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\title{An inverse scattering result for several\\ convex bodies}

\author{Latchezar Stoyanov}
\date{}

\begin{document}
\maketitle


\def\sn{S^{n-1}}
\def\ssn{S^{n-1}\times S^{n-1}}
\def\do{\partial \Omega}



\section{Introduction}
Let $K$ be a compact subset of ${\bf R}^n, n\geq 3, n$ odd, 
with $C^{\infty}$ boundary $\partial K$ such that
$$\Omega_K = \overline{{\bf R}^n\setminus K}$$ 
is connected.  
The {\it scattering operator} related  to the wave equation
in ${\bf R}\times \Omega$ with Dirichlet boundary condition on
${\bf R}\times \do$ can be represented as a unitary operator
$$S : L^2({\bf R}\times \sn ) \longrightarrow L^2({\bf R}\times \sn)$$
(see \cite{kn:LP1}).
The kernel of $S - $Id, which can be considered as a distribution
$$s_K (t,\theta,\omega) \in {\cal D}'({\bf R}\times \ssn ),$$
is called the {\it scattering kernel}.

\def\cc{C_0^{\infty}}
\def\ot{(\omega,\theta)}

It was shown by A.Majda \cite{kn:Ma} that for each obstacle $K$
the convex hull of $K$ can be recovered if one knows
sing supp $s_K (t,-\omega,\omega)$ for a dense set of $\omega$'s
in $\sn$ (see also \cite{kn:LP2} for a similar result). Consequently,
an obstacle $K$ is completely determined by the singularities of the
scattering kernel, provided we know in advance that $K$ is convex.
Moreover, one can distinguish between convex and strictly convex
obstacles by the same data (see \cite{kn:So} and \cite{kn:Y}). A general
problem arises: is an obstacle $K$
completely determined if we know sing supp $s_K(t,\theta,\omega)$
for all $\theta, \omega \in \sn$? 


In this paper we consider obstacles $K$ of the form
\begin{equation}
K = \bigcup_{j=1}^s K_j \:\:\: , \:\:\:K_i \cap K_j = \emptyset \:\:\:\:\: 
{\rm for}\:\:\: i\neq j,
\:\:\:\:\:\: K_j \:\: {\rm convex} \: {\rm for} \: {\rm all}\:\:
j = 1,\ldots,s,
\end{equation}
which satisfy the following condition
introduced by M.Ikawa \cite{kn:I}:
\begin{eqnarray*}
{\rm (H)} \:\:\:\:\:\:\:\:\:\:\:\:\: \:\:\:\:
(K\setminus (K_i \cup K_j)) \cap \:{\rm convex} \: 
{\rm hull}\: (K_i\cup K_j) = \emptyset \:\:\:\: {\rm for} \: {\rm all} \:\:\:
i,j = 1,\ldots,s.  
\end{eqnarray*}

We show that among the obstacles with these properties, $K$ can be completely
recovered if one knows sing supp $s_K (t,\theta,\omega)$ for almost
all $\ot \in \ssn$. 

To state precisely the corresponding result, consider 
another obstacle $L$ of the same type, namely
\begin{equation}
L = \bigcup_{j=1}^r L_j \:\:\:\: ,\:\:\: L_i \cap L_j = \emptyset \:\:\:\: 
{\rm for}\:\:\: i\neq j,
\:\:\:\:\:\: L_j \:\: {\rm convex} \: {\rm for} \: {\rm all}\:\:
j = 1,\ldots,r.
\end{equation}



{\bf 1.1. Theorem.} {\it Let $K$ and $L$ have the form {\rm (1)} and
{\rm (2)}, respectively, and let $K$ and $L$ satisfy the condition
{\rm (H)}. Assume that there exists a subset
${\cal R}$ of full Lebesgue measure in $\sn\times\sn$, 
such that for all $\ot \in {\cal R}$ we have 
\begin{equation}
{\rm sing}\:{\rm supp} s_K(t,\theta,\omega) \subset
{\rm sing}\:{\rm supp} s_L(t,\theta,\omega).
\end{equation}
Then for every $i = 1,\ldots,s$ there exists $j_i = 1,\ldots,r$
with $K_i = L_{j_i}$. In particular, $K \subset L$.}\\

As an immediate consequence one gets the following.\\

{\bf 1.2. Corollary.} {\it Let $K$ and $L$ have the form {\rm (1)} and
{\rm (2)}, respectively, and let both $K$ and $L$ satisfy the condition
{\rm (H)}. If there exists a subset
${\cal R}$ of full Lebesgue measure in $\sn\times\sn$ 
such that
$${\rm sing}\:{\rm supp} \; s_K(t,\theta,\omega) =
{\rm sing}\:{\rm supp} \; s_L(t,\theta,\omega)$$
holds for all $\ot \in {\cal R}$, then $K = L$.}\\

It is seen from the proof that the assertions of both Theorem 1.1 and
Corollary 1.2 remain true if $R$ is {\it residual } in $\ssn$ (instead
of having full Lebesgue measure), i.e. if $R$ contains a countable
inersection of open dense subsets of $\ssn$.\\

{\footnotesize
Acknowledgement. This paper has been written while the author was visiting
the University of Bordeaux I as chercheur associ\'{e} of C.N.R.S.}


\section{Proof of Theorem 1.1}

Throughout we assume that $K$ and $L$ 
satisfy the assumptions of Theorem 1.1. Let
 ${\cal R}$ be a subset of full Lebesgue measure of $\ssn$ such that
(3) holds for all $(\omega,\theta)\in {\cal R}$.
We are going to show that each convex component of $K$ coincides with some 
convex component
of $L$. To do this we shall use the connection between singularities 
of the scattering kernel and sojourn times of reflecting $\ot$-rays.

