Ann. Univ. Ferrara - Sez. VII - Sc. Mat.Vol. XLVII, 169-175 (2001)
Branch Points, Fourier Integrals and Pompeiu Problem.
FAUSTO SEGALA (*) - MATTEO TRIOSSI (**)
SUNTO - Si prova che se p è un polinomio e la mappa h(w)4kp(w) è univalente sul discounitario D del piano complesso, allora V4h(D) ha la proprietà di Pompeiu.
ABSTRACT - Let h be the square root of a polynomial and assume that h is univalent onthe unitary disk of the complex plane. Then the set V4h(D) has the Pompeiuproperty.
1. – Introduction.
In 1929 D. Pompeiu posed the following very hard problem. Let V be a sub-set of R 2 and fC 0 (R 2 ).
If
s(V)
f dx40 ( rigid motion s(1.1)
it is true that ff0?When the answer to this question is positive, V is called a Pompeiu
set (or V has the Pompeiu property). Many contributions towards thesolution of the problem were given since 1972-73 following the fundamentalsideas of the papers of Zalcman [3] and Brown-Schreiber-Taylor [1]. Ina recent paper of Garofalo-Segala [2], the Pompeiu property of a bounded
(*) Indirizzo dell’autore: Università di Ferrara, Italy.(**) Indirizzo dell’autore: Universidad Carlos III, Getafe, Spain.
FAUSTO SEGALA - MATTEO TRIOSSI170
simply connected open set V4h(D) is studied by analyzing the Riemannmap h (D is the unitary disk).
In this work we give a further contribution in that direction. Precisely weprove the following theorem.
THEOREM. Let p(w) be a polynomial and assume that h(w)4kp(w) isunivalent on D . Then V4h(D) is a Pompeiu set.
2. – Preliminaires.
From now on V will be a bounded simply connected open subset of theplane. By the celebrated theorem of Brown-Schreiber-Taylor [1], V is a Pom-peiu set iff xׯV (the Fourier-Laplace transform of the characteristic function of¯V) is not identically zero on every «circle» z 1
21z 224aD0 in C 2 . In [2] it is
shown that xׯV is identically zero on a «circle» of radius ka iff the function
rO ¯V
e rz2a
rS(z) dz(2.1)
is identically zero for every rD0, where S(z) is the Schwarz function of V .In this work we will prove that the Fourier integral (2.1) is not identically
zero, where V is under the hypotheses of Theorem, by analyzing its asymptot-ic expansion for rKQ .
We recall that the Schwarz function S(z) is given by
S(z)4 h u 1
h 21 (z)v .
If we denote by T the set of the singularities of h (obviously outside D), the set
of the singularities of S(z) is given by S4h g 1
Th (see picture below).
T1T S∂Ω
Figure 1
BRANCH POINTS, FOURIER INTEGRALS ETC. 171
3. – Polynomials of degree one.
We begin by proving theorem when h(w)4kw2a , with NaND1. By theinvariance of the Pompeiu property under rigid motions and dilations, we canassume
h(w)4o11w
221 .
We choose the branch kz4kre iw/2 , 2pEwGp and then we cut C on
11 w
242t , tF0, that is on T4 [2Q , 22]. It follows 1
T4 k2 1
2, 0l ,
S4 k k3
221, 0l . In this simple case we can give the explicit construction of
S(z). In fact we have
S(z)4o111
4z 218z21(3.1)
and we make the cut on 11 1
4z 218z42t , that is on k k3
221, 0l.
We must analyze the Fourier integral
J(r)4 ¯V
e rz2 (a/r) S(z) dz
with S(z) given by (3.1).We change the integration along ¯V with the integration along the path of
the picture below:
Gδ
Figure 2
FAUSTO SEGALA - MATTEO TRIOSSI172
Let d be the radius of the circle C . Denote by s(z) the jump of S and take intoaccount that
s (z)Ai
k2NzN, zK0 .(3.2)
If we require rkdKQ , for rKQ , we have a
rS(z)K0 on [2A , 2d] and
then e 2(a/r)S(z)A12 a
rS(z). Therefore we can write
J(r)AC
e rz2 (a/r) S(z) dz1a
r
2A
2d
e rz s(z) dz .
Now we make some calculations about the second Fourier integral. From (3.2)it follows
a
r
2A
2d
e rz s (z) dzAia
k2r
2A
2d
e rz dz
kNzN4
4ia
k2r
d
A
e 2rz dz
kz4
ia
k2r 3/2
dr
Ar
e 2u du
ku.
