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Cones with convoluted geometry that always scatter or radiate

We investigate fixed energy scattering from conical potentials having an irregular cross-section. The incident wave can be any arbitrary non-trivial Herglotz wave. We show that a large number of such local conical scatterers scatter all incident waves, meaning that the far-field will always be non-zero. In essence there are no incident waves for which these potentials would seem transparent at any given energy. We show more specifically that there is a large collection of star-shaped cones whose local geometries always produce a scattered wave. In fact, except for a countable set, all cones from a family of deformations between a circular and a star-shaped cone will always scatter any non-trivial incident Herglotz wave. Our methods are based on the use of spherical harmonics and a deformation argument. We also investigate the related problem for sources. In particular if the support of the source is locally a thin cone, with an arbitrary cross-section, then it will produce a non-zero far-field.


Introduction
When a wave of constant wavenumber encounters a penetrable obstruction, it typically produces a scattered wave. The scattered wave transmits information about the obstruction, which is a fact that is exploited by various imaging modalities [16,29]. However, an incident wave does not necessarily scatter even in the presence of a scatterer [12]. In the quantum mechanical context we call the energy of such a wave function a non-scattering energy. When dealing with with acoustic or electromagnetic scattering, we use the term non-scattering wavenumber instead. It might happen that a potential or an obstruction is e ectively transparent at certain fixed energies or wavenumbers, producing a zero far-field irrespective of the incident wave (see e.g. [24,32,37]). A striking converse to this type of phenomenon was first studied in [8], where the authors showed that an obstacle with a corner always scatters, i.e. it always produces a scattered wave that does not decay quickly, regardless of what the incident wave is or what energy it has. This type of obstacle is not transparent to any incident wave, i.e. it is always visible. The absence of non-scattering energies has in recent years been studied by a number of authors, see e.g. [4,5,6,8,14,15,18,19,27,36]. One of the themes of this paper is to study the absence of non-scattering-energies for potentials that are obtained from irregular cones.
It should also be pointed out that non-scattering energies are closely related to the so called interior transmission eigenvalues [13]. These are, in the acoustic setting, eigenvalues for a certain non elliptic and non self-adjoint spectral problem on the support of an inhomogenity of the refractive index. A non-scattering-energy is a always an interior transmission eigenvalue when the scatterer has compact support. The converse does not hold. It is well known that transmission eigenfunctions can be approximated by normalized Herglotz waves [45]. Such waves will produce arbitrarily small scattering. However this does not imply that there would be an incident wave producing no scattering. This only happens in the case of a non-scattering energy, and only if this sequence of Herglotz waves aproximates a non-scattering incident wave. The study of interior transmission eigenvalues has it roots in the analysis of certain numerical reconstruction methods in inverse scattering, namely the linear sampling method and the factorization method (for more on these see [16,17,29,30]). Energies which admit interior transmission eigenvalues cause challenges for these methods. For a survey on the topic of interior transmission eigenvalues see [13].
Another topic related to non-scattering are sources that produce no wave in the far field. A source f in acoustic scattering (∆ + k 2 )u = f produces a far-field u ∞ (θ) = C f (kθ), for details, see [6]. Therefore understanding source scattering at a fixed wavenumber is related to understanding the Fourier restriction problem [23]. By a nonradiating source, we mean a source that does not produce a wave in the far field at some wavenumber or energy. The topic of sources that radiate at all energies was first studied in [2], where it was shown that that nonradiating sources having a convex or nonconvex corner or edge on their boundary must vanish there. Subsequent studies of sources that always radiate include [3,6,7]. Here we extend the results in [2] by showing that sources obtained from very thin or very wide cones will always produce a radiating wave irrespective of their cross section.
We will describe very briefly and in a non technical wave the various concepts needed to understand our theorems. This is also to fix notation. Proper mathematical definitions are given in Section 2.
Recall that the scattering of an incident particle wave of energy λ > 0 by a potential V can be modeled by the equation Here u is the superposition of the wave function u i of an incident particle and a scattered wave u s , which is created by the interaction between u i and the scatterer V . The scattered wave will have an asymptotic expansion, and a detectable part of that is called the scattering amplitude or far-field pattern. By a non-scattering energy for a potential V , we mean a λ > 0 for which there is an incident wave which produces no scattering from V at energy λ. In other words, the scattering amplitude is zero even though the incident wave is not. A potential which always scatters is one that does not have any non-scattering energies.
We will analyze potentials whose support has a conical singularity on its boundary. We will consider cones C ⊂ R n , n = 3, that are determined by a compact cross section K ⊂ R n−1 , 0 ∈ int K in suitable coordinates by C = (tx , t) ∈ R n : x ∈ K, t ≥ 0 . (1.2) For technical reasons, we will also need some regularity assumptions. Definition 1.1. We call a cone C of the form (1.2) regular if the following conditions are satisfied.
(i) C is contained in a strictly convex closed circular cone, (ii) C has a connected exterior, and (iii) C has a bounded cross-section K ⊂ R 2 such that χ K ∈ H τ (R 2 ) for some τ ∈ (1/4, 1/2).
Such a cone is denoted by C σ and we usually take both σ and ρ 0 given by the context.
We will relate a conical potential to a cone as follows. Given a cone C with coordinates chosen as in (1.2), let V C be the potential where ϕ ∈ C 1/4+ε c (R 3 ), > 0, and Φ ∈ e −γ|·| L 2 (R 3 ), γ > 0, and supp Φ ⊂ {x 3 > 0}. Next, we state our main result for scattering from a potential. The theorem is a direct consequence of Theorem 5.3 and Proposition 7.6 (in the latter set C σ 1 = C σ , to obtain the theorem).
The main novelty of Theorem 1.3 is that it shows that there are a large number of conical potentials that always scatter, which have an irregular geometry, and the scattering happens for any incident Herglotz wave. Earlier results in the case of R 3 , have dealt with fairly regular geometries such as corners between hyperplanes [8] and circular cones or curvilinear polyhedra [18,19,36]). Where [36] uses a similar approach as we do here, utilizing complex geometrical optics solutions, and [18,19] use techniques from the theory of boundary value problems in corner domains. More recent results apply to more complicated geometries, such as [14] using stationary phase method and [38] using the methods from the theory of free boundary value problems. These results are not applicable to a general Herglotz incident wave and instead assume that it does not vanish on the boundary. Non vanishing is a major technical simplification. The proof in [8] would be much shorter if one would assume that u i (0) = 0 there. This is also demonstrated in this paper between Section 3 and sections 4, 5. The main di culty is as follows: if one assumes the source problem or a non vanishing incident wave, then one only needs to prove the non vanishing of an integral involving spherical harmonics of degree 2, as in Definition 3.1. With a general incident wave one needs instead to prove the non vanishing of the integral for all degrees, as in Definition 5.1.
Let us discuss briefly the proof of Theorem 1.3. Our starting point is the use of complex geometrical optics solutions and the theory of spherical harmonics. We use the latter to derive 1 Here we use the Hausdor distance between the sets A and B which is given by where d(x, A) := inf y∈A d(x, y), and d is induced by the Euclidean metric. a projection condition for cones which will guarantee potentials that always scatter. We call such cones admissible medium cones. See Definition 5.1.
The admissibility of a cone can be further verified by means of certain determinants being non zero. We use this determinant condition to prove that circular cones are admissible medium cones (and that they therefore always scatter) by computing explicit formulas for the determinants. This gives an alternative proof for the three dimensional result of [19,36] without the issue of a countable set of unhandled cones.
Testing for the determinant condition requires us to use certain results on the associated Legendre polynomials. We derive them in Section 8. In particular, we need a modification of the classical Christo el-Darboux formula in page 43 of [42]. Then to analyze potentials related to star-shaped cones, we perform a deformation argument, where we interpolate between the points of a circular cone and a star-shaped cone. Spherical harmonics are analytic, which can be used to show that the determinants of the deformed cones depend analytically on the deformation parameter. Analyticity then guarantees that certain critical integrals cannot vanish except in a countable set of points, and Theorem 1.3 follows.
We would like to further point out that even though our construction of complex geometrical optics solutions follows the argument of [36], we need to modify their argument to work with cones with irregular cross section. This is done in Section 4, where we also provide some explicit examples of cones with rather complex geometry to which our argument applies but previous arguments didn't.
We turn to the related problem of sources that radiate at every energy level next. Several types of source problems in wave propagation can be modeled by the Helmholtz or static Schrödinger equation. We will consider the following, in R n together with a radiation condition at infinity, the Sommerfeld radiation condition (2.3). It selects the outgoing wave from all possible solutions to (1.5). The above problem models a wave u created by the source f that is oscillating at energy λ > 0 and is radiating out to infinity. We will consider sources of the form where ϕ ∈ L ∞ (R n ) is compactly supported, Φ ∈ e −γ|·| L 2 (R n ) with γ > 0 is supported on a half-space, and C is a cone of the form (1.2) with vertex outside the above-mentioned half-space. It can be shown that the wave u has the asymptotic expansion The function α r ∈ L 2 (S n−1 ) is called the far-field pattern radiated by f . We say that a source term f always radiates if α r = 0 for all λ > 0. We will be interested in sources f determined by a fairly general family of cones that we call admissible source cones, Definition 3.1. These cones can have very irregular cross sections, e.g. with a fractal boundary. Our main result concerning these types of sources is a direct consequence of Theorem 3.9 in Section 3. Theorem 1.5. Assume that C ⊂ R 3 is a cone of the form (1.2), that is also an admissible source cone in the sense of Definition (3.1). Suppose furthermore that f is the source term (1.6), where ϕ is Hölder continuous at 0 and ϕ(0) = 0. Then for all λ > 0, we have that α r = 0 in (1.7). That is, f always radiates.
The first results on sources that always radiate were obtained in [2], where it was shown that sources having a convex or non-convex corner or edge on their boundary always radiate. The main novelty of Theorem 1.5 is that it shows that a source determined by a cone with a singularity at the vertex, will always radiate essentially regardless of the geometry of the cross section of the conical source. This follows because of propositions 3.6 and 3.7.
The paper is structured as follows. In Section 2 we review some preliminary results and notation that we use in the other sections. In particular, we set the notation for spherical coordinates. We then study the source problem in Section 3. After this, we construct complex geometrical optics solutions for potentials constructed from irregular cones in Section 4. In Section 5, we define the concept of an admissible medium cone and show that potentials related to these always scatter. Section 6 proves that circular cones are admissible medium cones. After this, in Section 7, we prove Theorem 1.3. Finally, in Section 8, we derive various formulas for associated Legendre polynomials that were used in earlier sections.

