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\topmatter
\title ON AN INTEGRO-DIFFERENTIAL TRANSFORM ON THE SPHERE\endtitle
\author E. Makai, Jr.$^*$, H\. Martini, T\. {\'O}dor$^{**}$ \endauthor
\thanks * Partially supported by Hungarian National Foundation for
Scientific
Research, Grant No\. T-031931 and FKFP Grant No. 0391/1997
\newline ** Partially supported by
Hungarian National Foundation for Scientific Research, Grant number 4427 
\endthanks



\rightheadtext\nofrills{\sevenrm On an integro-differential transform on the 
sphere}
\leftheadtext\nofrills{\sevenrm E. Makai, Jr.$^*$, H\. Martini, 
T\. {\'O}dor$^{**}$}
\address
$^*$, $^{**}$: Alfr\'ed R\'enyi Mathematical Institute of the Hungarian 
Academy of Sciences,
P.O.B.\ 127,
H-1364 Budapest, Hungary;
\newline Technische Universit\"at Chemnitz,
Fakult\"at f\"ur Mathematik,
D-09107 Chemnitz, Germany
\endaddress
\email
makai\@renyi.hu,
martini\@mathematik.tu-chemnitz.de,
odor\@renyi.hu
\endemail

%\date
%\today
%\enddate
\subjclass
1991 {\it Mathematics Subject Classification.} Primary:
44A12; Secondary: 33C55
\endsubjclass
\keywords
integro--differential 
operator, spherical harmonics, Funk--Hecke theorem 
\endkeywords


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\abstract In a recent paper the authors have proved that a convex body $K 
\subset \Bbb R ^d$, containing the origin $0$ in its interior, is symmetric 
with respect to $0$ if and only if $V_{d-1}(K \cap H') \ge V_{d-1}(K \cap
H)$ for all hyperplanes $H,H'$ such that $H$ and $H'$ are parallel and $H' 
\ni 0$ ($V_{d-1}$ is $(d-1)$--measure). For the proof the authors employed 
a new type of integro--differential transform, that let to correspond to a 
sufficently nice function $f$ on $S^{d-1}$ the function $R^{(1)}f$, where 
$(R^{(1)}f)(\xi )= \int\limits _{S^{d-1} \cap \xi ^\bot } ({\partial f}/
{\partial \psi }) d \eta $ --- where $\xi \in S^{d-1}$ and $\psi $ is 
geographic latitude --- and determined the null--space of the operator $R^
{(1)}$. In this paper we extend the definition to any integer $m \ge 1$, 
defining $(R^{(m)}f)(\xi )$ analogously as for $m=1$, but using $ 
{\partial ^m f}/{\partial \psi ^m}$ rather than $ 
{\partial f}/{\partial \psi }$. We investigate the null--space of the operator
$R^{(m)}$, determining it nearly.
\endabstract



\endtopmatter
\document

\vskip .4cm

%%%%%%%%%%%%%%INTRODUCTION%%%%%%%%%%%%%%%%%%%%%%

\heading
1.\ Introduction
\endheading
\vskip .4cm
In [Makai--Martini--\'Odor] we have proved that 
a convex body $K \subset \Bbb R ^d$, containing
the origin $0$ in its interior, has the property that for every 
hyperplane $H$ the hyperplane $H'$ parallel to $H$ and passing 
through 
$0$ satisfies $V_{d-1} (K \cap H') \ge V_{d-1} (K \cap H) $ ($V_{d-1}$ is 
$(d-1)$--volume), if and only if this convex body $K$ is symmetric with 
respect to $0$. 
The case $d=2$ was proved by
[Hammer, 1954]. (In fact he proved the analogous statement for 
$1$--dimensional sections for $K \subset \Bbb R ^d$, and we have proved 
the analogous statement for $k$--dimensional sections, $1 \le k \le d-1$.)

We have proved our result by using a new Radon--type transformation,
which can be considered as a common generalization of partial differential
operators and Radon--type transformations. 
Moreover we have proved our theorem by using spherical harmonics
and the Funk--Hecke theorem. 

Our integro--differential transformation let to correspond to a
sufficiently
nice function $f\colon S^{d-1}\to \Bbb R$ the function $R^{(1)}f\colon 
S^{d-1}\to \Bbb R$,
where $(R^{(1)}f)(\omega)$ 
is the integral over the large $S^{d-2}$ of $S^{d-1}$,
with pole $\omega$, of the derivative of $f$ in the direction $\omega$. To
prove our theorem we have established that the null--space of this operator $R$
consists of the even functions in the domain of $R$.

We investigate generalizations of this transformation, where we integrate
over the same large $S^{d-2}$
the
$m$'th derivative of $f$ with respect to the geographic latitude ($\omega$
considered as the north pole). The case $m=0$ is the spherical Radon
transformation
(Funk transformation), the case $m=1$ is the above transformation $R^{(1)}$. We
obtain that for $m \ge 1$ 
the null--space of this operator is the direct sum of the even
functions
(for $m$ odd) or odd functions (for $m$ even) in the domain of this
operator, respectively, and a
subspace of finite
dimension.

For general analytical background we refer to the books [Tricomi, 1957],
[Adams, 1975] and [Ziemer, 1989].
\vskip2truecm


%%%%%%%%PARAGRAPH 2%%%%%%%%%%%%%%

\heading
2. Preliminaries
\endheading
\vskip .4cm
As usual, $\Bbb R^d$ denotes the $d$--dimensional Euclidean space
which is endowed with the standard inner product and norm $|\,\cdot \,|$
structure. We will suppose $d \ge 2$. 
The origin is denoted by $0$ and
$V_{d-1}$ is the $(d-1)$--dimensional volume on the hyperplanes.

