Nonlinear Radon and Fourier Transforms

François Rouvière Université de Nice Laboratoire Dieudonné, UMR 7351

May 16, 2015

Abstract In this note we explain a generalization, due to Leon Ehrenpreis, of the classical Radon transform on hyperplanes. A function f on Rn can be reconstructed from nonlinear Radon transforms, obtained by integrating f and a …nite number of multiples x f over a family of algebraic hypersurfaces of degree m. This follows by solving a Cauchy problem for the nonlinear Fourier transform of f. We also give an inversion formula for this Radon transform.

1 Introduction

This expository note is an attempt at explaining the pages from Ehrenpreis’trea- tise [5] in which he develops the nonlinear Radon and Fourier transforms he had introduced in his previous papers [1][2][3][4]. The goal is to extend the classical hyperplane Radon transform R0f (integrals of a function f over all hyperplanes in Rn) to a family of algebraic submanifolds de…ned by higher degree polynomial equations. Is the generalized transform R still injective? Can we give an inversion formula? Unfortunately it is readily seen that R is no more injective (in general): reconstructing f from Radon transforms needs more than Rf alone. We shall explain here several results of the following type: there exists a …nite number of low-degree polynomial functions ak (with a1 = 1) such that f is determined by the Radon transforms R(akf). Besides, the restriction of the R(akf)’s to a certain subfamily of algebraic manifolds may even be su¢ cient, provided one increases the number of polynomials ak. After a brief reminder of the classical hyperplane transform (this Section) we shall introduce Ehrenpreis’nonlinear Radon transform and the related nonlinear Fourier transform, so as to get a projection slice theorem which plays a crucial role in this study (Section 2). The reconstruction problem boils down to a Cauchy problem for a system of partial di¤erential equations, solved in a naive way in Section 3 then, in Section 4, by the more sophisticated tools of harmonic polynomials. In Section 5 we discuss an inversion formula for the nonlinear Radon transform.

In order to motivate the forthcoming construction, let us brie‡yrecall a few facts n about the classical Radon transform R0. In the Euclidean space R it is given by integration of a compactly supported smooth function f (Rn) over the family of 2 D 1 all hyperplanes. A hyperplane being de…ned by the equation ! x = t where ! is a unit vector, t a real number and denotes the scalar product, we consider

R0f(t; !) := f; ! x=t Z an integral with respect to the measure induced on the hyperplane by the Euclidean measure dx of Rn. Note that (t; !) and ( t; !) de…ne the same hyperplane, thus R0f(t; !) = R0f( t; !). For any R we have 2 i! x i! x it e f(x)dx = dt e f(x) = e R0f(t; !)dt: Rn R ! x=t R Z Z Z Z This gives the projection slice theorem

f(!) = R0f(; !) (1)

n for R, ! R and ! = 1. b d Caution:2 on2 the left-handk k side of (1) the hat denotes the n-dimensional Fourier transform on x but on the right-hand side it denotes the 1-dimensional Fourier n 1 transform on t. Both sides are smooth functions on R S , rapidly decreasing with respect to . Knowing the integrals of f over all hyperplanes, i.e. R0f, the Fourier transform f is therefore known and R0 is easily inverted as follows. Writing the Fourier inversion formula for f in spherical coordinates we have b

n 1 i! x n 1 f(x) = (2) d! e R0f(; !) d ! =1 0 Zk k Z where d! is the Euclidean measure on the unit sphered of Rn. In order to use Fourier analysis in one variable we can replace 1 by : indeed R f(; !) = R f( ; !) 0 R 0 0 and, changing into then ! into !, we obtain R R d d i! x n 1 f(x) = C d! e R0f(; !) d ! =1 R j j Zk k Z 1 n d n 1 with C := 2 (2) . Let F (t; !) be a smooth function on R S , rapidly decreasing n 1 with respect to t, and let the operator @t be de…ned by j j n 1 n 1 @t F (; !) = F (; !) : j j j j n 1 k n 1 n 1 Thus @t = ( 1) @t if n = 2k+1 is odd; if n is even @t is the composition n j 1 j b b j j of @t and a Hilbert integral operator (see Helgason [7] p. 22). We infer the following inversion formula n 1 f = CR @t R0f (2) 0j j where the dual transform R0 is de…ned by

R0F (x) := F (! x; !)d! ! =1 Zk k (integration over the set of all hyperplanes containing x).

