coxeter system of lines and planes are sets of injectivity for the twisted spherical means

JID:YJFAN AID:6940 /FLA [m1L; v 1.133; Prn:13/05/2014; 10:54] P.1 (1-32)Journal of Functional Analysis ••• (••••) •••–•••

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Journal of Functional Analysis

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Coxeter system of lines and planes are sets ofinjectivity for the twisted spherical means

Rajesh K. SrivastavaSchool of Mathematics, Harish-Chandra Research Institute, Allahabad, 211019,India

a r t i c l e i n f o a b s t r a c t

Article history:Received 9 April 2012Accepted 7 March 2014Available online xxxxCommunicated by Alain Connes

Dedicated to Prof. E.M. Stein on theoccasion of his 80th birthday

Keywords:Coxeter groupHecke–Bochner identityHeisenberg groupLaguerre polynomialsSpherical harmonicsTwisted convolution

It is well known that a line in R2 is not a set of injectivityfor the spherical means for odd functions about that line.We prove that any line passing through the origin is a set ofinjectivity for the twisted spherical means (TSM) for functionsf ∈ L2(C), whose each of spectral projection e

14 |z|

2f × ϕk is

a polynomial. Then, we prove that any Coxeter system of evennumber of lines is a set of injectivity for the TSM for Lp(C),1 � p � 2. Further, we deduce that certain Coxeter systemof even number of planes is a set of injectivity for the TSMfor Lp(Cn), 1 � p � 2. We observe that a set S2n−1

R × C isa set of injectivity for the TSM for a certain class of functionson Cn+1.

© 2014 Elsevier Inc. All rights reserved.

1. Introduction

As an interesting result, Courant and Hilbert [14, p. 699] had proved that if the circularaverages of a function f , which is even with respect to a line L, vanish over all circlescentered at points of L, then f ≡ 0. As a consequence of this result, the circular averagesof a function f vanish over all circles centered at points of L if and only if f is odd with

E-mail address: [email protected].

http://dx.doi.org/10.1016/j.jfa.2014.03.0090022-1236/© 2014 Elsevier Inc. All rights reserved.

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respect to L (see [5, Lemma 6.3]). Hence, any line L in R2 is not a set of injectivity forthe spherical means for the odd functions about L.

However, this result does not continue to hold for injectivity of the twisted sphericalmeans on complex plane C, because of non-commutative nature of underlying geometryof the Heisenberg group (see [11–13,17]). The question, in general that any real analyticcurve can be a set of injectivity for the twisted spherical mean for L2(C), is still an openproblem. However, we are able to prove the following partial results for the TSM.

Let f ∈ L2(C) be such that for each k ∈ Z+ (set of non-negative integers), theprojection e

14 |z|

2f × ϕk is a polynomial. This space which consists of these functions f

is much larger than U(1)-finite functions in L2(C). Suppose the twisted spherical meanf × μr(x) of the function f vanishes ∀r > 0 and ∀x ∈ R. Then f = 0 a.e. That is, theX-axis is a set of injectivity for the TSM on C.

By rotation, it follows that any line passing through the origin is a set of injectivityfor the TSM. Since f × ϕk is a real analytic function, in the above case, we only needthe centers to be a sequence in R having a limit point.

With the same exponential condition, we observe that any curve γ(t) = (γ1(t), γ2(t)),which passes through the origin, where γj , j = 1, 2 are polynomials is also a set ofinjectivity for the TSM. It is an interesting question, whether a real analytic curve is aset of injectivity for the TSM for a smooth class of functions.

Further, to complete the arguments of our idea, we prove that any two perpendicularlines are a set of injectivity for the TSM on Lq(C). Moreover, this result implies thatany Coxeter system of even number of lines is a set of injectivity for the TSM on Lq(C).These results for the twisted spherical means are in sharp contrast to the well knownresult for injectivity of the Euclidean spherical means on R2, due to Agranovsky andQuinto [5].

On the basis of these striking results, it is therefore natural to ask, whether anyCoxeter system of odd number of lines can be a set of injectivity for the TSM. Webelieve, our techniques with slight modifications would continue to work for any Coxetersystem of lines to be a set of injectivity for the TSM, which we leave open for the timebeing.

Using Theorem 3.6, we prove that any Coxeter system of hyperplanes intersectingalong a line in Cn (n � 2) is a set of injectivity for the TSM for function in Lp(Cn),1 � p � 2. In order to know more about higher dimensional result, let us consider thefollowing Euclidean situation.

Let μr be the normalized surface measure on sphere Sr(x). Let F ⊆ L1loc(Rn). We

say that S ⊆ Rn is a set of injectivity for the spherical means for F if for f ∈ F withf ∗ μr(x) = 0, ∀r > 0 and ∀x ∈ S, implies f = 0.

In the work by Agranovsky and Quinto [5], it has been shown that sets of non-injectivity for the spherical means for Cc(Rn) (n � 2) are contained in the zero setof a certain harmonic polynomial. For non zero function f ∈ Cc(Rn), write S(f) =

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{x ∈ Rn: f ∗ μr(x) = 0, ∀r > 0}. Then they have proved that S(f) =⋂∞

k=0 Q−1k (0),

where

Qk(x) =∫Rn

f(y)|x− y|2k dy.

Since all of Qk cannot be identically zero, it follows that there exists the least positiveinteger ko such that Qko

�≡ 0. Hence ΔQko= 2ko(2ko + n − 1)Qko−1 = 0. That is,

S(f) ⊆ Q−1ko

(0). Since Qkois harmonic and a harmonic polynomial can vanish to a

Coxeter system of hyperplane intersecting along a line, it follows that for n > 2, anyCoxeter system of hyperplanes intersecting along a line may fail to be a set of injectivityfor the spherical means on Rn.

However, the above observation does not continue to hold for injectivity of the twistedspherical means on Cn (n � 2) (see [11–13,17]). The question, of any odd Coxeter systemof hyperplanes can be a set of injectivity for the twisted spherical mean for Lp(Cn), isstill unanswered.

In general, any real cone K ⊂ Rn (n > 2) is a set of injectivity for the spherical meansfor C(Rn) if and only if K is not contained in the zero set of any homogeneous harmonicpolynomial (see [9]). Hence it follows that any Coxeter system of hyperplanes intersectingalong a line which is contained in the zero set of some homogeneous harmonic must failto be a set of injectivity for the spherical means for the class of all continuous functionson Rn.

In a recent article, the author has proved an analogous result for the twisted sphericalmeans (TSM). A complex cone is a set of injectivity for the TSM for the class of allcontinuous functions on Cn as long as it does not completely lay on the level surface ofany bi-graded homogeneous harmonic polynomial on Cn. Further, he produces examplesof such level surfaces. That is, zero set of polynomial P (z) = az1z2 + |z|2, a is anynon-zero complex number and z ∈ Cn (see [33]).

Using Theorem 3.6 we prove that any pair of perpendicular planes in Cn (n � 2) is aset of injectivity for the TSM for function in Lp(Cn), 1 � p � 2. Suppose f × μr(z) = 0,∀r > 0 and ∀z ∈ Cn−1 × R ∪ Cn−1 × iR. Then f = 0. Using this, we conclude that anyCoxeter system of even number of planes intersecting along a line is a set of injectivityfor the TSM for Lq(Cn).

In a result of Narayanan and Thangavelu [22], it has been proved that the spherescentered at the origin are set of injectivity for the TSM on Cn. The author has generalizedtheir result for certain weighted twisted spherical mean (see [31]). In general, the questionof set of injectivity for the twisted spherical means (TSM) with real analytic weight isstill open. The author has proved that the spheres centered at the origin are sets ofinjectivity for the twisted spherical means with real analytic weight, for certain radialfunctions on Cn (see [32]).

Let z = (z′, zn+1) ∈ Cn+1 and S2n−1R = {z ∈ Cn: |z| = R}. In view of result

in [22] that S2n−1R is a set of injectivity for the TSM on Cn, we prove that the set


S2n−1R ×C is a set of injectivity for the twisted spherical means for the functions satisfying

e14 |z

′|2f(z) ∈ Lp(Cn+1) with 1 � p � ∞. However, this result can be generalized to setof the form S2n−1

R × Cm.On account of these results, we observe an “embedding property” of the sets of in-

jectivity in higher dimensions. In the Euclidean setup, the sets of injectivity for thespherical means on the unit sphere Sn−1 can be embedded into the sets of injectivityfor the spherical means on Rn (see [9]). It is feasible ask the following question aboutthe set of injectivity of embedded real cone. Let K be a real cone in Rn. Then, can theset K × Rm be a set of injectivity for the spherical means for the class of continuousfunctions on Rn+m is an interesting question.

Since Laguerre function ϕn−1k is an eigenfunction of the special Hermite operator

A = −Δz+ 14 |z|2, with eigenvalue 2k+n, the projection f×ϕn−1

k is also an eigenfunctionof A with eigenvalue 2k + n. As A is an elliptic operator and eigenfunction of an ellipticoperator is real analytic [19], the projection f × ϕn−1

k must be a real analytic functionon Cn. Therefore, any determining set for the real analytic functions is a set of injectivityfor the TSM on Lq(Cn) with 1 � q � ∞. By polar decomposition, the condition f ×μr(z) = 0, ∀r > 0 is equivalent to f×ϕn−1

k (z) = 0, ∀k ∈ Z+, set of non-negative integers.Therefore, any determining set for the real analytic functions is a set of injectivity forthe TSM on Lp(Cn) with 1 � p � ∞. For example, let γ(t) = r(t)eit, where r(t) bea non-periodic real analytic function on [0,∞) with limt→∞ r(t) = 0. Then Cn−1 ×{γ(t): t ∈ [0,∞)} is a set of injectivity for the TSM for Lp(Cn) with 1 � p � ∞. Fordetails on determining sets for real analytic functions, see [24,26].