Let $S^{d-1}$ denote the unit sphere with centre $0$;
its variable point will be denoted by $\omega, \,\, \xi$ or 
$\eta$. For $\omega \in S^{d-1}$ and $t \in \Bbb R$ let 
$H(\omega, t)$ be the hyperplane given by the equation $ \langle x,
\omega \rangle = t$. We write $\omega ^\bot$ for $H(\omega, 0)$. 
Often we will use a polar coordinate system on $S^{d-1}$, with pole 
some $\xi \in S^{d-1}$. That is, any $\omega \in S^{d-1}$ can 
be written as $\xi \sin \psi + \eta \cos \psi$,
where $\eta \in S^{d-1} \cap \omega ^\bot
$ and $- \pi /2 \le \psi \le \pi /2 $ (thus $\psi $ is the {\it geographic 
latitude}, which will be more convenient for us than the costumarily used 
$ \phi = \pi /2 - \psi $); 
then we write $\omega = (\eta, \psi)$. A real function 
defined on $S^{d-1}$ is called {\it {even (odd)}}, if for all $
\omega \in S^{d-1}$ we have $ f(- \omega )=f(\omega)\,\,\,\,(f(-\omega)
=-f(\omega))$. 


We turn to spherical harmonics, which are higher--dimensional
generalizations of
the trigonometric functions $\cos (nx)$, $\sin(nx)$ (these are
obtained for $d=2$). Standard references are [M{\"u}ller, 1966], [Seeley, 
1966], 
[Erd\'elyi et al., 1953] and,
for $d=3$
in more detail, [Sansone, 1959]; further references, with some geometrical
applications,
are e.g\. [Blaschke, 
1949], [Gardner, 1995], Appendix C, 
and also the survey paper [Groemer, 1993] and the books
[Schneider, 1993], pp\. 428--432, as well as [Groemer, 1996],
which contain ample further bibliography.

A polynomial $f\colon\Bbb R ^d\to\Bbb R$ is {\it harmonic\/}, if
$\sum\limits_{i=1}^d\left({\partial}/{\partial x_i}\right)^2f=0$. (This
is
invariant under the choice of an orthonormal base.) For an integer
$n\ge 0$ a
{\it spherical harmonic (of degree $n$)\/} in $d$ dimensions is
the restriction of a homogeneous harmonic polynomial $f\colon\Bbb R ^d
\to\Bbb R$
(of degree $n$) to $S^{d-1}$. (Since $d$ will be fixed, later we will not
refer to the dimension.) The spherical harmonics of degree $n$ form a
finite
dimensional vector space. Choosing from this subspace an orthonormal base
$\{Y_{ni}\,\,\vert\,\,1\le i\le N(d,n)\}$ (orthonormality meant in the
space
$L^2(S^{d-1})$, for Lebesgue measure on $S^{d-1}$), their union for each
$n\ge
0$ is a complete orthonormal system in $L^2(S^{d-1})$. Thus each $f\in
L^2(S^{d-1})$ has a Fourier expansion $\sum\limits_{n=0}^\infty\left(
\sum\limits_{i=1}^{N(d,n)} c_{ni}Y_{ni}\right)$. We will write
$\sum\limits_{i=1}^{N(d,n)}c_{ni}Y_{ni}=Y_n(f)$, thus the Fourier expansion
of
$f$ is $\sum\limits_{n=0}^\infty Y_n(f)$.

The spherical harmonics are the eigenfunctions of many linear operators
commuting with rotations. For example, 
the Funk--Hecke theorem ([Seeley, 1966], 
Theorem 3)
says
that if $F$ is measurable on $[-1,1]$, with $\int\limits_{-1}^1\vert
F(t)\vert(1-t^2)^{(d-3)/2}dt<\infty$, then any spherical harmonic $Y_n$ of
$n$--th degree is an eigenfunction of the integral operator $f\mapsto
g=g(\xi)=\int\limits_{S^{d-1}}F(\langle\xi,\eta\rangle)f(\eta)d\eta$, that
is
$$\int\limits_{S^{d-1}}
F(\langle\xi,\eta\rangle)Y_n(\eta)d\eta=\lambda_nY_n(\xi)\,,$$
where the eigenvalue $\lambda_n$ equals
$$
\lambda_n=V_{d-2}(S^{d-2})C_n(1)^{-1}\int\limits_{-1}^1F(t)C_n(t)
(1-t^2)^{(d-3)/2}dt\,.
$$
Here $V_{d-2}$ means $(d-2)$--dimensional volume, and $C_n(t)=
C_n^{(d-2)/2}(t)$ is the $n$'th Gegenbauer polynomial,
of order $(d-2)/2$, that is a non--zero polynomial of degree $n$,
satisfying for $0\le n<m$ the orthogonality relations
$$
\int\limits_{-1}^1C_n(t)C_m(t)(1-t^2)^{(d-3)/2}dt=0\,.
$$
There holds $C_n(1)\ne 0$, [Seeley, 1966], (3). For $n$ odd (even) $C_n$ 
is an odd
(even)
function [Erd\'elyi et al., 1953], \S 10.9, (16). References to Gegenbauer 
polynomials are
[Erd\'elyi et al., 1953] and [Tricomi, 1955].