Let
$\omega \in\sn, \theta\in\sn$. By a {\it reflecting $\ot$-ray} 
in $\Omega_K$, we mean an infinite ray in $\Omega_K$ with incoming direction
$\omega$ and outgoing direction $\theta$ making finitely many reflections
at $\do$ according to the law of the geometrical optics 
(see \cite{kn:PS1} or \cite{kn:PS2} for the precise definition). 
The {\it sojourn time} $T_{\gamma}$ of $\gamma$
is defined by 
$$T_{\gamma} = \langle \omega , x_1\rangle + \sum_{i=1}^{s-1} \| x_i -
x_{i+1} \| - \langle \theta , x_s \rangle,$$
where $x_1,\ldots,x_s$ are the successive reflection points of $\gamma$
(\cite{kn:G}). If $\gamma$ has no
segments tangent to $\partial K$, then it is called an {\it ordinary}
reflecting $\ot$-ray. By ${\cal L}_{\omega,\theta}(K)$ we denote the
{\it set of all reflecting $\ot$-rays in} $\Omega_K$.

It follows from Theorem 1.2 in \cite{kn:PS3} that there exists 
${\cal R}' \subset \ssn$ having full Lebesgue measure in $\ssn$
(and residual at the same time) such that for $\ot \in {\cal R}'$
we have
$${\rm sing} \: {\rm supp} \: s_K (t,\theta,\omega) =
\{ -T_{\gamma} : \gamma \in {\cal L}_{\omega,\theta}(K) \}$$
and
$${\rm sing} \: {\rm supp} \: s_L (t,\theta,\omega) =
\{ -T_{\gamma} : \gamma \in {\cal L}_{\omega,\theta}(L) \}.$$
Then 
$${\cal R}'' = {\cal R} \cap {\cal R}'$$
is a subset of full Lebesgue measure in $\ssn$, and according to
(3) and the above inclusions, one gets
\begin{equation}
\{ -T_{\gamma} : \gamma \in {\cal L}_{\omega,\theta}(K) \} \subset
\{ -T_{\gamma} : \gamma \in {\cal L}_{\omega,\theta}(L) \} \:\:\:\: ,
\:\:\:\: \ot \in {\cal R}''.
\end{equation}


\def\dk{\partial K}
\def\dl{\partial L}

Given $x \in \dk$, we denote by $N_K(x)$ the {\it unit normal vector} to
$\dk$ at $x$ pointing into $\Omega_K$. Thus
$$N_K : \dk \longrightarrow \sn$$
is the {\it Gauss map} of $\dk$. 

Theorem 1.1 will be easily derived by the following lemma.\\


{\bf 2.1. Lemma.} {\it For every $x \in \partial K$ there exists
$y \in \partial L$ such that} 
\begin{equation}
N_L(y) = \pm N_K(x) \:\:\:\: , \:\:\:\: \langle x-y, N_K(x)\rangle = 0.
\end{equation}

That is, the tangential hyperplane to $K$ at $x$ coincides with the
tangential hyperplane to $L$ at some $y \in \dl$.

The proof of Lemma 2.1 is rather long and will be devided into
several steps.\\


{\bf 2.2. Lemma.} {\it Let $Y$ and $Z$ be smooth convex (concave)
hypersurfaces in ${\bf R}^n$ with $Y \cap Z = \emptyset$. Let
$\Sigma(Y,Z)$ be the set of those $(y,z)\in Y\times Z$ such that the
tangent hyperplanes to $Y$ at $y$ and $Z$ at $z$ coincide and the Gauss
maps of $Y$ at $y$ and of $Z$ at $z$ are regular. Then
$\Sigma (Y,Z)$ is a smooth $(n-2)$-dimensional
submanifold of $Y\times Z$}.\\

{\it Proof.} Take smooth charts
$$\varphi : V \longrightarrow Y \:\:\: , \:\:\:
\psi : W \longrightarrow Z,$$
$V, W \subset {\bf R}^{n-1}$, around some points $y_0\in Y$ and
$z_0\in Z$ with $(y_0,z_0)\in \Sigma(Y,Z)$. For $v\in V$, denote by 
$N_Y(v)$ one of the unit normal vectors to $Y$ at $\varphi(y)$
so that $N_Y$ is continuos on $V$. In the same way we define
$N_Z$ on $W$. Since $(y_0,z_0)\in \Sigma(Y,Z)$, the normals to $Y$ at $y_0$
and to $Z$ at $z_0$ are collinear, and (choosing appropriately
the normal fields $N_Y$ and $N_Z$) we may assume that these normals
coincide. We may also assume that $N_Y$ is regular on $V$, $N_Z$ is regular
on $W$ and
$$N_Y^{(n)}(v) \neq 0 \:\:\: , \:\: v\in V \: \:\: , \:\:\:
N_Z^{(n)}(w)\neq 0 \:\:\: , \:\:\: w\in W.$$
Then $N_Y^{(n)}(v)$ and $N_Z^{(n)}(w)$ have one and the same sign for all
$v\in V$ and $w\in W$.  

