Journal of Mathematical Sciences, Vol. 198, No. 4, April, 2014

On the kernel of a hemispherical Funk transformation andits local analogs

Vitaliy V. Volchkov and Irina M. Savostyanova

Presented by R. M. Trigub

Abstract. Functions on a sphere with zero weighted means over all geodesic balls of a fixed radius arestudied. We obtain a description of such functions in the form of series in special functions.

Keywords. Legendre functions, spherical harmonics, Pompeiu transformation, hemispherical transforma-tion.

Introduction

Let R3 be a three-dimensional real Euclidean space with Euclidean norm |·|, S

2 = {x ∈ R3 : |x| = 1}.

The hemispherical transformation H on S2 puts the function f ∈ L(S2) in correspondence with its

integrals over all possible hemispheres from S2. P. Funk established [5] that the kernel of H consists

of even functions on S2 possessing the zero integral over the whole sphere. In addition, he reduced the

question about the inversion of H on a class of odd radial (i.e., depening only on x3) functions to theAbel integral equation. Other formulas of inversion for a hemispherical transformation and the imagesof various functional spaces under the action of H were found by B. Rubin [10] and S. Campi [4].

Generalizing the Funk transformation, P. Ungar [12] replaced the hemispheres in the definition ofH by the spherical caps with any fixed radius r and solved the problem of description of the kernel ofthe corresponding transformation Br. It turned out that, for some set of radii that is defined by thezeros of Gegenbauer polynomials, the transformation Br is injective. For the rest values of r (their setis countable and everywhere dense on (0, π)), the functions of the kernel of Br can be described withthe help of the expansion in a Laplace series in spherical harmonics on S

2. These results were furtherdeveloped and refined in works by R. Schneider, C. A. Berenstein, L. Zalcman, and E. Badertscher,where the generalizations for compact two-point-homogeneous spaces [3,11] and for locally symmetricspaces [1] were obtained.

In two last decades, a special attention was paid to the local analogs of the mentioned problemfor various homogeneous spaces X (see [15–17] and references therein). In addition to the essentialdifficulties related to the noninvariance of relevant kernels relative to the group of motions of the spaceX, some new effects arose. For example, as distinct from the global case, the functions on S

2\{ξ} (ξis any point of the sphere) possessing zero integrals over all balls from S

2\{ξ} with any fixed radiusform an infinite-dimensional space.

In the present work, we give description of the functions on a ball B ⊂ S2 that have zero weighted

means over all balls with fixed radius from B. The case under consideration is of interest, sincethe presence of a weight in integrals does not allows one to write the indicated intergal condition inthe form of a convolution equation with radial distribution whose theory was well developed recently

Translated from Ukrains’kiı Matematychnyı Visnyk, Vol. 10, No. 4, pp. 575–594, October–November, 2013.Original article submitted March 5, 2013

1072 – 3374/14/1984–0469 c© 2014 Springer Science+Business Media New York 469

(see [15–17]). In particular, the general theory of transmutative operators, which is a powerful tool forthe study of the properties of solutions of the convolution equations on various homogeneous spaces(see [16, Part 2]), is inapplicable in this case.

1. Formulation of the main result

As usual, by symbols N, Z, and Z+, we denote, respectively, the sets of natural, integer, and integernonnegative numbers. We introduce the spherical coordinates ϕ, θ on S

2 in the following way:

ξ1 = sin θ sinϕ, ξ2 = sin θ cos ϕ, ξ3 = cos θ, ϕ ∈ (0, 2π), θ ∈ (0, π), (1.1)

where ξ1, ξ2, and ξ3 are the Cartesian coordinates of the point ξ ∈ S2. Let R ∈ (0, π], and let BR be

an open geodesic ball with radius R centered at the point o = (0, 0, 1) ∈ S2, i.e.,

BR = {ξ ∈ S2 : ξ3 > cos R}.

Any function f ∈ Lloc(BR) corresponds to a Fourier series

f(ξ) ∼∞∑

k=−∞fk(θ)eikϕ, θ ∈ (0, R), (1.2)

where

fk(θ) =12π

2π∫0

f◦(ϕ, θ)e−ikϕdϕ, (1.3)

f◦(ϕ, θ) = f(sin θ sinϕ, sin θ cos ϕ, cos θ).

If f ∈ C∞(BR), then the series in (1.2) converges to f in the standard topology of the space C∞(BR)(see [16, Chapt. 11, i. 11.1]).