For suitable measures or distributions on $[-1,1]$ a formula similar to the
Funk--Hecke theorem holds, cf.\ for example Lemma 3.2.
\vskip2truecm

%%%%%%%%%%PARAGRAPH 3%%%%%%%%%%%

\heading 3. The operators $R^{(m)}_{\psi }$ and $R^P_{\psi }$\endheading
\vskip0.4truecm

Now we generalize the definition of the operator $R^{(1)}_{\psi }$ from
[Makai--Martini--\'Odor].
Let $d\ge 2$, and let $m\ge 1$ be an integer. We let
$Lip^{m-1}(S^{d-1}):=\{f\colon S^{d-1}\to\Bbb R\,\,\vert\,\, f$
has in some $C^\infty$--atlas of $S^{d-1}$ all partial derivatives of order
at most $m-1$, and all these are Lipschitz functions$\}$. We write
$Lip^0(S^{d-1})=Lip(S^{d-1})$. Let $-\pi/2\le\psi\le\pi/2$,
$\,f\in Lip^{m-1}(S^{d-1})$ and $\xi\in S^{d-1}$. Using polar coordinates
with
pole $\xi$, we define the integro--differential transform
$R_\psi^{(m)}f$
of $f$ by
$$(R_\psi^{(m)}f)(\xi)=
\int\limits_{S^{d-1}\cap\xi^\bot}
\frac{\partial^mf}{\partial\psi^m}(\eta,\psi)d\eta\,,$$
provided the
right--hand side exists. For 
$\psi=0$ we
drop the lower index.
The case $m=1$ of the following lemma is Lemma 3.6 of [Makai--Martini--\'Odor].

\proclaim{Lemma 3.1} Let $d\ge 2,\,\,m\ge 1$, $-\pi/2\le\psi\le\pi/2$
and
$f,g\in Lip^{m-1}(S^{d-1})$. Then, for almost all $\xi\in S^{d-1}$ we have
for almost all $\eta\in
S^{d-1}\cap\xi^\bot$ that $
({\partial^mf}/{\partial\psi^m})(\eta,\psi)$ exists, and,  
for almost all $\xi\in S^{d-1}$, the
integral defining $(R_\psi^{(m)}f)(\xi)$ exists. We have
$R_\psi^{(m)}f\in L^\infty(S^{d-1})$, and $R_\psi^{(m)}$ is
symmetric, i.e., $\int\limits_{S^{d-1}}
(R_\psi^{(m)}f)(\xi)g(\xi)d\xi=
\int\limits_{S^{d-1}}f(\xi)(R_\psi^{(m)}g)(\xi)d\xi$.
\endproclaim

\demo{Proof} The claimed existence of $\partial^mf/\partial\psi^m$ for
$\psi=\pm\pi/2$ follows from [Whitney, 1957], Ch.IX, Theorem 11A, by which
Lipschitz functions $S^{d-1} \to \Bbb R$ are almost everywhere 
differentiable, and 
otherwise
follows like
in Lemma 3.5 of [Makai--Martini--\'Odor], using the manifold 
$M_{\psi }=\{(\xi,\omega)\,\,\vert\,\,\xi,\omega\in S^{d-1}$,
$\langle\xi,\omega\rangle=\sin\psi\}$. 
Namely, $M_{\psi }$ is an $S^{d-2}$--bundle over $S^{d-1}$, via either 
projection $\pi _1:(\xi ,\omega ) \mapsto \xi, \,\,\pi _2 :(\xi ,\omega ) 
\mapsto \omega $. Applying locally the Fubini theorem to both $\pi _2 $ and 
$\pi _1$, we gain in turn, letting $\omega =(\eta , \psi )$, that $V_{2d-3}
(\{(\xi ,\omega )\in M_{\psi } \mid f(\omega )$ is not $m$ times 
differentiable at $\omega =(\eta , \psi )\})=0$, and $V_{d-1}(\{\xi \in 
S^{d-1} \mid \{ \omega \in S^{d-1} \mid \langle \xi ,\omega \rangle = \sin 
\psi ,\,\,f (\omega )$ is not $m$ times differentiable at $\omega =(\eta ,
\psi )\}$ does not have $(d-2)$--Lebesgue measure $0\})=0$ ($V_i$ denoting 
invariant Lebesgue type measure in the respective spaces). In other words: 
for almost all $\xi \in S^{d-1}$ we have for almost all $\omega \in 
\{\omega ' \in S^{d-1} \mid \langle \xi , \omega ' \rangle =\sin \psi \}$ 
that $f(\omega )$ is $m$ times differentiable at $\omega =(\eta ,\psi )$.
For almost all $\xi\in S^{d-1}$
we
have that the function $\eta\mapsto ({\partial^mf}/{\partial\psi^m})       
(\eta,\psi)$ is the almost everywhere limit of the continuous
functions
$$
\eta\mapsto\left(\frac{\partial^{m-1}f}{\partial\psi^{m-1}}
(\eta,\psi+\varepsilon)-
\frac{\partial^{m-1}f}{\partial\psi^{m-1}}
(\eta,\psi)\right)\biggm/\varepsilon
$$
with absolute value below a constant depending of $f$. Therefore the
function
$\eta\mapsto ({\partial^mf}/$ 
\newline ${\partial\psi^m})(\eta,\psi)$ is
integrable and $R_\psi^{(m)}f$ is bounded, for almost all $\xi$. Also
$R_\psi^{(m)}f$ is measurable, since by Lebesgue's dominated 
convergence theorem it is the almost everywhere limit
of
the (uniformly bounded set of) continuous functions
$$
\xi\mapsto\int\limits_{S^{d-1}\cap\xi^\bot}
\left(\frac{\partial^{m-1}f}{\partial\psi^{m-1}}
(\eta,\psi+\varepsilon)-
\frac{\partial^{m-1}f}{\partial\psi^{m-1}}(\eta,\psi)
\right)\varepsilon^{-1}d\eta\,.
$$
Hence $R_\psi^{(m)}f\in L^\infty(S^{d-1})$.

