Asymptotic Expansions of Finite Hankel Transforms and the Surjectivity of Convolution Operators

A compactly supported distribution is called invertible in the sense of Ehrenpreis-H\"ormander if the convolution with it induces a surjection from $\mathcal{C}^{\infty}(\mathbb{R}^{n})$ to itself. We give sufficient conditions for radial functions to be invertible. Our analysis is based on the asymptotic expansions of finite Hankel transforms. The dominant term may be the contribution from the origin or from the boundary of the support of the function. For the proof, we propose a new method to calculate the asymptotic expansions of finite Hankel transforms of functions with singularities at a point other than the origin.


Introduction
It is known that are both surjective, where P (D) ̸ = 0 is an arbitrary linear partial differential operator with constant coefficients [6,7,14].Since these mappings coincide with the convolution operator P (D)δ * on respective spaces, a natural question is to characterize compactly supported distributions that induce surjective convolution operators.Such a distribution is called invertible.Notice that δ(x − a) induces translation and hence is invertible.It was found that a compactly supported distribution is invertible if and only if its Fourier transform, an entire function, is slowly decreasing in a certain sense.This condition was found by Ehrenpreis [8] and was further studied by Hörmander [11].A self-contained account can be found in [12].Invertible distributions and slowly decreasing entire functions became and still are fundamental concepts in the research of convolution equations.Perturbation of invertible distributions are studied in [15], and a recent paper [5] investigates invertible distributions in an abstract setting.Convolution equations on symmetric spaces are discussed from the viewpoint of invertibility in [4].Slow decrease is a technical estimate from below and is not easy to grasp.The present authors expect that the notions of slow decrease and of invertibility will become less mysterious if many examples or sufficient conditions are found.The delta function δ S(0,r) supported on a sphere is invertible [13].Its normal derivatives are invertible as well [16].In [13,16], the asymptotic behavior of Bessel functions were used.As is well known [18], the Fourier transform of a radial function can be written in terms of an integral involving a Bessel function.The Fourier transforms of δ S(0,r) and its normal derivatives are limit cases.
In the present article, we prove the invertibility of some radial functions by using two methods of calculating asymptotic expansions of finite Hankel transforms 1 0 φ(s)J ν (rs)ds.One is a result in Wong [20], in which singularities at s = 0 determine the asymptotic behaviors.The other method, devised by the present authors, is useful to deal with singularities at s = 1.As far as they know, this is the first result about the asymptotic behaviors of Hankel transforms of functions singular at a point other than s = 0.
Notice that various additive formulas for supports and singular supports of convolutions and spaces of entire functions with a certain type of slow decrease are investigated in [3] and applied to study the surjectivity of convolution operators.Several examples are discussed in it and the characteristic function of an ellipsoid is studied by using a Bessel function.The tools developed in the present paper could be useful in that line of research.

Invertibility and slow decrease
In this article, we follow the convention for the definition of the Fourier transform of a compactly supported distribution ) is a radial function, its Fourier transform can be written in terms of an integral involving a Bessel function.
Theorem 2.1 ([18, p. 155]).Let f 0 (s), s > 0, be a function of a single variable.The Fourier transform of f (x) = f 0 (|x|), x ∈ R n , is written in terms of a Hankel transform.More precisely, Here J n/2−1 (z) denotes the Bessel function of the first kind of order n/2 − 1.
Remark 2.2.Notice that z −(n/2−1) J n/2−1 (sz) is an even entire function in a single variable z for each s > 0. If the support of f 0 (s) is bounded, then the Fourier transform of f (x) = f 0 (|x|) is an even entire function which is expressed by We employ the notions of invertibility and slow decrease originated by Ehrenpreis [8] and refined by Hörmander [11].(i) There is a constant A > 0 such that we have (ii) There is a constant A > 0 such that we have for any ξ ∈ R n .
