Задачи Радона для гиперболоидов

Аннотация

Мы предлагаем некоторый вариант преобразований Радона для пары X и Y гиперболоидов в R 3 , определенных уравнениями [x , x] = 1 and [ y, y] = -1 , y 1 ≥ 1 , соответственно, здесь [ x, y] = - x 1 y 1 + x 2 y 2 + x 3 y 3 . В качестве ядра этих преобразований мы берем δ([ x, y]) , где δ( t) - дельта-функция Дирака. Мы получаем два преобразования Радона D ( X ) → C ∞ ( Y ) и D ( Y ) → C ∞ ( X ) . Мы описываем ядра и образы этих преобразований. Для этого мы разлагаем полуторалинейную форму с ядром δ ([ x, y]) по скалярным произведениям компонент Фурье.

Полный текст

In this paper we offer a variant of Radon transforms for a pair of dual hyperboloids in R3 : the one-sheeted hyperboloid X : [x; x] = 1 ( [x; y] = x1y1+x2y2+x3y3 ) and the upper sheet of the two-sheeted hyperboloid Y : [y; y] = 1; y1 > 1 (the Lobachevsky plane). For a kernel of these transforms we take ([x; y]); x 2 X; y 2 Y; (t) being the Dirac delta function. This kernel gives two Radon transforms R : D(X) ! C1(Y) and R : D(Y) ! C1(X): We describe kernels and images of these transforms. For that we consider a sesqui-linear form with the kernel ([x; y]) and write the decomposition of this form into inner products of Fourier components. Results of this paper were announced in [4]. 1. Hyperboloids Let G be the group SO0(1; 2); it is a connected group of linear transformations of R3; preserving the form [x; y] = x1y1 + x2y2 + x3y3: We consider that G acts on R3 from the right. In accordance with this we write vectors in the row form. Let us take the following basis of the Lie algebra g of the group G: L0 = 0 @ 0 0 0 0 0 1 0 1 0 1 A; L1 = 0 @ 0 1 0 1 0 0 0 0 0 1 A; L2 = 0 @ 0 0 1 0 0 0 1 0 0 1 A: (1.1) The Casimir element in the universal enveloping algebra of g is (1=2) g; where g = L20 + L21 + L22 : (1.2) Consider subgroups K; H; A of the group G generating by elements L0; L1; L2; respectively. The subgroup K is a maximal compact subgroup of G: Denote by X the one-sheeted hyperboloid [x; x] = 1; and by Y the upper sheet of the two-sheeted hyperboloid [y; y] = 1; y1 > 1 (we consider that the x1 -axis goes up). These hyperboloids X and Y are homogeneous spaces of the group G with respect to translations x 7! xg; namely, X = G=H and Y = G=K: The subgroups H and K are stabilizers of points x0 = (0; 0; 1) 2 X and y0 = (1; 0; 0) 2 Y respectively. These hyperboloids have a G-invariant metric. It gives rise to the measures dx and dy and the Laplace-Beltrami operators X and Y respectively (all are G-invariant). As local coordinates on the hyperboloids we can take any two variables from x1; x2; x3: For Y it is natural to take y2; y3: Then we have dx = jx1j 1dx2dx3; dy = y 1 1 dy2dy3; X = @2 @x21 @2 @x22 + D2 1 + D1; D1 = x1 @ @x1 + x2 @ @x2 ; Y = @2 @y2 2 @2 @322 + D2 1 + D1; D1 = y2 @ @y2 + y3 @ @y3 : If M is a manifold, then D(M) denotes the space of compactly supported infinitely differentiable C-valued functions on M; with the usual topology, and D0(M) denotes 434 V. F. Molchanov the space of distributions on M of antilinear continuous functionals on D(M): For a differentiable representation of a Lie group, we retain the same symbol for the corresponding representations of its Lie algebra and of the universal enveloping algebra. Let us denote by UX and UY representations of our group G by translations on functions on X and X respectively (quasiregular representations): (UX (g)f)(x) = f(xg); (UX (g)f)(y) = f(yg): The