1 Introduction

It is well known[1,7,19]that any homogeneous polynomial of degree k,p ∈Pk,can be decomposed,in a unique fashion,as

where

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the notation Hk being used to denote the harmonic homogeneous polynomials of degree k.

One can easily find the projections πk(p)and χk(p).For example,if we apply the Laplacian to(1.1)we readily obtain so that

Interestingly,these projections appear in other,somewhat surprising places.Indeed,as explained in the section Spherical Harmonics via Differentiation of[1,Chapter 5],whenever a homogeneous differential operator of degree k is applied to r2−n in Rn one obtains an expression of the form u(x)r2−n−2k where u is not just homogeneous of degree k,but actually belongs to Hk.In fact,more is true,since u=(2−n)(−n)···(−n−2k+4)Y,that is,if p∈Pk and we denote(2−n)(−n)···(−n−2k+4)as An,k then

and in particular if Y∈Hk then

Let us start with the case whenHere the Funk-Hecke formula[3,(6.6)]yields,

Furthermore,we show that harmonic polynomials are also obtained when we take the derivatives of multipoles§Such harmonic multipoles have received increasing attention in recent years[2];see also[18].They play a fundamental role in the ideas of the late professor Stora on convergent Feyman amplitudes[17,21].of the form for some harmonic polynomial Y'∈Hk'.Indeed we obtain formulas for the derivativesof the principal value distribution p.v.and show that the ordinary part is a multipole of the form Z(x)/r2k'+2k+n−2 for some Z∈Hk+k'.

On the other hand,is there something special about the exponent 2−n?,that is,what happens if one computes p(∇)(rα)or more generallyAs we explain in the Section 6,for a general α the simplicity of(1.4)is just1Nevertheless,one can easily recover Y from the more complicated formula,as we explain in Section 6.,but(1.5)remains almost the same for any α.Actually if we replace r2−n by a more general function f(r)(1.5)remains basically the same but(1.4)is lost;this is a consequence of the Funk-Hecke formula[3,9,12].

2 Preliminaries and notation

We assume that we work in Rn with n≥3;results in the case n=2 are usually true too,but even if true the proofs sometimes require modifications.We shall employ the notations

由式(4)第1式、式(11)、式(12)、以及图1的△OA1B1、△OA2B2知，当φ为φ1、φ2时，φ为0、180，δ为δmax、δmin，α为α1、α2，注意到式(13)即知φ1、φ2是αmax可能出现位置。

Notice thatis the surface area of the unit sphere of Rn.

For results about distribution theory we refer to the textbooks [5,10,15,16,20],but give a summary of some important ideas below.The moment asymptotic expansion[5,Chapter 4]tell us that if g is a distribution defined in the whole Rn that decays very fast at 2Technically rapid decay means The expansion certainly holds if g has compact support.then g(λx)has the asymptotic expansion

where the constantsµα are the moments of g,

The notion of the finite part∗∗∗∗Hadamard introduced the notion of the finite parts,and the name,when considering the divergent integrals that appear in of the fundamental solutions of hyperbolic equations[11].of a limit[5,Section 2.4]is as follows.Suppose is a family of strictly positive functions defined for 0<ε<ε0 such that all of them tend to infinity at 0 and such that,given two different elementsthenis either 0 or ∞.

Definition 2.1.Let G(ε)be a function defined for 0<ε<ε0 withThe finite part of the limit of G(ε)as ε →0+with respect to F exists and equals A if we can write††Such a decomposition is unique since any finite number of elements of F is linearly independent.G(ε)=G1(ε)+G2(ε),where G1,the infinite part,is a linear combination of the basic functions and where G2,the finite part,has the property that the limit A=limε→0+G2(ε)exists.We then employ the notation

The Hadamard finite part limit corresponds to the case when is the family of functions ε−α|lnα|β,where α>0 and β≥0 or where α=0 and β>0.We then use the simpler notation F.p.limε→0+G(ε).

