On the Recovery Of a 2-D Function From the Modulus Of Its Fourier Transform

Transcription

1 On the Recovery Of a -D Function From the Modulus Of Its Fourier Transform Michael V. Klibanov Department of Mathematics and Statistics and Center for Optoelectronics University of North Carolina at Charlotte Charlotte, NC 83 mklibanv@ .uncc.edu October 3, 5 Abstract A uniqueness theorem is proven for the problem of the recovery of a complex valued compactly supported -D function from the modulus of its Fourier transform. An application to the phase problem in optics is discussed. 1 Introduction Let Ω R be a bounded domain and f ξ, η C Ω be a complex valued function. Consider its Fourier transform F x, y = f ξ, η e ixξ e iyη dξdη, x, y R. 1.1 Let Ω Gx, y = F x, y, x, y R. 1. We are interested in the question of the uniqueness of the following Problem. Given the function Gx, y, determine the function f ξ, η. The right hand side of 1.1 can often be interpreted as an optical signal whose amplitude and phase are F x, y and arg F x, y respectively, see, e.g., []. This problem is also called the phase problem in optics PPO meaning that only the amplitude of such an optical signal is measured. The latter reflects the fact that it is often impossible to measure the phase in optics, except of the case when the so-called reference signal is present e.g., the case of holography [18], see, e.g., [3]-[9], [11]-[16], [19], and []. 1

2 We assume that Ω =, 1, 1 is a square and the function f ξ, η has the form f ξ, η = exp [iϕ ξ, η], 1.3 where the real valued function ϕ C 4 Ω. be a small positive number. Denote Let Γ be the boundary of the square Ω and δ Ω δ Γ = {ξ, η Ω : dist [ξ, η, Γ] < δ}, where dist [ξ, η, Γ] is the Hausdorf distance between the point ξ, η and Γ. Hence, the subdomain Ω δ Γ Ω is a small neighborhood of the boundary Γ. The following uniqueness theorem is the main result of this paper Theorem 1. Assume that two functions f 1 ξ, η = exp [iϕ 1 ξ, η] and f ξ, η = exp [iϕ ξ, η] of the form 1.3 are solutions of the equation 1. with real valued functions ϕ 1, ϕ C 4 Ω satisfying conditions 1.4 and 1.5, where ϕ 1 ξ, η = ϕ ξ, η, ξ, η Ω, 1.4 Also, assume that either ϕ ξ, 1 η = ϕ ξ, η, ξ, η Ω. 1.5 ϕ jξ, > and ϕ jη, > for j = 1, 1.6a or ϕ jξ, < and ϕ jη, < for j = 1,. 1.6b In addition, let ϕ 1, = ϕ, = and both functions ϕ 1 ξ, η and ϕ ξ, η are analytic in a small neighborhood Ω δ Γ of the boundary Γ of the domain Ω as functions of two real variables ξ, η. Then ϕ 1 ξ, η = ϕ ξ, η in Ω. Remarks. a. We need conditions 1.4 and 1.5 for proofs of lemmata and 7. We need conditions 1.6a,b for the proof of Lemma, and, in a weaker form for the proof of Lemma 7. The condition of the analyticity of functions ϕ 1 ξ, η and ϕ ξ, η in Ω δ Γ can be replaced with the assumption that ϕ 1 ξ, η = ϕ ξ, η in Ω δ Γ. Such an assumption is often acceptable in the field of inverse problems. It should be pointed out that Lemma does not guarantee the uniqueness in the entire domain Ω. Now, if one would assume a priori that ϕ 1 ξ, η = ϕ ξ, η in Ω δ Γ thus focusing one the search of the function ϕ ξ, η in the major part Ω Ω δ Γ of the domain Ω, then it would be sufficient for the proof of Lemma 7 to replace 1.6a,b with ϕ ξ,, see 3.36 and b. To explain the assumption ϕ 1, = ϕ, =, we note that if the function f ξ, η is a solution of the equation 1., then functions f ξ, η e ic and f ξ, η e ic with an arbitrary real constant c are also solutions of this equation. Throughout the paper f denotes the complex conjugation. Hence, Theorem 1 can be reformulated by taking into account functions ϕ ξ, η + c and ϕ ξ, η + c, along with the function ϕ ξ, η.

3 Throughout the paper we assume that conditions of Theorem 1 are satisfied. Everywhere below j = 1,. Denote F j x, y = f j ξ, η e ixξ e iyη dξdη, x, y R, 1.7 Ω G j x, y = F j x, y, 1.8 where f j ξ, η = exp iϕ j ξ, η. So, we need to prove that the equality G 1 x, y = G x, y in R 1.9 implies that f 1 ξ, η = f ξ, η in Ω. The form 1.3 is chosen for two reasons. First of all, if both functions f ξ, η and arg [f ξ, η] would be unknown simultaneously, then 1. would be one equation with two unknown functions. It is unlikely that a uniqueness result might be proven for such an equation without some stringent additional assumptions. Another indication of this is an example of the non-uniqueness in section 3. Second, the representation 1.3 is quite acceptable in optics, see, e.g., [6], [1] and [13]. Derivations in these references are similar and, briefly, are as follows. Suppose that the plane {x 3 = } in the space R 3 = {x 1, x, x 3 } is an opaque sheet from which an aperture Ω is cut off. Suppose that a phase screen S is placed in the aperture Ω. The phase screen means a thin lens which changes only the phase of the optical signal transmitted through it, but it does not change its amplitude. For each point x 1, x, Ω consider the intersection of the straight line L x 1, x orthogonal to the plane {x 3 = } with S, i.e., consider L x 1, x S. Let ψ x 1, x and nx 1, x be respectively the thickness and the refraction index of this intersection. Suppose, a plane wave u = exp ikx 3 propagates in the half-space {x 3 < }. Consider the positive half-space {x 3 > }. Then the function F x, y of the form 1.1, 1.3 with ϕ x 1, x := k [nx 1, x 1] ψ x 1, x is approximately proportional to the wave field in the so-called Fraunhofer zone [], i.e., with kx 3 >> 1. Hence, our problem can be viewed as an inverse problem of the determination of the function ϕ x 1, x characterizing the phase screen from the amplitude of the scattered field measured far away from that screen. In addition, see, e.g., the paper [7], where the function ϕ x 1, x is called the aberration function phase errors and its reconstruction seems to be the subject of the main interest of [7]. The function F x, y can be continued in the complex plane C as an entire analytic function with respect to any of two variables x or y, while another one is kept real. It follows from the Paley-Wiener theorem [1] that the resulting function F z, y will be an entire analytic function of the first order of the variable z C. The major difference between 1-D and -D cases is that, unlike the 1-D case zeros of an analytic function of two or more complex variables are not necessarily isolated. For this reason, we consider below the analytic continuation of F x, y with respect to x only and keep y R. Thus, we consider the function F z, y, z C, y R. The example of the non-uniqueness in section 3 indicates that our main effort should be focused on the proof that complex zeros of the function F z, y can 3