Define the functions
$$f_i, g : V\times W \longrightarrow {\bf R}$$
by $f_i(v,w) = N_Y^{(i)}(v) - N_Z^{(i)}(w), i = 1,\ldots,n-1$ and
$g(v,w) = \langle \varphi(v) - \psi(w), N_Y(v)\rangle$, and
consider $F: V\times W \longrightarrow {\bf R}^n$ given by
$$F(v,w) = (f_1(v,w),\ldots,f_{n-1}(v,w),g(v,w)).$$
Then $F$ is smooth and we have
$$ \{ (v,w)\in V\times W : (\varphi(v),\psi(w))\in \Sigma(Y,Z) \} = F^{-1}(0).$$
Therefore the assertion will be proved if we show that $F^{-1}(0)$ is
a smooth $(n-2)$-dimensional submanifold of $V\times W$. To do this it is
sufficient to establish that $F$ is 
submersion on $F^{-1}(0)$ (cf. \cite{kn:GG}).

\def\dnw{\frac{\partial N_Z^{(i)}}{\partial w_j}(w)}
\def\dpw{\frac{\partial \psi}{\partial w_j}(w)}
\def\dnv{\frac{\partial N_Y^{(i)}}{\partial v_j}(v)}
\def\dvv{\frac{\partial \varphi}{\partial v_j}(v)}




Fix $(v,w)\in F^{-1}(0)$ and let
\begin{equation}
\sum_{i=1}^{n-1} A_i {\rm grad}_{v,w} f_i(v,w) + B {\rm grad}_{v,w}
g(v,w) = 0
\end{equation}
for some reals $A_i$, $B$. Considering in (6) the derivatives with respect
to $w_j$ ($1\leq j\leq n-1$), we get
\begin{equation}
- \sum_{i=1}^{n-1} A_i \dnw - B \langle \dpw , N_Y(v)\rangle = 0.
\end{equation}  
Set $A_n = 0$ and $A = (A_1,\ldots,A_n) \in {\bf R}^n$. Since
$F(v,w) = 0$, we have $f_i(v,w) = 0$ for all $i = 1,\ldots,n-1$, and
therefore $N_Y(v) = N_Z(w)$. Therefore (7) implies
$$0 = \langle A, \frac{\partial N_Z}{\partial w_j}(w)
 \rangle + B\langle \dpw, N_Z(w)\rangle 
= \langle A, \frac{\partial N_Z}{\partial w_j}(w) \rangle.$$
Since $N_Z$ is regular at $w$, we have that 
$\{ \frac{\partial N_Z}{\partial w_j}(w) \}_{j=1}^{n-1}$
is a linear basis in the tangent hyperplane $T_{\psi(w)}Z$. Hence
the latter equality implies $A = \lambda N_Z(w)$ for some
$\lambda \in {\bf R}$. Consequently, $0 = A_n = \lambda N_Z^{(n)}(w)$,
which shows that $\lambda = 0$. Thus, $A = 0$.

Next, consider in (6) the derivatives with respect to $v_j$
($ j = 1,\ldots,n-1$). We get
$$B(\langle \dvv , N_Y(v)\rangle + \langle \varphi(v) - \psi (w),
\frac{\partial N_Y}{\partial v_j}(v) \rangle ) = 0,$$
and consequently
$$B \langle \varphi(v) - \psi (w),
\frac{\partial N_Y}{\partial v_j}(v) \rangle  = 0.$$
Since $F(v,w) = 0$, we have that $N_Y(v) = N_Z(w)$ and 
$\varphi(v) - \psi(w)$ is orthogonal to $N_Y(v)$. On the other hand,
$\{ \frac{\partial N_Y}{\partial v_j}(v) \}_{j=1}^{n-1}$ is a 
linear basis of $T_{\varphi(v)}Y$, so
there exists $j = 1,\ldots,n-1$ with
$$\langle \varphi (v) - \psi(w), \frac{\partial N_Y}{\partial v_j}(v)
 \rangle \neq 0.$$
This yields $B = 0$ which proves the assertion.
$\spadesuit$\\


Now we begin with the {\it proof of Lemma} 2.1.

For $i = 1,\ldots, s$ denote by
${\cal N}_{K,i}$ the set of those $\omega \in \sn$ such that
$\omega$ is a regular value of the restriction of $N_K$ to $\dk_i$.
In the same way we indtroduce the sets ${\cal N}_{L,j}$ for
$j = 1,\ldots,r$. It follows by the Sard theorem that the sets
${\cal N}_{K,i}, {\cal N}_{L,j}$ have full Lebesgue measure in $\sn$.
Moreover, they are open and dense in $\sn$, so
$${\cal N} = (\bigcap_{i=1}^s {\cal N}_{K,i}) \bigcap (\bigcap_{j=1}^r
{\cal N}_{L,j})$$
is an open and dense subset of $\sn$ having full Lebesgue measure.