Consider now a function f defined in Rn that is probably not integrable over the whole space but which is integrable in the region|x|>ε for any ε>0.Then the radial finite part integral‡‡One should call the procedure(2.4)a radial finite part integral,since the use of the variable r means that f has been replaced by 0 inside a ball of radius ε.The results when solids of other shapes are removed could be very different[6,13,22].is defined as

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if the finite part limit exists.When the ordinary limit exists we call the integral a radial principal value integral and use the notationIf the finite part integralsfor all test functions φ that belong to a certain space such as or then we can define the distribution

We shall now consider how special is the exponent 2−n for obtaining spherical harmonics by differentiation.We shall see that if and we write in the form then in general To simplify our analysis we just consider the case when but the result remains true if α is not an even integer(a trivial case)unless α=2−n.

that is(5.4)with and But

Definition 2.2.Let be a distribution defined in the complement of the origin.Suppose the pseudofunction exists in Let be any regularization of f0.Then the delta part at 0 of f is the distribution whose support is the origin.We call the ordinary part of f.Notice that this delta part is in fact a spherical delta part.

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It must be emphasized that even though Pf(f0(x))is in a way the natural regularization of f0,other regularizations appear also very naturally.For example,consider the distribution Pfr−kin Rn;then the distributional derivativeis a regularization of the ordinary derivative of r−k;however[4,(3.16)]if k−n=2m is an even positive integer,then

where cm,n is given by(2.1),so thatis the delta part of the distributionSimilarly,in Rn for n≥2,and for m∈N,the distribution is a regularization of r−n−2m and in D'(Rn)its delta part is ln

3 The distributional derivative

Our aim in this section is to prove that the distributional derivative where Y∈Hk,does not have a delta part,that is,

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where An,k=(2−n)(−n)···(−n−2k+4).We shall give two proofs,both of which give important insight into the topic.Now we give our first proof,using induction;we give another proof in Section 5.

Let us recall[2]that if Y ∈Hk then the pseudofunction is actually a principal value,

an ordinary limit,not just a finite part limit. We emphasize[2]that in general this is not true if we replace Y by an arbitrary polynomial of p ∈Pk,nor it holds for exponents different from −n−2k+2 if the distribution is not regular. Notice also thatbutis not a principal value if k>0.

If we apply the moment asymptotic expansion(2.2)to a distribution of the type g(ω)δ(r−1),where being polar coordinates in Rn,we obtain the estimate .When multiplying with a harmonic polynomial the estimate is improved quite a bit.

其中In′为n采用an′策略的干扰用户集.为了简化公式推导，用表示集合X中用户满意度值改变量，则(12)式可简写为：

Lemma 3.1.Let Y∈Hk.Then as ε→0+

Proof.Indeed,the expansion of Y(ω)δ(r−ε)=ε−1 f(x/ε)as ε →0+,where f(x)=Y(ω)δ(r−ε),is obtained from(2.2)as

where are the moments.But since Y is a harmonic polynomial of degree k,we have[1]thatµα=0 whenever|α|

A different derivation of(5.2),that gives us extra information,is as follows,

Proposition 3.1.If Y∈Hk thendoes not have a delta part,that is,(3.1)holds.

Proof.If Y∈Hk then we can write

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where hY is a distribution concentrated at the origin,that depends on Y.Therefore,hY=for some polynomial qY.The Funk-Hecke formula[9,12],as presented in[3],shows that there are constants λk,that depend on k but not otherwise on Y,such that

We will prove by induction on k that λk=0 for all k.The result is of course true if k=0.Therefore,we assume that λk=0 and prove that λk+1=0.In order to do so it is enough to show that for just one harmonic polynomial Yk+1 ∈Hk+1 we have qYk+1=0;we will do it if where Y ∈Hk depends only on That there is at least one non zero Yk+1 ∈Hk+1 of this form is true because n≥3.We have,

and since distributional limits and distributional derivatives can be interchanged,

so that,remembering that Y(x)xn is harmonic and(3.3)becomes the limit as ε→0+of

and thus as required.

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4 The distributional derivative

We now consider the distributional derivative where p is a general homogeneous polynomial of degree k,p ∈Pk.We already know that the ordinary part ofdepends only on Y,the projection of p onto Hk.What we shall now show is that the delta part does not depend on Y at all,that is,it is a function of only p−Y.

Proposition 4.1.Let p∈Pk.Write p=Y+r2q,where Y=πk(p)∈Hk,and where q=χk(p)∈Pk−2.Then

Proof.Indeed,From the Proposition 3.1 we know that the first term equals Furthermore,it is well known[7]that|∇|2=∆,the Laplacian,and thatPutting these results together we immediately obtain(4.1).

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It is useful to evaluate(4.1)at a test function.Thus if φ∈S(Rn)we obtain the formula

Notice,in particular,the extreme cases,

valid if Y∈Hk,and

that holds for all q∈Pk−2.