4 be determined uniquely for each y a, b, where a, b R is a certain interval. This is achieved in two stages. First, we prove that asymptotic zeros can be determined uniquely Lemma 7. Next, it is shown that the rest of zeros can also be uniquely determined sections 4 and 5. The first uniqueness theorem for the PPO was proven by Calderon and Pepinsky [5]; also see the paper of Wolf [] for a similar result. These publications were concerned with the case of a real valued centro-symmetrical function f, which is different from our case of the complex valued centro-symmetrical function f satisfying conditions The latter causes a substantial difference in proofs of corresponding uniqueness results. Many publications discuss a variety of aspects of the PPO, see, e.g., above cited ones and references cited there; a recently published introduction to the PPO can be found in [8]. We also refer to the paper [17], which is concerned with the inverse problem of shape reconstruction from the modulus of the far field data; the mathematical statement of this problem is different from the above. A uniqueness result for the discrete case was proven in [3], where the function f is a linear combination of δ functions. For the continuous -D case, uniqueness theorems for the problem were proven in [1] and [13] assuming that ϕ C Ω. The goal of this publication is to replace the C with the C 4 via exploring some new ideas. The main new idea is presented in section 4. It an opinion of the author that the proof of Lemma 8 of this section is the most difficult element of this paper. The rest of the paper is devoted to the proof of Theorem 1. In section we prove that the function ϕx, y can be reconstructed uniquely near the boundary Γ of the domain Ω. In section 3 five lemmata are proven. In section 4 one more lemma is proven. We finalize the proof of Theorem 1 in section 5. Uniqueness In Ω δ Γ Results, similar with lemmata 1 and of this section were proven in [13] see lemmata.5-.7 and the proof of Theorem 3.1 in this reference. However, since the reference [13] is not easily available, it makes sense to present full proofs of lemmata 1 and here. In addition, these proofs are both significantly simplified and clarified compared with those of [13]. To prove that ϕ 1 = ϕ in Ω δ Γ, we need to analyze some integral equations. For any number ε, 1 denote P ε = { < x, y < ε}, a subdomain of the square Ω. Lemma 1. Let the number ε, 1. Let complex valued functions q, K s s = 1,, 3 be such that qx, y C P ε, K1 x, y, ξ, K x, y, η C P ε [, ε], and K 3 x, y, ξ, η C P ε P ε. Also, let K 1,, = K,, = 1.1 and q, = q x, = q y, =.. 4

5 Suppose that the complex valued function ux, y C P ε satisfies the integral equation α [αx + βy + qx, y] ux, y = K 1 x, y, ξ u ξ, y dξ + β y K x, y, η u x, η dη.3 + y K 3 x, y, ξ, η u ξ, η dηdξ, in P ε, and where α and β are two real numbers such that u, =,.4 αβ >..5 Then there exists such a number ε, ε] depending only on numbers α, β and functions q, K s s = 1,, 3 that ux, y = for x, y P ε. Note that because of the presence of the factor [αx + βy + qx, y] in the left hand side of the equation.3, this is not a standard Volterra equation. Indeed, we cannot simply divide both sides of.3 by [αx + βy + qx, y], since [αx + βy + qx, y] x=y= =. Proof of Lemma 1. We first prove that there exists a number ε, ε] such that ux, = and u, y =, for x, y, ε..6 Set in.3 y =. Denote vx = ux,, q x = qx, and K x, ξ = K 1 x,, ξ 1. Then [αx + q x] vx = α [1 + K x, ξ] v ξ dξ, x, ε. Differentiating this equation with respect to x, we obtain for x, ε [αx + q x] v x + q xvx = αk x, x vx + α By.1 and. we have for x, ε K x x, ξ v ξ dξ..7 K x, x = d [K ξ, ξ] x dξ, q x = q dξ ξ x ξ dξ, q x = q ξdξ. 5

6 In particular, this means that q x = ox as x. Hence, since by.5 α, then there exists a number ε 1, ε] depending on the number α and the function q x such that functions x αx + q x, K x, x Kx = αx + q x and qx = q x αx + q x are bounded in [, ε 1 ]. Divide.7 by the function αx + q x and integrate the resulting equality then. We obtain for x, ε 1 vx = [ ] K q ξ v ξ dξ + α dτ τ K x τ, ξ v ξ dξ..8 ατ + q τ Let αk x τ, ξ M for τ, ξ P ε1 P ε1, where M is a positive number. Then α dτ τ K x τ, ξ v ξ dξ ατ + q τ M dτ τ v ξ dξ.9 ατ + q τ where M 1 V x x, x, ε 1, V x = max v ξ.1 ξ x and the positive number M 1 depends only on M, α and q C[,ε1 ]. Hence,.8 and.9 imply that the following estimate takes place with another positive constant M depending only on M, α and norms K q, q C[,ε1 ] C[,ε1 ] Let t, ε 1 be an arbitrary number. By.11, vx M V x x, x, ε max vx max [M V x x]. x t x t Since the function V xx is monotonically increasing, then the latter inequality leads to V t M V t t, t, ε 1..1 Choose a number ε, ε 1 such that M ε < 1/. Then.1 leads to V t V t, t, ε. Hence V x = for x, ε. This and.1 imply that ux, := vx = in, ε, which is the first equality.6. The second equality.6 can be proven similarly. 6

7 Denote K 1 x, y, ξ = K 1 x, y, ξ 1, K x, y, η = K x, y, η Let in.3 x, y P ε, where the number ε, ε is the same as in.6. Apply the operator y x to both sides of.3. Using.1, we obtain αx + βy + qx, y u xy + q x u y + q y u x = [α K 1 x, y, x u y + β K ] x, y, y u x + α [ + α K 1y x, y, x + β K ] x x, y, y + K 3 x, y, x, y u K 1x x, y, ξ u y ξ, y dξ + β y K y x, y, η u x x, η dη.14 +α K 1xy x, y, ξ u ξ, y dξ + β y K xy x, y, η u x, η dη + K 3x x, y, ξ, y u ξ, y dξ + y K 3y x, y, x, η u x, η dη + y K 3xy x, y, ξ, η u ξ, η dηdξ, x, y P ε. The Taylor s formula and. imply that the function q 1 x, y = qx, y/ x + y is bounded in P ε. Hence,.5 implies that there exists such a number ε, ε] that functions x αx + βy + qx, y and y αx + βy + qx, y.15 are bounded in P ε. To see this, it is sufficient to introduce polar coordinates x = r cos θ and y = r sin θ with θ [, π/]. Further, the Taylor s formula,.1 and.13 imply that functions α K 1 x, y, x αx + βy + qx, y and β K x, y, y.16 αx + βy + qx, y are also bounded in P ε. In addition, by.4 u y, y = u x x, = and ux, y = y u y x, η dη = u x ξ, y dξ..17 For t, ε denote wt = max x,y t [ u xx, y + u y x, y ]..18 7