Note that to prove Lemma 2.1 it is sufficient to consider the case when 
$N_K(x)\in {\cal N}$. Indeed, assume that for every $x'\in \dk$
with $N_K(x')\in {\cal N}$ there exists $y'\in \dl$ with
$N_L(y') = \pm N_K(x')$, $\langle x' - y',N_K(x')\rangle = 0$.
Given an arbitrary $x\in \dk$, there exists a sequence
$\{ x_m \} \subset \dk$ with $N_K(x_m) \rightarrow N_K(x)$ and
$N_K(x_m)\in {\cal N}$ for each $m$. Taking a subsequence, we
may assume that all $x_m$ belong to one and the same convex component
$K_i$ of $K$ and 
$x_m \rightarrow x'' \in \dk$. Then
$N_K(x'') = N_K(x)$ and the convexity of $K_i$ implies 
$x\in T_{x''} K_i$. Since $N_K(x_m) \in {\cal N}$, by our assumption
there exists $y_m \in \dl$ with
$N_L(y_m) = \pm N_K(x_m)$, $\langle x_m - y_m,N_K(x_m)\rangle = 0$.
Passing again to an appropriate subsequence, we may assume 
$y_m \rightarrow y \in \dl$. Then we have
$N_L(y) = \pm N_K(x'')$, $\langle x'' - y,N_K(x'')\rangle = 0$.
Combining this with $N_K(x) = N_K(x'')$ and $x\in T_{x''}\dk$, we
see that (5) hold.

Fix an arbitrary $x^{(0)} \in \dk$ with $N_K(x^{(0)})\in {\cal N}$.
We want to find $y^{(0)}\in \dl$ such that 
$T_{x^{(0)}} \dk = T_{y^{(0)}} \dl$.  
We may assume that 
$x^{(0)} \in \dk_1$. It is sufficient to show that for each
sufficiently small 
neighbourhood $U$ of $x^{(0)}$ in $\dk_1$ there exist $x \in U$ and
$y \in \dl$ such that $T_x \dk = T_y \dl$. If this is so, then the
compactness of $\dl$ implies the existence of a point
$y^{(0)} \in \dl$ with $T_{x^{(0)}} \dk = T_{y^{(0)}} \dl$.

Let $U_0$ be a small neighbourhood of $x^{(0)}$ in $\dk_1$ such that
$N_K(x)\in {\cal N}$ for all $x\in U_0$. 
The first step in our proof will be to choose in a special way a
vector $\omega_0 \in \sn$ tangent to $\dk_1$ at some $x\in U_0$.

Given $i\neq j, i,j = 1,\ldots,r,$ denote by $\Sigma_{i,j}$ the set
of those $\omega \in \sn$ such that
$\omega = \frac{z-y}{\|z-y\|}$ for some $(y,z)\in \Sigma(\dl_i,\dl_j)$
(see the notation in Lemma 2.2). That is,
$$\Sigma_{i,j} = h_{i,j} (\Sigma(\dl_i,\dl_j)),$$
where
$$h_{i,j} : \dl_i \times \dl_j \longrightarrow \sn$$
is defined by
$$h_{i,j}(y,z) = \frac{z-y}{\|z-y\|}.$$
Since $h_{i,j}$ is smooth, it follows by the Sard theorem and Lemma 2.2
that $\Sigma_{i,j}$ is a finite union of compact subsets of measure
zero in $\sn$. Therefore
$${\cal S} = \sn \setminus ( \bigcup_{i\neq j} \Sigma_{i,j})$$
is a subset of full Lebesque measure in $\sn$. In particular, ${\cal S}$
is dense in $\sn$.

\def\tx{T_{x^{(0)}} \dk}

Next, using the fact that $K$ satisfies the condition (H), one can find
a line $l$ in ${\bf R}^n$ with $l\cap K = \{ x^{(0)}\}$. Indeed, for
each $i = 2,\ldots,s,$ $K'_i = K_i \cap T_{x^{(0)}} \dk$ is a convex
domain in $T_{x^{(0)}} \dk$ with $x^{(0)} \notin K'_i$. Let
$$C_i = \{ x^{(0)} + tz : t \geq 0, z \in K'_i \}.$$
Then $C_i$ is a cone in $\tx$ with vertex $x^{(0)}$. The condition (H)
implies $C_i \cap (\pm C_j) = \{ x^{(0)}\}$ for all 
$i\neq j, i,j = 2,\ldots,s$. It is then clear that there exists a line
$l \subset \tx$ through $x^{(0)}$ with $l\cap C_i = \{x^{(0)}\}$ for
all $i = 2,\ldots,s$.

Consider an arbitrary line $l$ with this property and denote by $\xi$
one of the directions of $l$, $\xi \in \sn$. The compactness of
$K\setminus K_1$ and the regularity of $N_K$ at $x^{(0)}$ 
show that for all $\omega\in\sn$ sufficiently close
to $\xi$ there exists $x\in U_0$ (not unique) so that $\omega$ is
tangent to $\dk_1$ at $x$ and the line $l(x,\omega)$, determined by
$x$ and $\omega$ has no common points with $K\setminus K_1$. Since
${\cal S}$ is dense in $\sn$, there exist $\omega_0 \in {\cal S}$
and $x\in U_0$ with
$$l(x,\omega_0)\cap (K\setminus K_1) = \emptyset,$$
$l(x,\omega_0)$ being tangent to $\dk_1$ at $x$. {\bf Fix $\omega_0$ 
and $x$ with these properties}.

