By V.V. Volchkov

ISBN-10: 9401000239

ISBN-13: 9789401000239

ISBN-10: 9401039992

ISBN-13: 9789401039994

Integral geometry bargains with the matter of settling on capabilities through their integrals over given households of units. those integrals de?ne the corresponding integraltransformandoneofthemainquestionsinintegralgeometryaskswhen this rework is injective. however, after we paintings with complicated measures or varieties, operators seem whose kernels are non-trivial yet which describe vital sessions of services. many of the questions bobbing up right here relate, in a single manner or one other, to the convolution equations. a number of the renowned guides during this ?eld contain the works by way of J. Radon, F. John, J. Delsarte, L. Zalcman, C. A. Berenstein, M. L. Agranovsky and up to date monographs by way of L. H¨ ormander and S. Helgason. until eventually lately examine during this sector used to be performed as a rule utilizing the means of the Fourier rework and corresponding equipment of advanced research. in recent times the current writer has labored out an primarily di?erent technique in keeping with the outline of varied functionality areas when it comes to - pansions in precise services, which has enabled him to set up very best leads to a number of popular problems.

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**Extra info for Integral Geometry and Convolution Equations**

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Since f ∈ Aϕ (U) we conclude that τ1 f ∗ τ2 ϕ = 0 for all τ1 , τ2 ∈ SO(n). 32) First, assume that ϕ ∈ D(Rn ). Let i = j and let τ2 be a rotation through angle θ in the plane (xi , xj ). 32) with respect to θ and by setting θ = 0, we obtain assertion (2) for ϕ ∈ D(Rn ), i = j. If i = j then Di,j f = 0 ∈ Aϕ (U). 19) we we obtain F k,l ∗ τ2 ϕ = 0. Analogously, for n have Fk,l,p ∗ τ2 ϕ = 0. 6), we obtain the same equality under assumptions of assertion (5). Since τ2 can be chosen arbitrarily, we have the assertions (2)–(5) in the case when ϕ ∈ D(Rn ).

7) we ﬁnd p(Δ)F = f which completes the proof. 2. 15) holds for all z ∈ C, where c, N, R > 0 are independent of z. 5 shows that f is a distribution of order at most N + n + k + 1. 5. Let E ⊂ C be a countable set. Assume that R > r > 0 and let f ∈ Dk,l (BR ). Then for each m ∈ Z+ there exists a function F ∈ (C m ∩ Ek,l )(Rn ) such that p(Δ)F = f in Br for some polynomial p : R1 → C. In addition, Z(q) ∩ (Z(Fk,l F ) ∪ E) = ∅, where q(z) = p(−z 2 ) and each zero of the polynomial q is simple. Proof.

59) where the constant c > 0 is independent of m. 60) 1 ⎪ 1 2 ⎪ 2 ⎪ tJν (ζm t)dt = Jν+1 (ζm ). 61) m=1 where cm = 1 2 2 (ζ ) Jν+1 m tf (t)Jν (ζm t)dt. 62) 0 Then the following equality holds N 1 t f (t) − lim N →+∞ 0 2 cm Jν (ζm t) dt = 0. m=1 In particular, if cm = 0 for all m ∈ N then f = 0. Assume now that f ∈ C k [0, 1], and let f (j) (0) = f (j) (1) = 0 for each j = 0, . . , k. 59) (1/2)−k we obtain the estimate cm = O ζm as m → ∞. 61) converges uniformly on [0, 1]. 1. Simplest properties of spherical harmonics Let Hk (Rn ), n 2 denotes the set of all homogeneous harmonic polynomials on Rn of degree k.

### Integral Geometry and Convolution Equations by V.V. Volchkov

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