(iii) If w ∈ E ′ (R n ) and ŵ/û is a holomorphic function, then ŵ/û is the Fourier transform of a distribution in Other classes of invertible distributions are discussed in [1].Let u be a compactly supported measure with an atom, v ∈ E ′ (R n ) have singular support disjoint from that of u and P (D) be a non-zero linear partial differential operator with constant coefficients.Then P (D)u + v is invertible.
Another useful fact in [1] is the following.
If f is real analytic in a neighborhood of singsupp u and f u is invertible, then u is invertible.In [13,16], the spherical mean value operator and its variants are discussed from the view point of invertibility.Proposition 2.7.
, where P j (D) is a non-zero linear partial differential operator with constant coefficients and Proof .The proofs of (i)-(iv) are easy.Theorem 2.3 (ii) and the Paley-Wiener-Schwartz theorem imply (v).Finally, (vi) follows from Theorem 2. .
On the other hand, we set S = {η ∈ R n ; |η| < B +A log(2+B)} ⊃ S(ξ 0 ).Here we may assume A is so large that A log(2 + B) > B. Then Proposition 2.11.Let p(z) be an even entire function of a single variable.Set q(x) = x α p(x) for x > 0, α ≥ 0. Assume there is a sufficiently small constant C > 0 and sufficiently large constants A, B > 0 such that we have Proof .The assertion is trivial when α = 0 (see Proposition 2.10).We have only to prove the case α > 0. By choosing a larger B if necessary, we may assume that On the other hand, B < A log(2 + x) holds if B/ log 2 ≤ A. We see that p(z) is slowly decreasing since (2.7) is valid if we adopt max{A+2α, B/ log 2} as a new value of A. Apply Proposition 2.10 to complete the proof.■
It is natural to assume c 0 ̸ = 0 as in (Φ 2 ) and we adopt it as a convention.See (4.3).If one wants to study the case φ(s) ∼ 0, one has only to consider the difference of two functions with the same nontrivial expansion.

Asymptotic expansion and invertibility
First, we utilize Wong's results reviewed in the previous section to consider the contribution of the singularities at s = 0 to Hankel transforms.We consider an infinitely differentiable function φ(s) and multiply it by a cut-off function.Therefore, m in (Φ 2 ) can be arbitrary and the conditions of uniform convergence are satisfied.The k-th coefficient in the right-hand side in (3.1) vanishes if and only if c k = 0 or 1 2 (µ + k − ν − 1) is a nonnegative integer.Notice that 1  2 (µ + k + ν + 1) cannot be a pole of the Gamma function since Re(µ + k + ν + 1) > k ≥ 0. This observation motivates us to introduce the set K in the proposition below.Proposition 4.1.Let µ, ν ∈ C and φ(s) be an infinitely differentiable function in (0, 1).We make the following assumptions: Moreover, let χ 0 (s) be an infinitely differentiable function such that χ 0 (s) = 1 in 0 < s ≤ ε and χ 0 (s) = 0 in 1 − ε ≤ s < 1, where 0 < ε < 1/3.We define the set where N 0 is the set of nonnegative integers.Then, we have the following.
as r → ∞, where A is an arbitrarily large real number.
where A is an arbitrarily large real number.
Proof .The set K is empty in this case, since we have µ = ν + 1 and k is even if c k ̸ = 0. ■ Next we consider the contribution of the singularities at s = 1 to finite Hankel transforms.