In 1996, Agranovsky and Quinto have proved a major breakthrough result in theintegral geometry, which completely characterizes the sets of injectivity for the sphericalmeans on the space of compactly supported continuous functions on R2. Their resultsays that the exceptional set for the sets of injectivity is a very thin set which consistsof a Coxeter system of lines union finitely many points.

Let μr be the normalized surface measure on sphere Sr(x). Let F ⊆ L1loc(Rn). We

say that S ⊆ Rn is a set of injectivity for the spherical means for F if for f ∈ F withf ∗ μr(x) = 0, ∀r > 0 and ∀x ∈ S, implies f = 0 a.e.

Theorem 1.1. (See [5].) A set S ⊂ R2 is a set of injectivity for the spherical meansfor Cc(R2) if and only if S � ω(ΣN ) ∪ F , where ω is a rigid motion of R2, ΣN =⋃N−1

l=0 {te iπlN : t ∈ R} is a Coxeter system of N lines and F is a finite set in R2.

In particular, any closed curve is a set of injectivity for Cc(R2). In fact, Agranovskyet al. [8] further prove that the boundary of any bounded domain in Rn (n � 2) is setof injectivity for the spherical means on Lp(Rn), with 1 � p � 2n

n−1 . For p > 2nn−1 , unit

sphere Sn−1 is an example of non-injectivity set in Rn. This result has been generalizedfor certain weighted spherical means, see [23]. In general, the question of set of injectivityfor the spherical means with real analytic weight is still open. In [23], it has been shown


that Sn−1 is a set of injectivity for the spherical means with real analytic weights forthe class of radial functions.

An analogue of Theorem 1.1 in the higher dimensions is still open and appeared asa conjecture in their work [5]. It says that the sets of non-injectivity for the Euclideanspherical means are contained in a certain algebraic variety. The following is their con-jecture.

Conjecture. (See [5].) A set S ⊂ Rn is a set of injectivity for the spherical meansfor Cc(R2) if and only if S � ω(Σ) ∪ F , where ω is a rigid motion of Rn, Σ is thezero set of a homogeneous harmonic polynomial and F is an algebraic variety in Rn ofco-dimension at most 2.

This conjecture remains unsolved, however a partial result related to this conjecturehas been proved by Kuchment et al. [10]. They also present a brief survey on the recentdevelopment towards the above conjecture. However, in this article, we observe that thisconjecture does not continue to hold for the spherical means on the Heisenberg groupHn = Cn × R. In fact result on H1 is an adverse to the Euclidean result, Theorem 1.1on R2.

In more general, let f ∈ L1loc(Cn) and write S(f) = {z ∈ Cn: f ×μr(z) = 0, ∀r > 0}.

Our main problem is to describe completely the geometrical structure of S(f) that wouldcharacterize which “sets” are set of injectivity for the TSM. For example, let f be a nonzero type function f(z) = a(|z|)P (z) ∈ L2(Cn) ∩ C(Cn), where P ∈ Hp,q. Here Hp,q isthe space of homogeneous harmonic polynomials on Cn of type

P (z) =∑

|α|=p, |β|=q

Cαβzαzβ .

Then S(f) = P−1(0)∪ F , where F is the union of finitely many spheres centered at theorigin. This means a set S ⊂ Cn is set of injectivity for twisted spherical means for typefunctions if and only if S � P−1(0). Since P is harmonic, by maximal principle P−1(0)cannot contain the boundary of any bounded domain in Cn. Hence the boundary of anybounded domain would be a possible candidate for set of injectivity for the TSM. Thequestion that the boundary of the bounded domain is a set of injectivity for the TSM hasbeen taken up by many authors (see [7,22,29]). Moreover, a substantial amount of workhas been done on the question of sets of injectivity for the spherical means in varioussetups for different class functions (see, [1–4,20,21,25,30,36–40]).

2. Notation and preliminaries

We define the twisted spherical means which arise in the study of spherical means onHeisenberg group. The group Hn, as a manifold, is Cn × R with the group law

(z, t)(w, s) =(z + w, t + s + 1 Im(z.w)

), z, w ∈ Cn and t, s ∈ R.
2


Let μs be the normalized surface measure on the sphere {(z, 0): |z| = s} ⊂ Hn. Thespherical means of a function f in L1(Hn) are defined by

f ∗ μs(z, t) =∫

|w|=s

f((z, t)(−w, 0)

)dμs(w). (2.1)

Thus the spherical means can be thought of as convolution operators. An importanttechnique in many problems on Hn is to take partial Fourier transform in the t-variableto reduce matters to Cn. Let

fλ(z) =∫R

f(z, t)eiλt dt

be the inverse Fourier transform of f in the t-variable. Then a simple calculation showsthat

(f ∗ μs)λ(z) =∞∫

−∞

f ∗ μs(z, t)eiλt dt

=∫

|w|=s

fλ(z − w)e iλ2 Im(z.w) dμs(w)

= fλ ×λ μs(z),

where μs is now being thought of as normalized surface measure on the sphere Ss(o) ={z ∈ Cn: |z| = s} in Cn. Thus the spherical mean f ∗μs on the Heisenberg group can bestudied using the λ-twisted spherical mean fλ ×λ μs on Cn. For λ �= 0, a further scalingargument shows that it is enough to study these means for the case of λ = 1.

Let F ⊆ L1loc(Cn). We say S ⊆ Cn is a set of injectivity for twisted spherical means

for F if for f ∈ F with f ×μr(z) = 0, ∀r > 0 and ∀z ∈ S, implies f = 0 a.e. The resultson set of injectivity differ in the choice of sets and the class of functions considered. Wewould like to refer to [7,22,31], for some results on the sets of injectivity for the TSM.

We need the following basic facts from the theory of bigraded spherical harmonics(see [35, p. 62] for details). We shall use the notation K = U(n) and M = U(n − 1).Then, S2n−1 ∼= K/M under the map kM → k.en, k ∈ U(n) and en = (0, . . . , 1) ∈ Cn.Let KM denote the set of all equivalence classes of irreducible unitary representationsof K which have a nonzero M -fixed vector. It is known that each representation in KM

has a unique nonzero M -fixed vector, up to a scalar multiple.For a δ ∈ KM , which is realized on Vδ, let {e1, . . . , ed(δ)} be an orthonormal basis

of Vδ with e1 as the M -fixed vector. Let tδij(k) = 〈ei, δ(k)ej〉, k ∈ K and 〈 , 〉 stand for theinner-product on Vδ. By Peter–Weyl theorem, it follows that {

√d(δ)tδj1: 1 � j � d(δ),

δ ∈ KM} is an orthonormal basis of L2(K/M) (see [35, p. 14] for details). Define Y δj (ω) =


√d(δ)tδj1(k), where ω = k.en ∈ S2n−1, k ∈ K. It then follows that {Y δ

j : 1 � j � d(δ),δ ∈ KM} forms an orthonormal basis for L2(S2n−1).

For our purpose, we need a concrete realization of the representations in KM , whichcan be done in the following way. See [27, p. 253], for details. For p, q ∈ Z+, let Pp,q

denote the space of all polynomials P in z and z of the form

P (z) =∑|α|=p

∑|β|=q

cαβzαzβ .

Let Hp,q = {P ∈ Pp,q: ΔP = 0}. The elements of Hp,q are called the bigraded solidharmonics on Cn. The group K acts on Hp,q in a natural way. It is easy to see thatthe space Hp,q is K-invariant. Let πp,q denote the corresponding representation of K

on Hp,q. Then representations in KM can be identified, up to unitary equivalence, withthe collection {πp,q: p, q ∈ Z+}.

Define the bigraded spherical harmonic by Y p,qj (ω) =

√d(p, q)tp,qj1 (k). Then {Y p,q

j :1 � j � d(p, q), p, q ∈ Z+} forms an orthonormal basis for L2(S2n−1). Therefore, fora continuous function f on Cn, writing z = ρω, where ρ > 0 and ω ∈ S2n−1, we canexpand the function f in terms of spherical harmonics as

f(ρω) =∑p,q�0

d(p,q)∑j=1

ap,qj (ρ)Y p,qj (ω), (2.2)

where the series on the right-hand side converges uniformly on every compact setK ⊆ Cn. The functions ap,qj are called the spherical harmonic coefficients of f and func-tion ap,q(ρ)Y p,q(ω) is known as the type function.

We also need an expansion of functions on Cn in terms of Laguerre functions ϕn−1k ’s,

which is know as special Hermite expansion. The special Hermite expansion is a use-ful tool in the study of convolution operators and is related to the spectral theoryof sub-Laplacian on the Heisenberg group Hn. However, more details can be foundin [35].

For λ ∈ R∗ = R \ {0}, let πλ be the unitary representation of Hn on L2(Rn) given by

πλ(z, t)ϕ(ξ) = eiλteiλ(x.ξ+ 12x.y)ϕ(ξ + y), ϕ ∈ L2(Rn

).

A celebrated theorem of Stone and von Neumann says that πλ is irreducible and upto unitary equivalence {πλ: λ ∈ R} are all the infinite dimensional unitary irreduciblerepresentations of Hn. Let

T = ∂

∂t, Xj = ∂

∂xj+ 1

2yj∂

∂t, Yj = ∂

∂yj− 1

2xj∂

∂t, j = 1, 2, . . . , n.

Then {T,Xj , Yj : j = 1, . . . , n} is a basis for the Lie Algebra hn of all left invariant vectorfields on Hn. Define L = −

∑nj=1(X2

j + Y 2j ), the second order differential operator


which is known as the sub-Laplacian of Hn. The representation πλ which induces arepresentation π∗

λ of hn, on the space of C∞ vectors in L2(Rn) is defined by

π∗λ(X)f = d

dt

∣∣∣∣t=0

πλ(exp tX)f.