Further we have
$$
\int\limits_{S^{d-1}}
(R_\psi^{(m)}f)(\xi)g(\xi)d\xi=$$
$$\lim_{\varepsilon\to 0}
\int\limits_{S^{d-1}}\left[
\int\limits_{S^{d-1}\cap\xi^\bot}\left(
\frac{\partial^{m-1}f}{\partial\psi^{m-1}}
(\eta,\psi+\varepsilon)-
\frac{\partial^{m-1}f}{\partial\psi^{m-1}}
(\eta,\psi)\right)\varepsilon^{-1}
d\eta\right]g(\xi)d\xi\,.
$$
An induction on $m$ shows that this equals
$\int\limits_{S^{d-1}}f(\xi)(R_\psi^{(m)}g)(\xi)d\xi$. The induction
basis is $m=1$, which follows from Lemma 3.6 of [Makai--Martini--\'Odor].
\qued
\enddemo

Defining $R_\psi^{(0)}$ in the analogous way, i.e., by
$$(R_\psi^{(0)}f)(\xi)=
\int\limits_{S^{d-1}\cap\xi^\bot}
f(\eta,\psi)d\eta\,,$$
we have
the following lemma, that generalizes Lemma 3.7 of [Makai--Martini--\'Odor].
Its statement in the case $m=0$ is due to [Radon, 1917] (for $d=3$), and to 
[Schneider, 1969], formula (5) (an alternative proof cf. in 
[Makai--Martini--\'Odor], Lemma 3.7), while in the case $m=1$ it is due to
[Makai--Martini--\'Odor], Lemma 3.7.

\proclaim{Lemma 3.2} Let $d\ge 2$, $m\ge 0$, $-\pi/2\le\psi\le\pi/2$, 
and
let $Y_n\colon S^{d-1}\to\Bbb R$ be a spherical harmonic of degree $n$. Then 
$Y_n$
is an eigenfunction of $R_\psi^{(m)}$, i.e\.,
$R_\psi^{(m)}Y_n=\lambda_nY_n$, with
$$
\lambda_n=V_{d-2}(S^{d-2})C_n^{(d-2)/2}(1)^{-1}
\left(\frac{d}{d\psi}\right)^mC_n^{(d-2)/2}(\sin\psi)\,\,.
$$
\endproclaim

\demo{Proof} By the case $m=0$ of our statement (referred to above) we have
$$
\int \limits _ {S ^{d-1} \cap \xi ^ \bot}
Y_n(\eta, \psi) d \eta =
V_{d-2}(S^{d-2})C_n(1)^{-1}C_n(\sin\psi)
\cdot Y_n(\xi)\,.
$$
From this case subsequent differentiations with respect to $\psi$
prove the
statement for any
$m\ge 1$. 
\qued
\enddemo

The following theorem is a generalization of Theorem 3.8 of 
[Makai--Martini--\'Odor].
The statement corresponding to the case $m=0$, $\psi=0$ in 
this theorem is the Funk integral
theorem (for $f\in 
C(S^{d-1})$),
cf.\ [Funk, 1913], Kap. 2, [Lifshitz--Pogorelov, 1954], 
[Schneider, 1969], [Helgason, 1980], Ch. 3, \S 1.B and [Helgason, 1984], while 
the case $m=1$ is contained in [Makai--Martini--\'Odor], Theorem 3.8.

\proclaim{Theorem 3.3} Let $d\ge 2$, $m\ge 1$ and
$-\pi/2\le\psi\le\pi/2$.
Then the null--space of the operator $R_\psi^{(m)}\colon
Lip^{m-1}(S^{d-1})\to
L^\infty(S^{d-1})$ equals $\{f\in Lip^{m-1}(S^{d-1})\,\,\vert$ the Fourier
expansion $\sum\limits_{n=0}^\infty Y_n(f)$ of $f$ satisfies that
$\left(d/d \psi \right)^mC_n^{(d-2)/2}(\sin\psi)\ne
0$
implies $Y_n(f)=0\}$. In particular, for $\psi=0$ and $m$ odd (even)
the
null--space of $R^{(m)}=R^{(m)}_0$ 
contains $\{f\in Lip^{m-1}(S^{d-1})\,\,\vert
\,\,f$
is even ($f$ is the sum of an odd function and a constant)$\}$.
\endproclaim

\demo{Proof} For the first statement we proceed analogously to 
[Alexandroff, 1937], [Petty, 1961], [Schneider, 1969], [Schneider, 1970], 
[Falconer, 1983].
Let $f\in Lip^{m-1}(S^{d-1})$. Then, by 3.1,
$R_\psi^{(m)}f\in L^\infty(S^{d-1})\subset L^2(S^{d-1})$, and, by
completeness of spherical harmonics, $R_\psi^{(m)}f=0$ holds a.e\. if
and
only if for each $n\ge 0$, and each spherical harmonic $Y_n$ of degree $n$
we
have $0=\langle R_\psi^{(m)}f,Y_n\rangle$, where $\langle\,\,,\,\,
\rangle$ now denotes 
scalar product in $L^2(S^{d-1})$. Letting $\sum\limits_{n=0}^\infty Y_n(f)$
be the
Fourier expansion of $f$, we have by 3.1 and 3.2 that $$\langle
R_\psi^{(m)}f,
Y_n\rangle=\langle f,R_\psi^{(m)}Y_n\rangle=\lambda_n\langle
f,Y_n\rangle=$$
$$V_{d-2}(S^{d-2})C_n(1)^{-1}
\left(\frac{d}{d\psi}\right)^mC_n(\sin\psi)\cdot\langle Y_n(f),
Y_n\rangle .$$ For fixed $n$ and $Y_n$ arbitrary this is $0$ if and only if
$\left(d/d \psi \right)^mC_n(\sin\psi)\cdot Y_n(f)=0$. This
implies the first statement.