representations UX and UY on the spaces L2(X; dx) and L2(Y; dy) are unitary with respect to the inner products hF; fiX = Z X F(x)f(x)dx; hF; fiY = Z Y F(y)f(y)dy: (1.3) We have UX( g) = X ; UY( g) = Y: (1.4) 2. Representations of the group SO0(1; 2) Recall some material about the principal non-unitary series of representations of the group G = SO0(1; 2); see, for example, [5]. Let C+ be the cone [x; x] = 0; x1 > 0: The group G acts transitively on it. For 2 C; let D (C+) be the space of C1 functions ' on C+ homogeneous of degree : '(tx) = t '(x); t > 0: Let T be the representation of G acting on this space by translations: (T (g)')(x) = '(xg): Take the section S of the cone C+ by the plane x1 = 1; it is a circle consisting of points s = (1; sin ; cos ): The Euclidean measure on S is ds = d : For a function ' on S; sometimes we write '( ) instead of '(s): The representation T can be realized on the space D(S) as follows (index 1 indicates the first coordinate of a vector): (T (g)')(s) = ' sg (sg)1 (sg) 1 : The element g; see (1.2), goes to a scalar operator: T ( g) = ( + 1)E: (2.1) The Hermitian form h ; 'iS = Z S (s)'(s)ds (2.2) is invariant with respect to the pair (T ; T 1); i. e. hT (g) ; 'iS = h ; T 1(g 1)'iS: (2.3) This formula follows from des = (sg) 1 1 ds; where es = (sg)=(sg)1: RADON PROBLEMS FOR HYPERBOLOIDS 435 Define an operator A in D(S) : (A ')(s) = Z S ( [s; u]) 1'(u)du: The integral converges absolutely for Re < 1=2 and can be continued as a meromorphic function to the whole -plane. It has simple poles at 2 1=2 + N: Here and further N = f0; 1; 2; :::g: For A we have hA ; 'iS = h ;A 'iS: (2.4) The operator A intertwines T and T 1; i. e. T 1(g)A = A T (g); g 2 G: A sesqui-linear form hA ; 'iS is invariant with respect to the pair (T ; T ): In particular, for 2 R; this form is an invariant Hermitian form for T : Take a basis m( ) = eim ; m 2 Z; in D(S): It consists of eigenfunctions of A : A m = a( ;m) m; (2.5) where a( ;m) = 2 +2 ( 1)m ( 2 1) ( + m) ( m) : (2.6) The composition A A 1 is a scalar operator: A A 1 = 1 8 !( ) E where !( ) is a “Plancherel measure” (see (5.2)): !( ) = 1 32 2 (2 + 1) cot : (2.7) The representation T can be extended to the space D0(S) of distributions on S by formula (2.3) where is a distribution and h ; 'iS is the value of the distribution at a test function ': It is an extension in fact, since D(S) can be embedded into D0(S) by means of the form (2.2), namely, we assign to a function 2 D(S) the functional ' 7! h ; 'iS in D0(S): Similarly the operator A can be extended to the space D0(S) by means of formula (2.4). If is not integer, then T is irreducible and T is equivalent to T 1 (by A or b A ). Let 2 Z; n 2 N: Subspaces V ;+ and V ;+ spanned by m for which m > and m 6 respectively are invariant. For < 0 they are irreducible and orthogonal to each other. For > 0 their intersection E is irreducible and has dimension 2 + 1: Let V d n = D(S)=En and V d n 1 = V n 1;+ + V n 1; : Let us denote by Td ; 2 Z; the representation on V d generated by T : The operator An vanishes on En and gives rise to the equivalence Td n Td n 1: 436 V. F. Molchanov There are four series of unitarizable irreducible representations: the continuous series consisting of representations T with = 1=2+i ; 2 R; the inner product is (2.2); the complementary series consisting of T with 1 < < 1; the inner product is hA ; 'iS with a factor; the holomorphic and antiholomorphic series. We need only their