8 Using.17, substitute ux, y = y u y x, η dη.19 in the right hand side of.14. Next, divide both sides of.14 by the function [αx + βy + qx, y] and apply the operator... dξ. to both sides of the resulting equality. Note that all kernels of integral operators in.14 are bounded. Also, since functions.15 are bounded, then and dξ u x ξ, η dη Qwt t, for x, y P t, αξ + βy + qξ, y t, ε.1 dξ u y τ, y dτ Qwt t, for x, y P t, αξ + βy + qξ, y t, ε.. Here and below in this proof Q denotes different positive constants independent on the parameter t, ε and functions u and w. Thus, using the fact that functions.15 and.16 are bounded and using also estimates.1 and., we conclude that the application of the operator. to the equality, which is obtained from.14 after the substitution.19 and division by the function [αx + βy + qx, y] leads to the following estimate u y x, y Qwt t, for x, y P t, t, ε..3 On the other hand, substituting in.14 ux, y = u x ξ, y dξ, dividing it then by the function [αx + βy + qx, y] and applying the operator we similarly obtain y...dη, u x x, y Qwt t, for x, y P t, t, ε..4 8

9 Summing up.3 and.4, we obtain u x x, y + u y x, y Qwt t, for x, y P t, t, ε. By.18, this is equivalent with u x x, y + u y x, y Qt max x,y t [ u xx, y + u y x, y ], for x, y P t, t, ε. Hence, max [ u xx, y + u y x, y ] Qt max [ u xx, y + u y x, y ], t, ε. x,y t x,y t The latter inequality and.18 lead to wt Qwt t, for t, ε. Choose the number ε, ε such that Qε < 1/. Then the latter inequality implies that wt wt, for t, ε. Hence, wt = for t, ε. This,.6 and.18 imply that ux, y = for x, y P ε. Lemma. ϕ 1 = ϕ in Ω δ Γ. Proof. For the sake of definiteness, we assume in this proof that the condition 1.6a is fulfilled. The proof in the case 1.6b is similar. Consider the function hx, y, Since ϕ 1 ϕ, =, then hx, y = sin [ ϕ1 ϕ x, y ] h, =..5 Since both functions ϕ 1 and ϕ are analytic in Ω δ Γ, it is sufficient to prove that hx, y = for x, y P σ for a number σ, 1. Let G x, y be the inverse Fourier transform of the function Gx, y defined in 1.. Then G x, y = fx + ξ, y + ηf ξ, η χ x + ξ, y + η χ ξ, η dξdη,.6 R where χ ξ, η is the characteristic function of the square Ω. Assuming that x, y Ω, we can rewrite the equality.6 in the form G x, y = 1 x 1 y fx + ξ, y + ηf ξ, η dξdη..7 9

10 Suppose that ϕ 1 ξ, η ϕ ξ, η in Ω δ Γ. Recall that f 1 ξ, η = exp [iϕ 1 ξ, η] and f ξ, η = exp [iϕ ξ, η]. Denote gξ, η = f 1 ξ, η f ξ, η. Then.7 leads to 1 x 1 y f 1 x + ξ, y + ηf 1 ξ, η dξdη 1 x 1 y f x + ξ, y + ηf ξ, η dξdη =..8 Since then.8 implies that 1 x 1 y f 1 x + ξ, y + η f 1 ξ, η f x + ξ, y + η f ξ, η = f 1 x + ξ, y + η g ξ, η + gx + ξ, y + η f ξ, η, f 1 x + ξ, y + ηg ξ, η dηdξ + 1 x 1 y gx + ξ, y + ηf ξ, η dηdξ =..9 Consider the second integral in.9. Changing variables, we obtain 1 x 1 y gx + ξ, y + ηf ξ, η dηdξ = 1 1 gξ, ηf ξ x, η y dηdξ..3 x y Note that by 1.4 and 1.5 gξ, η = g1 ξ, 1 η. Substituting this in the integral in the right hand side of.3 and changing variables ξ, η = 1 ξ, 1 η, we obtain 1 1 gξ, ηf ξ x, η y dηdξ =.31 x y 1 x 1 y g ξ, η f 1 ξ + x, 1 η + y dη dξ. Since by 1.4 and 1.5 f 1 ξ + x, 1 η + y = f ξ + x, η + y, then lead to 1 x 1 y f 1 x + ξ, y + ηg ξ, η dηdξ x 1 y f x + ξ, y + η gξ, ηdηdξ =, for x, y Ω. 1

11 It is convenient to make another change of variables x, y x, y = 1 x, 1 y and still keep the same notations for these new ones for brevity. Since by 1.4 and 1.5 f j 1 x + ξ, 1 y + η = f j x ξ, y η, j = 1,, then.3 becomes y [ f1 x ξ, y ηg ξ, η + f x ξ, y η gξ, η ] dηdξ =, for x, y Ω. Apply the operator y x to this equality. Note that f 1, = f, = 1. We obtain for x, y Ω We have Hence, Hence, Denote gx, y + gx, y = y y [ f1x x ξ, gξ, y + f x x ξ, gξ, y ] dξ [ f1y, y ηgx, η + f y, y ηgx, η ] dη.33 [ f1xy x ξ, y η gξ, η + f xy x ξ, y η gξ, η ] dηdξ. g = e iϕ 1 e iϕ = cos ϕ1 cos ϕ + i cos ϕ 1 cos ϕ [ ] ϕ ϕ = sin 1 ϕ1 + ϕ sin ϕ1 + ϕ i cos. [ ] ϕ1 + ϕ gx, y = hx, y sin ϕ1 + ϕ i cos..34 gx, y + gx, y = hx, y sin r 1 x, y = sin [ ] ϕ1 + ϕ x, y..35 [ ] ϕ1 + ϕ x, y. Since ϕ 1, = ϕ, =, then the Taylor s formula implies that the function r 1 x, y can be represented in the form r 1 x, y = αx + βy + qx, y,.36 where α = ϕ 1x, + ϕ x,, β = ϕ 1y, + ϕ y,.37 11