For the second statement observe that for $f$ constant we have
$R_\psi^{(m)}f=0$. Furthermore, for $\psi=0$, $m$ odd (even) and $f$
even
(odd) by 3.1 we have for almost all $\xi\in S^{d-1}$ that for almost all
$\eta\in S^{d-1}\cap\xi^\bot$ both
$({\partial^mf}/{\partial\psi^m})(\eta,0)$ and
$({\partial^mf}/{\partial\psi^m})(-\eta,0)$ exist and then have sum
0,
thus $(R^{(m)}f)(\xi)=0$ a.e\.
\qued
\enddemo

More generally, let $d\ge 2$, $m\ge 1$ an integer, $P$ a polynomial of
degree $m$, and $-\pi/2\le\psi\le\pi/2$. Then we define for $f\in
Lip^{m-1}(S^{d-1})$ and $\xi\in S^{d-1}$
$$
(R_\psi^{P}f)(\xi)=\int\limits_{S^{d-1}\cap\xi^\bot}
\left(P\left(\frac{\partial}{\partial\psi}\right)\right)
f(\eta,\psi)d\eta\,,$$
provided that the right--hand side exists. For $\psi =0$ we drop the lower 
index. Then, analogously like above, we have

\proclaim{Theorem 3.4} Let $d\ge 2$ be a fixed integer. Let $m\ge 1$ be an
integer, $P$ a real polynomial of degree $m$, and $-
\pi/2\le\psi\le\pi/2$. Furthermore, let 
$f,g\in Lip^{m-1}(S^{d-1})$. Then, for almost
all
$\xi\in S^{d-1}$ we have for almost all $\eta\in
S^{d-1}\cap\xi^\bot$ that
$\left(P\left({\partial}/{\partial\psi}\right)\right)f(
\eta,\psi)$ exists, and, for almost
all
$\xi\in S^{d-1}$, the integral defining 
$(R_\psi^{P}f)(\xi)$
exists. We have $R_\psi^{P}f\in L^\infty(S^{d-1})$, and $R_{\psi }^P$ is 
symmetric, i.e., $\int\limits_{S^{d-1}}
(R_\psi^Pf)(\xi)g(\xi)d\xi=
\int\limits_{S^{d-1}}f(\xi)(R_\psi^Pg)(\xi)d\xi$. The null--space
of the operator $R_\psi^{P}\colon Lip^{m-1}(S^{d-1})\to
L^\infty(S^{d-1})$ equals $\{ f\in Lip^{m-1}(S^{d-1})\,\,\vert\,\,$the
Fourier expansion
$\sum\limits_{n=0}^\infty Y_n(f)$ of $f$ satisfies that $\left(P
\left(d/d \psi \right)\right)C_n^{(d-2)/2}(\sin\psi)
\ne
0$ implies $Y_n(f)=0\}$.
\endproclaim 

\demo{Proof}
Let $P(t)=\sum_{r=0}^m c_r t^r$.
The existence statements, as well as 
the statement $R_{\psi}^{P}f \in L^{\infty}(S^{d-1})$
and symmetry of $R_{\psi}^{P}$, follow from 3.1. Then 
3.2 implies its
analogue for $R^{P}_{\psi}$, with eigenvalue 
$$ 
\lambda _n= V_{d-2} (S^{d-2}) C_n^{(d-2)/2} (1)^{-1} 
\(P \(
\frac {d} {d \psi }
\)\)
C_n^{(d-2)/2}(\sin \psi )\,. $$
Then, like in 3.3, we obtain that we have $(R_{\psi}^
{P} f)(\xi)=0$ a.e.\
if and only if
$$
\(P \(\frac {d} {d \psi }\)\)
C_n^{(d-2)/2}(\sin \psi) \ne  0
$$
implies $Y_n(f)=0$, where $\sum_{n=0}^{\infty}Y_n(f)$ is the
Fourier expansion of $f$.
\qued
\enddemo

{\bf {Remark.}} By Lemma 3.1 and Theorem 3.4 $R^{(m)}_{\psi }$ and 
$R^P_{\psi }$ are symmetric operators. Actually it is easily seen that their 
closures are self--adjoint operators in $L^2(S^{d-1})$, with domain
$\{\sum Y_n \mid Y_n$ is a spherical harmonic of degree $n,\,\,\sum (1+
\lambda _n ^2) \|Y_n\|^2 < \infty \}$, and are given by $\sum Y_n \mapsto \sum 
\lambda _nY_n$, where the $\lambda _n$ are the corresponding eigenvalues.

\vskip2truecm

%%%%%%%%%%%%%%%PARAGRAPH 4%%%%%%%%%%%%%%%%%%%%

\heading 4.\ The null--spaces of the operators $R^{(m)}$ and $R^P$\endheading
\vskip0.4truecm

Now, for $d\ge 2$, we turn to the case of general $m\ge 1$, and investigate
the
null--space of $R^{(m)}\colon Lip^{m-1}(S^{d-1})\to L^\infty(S^{d-1})$.

For $d=2$ the complete answer is given by

\proclaim{Proposition 4.1} Let $d=2$ and $m\ge 1$. Then the null--space of the
operator
$R^{(m)}\colon $ 
\newline
$Lip^{m-1}
(S^{d-1}) \to L^\infty(S^{d-1})$ equals for $m$ odd
(even) $\{f\in Lip^{m-1}(S^{d-1})\,\,\vert \,\,f$ is even ($f$ is the sum
of an odd
function and a constant)$\}$.
\endproclaim

 
\demo{Proof}
By 3.3, we have that $f\in \Lip^{m-1}(S^{d-1})$ satisfies $R^{(m)}f=0$ a.e.\
if and only if
$
\left. \(
d/{d \psi }
\)^m 
C^0_n(\sin \psi )\right|_{\psi=0}\allowbreak\ne 0
$
implies $Y_n(f)=0$. However, by [M\"uller, 1966], p. 33,
we have for some $a_n\ne 0$ that 
$
 C^0_n(\sin \psi)=C^0_n[\cos(\pi /2-\psi)]=a_n\cos [n(\pi /2-
\psi)].
$
A straightforward calculation gives then that
$$
\left. \(
\frac {d} {d \psi }
\)^m 
\cos [n(\pi /2-\psi )] \right|_{\psi=0}
$$
equals $0$ if $n=0$ or if $n-m$ is odd, and equals
$
 (-1)^{\lfloor n/2 \rfloor}(-1)^{\lfloor m/2 \rfloor} n^m\ne 0$
if $n-m$ is even and $n\ge 1$.
Then 3.3 implies our statement.
\qued