Preliminaries
We will review various results that are needed in the sequel. We start by specifying some function spaces after which we give a short review of basic scattering theory based on [1,26]. It is the same setup as in [35,36] for static scattering theory. The reader can find further details in these references.
Following [1,26,35,36] we use the spaces B(R n ) and B * (R n ). The space B(R n ) consists of those u ∈ L 2 (R n ) for which the norm is finite. See Section 14.1 in [26]. Here X 1 = {|x| < 1} and X j = {2 j−2 < |x| < 2 j−1 } for j ≥ 2. This is a Banach space whose dual B * (R n ) consists of all u ∈ L 2 loc (R n ) such that The set C ∞ c (R n ) of compactly supported smooth functions is dense in B(R n ) but not in B * (R n ). Their closure in B * (R n ) is denoted byB * (R n ), and u ∈ B * (R n ) belongs toB * (R n ) if and only if We will also use B * 2 andB * 2 , which take into account the derivatives up to second order of the function. These are defined via the norm Next, we give a short review of the relevant parts of scattering theory. In static or time harmonic scattering theory of energy λ > 0 one considers the equation in R n where u is a total wave function that is the superposition of an unperturbed incident wave function u i and a wave u s scattered by the potential function. That is, u = u i + u s where One further requires that the scattered wave u s satisfies the Sommerfeld radiation condition The potential V is a short range potential, by which we mean that V ∈ L ∞ (R n ) and that there are constant C, > 0, such that almost everywhere in R n . Note that this allows for potentials whose support is unbounded.
Will will furthermore assume that the incident wave is given by a Herglotz function, i.e.
It follows that u i ∈ B * 2 (R n ). See Proposition 2.1 in [35]. They also show that the incident particle u i has the asymptotics and moreover that the scattered wave has the asymptotics The function α s is called the scattering amplitude or the far field pattern. The relative scattering operator is the map For more technical details see [17] or more generally sections 7 and 8 in [1]. An α s ≡ 0 implies that u s decays quickly at infinity. We see thus define that λ is a non-scattering energy if the kernel of S λ is non-trivial. More precisely Definition 2.1. We call λ > 0 a non-scattering energy for a short range potential V if there is a g ∈ L 2 (S n−1 ) such that g = 0 and S λ g = 0.
The following Lemma provides another characterization of a non-scattering-energy.