Since $x$ will be fixed in the next arguments, without loss of generality
we may {\bf assume that $x = 0$}, i.e. $x$ coincides with the initial
point of the coordinate system in ${\bf R}^n$. We shall also assume that
$$N_K (0) = e = (0,\ldots,0,1).$$

\def\to{\theta(\omega)}
\def\ott{(\omega,\theta(\omega))}


For each
\begin{equation}
\omega \in \sn \:\:\: , \:\:\: \langle \omega,e\rangle < 0,
\end{equation}
define
$$\theta(\omega) = \omega - 2\langle \omega,e\rangle e \in \sn.$$
Then $\theta(\omega)$ is symmetric to $\omega$ with respect to
$T_0 \dk$. Consequently, for each $\omega$ with (8) there exists a
reflecting $\ott$-ray $\gamma(\omega)$ in $\Omega_{K_1}$ with exactly
one reflection point 0. It then follows by the choice of $\omega_0$
that if $\delta > 0$ is small enough and
\begin{equation}
\mid \omega - \omega_0 \mid < \delta,
\end{equation} 
then 
$\gamma(\omega)$ is contained in $\Omega_K$ and moreover $\gamma(\omega)$
is an ordinary reflecting $\ott$-ray in $\Omega_K$.

Fix $\delta > 0$ with the latter property. Note that 
$$T_{\gamma(\omega)} = \langle \omega, x\rangle - \langle \theta, 
x\rangle = 0,$$
since $x = 0$ by our assumption.\\

{\bf 2.3. Lemma.} {\it For each $\omega$ with {\rm (8)} and {\rm (9)}
there exists a reflecting $\ott$-ray $\gamma'(\omega)$ in $\Omega_L$
with $T_{\gamma'(\omega)} = 0$.}\\

\def\omtm{(\omega_m,\theta_m)}

{\it Proof of Lemma 2.3.} Fix $\omega$ with (8) and (9). Since 
${\cal R}''$ is dense in $\ssn$, $\ott$ can be approximated by elements
of ${\cal R}''$. Let $\{(\omega_m,\theta_m)\}\subset {\cal R}''$
be a sequence 
with $\omega_m \rightarrow \omega$, $\theta_m \rightarrow \to$. It is easy
to see that for sufficiently large $m$, there exists a (unique) reflecting
$\omtm$-ray $\gamma_m$ in $\Omega_K$ with one reflection point
$x_m \in \dk_1$. We may assume that there exists $\lim x_m = y \in \dk_1$.
Since
$$N_K(y) = \lim N_K(x_m) = \lim \frac{\theta_m - \omega_m}{\| \theta_m
- \omega_m \|} = \frac{\to - \omega}{\| \to - \omega \|} = N_K(0),$$
and $N_K$ is regular at $0 = x \in U_0$, 
we find that $y = 0$, i.e. $x_m \rightarrow 0$. So we may also assume
that $\| x_m \| \leq 1/2$ for all $m$.

It follows from $\omtm \in {\cal R}''$ and (4) that there exists a
reflecting $\omtm$-ray $\gamma'_m$ in $\Omega_L$ with
\begin{equation}
T_{\gamma'_m} = T_{\gamma_m} = \langle \omega_m, x_m\rangle -
\langle \theta_m, x_m \rangle = \langle \omega_m - \theta_m, x_m \rangle.
\end{equation}
In particular, $T_{\gamma'_m} \leq 1$ for all $m$.
This implies that the reflecting rays $\gamma'_m$ have bounded number of
reflection points. Indeed, let $d$ be the {\it minimal distance}
between convex components of $L$ and let $D$ be the {\it radius of a ball
with center $0$ containing} $L$. If $k_m$ is the number of reflection
points of $\gamma'_m$, then we clearly have
$$1 \geq T_{\gamma'_m} \geq (k_m - 1)d - 2D,$$
and so $k_m \leq \frac{1+2D}{d} + 1$.

Now passing to an appropriate subsequence, we may assume that each
$\gamma'_m$ has exactly $k$ reflection points ($k$ does not depend on $m$).
Moreover, we may assume that for each $m$ and each $i = 1,\ldots,k$
the $i$-th reflection point $x_i^{(m)}$ of $\gamma'_m$ belongs to one 
and the same convex component $L_{j_i}$ of $L$ and  there exists
$\lim_{m \rightarrow \infty} x_i^{(m)} = x_i \in \dl_{j_i}$. It is
then easy to see that $x_1,\ldots,x_k$ are the successive reflection
points of a reflecting (not necessarily ordinary) $\ott$-ray 
$\gamma'(\omega)$ in $\Omega_L$ with (cf (10))
$$T_{\gamma'(\omega)} = \lim_{m \rightarrow \infty} T_{\gamma'_m} = 0.$$
This proves the assertion. $\spadesuit$\\

Denote by $k(\omega)$ the number of reflection points of the reflecting
$\ott$-ray $\gamma'(\omega)$ from Lemma 2.3. The next step will be to
show that $k(\omega) \leq 2$ for all $\omega$ with (8) and (9), 
provided $\delta$ is chosen small enough. To do this we use the fact
that $L$ satisfies the condition (H).\\

{\bf 2.4. Lemma.} {\it  There exists $\varphi_0 \in (0,\frac{\pi}{2})$
such that if $X, Y, Z$ are three convex components of $L$ with 
$X \neq Y, Y \neq Z$ and
$\xi \in X, \eta \in Y, \zeta \in Z$, then the angle between the
vectors $\eta - \xi$ and $\zeta - \eta$ is not less than
$\varphi_0$.}\\

The proof of this assertion is an easy consequence of the condition (H)
and the compactness of $L$. We omit the details. $\spadesuit$.

