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Remarks on the general Funk-Radon transform and thermoacoustic tomography 7 0 0 2 V. P. Palamodov n a 07.01.2007 J 7 ] P 1 Introduction A . h ThegeneralizedFunktransformisaintegraltransformactingfromdensitiesonamanifold t a X to functions defined on a family Σ of hypersurfaces in X. The dual operator M◦ is m again a Funk transform which is defined for densities on the manifold Σ. We state two- [ side estimates for the operator M and a description of the range of the Funk transform 1 operatorM andapproximationtheoremforthekernelofthedualoperator. Thisoperator v 4 is similar to the integral transform related to a double fibration in the sense of Guillemin 0 [4] and some results can be extracted from his theory. Other results can be generalized 2 1 for the general double fibration. 0 7 0 2 Geometry / h t a Let X and Σ be smooth n-manifolds, n > 1 and F be a smooth closed hypersurface in m X ×Σ such that : v (i) the projections p : F → X and π : F → Σ have rank n. This condition implies i X that the sets F (σ) = π−1(σ),σ ∈ Σ and F (x) = p−1(x),x ∈ X are hypersurfaces in r X, respectively, in Σ. We call F incidence manifold. It can be defined locally by the a equation I(x,σ) = 0, where I is a smooth function in X × Σ such that dI 6= 0. If the hypersurface F is cooriented in X ×Σ, a function I can be chosen globally; we call it incidence function. If F is not cooriented, one can choose local incidence functions I ,α ∈ A such that I = ±I in the domain, where both functions I , I are defined. α β α α β The additional condition is (ii) the mapping q : F → Gn−1(X) is a local diffeomorphism, where Gn−1(X) = ∪ Gn−1 and Gn−1 means the manifold of n−1-subspaces in the tangent space T of X X x . x x at x; q(x,σ) = (x,H), where H denotes the tangent hyperplane to F (σ) at x. It follows that for any point x ∈ X and any tangent hyperplane H ⊂ T there is locally only one x hypersurface F (σ) that contains x and is tangent to H. 1 Proposition 2.1 The conditions (i-ii) are equivalent to the inequality detΦ 6= 0 in F, where ∂2I ... ∂2I ∂I ∂x1∂σ1 ∂x1∂σn ∂x1   ... ... ... ...   Φ =  ,    ∂2I ... ∂2I ∂I     ∂xn∂σ1 ∂xn∂σn ∂xn       ∂I ... ∂I I   ∂σ1 ∂σn  where x ,...,x and σ ,...,σ are local coordinates in X and in Σ, respectively. 1 n 1 n Proof. Suppose that detΦ 6= 0. Then also d I 6= 0 and d I 6= 0 which implies (i). x σ Choose a point (x ,σ ) and take a tangent vector θ to F (σ ) at x . The vector (θ,0) is 0 0 0 0 tangent to F at (x ,σ ) and the map q is well defined. Change the coordinates x and 0 0 i σ ,i = 1,...,n in such a way that ∂I(x ,σ )/∂σ = ∂I (x ,σ )/∂x = 0,i = 2,...,n at i 0 0 i 0 0 i this point. We have then ∂I ∂I . ∂2I n detΦ = detΨ, Ψ = . ∂x ∂σ (cid:26)∂x ∂σ (cid:27) 1 1 i j i,j=2 The inequality detΨ 6= 0 implies that the forms ∂d I(x ,σ)/∂σ ,...,∂d I(x ,σ)/∂σ x 0 2 x 0 n are independent. This means that the fields ∂/∂σ ,...,∂/∂σ do not move the point x 2 n 0 but rotate the tangent hyperplane to F (σ ) at x whereas the field ∂/∂σ move the point 0 0 1 x . This yields (ii). The inverse statement can be proved on the same lines. ◮ 0 It follows that the properties (i-ii) are symmetric with respect to X and Σ. The next condition is not symmetric: (iii) the projection p : F → X is proper. If F satisfies (ii) and (iii) and the hypersurface F (x) is not empty for a point x ∈ X, then the mapping q : F (x) → Gn−1 is x surjective, since the manifold Gn−1 is connected. x 3 The Funk transform Consider a hypersurface F as above defined by a incidence function I that fulfils (i). Define the Funk (or Minkowski-Funk-Radon) transform for densities f in X with compact support by means of the integral . 