Lemma 2.2.
Let V be a short-range potential. Then λ > 0 is a non-scattering energy for V if there exist functions v, w ∈ B * 2 (R n ) so that w ≡ 0 and Proof. Suppose that v and w are as in the claim. Now define Clearly u i and u s solve equations (2.2). The incident wave u i = w can be written as a Herglotz wave function by Theorem 14.3.3 in [26]. In other words there exists an g ∈ L 2 (S n−1 ) so that which would be impossible by (2.7) unless α s ≡ 0. Hence S λ g = 0.
We will also make use of the following Rellich-type theorem which is Theorem 4 in [44].
We use the spherical coordinate system given below. Denote by S n−1 the unit sphere in R n . Note that we will be a bit sloppy with notation, but this will not cause confusion. For example, given a function f : S 2 → C we will denote by f (x) its value at a point x ∈ S 2 . But we might also denote this value by f (ϑ, ϕ) and implicitly assume that x = (sin ϑ cos ϕ, sin ϑ sin ϕ, cos ϑ). For x ∈ R 3 we use the spherical coordintes (2.10) Much of our analysis will be based on the use of spherical harmonics. We denote the space of spherical harmonics of degree N by SH N . Moreover we will use the basis functions Y m N , given by for N ∈ N, m ∈ {−N, −N + 1, . . . , N − 1, N } and where P m N is the usual associated Legendre polynomial, which are defined by the Legendre polynomials P N in (8.1). For more information on these, see [22,31,40]. We shall have an opportunity to use the following classical result from the theory of spherical harmonics, known as Laplace's second integral representation for P m N . See Section 63 of Chapter III in [25]. Theorem 2.4. Let N ∈ N and let m be an integer such that |m| ≤ N . Then, for 0 ≤ ϑ < π/2, we have Remark 2.5. In fact, slightly more can be said: if m happens to be an integer with |m| > N , then the integral on the left-hand side vanishes for 0 ≤ ϑ < π/2.
Finally, we will be making use of the following lemma, which is essentially Lemma 3.6 in [8] or Lemma 4.3 in [36].
, and let C ⊂ R n be a closed cone with vertex at the origin such that inf x∈C\{0} x · Re ρ |x| · |Re ρ| > 0.
In particular, the value q = ∞ is acceptable, and then the exponent n/q − β − n reduces to −β − n as one would expect. The lemma is proven by the change of variables x = y/|ρ| and an application of Hölder's inequality.