sum Td : We shall call Td the representations of discrete series. For ' 2 D(S); denote by e' the coset of ' modulo En: Then the invariant inner product ( ; )n for Td n is ( e ; e')n = cnhAn ; 'iS; cn = a(n; n + 1) 1: (2.8) 3. Poisson and Fourier transforms First we determine distributions in D0(S) invariant with respect to the subgroup H under the representations T : We shall use the following notation for a character of the group R : t ;m = jtj (sgn t)m; where t 2 R = Rnf0g; 2 C; m 2 Z: In fact this character depends only on m modulo 2. Here and further the sign “ -” means the congruence modulo 2. It is easy to check that the distribution ;" = s ;" 3 = [x0; s] ;"; where 2 C; " = 0; 1; is H -invariant. Sometimes we write an integer instead of " with the same parity as ": As a function of ; ;" is a meromorphic function with simple poles at points 2 1 " 2N: Its residue at = n 1; n - "; is the distribution const (n)(s3) concentrated at two points s = (1; 1; 0): Here (t) is the Dirac delta function on the real line (a linear continuous functional on D(R) ). The space of H -invariants has dimension 2 for 6= n 1; n 2 N; and dimension 3 for = n 1: Every irreducible subfactor for T ; 2 Z; contains, up to a factor, precisely one H -invariant. In particular, n 1;n+1 and n;n+1 have non-zero projections into V 0 n 1; and D0(S)=V 0 n; respectively. The operator A carries ;" to 1;" with a factor: A ;" = j( ; ") 1;"; (3.1) where j( ; ") = 2 1=2 1 2 ( + 1) [1 ( 1)" cos ] : (3.2) It is easy to check that j( ; ")j( 1; ") = (8 !( )) 1; where !( ) is given by (2.7) or (5.2). The factor j( ; ") has simple poles at 2 1=2+N: By a general scheme [3], the H -invariant ;" gives rise to the Poisson kernel P ;"(x; s) = [x; s] ;"; x 2 X; s 2 S: This kernel gives rise to two transforms. The first of them, the Poisson transform P ;" : D(S) ! C1(X) is a linear continuous operator defined as follows: (P ;"')(x) = Z S [x; s] ;"'(s)ds: RADON PROBLEMS FOR HYPERBOLOIDS 437 It intertwines T 1 with UX ; therefore, its image consists of eigenfunctions of the Laplace- Beltrami operator: X P ;" = ( + 1)P ;" of parity " (see (2.1) and (1.4)). As a function in ; the Poisson transform behaves like ;" : it depends on meromorphically and has simple poles at 2 1 " 2N: Formula (3.1) gives P ;"A = j( ; ")P 1;": (3.3) Consider 2 Z: The transform P n 1;n+1 vanishes on En; it generates an operator on D(S)=En which intertwines Td n with UX : The Poisson transform Pn;n+1 considered on V d n 1 intertwines Td n 1 with UX : By (3.3), Pn;n+1 has the same image as P n 1;n+1: The second transform generated by the Poisson kernel is the Fourier transform F ;" : D(X) ! D(S) defined by (F ;"f) (s) = Z X [x; s] ;"f(x)dx: It is meromorphic in with simple poles at points 2 1 " 2N: It intertwines UX with T : It follows from (3.1) that A F ;" = j( ; ")F 1;": (3.4) For a function f 2 D(X); let us call two functions F ;"f; " = 0; 1; the Fourier components of f corresponding to the representation T : The Fourier and Poisson transforms are conjugate to each other with respect to forms (1.3) and (2.2): hP ;"'; fiX = h'; F ;"fiS: This relation allows to extend the Poisson transform to distributions on S: Consider the reducible case. The Fourier transform Fn corresponding to Td n is defined as the map of D(X) to D(S)=En which assigns to f 2 D(X) the corresponding coset of the function Fn;n+1f: By (2.8) and (3.4) we have (Fnf; Fnh)n = dnhF n 1;n+1f; Fn;n+1hiS; dn = 2n!2= (2n + 1)!: The