Reconstruction in the partial data Calder\'on problem on admissible manifolds

We consider the problem of developing a method to reconstruct a potential $q$ from the partial data Dirichlet-to-Neumann map for the Schr\"odinger equation $(-\Delta_g+q)u=0$ on a fixed admissible manifold $(M,g)$. If the part of the boundary that is inaccessible for measurements satisfies a flatness condition in one direction, then we reconstruct the local attenuated geodesic ray transform of the one-dimensional Fourier transform of the potential $q$. This allows us to reconstruct $q$ locally, if the local (unattenuated) geodesic ray transform is constructively invertible. We also reconstruct $q$ globally, if $M$ satisfies certain concavity condition and if the global geodesic ray transform can be inverted constructively. These are reconstruction procedures for the corresponding uniqueness results given by Kenig and Salo. Moreover, the global reconstruction extends and improves the constructive proof of Nachman and Street in Euclidean setting. We derive a certain boundary integral equation which involves the given partial data and describes the traces of complex geometrical optics solutions. For construction of complex geometrical optics solutions, following Nachman and Street and improving their arguments, we use a new family of Green's functions for the Laplace-Beltrami operator and the corresponding single layer potentials. The constructive inversion problem for local or global geodesic ray transforms is one of the major topics of interest in integral geometry.


Introduction
In 1980, Alberto Calderón [2] proposed the problem whether one can determine the electrical conductivity of a medium from voltage and current measurements at the boundary. In the mathematical literature, this problem is known as Calderón's inverse conductivity problem.
The Calderón's problem can be reduced to the problem of determining electric potential q from the Dirichlet-to-Neumann map associated to the Schrödinger operator −∆ + q. We will first discuss the case of Euclidean space in dimension n ≥ 3. In the fundamental paper by Sylvester and Uhlmann [20] it was shown that bounded potential in a bounded domain of Euclidean space can be uniquely determined from the knowledge of the Dirichlet-to-Neumann map. Since then, substantial progress has been achieved on Calderón's problem. Then corresponding reconstruction procedure was given by Nachman [14] and independently by Novikov [16]. The reader is referred to recent expository paper by Uhlmann [22] for a survey of progress made on Calderón's problem.
In the current paper we are interested in the case when the Dirichlet-to-Neumann map is known only on part of the boundary. Let Γ + and Γ − be the open subsets of the boundary where Dirichlet data inputs are prescribed and Neumann data measurements are made. The first result is due to Bukhgeim and Uhlmann [1]. They prove unique determination result if Γ + and Γ − are roughly complementary and slightly more than half of the boundary. This result has been improved significantly by Kenig, Sjöstrand and Uhlmann [10] where they show that bounded potential can be uniquely recovered if Γ − possibly very small open subset of the boundary, but Γ + must be slightly larger than the complement of Γ − in the boundary. Constructive proof of this result is given by Nachman and Street [15]. For recent results on Calderón's inverse problem with partial data, see [7]. The approaches of [1,10,15] are based on Carleman estimates with boundary terms.
There is a result by Isakov [6] where he gives uniqueness result when Γ − = Γ + = Γ and the inaccessible part of the boundary for measurements is either part of a hyperplane or part of a sphere. This work is based on a reflection argument.
In the current paper we consider partial data Calderón's problem on manifolds. The methods of [10,6] were unified and extended to so-called admissible manifolds (which will be described below) by Kenig and Salo [8] obtaining improved results. To appreciate these improvements, the reader is referred to [8,Section 3] for detailed corresponding discussions. The goal of this paper, is to give the reconstruction procedures to the corresponding results of [8].
By the results of Section 2, Λ g,q is a bounded linear operator Λ g,q : H g (∂M ) → H −3/2 (∂M ). Given two open subsets Γ − , Γ + ⊂ ∂M . The partial data inverse problem is to determine q from the knowledge of Λ g,q f on Γ − for all f ∈ H g (∂M ) supported in Γ + .
We need to introduce the notion of admissible manifolds.