The Source Problem
In this section we prove Theorem 1.5. We begin with a few definitions.
Definition 3.1. We call a closed cone C ⊂ R 3 with vertex at the origin an admissible source cone if it is contained in some closed strictly convex circular cone, if its exterior is connected, and if its spherical cross section C ∩ S 2 has the property that Here Y m 2 denotes a spherical harmonic of degree two (2.11). Remark 3.2. We point out that Definition 3.1 is invariant with respect to rotation. That is, if C satisfies all the conditions of the definition, then the rotated cone also satisfies the conditions for any O ∈ SO(3, R).
We will show that smooth enough sources that have a jump on the boundary near a vertex of an admissible cone produce a non-zero far-field, see Theorem 3.9. But before that, we will give some examples of admissible source cones. In particular, cones whose vertex angle is small are admissible no matter the shape of their cross-section. Definition 3.3. Let γ ∈ (0, π/2). Then we define C γ to be the closed strictly convex circular cone . We also define a "magic angle" ϑ 0 to be the unique angle ϑ 0 ∈ (0, π/2) for which cos ϑ 0 = 1/ √ 3.
Proof. The cone C γ is contained in a closed strictly convex circular cone, namely C γ itself. It is also clear that the exterior R 3 \ C γ is connected. To prove the integral condition, we shall prove that After moving to spherical coordinates, writing Y 0 2 in terms of the Legendre polynomial P 2 as in (2.11), and observing that the integrand is independent of the variable ϕ, we only have to show that γ 0 3 cos 2 ϑ − 1 sin ϑ dϑ = 0.
But we can compute that γ 0 3 cos 2 ϑ − 1 sin ϑ dϑ = cos γ sin 2 γ, and clearly the product is non-zero. Proposition 3.6. If C ⊂ R 3 is a closed strictly convex cone with vertex at the origin, if C is not a set of measure zero, if C has connected exterior, and if C ⊆ C ϑ 0 , then C is an admissible source cone.
Proof. This follows directly from the observation that, in spherical coordinates, the function Y 0 2 is strictly positive when ϑ > ϑ 0 . Thus, in the integral the integrand is strictly positive for all x, except possibly for x lying in a set of measure zero, and therefore the integral must also be strictly positive.
Proof. This case can be handled in exactly the same way as the previous example, except that now Y 0 2 is strictly negative almost everywhere in C ∩ S 2 .
then C is an admissible source corner.
Proof. Again, the point is that, in spherical polar coordinates, Y 0 2 is strictly positive when ϑ < ϑ 0 and strictly negative when ϑ 0 < ϑ < π/2, and so After moving to spherical coordinates, writing Y 0 2 in terms of P 2 and simplifying, we see that the original integral is strictly positive if cos γ 1 sin 2 γ 1 + cos γ 2 sin 2 γ 2 − cos ϑ 0 sin 2 ϑ 0 > 0, and this is precisely the hypothesis we made.
We can prove the main theorem about always scattering sources. Note in the proof that since the cone C might have a very rough boundary, we cannot use the simple integration by parts formula from [2].
be an admissible source cone with vertex at the origin. Finally, assume that u ∈B * 2 solves the equation Proof. Let r > 0 be so small that the ball B(0, 2r) is contained in the component of R 3 \supp Φ containing H. By Theorem 2.3, unique continuation and the connectedness of the exterior of C from Definition 3.1, we must have u| B(0,2r)\C ≡ 0. We also observe that, by rotating the entire configuration, if necessary, we may assume that C is oriented so that C \{0} ⊂ R 2 ×R + . Next, let √ λ ≤ τ < ∞ and ψ ∈ R. We shall employ the complex geometrical optics In particular, we have ρ · ρ = −λ, so that We study the limit τ → ∞, and so all the implicit constants below will be independent of τ , even though they are allowed to depend on ψ, ϕ, α, r, ϕ 0 , C, u, the implicit constant in the theorem statement, and the cut-o function χ chosen below. In the We choose a cut-o function χ ∈ C ∞ c (R 3 ) so that χ| B(0,r) ≡ 1 and that χ| R 3 \B(0,2r) ≡ 0, and we write A for the annulus B(0, 2r) \ B(0, r). Then, using our knowledge of the supports of χ, ∇χ, ∆χ, Φ and u, we may argue that In C, the factor e −ρ·x is exponentially decaying, and we may estimate, for some constant δ > 0 depending only on C and r, that since the constants are independent of τ .
We may now continue with We may also estimate Thus, we may further continue with a change of variables to get .
we get by the dominated convergence theorem, after taking τ → ∞, Moving next to polar coordinates, we obtain Let m ∈ {−2, −1, 0, 1, 2}. Moving to spherical coordinates, multiplying both sides with e imψ , and integrating with respect to ψ over the interval (0, 2π) gives Using Laplace's second representation formula, i.e. Theorem 2.4, and multiplying by appropriate constant factors gives Since, by Definition 3.1, the last integral must be nonzero for some choice of m, we conclude that ϕ 0 = 0.

Complex geometrical optics solutions
Here we construct so called complex geometrical optics solutions (CGO-solutions for short) for the Schrödinger equation with a cone like potential. To this end we first show that the admissible cones give continuous multiplication operators between certain Sobolev spaces. This argument closely follows the argument in [36]. We then construct the relevant CGOsolutions, as in that paper, as a consequence of the L p -resolvent estimates of [28] and the multiplier properties of the cone like potential. We end the section by giving some example cones with fairly complicated structure that admit CGO-solutions.
We need to show that V is a pointwise multiplier. The idea is to first establish that if the cross-section of the cone is such that the characteristic function of the cross-section has suitable properties, then also a cylinder with the same cross-section has those properties. This can then be used to cut a infinite cone in to finite pieces with the relevant property, and using scaling properties of the Sobolev norms we get the relevant property for the cone itself.
Our starting point is a slightly reformulated version of Proposition 3.9 in [36]. Then Proof. Let us prove the claim for p = 2. The claim for p < 2 follows from this case because χ C has compact support. Set x := (x , x 3 ), and I := [−1, 1]. Note that so that For χ I one has the estimate One has ξ 2τ ≤ C ξ 2τ ξ 3 2τ . Using these facts gives then that The previous lemma can be used as in [36] to deduce the following proposition.
Proof. This lemma is the analogue of Proposition 3.7 in [36]. Its proof can be easily modified to work in our case. We only need to use our Lemma 4.1 instead of Proposition 3.9 in [36].
We consider a potential V defined using the characteristic function of the previous lemma next.