1 f Mf(σ) = lim f = , σ ∈ Σ, (1) ε→0 2ε Z|I(·,σ)|≤ε ZF(σ) dxI where ω = f/d I is a n−1-form such that d I∧ω = f. This form is defined up to a term x x d I ∧χ, where χ is a n−2-form. Therefore the restriction of ω to the curve F (σ) is a x well-defined density. This density does not change, if we replace I by −I. Therefore the 2 Funk transform is well defined. The function Mf is also continuous. Suppose that the condition (iii) is fulfilled. If a density f is supported in a compact set K ⋐ X, then Mf . is supported in the compact set Λ = π(p−1(K)) ⋐ Σ. Example 1. Let X andΣ be unit spheres in Euclidean 3-spaces and the hypersurface F be defined by the global incidence function I(x,σ) = x σ +x σ +x σ . The operator 1 1 2 2 3 3 M coincides with the classical Minkowski-Funk transform, [3]. The Funk transform can be also defined on projective planes X = P2 = S2/Z ,Σ = P2, if we take ±I as 2 local incidence functions. The dual operator M◦ coincides with M through the natural ∼ isomorphism X = Σ. Example 2. Let (X,g) be a Riemannian 2-manifold with boundary and Σ be the family of closed geodesic curves γ. Take a density f = fdS, where dS is the Riemannian area form and f is a continuous function with compact support. Then we can write the geodesic integral transform as the Funk transform Mf(γ) = fds, Z γ if we take an incidence function I for the family F such that |∇ I| = 1. Any smooth x weight function w = w(x,σ) can be included, by replacing the function I to w−1I. 4 Above estimates The scale of Sobolev L -norms k·kα,α ∈ R is defined for functions supported in an 2 arbitrary compact set K ⊂ X. Fix a volume form dX in X and define kfkα = kf kga for a 0 density f = f dX withsupportinK.DenotebyHα (X,Ω), Hα (X)thespaceofdensities 0 K K (distributions), respectively, of (generalized) functions supported in K with finite norm k·k . For an arbitrary compact set Λ ⊂ Σ we define the spaces Hα(Σ,Ω), Hα(Σ) in the α Λ Λ same way. Proposition 4.1 For any family F that fulfils (i-ii), an arbitrary compact set K ⋐ X, any smooth function ε in Σ with compact support and any real α the inequality holds kεMfkα+(n−1)/2 ≤ C kfkα α for f ∈ Hα (X,Ω). K Proof. The Funk transform can be expressed as an oscillatory integral Mf(σ) = exp(2πıτI(x,σ))f(x)dτ. ZK ZR The critical set of the phase function τI(x,σ) is the hypersurface F (σ) and the condition d I 6= 0 implies that the phase function is non-degenerate. The corresponding conic x Lagrange manifold is L = {(x,σ,ξ,ρ) ∈ T∗(X ×Σ),I(x,σ) = 0,ρ = τd I,ξ = τd I,τ 6= 0}. σ x 3 Lemma 4.2 Rank of the matrix ∂(x,ξ) is equal to 2n in any point of Λ. ∂(σ,ρ) Proof of Lemma. Suppose that the rank is less 2n. Then there exists a vector t = (δx,δσ,δξ,δρ) in T∗(X ×Σ) tangent to L such that δσ = 0,δρ = 0. This yields dI(δx) = 0,δρ = δτd I +τd d I(δx), σ x σ δξ = δτd I +τd2I(δx) x x for a tangent vector δτ to R. The first line implies that the vector (τδx,δτ) fulfils the equation (τδx,δτ)Φ = 0. By Proposition 2.1 this vector vanishes, that is δx = 0,δτ = 0. The second line gives δξ = 0. ◮ Bythis Lemma theprojections ofL to T∗(X)andto T∗(Σ) aresubmersions. Inother terms, L is