12 and the function qx, y C Ω satisfies conditions.. By 1.6a and.37 α, β >..38 Hence.35 and.36 imply that the function gx, y + gx, y can be represented as gx, y + gx, y = [αx + βy + qx, y] hx, y..39 Consider first two integrals in the right hand side of.33. Using the Taylor s formula,.34 and.37, we obtain [ f 1x x ξ, gξ, y + f x x ξ, gξ, y ] [ sin = iϕ 1x, [1 + r x ξ] ϕ1 + ϕ ϕ1 + ϕ ξ, y + i cos ξ, y ] hξ, y+ iϕ x, [1 + r 3 x ξ].4 [ ] ϕ1 + ϕ sin ϕ1 + ϕ ξ, y i cos ξ, y hξ, y = [ϕ 1x, + ϕ x, ] [1 + G 1 x, y, ξ] hξ, y = α [1 + G 1 x, y, ξ] hξ, y, where functions r x, r 3 x C [, 1], the function G 1 x, y, ξ C Ω [, 1] and r = r 3 = G 1,, =. Similarly [ f 1y, y η gx, η + f y, y η gx, η ] = β [1 + G x, y, η] hx, η,.41 where the function G x, y, η C Ω [, 1] and G,, =. Denote K 1 x, y, ξ := [1 + G 1 x, y, ξ], K x, y, η := 1 + G x, y, η, Thus, K 3 x, y, ξ, η := f 1xy x ξ, y η +f xy x ξ, y η [ sin [ sin ϕ1 + ϕ ϕ1 + ϕ ξ, η i cos ξ, η + i cos ϕ1 + ϕ K 1, K C Ω [, 1], K 3 C Ω Ω. ϕ1 + ϕ ξ, η ] ξ, η Further,.34 and imply that the function hx, y satisfies the following integral equation [αx + βy + qx, y] hx, y = α K 1 x, y, ξ u ξ, y dξ + β y ]. K x, y, η u x, η dη.4 1

13 + y K 3 x, y, ξ, η u ξ, η dηdξ, in Ω, where numbers α, β and functions q, K 1, K and K 3 satisfy conditions of Lemma 1. Thus, by Lemma 1 there exists a number σ, 1 such that hx, y = for x, y P σ. Note, however that it does not follow from the proof of Lemma that the function [αx + βy + qx, y] 1 has no singularities at points x, y Ω, which are located far from the boundary Γ. Hence, it is unclear what does the equation.4 imply for these points. Thus, we should proceed with the proof of Theorem 1. 3 Lemmata By 1. Gx, y = F x, y F x, y, x, y R. 3.1 Hence, the analytic continuation Gz, y of the function Gx, y is Gz, y = fξ, ηe izξ e iyη dξdη fξ, ηe izξ e iyη dξdη. 3. Ω Ω Denote F z, y = F z, y. Then one can rewrite 3. as Gz, y = F z, y F z, y. 3.3 Hence, Gz, y and F z, y are entire analytic functions of the first order of the variable z C for every y R. Since functions F 1 z, y and F z, y are analytic with respect to y R as functions of the real variable, it is sufficient to prove that F 1 z, y = F z, y for z C and for every y a, b for an interval a, b R. And this is what is done in this paper below. Consider an example of the non-uniqueness, which is sometimes called the complex zero-flipping in the physics literature, see, e.g., [8]. The function F z, y can be represented in the form [1] F z, y = z ky e gz,y n=1 1 z z exp, y R, 3.4 a n y a n y where ky is an integer, gz, y is a linear function of z and {a n y} n=1 is the set of zeros of the function F z, y. Each zero is counted as many times as its multiplicity is. The integer ky, the function gz, y and zeros {a n y} n=1 depend on y as on a parameter. Specific types of such dependencies e.g., analytic, continuous, etc. do not affect the rest of the proof of Theorem 1. Thus, for brevity we will not indicate dependencies of these on 13

14 y in some but not all formulas below. Suppose, for example that Im a 1. Consider the function F x, y, Note that F x, y = x a 1 x a 1 F x, y. x b x b = 1, x R, b C. 3.5 Hence, by 3.5 F x, y = F x, y for all x, y R. In addition, it can be easily shown that the inverse Fourier transform f ξ, η of F x, y has its support in Ω. Thus, the most difficult aspect of the PPO is to determine complex zeros in 3.4. Lemma 3. For an y, let {a n } n=1 C be the set of all zeros of the function F z, y as indicated in 3.4. Then {a n } n=1 is the set of all zeros of the function F z, y. Thus, F a, y = F a, y =. The multiplicity of each zero z = a of the function F z, y equals the multiplicity of the zero z = a of the function F z, y. The set of zeros of the function Gz, y is {a n } n=1 {a n} n=1. Proof. Let F a, y =. This means that fξ, ηe izξ e iyη dξdη z=a =. Consider the complex conjugate of both sides, = fξ, ηe izξ e iyη dξdη z=a = Ω fξ, ηe izξ e iyη dξdη z=a = F a, y = F a, y. S S Further, let z = a be a zero of the multiplicity s. Then differentiating last two formulas k 1 k s times with respect to z, we obtain the statement of this lemma about the multiplicity. Lemma 4. For each y R real zeros of functions F 1 z, y and F z, y coincide. Proof. By 1.9 G 1 z, y = G z, y, z C, y R. Hence, for any fixed y R all zeros real and complex of functions G 1 z, y and G z, y coincide. By 3.3 and Lemma 3 the multiplicity of each real zero x = a of the function G j x, y is twice the multiplicity of the zero x = a of the function F j x, y. First, consider the problem of the determination of the number ky and the function gz, y in 3.4. For a positive integer m denote I m = mπ + π/, mπ + 3π/. Lemma 5. Suppose that there exists a positive number N such that for every integer m > N sets of zeros of functions F 1 z, y and F z, y coincide for every y I m. Then there exists a number N > N such that for every integer m > N and for every number y I m corresponding numbers k 1 y and k y and functions g 1 z, y and g z, y in products 3.4 for functions F 1 and F coincide. 14

15 Proof. Denote p j ξ, y = 1 e iyη f j ξ, η dη By Lemma p 1 ξ, y = p ξ, y for ξ [, δ 1 δ, 1]. Hence, we can denote By 1.4 and p ξ, y := p 1 ξ, y = p ξ, y for ξ [, δ 1 δ, 1]. 3.1 p1, y = p, y 3.13 p ξ 1, y = p ξ, y Using 1.7, 3.11 and 3.13 we obtain F j z, y = 1 e iz 1 1 p1, y e izξ p jξ ξ, ydξ iz Hence, F j z, y = e iz [p1, y + o1], for Im z iz 3.16a and F j z, y = 1 [p1, y + o1], for Im z. 3.16b iz Setting in 3.11 ξ := 1, integrating by parts and recalling that f1, 1 = f, = 1, we obtain p1, y = 1 e iy 1 1 e iyη f ξ 1, η dη iy Because of the choice of intervals I m, we have e iy 1, y Im, m = 1,, The Riemann-Lebesgue lemma implies that one can choose a positive integer N > N so large that 1 e iyη f ξ 1, η dη.1, y {y > N} Hence, imply that p1, y 1 y, m > N, y I m