Proposition 4.3. Let C δ be as in Lemma 4.2. Consider the potential
for any g ∈ H τ,4 (R 3 ) and Proof. Use the notation follows. In more detail, choosep = 2 in Proposition 3.5 in [36], which gives p = 4 and the Hölder conjugate p = 4 3 . Furthermore by the results on p. 205 in [43], we see that Lemma 4.2 with p = 4/3 and this same result in [43] shows (4.2) immediately.
The previous proposition shows that potentials of the form ϕ x −α χ C δ give continuous multiplier operators in the sense of (4.1). The uniform Sobolev estimates of [28] and multiplier property (4.1) can be used to deduce the following existence result for a inhomogeneous equation. Then, the norm bound (4.2) implies the existence of complex geometrical optics solutions for the homogeneous partial di erential equation (−∆ + V − λ)ψ = 0. This is a standard argument, done for potentials supported on a circular cone in Proposition 3.3 of [36]. We nevertheless repeat parts of the proof here for the convenience of the reader.
for all g ∈ H τ,4 (R 3 ). Then there exist constants C > 0 and R > 0 such that whenever ζ ∈ C n satisfies | Re ζ| ≥ R and f ∈ H τ,4/3 (R 3 ) the equation Proof. Here we only repeat the case involving a nonzero V . The first part of the proof of Proposition 3.3 in [36], gives an solution operator G ζ for the case V = 0, for which and (−∆ + 2ζ · D)G ζ g = g for any g ∈ H τ,4/3 (R 3 ). Furthermore G ζ obeys the norm estimate for all g in that space. Proposition 4.3 implies that again for all g as before. We can now choose an R > 0, so that CR −1/2 = 1 2 , and then assuming | Re(ζ)| ≥ R we have for any g. Note that a solution to the equation then we get a solution ψ to (4.3) by setting ψ = G ζ v. Estimate (4.4) shows that (I + V G ζ ) −1 can be expressed as a Neumann series converging in H τ,4/3 (R 3 ). By using the norm estimate for G ζ and (4.4) we have moreover that The solvability result gives us then finally the existence of suitable CGO solutions. Theorem 4.5. Let λ ∈ R + , let ρ ∈ C 3 satisfy ρ·ρ = −λ, and assume that |Im ρ| is large enough.
The next example is a cone with a porous structure. We will use some further results from [21] to analyze this. We use in particular the fact that we can determine what Sobolev space the characteristic function of a set is in by looking at the box-counting dimension of the boundary of the set.  Figure 2. The set A can be constructed iteratively by starting with the four largest circles. One then adds all the circles that are mutually tangent to any three of the initial four circles. In this way, one obtains four new circles, or eight circles in total. One then adds the circles that are mutually tangent to the eight circles similarly, and so forth. Now consider the setÃ ⊂ R 2 , which is constructed as the Apolloninan gasket, but so thatÃ contains the entire interior disk of the circles, but only of those added on every other iteration in the construction of A. See Figure 2. Now we consider the conic set The setÃ is a porous set with non zero Lebesgue measure, which can be analyzed in a straightforward manner. Clearly we have that By the monotonicity of the box-counting dimension (see p.48 in [20]) we obtain that dim M (∂Ã) ≤ dim M (A) ≤ 1.32. We can now apply the results in [21] (see also [39]). Let D ⊂ R n , then in general we have that for 1 ≤ p < ∞. See Theorem 1.3 in [21] and the related comments. For ∂Ã and p = n = 2, the condition becomes dim M (∂Ã) < 2 − 2τ, which follows if τ < 1 − 0.5 · 1.32 = 0.34. Again, we see that Theorem 4.5 implies that we can construct CGO solutions for the potential V = x −α χ C with the desirable remainder estimates if α > 9/4, since we can choose τ so that 1/4 < τ < 0.34.
Remark 4.8. Condition (4.6) has a partial converse that states that for sets D ⊂ R n , we have that for 1 ≤ p < ∞. See [21]. This sets some limits on the applicability of the above constructions. It follows that we cannot obtain CGO-solutions with the above method for potentials supported on cones whose cross-section D ⊂ R 2 has a fractal boundary with dim M (∂D) large enough, since the CGO construction demands that τ > 1/4.

Admissible cones always scatter
In this section we define the notion of an admissible medium cone. Admissibility amounts essentially to an orthogonality condition and certain regularity assumptions. We then prove that admissible medium cones result in potentials that always scatter. After this we formulate a simple determinant condition for medium cones that are admissible.
Definition 5.1. We call a closed cone C ⊂ R 3 an admissible medium cone, if (i) C is contained in a strictly convex closed circular cone, (ii) C has a connected exterior, (iii) C has a bounded cross-section D ⊂ R 2 such that χ D ∈ H τ (R 2 ) for some τ ∈ (1/4, 1/2), and (iv) for any spherical harmonic H ≡ 0 of arbitrary degree N , there exists an index m ∈ {−N − 2, . . . , N + 2} so that for some ε ∈ R + , Φ ∈ e −γ|·| L 2 (R 3 ) for some γ ∈ R + so that Φ| H ≡ 0 for some open half-space H ⊂ R 3 , and so that the origin belongs to the component of R 3 \ supp Φ containing H, and where C is an admissible medium cone with vertex at the origin. Finally, assume that λ is a non-scattering energy for the potential V . Then ϕ(0) = 0.

Proof. By Lemma 2.2, there exist solutions
and w ≡ 0. Let r ∈ R + be so small that the ball B(0, 2r) is contained in the component of R 3 \ supp Φ containing H. By Theorem 2.3, unique continuation and the connectedness of the exterior of C from Definition 5.1, we have u| B(0,2r)\C ≡ 0.
In particular, we have ρ · ρ = −λ, and, assuming that τ is large enough, Theorem 4.5 gives us a solution u 0 = e −ρ·x (1 + ψ) to the equation for some q ∈ (6, 12), and satisfying the estimate where δ ∈ R + is fixed. We shall study the limit τ → ∞, and so all the implicit constants will be independent of τ , but they are allowed to depend on everything else. Since w is real-analytic and w ≡ 0, there exists a homogeneous complex polynomial H(x) of some degree N ∈ {0} ∪ Z + such that H(x) ≡ 0 and for all x ∈ B(0, 2r). The equation (−∆ − λ) w = 0 implies that H is harmonic. This can be seen by applying the di erential operator to the Taylor expansion of w. The argument is essentially the same as in Lemma 17 of [9].
We choose a cut-o function χ ∈ C ∞ c (R 3 ) so that χ| B(0,r) ≡ 1 and χ| R 3 \B(0,2r) ≡ 0, and write A for the annular domain B(0, 2r) \ B(0, r). We may then argue, remembering that V and ϕ χ C coincide in the support of χ, that Since we may estimate |e −ρ·x | e −δ τ in the domain A for some constant δ ∈ R + not depending on τ , we have Now, we may expand the integrand u 0 χϕw in the remaining integral in steps to get Using Lemma 2.6 and the remainder estimate of Theorem 4.5 for ψ, we have for the first of the four integrals on the right-hand side that For the second integral we use that Hölder continuity which implies that |ϕ(x) − ϕ(0)| ≤ C|x| 1/4+ and Lemma 2.6 with q = ∞, so that have that For the third integral we use Lemma 2.6 similarly and obtain that In particular, all three are o(τ −N −3 ), and we may continue The last integral may be estimated to be
And Since e −ρ/|ρ|·x e −x 3 / √ 3 in C, the dominated convergence Theorem gives, when taking τ → ∞, that Writing ω for the vector √ 2ρ 0 / |ρ 0 | = i cos ψ, i sin ψ, 1 , and remembering that ω · y has a positive real part for y ∈ C ∩ S 2 , we may compute through the use of polar coordinates that Since C was assumed to be an admissible medium cone, the last integral is non-zero for some choice of m, and thus ϕ(0) = 0.
In the rest of this section we show that the admissibility of a medium cone can be ascertained by studying certain determinants, which we will shortly define.
We are particularly interested in families of cones {C ρ : ρ ∈ I ⊂ R} depending on the parameter ρ. Consider the (2N + 1)-tuple Ψ of basis functions of SH N +2 , such that These give us information of the projections of the basis functions Y l N on the basis functions in Ψ N in the L 2 (K ρ )-space. Define the matrices And let the corresponding determinants be Note that we are mainly interested in a specific choice of Ψ, which is , as this choice of Ψ is particularly useful for analyzing circular cones. Moreover we will use the abbreviations Proof. Assume that C ρ is not an admissible medium cone, so that there exists an N and for all Y ∈ SH N +2 . We can write H, as