\proclaim{Lemma 4.2} 
Let $d \ge 3$ and $ m\ge 1$. Then there exists a polynomial 
$\varGamma _{d,m,n}$ of $n$, of degree $m$, such that for 
$n-m$ even we have  
$$
\left. \(\frac {d} {d \psi }\)^m 
C^{(d-2)/2}_n(\sin \psi)\right|_{\psi=0}=0
$$
if and only if $\varGamma _{d,m,n}=0$. We have $\varGamma_{d,0,n}=1$,
$\varGamma_{d,1,n}=n+d-3$, $\varGamma_{d,2,n}=n(n+d-3)$ and 
$\varGamma_{d,3,n}=(n+d-3)(n^2 +(d-2)n-(d-2))$.
\endproclaim

\demo{Proof} By [Tricomi, 1955], p. 182, for $d\ge 3$ we have
$$
C^{(d-2)/2}_n(\sin \psi)=C^{(d-2)/2}_n[\cos (\pi /2-
\psi)]=$$
$$(-1)^n \sum_{i=0}^n
\binom{-d/2+1}{i}
\binom{-d/2+1}{n-i}
\cos [(n-2i)(\pi /2-\psi)]\,.
$$
Using 
$$
\(\frac {d} {d \psi }\)^m \cos \psi =
\cos (\psi+m\pi /2)\,,
$$
we have for $n-m$ even that
$$
\gamma_{d,m,n}:=
\left. 
\(\frac {d} {d \psi }\)^m 
C^{(d-2)/2}_n(\sin \psi)
\right|_{\psi=0}=
$$
$$
(-1)^n \sum_{i=0}^n
\binom{-d/2+1}{i}
\binom{-d/2+1}{n-i}[-(n-2i)]^m
\cos [(n-2i)\pi /2+m\pi /2)]=
$$
$$
(-1)^n(-1)^{(n+m)/2}
\sum_{i=0}^n (-1)^i
\binom{-d/2+1}{i} \binom{-d/2+1}{n-i}(2i-n)^m.
$$
Here $(2i-n)^m=\sum\limits_{p=0}^m(-1)^p\binom{m}{p}2^{m-p}n^pi^{m-p}$,
and 
$$
i^{m-p}=\sum_{q=0}^{m-p}c_{m-p, q}\binom{i}{q}q!\,,
$$
for some constants $c_{m-p, q}$, where $c_{m-p, m-p}=1$. Hence
$$
(2i-n)^m=\sum_{q=0}^m f_{m, q}(n)\binom{i}{q}q!\,,
$$
where 
$$
   f_{m,q}(n)=\sum_{p=0}^{m-q}(-1)^p\binom{m}{p} 2^{m-p} c_{m-p, q} n^p.
$$
Therefore 
$$
\gamma _{d,m,n}=(-1)^n (-1)^{(n+m)/2}\sum_{q=0}^m	f_{m,q}(n) \delta 
_{d,n,q}\,,
$$
where
$$
\delta _{d,n,q}=
\sum_{i=0}^n (-1)^i
\binom{-d/2+1}{i} \binom{-d/2+1}{n-i}
\binom{i}{q}q!\,.
$$

Here $\delta _{d,n,q}$ is the  coefficient of $x^n$ in the 
power series expansion of
$$
x^q\left[
\(\frac{
d} {dx
}\)^q (1-x)^{(2-d)/2}
\right] (1+x)^{(2-d)/2}=
$$
$$
(-1)^q \(-\frac{d}{2}+1\)\(-\frac{d}{2}\)\dots 
\(-\frac{d}{2}-q+2\)x^q
(1-x^2)^{-{d/2}-q+1}(1+x)^q =
$$
$$
(-1)^q \(-\frac{d}{2}+1\)\(-\frac{d}{2}\)\dots 
\(-\frac{d}{2}-q+2\)x^q
\sum_{k=0}^\infty (-1)^k \binom{-d/2-q+1}{k}
x^{2k}\sum_{l=0}^q \binom{q}{l}x^l,
$$
i.e., 
$$
\delta _{d,n,q}=
(-1)^q \(-\frac{d}{2}+1\)\(-\frac{d}{2}\)\dots 
\(-\frac{d}{2}-q+2\)
\times
$$
$$
\sum_{k\ge 0,\, 0\le l\le q,\, 2k+l=n-q} (-1)^k
\binom{-d/2-q+1}{k}\binom{q}{l}\,.
$$
Here in the last summation we have $2k=n-q-l\in [n-2q, n-q]$,
thus 
$\lceil n/2 \rceil-q\le k\le \lfloor (n-q)/2 \rfloor\le 
\lfloor n/2 \rfloor$.
Therefore we introduce the notation $k=\lfloor n/2 \rfloor-j$;
then $0\le \lfloor n/2 \rfloor-\lfloor (n-q)/2 \rfloor=
q/2+\delta \le j\le q-\varepsilon $. Here $\delta =0$ for $q$ 
even, $\delta =1/2$ or
$-1/2$ for $q$ odd and $n$ even or odd,
respectively, and
$\varepsilon =0$ for $n$ even, $\varepsilon =1$ for $n$ odd. Furthermore,
$l=n-2k-q=\varepsilon +2j-q$.