An easy calculation shows that π∗(Xj) = iλxj , π∗(Yj) = ∂∂xj

, j = 1, 2, . . . , n. Therefore,π∗λ(L) = −Δx +λ2|x|2 =: H(λ), the scaled Hermite operator. The eigenfunction of H(λ)

is given by φλα(x) = |λ|n4 φα(

√|λ|x), α ∈ Zn

+, where φα are the Hermite functions on Rn.Since H(λ)φλ

α = (2|λ| + n)|λ|φλα. Therefore,

L(πλ(z, t)φλ

α, φλβ

)=

(2|λ| + n

)|λ|

(πλ(z, t)φλ

α, φλβ

).

Thus the entry functions (πλ(z, t)φλα, φ

λβ), α, β ∈ Zn

+ are eigenfunctions for L. As(πλ(z, t)φλ

α, φλβ) = eiλt(πλ(z)φλ

α, φλβ), these eigenfunctions are not in L2(Hn). However

for a fix t, they are in L2(Cn). Define Lλ by L(eiλtf(z)) = eiλtLλf(z). Then the functions

φλαβ(z) = (2π)−n

2(πλ(z)φλ

α, φλβ

),

are eigenfunction of the operator Lλ with eigenvalue 2|λ| + n. The functions φλαβ ’s are

called the special Hermite functions and they form an orthonormal basis for L2(Cn) (see[35, Theorem 2.3.1, p. 54]). Thus, for g ∈ L2(Cn), we have the expansion

g =∑α,β

⟨g, φλ

αβ

⟩φλαβ .

To further simplify this expansion, let ϕn−1k,λ (z) = ϕn−1

k (√

|λ|z), the Leguerre function ofdegree k and order n− 1. The special Hermite functions φλ

αα satisfy the relation

∑|α|=k

φλα,α(z) = (2π)−n

2 |λ|n2 ϕn−1k,λ (z). (2.3)

Let g be a function in L2(Cn). Then g can be expressed as

g(z) = (2π)−n|λ|n∞∑k=0

g ×λ ϕn−1k,λ (z),

whenever λ ∈ R∗ (see [35, p. 58]). In particular, for λ = 1, we have

g(z) = (2π)−n∞∑k=0

g × ϕn−1k (z), (2.4)

which is called the special Hermite expansion for g.


We need the Hecke–Bochner identity for the spectral projections f×ϕn−1k , for function

f ∈ L2(Cn). See [35, p. 70]. For k ∈ Z+, the Laguerre function ϕn−1k is defined by

ϕn−1k (z) = Ln−1

k (12 |z|2)e−

14 |z|

2 , where

Ln−1k (x) =

k∑j=0

(−1)j(k + n− 1k − j

)xj

j! ,

is the Laguerre polynomial of degree k and order n− 1.

Lemma 2.1. Let aP ∈ L2(Cn), where a is a radial function and P ∈ Hp,q. Then

aP × ϕn−1j (z) = (2π)n

⟨a, ϕn+p+q−1

k−p

⟩P (z)ϕn+p+q−1

k−p (z)

= (2π)−nP (z)a× ϕn+p+q−1k−p (z), (2.5)

if k � p and 0 otherwise. The convolution in the right hand side is on the space Cn+p+q.

3. Sets of injectivity for the twisted spherical means

In this section, we prove that the X-axis is a set of injectivity for the TSM for acertain class of functions in L2(C). Then, we replicate the method to prove that X-axistogether with Y -axis is a set of injectivity for the TSM for Lq(C). In the later case, wededuce a density result for Lp(C), 2 � p < ∞.

Next, we find an expansion for f × ϕ0k with help of Hecke–Bochner identities for

spectral projection. The real analyticity of f × ϕn−1k can also be understood by the fact

that f × ϕn−1k can be extended to a holomorphic function on C2n.

Proposition 3.1. Let f ∈ L2(C). Then the real analytic expansion of Qk(z) = f × ϕ0k(z)

can be written as

Qk(z) =k∑

p=0Cp0

k−pzpϕp

k−p(z) +∞∑q=0

C0qk zqϕq

k(z). (3.1)

Proof. We know that

f(z) =∞∑p=0

ap0(|z|

)zp +

∞∑q=0

a0q(|z|)zq, for z ∈ C.

Since f ∈ L2(C), using the Hecke–Bochner identity for the spectral projections as inLemma 2.1, we can express f × ϕ0

k(z) as

Qk(z) = f × ϕ0k(z) =

k∑Cp0

k−pzpϕp

k−p(z) +∞∑

C0qk zqϕq

k(z),
p=0 q=0


where the series on the right-hand side converges to Qk in L2(C). In order to show thatthe series converges uniformly on every compact set K ⊆ C, it is enough to show thatthe series

h(z) =∞∑

q=k+1

C0qk zqϕq

k(z),

converges uniformly on every ball BR(o) in C. Since Qk ∈ L2(C), it follows that h ∈L2(C) and

‖h‖2L2(C) =

∞∑q=k+1

∣∣C0qk

∣∣2∥∥zqϕqk

∥∥2L2(C) < ∞.

Since

∥∥zqϕqk

∥∥2L2(C) =

∞∫0

∫S1

|rω|2q(ϕqk

)2r dr dω

= 2π∞∫0

(ϕq+1−1k

)2r2(q+1)−1 dr

= 2π2q (k + q)!k! . (3.2)

Therefore, the coefficients C0qk ’s must satisfy an estimate of type

∣∣C0qk

∣∣ � C

(k!

2q+1(k + q)!

) 12

, (3.3)

where C is a constant and independent of q. Now, let |z| � R. Then, we have

∣∣h(z)∣∣ � e−

14 |z|

2∞∑

q=k+1

∣∣C0qk

∣∣|z|q∣∣∣∣∣

k∑j=0

(−1)j(q + k

k − j

) (12 |z|2)

j

j!

∣∣∣∣∣

� Ce−14 |z|

2∞∑

q=k+1

(k!

2q+1(k + q)!

) 12

|z|q (q + k)!k!q!

k∑j=0

(12 |z|2)

j

j!

� Ce−14 |z|

2∞∑

q=k+1

((q + k)!2q+1k!q!

) 12 |z|q

(q!) 12e

12 |z|

2

� Ce14R

2∞∑ (

(q + k)!2q+1k!q!

) 12 Rq

(q!) 12< ∞.

q=k+1


Thus the function h is real analytic on C. That is, the right-hand side of (3.1) is a realanalytic function which agrees to the real analytic function Qk a.e. on C. Hence (3.1) isa real analytic expansion of Qk. �

We would like to call (3.1) the Hecke–Bochner–Laguerre series for the spectral pro-jections. We study this series carefully and use it to prove the most striking results,Theorems 3.4, 3.6 of this article.

Theorem 3.2. Let f ∈ L2(C). Suppose f × μr(x) = 0, ∀r > 0 and ∀x ∈ R. Thenf × ϕ0

0 ≡ f × ϕ01 ≡ 0 on C.

Proof. Since f × μr(x) = 0, ∀r > 0 and ∀x ∈ R, by polar decomposition, it follows thatQk(x) = f × ϕ0

k(x) = 0, ∀x ∈ R and ∀k ∈ Z+. For k = 0, we have

Q0(x) = C000 ϕ0

0(x) + C000 ϕ0

0(x) + C010 xϕ1

0(x) + C020 x2ϕ2

0(x) + · · · = 0, ∀x ∈ R.

On equating the coefficients of 1, x, x2, . . . to zero, we get C0q0 = 0, ∀q � 0. Hence Q0 ≡ 0

on C. From Eq. (3.1), we have

Q1(x) = C001 ϕ0

1(x) + C100 xϕ1

0(x) +∞∑q=0

C0q1 xqϕq

1(x) = 0, ∀x ∈ R.

Using the argument x → −x, it follows that

2C001 ϕ0

1(x) +∞∑

m=0C0,2m

1 x2mϕ2m1 (x) = 0.

By equating coefficient of 1, x2, x4, . . . , we get C0,2m1 = 0, for m = 0, 1, 2, . . . . Hence the

series of Q1(x) reduces to

Q1(x) = C1,00 xϕ1

0(x) +∞∑

m=0C0,2m+1

1 x2m+1ϕ2m+11 (x) = 0.

By canceling e−14x

2 in the above series, we have

C1,00 x +

∞∑m=0

C0,2m+11 x2m+1

(2m + 2 − 1

2x2)

= 0.

On equation the coefficients of x, x3, x5, . . . , we get the following recursion relations

C1,00 = −2C0,1

1 and C0,2m+11 = C0,1

12m ; for m = 1, 2, 3, . . . . (3.4)
2 (m + 1)!


Now, we can write Q1(z) = C1,00 zϕ1

0(z) + h(z), where the series

h(z) =∞∑

m=0C0,2m+1

1 z2m+1ϕ2m+11 (z)

converges in L2(C). We claim that all the coefficients C0,2m+11 ; m = 0, 1, 2, . . . are zero.

Here two cases arise. If h1(z) has finitely many non-zero coefficients, then Q1(z) ispolynomial times Gaussian and hence the condition Q1(x) = 0, ∀x ∈ R, by equating thecoefficient of highest degree term to zero, implies that each coefficient has to be zero.(Please see the proof of Theorem 3.4.) On the other hand, suppose infinitely many of thesecoefficients are non-zero. Then, by the estimate (3.2) and the recursion relations (3.4),we have

‖h‖2L2(C) =

∞∑m=0

∣∣C0,2m+11

∣∣2∥∥z2m+1ϕ2m+11

∥∥2L2(C)

= 2π∣∣C0,1

1∣∣2 ∞∑

m=0

22m+1(2m + 2)!(22m(m + 1)!)2

= 4π∣∣C0,1

1∣∣2 ∞∑

m=0

(2m + 2)!22m((m + 1)!)2 =

∞∑m=0

bm = ∞.

The series on the right-hand side diverges by Raabe’s test (see, [18, p. 36]). Since

limm→∞

{m

(bm

bm+1− 1

)}= −1

2 < 1.