Then in particular
So that for any choice of Ψ, the vectorᾱ := (α −N , .., α N ) = 0 is such that This implies that all the matrix C Ψ,N (ρ) is singular, for any choice of Ψ. In particular we have that D Ψ,N (ρ) = 0, for all Ψ which proves the claim.

Circular cones
Here we use the properties of spherical harmonics to prove that circular cones are admissible medium cones, utilizing the determinant condition of Lemma 5.4. We also use some results on associated Legendre polynomials, which we prove later in Section 8. In this and the following sections, we use the notation below for circular cones. Compare with Definition 3.3. Recall also the notational convention mentioned before (2.10) that allows us to write (ϑ, ϕ) ∈ S 2 instead of x ∈ S 2 , x = (sin ϑ cos ϕ, sin ϑ sin ϕ, cos ϑ). Definition 6.1. By a circular cone C ρ we denote a cone that can be represented in a spherical coordinate system as where ρ ∈ (0, π/2). Also we let K ρ := C ρ ∩ S 2 . Note that a half space is not a spherical cone according to this definition.
Recall that the elements of the matrix C N (ρ) are given by and moreover that The next proposition shows that circular cones are admissible medium cones, and taken together with Theorem 5.3 we see that circular cones always scatter.

Proposition 6.2. A circular cone C ρ is an admissible medium cone.
Proof. By Lemma 5.4 it is clear that it is enough to show that D N (ρ) = 0. In the case of a circular cone the matrix C N (ρ) is an diagonal matrix, since where x 0 = cos ρ, and where we used the definition in (8.1) to deduce that P −m M = cP m M , for some constants c. We will evaluate the integral on the last line. Clearly we can assume that m ≥ 0. By lemmas 8.1, 8.3, 8.4 and 8.5, we know that In particular we have that since P m m and P m m+1 have no zeros on the interval x 0 ∈ (0, 1), and since x 0 ∈ (0, 1), because ρ ∈ (0, π/2). It follows that all the diagonal elements of C N (ρ) are bounded away from zero, when ρ ∈ (0, π/2), and hence D N (ρ) = 0. A circular cone is thus an admissible medium cone.

A density argument for star-shaped cones
In this section we finish the proof of Theorem 1.3 by proving Proposition 7.6. The latter shows that all star-shaped cones have admissible medium cones arbitrarily close to them. Theorem 5.3 imples that these admissible ones always scatter.
Note that C σ ρ is a circular cone when ρ = 0, in particular C σ 0 = C ρ 0 . For ρ = 1, we have C σ ρ = C σ 1 = C σ . The star-shaped deformation is thus essentially given by an interpolation between the points in the circular cone C ρ 0 and the star-shaped cone C σ .
As in the previous sections, we are interested in the functions this time integrated over the star-shaped deformation cap instead of a circular cap. Furthermore recall that and that We will also be dealing with the associated Legendre polynomials 2 which are defined in (8.1) in terms of the Legendre polynomials P n , but we reproduce the equation here for convenience.
where m = 0, .., n. Moreover we set P m n := 0, for m > n. It will be convenient to extend the definition of the associated Legendre polynomials to C. Note that extending the factor (1 − x 2 ) 1/2 can be done in several ways depending on which branch of the square root we choose. where we choose the square root so that it has the branch cut along (−∞, 0), that is √ z = √ re iϕ = √ re iϕ/2 , when ϕ ∈ (−π, π). Lemma 7.3. The functions P m n are complex analytic in {x + iy ∈ C : −1 < x < 1, y ∈ R}. Proof. Since ∂ m z P n (z) is a polynomial and (1 − z) m/2 is a product, it will be enough to show that (1 − z 2 ) 1/2 is complex analytic on {x + iy ∈ C : −1 < x < 1, y ∈ R}. Now Since Re z 2 = x 2 − y 2 ∈ (−∞, 1) when −1 < x < 1, we have is analytic and is analytic. Finally √ z : C \ (−∞, 0] → C is analytic since we choose the branch cut in Definition 7.2 to be (−∞, 0). The range of (1 − z) is contained in the former's domain and so the restriction of √ z to the range of (1 − z) is analytic too. Thus (1 − z 2 ) 1/2 is complex analytic on {x + iy ∈ C : −1 < x < 1, y ∈ R}, which proves the claim.
Next we will shows that D N,σ are analytic functions in the parameter ρ. For this it is convenient to make use of the following elementary lemma. But before that, let us recall some notation. For two sets of complex numbers A, B ⊂ C we define This notation is particularly useful for talking about rectangles.  g[(a, b)] is the image of the interval (a, b) under g. Assume furthermore that (ii) for any z 0 ∈ R , one has the estimate b a |f (z 0 g(t))h(t)| dt < ∞.