It is also interesting to observe that if we write the general distributional derivativeas a sum of two parts,ford(x;ε),its locally integrable part,and fsing(x;ε),its singular part,a distribution concentrated on the sphere r=ε,then the limit of ford(x;ε)as ε→0+exists,but—unless p=Y—it does not equal

5 Another proof and a generalization

We shall write for either the regular distribution rα,when locally integrable,or the pseudofunction if not.Since all distributional derivatives ofare also pseudofunctions,or regular,so that there are no delta parts[2].In a sense the formulas for the ordinary and for the distributional derivatives are the same!

The constant Wn,k,0 equals 1 if k=0 and the product n···(n+2k−2)if k>1.

Notice that the convolution of the multipolesand r2−n is well defined as a distribution because of the behavior of the functions at infinity.We can then compute the distributional Laplacian of this convolution in two different ways,namely

and

Comparison of the two results thus gives

giving another proof of(3.1)since(−1)k+1(2−n)Wn,k,0/(n+2k−2)equals the product(2−n)(−n)···(−n−2k+4),that is,An,k.

These ideas actually allow us to generalize(3.1).

Proposition 5.1.Ifand then

Proof.We compute the distributional Laplacian of the convolution of the two multipolesand p.v.in two ways—using(5.1)—as we did above,equate the results,and simplify.

Notice that formula(3.1)corresponds to the case when

We can now give our first proof of(3.1).

whereandIn generalis not a pseudofunction,only when YY' is harmonic.

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We can also give a generalization of the Proposition 4.1.

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Proposition 5.2.Letwith and let .If and q=then

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Proof.Write p=Y+r2q1.Then,employing(5.3),with and and(5.1)we obtain

In general it is not possible to separate the contribution to a distribution from a given point;to talk about the delta part at x0 of all distributions does not make sense(The delta part will not exist,in general.For instance,the function sinr−k is locally integrable in Rn,and thus it gives a well defined regular distribution in D'(Rn).If k>n,the distributional derivativeis another well defnied distribution,but its delta part at the origin is not defined,since does not exist.).However,sometimes,we can actually separate the delta part[2].

so that and ,as required.

We have therefore encountered another instance where spherical harmonics are obtained by differentiation.Indeed,whenever and the ordinary derivativeis always of the form for a harmonic polynomial Z and some constant c that depends on n,k,and Will this be the case for derivatives of the type for The answer is no:Just computing simple cases of this derivative for a that is not harmonic will convince the reader that if we write the result as for some then in general

6 The derivatives p(∇)(rα)

The notation Pf(f(x))was introduced by Schwartz[20],who called it a pseudofunction.

We now present an alternative proof of the Proposition 3.1.It is based on the formula[2,Prop.5.5]

Several further questions arise,however.First,since the function r2−n is singular at the origin,these formulas hold in Rn{0}but not in all Rn,so what are the corresponding formulas for the distributional derivatives†Following Farassat[6]we denote distributional derivatives with an overbar,namely,and so on.and that is,the corresponding formulas in the whole space‡Distributional derivatives of this kind play an important role in Physics;the distributional derivativeswere given by Frahm[8],and can be found in the textbooks[14]..Curiously,while in general will contain extra terms,namely a delta part,the distributional expression remains basically equal to(1.5)since does not have a delta part;delta parts and ordinary parts of a distribution are explained in Section 2. We give two different proofs of the formula forone by induction in Section 3 and another in Section 5.We also consider the distributional derivativein Section 4,showing that in general the ordinary part of this derivative depends only on Y,while the delta part depends only on q.

so that as in the case of the exponent 2−n,the derivative gives us a constant multiple of times Actually this is the case not only for powers of r,since for any radial tempered distribution f0,f0(x)=f(r)for f a distribution of one variable,the Funk-Hecke formula[3,Theorem 6.4]gives

where is the distributional operator given by

where Again is a multiple of Yk(x),the same for all Y∈Hk.

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On the other hand[3,(6.9)]

where

Therefore if p ∈PkHk and we writein the formthen in general

Clearly,even though is not a multiple of it is very easy to find Y from

for some constant cα.However,the simplicity of the formula(1.4)is lost.

We finish with a generalization of the Proposition 5.1.

Proposition 6.1.If Y∈Hk and then

Proof.Indeed,if we use(6.2)we see that both sides equal

Notice that,in particular,if both harmonic polynomials are of the same degree,then replacing α−2k by α we obtain

if

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