By Pμν (μ, ν ∈ C), we denote Legendre functions of the first kind on (-1,1), i.e.,

Pμν (x) =

1Γ(1 − μ)

(1 + x

1 − x

)μ2

F

(−ν, ν + 1; 1 − μ;

1 − x

2

), μ �∈ N, (1.4)

Pμν (x) = (−1)μ(1 − x2)

μ2

(d

dx

)μ

Pν(x), μ ∈ N,

where F is the Gauss hypergeometric function, Γ is the gamma-function, and Pν = P 0ν (see [2]). We

setpν,k(θ) = P−k

ν (cos θ), ν ∈ C, k ∈ Z. (1.5)

Lemma 2.4 will show that, at fixed k ∈ Z+, θ ∈ (0; π), the function h(ν) = pν,k(θ) has the infinitenumber of zeros. In addition, all zeros of h are real and prime, are located symmetrically relative to−1

2 , and lie outside the interval [−k−1, k]. We denote the set of zeros of this function from the interval(k; +∞) by Nk(θ).

Let O(3) be the orthogonal group in R3. In what follows, we will consider that r is a fixed number

from (0; R), and Br is the closure of the ball Br. For M ∈ Z+, s ∈ Z+ ∪ {∞}, we set

Vr,M (BR) =

{f ∈ Lloc(BR) :

∫Br

f(τξ)(ξ1 + iξ2)Mdξ = 0 ∀ τ ∈ O(3) : τBr ⊂ BR

},

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V sr,M (BR) = Vr,M (BR) ∩ Cs(BR),

where dξ is an element of the area on S2. For M = 0, the class Vr,M (BR) coincides with the class

of functions f ∈ Lloc(BR) satisfying the convolution equation f ∗ χr = 0 in BR−r, where χr is theindicator of the ball Br. As was mentioned above, the equations of such type were studied by manyresearchers for various homogeneous spaces (see [15–17]). We note that the function χr arising in thedefinition of the class Vr,0(BR) depends only on θ. This corresponds to the zero coefficient in expansion(1.2). In this connection, the form of a weight in the definition of the class Vr,M (BR) is quite natural.

The main result of our work is given by the following proposition.

Theorem 1.1. Let f ∈ C∞(BR). Then, in order that f belong to the class Vr,M (BR), it is necessaryand sufficient that, for any k ∈ Z, the expansion

fk(θ) =∑

ν∈NM+1(r)

cνpν,|k|(θ) + (sin θ)|k|M−|k|−1∑

j=0

γj(cos θ)j (1.6)

be valid. Here, 0 ≤ θ < R, cν , γj ∈ C, cν = O(ν−a) as ν → +∞ for any a > 0, and the second sum in(1.6) equals zero for |k| ≥ M .

We note that an analogous result for a real Euclidean space was obtained in [14]. However, themethods in [14] used essentially a vector structure of R

n and require the significant development inthe spherical case.

2. Auxiliary assertions about the functions pν,k

By Dk, we denote a differential operator defined on the space C1(0, π) as follows:

(Dku)(θ) = (sin θ)k d

dθ

(u(θ)

(sin θ)k

), u ∈ C1(0, π).

Let also L be the Laplace operator on S2, i.e.,

L =∂2

∂θ2+ ctg θ

∂

∂θ+

1sin2 θ

∂2

∂ϕ2.

Lemma 2.1. The equalities

Dk pν,k = (k − ν)(k + ν + 1)pν,k+1, D−k pν,k = pν,k−1, (2.1)

(L + ν(ν + 1)Id)(pν,k(θ)eikϕ) = 0, (2.2)

where Id is the identity operator, hold.

Proof. Using the formula

(1 − x2)dPμ

ν (x)dx

= −νxPμν (x) + (ν + μ)Pμ

ν−1(x)

(see [2]), we obtain

p′ν,k(θ) = ν ctg θ pν,k(θ) +(k − ν)sin θ

pν−1,k(θ).

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From whence, we have

Dk pν,k(θ) =(k − ν)sin θ

(pν−1,k(θ) − cos θ pν,k(θ)), (2.3)

D−k pν,k(θ) =1

sin θ((ν + k) cos θ pν,k(θ) − (ν − k)pν−1,k(θ)). (2.4)

SincePμ

ν−1(x) − xPμν (x) = (ν − μ + 1)

√1 − x2Pμ−1

ν (x),

(ν − μ)xPμν (x) − (ν + μ)Pμ

ν−1(x) =√

1 − x2Pμ+1ν (x)

(see [2]), relations (2.3) and (2.4) yield (2.1).The operator L acts on the function u(ξ) = v(θ)eikϕ by the rule

(Lu)(ξ) = (�kv)(θ)eikϕ,

where

�k =d2

dθ2+ ctg θ

d

dθ− k2

sin2 θId.