Combining this formula and integration by parts, we get if k ≤ N − 1.We obtain for N ≥ 1.Notice that (4.7) holds for N = 0 as well.On the other hand, it is well known that as R ∋ z → ∞ (see, for example, [17, formula (10.17.3)]).By (4.8) and the boundedness of J ν+N (z) as z → +0, there exists a bounded function R(z) (0 < z < ∞) such that We have The first term in the right hand side is of order o r −1/2 as r → ∞ by the Riemann-Lebesgue lemma.The second term is of order O r −3/2 by the boundedness of R. We have shown and the combination of it and (4.7) gives I 0 = o r −(N +1/2) .■ Proposition 4.4.Let N be a nonnegative integer and ν, λ 0 , . . ., λ m , Λ, a 0 , . . ., a m be complex numbers.Assume where a 0 ̸ = 0 and ψ(t) is an infinitely differentiable function in Then as r → ∞, Proof .By [9, formula (26(33)a)] or [10, formula (6.567.1)],we have if r > 0, Re ν > −1, Re α > −1. 3 We employ this formula and Proposition 4.3 to obtain We finish the proof by using (4.8).■ Now we give our main results.
for some A ℓ .The second sum, which corresponds to s n/2 m k=0 a k 1 − s 2 λ k , does not contribute to the finite Hankel transform because of Proposition 4.2.
Let K n = K(µ, ν(n), {c k } k ) be the set defined by (4.4) with ν = ν(n).If K n ̸ = ∅, we have and if K n = ∅, H 0 decreases rapidly of arbitrary order by (4.5).On the other hand, for φ defined in Proposition 4.4, we have We can prove that H 0 + H 1 satisfies (2.8) in the following way.
Notice that φ(s) − φ(s) − s n/2 φ s 2 vanishes near s = 0, 1 and The finite Hankel transform 1 0 s λ 1 − s 2 ρ−1 J ν (rs)ds can be written in terms of the generalized hypergeometric function 2 F 3 by [10, formula (6.569)] and it is good enough to prove invertibility.The advantage of our method is that it is stable under small perturbations and works even if no closed form expression is available.Remark 4.7.In (4.2), j can be arbitrarily large.This assumption can be relaxed in some cases.We give such an example.Assume We do not assume that term by term differentiation is possible:4 m in (Φ 1 ) is 0.Moreover, we have removed the assumption c 0 ̸ = 0 and do not need the set K n .We keep all the other assumptions of Theorem 4.5 unchanged.By (3.1) and Remark 3.2, we have H 0 = o(1).If Re λ 0 ≤ −3/2, H 1 is dominant and f (x) defined by (4.10) is invertible.
, where P (D) ̸ = 0 is a linear partial differential operator with constant coefficients.It give rise to P (D)δ(x) * , which is nothing but P (D).So P (D) : for any ξ ∈ R n satisfying |ξ| ≥ B, if and only if p(ζ) is slowly decreasing in the sense of Definition 2.8.Proof .The "if" part is trivial.We prove the "only if" part.If |ξ 0 | = B, we set 3 (ii).Indeed, if u and v satisfy (2.1) with a common constant A, then u ⊗ v satisfies (2.1) with 2A instead of A. ■ Definition 2.8.In the present paper, we extend the terminology of Definition 2.5 slightly: an entire function p(ζ), not necessarily the Fourier transform of a compactly supported distribution, is called slowly decreasing if there is a constant A > 0 such that we have sup |p(η)|; η ∈ R n , |η − ξ| < A log(2 + |ξ|) > (A + |ξ|) −A (2.2) for any ξ ∈ R n .Proposition 2.9.Let p(ζ) be an entire function.There are constants A, B > 0 such that sup |p(η)|; η ∈ R n , |η − ξ| < A log(2 + |ξ|) > (A + |ξ|) −A (2.3) For any ξ ∈ R n , set x = |ξ|.The radial function p(|η|), η ∈ R n , satisfies the condition in Proposition 2.9, since the combination of then (2.2) holds if we replace A with Ã. ■ Proposition 2.10.Let p(z) be an even entire function of a single variable.If p(z) is slowly decreasing, then p ζ 2 is a slowly decreasing function in ζ ∈ C n .Proof .Notice that p ζ 2 is well-defined since p(z) is even.There are constants A, B > 0 such that we have sup |p(y)|; y ∈ R, |y − x| < A log(2 + x) > (A + x) −A (2.7) for any x ∈ R satisfying x ≥ B.