Suppose that (i) a complex valued function f is complex analytic on
Then F is complex analytic on R . In particular F is real analytic on (c, d).
Proof. Firstly notice that the first part of condition (ii) guarantees that F is well defined on the rectangle R . Secondly, it su ces to compute the derivative F (z 0 ) when z 0 ∈ R to prove the claim. Choose a sequence (z m ), and we are going to show that F D m converges as m → ∞ and the limit will be F (z 0 ). Note that z 0 ∈ R , which is an open set, and since we are interested in the limit only, we may assume that for all m the whole segment from z 0 to z 0 + z m is in a compact subset K R , which depends only on z 0 and R . This implies that for any t ∈ (a, b) and ξ on that segment, i.e. that ξg(t) will stay a positive distance from ∂R. This will be used later in the proof after a mean value theorem.
To study the limit of F D m (z) define Let us study the limit of these as m → ∞. Firstly, the pointwise limit. If g(t) = 0 then D m (z 0 , t) = 0 so the pointwise limit exists and is 0. If g(t) = 0 then the pointwise limit of D m (z 0 , t) is f (z 0 g(t))g(t) a formula which also applies to the case g(t) = 0. Next, let us show that the |D m (z 0 , ·)| have an integrable upper bound in the interval (a, b). If g(t) = 0, we have by the mean value theorem for some ξ ∈ C which lies on the segment connecting z 0 to z 0 + z m . This also holds when g(t) = 0. The supremum above is a finite number because f is complex analytic in R by (i), and so its derivative f is bounded on its compact subsets. The set Kg(a, b) is such a set according to (7.2). This means that for any fixed z 0 there is a finite constant C depending on a, b, c, d, δ, z 0 and f such that |D m (z 0 , t)| ≤ C|g(t)| ≤ C g ∞ for all t ∈ (a, b), and this is integrable. By the dominated convergence Theorem and since h ∈ L ∞ (a, b), we have that for each z 0 ∈ R the following holds and that the latter is a finite complex number. This gives the existence of F (z 0 ). Hence F is complex analytic in R . Proof. Using spherical coordinates and referring to (2.11) we have that Changing the variable symbols in preparation for Lemma 7.4, we can write We want to apply Lemma 7.4 to show that I k,l N,σ is real analytic in (− 0 , 1) for some 0 > 0. In order to do this we need to check that the conditions (i) and (ii) of the Lemma 7.4 can be satisfied. For condition (i), we need to find 0 > 0 and δ 0 > 0 so that f is complex analytic on the rectangle and 0 ≤ so that C σ ρ is defined for ρ ∈ (− 0 , 1). Choose 0 > 0 by (7.5) 0 := min Since g(t) ∈ (0, π/2 − ρ 0 ) for all t, and by our choice of 0 , we have for x 0 ∈ (− 0 , 1), that Recal also that cos(x + iy) = cos(x) cosh(y) − i sin(x) sinh(y), where each of cos, cosh, sin, sinh are real-valued when their arguments are real. Let δ 2 > 0 be arbitrary and let (7.6) δ 0 := δ 1 g L ∞ , δ 1 = 1 2 min arsinh δ 2 , arcosh 1 cos(ρ 0 /2) both of which are positive, and they give δ 0 g(t) ∈ (0, δ 1 ) for all t. Then for − 0 < x 0 < 1 and −δ 0 < y 0 < δ 0 , we have that Denote α = 0, β = cos(ρ 0 /2) cosh δ 1 . Then by our choice of δ 1 we see that α < β < 1 and sinh δ 1 < δ 2 , and so 2π). In other words, is complex analytic. In particular the associated Legendre polynomials P m n are analytic in the range of f 2 . In fact, R is a positive distance 1 − β away from any point where we do not know them being analytic. See Lemma 7.3.
Let us prove next that f is complex analytic on the rectangle R given in (7.3). We know that P k N +2 P l N is analytic on the vertical strip {x + iy ∈ C : −1 < x < 1, y ∈ R} by Lemma 7.3. Hence dw is likewise analytic in the strip {x + iy ∈ C : −1 < x < 1, y ∈ R}, which happens to contain R . Thus f = f 1 • f 2 is analytic on R and so condition (i) of Lemma 7.4 is satisfied.
Next, we show that condition (ii) is satisfied, i.e. that we have 2π 0 |f (z 0 g(t))h(t)|dt < ∞ for any z 0 ∈ R . Recall that |h(t)| = 1 in our case. Notice that f 1 is in fact uniformly bounded on R . This is because f 1 is analytic in the vertical strip mentioned above, and R is contained in it with distance 1 − β > 0 from its boundary. Since f 2 : R → R , and f 1 : R → C is uniformly bounded, we have f = f 1 • f 2 : R → C uniformly bounded. Given z 0 ∈ R and t ∈ (0, 2π) we have z 0 g(t) ∈ R and so we have a uniform bound for f (z 0 g(t)). Hence the integral is finite. Lemma 7.4 imples that the maps ρ → I k,l N,σ (ρ) are real analytic on (− 0 , 1) where 0 is given by (7.5). The determinant functions D N,σ are obtained from matrices having elements I k,l N,σ as their components. The determinant is a sum of products of these elements and is therefore also real analytic in (− 0 , 1).
Finally, by the proof of Proposition 6.2 we know that D N,σ (0) = 0, since C σ 0 is a circular cone. The functions D N,σ are thus not identically zero in ρ. This completes the proof.
The set Z is countable as a countable union of countable sets. This follows because D N,σ has at most a countable number of zeros in (− 0 , 1) since it is analytic and not identically zero there by Lemma 7.5.