We continue with the proof of Lemma 2.1. Assume that
\begin{equation}
\delta < \frac{d}{D} \sin^2 \frac{\varphi_0}{4},
\end{equation}
and fix $\omega$ with (8) and (9). Let $y_1,\ldots, y_k \:\: (k = k(\omega))$
be the successive reflection points of the reflecting $\ott$-ray
$\gamma'(\omega)$ in $\Omega_L$ (see Lemma 2.3). Then we have
\begin{eqnarray*}
0 = T_{\gamma'(\omega)} 
& = & \langle \omega, y_1\rangle + \sum_{i=1}^{k-1} \| y_i - y_{i+1}\|
- \langle \to , y_k\rangle \cr
& = & \langle \omega - \to, y_k\rangle + \sum_{i=1}^{k-1} {[\| y_{i+1}
- y_i \| - \langle \omega, y_{i+1} - y_i\rangle ]}.\cr
\end{eqnarray*}
Since $\mid \langle \omega - \to, y_k \rangle \mid \leq 2D\delta$ and
$$\| y_{i+1} - y_i\| - \langle \omega, y_{i+1} - y_i\rangle =
(1- \cos \varphi_i ) \| y_{i+1} - y_i\|,$$
$\varphi_i \in {[0,\frac{\pi}{2}]}$ being the angle between the vectors
$\omega$ and $y_{i+1} - y_i$, we find
$$2D\delta \geq 2\sum_{i=1}^{k-1} \|y_{i+1} - y_i\| \sin^2\frac{\varphi_i}
{2} \geq 2d \sum_{i=1}^{k-1} \sin^2\frac{\varphi_i}{2}.$$
Combining this with (11), one obtains
$$\sin^2\frac{\varphi_i}{2} \leq \frac{D}{d}\delta < 
\sin^2\frac{\varphi_0}{4}$$
and therefore $\varphi_i < \frac{\varphi_0}{2}$ for each 
$i = 1,\ldots,k-1$.

Assume that $k > 2$ and denote by $\varphi$ the angle
between the vectors $y_2 - y_1$ and $y_3 - y_2$. Since
$\varphi \leq \varphi_1 + \varphi_2$, we get $\varphi < \varphi_0$,
which is a contradiction with the
choice of $\varphi_0$ (see Lemma 2.4). Thus, $k = k(\omega) \leq 2$.

In this way we have shown that, if $\delta$ satisfies (11), then
$k(\omega) \leq 2$ for each $\omega$ with (8) and (9).

Next, we assume that $\delta$ satisfies (11). Given $\omega$ with (8)
and (9), we have $\omega \neq \to$, and so $\gamma'(\omega)$ has at
least one {\it transversal reflection point} $y$, that is 
$\gamma'(\omega)$ is not tangent to $\dl$ at $y$. The final step in the
proof of Lemma 2.1 will be to show that there exists $\omega$ with
(8) and (9) so that $\gamma'(\omega)$ has exactly one transversal
reflection point. This will be derived by the following lemma.\\

{\bf 2.5. Lemma.} {\it Let $P\neq Q$ be two convex components of $L$
and let ${\cal T}$ be the set of those $\omega$ with {\rm (8)} and
{\rm (9)} such that $\gamma'(\omega)$ has two transversal reflection
points, the first of which belongs to $P$ and the second to $Q$. 
Assume that $\omega_0$ is a point of Lebesgue measure density of
${\cal T}$ in $\sn$. Then there exist $y\in P, z\in Q$ such that}
$$N_L(y) = \pm N_L(z) = \pm e \:\:\: , \:\:\: 
\frac{z-y}{\| z-y\|} = \omega_0 .$$

{\it Proof of Lemma 2.5.} Recall that by our assumptions we have
$x = 0$ and $N_K (0) = e = (0,\ldots,0,1)$. Consequently,
$\omega_0 = (u_0;0)$ for some $u_0 \in {\bf R}^{n-1}$ with
$\| u_0\| = 1$. Set
$$B_0 = \{ u\in {\bf R}^{n-1} : \| u\| < 1 \}.$$
For $u = (u_1,\ldots,u_{n-1})\in B_0$ define
$$\omega(u) = (u; -\sqrt{1-\| u\|^2}) \:\: , \:\: \theta(u) = 
\theta(\omega(u)) = (u;\sqrt{1-\| u\|^2}).$$
Clearly, if $u$ is sufficiently close to $u_0$, then $\omega(u)$
satisfies (8) and (9). Therefore $u_0$ is a point of positive Lebesgue
measure density of 
$$U' = \{ u \in B_0 : \omega(u) \in {\cal T} \: \}$$
in ${\bf R}^{n-1}$.