This contradicts the fact that the series h is L2(C) summable. Thus, we get C0,2m+11 = 0,

for m = 0, 1, 2, . . . . Hence, we conclude that Q1 ≡ 0. �Remark 3.3. Under the same assumptions as in Theorem 3.2, it would be interesting toknow, whether Qk ≡ 0 for k � 2. The argument used to show Q1 ≡ 0 does not seemto work in this case. In another attempt, using the recursion relations Ln

k = Ln−1k +

· · · + Ln−10 , Ln

k − Lnk−1 = Ln−1

k and the result that f × ϕ00 = f × ϕ0

1 = 0, we can easilydeduce that f × ϕ0

2 = f × ϕ12. But, we are not able to conclude any thing more on

account of the facts that f × ϕ02 is an eigenfunction of the special Hermite operator A

and ϕ12 = ϕ0

2 + ϕ01 + ϕ0

0.However, we prove the following partial result that any line passing through the origin

is a set of injectivity for the TSM for a certain class of functions in L2(C). Since for anyσ ∈ U(n), we have f × μr(σ.z) = (π(σ)f)× μr(z). It follows that a set S ⊂ C is a set ofinjectivity for the TSM if and only if for each σ ∈ U(n), the set σ.S is a set of injectivityfor the TSM. In view of this, it is enough to prove that the X-axis is a set of injectivityfor the TSM.


Theorem 3.4. Let f ∈ L2(C) and for each k ∈ Z+ the projection e14 |z|

2f × ϕ0

k is apolynomial. Suppose f × μr(x) = 0, ∀r > 0 and ∀x ∈ R. Then f = 0 a.e.

Proof. Since f ∈ L2(C), by polar decomposition, the condition f × μr(x) = 0, ∀r > 0and ∀x ∈ R is equivalent to f ×ϕ0

k(x) = 0, ∀k � 0 and ∀x ∈ R. From Eq. (3.1), we have

f × ϕ0k(z) =

k∑p=0

Cp0k−pz

pϕpk−p(z) +

∞∑q=0

C0qk zqϕq

k(z).

By the given exponential condition, we can write f×ϕ0k(z) = P (z, z)e− 1

4 |z|2 . Let z = teiθ.

Then for each fixed t, the function f ×ϕ0k(teiθ) is a trigonometric polynomial. Using the

orthogonality of einθ it follows that there exist m = m(k) ∈ Z+ such that

f × ϕ0k(z) =

k∑p=0

Cp0k−pz

pϕpk−p(z) +

m∑q=0

C0qk zqϕq

k(z). (3.5)

Therefore, for each k ∈ Z+, we have

k∑p=0

Cp0k−px

pϕpk−p(x) +

m∑q=0

C0qk xqϕq

k(x) = 0, ∀x ∈ R.

The constant term in the above expansion 2C00k = 0, hence

k∑p=1

Cp0k−px

pϕpk−p(x) +

m∑q=1

C0qk xqϕq

k(x) = 0, ∀x ∈ R.

On equating the coefficient of the highest degree term xm+2k to zero, we get C0mk = 0.

Similarly, continuing this argument up to x2k, we obtained C0mk = C

0(m−1)k = · · · =

C01k = 0. Then equate the coefficients of x, x2, . . . , x2k−1 to zero, we find C10

k−1 = C20k−2 =

· · · = Cp00 = 0. Thus f ×ϕ0

k ≡ 0, ∀k � 0. Hence f = 0 a.e. This completes the proof. �Remark 3.5. (a) Since f ×ϕk is real analytic and the zero set of a real analytic functionis isolated, in this case, we only need the centres to be a sequence in R having a limitpoint. It is clear from (3.5) that any curve γ := {(γ1(t), γ2(t)): t ∈ R}, where γj , j = 1, 2are polynomials is also a set of injectivity for the TSM. A natural question is thatγ := (γ1, γ2) with γj ’s are real analytic is a set of injectivity for the TSM. We believethat this will help in characterizing non-injectivity sets for TSM.

(b) Let us consider the functions

fm(z) =∞∑

ap0(|z|

)zp +

m∑a0q(|z|)zq.

p=0 q=0


Then by the similar argument as in the proof of Theorem 3.2, we can easily deduce thate

14 |z|

2fm×ϕ0

k is a polynomial. In fact these are the only functions for which e14 |z|

2f ×ϕ0

k

is a polynomial. This can be seen from Eq. (3.5), when we reverse the process using theHecke–Bochner identities. Thus the space considered is just not empty, it includes allthe sequence (fm) which converges to

f(z) =∞∑p=0

ap0(|z|

)zp +

∞∑q=0

a0q(|z|)zq

in L2(C).We believe that the proof of Theorem 3.4, without exponential condition on spectral

projection would need a finer argument and hence we prefer to return to this questionlater.

Next, we prove the stronger result that X-axis together with Y -axis is a set of injec-tivity for the TSM for any function in Lq(C).

For η ∈ C, define the left twisted translate by

τηf(ξ) = f(ξ − η)e i2 Im(η.ξ).

Then τη(f×μr) = τηf ×μr. Let S be a set of injectivity for the TSM on Lq(C). Supposef × μr(z − η) = 0, ∀r > 0 and ∀z ∈ S. Then

τηf × μr(z) = τη(f × μr)(z) = ei2 Im(η.z)f × μr(z − η) = 0,

for all r > 0 and ∀z ∈ S. Since the space Lq(C) is twisted translations invariant, itfollows that a set S ⊂ C is set of injectivity for the TSM if and only if for each η ∈ C,the set S − η is a set of injectivity for the TSM. That is, the Euclidean translate of theset S is also a set of injectivity for the TSM on Lq(C). By rotation and translation, it isobvious that any two perpendicular lines can be set of injectivity for the TSM, providedX-axis and Y -axis together is a set of injectivity for the TSM.

Theorem 3.6. Let f ∈ Lq(C), for 1 � q � 2. Suppose f ×μr(x) = f ×μr(ix) = 0, ∀r > 0and ∀x ∈ R. Then f = 0 a.e.

Let f ∈ Lq(C). Then by convolving f with a right and radial compactly supportedsmooth approximate identity, we can assume f ∈ L2(C). Let us consider the Hecke–Bochner–Laguerre series

Qk(z) =k∑(

Cp0k−pz

pϕpk−p(z) + C0p

k zpϕpk(z)

)+

∞∑C0p

k zpϕpk(z). (3.6)

p=1 p=k+1


The proof is now based on symmetries and then cancellations. We decompose the aboveseries into four (disjoint) series, each of which after equating its coefficients to zero,gives a system of solvable recursion relations. Using these recursion relations togetherwith some basic properties of Laguerre polynomials, we show that all the coefficientsappeared in the series (3.6) are zero.

Let E+ and O+ denote the sets of even and odd positive integers respectively. LetEk = E+ ∩ {1, 2, . . . , k}, Fk = E+ ∩ {k + 1, k + 2, . . .}, Gk = O+ ∩ {1, 2, . . . , k} andHk = O+ ∩ {k + 1, k + 2, . . .}. Then, we can decompose the above series as Qk(z) =Uk(z) + Vk(z2), where

Uk(z) =∑p∈Gk

(Cp0

k−pzpϕp

k−p(z) + C0pk zpϕp

k(z))

+∑p∈Hk

C0pk zpϕp

k

and

Vk

(z2) =

∑p∈Ek

(Cp0

k−pzpϕp

k−p(z) + C0pk zpϕp

k(z))

+∑p∈Fk

C0pk zpϕp

k.

We shall call Uk and Vk as odd and even series respectively. For x ∈ R, it is giventhat Uk(x) + Vk(x2) = 0. Using the argument x → −x, we have Uk(x) = Vk(x2) = 0.Similarly, Uk(ix) + Vk((ix)2) = 0, implies Uk(ix) = Vk((ix)2) = 0. Indeed, we can putthese conditions as follows.

(A) Uk(x) = Uk(ix) = 0, and(B) Vk(x2) = Vk((ix)2) = 0.

In order to prove Theorem 3.6, we prove that Qk ≡ 0, ∀k � 0. Now, we divide the proofinto two parts: 0 � k � 3 and k � 4.

Lemma 3.7. Let f ∈ L2(C) and 0 � k � 3. Suppose Qk(x) = Qk(ix) = 0, ∀x ∈ R. ThenQk ≡ 0.

Proof. Since, we have shown in Theorem 3.2 that Q0 ≡ Q1 ≡ 0, we only need to proveQk ≡ 0 for k = 2, 3. For k = 2, by condition (A), we get a pair of equations

x(C10

1 ϕ11(x) + C01

2 ϕ12(x)

)+

∞∑m=2

C0,2m−12 x2m−1ϕ2m−1

2 (x) = 0,

x(C10

1 ϕ11(x) − C01

2 ϕ12(x)

)+

∞∑m=2

(−1)mC0,2m−12 x2m−1ϕ2m−1

2 (x) = 0.

On adding these two equations, we have

xC101 ϕ1

1(x) +∞∑

C0,4m−52 x4m−5ϕ4m−5

2 (x) = 0.
m=2


By equating the coefficients of x, x3, x7, . . . to zero, we get C101 = 0 and C0,4m−5

2 = 0, form = 2, 3, . . . . Similarly by subtracting and then equating the coefficients of x, x5, x9, . . .

to zero, we obtain C012 = 0 and C0,4m−7

2 = 0, for m = 3, 4, . . . . Hence U2 ≡ 0. Bycondition (B), we have

x2(C200 ϕ2

0 + C022 ϕ2

2)

+∞∑

m=2C0,2m

2 x2mϕ2m2 (x) = 0,

−x2(C200 ϕ2

0 + C022 ϕ2

2)

+∞∑

m=2(−1)mC0,2m

2 x2mϕ2m2 (x) = 0.