We have 
$$
\delta _{d,n,q}\binom{-d/2+1}{\lfloor n/2 \rfloor}^{-1}=
\sum_{q/2+\delta \le j\le q-\varepsilon ,\,\, j\le 
\lfloor n/2 \rfloor}
\delta _{d, n, q, j}\,,
$$
where 
$$
\delta _{d, n, q, j}=
(-1)^q 
\(-\frac{d}{2}+1\)\(-\frac{d}{2}\)\dots 
\(-\frac{d}{2}-q+2\)\times
$$
$$
(-1)^{\lfloor {n/2} \rfloor-j}
\binom{-d/2-q+1}{\lfloor n/2 \rfloor-j}
\binom{-d/2+1}{\lfloor n/2 \rfloor}^{-1}
\binom{q}{\varepsilon +2j-q}=
$$
$$
(-1)^{\lfloor {n/2} \rfloor}(-1)^{q-j}
\(-\frac{d}{2}+1\)\(-\frac{d}{2}\)\dots\(-\frac{d}{2}-q+2\)
\times
$$
$$
\(-\frac{d}{2}-q+1\)\dots
\(-\frac{d}{2}-q-\left\lfloor\frac{n}{2}\right\rfloor+j+2\)
\times
$$
$$
\(\left\lfloor\frac{n}{2}\right\rfloor-j\)!^{-1}
\(-\frac{d}{2}+1\)^{-1}\dots
\(-\frac{d}{2}-\left\lfloor\frac{n}{2}\right\rfloor+2\)^{-1}
\left\lfloor\frac{n}{2}\right\rfloor!
\binom{q}{\varepsilon +2j-q}=
$$
$$
(-1)^{\lfloor n/2 \rfloor}(-1)^{q-j}
\(-\frac{d}{2}-\left\lfloor\frac{n}{2}\right\rfloor+1\)\dots
\(-\frac{d}{2}-\left\lfloor\frac{n}{2}\right\rfloor-q+j+2\)
\times
$$
$$
\left\lfloor\frac{n}{2}\right\rfloor\dots
\(\left\lfloor\frac{n}{2}\right\rfloor-j+1\)
\binom{q}{\varepsilon +2j-q}=
(-1)^{\lfloor n/2 \rfloor} g_{d , q, j}(n)\,,
$$
where $g_{d , q, j}(n)$ equals, for $n$ having the parity of $m$,
a polynomial of $n$, of degree 
$(q-j)+j=q$, and of leading coefficient $2^{-q}\binom{q}{\varepsilon +2j-q}$.
However, for $0\le \lfloor n/2 \rfloor<j$ we have
$g_{d,q,j}(n)=0$, thus
$$
\delta _{d,n,q}\binom{-d/2+1}{\lfloor n/2 \rfloor}^{-1}=
(-1)^{\lfloor n/2 \rfloor}\sum_{j=q/2+\delta }^{q-\varepsilon }
g_{d, q, j}(n)\,.
$$
Collecting these results, we have 
$$
(-1)^n(-1)^{(n+m)/2}(-1)^{\lfloor n/2 \rfloor}
\gamma _{d, m, n}\binom{-d/2+1}{\lfloor n/2 \rfloor}^{-1}=$$
$$(-1)^{\lfloor n/2 \rfloor}
\sum_{q=0}^m f_{m, q}(n)\delta _{d, n, q}
\binom{-d/2+1}{\lfloor n/2 \rfloor}^{-1}=
$$
$$
\sum_{q=0}^m f_{m, q}(n)
\sum_{j=q/2+\delta }^{q-\varepsilon } g_{d, q, j}(n):=
\varGamma _{d, m, n}\,.
$$
Here each $f_{m, q}(n)$ or $g_{d, q, j}(n)$ equals, for
$n$ having the parity of $m$, a polynomial of $n$, of degree $m-q$
or $q$, respectively.
Moreover, $\delta $ and 
$\varepsilon $ in the bounds of the inner summation only
depend on the parities of $n$ and $q$.
Therefore $\varGamma _{d,m,n}$ equals, for $n$ having the parity of $m$,
a polynomial 
of $n$, of degree $\le m$.
Furthermore, we have $\gamma _{d,m,n}=0$ if and only if 
$\varGamma _{d, m, n}=0$.

Now we show that $\varGamma _{d, m, n}$ equals, for $n$ having the parity
of $m$, a 
polynomial of $n$, of degree exactly $m$. More exactly, we show
that its
leading coefficient is $1$.
We have 
$$
\varGamma _{d, m, n}=\sum_{q=0}^m f_{m, q}(n)\sum_{j=q/2+\delta }^
{q-\varepsilon }
g_{d, q, j}(n)\,.
$$
Here, by $c_{q, q}=1$, the leading term of $f_{m, q}(n)$ is
$(-1)^{m-q}\binom{m}{q}2^q n^{m-q}$.
Similarly, the leading term of $g_{d, q, j}(n)$ is
$2^{-q}\binom{q}{\varepsilon +2j-q}n^q$. Hence the coefficient of $n^m$
in $\varGamma _{d, m, n}$ is
$$
\sum_{q=0}^m(-1)^{m-q}\binom{m}{q}
\sum_{j=q/2+\delta }^{q-\varepsilon }\binom{q}{\varepsilon +2j-q}\,.
$$
A small discussion, taking into consideration the parities of $m$ and $q$,
shows that this equals
$$
\sum_{q=0}^m(-1)^{m-q}\binom{m}{q}
\sum_{0\le \varepsilon +2j-q\le q}\binom{q}{\varepsilon +2j-q}=
$$
$$
\sum_{q=1}^m(-1)^{m-q}\binom{m}{q}2^{q-1}+\frac{1}{2}+(-1)^m \frac{1}{2}=
$$
$$
(2-1)^m\frac{1}{2}- 
(-1)^m\frac{1}{2}+\frac{1}{2}+(-1)^m\frac{1}{2}=1.
$$
The above considerations show that, for any given $m$, 
we can evaluate $\gamma _{d,m, n}$ for all $n$. Performing the above
calculations for $m \le 3$, we obtain the formulas for 
$\varGamma _{d,m,n}$, given in the lemma. 
\qued
\enddemo

Using the above considerations we prove the following theorem, that is a 
generalization of the last statement of Theorem 3.8 of 
[Makai--Martini--\'Odor].