2 2 A Nonlinear Radon Transform

2.1 Integration on Hypersurfaces

Let ' : R be a smooth function on an open subset of the Euclidean space ! Rn. A convenient way to introduce our Radon transform is to consider …rst, for f ( ) (a smooth function with compact support contained in ) and t R , 2 D 2

f'(t) := f(x)dx Z'(x)

R'f(t) := (f')0 (t) = @t f(x)dx: (3) Z'(x)

u('(x))f(x)dx = u(t)R'f(t)dt: (4) n ZR ZR (iii) Let dSt be the Euclidean measure on the hypersurface St := x '(x) = t . Then f 2 j g 1 R f(t) = f(x) dS (x): (5) ' ' (x) t ZSt k 0 k Formula (5) gives the geometrical meaning of R'f as an integral of f over the level hypersurface '(x) = t; we may write it for short as

R'f(t) = f: (6) Z'(x)=t According to (4) it may also be viewed as R'f(t) = ' t; f where ' t is the h i pullback by ' of the Dirac measure t of R at t (see Friedlander [6] Section 7.2 or Hörmander [8] Section 6.1). Proof. (i) and (iii) Given a we have '0(a) = 0 thus (for instance) @n'(a) = 0. By the inverse function theorem2 there exists an open6 neighborhood U of a such6 that the map x = (x0; xn) y = (x0;'(x)) is a di¤eomorphism of U onto V I, where 7! n 1 x0 = (x1; :::; xn 1), V is an open neighborhood of (a1; :::; an 1) in R and I is an open interval containing '(a). Let y = (y ; yn) x = (y ; (y ; yn)) denote the 0 7! 0 0 inverse map. Then dy = @n'(x) dx and, assuming supp f U, we have j j f f'(t) = f(x)dx = (y0; (y0; yn))dy0dyn: @n' Z'(x)

@i'(y0; (y0; t)) + @n'(y0; (y0; t))@i (y0; t) = 0

1=2 n 1 2 for i = 1; :::; n 1. It follows that ' = @n' 1 + (@i ) and, for t I, k 0k j j 1 2 P n 1 1=2 f R f(t) = (y ; (y ; t)) 1 + @ (y ; t) 2 dy ' ' 0 0 i 0 0 V 0 1 ! Z k k X f = (x)dS (x); ' t ZSt k 0k the hypersurface integral being computed by means of the parameters y0. The latter equality also holds for t = I (both sides vanish) and this proves (i) and (iii) for supp f U. The general2 case follows by partition of unity. (ii) Since supp R'f [m; M] we have M M M u(t)R'f(t)dt = u(t)(f')0 (t)dt = [u(t)f'(t)] u0(t)f'(t)dt m ZR Zm Zm = u(M) f(x)dx u0(t)f(x)dtdx: Z Z'(x)

M f(x)dx u0(t)dt = f(x)(u(M) u('(x)))dx Z Z'(x) Z and (4) follows.

2.2 Nonlinear Radon and Fourier Transforms We now wish to extend the classical Radon transform of Section 1, replacing the hyperplanes ! x = t by level hypersurfaces of homogeneous polynomials of given degree m 1 in Rn. We write such polynomials as

p(x) := x =m j Xj n n n where x R and, in multi-index notation, = ( 1; :::; n) N , = 1 i, 1 2 n 2 j j x = x xn and R. 1 2 P 4 It is easily checked that the number of terms in =m is the binomial coe¢ cient (m+n 1)! j j N = N(m; n) = m!(n 1)! . Indeed let us consider P n n 1 2 2 = 1 + txj + t xj + : 1 txj j=1 j=1 Y Y m Expanding the product we see that the coe¢ cient of t is =m x , therefore j j m equals N(m; n) when all xi’s are 1. Thus N(m; n) is the coe¢ cient of t in the expansion of (1 t) n and the result follows. Note that N > nPfor n 2 and m 2. Let RN , = 0, and := x p(x) = 0 . By Euler’s identity for the 2 6 f j n 6 g homogeneous function '(x) = p(x) on R the gradient '0 does not vanish on . The level surface p(x) = t is thus a smooth hypersurface of Rn for t R, t = 0. The nonlinear Radon transform of a test function f ( ) is then2 de…ned,6 in the notation of (6), by 2 D

Rf(t; ) := R'f(t) = f: (7) p(x)=t Z

For m = 1 we have N = n and R is the classical hyperplane Radon transform R0.