In a quite similar way, we find V2 ≡ 0 and hence Q2 ≡ 0. Here, by adding and subtracting,we get the coefficients to be more disjoint, which is the only difficulty we need to resolve.The method used in this case will be repeated for k � 3.

For k = 3, by condition (A), we have

x(C10

2 ϕ12(x) + C01

3 ϕ13(x)

)+

∞∑m=3

C0,2m−13 x2m−1ϕ2m−1

3 (x) = 0,

x(C10

2 ϕ12(x) − C01

3 ϕ13(x)

)+

∞∑m=3

(−1)mC0,2m−13 x2m−1ϕ2m−1

3 (x) = 0.

As very similar to above, the pair of equations implies that U3 ≡ 0. By condition (B),we have

x2(C201 ϕ2

1(x) + C023 ϕ2

3(x))

+∞∑

m=2C0,2m

3 x2mϕ2m3 (x) = 0,

−x2(C201 ϕ2

1(x) + C023 ϕ2

3(x))

+∞∑

m=2(−1)mC0,2m

3 x2mϕ2m3 (x) = 0.

This shows that V3 ≡ 0 and hence Q3 ≡ 0. �Next, we prove the following lemma for the case k � 4, which completes the proof of

Theorem 3.6.

Lemma 3.8. Let f ∈ L2(C) and k � 4. Suppose Qk(x) = Qk(ix) = 0, ∀x ∈ R. ThenQk ≡ 0.

Proof. First, we show that the odd series Uk ≡ 0, ∀k � 4. By condition (A), we canwrite


∑p∈Gk

xp(Cp0

k−pϕpk−p(x) + C0p

k ϕpk(x)

)+

∑p∈Hk

C0pk xpϕp

k(x) = 0,

∑p∈Gk

(−1)p−12 xp

(Cp0

k−pϕpk−p(x) − C0p

k ϕpk(x)

)+

∑p∈Hk

(−1)p+12 C0p

k xpϕpk(x) = 0.

By adding and subtracting, we get two series in which there are no common coefficients.On equating the coefficients of x, x3, x5, . . . , in both the new series to zero, we get allthe coefficients of odd series Uk are zero. Hence Uk ≡ 0, ∀k � 4. Since i4 = 1, it showsthat there must occur some change in the pattern of the even series Vk for k � 4. Forinstance consider V4. By condition (B), we have

x2(C202 ϕ2

2(x) + C024 ϕ2

4(x))

+ x4(C400 ϕ4

0(x) + C044 ϕ4

4(x))

+∞∑

m=3C0,2m

4 x2mϕ2m4 (x) = 0,

−x2(C202 ϕ2

2(x) + C024 ϕ2

4(x))

+ x4(C400 ϕ4

0(x) + C044 ϕ4

4(x))

+∞∑

m=3(−1)mC0,2m

4 x2mϕ2m4 (x) = 0.

On adding and subtracting, we get the two series

x4(C400 ϕ4

0(x) + C044 ϕ4

4(x))

+∞∑

m=4C

0,4(m−2)4 x2mϕ

4(m−2)4 (x) = 0, (3.7)

x2(C202 ϕ2

2(x) + C024 ϕ2

4(x))

+∞∑

m=3C0,4m−6

4 x4m−6ϕ4m−64 (x) = 0. (3.8)

By equating the coefficients of x4, x6, x8, . . . , in Eq. (3.7) to zero, we get, all the co-efficients in (3.7) are zero. By canceling e−

14x

2 in series (3.8) and using x2 → 2x, wehave

2x(C20

2 L22(x) + C02

4 L24(x)

)+

∞∑m=3

C0,4m−64 (2x)2m−3L4m−6

4 (x) = 0.

On equating the coefficients of x and x2 to zero, we get

(L2

2(0) L24(0)

(L22)′(0) (L2

4)′(0)

)(C20

2C02

4

)=

(6 6−4 −20

)(C20

2C02

4

)= 0. (3.9)

Thus C202 = C02

4 = 0 and hence we find V4 ≡ 0. Equivalently, we can use onwards towrite: on equating the coefficients of x2 and x4 in Eq. (3.8) to zero, we get Eq. (3.9).Now, it only remains to show that Vk ≡ 0, ∀k � 5. In this case, by condition (B), wehave


∑p∈Ek

xp(Cp0

k−pϕpk−p + C0p

k ϕpk

)+

∑p∈Fk

C0pk xpϕp

k = 0,

∑p∈Ek

(−1)p2 xp

(Cp0

k−pϕpk−p + C0p

k ϕpk

)+

∑p∈Fk

(−1)p2C0p

k xpϕpk = 0.

We further require a partition of the set E = A1 ∪ A2, where A1 = {4t− 2: t ∈ N} andA2 = {4t: t ∈ N}. By adding and subtracting, we will get the following pair of serieshaving brackets

∑p∈Ek∩A2

xp(Cp0

k−pϕpk−p + C0p

k ϕpk

)+

∑p∈Fk∩A2

C0pk xpϕp

k = 0, (3.10)

∑p∈Ek∩A1

xp(Cp0

k−pϕpk−p + C0p

k ϕpk

)+

∑p∈Fk∩A1

C0pk xpϕp

k = 0. (3.11)

Since the matrix(

Lpk−p(0) Lp

k(0)(Lp

k−p)′(0) (Lpk)′(0)

)

is non-singular and p ∈ A2, the brackets in (3.10) are not a problem. Hence on equatingthe coefficients of x2, x4, x6, . . . to zero in (3.10), it follows that all the coefficients in (3.10)are zero. Similarly, in (3.11), as p in A1, it also follows that all the coefficients in (3.11)are zero. Thus we find Vk ≡ 0, ∀k � 5 and hence Qk ≡ 0, ∀k � 0. This completes theproof. �Remark 3.9. (a) We can also prove the general case by calculating case-wise, whenk = 2m, 2m + 1 and p = 4t − 2, 4t, but it would only make the calculation to be morecomplicated. In another attempt, to get a more transparent proof of Theorem 3.6, keepapplying the right invariant operator A = ∂

∂z + 14 z to Qk(z). Then a straightforward

calculation shows that

ApQk(0) = p!ϕpk−p(0)Cp0

k−p,

if p � k and 0 otherwise. By the condition Qk(x) = Qk(ix) = 0, ∀x ∈ R, it followsthat AQk(0) = 0. This implies C10

k−1 = 0 and in turn C01k = 0. In view of this, we can

immediately conclude that Q1 ≡ 0. But for k � 2, the conditions on Qk do not implyApQk(0) = 0, when p � 2, otherwise this would lead to a more transparent proof ofTheorem 3.6.

(b) In Theorem 3.6, we have shown that any two lines having angle π/2 are a set ofinjectivity for the TSM on C. However, the question that any two lines having positiveangle less than π/2 can be a set of injectivity for the TSM on C is still unanswered.

(c) Consider Coxeter system of N -lines ΣN =⋃N−1

l=0 {teiθl : θl = πlN , t ∈ R}. Suppose

θl = π/2. Then l = N/2. By Theorem 3.6, it follows that any Coxeter system of evennumber of lines is also a set of injectivity for the TSM on Lq(C).


Next, we set to describe the problem for any Coxeter system of odd lines. Let1, ω, ω2, . . . , ωN−1 be the N roots of unity. Then ΣN =

⋃N−1l=0 {ωlx: x ∈ R}. We have

formulated this problem in the following way.

Conjecture. Let f ∈ L2(C). Suppose f × μr(ωlx) = 0, ∀r > 0, and l = 0, 1, . . . , N − 1and ∀x ∈ R. Then f = 0 a.e.

We would like to produce some evidence about the feasibility of this problem. In thiscase, we get a system of solvable recursion relations by decomposing the series usingsymmetries, however those recursion relations for higher values of k give rise to a higherorder square matrix, which is needed to show non-singular. This is the only difficulty ingetting a solution to this problem. Since Qk(ωlx) = 0, ∀l; l = 0, 1, . . . , N−1 and ∀x ∈ R,as similar to Theorem 3.6, we will have the following conditions.

(A) Uk(ωlx) = 0, ∀l; l = 0, 1, . . . , N − 1, and(B) Vk((ωlx)2) = 0, ∀l; l = 0, 1, . . . , N − 1.

For N = 3, by a simple argument that finds all odd positive integers which are notdivisible by 3, we can find a partition of the set of natural numbers as N =

⋃2i=0 Ai, where

Ao = {1, 2, 3, 4}, A1 = {6t− 1, 6t: t ∈ N} and A2 = {6t+ 1, 6t+ 2, 6t+ 3, 6t+ 4: t ∈ N}.For k ∈ Ao, the condition (A) together with the facts 1 + ω + ω2 = 0 and ω3 = 1,

implies that

C03k x3ϕ3

k + C09k x9ϕ9

k + · · · = 0.

On equating the coefficient of x3, x9, . . . to zero, we get C03k = C09

k = · · · = 0. Hence theodd series Uk reduces to

x(C10

k−1ϕ1k−1 + C01

k ϕ1k

)+ C05

k x5ϕ5k + C07

k x7ϕ7k + · · · = 0. (3.12)

On equating the coefficients of x and x3 in Eq. (3.12) to zero, we get

(L1k−1(0) L1

k(0)(L1

k−1)′(0) (L1k)′(0)

)(C10

k−1C01

k

)=

(k k + 1

−k(k−1)2

−k2

)(C10

k−1C01

k

)= 0. (3.13)

Thus C10k−1 = C01

k = 0 and hence we find Uk ≡ 0, ∀k ∈ Ao. For k ∈ {1, 2, 3}, bycondition (B) and the facts 1 + ω + ω2 = 0 and w3 = 1, we have

C06k x6ϕ6

k + C0,12k x12ϕ12

k + · · · = 0.