\proclaim{Theorem 4.3}
Let $d\ge 3$  be a fixed integer. Then, for any integer  $m\ge 1$,
there exists a set $A_m$ of non--negative integers of the same
parity as $m$, with cardinality $|A_m|\le m$, such that the following holds.
The null--space of the operator $R^{(m)} \colon Lip^{m-1}(S^{d-1})
\to L^\infty (S^{d-1})$ equals $\{ f \in Lip^{m-1}(S^{d-1})\,\,|
\,\,f$ is of the form $f=g+\sum\limits_{n\in A_m}Y_n$,
where $g$ is even (odd) for $m$ odd (even), and $Y_n$
is a spherical harmonic of degree $n\}$. In particular, for
$m=2$, $3$ we have $A_2=\{0\}$, $A_3=\emptyset$.
\endproclaim

\demo{Proof}
By 3.3, we have that $f \in Lip^{m-1} (S^{d-1})$ satisfies 
$R^{(m)}f=0$ a.e.\
if and only if, for the Fourier expansion   
$\sum_{n=0}^\infty Y_n(f) $ of $f$, we have that
$\gamma _{d, m, n}:=\left.\(
d/{d \psi }
\)^m C_n^{(d-2)/2}(\sin 
\psi )
\right|_{\psi =0}\ne 0$
implies $Y_n(f)=0$. Since $C_n$ is odd (even) for $n$ odd
(even), therefore
for $n-m$ odd we have
$\gamma _{d, m, n}=0$. So we only need to consider the case $n-m$ even.
Then by 4.2 we have $\gamma _{d, m, n}=0$ if and only if 
$\varGamma _{d, m, n}=0$,
and $\varGamma _{d, m, n}$ equals, for $n$ having the parity
of $m$, a polynomial of $n$, of degree $m$.

We let $A_m=\{n\,\,|\,\, \text{$n\ge 0$ is an integer, $n-m$ is even,
$\varGamma _{d,m,n}=0$}\}$. Then $\{ f \in Lip^{m-1} $
$(S^{d-1})\,\,|\,\,
R^{(m)}f=0$ a.e\.$\}=\bigl \{ f \in Lip^{m-1}(S^{d-1})\,\,|\,\,f= \sum
\{Y_n(f)\,\,|\,\, n \ge 0$ is an integer, and either $2 \nmid 
(n-m),$ or $(2 \mid (n-m)$ and $n \in A_m)\}\bigr \} $.

For the cases $m=2$, $3$, we consider the equation
$\varGamma _{d,m,n}=0$ from 4.2. For $m=2$ its only non--negative  even root
is $n=0$. Now let $m=3$ and $n\ge 0$ odd. Then we have $n+d-3\ge 1$. 
Furthermore, 
the discriminant of $n^2+(d-2)n-(d-2)$ is $d^2-4$, that is not
a perfect square for $d\ge 3$, thus the roots of this polynomial 
are irrational.
\qued
\enddemo

\proclaim{Theorem 4.4}
Let $d \ge 2$ be a fixed integer. Let $m \ge 1$ be an integer, $P$ a 
polynomial of degree $m$, and $\psi =0$. If 
$P$ is odd
(even) for $m$ odd (even), then there exists a set $A_P$ of
non--negative
integers of the same parity as $m$, with cardinality $\vert A_P\vert\le m$,
such that the following holds. The null--space of the operator 
$R^P=R_0^{P}$ equals $\{f\in
Lip^{m-1}(S^{d-1})\,\,\vert \,\,f$ is of the form $f=g+
\sum\limits_{n\in A_P}Y_n$, where $g$ is even (odd)
for $m$ odd (even), and $Y_n$ is a spherical harmonic of
degree $n\}$.
\endproclaim

\demo{Proof} Suppose that $P$ is such as given in the theorem. Then we
 have 
$
%\left .
\(P \(
d / {d \psi }
\)\) C_n^{(d-2)/2}$
\newline $\left . (\sin \psi )
\right|_{\psi =0}=0
$
for $n-m$ odd. 
Now let us suppose that $n-m$ is even.
First we suppose that $d\ge 3$. We have, like at 4.3 and 4.2,
$$
\left .
\(P \(\frac {d} {d \psi }\)\)
C_n^{(d-2)/2}(\sin \psi )
\right|_{\psi =0}=
\sum_{r=0}^m c_r \gamma _{d, r, n}=
$$
$$
(-1)^n (-1)^{\lfloor {n/2} \rfloor}
\binom{-d/2+1}{\lfloor n/2 \rfloor}
\sum_{r=0}^m c_r (-1)^{(n+r)/2} \varGamma _{d, r, n}
$$
(observe that $n \equiv r\,\, ({\text {mod} }\,\, 2) $).
This expression equals $0$ if and only if the last sum equals $0$.
Furthermore, this last sum equals an $m$--th degree  polynomial of $n$,
if $n-m$ is even.
Now let $A_{P}$ be the set of those non--negative integer
roots $n$
of this polynomial, for which $n-m$ is even.

Now let $d=2$. Then we have from the proof of 4.1, with the same $a_n\ne 0$,
$$
\left. \(P \(\frac {d} {d \psi }\)\)
C_n^0(\sin \psi )\right|_{\psi =0}=
a_n (-1)^{\lfloor n/2 \rfloor}
\sum_{r=0}^m c_r (-1)^{\lfloor r/2 \rfloor} n^r .
$$
Then we define $A_{P}$ using this last sum, like above.
\qued
\enddemo

\vskip1truecm

{\bf Acknowledgment.} The authors express their gratitude to R. Gardner 
for his several valuable remarks concerning this paper. 


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\enddocument 

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