Properties of R. (i) By Proposition 1, for f ( ) and = 0, Rf(:; ) is a compactly supported 2 D 6 smooth function of t on R. By (4)

F ( p(x); ) f(x)dx = F (t; )Rf(t; )dt n ZR ZR for = 0 and any F continuous on R RN . In particular, for R, 6 2 i p(x) it e f(x)dx = e Rf(t; )dt = Rf(; ) = Rf(1; ) (8) n ZR ZR is the one-dimensional Fourier transform of Rf withc respect toc the variable t. This extends the projection slice theorem (1). (ii) The left-hand side of (8) is well-de…ned for all f (Rn) (without assuming 2 D supp f ), and extends to an entire function of (; ) on C CN . This suggests de…ning Rf(; 0) = f, that is Rf(t; 0) = f(x)dx (t) where is the Dirac Rn measure at the origin of . R R R Actually,c the restrictive assumptions supp f , t = 0, = 0 may be left out in the sequel, as we shall work with Rf rather than Rf6 . 6 (iii) From (8) it follows that c i p(x) @ Rf(; ) = i e x f(x)dx = iR\(x f)(; ); (9) n ZR therefore c

@ Rf(t; ) = @tR (x f)(t; ) (10) for f ( ), = 0 and Nn, = m. 2 D 6 2 j j 5 (iv) Note that, for m even, Rf = 0 whenever f is an odd function: R is not an injective map and, in this case, f cannot be reconstructed from Rf alone. We shall see in the next sections how to circumvent this di¢ culty.

Let us introduce the nonlinear Fourier transform of f de…ned, for all f 2 (Rn), by D i( x+ p(x)) n N f(; ) := e f(x)dx , R , R : (11) n 2 2 ZR It extends to ane entire function of (; ) Cn CN . As a function on Rn RN it is bounded by f(x) dx and, for …xed 2, it is rapidly decreasing with respect to . Rn j j On the one hand f(; 0) = f() is the classical n-dimensional Fourier transform R of f; on the other hand f(0; ) = Rf(; ) is the 1-dimensional Fourier transform of Rf: e b e c f(; )

.&e f(; 0) = f() f(0; ) = Rf(1; ):

Reconstructing f(; ) frome f(0; b) would thereforee allowc to reconstruct f from Rf. For this we shall consider partial di¤erential equations satis…ed by f. e e 2.3 Partial Di¤erential Equations e

Taking derivatives of (11) under the integral sign we get, for j = 1; :::; n and Nn, = m, 2 j j i( x+ p(x)) @j f(; ) = i e xjf(x)dx = i(^xjf)(; ) (12) n ZR e i( x+ p(x)) @ f(; ) = i e x f(x)dx = i(^x f)(; ): (13) n ZR e n N Thus f satis…es the system of N linear partial di¤erential equations on R R m 1 n e i @ f = @ f for N ; = m: (14) 2 j j

For any ; ; ; Nn of lengthe m suche that x x = x x we infer that, as a function of , f satis…es2 the Plücker equations

@ @ @ @ f = 0: (15) e Given ; , all such multi-indices ; are obtainede as = ", = + ", where n n " = ("1; :::; "n) Z satis…es j "j j for j = 1; :::; n and "j = 0. 2 1 2 2 P Example. For m = n = 2 we have p(x) = 1x1 + 2x2 + 3x1x2 (here N = 3) and 2 2 i@1 f = @1 f , i@2 f = @2 f , i@3 f = @1 @2 f: 2 2 2 2 The identity (x1x2) = x x leads to he hyperbolic equation @ f = @ @ f. e 1 2 e e e e e3 1 2