16 Choose an arbitrary integer m > N. Then zeros of functions F 1 z, y and F z, y coincide for all y I m. Also, 3.16a,b and 3. imply that and Let F j z, y = e iz iz p 1, y 1 + o1, for Im z, y I m, 3.1a F j z, y = 1 iz p 1, y 1 + o1, for Im z, y I m, 3.1b F j z, y = z k jy e g jz,y p 1, y, y I m. 3. n=1 1 z z exp. a n y a n y Then for all y I m log [ ] F1 z, y = k 1 y k y log z + g 1 z, y g z, y. 3.3 F z, y On the other hand, by 3.1a and 3. [ ] F1 z, y log = o1, F z, y for Im z, y I m Since g 1 z, y and g z, y are linear functions of the variable z for every y R, then comparison of 3.3 and 3.4 shows that k 1 y k y = and g 1 z, y g z, y = for all y I m. Lemma 6. There exists a positive number N = NF 1, F and a positive number T = T N such that for every integer m > N and for every y I m all zeros of functions F 1 z, y and F z, y are located in the strip { Im z < T }. Proof. Choose a number N = NF 1, F such that 3.19 and 3. hold. Let m > N be an integer. By 3.1a,b and 3. there exists a positive number T = T N independent on m as long as m > N such that F j z, y expim z z p1, y for Im z T, y I m and F j z, y 1 z p1, y for Im z T, y I m. Lemma 7. There exists a positive number N = NF 1, F such that there exists a positive number M = MN such that for all integers m > N and for every y I m zeros of functions F 1 z, y and F z, y coincide in { z > M}. 16

17 Proof. We use notations of the proof of Lemma 5. Choose a positive number N 1 = N 1 F 1, F such that 3.19 and 3. are fulfilled. Let T = T N 1 be the number of Lemma 6. First, we prove that for every integer m > N 1 and for every y I m both functions F 1 z, y and F z, y have infinitely many zeros. Fix an y I m. Let, for example the function F 1 z = F 1 z, y has only a finite number s zeros in C. Then 3.4 implies that F 1 z = P s ze γz, 3.5 where γ is a complex number and P s z is a polynomial of the degree s. However, by 3.15 and the Riemann-Lebesgue lemma hx = 1 [ e ix 1 ] 1 p1, y + o for x, x R. 3.6 ix x Since by 3. p1, y, then 3.6 contradicts with 3.5. The latter and Lemma 6 imply that for every integer m > N 1, for every y I m and for each positive number K there exists a zero z j K { z > K} { Im z < T } of the function F j z, y. Integrating by parts in 3.15 and using 3.13 and 3.14, we obtain for every integer m > N 1 and for all y I m iz F j z, y = e iz 1 p1, y + 1 e iz + 1 p ξ 1, y 3.7 iz 1 z e iz 1 p ξξ 1, y + 1 z 1 e izξ 3 ξ p j ξ, ydξ. Lemma 6 tells one that in order to find the asymptotic behavior of zeros of functions F j z, y, one should investigate the behavior of these functions at Re z with Im z < T. Integrating by parts in 3.11 and using 1.5, we obtain ξ k p j ξ, y = 1 e iy 1 1 ξ k f j ξ, 1 + e iyη ξ k η f j ξ, η dη, k =, 1,, iy It follows from 3.1, 3. and 3.8 that one can the choose a number M 1 = M 1 N 1 so large that 1 p1, y + iz p ξ1, y 1 y, for m > N 1, y I m, z > M Also, 3. implies that 1 p1, y + p ξ 1, y/iz = 1 p1, y 1 p ξ1, y pz, y +, 3.3 izp1, y z for m > N 1, y I m, z > M 1, where pz, y C. Here and below in this proof C denotes different positive numbers which are independent on z { ż > 1} { Im z < T }, N 1, N, M k 17

18 k = 1,..., 5, complex numbers z j which are chosen below and the parameter y I m, as long as the integer m > N 1. Although, in principle at least each function F j, j = 1, has its own constant C j, but we always choose C = maxc 1, C. By 3.7 and 3.8 we have iz F j z, y = e iz p1, y + 1 iz p ξ1, y p1, y 1 iz p ξ1, y + B j z, y z { z > M 4 } { Im z < T }, m > N 1 and y I m, where the function B j can be estimated as, 3.31 B j z, y C y z. 3.3 Dividing 3.31 by the function [p1, y + p ξ 1, y/iz], using 3.9, 3.3 and 3.3, we obtain for m > N 1, y I m, Im z < T iz F j z, y p1, y iz p ξ1, y = e iz 1 + p ξ1, y izp1, y + B j z, y, 3.33 z { z > M 4 } { Im z < T }, m > N 1 and y I m, where the function B j z, y satisfies the estimate B j z, y C z, 3.34 z { z > M 4 } { Im z < T }, m > N 1 and y I m. Choose an integer m > N 1 and fix a number y I m. Let z j { z > M 1 } { Im z < T } be a zero of the function F j z, y. Then 3.33 implies that exp iz j = 1 p ξ1, y iz j p1, y B j z j, y. Since exp iz j = exp iz j + iπs for any integer s, then there exists an integer n z j such that iz j + iπn z j = log 1 p ξ1, y izp1, y B j z, y By 1.6a,b Hence, 1.4, 1.5 and 3.36 imply that ϕ ξ, 3.36 ϕ ξ, = ϕ ξ 1,

19 Thus, 1.3, 3.17, 3.18, 3.8, 3.37 and the Riemann-Lebesgue lemma imply that one can choose the number N = N F 1, F N 1 and where gy is a complex valued function such that p ξ 1, y ip1, y = ϕ ξ1, g y, 3.38 gy < 1 8, m > N, y I m Without loss of generality we assume from now on in this proof that the integer m, which was chosen after 3.34 is so large that m > N and the fixed number y I m. By 3.34, 3.38 and 3.39 one can choose the number M = M N M 1 so large that log 1 p ξ1, y izp1, y B j z, y = ϕ ξ1, g j z, y, 3.4 z where the function g j z, y is such that g j z, y < 1 4, 3.41 z { z > M } { Im z < T }, m > N, y I m. Substituting the right hand side of 3.4 in the right hand side of 3.35 and setting y := y, we obtain z j = πn z j i ϕ ξ1, 1 z j 1 + g j z j, y. 3.4 Since we are concerned in this lemma with the asymptotic behavior of zeros of functions F j z, y, then we can assume now that the zero z j { z > M } { Im z < T }. Choose the number M 3 = M 3 N M so large that C/M 3 < M 3 /8. Similarly with the above we now assume that z j { z > M 3 } { Im z < T }. Hence, 3.41 and 3.4 lead to ϕ Imz j ξ 1, g j z j, y z j < C < M 3 M On the other hand, since z j > M 3, then 3.4 and 3.43 imply that M 3 < z j < πn z j + M 3 /8. Hence, Combining 3.44 with 3.4 and 3.43, we obtain πn z j > 7 8 M z j = πn z j [1 + λ j z j, y ],