The operator �k can be represented as

�k = D−k−1Dk − k(k + 1)Id = Dk−1D−k − k(k − 1)Id. (2.5)

Relation (2.2) follows now from (2.5) and (2.1).

Lemma 2.2. (i) Let ε, θ ∈ (0, π), k ∈ Z+. Then, as ν → ∞ so that | arg ν| < π− ε, the asymptoticequality

pν,k(θ) =

√2

π sin θ

cos((ν + 12)θ − π

4 (2k + 1))

(ν + 12)k+ 1

2

+ O

(eθ|Im ν|

|ν|k+ 32

)(2.6)

holds uniformly in θ on any interval [α, β] ⊂ (0, π).

(ii) If ν ∈ C, θ ∈ (0, π), k ∈ Z+, then

|pν,k(θ)| ≤ 1k!

(sin

θ

2

)k (cos

θ

2

)−k−1

eθ|Im ν|. (2.7)

(iii) Let 0 < a < π, s, k ∈ Z+. Then

maxθ∈[0,a]

∣∣∣∣dspν,k(θ)dθs

∣∣∣∣ = O(νs−k), ν → +∞. (2.8)

Proof. By the Mehler–Dirichlet formula,

pν,k(θ) =(sin θ)−k

√2πΓ

(k + 1

2

)θ∫

−θ

(cos t − cos θ)k− 12 ei(ν+ 1

2)tdt (2.9)

(see [2]). Relation (2.9) and the asymptotic expansion of Fourier integrals (see [9]) yield (2.6).To prove (2.7), we again use (2.9). Then

|pν,k(θ)| ≤ (sin θ)−k

√2πΓ

(k + 1

2

)θ∫

−θ

(cos t − cos θ)k− 12 dt eθ|Im ν|.

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Since

θ∫0

(cos t − cos θ)k− 12 dt =

1∫cos θ

(x − cos θ)k− 12

dx√1 − x2

≤ 1√1 + cos θ

1∫cos θ

(x − cos θ)k− 12 (1 − x)−

12 dx

=√

π2k− 12 Γ

(k + 1

2

)k!

(sin

θ

2

)2k (cos

θ

2

)−1

,

we obtain estimate (2.7).Finally, we will prove (2.8). For a < π/2, estimate (2.8) follows from the integral representation

pν,−k(θ)eikϕ = ikΓ(ν + k + 1)2πΓ(ν + 1)

π∫−π

(cos θ + i sin θ cos(ψ − ϕ))νeikψdψ, θ ∈ (0, π/2)

and the equality

pν,−k(θ) = (−1)k Γ(ν + k + 1)Γ(ν − k + 1)

pν,k(θ) (2.10)

(see [2]). On the other hand, the asymptotic expansion (2.6) and the second relation in (2.1) showthat

max0<α≤θ≤β<π

∣∣∣∣dspν,k(θ)dθs

∣∣∣∣ = O(νs−k−1/2), ν → +∞.

Combining these two cases, we obtain statement (iii).

Further, we require a generalization of the classical formula of multiplication for Legendre polyno-mials (see [13]). To formulate and to prove the appropriate assertion, we introduce some notation.

Let k, m ∈ Z, ν ∈ C, and let the real numbers θ1, θ2 satisfy the conditions: 1) sin2(θ1/2) �= 1,cos θ2 �= −1; 2) sin θ1 sin θ2 ≥ 0, cos(θ1 + θ2) > −1 or sin θ1 sin θ2 ≤ 0, cos(θ1 − θ2) > −1. We definethe integral Ik,m

ν (θ1, θ2) in the following way:

Ik,mν (θ1, θ2) =

12π

π∫−π

P−kν (cos θ1 cos θ2 − sin θ1 sin θ2 cos ψ)eimψ

× (cos θ2 sin θ1 + sin θ2 cos θ1 cos ψ − i sin θ2 sin ψ)k

(1 − (cos θ1 cos θ2 − sin θ1 sin θ2 cos ψ)2)k/2dψ.

For l ∈ Z+, l ≥ max{|k|, |m|}, we introduce the functions P lkm on (-1,1), following [13], with the help

of the equalities

P lkm(x) =

ik−m

2k(k − m)!

√(l − m)!(l + k)!(l − k)!(l + m)!

(1 + x)k+m

2 (1 − x)k−m

2

× F

(l + k + 1, k − l; k − m + 1;

1 − x

2

), (2.11)

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if k ≥ m, and

P lkm(x) =

im−k

2m(m − k)!

√(l + m)!(l − k)!(l + k)!(l − m)!