6 e e 3 A Cauchy Problem

Given f (Rn) let us now try to reconstruct f(; ) from f(0; ) = Rf(1; ) by solving a2 Cauchy D problem for the system (14) with data on = 0. In order to achieve this goal we shal l of course need moree than Rf(1; e) : let usc recall that f(0; ) = 0 for m even and f odd, though f may be not identically zero. It should be noted that f(0; ) satis…es the Plücker equations (15),c but this fact will not be takene into account here (see Remark belowe however). Since f is ane entire function we have e f(; ) = @ f(0; ) ; ! Nn X2 e e an absolutely convergent series for all Cn and CN . To work it out we shall only need the2 derivatives2 @ f(0; ) for < m; the j j higher order derivatives will be given by (14). More precisely, @ f = ij jx f for all m 1 e n by (12), and equals i @ f by (14) if = m. For any N we may write j j 2e g = qm + r with q; r N, 0 r < m, and factorize @ as j j 2 e @ = @ 1 @ q @ n with 1; ::: q; N , 1 = = q = m and = r; this factorization is not unique. It follows2 that j j j j j j

q @ f = ij j @ @ (^x f) 1 q and e q f(; ) = ij j @ @ (^x f)(0; ) 1 q ! Nn X2 (with q; 1; :::; q; edepending on in the sum). Remembering (^x f)(0; ) = R\(x f)(1; ) for = 0, we see that f is determined 6 by the nonlinear Radon transforms of all functions x f for Nn and < m. m 1 m 2 j j Their number is k=0 N(k; n) = N(m 1; n + 1) = n N(m; n) (inductione on m). In particular if R(x f) = 0 for all with < m, then f = 0. P j j Example. For m = n = 2 (Section 2.3), @ 1 @ 2 factorizes as powers of @2 and @2 , 1 2 1 2 possibly composed with @1 or @2 or @1 @2 according to the parity of 1 and 2. Gathering together similar terms the above result reads

f(; ) = C(D1)C(D2)f + S(D1)S(D2)D3f+ + i S(D )C(D )(^x f) + i C(D )S(D )(^x f) (16) e e 1 1 e 2 1 2 1 2 2 where

2 2 D1 = i1@1 , D2 = i2@2 , D3 = i12@3 1 zk 1 zk C(z) = , S(z) = (2k)! (2k + 1)! k=0 k=0 X X

7 and, in the right-hand side of (16), f, (^x1f), (^x2f) are evaluated at (0; ). Thus the knowledge of the three Radon transforms Rf, R (x1f) and R (x2f) determines f. Remark. The Plücker equations (15),e here @2 f = @ @ f, haven’t been taken 3 1 2 into account. They imply @2kf = (@ @ )k f, @2k+1f = (@ @ )k @ f for k e , 3 1 2 3 1 2 3 N hence the Taylor expansion e e 2 e e e e k f(0; ; ; ) = @k f(0; ; ; 0) 3 1 2 3 3 1 2 k! k N X2 e e = C (E) f(0; 1; 2; 0) + 3S (E)(@3 f)(0; 1; 2; 0) (17)

2 where E = 3@1 @2 , and similarlye e

@3 f(0; 1; 2; 3) = 3@1 @2 S (E) f(0; 1; 2; 0) + C (E)(@3 f)(0; 1; 2; 0): (18)

Combininge (16) (17) and (18) it followse that f can be reconstructede from f, @3 f,

(^x1f), @3 (^x1f), (^x2f) and @3 (^x2f) at (0; 1; 2; 0) only. e e e Remembering (13) @3 f = i(x^1x2f), these 6 functions can be replaced by f, (^x1f), ^2 ^2 \2 (^x2f), (x^1x2f), x1x2f and x1x2f , that is Rf, R\(x1f),..., R(x1x2f) evaluated e 2 e at (1; 1; 2; 0). In other words the integrals of f, x1f,..., x1x2f over the conics 2 2 1x1 + 2x2 = t will determine f. A stronger (andc more general) result is given in the next section.