Fourier transform corresponding to Td n 1 is F n 1;n+1: The representation T has one up to a factor K -invariant, it is the function equal to 1 identically on S : (s) = [y0; s] = 1: The representations of the discrete series have no K -invariants. The corresponding Poisson transform Q : D(S) ! C1(Y) and Fourier transform D(Y) ! D(S) are defined by (Q ')(y) = Z S [ y; s] '(s)ds; (G h) (s) = Z Y [ y; s] h(y)dy: 438 V. F. Molchanov Notice that [ y; s] > 0 for all y 2 Y and s 2 S: The Poisson transform Q intertwines T 1 with UY; therefore, its image consists of eigenfunctions of the Laplace-Beltrami operator: Y Q ;" = ( + 1)Q ;": (3.5) 4. Spherical functions Let 2 C; " = 0; 1: Let us define a spherical function ;" on the hyperboloid Y as follows ;"(y) = hT (g) ;"; 1iS (4.1) = h ;"; T 1(g 1) 1iS = Z S ;"[ y; s] 1ds; (4.2) where g 2 G is such that y0g = y: As the distribution ;" does, the spherical function ;" is given by an integral absolutely convergent for Re > 1 and can be continued analytically in to a meromorphic function. It has poles where ;" has and of the same (the first) order. The function ;"(y) is a function of class C1 on Y invariant with respect to H : ;"(yh) = ;"(y); h 2 H: Therefore, it depends on y3 = [x0; y] only: ;"(y) = ;"(y3); (4.3) where ;"(c) is a function in C1(R). Lemma 4.1. The function ;" has the following integral representation: ;"(c) = Z 2 0 c + p c2 + 1 cos - ;" d : (4.4) P r o o f. Let us take in (4.1) as g the matrix a = exp t L2; see (1.1), in A: a = 0 @ cosh t 0 sinh t 0 1 0 sinh t 0 cosh t 1 A: We have (T (a) ;") (s) = [x0; sa] ;" = (sinh t + s3 cosh t) ;" : By (4.1), the value of ;" at the point y0a = (cosh t; 0; sinh t) is just (4.4) with c = sinh t: It follows from (4.4) that the function ;" has parity " : ;"( c) = ( 1)" ;"(c): RADON PROBLEMS FOR HYPERBOLOIDS 439 Equality (4.2) shows that the spherical function ;" is the Poisson transform of the H -invariant: ;" = Q 1 ;": (4.5) Consider ;" as a distribution on Y : h ;"; fiY = Z Y ;"(y)f(y)dy; (4.6) where f 2 D(Y): The right hand side in (4.6) can be rewritten as an iterated integral, then we obtain: h ;"; fiY = Z 1 1 ;"(c)(Mf)(c)dc; (4.7) where (Mf)(c) = Z Y (y3 c)f(y)dy; The map M assigns to a function f its integrals over H -orbits. It is a continuous operator from D(Y) onto D(R): Lemma 4.2. The value (4.6) is expressed in terms of Fourier components: h ;"; fiY = h ;";G 1fiS: (4.8) P r o o f. Let h(y) be a majorant of the function f(y); depending on y1 only. Then for Re > 1 the right hand side in (4.8) is majorized by the integral Z 2 0 j cos j d Z Y [y; s] 1 h(y)dy; (4.9) where = Re : In fact, the integral over Y here does not depend on s: Therefore, integral (4.9) converges absolutely and the order of integration can be inverted. So we get equality (4.8) for Re > 1: To other this equality is extended by analycity. Let be a distribution on Y invariant with respect to H: Assign to it two things: a convolution with of functions f in D(Y) and a sesqui-linear functional K on the pair (D(X); D(Y) ). The convolution ? f is the following function on X : ( ? f)(x) = h ; UY(g)fiY = Z Y (y)f(yg)dy; the functional is: K( jh; f) = hh; ? fiX = Z X h(x)h ; UY(g)fiY dx = Z X-Y (yg 1)h(x)f(y)dx dy; 440 V. F. Molchanov where h 2 D(X); f 2 D(Y) and g is an arbitrary element in G carrying x0 to x: Since is H -invariant, these formulae do not depend on the choice of g for given x: The convolution is a linear map D(Y) ! C1(X); intertwining UY