20 where the function λ j z, y satisfies the following estimate λ j z, y < 1 7, 3.46 z { z > M 3 } { Im z < T }, m > N, y I m. It follows from 3.45 and 3.46 that Re z j 3.47a and sgn [Re z j ] = sgn n z j, 3.47b where sgnx = 1 if x > and sgnx = 1 if x < for x R. Note that for each above zero z j of the function F j there exists only one integer n z j. Indeed, if there exists a second one n z j, then 3.4 implies that z j = πn z j i ϕ ξ1, 1 z j 1 + g j z j, y. Subtracting this formula from 3.4, we obtain π [n z j nz j ] =. Consider now sgnimz j. It follows from that one can choose a number M 4 = M 4 N M 3 so large that for any zero z j { z > M 4 } { Im z < T } of the function F j z, y the following equality is true Imz j = ϕ ξ1, 1 πn z j along with 3.47, where the function µ j z, y satisfies the estimate 1 + µj z j, y, 3.48 µj z, y 1, 3.49 z { z > M 4 } { Im z < T }, m > N, y I m. Since ϕ ξ 1, 1, then 3.47a,b-3.49 lead to and Imz j 3.5 sgnimz j = sgn [ ϕ ξ 1, 1 ] sgn [Re z j ] = sgn [ ϕ ξ 1, 1 ] sgn n z j We show now that the multiplicity of the zero z j is one, as long as z j { z > M 5 } { Im z < T }, where the number M 5 M 4 is chosen below. Since the formula 3.31 was derived from formulas 3.7 and 3.8, then 3.31 can be rewritten in the form iz F j z, y =

21 e iz p1, y + 1 iz p ξ1, y p1, y 1 iz p ξ1, y z e iz 1 p ξξ 1, y + 1 z 1 e izξ 3 ξ p j ξ, ydξ, z { z > M 4 } { Im z < T }, m > N, y I m. Differentiating both sides of the formula 3.5 with respect to z, setting then m := m, y := y, z := z j { z > M 4 } { Im z < T } and assuming that F j z j, y = z F j z j, y =, we obtain exp iz j p1, y + 1 p ξ 1, y = H jz j, y, 3.53 iz j y zj where the function H j z, y satisfies the estimate H j z, y C, 3.54 z { z > M 4 } { Im z < T }, m > N, y I m. Hence, dividing 3.53 by p1, y + p ξ 1, y /iz j and using 3.9 and 3.54, we obtain exp iz j = H j z j, y, 3.55 zj where the function H j z, y satisfies the following estimate H j z, y C z { z > M 4 } { Im z < T }, m > N, y I m. Since exp iz j exp Im z j and z j > M 4, then replacing in 3.43 M 3 with M 4, we obtain exp iz j exp C/M 4. Hence, 3.55 and 3.56 imply that C exp CM M 4 Choose a number M 5 = M 5 N M 4 so large that C M 5 < 1 exp C M Again, we can assume similarly with the above that z j { z > M 5 } { Im z < T }. On the other hand, it follows from 3.57 that if the multiplicity of the zero z j is greater than 1, then one should have C exp CM M 5 1

22 Inequalities 3.58 and 3.59 contradict with each other. This contradiction proves that the multiplicity of the zero z j is 1, as long as z j { z > M 5 } { Im z < T }. Set now N = NF 1, F := N and M = MN := M 5. For the sake of definiteness, let j = 1. Consider the zero z 1 { z > M} { Im z < T } of the function F 1 z, y, i.e., F 1 z 1, y =. We are going to prove now that F z 1, y =, which would be sufficient for the validity of Lemma 7. By 1.9 and 3.3 F 1 z 1, y F 1 z 1, y = F z 1, y F z 1, y =. Suppose that F z 1, y. Then F z 1, y =. Hence, Lemma 3 implies that F z 1, y =. Since Re z 1 = Re z 1, then 3.47a implies that Re z 1 = Re z 1. Formulas 3.5 and 3.51 are valid for any zero z j { z > M} { Im z < T } of the function F j for j = 1,. Denote z := z 1. Then F z, y =. Hence, using 3.5 and 3.51, we obtain Imz 1 and sgnimz 1 = sgn [ ϕ ξ 1, 1 ] sgn Re z 1. But since F 1 z 1, y =, then formulas 3.5 and 3.51 are also true for z 1, i.e., Imz 1 and signimz 1 = sgn [ ϕ ξ 1, 1 ] sgn Re z 1. Thus, we have obtained that Imz 1 Imz 1 and signimz 1 = sgnimz 1, which is impossible, since Imz 1 = Imz 1. This proves that F z 1, y =. 4 Zeros In { z < M} Both in this and next sections numbers N = N F 1, F and M = MN are those, which were chosen in Lemma 7. Let m > N be an integer. Fix an arbitrary number y 1 I m. So, in both sections 4 and 5 we assume that y = y 1 and do not indicate the dependence on the parameter y for brevity, keeping in mind, however that this dependence exists. Recall that we assume the existence of two functions f 1 ξ, η and f ξ, η, which correspond to the same function G in 1.. Hence, 1.9 and 3.3 imply that F 1 z F 1 z = F z F z, z C. 4.1 Let Φz, z C be an entire analytic function. Denote ZΦ the set of all zeros of this function. Also, denote ZM, Φ = ZΦ { z < M}, and Using Lemma 4, we obtain Z Φ = ZΦ {Im z = }, Z + M, Φ = ZM, Φ {Im z > } Z M, Φ = ZM, Φ {Im z < }. Z F 1 = Z F. 4.