4 Harmonic Polynomials and the Cauchy Problem

Two chapters of [5] are devoted to a general theory of harmonic polynomials which, when applied to nonlinear Radon transforms, leads to a re…ned version of the results of Section 3. We shall only present here a simpli…ed approach to the harmonic polynomials relevant to our problem. Notation. All polynomials considered here have complex coe¢ cients. Let us order m m the N monomials (x ) =m as x1 ; :::; xn …rst, then x B where B is the set of the N n remaining multi-indicesj j of length m. In accordance2 with this we replace our N n N n previous notation = ( ) =m R by (; ) R R with = (1; :::; n) j j 2 2 n m and = ( ) B ; the former x is replaced by j=1 jxj + B x . 2 n+N 2 n Let (x; p; q) R denote dual variables to (; ; ), with x = (x1; :::; xn) R , 2 n P N n P P 2 p = (p1; :::; pn) R and q = (q ) R . 2 B 2 In this new notation the partial2 di¤erential equations (14) become

m i@ f = i@ f , ( i@) f = i@ f for j = 1; :::; n and B: (19) j j 2 They are duale to e e e

m x F = pjF , (x q )F = 0 for j = 1; :::; n and B; (20) j 2 where F is the tempered distribution on Rn+N corresponding to f via the Fourier transform on Rn+N (being smooth and bounded, f is tempered on Rn+N ). e 8 e n N n N Let us introduce the following N polynomials on R R = R : m uj(x; q) := x , u (x; q) := x q for j = 1; :::; n and B: (21) j 2 The system (20) implies that the support of F is contained in the closed set V of Rn+N de…ned by the N equations

n n N n V = (x; p; q) R R R uj(x; q) = pj ; u (x; q) = 0 ; 1 j n; B : 2 j 2 n+N Being the graph of a map x (p; q), V is a n-dimensional submanifold of R . 7! n N n De…nition 2 A polynomial function h(x; q) on R R is called harmonic if

uj(@x;@q)h = 0 , u (@x;@q)h = 0 for j = 1; :::; n and B: 2 It is called homogeneous of degree d if h(tx; tmq) = tdh(x; q) for all t R (thus 2 each xj has degree 1 and each q has degree m).

D D Proposition 3 Let D := B q @x . Then u (@x;@q) = e @q e . The space of harmonic polynomials2 is mn-dimensional. Its elements are given by P h = eDf where f is an arbitrary polynomial of the following form

f(x) = a x with 0 j m 1 for j = 1; :::; n and a C: 2 Nn X2 Besides h = eDf is homogeneous of degree d (in the sense of De…nition 4) if and only if f is homogeneous of degree d.

Proof. Since u (@x;@q) = @x @q we have [D; u (@x;@q)] = @x and D;@x = 0, 2 thus (ad D) u (@x;@q) = 0 and h i

D D ad D e u (@x;@q)e = e u (@x;@q) = (1 ad D) u (@x;@q)

= u (@x;@q) @ = @q : x [This proof may also be written without any Lie formalism, by computing the deriv- tD tD ative with respect to t of e u (@x;@q)e .] Since eD is a linear isomorphism of the space of polynomials onto itself, a polynomial h(x; q) is harmonic if and only if

m D @ h = 0 , @q e h = 0 for j = 1; :::; n and B: xj 2 The latter equations imply h = eDf for some polynomial f in the x variables. Since m m D;@xj = 0 the former equations imply @xj f = 0 for j = 1; :::; n whence our claim habout fi. The operator D preserves homogeneity in (x; q) and the last statement follows.

9 Examples. Let us write down, as an example, a basis of homogeneous harmonic polynomials for n = 2 and m = 4. Here N = 5, = ( 1; 2) with 0 j 3, 1 2 1 + 2 = 4, q = (q13; q22; q31) and D = q 1 2 @x1 @x2 . The 16 monomials f(x) = a b x1x2, 0 a 3, 0 b 3, make up a basis of the relevant polynomials f. Since the degree off is 6at most we have D2Pf = 0 and the 16 corresponding harmonic polynomials are h = f + Df, that is1

2 2 3 2 2 3 1 , x1 , x2 , x1 , x1x2 , x2 , x1 , x1x2 , x1x2 , x2 , 3 2 2 3 x1x2 + 6q31 , x1x2 + 4q22 , x1x2 + 6q13 , 3 2 2 3 x1x2 + 12q22x1 + 12q31x2 , x1x2 + 12q13x1 + 12q22x2 , 3 3 2 2 x1x2 + 18q13x1 + 36q22x1x2 + 18q31x2 .