\def\ou{\omega(u)}
\def\tu{\theta(u)}
\def\du{\partial u_i}
\def\dtu{\frac{\partial T}{\partial u_i}}
\def\dou{\frac{\partial \omega}{\partial u_i}}
\def\dttu{\frac{\partial \theta}{\partial u_i}}
\def\dyu{\frac{\partial y}{\partial u_i}}
\def\dzu{\frac{\partial z}{\partial u_i}}
\def\zy{\frac{z(u) - y(u)}{\|z(u)- y(u)\|}}



Let $u' \in U'$ be sufficiently close to $u_0$. Then there exist
$y(u') \in P$ and $z(u') \in Q$ which are the successive reflection
points of a reflecting $(\ou,\tu)$-ray in $\Omega_L$. This is in
fact the ray $\gamma'(\ou)$, and so $T_{\gamma'(\ou)} = 0$.

Set $L' = P\cup Q$. It is clear that there exists an open $U$ with
$U' \subset U\subset B_0$ such that for each $u\in U$ there exist
unique $y(u) \in \partial P, z(u)\in \partial Q$ which are the
successive reflection points of a reflecting $(\ou,\tu)$-ray 
$\gamma'(u)$ in $\Omega_{L'}$. However, the sojourn time
$T(u) = T_{\gamma'(u)}$ might be non-zero. It is easy to see that
$y(u)$ and $z(u)$ are smooth maps of $u\in U$ (cf. for example
Appendix b in \cite{kn:Sj} or Chapter 10 in \cite{kn:PS2}). Hence
$$T(u) = \langle \ou, y(u)\rangle + \|y(u) - z(u)\| -
\langle \tu, z(u)\rangle $$
is a smooth function of $u$. Let us calculate $\dtu$ for $u\in U$
and $i = 1,\ldots,n-1$. We have
\begin{eqnarray*}
\dtu (u) 
& = & \langle \dou (u), y(u)\rangle + \langle \ou, \dyu (u)\rangle
+ \langle \zy, \dzu (u) - \dyu (u)\rangle \cr
&   & - \langle \dttu (u), z(u)\rangle
- \langle \tu ,\dzu (u)\rangle.\cr
\end{eqnarray*}
Since $\dyu (u)$ is tangent to $\partial P$ at $y(u)$, it follows by the
reflection law at $y(u)$ that
$$\langle \zy , \dyu (u) \rangle = \langle \ou ,\dyu (u)\rangle.$$
Similarly, one finds
$$\langle \zy , \dzu (u) \rangle = \langle \tu ,\dzu (u)\rangle.$$
Using the latter two equalities, we deduce from above that
\begin{equation}
\dtu (u) = \langle \dou (u), y(u) \rangle - \langle \dttu (u),
z(u) \rangle.
\end{equation}

\def\tiu{\tilde{u}}



Fix $\epsilon \in (0,1)$ such that 
$u\in U, \|u- u_0\| < \epsilon$ imply $\|\ou - \omega_0\| < \delta$.
Clearly, there exist points $u \in U$ with $\|u - u_0\| < \epsilon$
which are points of positive Lebesgue measure density of $U'$ (otherwise $U'$
would have measure zero). Fix a point $\tilde{u}$ with these properties.
We claim that
\begin{equation}
\dtu (\tiu) = 0 \:\:\:\: , \:\:\: i = 1,\ldots,n-1.
\end{equation}
Consider an arbitrary $i =1,\ldots,n-1$. It is sufficient to show that
each neighbourhood $V$ of $\tiu$ in ${\bf R}^{n-1}$ contains a point $v$
with $\dtu (v) = 0$. Fix a cubic neighbourhood $V$ of $\tiu$
and let $e_i = (0,\ldots,0,1,0,\ldots,0)$ be the $i$-th basis vector
in ${\bf R}^{n-1}$. Since 
$V\cap U'$ has positive measure in ${\bf R}^{n-1}$, there exists
$v = (v_1,\ldots,v_{n-1}) \in V$ such that $v+se_i \in V\cap U'$ for at 
least two different values of $s \in {\bf R}$. For such values of $s$
the definition of $U'$ implies $T(v+ s e_i) = 0$. Therefore there exists
$v' = v+ s' e_i \in V$ with $\dtu (v') = 0$. This proves (13).