and UX : ? (UY(g)f) = UX (g) ( ? f) : For the spherical function ;"; the convolution and the functional are expressed in terms of the Poisson and Fourier transforms: ( ;" ? f) (x) = (P ;"G 1f) (x); K( ;"jh; f) = hF ;"h;G 1fiS: (4.10) The kernel K ;"(x; y) of the functional (4.10) is K ;"(x; y) = Z S [x; s] ;"[ y; s] 1ds: Lemma 4.3. The function ;" has the following property of symmetry in : 1;" = 1 + ( 1)" cos sin ;": (4.11) P r o o f. By Lemma 4.2, (3.1), (2.4), (2.5) and Lemma 4.2 again we have: h 1;"; fiY = h 1;";G fiS = j( ; ") 1hA ;";G fiS = j( ; ") 1h ;";A G fiS = a( ; 0)j( ; ") 1h ;";G 1fiS = a( ; 0)j( ; ") 1h ;"; fiY: Substituting here values of a( ; 0) and j( ; ") from (2.6) and (3.2), we get (4.11). Lemma 4.4. The spherical function ;" is an eigenfunction of the Laplace-Beltrami operator: Y ;" = ( + 1) ;": (4.12) P r o o f. The function ;" is the Poisson transform of the function ;"; see (4.5). It remains to remember (3.5). On functions depending on y3 = c only, the operator Y becomes to the following differential operator (the H -radial part of Y ): L = (c2 + 1) @2 @c2 + 2c @ @c : (4.13) Lemma 4.5. The function ;"; see (4.3) and (4.4), is an eigenfunction of L : L ;" = ( + 1) ;": RADON PROBLEMS FOR HYPERBOLOIDS 441 The lemma follows immediately from Lemma 4.4. Theorem 4.1. The spherical function ;"(y) is expressed in terms of the Legendre functions P (see [2, Ch. III]) of the imaginary argument: ;"(y) = 2 ei =2 + ( 1)"e i =2 n P (iy3) + ( 1)"P ( iy3) o : (4.14) P r o o f. Denote for brevity: P (it) = B (t); (4.15) also for a function '(t) on R we shall denote b'(t) = '( t): Equality (4.14) is equivalent to the following expression of the function ;" : ;"(c) = 2 ei =2 + ( 1)"e i =2 n B (c) + ( 1)" b B (c) o : (4.16) So we have to prove (4.16). The Legendre function P (z) is analytic in the z -plane with the cut ( 1; 1]; satisfies the equation: (z2 1) @2 @z2 + 2z @ @z w = ( + 1)w (4.17) and has the integral representation P (z) = 1 2 Z 2 0 z + p z2 1 cos - d : (4.18) Let be not integer. Then the functions P (z) and b P (z) form a basis of solutions of equation (4.17). For z = ic equation (4.17) becomes the equation: Lw = ( + 1)w: In virtue of Lemma 4.5 the function ;" is a linear combination of functions B and b B : Coefficients of this linear combination could be found out by computing values of functions ;"; B and b B and their derivatives at the point c = 0; using (4.6) and explicit expressions [2, 3.4(20),(23)]. But it is more convenient for us to find them in another way. Let z tend to ic; c 2 R; in (4.18) such that Re z > 0: We get: B (c) = 1 2 ei =2 Z 2 0 c + p c2 + 1 cos i0 - d : (4.19) Denote Z (c) = Z 2 0 c + p c2 + 1 cos i0 - + d : Then b Z (c) = Z 2 0 c + p c2 + 1 cos i0 - d : 442 V. F. Molchanov Applying to (4.19) the formula: (t i0) = t + + e i t ; we obtain B = 1 2 h ei =2Z + e i =2 b Z i ; (4.20) whence b B = 1 2 h e i =2Z + ei =2 b Z i : (4.21) From (4.20) and (4.21) we have Z = i sin h ei =2B e i =2 b B i ; (4.22) b Z = i sin h e i =2B + ei =2 b B i : (4.23) Since ;" = Z + ( 1)" b Z : we obtain (4.16) by (4.22) and (4.22). Let us establish some estimates for spherical functions of the continuous series ( = 1=2 + i ). They show that values of these spherical functions at f decrease rapidly when their parameter tends to infinity. Theorem 4.2. Let = 1=2+i ; 2 R: For any compact set W Y; there exists a number C > 0 such that for any f 2 D(Y) with the support in W the following estimate