For m = n = 2 (already considered) we have N = 3, q R, and the corresponding basis of harmonic polynomials is 2

1 , x1 , x2 , x1x2 + q:

n More generally, let A denote the set of all N such that 0 j m 1 for D 2 j = 1; :::; n. By Proposition 5 the h := e x , A, make up a basis of the space of harmonic polynomials. 2

n N n Proposition 4 For any polynomial P (x; q) on R R there exists a family of n N m polynomials Q , A, on R such that 2

P (x; q) = Q (u1(x; q); :::; uN (x; q))h (x; q); A X2 where u1; :::; uN denote the polynomials de…ned by (21).

Proof. Let a; b = a(@)b(0) be the Fischer inner product on the space of polynomi- n hN ni als on R R . Then h is harmonic if and only if uk(@x;@q)h = 0 for k = 1; :::; N, i.e. auk; h = 0 for all polynomials a. The space of harmonic polynomials is thus h i N the orthogonal complement of the ideal k=1 ak(x; q)uk(x; q) generated by the u ’s(where the a ’sare arbitrary polynomials). k k nP o A given P (x; q) now has a unique decomposition as

N P = h + akuk k=1 X with h harmonic. Separating homogeneous components we may assume P is homogeneous of degree d (in the sense of De…nition 2). Since uk is homogeneous, each homogeneous component of a harmonic polynomial is harmonic. We may therefore assume h and all akuk homogeneous of degree d, therefore ak is homogeneous of degree d m. Writing similar decompositions for each ak the result easily follows. 1 Cf. [5] p. 312, where the coe¢ cients 16 should be replaced, I think, by 18.

10 Example. For m = n = 2 the generators and harmonic polynomials are respectively

2 2 u1 = x , u2 = x , u3 = x1x2 q 1 2 h0 = 1 , h1 = x1 , h2 = x2 , h3 = x1x2 + q and the …rst non-trivial examples of decomposition in Proposition 4 are:

2x1x2 = u3h0 + h3 , 2q = u3h0 + h3 x1q = u3h1 + u1h2 , x2q = u3h2 + u2h1 2 2 q = u1u2h0 u3h3 , 2x1x2q = (2u1u2 u )h0 u3h3: 3

Replacing xj by i@ and q by i@ we infer from Proposition 4 an equality j of di¤erential operators. Applying them to f we obtain

m P ( i@; i@)f = Q i@ ; ( ei@) ( i@ ) h ( i@; i@) f j A X2 e e = Q ( i@; 0) h ( i@; i@) f A X2 e in view of (19) and the commutativity of di¤erential operators. In particular all derivatives @ @ f may be written in this form with polynomials Q depending on ; whence, by Taylor’sformula on the variables (; ), e f(; ; ) = Q ( i@; 0)h ( i@; i@) f(0; ; 0) (22) ! ! X e n N n e where runs over all N , N and A. Remembering (12)(13) 2 2 2 i@ f = x f, i@ f = x f we have h ( i@ ; i@ ) f = (h (x; q)f) with q = x j P j for B. 2 e g e g e e Lemma 5 For all there exists a positive integer C such that, when replacing each q by x for B, 2

h (x; q) = h (x; (x ) B) = C x : 2

Proof. For Nn we have 2 ! Dx = q @ x = q x x ( )! B B X2 X2 ! D2x = q q x ( )! ; B X2 etc (the coe¢ cients being 0 unless , resp. + ). When replacing q by 2 x , q by x etc, the polynomials Dx , D x etc thus become x times a positive D integer coe¢ cient. The same holds for h = e x , whence the lemma.