23 By Lemma 7 ZF 1 ZM, F 1 = ZF ZM, F. 4.3 Hence, Lemma 5 and 4. imply that in order to establish Theorem 1, it is sufficient to prove that Z + M, F 1 = Z + M, F 4.4 and Let Z M, F 1 = Z M, F 4.5 {a k } n 1 k=1 = Z +M, F 1 and {b k } n k=1 = Z +M, F. 4.6 In both cases each zero is counted as many times as its multiplicity is. Consider functions B 1 z and B z, n 1 z a k B 1 z = F 1 z, 4.7 z a k The main result of this section is Lemma 8. Z B 1 = Z B. Proof. By 4. and and B z = F z Also, it follows from 4.3 and that k=1 n k=1 z b k z b k. 4.8 Z B 1 = Z B 4.9 Z + M, B 1 = Z + M, B =. 4.1 ZB 1 ZM, B 1 = ZB ZM, B Thus, imply that all what we need to prove in this lemma is that Let Z M, B 1 = Z M, B. 4.1 {c k } s k=1 = Z M, B In 4.13 we count each zero c of the function B 1 as many times as its multiplicity is. The main idea of this proof is to show that a combination of Lemma 3 and 4.1 with 4.7 and 4.8 leads to {c k } s k=1 Z M, B To establish 4.14, we should consider several possible cases for zeros {c k } s k=1. Consider the zero c 1 Z M, B 1. Either F 1 c 1 = or F 1 c 1. We consider both these cases. 3

24 Case 1. Assume first that F 1 c 1 = 4.15 By 4.1 and 4.15 at least one of the two equalities 4.16, 4.17 takes place, F c 1 =, 4.16 F c 1 = Assuming that 4.15 is true, consider cases 4.16 and 4.17 separately. We denote them C 11 and C 1 respectively. Case C 11. Suppose that 4.16 is true. Since by 4.13 Im c 1 >, then c 1 Z + M, F. Let, for example c 1 = b 1. Then c 1 = b 1. Therefore, c 1 is present in the nominator of the first term of the product in 4.8, which implies that B c 1 =. Hence, Case C 1. Assume that 4.16 is invalid, i.e., c 1 Z M, B F c Then 4.17 holds. Because of 4.19, the number c 1 is not present in denominators of the product in 4.8. On the other hand, 4.1, 4.15, 4.19 and Lemma 3 imply that F c 1 =. Hence, by 4.8 B c 1 =, which implies Thus, the assumption 4.15 led us to 4.18 in both possible cases C 11 and C 1. Consider now the Case, which is opposite to the Case 1. Case. Suppose that F 1 c By 4.13 B 1 c 1 =. 4.1 Since Im c 1 >, then 4. implies that c 1 / Z + M, F 1, which means that c 1 is not present in nominators of the product in 4.7. Hence, 4.7 and 4.1 lead to F 1 c 1 =. Hence, by 4.1 at least one of the two equalities 4., 4.3 takes place F c 1 =, 4. F c 1 =. 4.3 Assuming that 4. is true, consider cases 4. and 4.3 separately. We denote these two cases C 1 and C respectively. Case C 1. Suppose that 4. is true. Since Im c 1 <, then c 1 is not present in denominators of the product in 4.8. Hence, B c 1 =. This means that 4.18 is true. Case C. Assume now that 4. is invalid. Hence, 4.3 holds. Hence, Lemma 3 implies that 4.16 holds note that we cannot now refer to the above case C 11, because being inside of Case, we do not assume that 4.15 is valid. Since Im c 1 >, then 4

25 4.16 means that c 1 Z + M, F. Let, for example c 1 = b 1. Hence, c 1 = b 1. Therefore, c 1 is present in the nominator of the first term of the product in 4.8, which implies that B c 1 =. This means that 4.18 is true. Thus, the conclusion from Cases 1 and is that 4.18 holds. We show now that c Z M, B. 4.4 Hence, using 4.13 and 4.18, we obtain that B 11 z = B 1z z c 1 and B 1 z = B z z c are entire analytic functions. Also, by 4.13 and 4.5 {c k } s k= = Z M, B It follows from 4.5 that in order to prove 4.4, it is sufficient to prove that c Z M, B We again consider two possible cases. Case 3. Suppose that 4.15 is true. Because of 4.1 and 4.15, at least one of equalities 4.16 or 4.17 holds. We again consider cases 4.16 and 4.17 separately and denote them C 31 and C 3 respectively. Since 4.15 holds, we can assume that c 1 = a 1. Case C 31. Suppose that 4.16 holds. Let, for example c 1 = a 1 = b 1. Introduce functions F 11 z and F 1 z by F 11 z = F 1z z a 1 and F 1 z = F z z a Since F 1 a 1 = F a 1 =, then F 11 z and F 1 z are entire analytic functions. Hence, Z + M, F 11 = {a k } n 1 k= and Z + M, F 1 = {b k } n k=. It follows from 4.7, 4.8, 4.15, 4.5, and 4.8 that formulas for functions B 11 z and B 1 z can be written as n 1 z a k B 11 z = F 11 z z a k k= n z b k and B 1 z = F 1 z. 4.9 z b k k= Note that by 4.8 F 11 z = F 11 z = F 1 z z a 1 and F 1 z = F 1 z = F z z a 1. Hence, F j1 z F j1 z = F jz F j z z a 1 z a 1. 5

26 Hence, 4.1 leads to F 11 z F 11 z F 1 z F 1 z. 4.3 Relations 4.15, 4.5, 4.6 and enable us to repeat arguments of the above Case 1 replacing c 1 with c, B 1 z with B 11 z, B z with B 1 z, F 1 z with F 11 z, and F z with F 1 z. Thus, we obtain 4.7, which, in turn leads to 4.4. Case C 3. Assume now that 4.16 is invalid. Then 4.17 holds. Since we are still within Case 3, then c 1 = a 1. Hence, 4.17 and Lemma 3 imply that F a 1 =. Introduce entire analytic functions F 1 z and F z as Then Hence, 4.1 implies that F 1 z = F 1z z a 1, F z = F z z a F j z F j z = F jz F j z z a 1 z a 1. F 1 z F 1 z F z F z. 4.3 Hence, using 4.7, 4.8, 4.15, 4.5, and 4.31, we conclude that formulas for functions B 11 z and B 1 z can be written as n 1 z a k B 11 z = F 1 z z a k k= n z b k and B 1 z = F z z b k Therefore, relations 4.15, 4.5, 4.6, and enable us to repeat arguments of the above Case 1 replacing c 1 with c, B 1 z with B 11 z,b z with B 1 z, F 1 z with F 1 z, and F z with F z. This leads to 4.7, which, in turn implies 4.4. Thus, both cases C 31 and C 3 led us to 4.4. This proves that if F 1 c 1 =, then c Z M, B.The alternative to the Case 3 Case 4 with F 1 c 1 is considered similarly. The only difference is that instead of Case 1 we should refer to Case for the repetition of the arguments. Thus, we have established that both zeros c 1, c Z M, B. To prove that c 3 Z M, B, we need to consider entire analytic functions B 1 z = B 11z z c and B 1 z = B 1z z c and repeat the above. Therefore, repeating this process, we obtain Hence, 4.13 and 4.14 lead to Z M, B 1 Z M, B. Similarly, Z M, B 1 Z M, B. Thus, 4.1 is true. Consider now zeros of functions F 1 z and F z in {Im z < } { z < M}. Let {a k }n 3 k=1 = Z M, F 1 and {b k }n 4 k=1 = Z M, F. Similarly with 4.7 and 4.8 we introduce functions B1 z and B z by B 1 z = F 1 z n 3 k=1 z a k z a k and B z = F z 6 k=1 n 4 k=1 z b k. z b k