At this point we choose dr41. Then
a
r
2A
2d
e rz s(z) dzAiak2
r 3/2
1
Q
e 2u 2du .
Now we analyze the integral over C .
C
e rz2 (a/r) S(z) dzAi
r
2p
p
exp ye iw2a
2 k2kre 2iw/2z e iw dw4
4ia
2 k2r 3/2
2p
p
exp (e iw ) e iw/2 dw4ia
k2r 3/2
0
Q
exp (2e 2y ) e 2y/2 dy4
4ia
k2r 3/2
0
1e 2u
kudu4
iak2
r 3/2
0
1
e 2u 2du .
BRANCH POINTS, FOURIER INTEGRALS ETC. 173
Summing up, we conclude that
J(r)Aiakp
k2r 23/2 , rK1Q .
4. – Polynomials of general degree.
Let p(w) be a polynomial with even degree N42p12 with N distinctroots. Then the Riemann surface of kp(w) is a sphere with p handles, i.e. asurface of genus p (picture 3), since the roots x1 , R , xN are ramification pointsand we make p cuts along p lines joining p couples of these roots (see figure 3).
Figure 3
When N42p11, the Riemann surface is of the same type, if Q is regard-ed as a ramification point. Therefore in the case N even, we may choose a
branch of kp(w) which is holomorphic outside N
2cuts and which has a pole of
order N
2at Q . In the case N odd we may choose a branch of kp(w) holomor-
phic outside N11
2cuts, with a cut joining a finite root with Q . In the picture 4
we illustrate the set T of the singularities of h(w).We give also a picture which illustrates the corresponding set S of the sin-
gularities of the Schwarz function.A cut T relative to h(w) on the plane w induces a cut S relative to S(z) on theplane z . In general it is extremely complicated to find the principal contribu-tion to
g
e rz2 (a/r) S(z) dz where g is a contour around S .
To solve the problem we explain our idea.
Let A , B the extreme points of T and h g 1
Ah , h g 1
Bh the corresponding
FAUSTO SEGALA - MATTEO TRIOSSI174
D D
∞
N oddN even
Figure 4
points of S . We denote by L the segment line joining h g 1
Ah and h g 1
Bh and
choose the cut T in the following way:
T41
h 21 (L).
Then the corresponding set S is given by S4L! So we can assume that thecuts relative to S(z) are segments line. In the case N42p even, we have toconsider two types of Fourier integrals:
H(r)4g
e rz2 (a/r) S(z) dz(4.1)
N oddN even
Figure 5
BRANCH POINTS, FOURIER INTEGRALS ETC. 175
where g is a contour around the cut L ,
K(r)4C
e rz2 (a/r) S(z) dz(4.2)
where C is a circle around zero and S(z) has a pole of order p in 0 .The integral (4.2) is studied in Garofalo-Segala [2] by using the method of
the steepest descent and its principal contribution for rKQ is given by
const . r2
2
p11 exp gCrp21
p11 hfor some constant C .
In order to find the principal contribution to H(r), we observe that L is farfrom zero and therefore
H(r)Aa
r
L
e rz s(z) dz
where s is the jump of S(z) over L .We write L4]A1 tB , t [0 , 1 ]( with Re BD0 and make some calcula-
tions:
H(r)Aa
re rA
1/2
1
e rtBk12t dt4a
re r(A1B)
0
1/2 e 2rBu kuduAakp
2B 3/2
1
r 3/2e r(A1B) .
In the case N odd, we have to integrate along a segment line whose zero is anextreme point. By using the invariance of the Pompeiu property under rota-tions, we may assume L4 [2z 0 , 0 ] with z 0D0. Then we integrate along thepath of the picture 2 and the radius d of the circle is forced by the conditionrd N/2KQ .
Then the calculations are of the same type of those developed in section 3.The proof of theorem is complete.
R E F E R E N C E S
[1] L. BROWN - B. M. SCHREIBER - B. A. TAYLOR; Spectral synthesis and the Pompeiuproblem, Ann. Inst. Fourier, 23 (1973), pp. 125-154.
[2] N. GAROFALO - F. SEGALA, Univalent functions and Pompeiu problem, Trans.Amer. Math. Soc., 346 (1) (1994), pp. 137-146.
[3] L. ZALCMAN, Analyticity and the Pompeiu problem, Arch. Rat. Anal. Mech., 47(1972), pp. 237-254.
Pervenuto in Redazione il 27 marzo 2001.
Top Related