Going back to (22) we have h ( i@; i@) f = C x f and we conclude that, n n N n for (; ; ) R R R , 2 e g f(; ; ) = C Q ( i@ ; 0)(^x f)(0; ; 0) : ! ! ;; X e Therefore the restriction to all (0; ; 0) of the mn functions x f, A, determines f. In other words, the Cauchy problem for (19) is well-posed with the2 Cauchy data n+N h ( i@; i@) f = C x f on the n-plane of R de…nedg by = = 0. In terms of Radon transforms we obtain the following result. e g Theorem 6 A function f (Rn) is uniquely determined by the mn nonlinear 2 D n n Radon transforms R (x f)(t; ; 0) (with N , 0 j < m, t R, R 0 ), that is by the integrals of each x f on the hypersurfaces2 2 2 n f g

m m 1x + + nx = t: 1 n 5 Inversion Formulas

Let us now look for an inversion formula for the nonlinear Radon transform. The nonlinear Fourier transform f is greatly overdetermined, with n + N variables (; ) instead of n for f. As in Section 3 we shall restrict f to = 0 and, assuming m m the monomials x are orderede as x1 ; :::; xn …rst, followed by the other x ’s, it turns out that (as in the …nal remark of Section 3) wee can also restrict to = n (1; :::; n; 0; :::; 0), written as R for short. Then 2 i n xm n f(0; ) = e 1 j j f(x)dx = Rf(; ) with R; R : (23) n 2 2 ZR P e c 5.1 First Case: m odd n Let U denote the dense open subset of R de…ned by xj = 0 for all j. For m odd the map : x y = xm := (xm; :::; xm) is a di¤eomorphism6 of U onto itself. Then 7! 1 n i y Rf(; ) = e g(y)dy = g() (24) n ZR n with y = jyj andc b 1 P n (1=m) 1 1=m n g(y) := m y1 yn f(y ) , y R : j j 2 As above g denotes the classical n-dimensional Fourier transform and Rf is the 1-dimensional Fourier transform with respect to t. The change x y thus reduces the nonlinear Radon transform R to thec linear b 7! one considered in the introduction: Rf(t; ) = R0g(t; ). But g is not necessarily smooth, g() = Rf(1; ) is not necessarily rapidly decreasing and the inversion formula (2) may become invalid here. However g is integrable on Rn and vanishes outside a compactc set, therefore de…nes a tempered distribution. Denoting by the b F 12 inverse Fourier transform for tempered distributions on Rn we have g = g hence, for any u (U), F 2 D b f(x)u(xm)dx = g(y)u(y)dy = g(y); u(y) hF i ZU ZU = ( g)(x); det 0(x) u( (x) F j b j n m 1 m = m (x1 xn) ( g)(x); u(x ) ; b F using the pullback by of the distribution g on U (cf. [6] p. 80). The absolute F b value may be skipped here since m 1 is even and det 0 > 0. Therefore, for n f (R ), b 2 D n m 1 f(x) = m (x1 xn) ( Rf(1; ))(x); (25) F an equality of distributions on U. c

5.2 Second Case: m even

The above map : x y is no more a bijection: given y with all yj > 0, the m 7! n 1=m 1=m equations y = x now have 2 solutions x = y ; :::; yn . 1 n n For x; y R we write xy := (x1y1; :::; xnyn). Let E := 1; 1 denote the set 2 f g of all " = ("1; :::; "n) with "j = 1 and n n R+ := x R xj > 0 for 1 j n : f 2 j g n Viewing the integral (23) as a sum of integrals over the quadrants "R+, " E, we 1=m n 2 obtain, by the change of variables x y with xj = "jy , yj > 0, on "R+, 7! j

i y Rf(; ) = f(0; ) = e g(y)dy n ZR+ n c en with R, R and, for y R+, 2 2 2 n (1=m) 1 1=m g(y) := m (y1 yn) f("y ): " E X2 n Let H denote the Heaviside function H(y) = 1 if y R+, H(y) = 0 otherwise. Equation (24) is now replaced by 2