Now it follows from (12) and (13) that
$$\langle \dou (\tiu), y(\tiu)\rangle - \langle \dttu (\tiu),
z(\tiu) \rangle = 0 \:\:\:\: , \:\: i =1,\ldots,n-1.$$
On the other hand,
$$\dou (\tiu) = e_i + \frac{\tiu_i}{\sqrt{1-\| \tiu \|^2}} e \:\:\: ,
\dttu (\tiu) = e_i - \frac{\tiu_i}{\sqrt{1-\| \tiu \|^2}} e,$$
therefore
$$y_i(\tiu ) + \frac{\tiu_i}{\sqrt{1-\| \tiu \|^2}} y_n(\tiu) -
z_i(\tiu) + \frac{\tiu_i}{\sqrt{1-\| \tiu \|^2}} z_n(\tiu) = 0.$$
That is
$$z_i(\tiu) - y_i (\tiu) = \lambda \tiu_i \:\:\: , \:\: i =1,\ldots,n-1$$
with $\lambda = \frac{z_n(\tiu) + y_n(\tiu)}{\sqrt{1 - \| \tiu \|^2}}$.
Consequently, there exists $\mu \in {\bf R}$ with
$$z(\tiu) - y(\tiu) = \lambda \tiu + \mu e.$$
In particular, we see that the vector $z(\tiu) - y(\tiu)$ is collinear
with the two-dimensional plane $\Pi(\tiu)$, determined by the point
$y(\tiu)$ and the vectors $\tiu$ and $e$. Since the vector $\omega(\tiu)$
is collinear with this plane, the law of reflection at $y(\tiu)$ yields
that $N_L(y(\tiu))$ is also collinear with $\Pi(\tiu)$. Using the same
argument for $\theta(\tiu)$ and the law of reflection at $z(\tiu)$, we
find that $N_L(z(\tiu))$ is collinear with $\Pi(\tiu)$, too. In this way
we have shown that the vectors 
$\tiu, \omega(\tiu), \theta(\tiu), N_L(y(\tiu)), N_L(z(\tiu))$
are collinear. 

Since $u_0$ can be approximated by vectors $\tiu \in U$ having the
latter property, using the compactness of $\partial P$ and $\partial Q$,
we deduce that there exist $y\in \partial P$ and $z\in \partial Q$ such
that $\zy = \omega_0$, $\langle z-y, N_L(y)\rangle = 0$,
$\langle z-y,N_L(z)\rangle = 0$ and the vectors $N_L(y)$ and $N_L(z)$
are collinear with the two-dimensional plane, determined by 
$\omega_0$ and $e$. This clearly implies that both $N_L(y)$ and
$N_L(z)$ are collinear with $e$, i.e. $N_L(y) = \pm N_L(z) = \pm e $.
$\spadesuit$\\

Now assume that for each $\omega$ with (8) and (9), 
$\gamma'(\omega)$ has two transversal reflection points. For
$i,j = 1,\ldots,r$ denote by ${\cal T}_{i,j}$ the set of those
$\omega$ with (8) and (9) such that the first reflection point of
$\gamma'(\omega)$ belongs to $L_i$ and the second one to $L_j$.
Since $\cup_{i,j=1}^r {\cal T}_{i,j}$ is an open subset of $\sn$
(so it has positive measure) and $\omega_0$ belongs to its closure,
there exist $i \neq j$ such that $\omega_0$ is a point of positive
Lebesgue measure density of ${\cal T}_{i,j}$. Fix $i$ and $j$ with
this property and apply Lemma 2.5 to $P = L_i$, $Q = L_j$ and
${\cal T} = {\cal T}_{i,j}$. We get that there exist $y\in\dl_i$
and $z\in\dl_j$ such that $N_L(y) = \pm N_L(z) = \pm e$ and  
$\omega_0 = \frac{z-y}{\|z-y\|}$. Since $e = N_K(0)\in {\cal N}$,
it follows by the definition of ${\cal N}$ that $N_L$ is regular at 
both $y$ and $z$. Consequently,
$(y,z) \in \Sigma(\dl_i , \dl_j)$ and 
$\omega_0 = h_{i,j}(y,z) \in \Sigma_{i,j}$ in contradiction with
$\omega_0 \in {\cal S}$. 

Therefore there exists $\omega$ with (8) and (9) such that 
$\gamma'(\omega)$ has exactly one transversal reflection point
$y(\omega)$. Fix $\omega$ with this property and set 
$y = y(\omega)$. Then $\omega - \theta(\omega)$ is a non-zero
vector collinear with both $N_K(0) = N_K(x)$ (recall that $x = 0$ by
our assumption) and $N_L(y)$. Hence $N_K(x) = \pm N_L(y)$. On the
other hand, 
$$0 = T_{\gamma'(\omega)} = \langle \omega - \theta(\omega), y
\rangle$$
shows that $\langle x - y, N_K(x)\rangle = 0$ which implies that
$T_x \dk = T_y \dl$. This completes the proof of Lemma 2.1.
$\spadesuit$\\

{\it Proof of Theorem 1.1}. It is sufficient to show that $K_1$
coincides with some convex component of $L$. 

Fix an element $x$ of $\dk_1$. It follows from
Lemma 2.1 that there exists $y\in \dl$ with $T_x \dk = T_y \dl$.
Fix $y$ with this property and let $y \in L_i$ for some
$i = 1,\ldots,r$. We will show that $K_1 = L_i$. According to
the condition (H) and Lemma 2.1, one gets easily that the set $W$ of those
$x'\in \dk_1$ for which there exists $y'\in \dl_i$ with
$T_{x'} \dk = T_{y'} \dl$ is open in $\dk_1$. On the other hand, $W$
is also closed in $\dk_1$, and since $W \neq \emptyset$ and
$\dk_1$ is connected, we find $W = \dk_1$. Consequently, any
supporting hyperplane to $K_1$ is a supporting hyperplane to $L_i$ as well.
Now using the convexity of $K_1$ and $L_i$ we conclude that
$K_1 = L_i$. 




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\end{thebibliography}


$\:\:\:\:$
Department of Mathematics, University of Western Australia,
Nedlands, Perth 6009, Western Australia; stoyanov@maths.uwa.edu.au





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