This shows that C062 = C0,12

2 = · · · = 0, and in turn the even series Vk reduces to

x2(C20k−2ϕ

kk−2 + C02

k ϕ2k

)+ C04

k x4ϕ4k + C08

k x8ϕ8k + · · · = 0.


Let Fnk = Ln

k (0). On equating the coefficients of x2, x4 and x6 to zero, we get⎛⎝F 2

k−2 F 2k 0

F 20 −F 3

k−1 F 4k

0 F 4k−2 −F 5

k−1

⎞⎠

⎛⎝C20

k−2C02

k

C04k

⎞⎠ = 0.

This implies C20k−2 = C02

k = C04k = 0. Thus, it follows that Vk ≡ 0 and hence Qk ≡ 0,

for k ∈ {0, 1, 2, 3}. This gives a strong evidence about the existence of the problem. Wewould also like to focus on to the proof, for higher values of k. Let k ∈ A1 ∪A2. Then bythe condition (A) together with the facts 1 + ω + ω2 = 0 and ω3 = 1, the odd series Uk

reduces to

x(C10

k−1ϕ1k−1 + C01

k ϕ1k

)+ x5(C50

k−5ϕ5k−5 + C05

k ϕ5k

)+ x7(C70

k−7ϕ7k−7 + C07

k ϕ7k

)+ · · · + xr

(Cr0

k−rϕrk−r + C0r

k ϕrk

)+ · · · + C0j

k xjϕjk + · · · = 0.

On equating the coefficients of x and x3 to zero, we get C10k−1 = C01

k = 0. Thus

x5(C50k−5ϕ

5k−5 + C05

k ϕ5k

)+ x7(C70

k−7ϕ7k−7 + C07

k ϕ7k

)+ · · ·

+ xr(Cr0

k−rϕrk−r + C0r

k ϕrk

)+ · · · + C0j

k xjϕjk + · · · = 0. (3.14)

The main problem here is to remove the brackets (mixed term) in this series. Then, itis easy to show that rest of coefficients are zero. To remove the brackets, we need toidentify the indices r, j, appeared in (3.14) and the number of equations m required. ByEq. (3.14), we can write the following table.

k r j m = j+12 − 2 k − r

k ∈ A1 6t − 1 2k − 1 k − 2 0, 1k ∈ A2 6t + 1 2k + 1 k − 1 0, 1, 2, 3

Let k ∈ A1. Equate the coefficients of x5, x7, . . . , xj to zero. In order to show thatcoefficients in the brackets are zero, we need to show that the following matrices arenon-singular. For k = 5, we have the matrix

⎛⎝F 5

0 F 55 0

0 −F 64 F 7

50 F 7

3 −F 84

⎞⎠ ,

which is non-singular. For higher values of k, we get the higher order matrices whichshould be non-singular. This is the only difficulty in the above arguments. Hence, weleave this problem open for future research.

Remark 3.10. In the case, when N = 1 (without exponential decay), we cannot obtainthe m×m matrices. However for N � 3, we have the m×m matrices which are neededto be non-singular.


From Remark 3.9(b), it is clear that the set Σ2N is a set of injectivity for the TSMfor Lq(C), with 1 � q � 2. As a dual problem, it is natural to ask that Σ2N is a set ofdensity for Lp(C), for 2 � p < ∞. The following result would emerge as the first resultabout the sets of density in terms of the TSM. Let C�

c(C) denote the space of radialcompactly supported continuous functions on C. Let τzf(w) = f(z − w)e i

2 Im(z.w).

Proposition 3.11. The subspace F (Σ2N ) = Span{τzf : z ∈ Σ2N , f ∈ C�c(C)} is dense

in Lp(C), for 2 � p < ∞.

Proof. Let 1p + 1

q = 1. Then 1 � q � 2. By Hahn–Banach theorem, it is enough to showthat F (Σ2N )⊥ = {0}. Let g ∈ Lq(C) be such that

∫C

τzf(w)g(w) dw = 0, z ∈ Σ2N , ∀f ∈ C�c(C).

That is,

g × f(z) = f × g(z) = 0.

Let the support of f be contained in [0, t]. Then by passing to the polar decomposition,we get

t∫r=0

g × μr(z)f(r)r2n−1 dr = 0.

By differentiating the above equation, it follows that g×μt(z) = 0, ∀t > 0 and ∀z ∈ Σ2N .Thus by Theorem 3.6, we conclude that g = 0 a.e. on Cn. �4. Discussion on sets of injectivity in higher dimension

More generally, similar to the work of Agranovsky and Quinto [5], let f ∈ L1loc(Cn)

and write S(f) = {z ∈ Cn: f × μr(z) = 0, ∀r > 0}. Our main problem is to describethe complete geometrical structure of S(f) that would ensure which “sets” are sets ofinjectivity for the TSM. There is one such result.

Lemma 4.1. Let f ∈ Lp(Cn) ∩ C(Cn), for 1 � p � ∞. Then

S(f) =∞⋂k=0

Q−1k (0).


Proof. Let z ∈ S(f). Then by polar decomposition, it follows that Qk(z) = 0, ∀k � 0.Hence S(f) ⊆

⋂∞k=0 Q

−1k (0). Conversely, let Qk(z) = 0, ∀k � 0. Then

∞∫r=0

f × μr(z)ϕn−1k (r)r2n−1 dr = 0, ∀k � 0.

Since the set {ϕn−1k : k = 0, 1, 2, . . .} is an orthonormal set for L2(R+, r

2n−1 dr) andf×μr(z) is continuous in r, it follows that f×μr(z) = 0, ∀r > 0 and hence z ∈ S(f). �

Next, we find out S(f) for the type function f(z) = a(|z|)P (z) on Cn. For this, weneed the following result of Filaseta and Lam [15], about the irreducibility of Laguerrepolynomials. Define the Laguerre polynomials by

Lαk (x) =

k∑i=0

(−1)i(α + k

k − i

)xi

i! ,

where k ∈ Z+ and α ∈ C.

Theorem 4.2. (See [15].) Let α be a rational number, which is not a negative integer.Then for all but finitely many k ∈ Z+, the polynomial Lα

k (x) is irreducible over therationals.

Using Theorem 4.2, we have obtained the following corollary about the zeros of La-guerre polynomials.

Corollary 4.3. Let k ∈ Z+. Then for all but finitely many k, the Laguerre polynomialsLn−1k (x)’s have distinct zeros over the reals.

Proof. By Theorem 4.2, there exists ko ∈ Z+ such that Ln−1k ’s are irreducible over Q

whenever k � ko. Therefore, we can find polynomials P1, P2 ∈ Q[x] such that P1Ln−1k1

+P2L

n−1k2

= 1, over Q with k1, k2 � ko. Since this identity continues to hold on R, itfollows that Ln−1

k1and Ln−1

k2have no common zero over R. �

Proposition 4.4. Let f be a non-zero type function f = aP ∈ L2(Cn), where P ∈ Hp,q.Then S(f) = P−1(0) ∪ F , where F is a finite union of spheres in Cn.

Proof. Since f �≡ 0, there exists at least some k ∈ Z+ for which Q−1k (0) �= Cn. Therefore,

Q−1k (0) = P−1(0) ∪

(ϕn+p+q−1k−p

)−1(0),

for some k ∈ Z+. Hence S(f) = P−1(0) ∪ F . �


Proposition 4.5. Let f = aP ∈ L2(Cn), where P ∈ Hp,q. Suppose Qk is not identicallyzero on Cn, for all but finitely many k. Then S(f) = P−1(0).

Proof. Since Qk(z) = f ×ϕn−1k . Then by Lemma 4.1, we have S(f) =

⋂∞k=0 Q

−1k (0). By

Hecke–Bochner identity, we can write

Qk(z) = (2π)n⟨a, ϕn+p+q−1

k−p

⟩P (z)ϕn+p+q−1

k−p (z).

Since Qk �≡ 0 for infinitely many k ∈ Z+. Therefore,

Q−1k (0) = P−1(0) ∪

(ϕn+p+q−1k−p

)−1(0) �= Cn,

for infinitely many k. In view of Corollary 4.3, the functions ϕn+p+q−1k−p ’s cannot have

a common zero except for finitely many k ∈ Z+ with k � p. Hence, we conclude thatS(f) = P−1(0). �

Now, we would like to address the problem in the higher dimensional space Cn withn � 2. Let f ∈ L2(Cn). Consider the spherical harmonic decomposition of f as

f(z) =∞∑p=0

∞∑q=0

d(p,q)∑j=1

apqj(|z|

)P jpq(z). (4.1)

In view of the Hecke–Bochner identities (2.5), we conclude that

f × ϕn−1k0

=k0∑p=0

∞∑q=0

d(p,q)∑j=1

Cpqk0−p,jP

jpqϕ

n+p+q−1k0−p

=k0∑p=0

∞∑q=0

P k0pq ϕ

n+p+q−1k0−p , where P k0

pq ∈ Hp,q.

Now look at the following concrete expression for the spectral projections

Qk(z) =k∑

p=0

∞∑q=0

P kpq(z)ϕ

n+p+q−1k−p (z), P k

pq ∈ Hp,q. (4.2)

Remark 4.6. (a) As very much similar to complex plane C, our believe suggests thatany Coxeter system of hyperplanes can be a set of injectivity for the TSM on Cn. Forinstance on C2, suppose the function

Q(z1, z2) =2∑(

apzp1 + bpz

p2)

+2∑(

cq zq1 + dq z

q2)

p=1 q=1


vanishes on each of co-ordinate axis, i.e., Q(x, 0) = Q(ix, 0) = Q(0, x) = Q(0, ix) = 0,∀x ∈ R. Then Q ≡ 0. As another example, consider a typical polynomial P (z1, z2) =czp1 z

q2 ∈ Hp,q. Suppose P (z1, x2) = 0, ∀z1 ∈ C and ∀x2 ∈ R. Then P ≡ 0. In view of

these arguments, write

S = (C× R) ∪ (C× iR) ∪ (R× C) ∪ (iR× C).