Results on associated Legendre polynomials
In this section we derive some results on associated Legendre polynomials, that are utilized in computing explicitly the determinants D N in the case of circular cones, which is done in Section 6. First we extend some well known formulas for certain inner products of Legendre polynomials to the case of the associated Legendre polynomials, after which we derive a modification of the Christo el-Darboux formula, which can be used to analyze these inner products.
Recall firstly that the associated Legendre polynomials are defined in terms of the Legendre polynomials P n , as where m = 0, .., n. Moreover we set P m n := 0, for m > n. The Legendre polynomials are defined in [11] equation (9), p.10.
We will first derive an integral formula for the special case of the associated Legendre polynomials of the form P 0 n , i.e. in the case when they coincide with the Legendre polynomials P n . This formula can be found in [11] equation (5), p.172. We however give the proof as a convenience to the reader. Lemma 8.1. Let x 0 ∈ (−1, 1). Then we have the following formula where C = 1/(4n + 6).
Proof. Note firstly that P 0 n = P n . The Legendre polynomials in the claim solve the ODEs ∂ x (1 − x 2 )∂ x P n + n(n + 1)P n = 0, Multiplying by P n and P n+2 , and subtracting the resulting equations gives, then that which proves the claim.
Proof. By taking repeated derivatives of the Legendre ODE (1 − x 2 )∂ 2 x P n − 2x∂ x P n + n(n + 1)P n = 0, one arrives by an induction argument at the formula Multiplying by (1 − x 2 ) m−1 , gives then that We are now ready to generalize the result of Lemma 8.1 to the case m = 1, .., n − 1.
Proof. Assume that 1 ≤ m ≤ n − 1. By integration by parts we get that Then using the formula of Lemma 8.2 we have that where C 1 = (n + 2)(n + 3) − m(m − 1). Likewise we can use integration by parts to deduce that where C 2 = n(n + 1) − m(m − 1). Subtracting these we obtain that where the constant is given by C 3 := C 1 − C 2 = 4n + 6. This can be further simplified, since by a straight forward computation Putting this together gives then that This proves the claim for the cases m = 1, .., n − 1.
Finally we prove the results of lemmas 8.1 and 8.3, for the case m = n.
Proof. Note that ∂ n+1 x P n = 0. By integration by parts we have that Let's rewrite the first term on the r.h.s. of the last line. Using Lemma (8.2) we have that x P n+2 ) = −(4n + 6)(1 − x 2 ) n ∂ n x P n+2 .
Notice that by inserting this back into to (8.2) we get the term on the l.h.s. of the claim. It is thus enough to show that the boundary term in (8.2) gives the expression on the r.h.s. of the claim. For this we use the fact that Now we have that ∂ x P n n = ∂ x (1 − x 2 ) n/2 ∂ n x P n , so the previous equations yields by a short computation that x P n+2 ∂ n x P n = (1 − x 2 ) P n n ∂ x P n n+2 − P n n+2 ∂ x P n n .
Going back to (8.2) we see that and that the claim holds.
Next we prove the following Christo el-Darboux type formula, which is a slight modification of the usual formula (see p.43 in [42]). Note that the r.h.s. of the claim has a sign determined by the sign of x.  implies that (8.3) P m n+1 = a n xP m n − b n−1 P m n−1 , P m n+2 = a n+1 xP m n+1 − b n P m n . Using the second of these equalities gives that P m n (x)P m n+2 (y) − P m n (y)P m n+2 (x) = a n+1 yP m n+1 (y)P m n (x) − xP m n+1 (x)P m n (y) . (8.4) Lets evaluate the two terms on the r.h.s. of the equality. By successively applying the recurrence relation in (8.3) we get the equations P m n+1 (y)P m n (x) = a n yP m n (y)P m n (x) − b n−1 P m n−1 (y)P m n (x), P m n−1 (y)P m n (x) = a n−1 xP m n−1 (x)P m n−1 (y) − b n−2 P m n−2 (x)P m n−1 (y), P m n−1 (y)P m n−2 (x) = a n−2 yP m n−2 (x)P m n−2 (y) − b n−3 P m n−2 (x)P m n−3 (y), Using these equations to rewrite the first term on the r.h.s of (8.4), we get that yP m n+1 (y)P m n (x) =a n y 2 P m n (y)P m n (x) − a n−1 b n−1 xyP m n−1 (x)P m n−1 (y) + a n−2 b n−1 b n−2 y 2 P m n−2 (x)P m n−2 (y) − a n−3 b n−1 b n−2 b n−3 xyP m n−3 (x)P m n−3 (y) . . . − a m b n−1 . . . b m z 1 yP m m (y)P m m (x). For the second term on the r.h.s of (8.4), we get similarly the expression xP m n+1 (x)P m n (y) =a n x 2 P m n (x)P m n (y) − a n−1 b n−1 yxP m n−1 (y)P m n−1 (x) + a n−2 b n−1 b n−2 x 2 P m n−2 (y)P m n−2 (x) − a n−3 b n−1 b n−2 b n−3 yxP m n−3 (y)P m n−3 (x) . . . Subtracting the two equalities gives then that yP m n+1 (y)P m n (x) − xP m n+1 (x)P m n (y) =a n (y 2 − x 2 )P m n (y)P m n (x) + a n−2 b n−1 b n−2 (y 2 − x 2 )P m n−2 (x)P m n−2 (y) + . . . + a m b n−1 . . . b m (xz 2 − yz 1 )P m m (x)P m m (y). Notice that the last term is non zero only if n − m is even, and in this case it equals to y 2 − x 2 . Going back to (8.4), we get that P m n (x)P m n+2 (y) − P m n (y)P m n+2 (x) x − y = −a n+1 a n (y + x)P m n (y)P m n (x) + a n−2 b n−1 b n−2 (y + x)P m n−2 (x)P m n−2 (y) + . . .
Adding ±P m n+2 (y)P m n (y) in the denominator and taking the limit y → x, gives that if n − m is even