holds: jh ;"; fiYj 6 C max y mY f (y) ( 2 + 1=4) m; m 2 N: ( 4.24) P r o o f. Take h 2 D(Y) depending on y1 only, such that h(y) > 0; h(y) = 1 on W: Then h; where = max jf(y)j; is a majorant for f depending on y1 only. Arguing as in the proof of Lemma 4.2, we obtain jh ;"; fiYj 6 C ; (4.25) where C is the number C = Z 2 0 j cos j 1=2d Z Y [ y; s] 1=2h(y) dy: Now apply the estimate (4.25) to the function mY f; m 2 N; transfer the operator Y to the function ;"; since it is self-adjoint, and use (4.12). Since j ( + 1)j = 2 + 1=4 for = 1=2 + i ; we get (4.24). Let us write expressions of ;" for integer. In the notation ;" sometimes it is convenient to write an integer instead of " with the same parity as ": RADON PROBLEMS FOR HYPERBOLOIDS 443 Let n 2 N: Let first = n: For n;n+1 we have to evaluate an indeterminacy in (4.14). We have n;n(y) = 2 i nPn(iy3); n;n+1 = 4i1 nQ n(iy3); where Pn(z) is the Legendre polynomial and Q n(z) is the Legendre function which differs from the Legendre function of the second kind Qn(z) by the cut on the z -plane: for Qn(z) one takes the cut [ 1; 1]; but for Q n(z) one has to take the cut ( 1; 1][[1;1) ; therefore, we have: Q n(z) = 1 2 Pn(z) ln 1 + z 1 z Wn 1(z); cf. [2, 3.6(24)], where the principal branch of the logarithm is taken and Wn 1(z) is a polynomial of degree n 1 indicated in [2, 3.6.2]. For = n 1 we use the relation (4.11). For " - n the function ;" has a pole at = n 1 because of ;": We have n 1;n+1 = 0; Res = n 1 ;n = ( 1)n+1 (2= ) n;n: 5. Eigenfunction decomposition of the radial part of the Laplace-Beltrami operator In this Section we obtain the eigenfunction decomposition of the operator (see (4.13)) L = (c2 + 1) @2 @c2 + 2c @ @c defined on the real line R: We use the function ;"(c); see (4.3) and (4.4). Recall that it has parity " and satisfies the equation: Lw = ( + 1)w: Let us denote by ('; ) the L2(R) inner product of functions '; : ('; ) = Z 1 1 '(c) (c) dc: Theorem 5.1. There is the following eigenfunction decomposition of the operator L : ('; ) = Z 1 1 !( ) X " ('; ;")( ;"; ) = 1=2+i d ; (5.1) where !( ) = 1 32 2 (2 + 1) cot (5.2) so that ! 1 2 + i = 1 16 2 tanh : 444 V. F. Molchanov P r o o f. Let us write the resolvent R = ( E L) 1 of the operator L: Let h 2 L2(R) and R h = f; then h = ( E L)f; so that Lf f = h: (5.3) Let us take in the form = ( + 1): The correspondence 7! maps the half plane Re > 1=2 onto the -plane with the cut ( 1; 1=4 ] one-to-one. Let f1; f2 be eigenfunctions of the operator L with the eigenvalue = ( + 1) with Re > 1=2: They behave at infinity ( 1) as Ajcj + Bjcj 1: Let us take them such that they are square integrable at +1 and 1 respectively. Then for c ! +1: f1(c) B1c 1; f2(c) A2c + B2c 1; and for c ! 1: f1(c) C1jcj + D1jcj 1; f2(c) D2jcj 1: The wronskian W of these functions is W = W0 c2 + 1 ; W0 = (2c + 1)B1A2: We have already several eigenfunctions: P (ic); P ( ic); Z (c); b Z (c); ;"(c); " = 0; 1: By [2, 3.2(18)] the Legendre functions behave when c ! +1 as follows: P (ic) p( ) ei =2 c + p( 1) ei( 1) =2 c 1; P ( ic) p( ) e i =2 c + p( 1) ei( +1) =2 c 1; where p( ) = 2 1B + 1 2 ; 1 2 ; B(a; b) being the Euler beta function. By (4.22), (4.22) it gives that when c ! +1 we have Z (c) 2 p( ) c 2 sin p( 1) c 1; b Z (c) 2 cot p( 1) c 1: Therefore, as a mentioned-above basis f1; f2 of solutions of the equation Lw = w; = ( + 1); we can take the pair b Z ; Z : Then W0 = (2 + 1) 2 p( ) 2 cot p( 1) = 4 : Therefore, the solution f of equation (5.3) is f(c) = 1 4 n b Z (c) Z c 1 Z (t)h(t)dt + Z (c) Z 1 c b Z (t)h(t)dt o : RADON PROBLEMS FOR HYPERBOLOIDS 445 Thus, for Im 6= 0; the resolvent R is an integral operator with the kernel K (c; t) = ( (1=4 ) b Z (c)Z (t); c > t; (1=4 )Z (c) b Z (t); c < t; (5.4) here = ( + 1) and belongs to the half plane Re > 1=2 with the cut along the real axis. Let '; 2 L2(R): By the Titchmarsh-Kodaira theorem [1, XIII] we have ('; ) = lim "!