(1 + x)k+m

2 (1 − x)m−k

2

× F

(l + m + 1, m − l; m − k + 1;

1 − x

2

), (2.12)

if k < m. In addition, if α, β ∈ C, α �= −1,−2, . . ., we set

Φν,α,β(θ1) = F

(α + β + 1 + ν

2,α + β + 1 − ν

2; α + 1; sin2 θ1

2

). (2.13)

We note that functions (2.11)–(2.13) are closely related to the Jacobi classical polynomials (see [13]).

Lemma 2.3. Let the numbers θ1, θ2, and θ1 +θ2 belong to the interval [0, π), k, m ∈ Z, k ≥ m, ν ∈ C.Then

Ik,mν (θ1, θ2) = Ik,−m

ν (θ1 − π, θ2 − π)

=1

(k − m)!

(sin

θ1

2

)k−m (cos

θ1

2

)k+m

Φ2ν+1,k−m,k+m(θ1)pν,m(θ2), (2.14)

Ik,mν (−θ1, θ2) = Ik,−m

ν (π − θ1, θ2 − π)

=(−1)m+k

(k − m)!

(sin

θ1

2

)k−m (cos

θ1

2

)k+m

Φ2ν+1,k−m,k+m(θ1)pν,m(θ2).

Proof. By the formula of multiplication for the functions P lkm, we have (see [13])

P lkm(cos θ1)P l

m0(cos θ2) =12π

π∫−π

P lk0(cos θ1 cos θ2 − sin θ1 sin θ2 cos ψ)eimψ

× (cos θ2 sin θ1 + sin θ2 cos θ1 cos ψ − i sin θ2 sinψ)k

(1 − (cos θ1 cos θ2 − sin θ1 sin θ2 cos ψ)2)k/2dψ. (2.15)

According to [13],

P lm0(x) = im

√(l + m)!(l − m)!

P−ml (x). (2.16)

Therefore, relation (2.15) can be presented as

Ik,ml (θ1, θ2) = im−k

√(l + m)!(l − k)!(l + k)!(l − m)!

P lkm(cos θ1)pl,m(θ2). (2.17)

Since k ≥ m, relations (2.11), (2.13), and (2.17) yield

Ik,ml (θ1, θ2) =

1(k − m)!

(sin

θ1

2

)k−m (cos

θ1

2

)k+m

Φ2l+1,k−m,k+m(θ1)pl,m(θ2). (2.18)

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Consider now the entire function

f(z) = z(z − 1) · · · (z − a + 1)(Ik,mz (θ1, θ2) − 1

(k − m)!

(sin

θ1

2

)k−m (cos

θ1

2

)k+m

× Φ2z+1,k−m,k+m(θ1)pz,m(θ2)),

where a = max{|m|, |k|}. Then f(n) = 0 for n ∈ Z+ (see (2.18)). In addition, f satisfies the estimate

|f(z)| ≤ c(1 + |z|)3ae(θ1+θ1)|Im z|, z ∈ C,

where c is independent of z (see (2.7), (2.10), and [16, Chapt. 7, Corollary 7.2]). From whence, weconclude(see [16, Chapt. 7, Lemma 7.5]) that f ≡ 0. This gives the coincidence of the extreme partsin (2.14). The remaining equalities in the lemma can be deduced from the proved relation with thehelp of simple changes of the variable in the integral.

Concluding this section, we present some properties of zeros ν of function (1.5).

Lemma 2.4. Let r ∈ (0; π), k ∈ Z+. Then

(i) the function ν → pν,k(r) (ν ∈ C) has the infinite number of zeros;

(ii) all zeros of this function are real, prime, and are located symmetrically relative to the pointν = −1

2 ;

(iii) if ν ∈ [−k − 1, k], then pν,k(r) > 0;

(iv) for any ε > 0, ∑ν∈Nk(r)

1ν1+ε

< ∞, (2.19)

where Nk(r) = {ν > k : pν,k(r) = 0} as above.

Proof. Consider the even entire function

qr(λ) = pλ− 12,k(r), λ ∈ C. (2.20)

It follows from (2.9) that qr(λ) = qr(λ) and, in particular, qr is real-valued on R1. In addition, qr > 0

on the imaginary axis. Moreover, qr → 0 as λ → ∞ along the real axis and is a function of r of theexponential type (see (2.7)). From whence and by the Hadamard theorem of factorization [7, Chapt. 1,Sect. 3, i. 8] we make conclusion that qr has the infinite number of zeros. This yields statement (i).