i y Rf(; ) = e H(y)g(y)dy = Hg(): n ZR c c n Again Hg is integrable and vanishes outside a compact set, hence tempered on R , and as above the Fourier inversion Hg = Hg implies the following equality of n F distributions on R+ c n m 1 f("x) = m (x1 xn) ( Rf(1; ))(x): (26) F " E X2 c 13 n This gives f if its support is contained in some quadrant "R+. Otherwise we must separate the components f("x), which can be achieved by replacing f with x f for suitably chosen ’sas follows. With each " = ("1; :::; "n) E we associate the monomial 2 p"(x) := xi xi 1 k where 1 i1 < < ik n is the (ordered) set of indices i such that "i = 1; for instance, n = 4 and " = ( 1; 1; 1; 1) yield p"(x) = x1x3. The map " p" is a 7! bijection of E onto the set of divisors of x1 xn. Let "; E. A minus sign occurs in p"(x) = p"(1x1; :::; nxn) each time there 2 is a factor xi, that is "i = 1, and the corresponding i is 1. Therefore k(";) p"(x) = a";p"(x) with a"; := ( 1) ; (27) where k("; ) denotes the number of indices i such that "i = i = 1. Example. For n = 2 the matrix (a";) is given by the table:

p" 1 x1 x2 x1x2 " ++ + + ++ 1 1 1 1 + 1 1 1 1 + 1 1 1 1 1 1 1 1 Our inversion formula for R will be inferred from the following combinatorial lemma.

Lemma 7 The set E = 1; 1 n being provided with some ordering, the 2n 2n f g 2 n matrix A = (a";)"; E is symmetric and A = 2 I (where I is the unit matrix). 2 Proof. The symmetry is clear by the de…nition of k("; ). For "; ; E we have k("; ) = k("; ) + k("; ) since "i = ii = 1 is equivalent 2 to "i = 1 and i = 1, i = 1 or (exclusive or) "i = 1 and i = 1, i = 1. Therefore a";a"; = a"; : (28) Besides, for …xed E, 2 n (1 + ixi) = 1 + ixi + ijxixj + + 1 nx1 xn i=1 i i

Let us consider Sf(x) := E f(x). Replacing f by p"f we obtain, in view of (27), 2 P S(p"f)(x) = (p"f)(x) = p"(x) a";f(x); E X2 X 1 n which can be inverted by A = 2 A (Lemma 7) as

n 1 f(x) = 2 a";p"(x) S(p"f)(x) " E X2 for each E. By (26) applied to each p"f we have 2 n m 1 S(p"f)(x) = m (x1 xn) ( Rp["f(1; ))(x) F n on R+ and the latter equations show that f can be reconstructed in each quadrant of n n R from the 2 nonlinear Radon transforms Rf, R(xif), R(xixjf),..., R(x1 xnf). Summarizing we have proved the following theorem. Let us recall our notation: Rf = Rf(1; ) is given by (23) with Rn, is the inverse Fourier transform n 2 F of tempered distributions on R , is the pullback of distributions by (x) = m m n (cx ; :::;c x ), E = 1; 1 and p", a"; are de…ned before Lemma 7. 1 n f g Theorem 8 The nonlinear Radon transform (7) is inverted by the following formulas, where f (Rn). 2 D (i) if m is odd n m 1 f(x) = m (x1 xn) ( Rf)(x) F (equality of distributions on the open set x1 = 0; :::; xn = 0); 6 c6 (ii) if m is even: for E, 2 n m 1 m 1 f(x) = a";p"(x) (x1 xn) ( Rp["f)(x) 2 F " E X2

(equality of distributions on the open set x1 > 0; :::; xn > 0).

References

[1] Ehrenpreis, Leon, The Radon Transform and Tensor Products, Contemp. Math. 113 (1990), 57-63.

[2] Ehrenpreis, Leon, Nonlinear Fourier Transform, Contemp. Math. 140 (1992), 39-48.

15 [3] Ehrenpreis, Leon, Parametric and Non Parametric Radon Transform, in 75 Years of Radon Transform, International Press 1994.

[4] Ehrenpreis, Leon, Some Nonlinear Aspects of the Radon Transform, Lectures in Applied Math. 30 (1994), 69-81.

[5] Ehrenpreis, Leon, The Universality of the Radon Transform, Oxford University Press 2003.

[6] Friedlander, F., Introduction to the Theory of Distributions, Cambridge Univer- sity Press 1982.

[7] Helgason, Sigur†ur, Integral Geometry and Radon transforms, Springer 2011.

[8] Hörmander, Lars, The Analysis of Linear Partial Di¤erential Operators I, Springer-Verlag 1983.