+0 1 2 i Z 1 1 (R i"'; ) d Z 1 1 (R +i"'; ) d : Let us pass to : Then d = (2 + 1)d and we denote S = R : The operator function S is analytic in the half plane Re > 1=2: Therefore, ('; ) = 1 2 Z 1 1 (2 + 1)(S '; ) = 1=2+i d : We can keep here only the even part in of the integrand: ('; ) = 1 4 Z 1 1 (2 + 1) ((S S 1)'; ) = 1=2+i d : Let us compute the kernel M (c; t) of the operator S S 1: Let c > t: By (5.4) we have M (c; t) = 1 4 n b Z (c)Z (t) b Z 1(c)Z 1(t) o Let us insert here (4.22) and (4.22) and use that the Legendre function P is unchanged under 7! 1: We obtain (recall notation (4.15)): M (c; t) = cos 2 sin2 n B (c) b B (t) + b B (c)B (t) o : (5.5) For c < t; we obtain the same expression. Further, if = 1=2 + i ; then for the Legendre function P on the imaginary axis we have P (ic) = P ( ic) = P 1( ic) = P ( ic); or, in terms of B : B (c) = b B (c) = b B 1(c) = b B (c): Therefore, equality (5.5) gives ('; ) = Z 1 1 (2 + 1) cos 8 sin2 n (';B )(B ; ) + ('; b B )( b B ; ) o = 1=2+i d : (5.6) This formula is the desired eigenfunction decomposition - in the basis B ; b B : Now let us pass in (5.6) from B ; b B to ;"; " = 0; 1; by B = 1 2 cos 2 ;0 + i sin 2 ;1 - ; b B = 1 2 cos 2 ;0 + i sin 2 ;1 - ; then we obtain (5.1). 446 V. F. Molchanov 6. Decomposition of a sesqui-linear form on the pair of hyperboloids Let us consider the following sesqui-linear form A(h; f) defined on the pair (D(X); D(Y) : A(h; f) = Z X-Y ([x; y])h(x)f(y)dxdy: The main result of our work consists of Theorem 6.1, which gives the decomposition of this form in terms of Fourier components of functions h and f: The decomposition contains Fourier components of the continuous series ( = 1=2 + i ). Theorem 6.1. The sesqui-linear form A(h; f) decomposes into Fourier components of the continuous series F ;0f and G h; = 1=2 + i ; 2 R; as follows: A(h; f) = Z 1 1 ( )hF ;0h;G fiS = 1=2+i d ; (6.1) where ( ) = 2!( )B 2 ; 1 2 (6.2) = 1 16 2 (2 + 1) cot B 2 ; 1 2 ; (6.3) the factor !( ) is given by (5.2), so that 1 2 + i = 1 8 5=2 tanh sin 1 4 + i 2 1 4 + i 2 2 P r o o f. Let us take in (5.1) as ' the characteristic function of the interval [0; a] divided by a and as the function Mf; f 2 D(Y): We can consider that a 2 [0; 1]: We obtain 1 a Z a 0 Mf(c) dc = X " Z 1 1 "( ) h 1 a Z a 0 1=2+i ;"(c)dc i d ; (6.4) where we denoted "( ) = !( ) ( ;";Mf) = 1=2+i = h ;"; fiY = 1=2+i ; see (4.7). Let a tend to 0. Then the left hand side of (6.4) goes to Mf(0): Let us prove that we can pass to the limit under the integral over in the right hand side of (6.4). By the mean value theorem, the integral in the right hand side of (6.4) is equal to F"(a) = Z 1 1 "( ) 1=2+i ;"(-)d ; (6.5) where - is a number in [ 0; a] (depending on a; and " ). We have to prove that F"(a) ! F"(0) (6.6) RADON PROBLEMS FOR HYPERBOLOIDS 447 when a ! 0; where F"(0) = Z 1 1 "( ) 1=2+i ;"(0)d : (6.7) Let us take an arbitrary number > 0: In virtue of Theorem 4.2 both functions "( ); " = 0; 1; decrease rapidly when j j ! 