To prove (ii), we use the relation

(μ − ν)(μ + ν + 1)

r∫0

pν,k(θ)pμ,k(θ) sin θ dθ = sin r(pμ,k(r)p′ν,k(r) − pν,k(r)p′μ,k(r)) (2.21)

(see [2]). Let qr(λ) = 0. Assume that λ �∈ R1. Then λ2 �= λ

2, since iλ �∈ R1. Setting μ = ν = λ − 1

2 in(2.21) and taking into account that qr(λ) = 0, we obtain

r∫0

|pλ− 12,k(θ)|2 sin θ dθ = 0, (2.22)

475

which is impossible. Assume now that qr(λ) = q′r(λ) = 0. Setting μ = λ− 12 in (2.21), we obtain again

(2.22) as ν → μ. Thus, all zeros of qr are real, prime, and symmetric relative to the point λ = 0. Thisand relation (2.20) yield statement (ii).

Statement (iii) follows from the equality

pν,k(r) =(sin r)k

2kk!F

(k − ν, ν + k + 1; k + 1; sin2 r

2

)(see [2]) and the definition of a hypergeometric function. Finally, using the connection between theorder of an entire function and the exponent of convergence of the sequence of its zeros (see [7, Chapt. 1,§2, i.2]), we obtain (iv).

Lemma 2.5. Let r ∈ (0; π), k ∈ Z+,

δ(μ, ν) =

r∫0

pν,k(θ)pμ,k(θ) sin θ dθ, μ, ν ∈ Nk(r).

Then δ(μ, ν) = 0 for μ �= ν, andδ(ν, ν) >

c

ν2k+2, (2.23)

where the constant c > 0 is independent of ν.

Proof. For μ �= ν, the statement follows from (2.21). It is sufficient to prove inequality (2.23) for allsufficiently large ν ∈ Nk(r). Let ν > π

4r − 12 . We set

gk(θ, t) = (cos t − cos θ)k− 12 , 0 ≤ t ≤ θ ≤ π. (2.24)

Then relation (2.9) yields

δ(ν, ν) =

r∫0

(pν,k(θ))2 sin θ dθ

=2

πΓ2(k + 1/2)

r∫0

(sin θ)1−2k

( θ∫0

gk(θ, t) cos(ν +

12

)t dt

)2

dθ

≥ 2πΓ2(k + 1/2)

π4(ν+1/2)∫

0

(sin θ)1−2k

( θ∫0

gk(θ, t) cos(ν +

12

)t dt

)2

dθ

≥ 1πΓ2(k + 1/2)

π4(ν+1/2)∫

0

(sin θ)1−2k

( θ∫θ2

gk(θ, t) dt

)2

dθ. (2.25)

We estimate the internal integral in (2.25) as follows:

θ∫θ2

gk(θ, t) dt =

cos θ2∫

cos θ

(x − cos θ)k− 12

dx√1 − x2

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≥ 1sin θ

cos θ2∫

cos θ

(x − cos θ)k− 12 dx =

22k + 1

(cos θ

2 − cos θ)k+ 1

2

sin θ. (2.26)

With regard for the relation

cos θ2 − cos θ

sin θ=

sin 3θ4

2 cos θ2 cos θ

4

≥ 12

sin3θ

4≥ 3θ

4π

for 0 < θ < π4(ν+1/2) , (2.25), and (2.26), we obtain

δ(ν, ν) ≥ 1πΓ2(k + 3/2)

π4(ν+1/2)∫

0

(3θ

4π

)2k+1

dθ,

which yields (2.23).

Lemma 2.6. Let k ∈ Z+, u ∈ L[0, r], and let

v(z) =

r∫0

u(θ)pz,k(θ) sin θ dθ. (2.27)

If v(z) = 0 for all z ∈ Nk(r), then u = 0.

Proof. Relations (2.27), (2.9), and (2.24) yield

v(z) =1√

2πΓ(12 + k)

r∫0

u(θ)K(z, θ)(sin θ)1−kdθ, (2.28)

where

K(z, θ) =

θ∫−θ

ei(z+ 12)tgk(θ, t)dt. (2.29)

Integrating (2.29) by parts, we obtain

K(z, θ) =( i

z + 12

)kθ∫

−θ

ei(z+ 12)t

( d

dt

)k(gk(θ, t)) dt. (2.30)

It is seen from (2.28), (2.30), and (2.7) that

|v(z)| ≤ c

(e

r2|Im z| +

er|Im z|

(1 + |z|)k

), z ∈ C, (2.31)

where c is independent of z. Consider the function w(z) = v(z)/pz,k(r). Since pz,k(θ) = p−z−1,k(θ)(see [2]), the condition, Lemma 2.4, and estimates (2.31) and (2.7) imply that w is an entire functionof at most the first order, and w(z) = w(−z − 1), z ∈ C. In addition, w(z) = O(|z|) as z → ∞ alongthe straight lines Im z = ±Re z (see (2.6)). From whence and from the Phragmen–Lindelof principle,

477

we conclude that w is a polynomial of at most the first power. By virtue of the evenness of w relativeto −1

2 , we have w(z) = w(0) for all z. This means (see (2.28), (2.29), and (2.9)) that

r∫−r

ei(z+ 12)t

r∫|t|

u(θ)(sin θ)1−kgk(θ, t) dθ dt = w(0)(sin r)−kK(z, r).

Thenr∫

t

u(θ)(sin θ)1−k gk(θ, t)gk(r, t)

dθ = w(0)(sin r)−k

for all t ∈ (0, r). Passing to the limit as t → r, we conclude that w(0) = 0. Thus, we obtain theAbel-type integral equation

r∫t

u(θ)(sin θ)1−kgk(θ, t) dθ = 0, t ∈ (0, r)

whose solution is the zero function (see [6, Chapt. 1, §2, the proof of Theorem 2.6]).

3. Properties of the classes V sr,M(BR)

Let 0 < R ≤ π, f ∈ Lloc(BR), fk(θ) be the Fourier coefficients defined by equality (1.3). Forbrevity, we set

fk(ξ) = fk(θ)eikϕ, k ∈ Z.

Further, all functions from the class C(BR \ {o}) that admit a continuous extension to the point {o}are considered additionally defined at this point by continuity.

Lemma 3.1. Let f ∈ V sr,M (BR). Then

(i) f ∈ V sr,M (BR);

(ii) fk ∈ V sr,M (BR) for any k ∈ Z;

(iii) if s ≥ 1 and f(ξ) = u(θ)eikϕ, then the functions (Dku)(θ)ei(k+1)ϕ and (D−ku)(θ)ei(k−1)ϕ belongto the class V s−1

r,M (BR);

(iv) if s ≥ 2 and f(ξ) = u(θ)eikϕ, then (�ku)(θ)eikϕ ∈ V s−2r,M (BR);

(v) if s ≥ 1 and f(ξ) = u(θ)ei(M+1)ϕ, then u(r) = 0.

Proof. In the course of the proof, we consider that τ is an element of the group O(3) such thatτBr ⊂ BR.

(i) Let κα be an orthogonal transformation in R3 that acts by the rule

καξ = (ξ1 cos α + ξ2 sinα, ξ1 sinα − ξ2 cos α, ξ3), α ∈ R. (3.1)

Then κα = κ−1α and τκαBr ⊂ BR. Therefore,

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∫Br

f(τξ)(ξ1 + iξ2)Mdξ =∫Br

f(τκακαξ)(ξ1 + iξ2)Mdξ

= eiαM

∫Br

f(τκαη)(η1 − iη2)Mdη = eiαM

∫Br

f(τκαη)(η1 + iη2)Mdη = 0,

which was required in the first assertion.(ii) By τα, we denote a rotation of R

3 in the plane (x1, x2) by an angle of α, i.e.,

ταξ = (ξ1 cos α − ξ2 sin α, ξ1 sin α + ξ2 cos α, ξ3). (3.2)

Relation (1.2) yields the formula

fk(ξ) =12π

2π∫0

f(ταξ)eikαdα. (3.3)

In particular, fk ∈ Cs(BR). Then, according to (3.3), we have

∫Br

fk(τξ)(ξ1 + iξ2)Mdξ =12π

2π∫0

∫Br

f(τατξ)(ξ1 + iξ2)Mdξeikαdα.

Since τατBr ⊂ BR for any α ∈ [0, 2π], we obtain statement (ii).(iii), (iv) Let at be a rotation of R

3 in the plane (x2, x3) by an angle of (−t), i.e.,

atξ = (ξ1, ξ2 cos t + ξ3 sin t,−ξ2 sin t + ξ3 cos t). (3.4)

For sufficiently small |t|, the condition yields∫τBr

F (atξ)PM (τ−1ξ) dξ = 0, (3.5)

where F (x) = f(x/|x|), PM (ξ) = (ξ1 + iξ2)M . Differentiating (3.5) with respect to t and setting t = 0,we obtain ∫

τBr

h(ξ)PM (τ−1ξ) dξ = 0, (3.6)

where h(ξ) = ξ3∂F∂x2

(ξ) − ξ2∂F∂x3

(ξ). The direct calculation indicates that

h◦(ϕ, θ) = cos ϕ∂f◦

∂θ− sinϕ ctg θ

∂f◦

∂ϕ=

12(Dku)(θ)ei(k+1)ϕ +

12(D−ku)(θ)ei(k−1)ϕ.

Statement (iii) follows now from (3.6) and assertion (ii), and (iv) follows from (iii) and (2.5).(v) Using statement (iii), we have

(D−M−1u)(θ)eiMϕ ∈ Vr,M (BR).

Integrating this function over the ball Br, we obtain the equality u(r) = 0.

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Lemma 3.2. (i) Let ν ∈ NM+1(r), k ∈ Z. Then the function sν,k(ξ) = pν,|k|(θ)eikϕ belongs to theclass V ∞

r,M (Bπ).

(ii) Let M ≥ 1, |k| ≤ M − 1. Then the function

f(ξ) = (sin θ)|k|M−|k|−1∑

j=0

γj(cos θ)jeikϕ, γj ∈ C

belongs to the class V ∞r,M (S2) for any r ∈ (0, π).

Proof. (i) The smoothness of the functions sν,k follows from the equalities

sν,k(ξ) =1k!

F

(−ν, ν + 1; 1 + k;

1 − ξ3

2

)(ξ2 + iξ1)k

(1 + ξ3)k, k ≥ 0,

sν,k(ξ) =1

(−k)!F

(−ν, ν + 1; 1 − k;

1 − ξ3

2

)(ξ2 − iξ1)−k

(1 + ξ3)−k, k < 0

(see (1.4) and (1.1)). We now prove that sν,k ∈ Vr,M (Bπ).Let k ≥ M . Using Lemma 2.3, relation (2.10), and the equality

r∫0

(sin θ)M+1P−Mν (cos θ) dθ = (sin r)M+1P−(M+1)

ν (cos r) (3.7)

(see [8, Section 1.12.1, formula (8)]), for |t| < π − r, α, β ∈ R, we have

∫Br

sν,k(τβatταξ)(ξ1 + iξ2)Mdξ =2π

(k − M)!iMe−iMαe−ikβ

× (sin r)M+1

(sin

t

2

)k−M (cos

t

2

)k+M

Φ2ν+1,k−M,k+M (t)pν,M+1(r), (3.8)

∫Br

sν,k(κβatταξ)(ξ1 + iξ2)Mdξ =2π

(k + M)!(−1)k+M iMe−iMαe−ikβ

× (sin r)M+1

(sin

t

2

)k+M (cos

t

2

)k−M Γ(ν + M + 1)Γ(ν − M + 1)

Φ2ν+1,k+M,k−M (t)pν,M+1(r), (3.9)

where τα, κβ , and at are in (3.1), (3.2), and (3.4). Since pν,M+1(r) = 0, relations (3.8) and (3.9) andthe Euler theorem of expansion of proper orthogonal matrices in a composition of rotations aroundthe coordinate axes yield the required statement for k ≥ M . The general case can be easily obtainedwith the help of Lemma 3.1, (i), (iii), and (2.1).

(ii) In view of Lemma 3.1, (i), we may consider that k ≥ 0. Then

f(ξ) = (ξ2 + iξ1)kM−k−1∑

j=0

γjξj3.

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To prove (ii), it is sufficient to prove the equality∫Br

Y (ξ)(ξ1 + iξ2)Mdξ = 0, (3.10)

where Y is any spherical harmonics on S2 of degree of at most M −1 (see [13, Chapt. 3, Sect. 6, i. 5]).

A spherical harmonic of order m on S2 takes the formm∑

n=−m

cnpm,n(θ)einϕ, cn ∈ C.

Therefore, relation (3.10) follows from the orthogonality of the trigonometric system {einϕ}∞n=−∞ inL2[0, 2π].

Lemma 3.3. Let k ∈ Z, f(ξ) = u(θ)eikϕ ∈ VM+|k|+4r,M (BR), and f = 0 in Br. Then f = 0 in BR.

Proof. Without any loss of generality, we may consider that k = 0, i.e., f(ξ) = u(θ) (see Lemma 3.1,(iii)). Let 0 < ε < R − r. Consider the function wε satisfying the following conditions: 1) wε ∈C∞[0, π]; 2) wε = 1 on [0, R − ε] and wε = 0 on [R − ε/2, π]. For θ ∈ [0, π], we set Φ(θ) = u(θ)wε(θ),where u = 0 on [R, π]. Then Φ ∈ CM+4[0, π], and

Φ(θ) =∞∑

j=0

αjPj(cos θ), (3.11)

where

αj =2j + 1

2

π∫0

Φ(θ)Pj(cos θ) sin θ dθ

(see [13, Chapt. 3, Sect. 6, i. 4, formulas (21), (22)]). In this case, αj = O(j−M−2) as j → +∞,and series (3.11) converges absolutely and uniformly on [0, π]. Then we use mapping (3.4), where0 < t < R − r − ε. The condition yields∫

Br

F (atξ)(ξ1 + iξ2)Mdξ = 0, (3.12)

where F (ξ) = Φ(arccos ξ3). Expansion (3.11) indicates that

F (atξ) =∞∑

j=0

αjPj(ξ3 cos t − ξ2 sin t).

Therefore, the transition to the spherical coordinates in (3.12) gives the relation

∞∑j=0

αj

r∫0

( 2π∫0

Pj(cos θ cos t − sin θ sin t cos ϕ)e−iMϕdϕ

)(sin θ)M+1dθ = 0. (3.13)

Let us set l = j, k = 0, m = −M , θ1 = r, and θ2 = t in (2.15). Then relations (2.16), (2.10), (3.13),and (3.7) yield

∞∑j=0

αjpj,M+1(r)pj,−M (t) = 0. (3.14)

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Let us apply the operator DM to (3.14). Then relation (2.1) yields

∞∑j=0

αjpj,M+1(r)pj,−M−1(t) = 0. (3.15)

The product of Legendre functions in (3.15) can be presented in the form

pj,M+1(r)pj,−M−1(t) =1π

t+r∫|t−r|

Pj(cos θ)TM+1

(cos r cos t − cos θ

sin r sin t

)

× sin θdθ√(cos θ − cos(t + r))(cos(t − r) − cos θ)

,

where TM+1 is a Chebyshev polynomial of the first kind (see [13, Chapt. 3, Sect. 4, i. 3, formula (7)]).In view of (3.11), we arrive at the equation

t+r∫|t−r|

Φ(θ)TM+1

(cos r cos t − cos θ

sin r sin t

)sin θdθ√

(cos θ − cos(t + r))(cos(t − r) − cos θ)= 0,

0 < t < R − r − ε.

From whence, we conclude (see [18, Lemma 4]) that Φ = 0 on [0, R− ε]. By virtue of the arbitrarinessof ε ∈ (0, R − r), f = 0 in BR.

4. Proof of Theorem 1.1

The sufficiency of the conditions in Theorem 1.1 follows easily from Lemma 3.2 (see (1.2)). Wenow prove the necessity. By Lemma 3.1, (ii), fk ∈ V ∞

r,M (BR) for any k ∈ Z. First, we consider the casewhere k = M + 1. We set

cν =1

δ(ν, ν)

r∫0

fM+1(θ)pν,M+1(θ) sin θ dθ, ν ∈ NM+1(r), (4.1)

where

δ(ν, ν) =

r∫0

(pν,M+1(θ))2 sin θ dθ.

Integrating (4.1) by parts with regard for (2.1), (2.5), and Lemma 3.1, (iv), (v), we obtain

cν =1

δ(ν, ν)(−1)m

(ν(ν + 1))m

r∫0

(�mM+1fM+1)(θ)pν,M+1(θ) sin θ dθ

for any m ∈ Z+. Therefore, relations (2.23) and (2.8) yield

cν = O

(1

ν2m−M−3

), ν → +∞. (4.2)

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We now define a smooth function F on [0, R) by the equality

F (θ) =∑

ν∈NM+1(r)

cνpν,M+1(θ) (4.3)

(see (4.2), (2.8), and (2.19)). Using Lemma 2.5, we obtain

cν =1

δ(ν, ν)

r∫0

F (θ)pν,M+1(θ) sin θ dθ, ν ∈ NM+1(r). (4.4)

Comparing (4.1) with (4.4) and taking Lemma 2.6 into account, we conclude that fM+1 = F on [0, r].Then the function (fM+1(θ) − F (θ))ei(M+1)ϕ is equal to zero in Br and belongs to V ∞

r,M (BR) (seeLemma 3.2, (i)). The required expansion (1.6) for k = M + 1 follows now from Lemma 3.3 and (4.3).The case where k ≥ 0 is reduced to the above-considered one by induction on k with the use of (2.1)and Lemma 3.1 (iii). In this case for k = M, we have to consider that the function g(ξ) = (sin θ)MeiMϕ

does not belong to the class Vr,M (BR), since∫Br

g(ξ)(ξ1 + iξ2)Mdξ �= 0.

From whence, we obtain representation (1.6) for k < 0 in view of Lemma 3.1, (i). Thus, Theorem 1.1is completely proved.

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Translated from Russian by V. V. Kukhtin

Vitaliy Vladimirovich Volchkov and Irina Mikhailovna SavostyanovaDonetsk National University,24, Universitetskaya Str., Donetsk 83001, UkraineE-Mail: v.volchkov@donnu.edu.ua,cavost@mail.ru

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