It is natural to ask, whether the set S can be a set of injectivity for the TSM on Lq(C2),for 1 � q � 2. We follow this question in the next section.

5. Sets of injectivity for the twisted spherical means in CCCn

In this section, we prove that any Coxeter system of even number of hyperplanesintersecting along a line is a set of injectivity for the TSM for Lp(Cn) (n � 2), for1 � p � 2. Then, we prove that the set S2n−1

R × C is a set of injectivity for the TSMfor a certain class of functions on Cn+1. In the first case, we deduce a density result forLp(Cn) with 2 � p < ∞.

As the space Lp(Cn) is twisted translations invariant, to prove any Coxeter system ofeven number of hyperplanes intersecting along a line is a set of injectivity for the TSMfor Lp(Cn), it is enough to prove that the set Cn−1 ×Σ2N , where

⋃2N−1l=0 {te iπl

2N : t ∈ R},is a set of injectivity for the TSM for Lp(Cn).

Using Theorem 3.6, we prove the following result. Let Sn = Cn−1 ×Σ2N .

Theorem 5.1. Let f ∈ Lp(Cn), for 1 � p � 2. Suppose f × μr(z) = 0, ∀r > 0 and∀z ∈ Sn. Then f = 0 a.e. on Cn.

Proof. Since f × μr(z) = 0, ∀r > 0, by polar decomposition, it follows thatf × ϕn−1

k (z) = 0, ∀k ∈ Z+. Given that f ∈ Lp(Cn). By convolving f with a right andradial compactly supported smooth approximate identity, we can assume f ∈ L2(Cn).In order to prove the result on Cn, we first prove the result on C2 and then by inductionhypothesis on n, we deduce it for Cn. Since

ϕ1k(z1, z2) =

∑β1+β2=k

ϕ0β1

(z1)ϕ0β2

(z2).

Therefore, we can write

f × ϕ1k(z1, z2) =

∑β1+β2=k

∫C2

f(z1 − w1, z2 − w2)ϕ0β1

(w1)ϕ0β2

(w2)

× ei2 Im(z1.w1+z2.w2) dw1 dw2

=∑

β +β =k

∫f ×2 ϕβ2(z1 − w1, z2)ϕ0

β1(w1)e

i2 Im(z1.w1) dw1

1 2 C


=∑

β1+β2=k

∫C

Fz2,β2(z1 − w1)ϕ0β1

(w1)ei2 Im(z1.w1) dw1

=∑

β1+β2=k

Fz2,β2 ×1 ϕ0β1

(z1), (5.1)

where the function Fz2,β2 is defined by

Fz2,β2(z1) =∫C

f(z1, z2 − w2)ϕ0β2

(w2)ei2 Im(z2.w2) dw2.

For fixed z2, using the Minkowski integral inequality, it can show that the functionFz2,β2 ∈ L2(C). By the hypothesis, f × ϕ1

k(z1, z2) = 0, ∀(z1, z2) ∈ S2 and ∀k ∈ Z+.Therefore, from Eq. (5.1), we can write

∑β1+β2=k

Fz2,β2 ×1 ϕ0β1

(z1) = 0, ∀(z1, z2) ∈ S2 and ∀k ∈ Z+.

As the above equation is valid for each k ∈ Z+, it follows that the sum over each ofthe diagonal β1 + β2 = k is zero. Using the facts that set {ϕ0

β1: β1 ∈ Z+} forms an

orthogonal basis for L2(C) and S2 = C×Σ2N , it follows that

Fz2,β2 ×1 ϕ0β1

(z1) = 0, ∀β1, β2 ∈ Z+.

By Eq. (5.1), we have

f ×(ϕ0β1ϕ0β2

)(z1, z2) = 0, ∀(z1, z2) ∈ S and ∀β1, β2 ∈ Z+.

Now, we can write,

f ×(ϕ0β1ϕ0β2

)(z1, z2) =

∫C

Gz1,β1(z2 − w2)ϕ0β1

(w2)ei2 Im(z2.w2) dw1

= Gz1,β1 ×2 ϕ0β2

(z2).

By the given condition, we have Gz1,β1 ×2 ϕ0β2

(z2) = 0, ∀β1, β2 ∈ Z+. For each fixed β1,in view of Theorem 3.6, we can conclude that Gz1,β1(z2) = 0, ∀(z1, z2) ∈ C2. That is,f ×1 ϕβ1(z1, z2) = 0, ∀β1 ∈ Z+. Therefore,

f × ϕ1k(z1, z2) =

∑β1+β2=k

∫C

f ×1 ϕβ1(z1, z2 − w2)ϕ0β1

(w2)ei2 Im(z2.w2) dw = 0,

for all k ∈ Z+. Hence, we conclude that f = 0, a.e. C2. In order to prove the resultfor n > 2, we use the induction hypothesis on n. Suppose the result is true for n − 1with n > 2. That is, the set Sn−1 = Cn−2 ×Σ2N is a set of injectivity for the TSM for


L2(Cn−1). Let k = β1 +β2 + · · ·+βn = β1 + |γ| and z = (z1, z2, . . . , zn) = (z1, z′). Then,

as similar to C2 case, we can write

f × ϕn−1k

(z1, z

′) =∑

β1+|γ|=k

Fz′,γ ×1 ϕ0β1

(z1).

Given that

f × ϕn−1k

(z1, z

′) =∑

β1+|γ|=k

Fz′,γ ×1 ϕ0β1

(z1) = 0, ∀(z1, z

′) ∈ Sn and ∀k ∈ Z+.

Since, the set {ϕ0β1

: β1 ∈ Z+} forms an orthogonal basis for L2(C) and Sn = Cn−1×Σ2N ,it follows that Fz′,γ ×1 ϕ

0β1

(z1) = 0, ∀β1 ∈ Z+ and ∀|γ| ∈ Z+. This in turn implies thatfor each fixed γ ∈ Zn−1

+ , we get Fz′,γ ×1 ϕ0β1

(z1) = 0, ∀β1 ∈ Z+ and ∀(z1, z′) ∈ C×Sn−1.

Once again using the orthogonality of the set {ϕ0β1

: β1 ∈ Z+} in L2(C), we concludethat Fz′,γ(z1) = 0, ∀(z1, z

′) ∈ C× Sn−1 and ∀γ ∈ Zn−1+ . Therefore, we have

∑|γ|=k

Fz′,γ(z1) =∑|γ|=k

∫Cn−1

f(z1, z

′ − w′) n∏j=2

ϕ0βj

(wj)ei2 Im(z′.w′) dw′ = 0,

∀k ∈ Z+ and ∀z′ ∈ Sn−1. By the assumption that Sn−1 is a set of injectivity for theTSM for L2(Cn−1), we infer that f = 0 a.e. on Cn. �Remark 5.2. (a) By the proof of Theorem 5.1, it reveals that if S is a set of injectivityfor the TSM for Lp(C), then the set Cn−1 × S will be a set of injectivity for the TSMfor Lp(Cn). That is, the sets of injectivity for the TSM on Cn can be embedded into thesets injectivity for the TSM on Cn+1.

(b) We would like to mention that the question of Coxeter system of odd numberof hyperplanes intersecting along a line is a set of injectivity for the TSM for Lp(Cn)(n � 2) can be answered, once we know that any Coxeter system of odd number of linescan be a set of injectivity for the TSM for Lp(C). However, the later question on thecomplex place C is itself an open problem.

Next, we prove that the set Sn+1 = S2n−1R × C is a set of injectivity for the TSM for

a certain class of functions on Cn+1. Let z = (z′, zn+1) ∈ Cn+1. In order to prove thisresult, we need the following result from [22].

Theorem 5.3. (See [22].) Let f be a function on Cn such that e14 |z|

2f(z) ∈ Lp(Cn), for

1 � p � ∞. Suppose f × μr(z) = 0, ∀r > 0 and ∀z ∈ S2n−1R . Then f = 0 a.e. on Cn.

Remark 5.4. Since τη(f × μr) = τηf × μr. Therefore, the function space considered asin the above Theorem 5.3 is not twisted translation invariant, it follows that a spherecentered off the origin is not set of injectivity for the TSM on Cn.


Theorem 5.5. Let f be a function on Cn+1 such that e14 |z

′|2f(z) ∈ Lp(Cn+1) for1 � p � ∞. Suppose f × μr(z) = 0, ∀r > 0 and ∀z ∈ Sn+1. Then f = 0 a.e. on Cn+1.

Proof. Let k = β1 + β2 + · · · + βn+1 = |γ| + βn+1. Then, we can write

f × ϕnk

(z′, zn+1

)=

∑|γ|+βn+1=k

Fz′,γ ×(n+1) ϕ0βn+1

(zn+1).

By the given conditions, we have

f × ϕnk

(z′, zn+1

)=

∑|γ|+βn+1=k

Fz′,γ ×(n+1) ϕ0βn+1

(zn+1) = 0,

for all (z′, zn+1) ∈ Sn+1 and ∀k ∈ Z+. Since, the set {ϕ0β1

: βn+1 ∈ Z+} forms anorthogonal basis for L2(C), it follows that Fz′,γ ×(n+1) ϕ

0β1

(zn+1) = 0, ∀βn+1 ∈ Z+ and∀|γ| ∈ Z+. This in turn implies that for each fixed γ ∈ Zn

+, Fz′,γ ×(n+1) ϕ0β1

(zn+1) = 0,∀βn+1 ∈ Z+ and ∀(z′, zn+1) ∈ Sn+1. Once again using the orthogonality of the set{ϕ0

βn+1: βn+1 ∈ Z+} in L2(C), we conclude that Fz′,γ(zn+1) = 0, ∀(z′, zn+1) ∈ Sn+1 and

∀γ ∈ Zn+. Hence, we can write

∑|γ|=k

Fz′,γ(zn+1) =∑|γ|=k

∫Cn

f(z′ − w′, zn+1

) n∏j=1

ϕ0βj

(wj)ei2 Im(z′.w′) dw′ = 0,

∀k ∈ Z+ and ∀z′ ∈ S2n−1R . Therefore, in view of Theorem 5.3, we infer that f = 0

a.e. Cn. �Since the set Sn = Cn−1 × Σ2N is a set of injectivity for the TSM for Lp(Cn), with

1 � p � 2. As a dual problem, it is natural to ask that Sn is a set of density for Lq(Cn),for 2 � q < ∞. Let C�

c(C) denote the space of radial compactly supported continuousfunctions on C. Let τzf(w) = f(z − w)e i

2 Im(z.w).

Proposition 5.6. The subspace F (Sn) = Span{τzf : z ∈ Sn, f ∈ C�c(Cn)} is dense in

Lq(C), for 2 � q < ∞.