0: Therefore, there exists a number A such that Z j j>A "( ) d < : (6.8) It follows from formula (4.4) that the function 1=2+i ;"(c) is bounded, i. e. 1=2+i ;"(c) 6 N; (6.9) N being some number, for all 2 R; " = 0; 1; and all c from some finite interval, for example, [0; 1]: Indeed, formula (4.4) implies the inequality ;"(c) 6 Z 2 0 c + p c2 + 1 cos 1=2 d (6.10) since the function of c in the right hand side of (6.10) (it is the function 1=2;0 ) is continuous with respect to c: On the other hand, since the function 1=2+i ;"(c) is continuous with respect to and c; there exists a number > 0 such that 1=2+i ;"(-) 1=2+i ;"(0) < (6.11) for j j 6 A and 0 6 - < : Then for 0 < a < we have F"(a) F"(0) 6 Z 1 1 "( ) 1=2+i ;"(-) 1=2+i ;"(0) d = Z A A + Z j j>A 6 Z A A "( ) d + 2N Z j j>A "( ) d 6 (C" + 2N) ; (6.12) where C" Z 1 1 "( ) d ; here we used (6.5), (6.7)-(6.9), (6.11). Inequality (6.12) proves (6.6). Now we may pass to the limit in (6.4) when a ! 0: We obtain Mf(0) = Z 1 1 !( ) X " ;"(0) h ;"; fiY = 1=2+i d : (6.13) By (4.4) we have ;"(0) = Z 2 0 (cos ') ;"d' = [ 1 + ( 1)" ]B + 1 2 ; 1 2 : 448 V. F. Molchanov We see that ;"(0) is equal to zero for " = 1; so that only one summand in (6.13) remains - with " = 0: Since = 1 for = 1=2 + i ; (6.14) equality (6.13) is Mf(0) = Z 1 1 ( )h ;0; fiY = 1=2+i d ; (6.15) where ( ) is given by (6.3), (6.4). The left hand side in (6.15) is Mf(0) = Z Y [ x0; y ] f(y) dy: (6.16) Taking into account (6.16) let us apply (6.15) to a shifted function (UY(g)f) (y) = f(yg); g 2 G: We get Z Y ([ x; y ]) f(y)dy = Z 1 1 ( )h ;0; UY(g)fiY = 1=2+i d ; (6.17) where x = x0g: Now multiply both sides of (6.17) by a function h(x) in D(X) and integrate over x 2 X: In the right hand side we may invert the order of integrations - in virtue of Lemma 6.1, see below. We obtain: A(h; f) = Z 1 1 ( ) Z X h ;0; UY(g)fiY h(x)dx = 1=2+i d : The integral over X is nothing but the functional K( ;0jh; f): Substituting its expression (4.10) in terms of Fourier components and taking into account (6.14), we get (6.1). Lemma 6.1. For any function f(y) in DY the integral in the right hand side of (6.17) converges absolutely and uniformly with respect to x = x0g on any compact V X: P r o o f. The hyperboloid X can be embedded into the group G as the product AK of subgroups A and K: By continuity of UY; the union of supports of all functions UY(g)f; where g = ak is such that x0g 2 V is some compact W in Y: By Theorem 4.1 there exists C > 0 such that for any g = ak; x0g 2 V; the following inequality holds h ;"; UY(g)fiY 6 C max y mY UY(g)f - (y) ( 2 + 1=4) m: Since Y commutes with translations, we have max y mY UY(g)f (y) = max y UY(g) mY f - (y) = max y mY f(y) ; so that there exist numbers Cm; m 2 N; such that h ;"; UY(g)fiY 6 Cm ( 2 + 1=4) m; for all x = x0g 2 V and all m 2 N; whence the lemma. RADON PROBLEMS FOR HYPERBOLOIDS 449 The quasiregular representation of G = SO0(1; 2) on X contains representations of the continuous series with multiplicity two and the analytic and antianalytic series with multiplicity one, and the quasiregular representation of G on Y contains representations of the continuous series with multiplicity one. Theorem 6.1 gives Theorem 6.2. The kernel of the Radon transform R consists of functions belonging to the discrete spectrum and to the odd part of the continuous spectrum on X; its image goes in C1(Y): The kernel of the Radon transform R is f0g; its image consists of functions belonging to the even part of the continuous spectrum X: