^2 d u \ . \label{causch}
\end{equation}

By integrating (\ref{causch}) and using (\ref{dado}) we find

$$\big\Vert  r  \big\Vert ^2_{L_2} \leq  \frac{1}{(n - 1)!} 
\big\Vert P \big\Vert ^{2 n}_{L_2} \big\Vert r \big\Vert ^2_{L_2}\ 
{\buildrel n \rightarrow \infty \over \longrightarrow }\ 0 \ .$$

It follows then that

$$ r (x) = \varphi (x) - \psi (x) = 0\ . $$

\bull %%%q.e.d.

\subsection{\it Proofs of Theorems 1 and 2}

Theorem 1 is a direct consequence of Proposition 8 and Proposition 9.

\noindent The proof of Theorem 2 is similar;  we only have to make the following changes:

\begin{itemize}
\item We replace $f (\lambda , \varphi , \psi )$ by
$$ \tilde f (\lambda , \varphi , \psi ) = f (- \lambda , \varphi , \psi ) 
f (\lambda , \varphi , \psi )$$
Thus, since $\mu_{l n} (\varphi , \alpha ) = \mu_{l n} (\psi , \alpha )$ and
 $\mu_{k n} (\varphi , \beta ) =
\mu_{k n} (\psi , \beta ) $ for all $n \in  \ENE$ one has by Lemma 3 that for any $n \in  \ENE$
 $$\tilde f \big( \mu_{l n} (\varphi , \alpha ) \big) =
  \tilde f \big( \mu_{k n} (\varphi , \beta ) \big) = 0$$
  and 
  $$ \tilde f \big( - \mu_{l n} (\varphi , \alpha ) \big) =
    \tilde f \big( - \mu_{k n} (\varphi , \beta ) \big) = 0\ .$$
\item By Lemma 4 one has
$$ \tilde f (\lambda) = o (e^{\mid Im \lambda \mid  4 a}) 
\mbox{ as }\mid \lambda \mid  \rightarrow + \infty .$$

\item In Proposition 8 we replace $h$ by $\tilde h$ where
$$\tilde h (x) = (\lambda  - \mu  _0) (\lambda - \nu _0) \prod\limits_{n > 0}
\frac{(\mu_{l n}^2 - \lambda^2 ) (\nu_{k n}^2 - \lambda^2 )}{(l n \pi)^2 (k n \pi)^2 }$$
with $\mu _j = \mu _j (\varphi , \alpha )$ and $ \nu _j = \mu _j (\varphi, \beta ) (j \in \ENE)$.
 
The function $\lambda  \longmapsto \frac{\tilde f (\lambda)}{\tilde h (\lambda)}$ is still entire.

\item By Lemma 7 there exist $C >0$ and $ (\gamma _p)_{p \geq 1}$ with 
$\gamma _p {\longrightarrow \atop p \rightarrow \infty } \infty $ such that uniformly on 
$\mid \lambda \mid = \gamma _p \hfill \break
\mid  h (x) \mid  \geq  C \exp \Big( \frac{1}{l} + \frac{1}{k}  \Big) \mid Im \lambda \mid $.
\item As in Proposition 8 we obtain, using 
$\frac{1}{l} + \frac{1}{k}  \geq  4 a$, that 
$\Big\vert \frac{\tilde f ( \lambda )}{\tilde h (\lambda )} \Big\vert = o (1)$ 
for $p \rightarrow \infty $ uniformly on $\mid \lambda \mid  = \gamma _p$. 
Hence by the maximum principle $\tilde f \equiv 0$, i. e. $f \equiv 0$.

\item We apply Proposition 9 to conclude to $\varphi  = \psi $.
\end{itemize}

%    \bull %%%q.e.d.
\medskip

{\bf Acknowledgement:} B. G. would like to acknowledge the support of the ACI project (French Government)
and the hospitality of the IIMAS-UNAM institute. R. del R. gratefully acknowledges support by projects IN-102998 PAPIIT-UNAM and 27487E CONACyT (Mexican Goverment) and the hospitality of the Dept. of Mathematics of the University of Nantes.

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\end{thebibliography}


\end{document}
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