locally the graph of a canonical transformation. The symbol a(x,σ,ξ,ρ) = 1 is a homogeneous function of ξ,ρ of order 0. The order m of the Fourier integral operator M satisfies the equation We have m+dimX ×Σ/4−N/2 = 0, where dimX ×Σ = 2n and N = 1 is the number of variables τ. This yields m = (1−n)/2, which means that the functional . I(ψ) = exp(2πıτI(x,σ))ψ(x,σ)dτ ZΣZX ZR defined for a smooth densities ψ in X ×Σ with compact support, is a distribution of the class I(1−n)/2(X ×Σ,L) in the sense of Definition 25.4.9 of [6]. By Corollary 25.3.2 the operator εM defines a continuous map Hα (X,Ω) → Hα+(n−1)/2(Σ) for any real α, where K Λ Λ = suppe. ◮ Corollary 4.3 If F fulfils (i-iii), the Funk transform M can be extended to a bounded operator Hα (X,Ω) → Hα+(n−1)/2(Σ) for any α,K and Λ = π(p−1(K)). K Λ Due to (iii), we have suppMf ⊂ Λ and the cutoff factor ε can be dropped out. 5 Dual Funk transform Let ϕ be a density in Σ with compact support. Define the dual Funk transform as follows 1 ϕ M◦ϕ(x) = lim ϕ = . (2) ε→0 2ε Z|I(x,·)|≤ε ZF(x) dσI If F satisfies (iii), then the manifold F (x) is compact for any x ∈ X and the integral (2) is well defined for any continuous density ϕ in Σ. The natural pairing . (f,φ) 7→ hf,φi = fφ Z X is well defined for a densities f and functions φ on X provided one of them has compact support. 4 Proposition 5.1 The operator −M◦ is dual to M. Proof. We have f f ∧ϕ¯ ϕ¯ hMf,ϕi = M (f)ϕ¯ = ϕ¯ = = − f ∧ , Z Z Z d I Z d I Z d I Σ Σ F(σ) x F x F σ since dI = d I +d I = 0 on F. The right-hand side equals x σ ϕ¯ − f = − fM◦(ϕ¯) = −hf,M◦ϕi. ◮ Z Z d I Z X F(x) σ X 6 Backprojection and two side estimates Definition. Fix some area forms dX in X and dΣ in Σ. The back projection operator M∗ : g 7→ M◦(gdΣ)dX transforms functions defined in Σ to densities in X. Definition. We say that points x,y ∈ X are conjugate with respect to F, if x 6= y and the form d I(x,σ)∧d I(y,σ) defined in Σ vanishes. σ σ Theorem 6.1 If a family F has no conjugate points, fulfils (i-ii) and the condition: (iv) the projection q : π−1(Λ) → Gn−1(K) is surjective for some sets K ⋐ X,Λ ⋐ Σ, then for arbitrary cutoff function ε such that ε = 1 in Λ and any α > β the estimate kfkα ≤ C kεMfkα+(n−1)/2 +C kfkβ (3) α β holds for the Funk transform of densities f supported in K, where C and C do not α β depend on f. Lemma 6.2 The composition M∗εM is an elliptic PDO in K of order 1−n. Proof of Lemma. Write f = f dX, where f is a function supported in K and calculate 0 0 M∗εMf ε(σ)dΣ f(x) (y) = dX Z d I Z d I(x,σ) F(y) σ F(σ) y ε(σ)dΣ dX = f (x) 0 Z d I(y,σ) Z d I(x,σ) I(y,σ)=0 σ I(x,σ)=0 x ε(σ)dΣ = − f (x)dX, 0 Z Z d I(y,σ)∧d I(x,σ) I(y,σ)=0 I(x,σ)=0 σ σ since dI = d I+d I = 0 in F. We can write the right-hand side as A(y,x)f(x), where x σ R ε(σ)dΣ A(y,x) = − . (4) Z d I(y,σ)∧d I(x,σ) F(x)∩F(y) σ σ 5 The dominator does not vanishes for y 6= x, since there is no conjugate points. The quotient Q in (4) is well defined as n − 2-form up to an additive term d I(y,σ) ∧ σ S + d I(x,σ) ∧ R, where S and R are some n − 3-forms. The integral of this term σ along the smooth manifold F (x) ∩ F (y) vanishes and the function A(y,x) is a well defined and smooth, except for the diagonal. Near the diagonal we can write I(y,σ) = I(x,σ)+ (y −x )∂I (x,σ)/∂x +O |y −x|2 and i i i P (cid:0) (cid:1) ∂I(x,σ) d I(x,σ)∧d I(y,σ) = (y −x )d I(x,σ)∧d +O |y −x|2 . σ σ i i σ σ ∂x Xi i (cid:0) (cid:1) The forms d I(x,σ)∧d ∂I(x,σ)/∂x ,i = 1,...,n do not vanish and are linearly indepen- σ σ i dent, since of Proposition 2.1. Therefore the product d I(x,σ) ∧d I(y,σ) is bounded σ σ by c|x−y|frombelowasy → x.Therefore we have A(y,x) = a(y)|x−y|−1+O(1) near thediagonal, where aisasmoothpositive function. Thisimplies thatM∗εM isaclassical integral operator on K with weak singularity, moreover it is a pseudodifferential operator of order 1−n. It is an elliptic operator, since the symbol a is positive. ◮ Proof of Theorem. The support of the function M∗εMf is contained in the compact set p(suppε) ⊂ X. By Proposition 4.1 M∗ is (n−1)/2-smoothing operator, which yields kM∗εMfkα+n−1 ≤ CkεMfkα+(n−1)/2. (5) By Lemma 6.2 the operator M∗εM is elliptic of order n − 1, therefore the standard inequality holds kfkα ≤ C kM∗εMfkα+n−1 +C kfkβ α β for an arbitrary β and some constants C ,C . Taking in account (5) yields (3). ◮ α β Corollary 6.3 The eigenvalues λ of the operator M∗εM numbered in decreasing order k satisfy the estimate ck(1−n)/2 ≤ λ ≤ Ck(1−n)/2, k ≥ k . k 0 For the Radon transform the eigenvalues are calculated in [10]. Corollary 6.4 Suppose that for some β < α the equation Mf = 0,f ∈ Hβ (X,Ω) implies K f = 0. Then the two-side estimate holds: c kfkα ≤ kεMfkα+(n−1)/2 ≤ C kfkα. (6) α α Proof. The right-hand side inequality follows from Proposition 4.1. Suppose that the left-handsideestimatedoesholdfornoc .Thenthereexistsasequence {f } ⊂ Hα (X,Ω) α k K such that kf k ≥ kkMf k ,kf k = 1,k = 1,2,... (7) k α N α+1/2 k β The inequality (3) implies that kf k ≤ 2C for k > 2C and kMf k → 0. Because k α β α k α+1/2 the imbedding Hα (X,Ω) → Hβ (X,Ω) is compact, we can choose a subsequence (denote K K 6 itagain{f })suchthatf → ginHβ (X,Ω).ByProposition4.1kMf → Mgk → k k K k β+(n−1)/2 0, which implies Mg = 0. By the condition g = 0; it follows that kf k → 0 in contradic- k β tion with (7). ◮ Remarks. Mukhometov’s result [9] implies the estimate kfk0 ≤ CkMfk1 for the case n = 2. An estimate of this kind for more general situation was obtained by Sharafutdiniv [15], Ch. IV. Inequalities for Sobolev norms are well known for the Radon transform. Estimates for shift derivatives of order α+1/2 (n = 2) were obtained by several authors. Natterer [10] has shown that (6) holds also for angular derivatives. For the attenuated Radon transform see Rullgard [14]. OurapproachissimilartothatofLavrent’evandBukhgeim[7],wherethecomposition M∗M was described asanintegral operator inthe localcase. Guillemin [4],[5]has defined the ‘generalized Radon transform’ R for an arbitrary double fibration. This transfrom is treatedasanellipticFourierintegraloperatorandR∗Risshowntobeapseudodifferential elliptic operator under the ‘Bolker condition’. This condition is equivalent to absence of conjugate points in our situation. More details are given in the paper of T. Quinto [13]. 7 Range conditions and approximation Let K be a compact set in X and α ∈ R. We define Hα(K,Ω)to be the dual space of H−α(X) and use the notation k·kα for the norm in Hα(K,Ω). The trace operator K Hα(X,Ω) → Hα(K,Ω) is well defined and is open for an arbitrary compact set L ⊂ X L such that K ⋐ L, since H−α(X) is a subspace of H−α(X). Therefore Hα(K,Ω) can be K L realized as the quotient space of Hα(X,Ω) modulo the kernel of the trace operator. The L last one consists of densities f supported in L\K. For any β > α we have the operator η′ : Hβ(K,Ω) → Hα(K,Ω), which is dual to the natural imbedding η : H−α(X) → K H−β(X). If the boundary of K is smooth, the imbedding η has dense image and η′ is K . injective. Thenwecandefinetheintersection H∞(K,Ω) = ∩ Hα(K,Ω);anydensity f ∈ H∞(K,Ω) is smooth in the interior of K. Similarly, we definαe Hα(K) =. H−α(X,Ω) ′. K Suppose that the incidence manifold F fulfils (i-iii). If a density f is(cid:0)supported in(cid:1)a compact set K, then the support of Mf is contained in the compact set Λ = π(p−1(K)). The hypersurface F (x) is compact for any pointx ∈ X and the dualtransform M◦ is well defined for all continuous densities in Σ. Moreover, it can be extended to a continuous operator M◦ : Hα(Λ,Ω) → Hα+(n−1)/2(K) for any α by means of the duality hM◦g,fi = −hg,Mfi, f ∈ H−α−(n−1)/2(X,Ω), g ∈ Hα(Λ,Ω). K By Proposition 4.1, we have Mf ∈ H−α(Σ), hence the right-hand side is well defined. If Λ α ≥ 0, the function M◦g defined by this formula is equal to the integral (2), which has sense, at least, for almost all x ∈ K. Theorem 7.1 Suppose that F satisfies (i-iii) and has no conjugate points. Then for any K ⋐ X and arbitrary α ∈ R∪{∞} the image of the Funk operator M : Hα (X,Ω) → Hα+(n−1)/2(Σ), Λ = π p−1(K) K Λ (cid:0) (cid:1) 7 α+(n−1)/2 is closed and coincides with the subspace of functions ϕ ∈ H (Σ) such that gϕ = Λ 0 for any solution g ∈ H−α−(n−1)/2(Λ,Ω) of the equation R M◦g(x) = 0,x ∈ K. (8) Proof. The image of M is closed by Theorem 6.1, thereby it coincides with the polar of the kernel of the dual operator M◦. ◮ Theorem 7.2 If F fulfils (i-iii) and has no conjugate points. Then for any set K ⋐ X with smooth boundary and arbitrary real α ∈ R any density g ∈ Hα(Λ,Ω),Λ = . π(p−1(K)) thatfulfils (8) canbe approximatedby solutionsh ∈ H∞(Λ,Ω) = ∩ Hα(Λ,Ω). α Proof. Let Solβ denote the space of solutions of (8) in the class Hβ(Λ,Ω). We show first that g can be approximated by elements of Solβ for any β > α. It is sufficient to check that any functional φ on Hα(Λ,Ω) that is equal to zero on Solβ also vanishes on g. The dual space is isomorphic to H−α(Σ), which implies φ ∈ H−α(Σ). By Corollary Λ Λ 4.3 the Funk transform defines the continuous operator M : H−β−(n−1)/2(X,Ω) → H−β(Σ). β K Λ By Theorem 7.1, the range of this operator is closed and coincides with the polar set of Solβ. It follows that, φ = Mψ for a density ψ ∈ H−β−(n−1)/2(X,Ω). By Theorem 6.1 we K −α−(n−1)/2 have ψ ∈ H (X,Ω) and can write K hφ,gi = hMψ,gi = −hψ,M◦gi = 0, since the function M◦g is well defined as element of Hα+(n−1)/2(X). This yields that g K is contained in the closure of the space Solβ. Now we approximate g by elements of the space H∞(Λ,Ω). Let k·kα be the norm Λ in the space Hα(Λ,Ω), which is dual to the norm k·k−α. We may assume that the norm k·kα is monotone increasing function of α. Take an arbitrary ε and choose a Λ function h ∈ Hα+1(Λ,Ω) such kh −gkα < ε/2, then we choose a function h ∈ 1 1 Λ 2 Hα+2(Λ,Ω) such that kh −h kα+1 < ε/4 and so on. We obtain a sequence {h } such 2 1 Λ k that kh −h kα+k < 2−k−1ε for k = 1,2,... This sequence converges to an element h in k+1 k Λ any space Hβ(Λ,Ω),β > α. It follows that h ∈ H∞(Λ,Ω) and kh−gkα ≤ ε. ◮ Λ Corollary 7.3 Under conditions of Theorem 7.1, it is sufficient to check the equation gϕ = 0 for densities g ∈ H∞(Λ,Ω) that satisfies (8). R 8 Thermoacoustic tomography We apply the above results for the thermo/opto/photoacoustic geometry. First, consider the case of complete acquisition geometry. Let X be the open unit ball in an Euclidean 8 space E, S be the sphere of radius R > 1 and Σ = S ×R. The manifold F ⊂ X ×Σ R R is given by the equation I(x;y,r) = |y −x| − r = 0,y ∈ S ,0 < r. The manifold F R obviously fulfils (i),(iii) and has no conjugate points. Check that the condition (ii) is also satisfied. It is sufficient to check that, the sphere F (y,r) can not be tangent to F (z,s) at a point x ∈ X, if the points (y,r) and (z,s) are sufficiently close in Σ. The condition |y +z| > 2 is sufficient for this. The Funk operator Mf(y,r) = f (x)dS, f = f dx (9) 0 0 Z |x−y|=r isthesphericalintegraltransform,wheredS istheEuclideansurfaceareaformonspheres. The kernel of dual transform M◦ consists of densities ϕ = φdSdr in Σ such that ϕ φdSdr 0 = = = − φdS. Z d I Z d(|x−y|−r) Z F(x) σ F(x) for any x ∈ X. Theorem 7.2 yields Corollary 8.1 For any compact set K ⊂ X with smooth boundary and arbitrary α ∈ R ∪{∞} the range of the Funk operator M : Hα (X,Ω) → Hα+(n−1)/2(Σ) coincides with K Λ the set of functions g in Σ such that gϕ = 0 (10) Z Σ for any density ϕ = φdSdr such that φ ∈ C∞(Σ) and φdS = 0, x ∈ K. (11) Z F(x) Remark. Forthe operatorM acting onC∞-densities therangeconditions were given in the papers [11], [1], [2]. The conditions of [1] and [2] give full description of the range of M, but have implicit form. Weextract some explicit range conditions fromCorollary 8.1. For anarbitraryx ∈ K, the manifold F (x) is the intersection of the cone surface |y −x| = r with the cylinder Σ. This intersection is contained in the hyperplane P (x) = y,s;2hx,yi+s = |x|2 +R2 ⊂ E×R, (cid:8) (cid:9) where we set s = r2. Thus, the condition (11) means vanishing of integrals of φdS over intersections of Σ with the hyperplanes P (x),x ∈ K. Suppose that φ is a polynomial in s : φ(y,s) = φ (y)(s−R2)k and have k P φdS = φ (y) |x|2 −2hx,yi kdS(y). k Z Z F(x) Xk S (cid:0) (cid:1) 9 Set x = tz for |z| = 1 and 0 ≤ t < 1 and develop the right-hand side in powers of t : φdS = φ (y) t2 −2thz,yi kdS k Z Z F(x) X S (cid:0) (cid:1) = φ (y)dS −t 2φ (y)hz,yidS 0 1 Z Z +t2 φ (y)+4φ (y)hz,yi2 dS 1 2 Z (cid:2) (cid:3) −t3 4φ (y)hz,yi+8φ (y)hz,yi3 dS 2 3 Z (cid:2) (cid:3) +t4 φ (y)+12φ (y)hz,yi2 +16φ (y)hz,yi4 dS 2 3 4 Z (cid:2) (cid:3) +.. = 0 The right-hand side vanishes for all t, which yields the system of equations φ (y)dS = 0, 0 Z φ (y)hz,yidS = 0, 1 Z 4φ (y)hz,yi2 +φ (y) dS = 0, 2 1 Z (cid:2) (cid:3) 2φ (y)hz,yi3 +φ (y)hz,yi dS = 0, 3 2 Z (cid:2) (cid:3) 16φ (y)hz,yi4 +12φ (y)hz,yi2 +φ (y) dS = 0, 4 3 2 Z (cid:2) (cid:3) ... Corollary 8.2 Any solution (φ ,φ ,φ ,...) of this system such that φ = 0 for all j > k 0 1 2 j for some k yields a function φ(y,s) = kφ (y)(s−R2)j that is a polynomial in s of 0 j order k, fulfils (11) and is orthogonal toPthe range of M. Therearemanysolutionsofthisform,sincethesystemhastriangleformwithdiagonal terms φ (y)hz,yijdS, |z| = 1,j = 0,1,...,k. j Z To solve these equation we only need to fix the moments of φ of degree j. There are only j n+j−1 linearly independent j-moments, hence one can find infinitely many independent n−1 s(cid:0)olutio(cid:1)nswhicharefinitesumsofharmonics. Inparticular,wecantakeforφ anyfunction 0 on the sphere with zero average, an arbitrary function φ with zero linear moments of 1 φ and set φ = 0 for k > 1 etc. The range conditions of S. Patch [11] are apparently 1 k contained in (10) for polynomial φ. 10