Proof. Let 1p + 1

q = 1. Then 1 � p � 2. By Hahn–Banach theorem, it is enough to showthat F (Sn)⊥ = {0}. Let g ∈ Lp(Cn) be such that

∫C

τzf(w)g(w) dw = 0, z ∈ Σ2N , ∀f ∈ C�c

(Cn

).

That is,

g × f(z) = f × g(z) = 0.


Let the support of f be contained in [0, t]. Then by passing to the polar decomposition,we get

t∫r=0

g × μr(z)f(r)r2n−1 dr = 0.

By differentiating the above equation, it follows that g× μt(z) = 0, ∀t > 0 and ∀z ∈ Sn.Thus by Theorem 5.1, we conclude that g = 0 a.e. on Cn. �6. Some remarks and related open problems

Remark 6.1. (a) In the article by Agranovsky et al. [7], it has been shown that boundaryof any bounded domain in Cn is a set of injectivity for the TSM for a class of functions fsatisfying f(z)e( 1

4+ε)|z|2 ∈ Lp(Cn), for some ε > 0 and 1 � p � ∞. However to provethis result for ε = 0 is an open problem. The sphere S2n−1

R is an example with ε = 0, asmentioned in Theorem 5.3.

We are thinking to do away with exponential condition. Though it is not possiblefor sphere, because of the functional relation ϕn−1

k × μr(z) = B(n, k)ϕn−1k (r)ϕn−1

k (|z|).We are working for the real analytic curves γ having non-constant curvature can be thesets of injectivity for the TSM for Lq(C) with 1 � q � 2. We know that the spectralprojections Qk = f × ϕn−1

k have a real analytic expansion for Qk as

Qk(z) =k∑

p=0Cp0

k−pzpϕp

k−p(z) +∞∑q=0

C0qk zqϕq

k(z).

Suppose Qk(z) = 0, ∀k ∈ Z+, and ∀z ∈ γ. For a curve γ(t) = r(t)eit of non-constantcurvature, radius r(t) will vary in an interval. The fact that Qk is a real analytic function,Qk(γ(t)) must be a real analytic function, which vanishes over an interval. It is interestingto know that can Qk(γ(t)) will vanish for all t ∈ R. In particular, it can be possiblewhen r(t) is a non-periodic function. For example, spiral γ(t) = {(et cos t, et sin t): t ∈(−∞, 0]}. Hence spiral is a set of injectivity for the TSM for Lq(C) with 1 � q � ∞.For a more general example, let γ(t) = r(t)eit, where r(t) be a non-periodic real analyticfunction on [0,∞) with limt→∞ r(t) = 0. Then γ(t) is a set of injectivity for the TSMfor Lq(C) with 1 � q � ∞. Moreover, γ(t) is a determining curve for any real analyticfunction on C.

(b) Let K be a non-trivial connect subgroup of U(n). For non-zero vector z0 in Cn,the K-orbit of z0 is denoted by Kz0 = {k.z0: k ∈ U(n)}. We know that the sphere S2n−1

which orbit is unit vector, is a set of injectivity for the TSM for certain class functionsin Lp(Cn). Suppose f × μr(z) = 0, for all r > 0 and for all z in Kz0 . Whether it impliesf = 0 is an interesting question. In an article by Ratnakumar et al. [28], they havegeneralized the idea of two radii theorem for the K-orbit.


(c) We are working for the set S = R×R∪R×iR can be set of injectivity for the TSMfor Lq(C2), 1 � q � 2. We start with a small class of functions f(z1, z2) = zp1ϕ

0α1

(z1)h(z2).Suppose f × μr(z1, z2) = 0, ∀r > 0 and ∀(z1, z2) ∈ R × R ∪ R × iR. Then f = 0 a.e.on C2. If it goes to happen that S is a set of injectivity for the TSM for Lq(C2) then itwould be a surprise contrast to the sets of injectivity for the Euclidean spherical meanson R4, where the minimal dimension of a set of injectivity is three.

(d) We are working to locate the sets of non-injectivity for the spherical means onreal hyperbolic spaces Bn (n � 2). To characterize all sets of non-injectivity for thespherical means on Bn, is an open problem. For nonzero function f ∈ Cc(Bn), defineS(f) = {x ∈ Bn: Msf(x) = 0, ∀s > 0}. Then S(f) =

⋂∞k=0 Q

−1k (0), where

Qk(x) =∞∫

s=0

(sinh s)2k+n−1Msf(x) ds.

Using the fact that spherical mean satisfies the Darboux equation, we derive a recursionrelation LkQk = (2k + n − 2)Qk−1, where Lk = Lx − 2k(2k + n − 1)Id. Since all ofQk cannot be identically zero, therefore, it follows that there exists the least positiveinteger ko such that Qko

�≡ 0. Hence LkoQko

= 2ko(2ko + n − 1)Qko−1 = 0. Thus, weinfer that LxQko

= 2k(2k + n − 1)Qkoand S(f) ⊆ Q−1

ko(0). As the Laplace–Beltrami

operator L is an elliptic operator, therefore, its eigenfunction Qkois a real analytic

function (see [19]). Since the zero set of a real analytic function is a very thin set, it followsthat the sets of non-injectivity for the spherical means on Bn lie on some real analyticsurface. However, it remains difficult to locate exactly the sets of non-injectivity for thespherical means. We are further analyzing the function Qko

to reach on a conclusion.In general, a conjecture by Zalcman et al. [9], “sets of non-injectivity for the Euclideanspherical means are contained in the zero set of a certain eigenfunction of the Laplacian”,is still unsolved for the spherical means for L1

loc(Bn).(e) It is interesting and long standing open problem to locate the sets of non-injectivity

for the spherical means on Rn (n � 2). To characterize all sets of non-injectivity for thespherical means on Rn, is an open problem. Consider a nonzero function f ∈ L2(Rn) ∩C(Rn) and define S(f) = {x ∈ Rn: Rf(x, s) = 0, ∀s > 0}. Then S(f) can have animplicit description like S(f) =

⋂∞k=0 Q

−1k (0), where

Qk(x) =∞∫

s=0

Rf(x, s)s2k+n−1e−s22 ds.

However, knowing the geometry of Q−1k (0) is a difficult problem. In fact, there are some

other implicit descriptions for S(f) to exist (see [6]).(f) Let us rewrite Qkf = f ×ϕn−1

k . From the explicit expression of Qk, given by (4.2),it follows that


‖Qkf‖22 =

k∑p=0

∞∑q=0

∥∥Y kpq(f)

∥∥22

∥∥ϕn+p+q−1k−p

∥∥22, (6.1)

where Y kpq(f)’s are spherical harmonics depending upon f . In the work [34], Stempak and

Zienkiewicz have established that for f ∈ Lr(Cn), with 1 � r < 2(2n+1)2n+3 , the operators

Qk satisfy the estimate ‖Qkf‖2 � Ck‖f‖r. On the basis of the equality (6.1), it is naturalto ask, whether the map f → f × ϕn−1

k would satisfy an end point estimate.(g) Let μ be a finite Borel measure which is supported on a curve γ and S be a

non-empty set in R2. Then the pair (γ, S) is called a Heisenberg Uniqueness pairs (HUP)for μ if its Fourier transform μ(x, y) = 0, ∀(x, y) ∈ S, implies μ = 0. In a recent work [16],Hedenmalm et al. prove the following result. Suppose μ is supported on the hyperbolaγ = {(x, y): xy = 1} and μ vanishes on the lattice-cross S = αZ × {0} ∪ {0} × βZ.Then μ = 0 if and only if αβ � 1, where α, β ∈ R+. This is a variance of uncertaintyprinciple for Fourier transform. In view of this and the fact that ϕ0

k × μ is real analytic,it is natural to ask the following question. Let μ be a finite measure supported on a realanalytic curve γ and S be a non-empty set in C. Then find all those non-trivial pairs(γ, S) such that ϕ0

k × μ(z) = 0, ∀z ∈ S and ∀k � 0. It implies μ = 0. Here, we skip towrite further details about these ideas and they might be appear in the successive work.

Concluding remarks. We would like to point out the key motivation behind X-axistogether with Y -axis is a set of injectivity for the TSM on C. Consider the functionQ(z) = c0z + c1z + c2z

3 + c3z3. Suppose Q(x) = Q(ix) = 0, ∀x ∈ R. Then Q ≡ 0. This

result can also be interpreted on the Heisenberg group for the spherical means givenby (2.1). The set Σ2N =

⋃2N−1l=0 {(ωlx, t): x, t ∈ R} is a set of injectivity for the spherical

means on H1.

Acknowledgments

The author wishes to thank E.K. Narayanan and S. Thangavelu for several fruitfuldiscussions, specially, during my short visit to IISc, Bangalore. The author would liketo extend a sincere thank to M.L. Agranovsky for his reasonable suggestion and re-marks. The author would also like to gratefully acknowledge the support provided bythe Department of Atomic Energy, Government of India (HRI/4013/1866).

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coxeter system of lines and planes are sets of injectivity for the twisted spherical means

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