27 Lemma 9. Z B1 = Z B. We omit the proof of this lemma, because it is quite similar with the proof of Lemma 8. 5 Proof of Theorem 1 We recall that in this section numbers N = N F 1, F and M = MN are those, which were chosen in Lemma 7, m > N is an integer, an arbitrary number y 1 I m is fixed, and we set y := y 1. So, for brevity we do not indicate in this section the dependence on the parameter y. It was established in the beginning of section 4 that in order to prove Theorem 1, it is sufficient to proof 4.4 and 4.5. By 1.8, 1.9, 3.5, 4.7 and 4.8 B 1 x = B x, x R. Hence, lemmata 5 and 8 imply that the function gz and the integer k in analogs of infinite products 3.4 for functions B 1 and B are the same for both these functions. Hence, Lemma 8 and 3.4 imply that B 1 z = B z, z C. Hence, 4.7 and 4.8 lead to n1 z a k F 1 z + 1 F 1 z = F z + z a k k=1 n k=1 z b k z b k 1 F z, z C. 5.1 Set in 5.1 z := x, and apply the inverse Fourier transform with respect to x, 1 π... e ixξ dx. We can write each of functions [ n1 ] x a k Q 1 x = 1 x a k k=1 and Q x = [ n ] x b k 1 x b k k=1 as a sum of partial fractions, i.e., as a sum of D sk x d k s, Im d k >, j = 1,..., t, t maxn 1, n with certain constants D sk, where d k Z + M, F 1 Z + M, F. The theory of residuals [1] implies that 1 x d k s e ixξ dx = H ξ P s 1 ξ e id kξ, 5. where H ξ = 1 for ξ > and H ξ = for ξ < is the Heaviside function and P s 1 ξ is a polynomial of the degree s 1. 7

28 Let V 1 ξ and V ξ be the inverse Fourier transforms of functions Q 1 x and Q x respectively. By 5. V 1 ξ = V ξ = for ξ <. Thus, 5.1 and 5. imply that p 1 ξ + ξ p 1 ξ θ V 1 θ dθ = p ξ + ξ p ξ θ V θ dθ, 5.3 where functions p 1 ξ and p ξ are defined as p 1 ξ := p 1 ξ, y 1, p ξ := p ξ, y 1 and functions p 1 ξ, y and p ξ, y were defined in By 3.1 p 1 ξ = p ξ := p ξ for ξ, δ. Denote W ξ = V 1 ξ V ξ. Hence, 5.3 leads to ξ p ξ θ W θ dθ =, ξ, δ. 5.4 Differentiate 5.4 with respect to ξ and note that by 3.13 p1 = p and by 3. p. We obtain the following Volterra integral equation with respect to the function W ξ W ξ + 1 ξ p ξ θ W θ dθ =, p ξ, δ. 5.5 Hence, W ξ = for ξ, δ. Since the function W ξ is a linear combination of functions P s 1 ξ e idkξ for ξ >, then W ξ is analytic with respect to ξ, and can be continued in C as an entire analytic function W z. Therefore, W z = for all z C. Hence, Z + M, F 1 = Z + M, F, which proves 4.4. We omit the proof of 4.5, since it can be carried out quite similarly via the use of Lemma 9 instead of Lemma 8. Acknowledgments This work was supported by the grant W911NF from the US Army Research Office and by the research grant from the University of North Carolina at Charlotte. The author is grateful to Michael Fiddy and Paul Sacks for many useful discussions and for pointing him to right references. The author also appreciates an anonymous referee for quite useful comments, which helped to improve the quality of presentation. References [1] L.V. Ahlfors, Complex Analysis, McGraw-Hill, Inc., New York, 1979 [] M. Born and E. Wolf, Principles of Optics, Pergamon Press, Oxford, 198. [3] B.J. Brames, Unique phase retrieval with explicit support information, Optics Letters, , [4] R.E. Burge, M.A. Fiddy, A.H. Greenway and G. Ross, The phase problem, Proc. Royal Soc. A, ,

29 [5] A. Calderon and R. Pepinsky, On the question of the uniqueness of solutions of the phase problem in crystal analysis, in Computing Methods And the Phase Problem In X- ray Crystal Analysis, R. Pepinsky editor, X-Ray Crystal Analysis Laboratory, Pennsylvania State College 195, [6] D. Dobson, Phase reconstruction via nonlinear least squares, Inverse Problems, 8 199, [7] J.R. Fienup, J.C. Marron, T.J. Schulz and J.H. Seldin, Hubble space telescope characterization by using phase retrieval algorithms, Applied Optics, , [8] M.A. Fiddy and H.J. Caulfield, The significance of phase and information, in Tribute to Emil Wolf. Science and Engineering Legacy of Physical Optics, T.P. Jannson editor, SPIE Press, Bellingham, Washington 5, [9] H. Hauptman, The phase problem of x-ray crystallography, in Signal Processing, Part II: Control Theory and Applications, Institute of Mathematics and Applications of the University of Minnesota, Volume 3 199, [1] L. Hörmander, The Analysis of Linear Partial Differential Operators, I. Distribution Theory and Fourier Analysis, Springer, Berlin, [11] N.E. Hurt, Phase Retrieval and Zero Crossings: Mathematical Methods in Image Reconstruction, Kluwer, Dordrecht,. [1] M.V. Klibanov, Determination of a function with compact support from the absolute value of its Fourier transform and an inverse scattering problem, Differential Equations, 1987, [13] M.V. Klibanov, Inverse scattering problems and restoration of a function from the modulus of its Fourier transform, Siberian Math. J., , [14] M.V. Klibanov and P.E. Sacks, Phaseless inverse scattering and the inverse problem in optics, J. Math. Phys., , [15] M.V. Klibanov and P.E. Sacks, Use of partial knowledge of the potential in the phase problem of inverse scattering, J. Comput. Phys., , [16] M.V. Klibanov, P.E. Sacks and A.V. Tikhonravov, The phase retrieval problem, Inverse Problems, , 1-8. [17] R. Kress and W. Rundell, Inverse obstacle scattering with modulus of the far field pattern as data. In Inverse Problems in Medical Imaging and Nondestructive Testing Oberwolfach, 1996, , Springer, Vienna. [18] G. Stroke, An Introduction to Coherent Optics and Holography, Academic Press, New York, [19] V.V. Volkov, Y. Zhu and M. De Garef, A new symmetrized solution of phase retrieval using the transport of intensity equation, Micron, 33, [] E. Wolf, Is a complete determination of the energy spectrum of light possible from measurements of the degree of coherence?, Proc. Phys. Soc., 8 196,