Definition. A compact Riemannian manifold (M, g) with boundary of dimension n ≥ 3, is said to be admissible if it is conformal to a submanifold with boundary of R × (M 0 , g 0 ) where (M 0 , g 0 ) is simple (n − 1)-dimensional manifold. By simplicity of (M 0 , g 0 ) we mean that the boundary ∂M 0 is strictly convex, and for any point x ∈ M 0 the exponential map exp x is a diffeomorphism from its maximal domain in where ϕ(x) = x 1 . The function ϕ is a natural limiting Carleman weight in (M, g); see [3]. In the results below we assume that there is a part which inaccessible for measurements Γ i ⊂ ∂M tan , and the accessible part will be denoted by Γ a = ∂M tan \ Γ i .
We say that a unit speed geodesic γ : [0, T ] → M 0 , on transversal simple manifold (M 0 , g 0 ), is nontangential if γ(0), γ(T ) ∈ ∂M 0 and γ(t) ∈ M int 0 if t ∈ (0, T ). The first main result of our paper, says that one can reconstruct the local attenuated geodesic ray transform of the one-dimensional Fourier transform (with respect to x 1variable) of the potential q from the partial knowledge of the Dirichlet-to-Neumann map with Γ + ⊃ ∂M + ∪ Γ a and Γ − ⊃ ∂M − ∪ Γ a . Theorem 1.1. Let (M, g) be an admissible manifold, and suppose that q ∈ C(M ) such that 0 is not a Dirichlet eigenvalue of −∆ g + q. Let Γ i ⊂ ∂M tan be closed such that for some open E ⊂ ∂M 0 one has Let Γ a = ∂M tan \ Γ i and let Γ ± ⊂ ∂M be a neighborhood of ∂M ± ∪ Γ a . Then for any given nontangential geodesic γ : [0, T ] → M 0 with endpoints on E and for any λ ∈ R, the integral T 0 e −2λt (cq)(2λ, γ(t)) dt can be constructively recovered from the knowledge of Λ g,q (f ) on Γ − for all f ∈ H g (∂M ) supported in Γ + . Here q is extended outside of M by zero, and (cq) is the one-dimensional Fourier transform of q with respect to x 1 -variable. This is a constructive version of the corresponding uniqueness result by Kenig and Salo [8,Theorem 2.1].
In the next result, we consider the local geodesic ray transform I O in an open subset O of the transversal simple manifold (M 0 , g 0 ) which is defined for f ∈ C(M 0 ) as Using Theorem 1.1 one can constructively recover potentials in the set where the local geodesic ray transform is invertible.
This result gives a constructive proof of the corresponding uniqueness result by Kenig and Salo [8,Theorem 2.2]; the latter is the above mentioned generalization of the result of Isakov [6].
Constructive invertibility of the local ray transform, to the best of author's knowledge, is known in the following case: if M 0 has dimension n ≥ 3 and if p ∈ ∂M 0 is such that ∂M 0 is strictly convex near p, then there is an open O ⊂ M 0 containing p on which I O is constructively invertible; this result is due to Uhlmann and Vasy [23]. In two dimensions, no such result is known. Even injectivity of the local geodesic ray transform is an open question.
If ∂M tan has zero measure in ∂M , we give the reconstruction procedure to determine potentials globally. The problem is reduced to the constructive invertibility of the global geodesic ray transform on the transversal simple manifold M 0 . Theorem 1.3. Let (M, g) be an admissible manifold, and suppose that q ∈ C(M ) such that 0 is not a Dirichlet eigenvalue of −∆ g + q. Suppose that ∂M tan is of zero measure in ∂M . If the global geodesic ray transform is constructively invertible in M 0 , then q can be constructively determined in M from the knowledge of This is a generalization with refinements to admissible manifolds of the corresponding result by Nachman and Street [15] in Euclidean setting. More precisely, comparing to [15], we do not assume that the subsets of Dirichlet data inputs overlap with the subsets of Neumann data measurements. So our reconstruction procedure is new even in Eulidean space. The version of Theorem 1.3 was given by Kenig, Salo and Uhlmann [9] for full data case on admissible manifolds of dimension three.
Constructive invertibility of the global ray transform is known in the following cases: where Ω ⊂ R n is open and bounded with C ∞ boundary, and e is the Euclidean metric. In this case inversion formula is given in the book of Sharafutdinov [19,Section 2.12]. • (M 0 , g 0 ) of dimension n ≥ 3, have strictly convex boundary and is globally foliated by strictly convex hypersurfaces. For such case, there is a layer stripping type algorithm for reconstruction developed by Uhlmann and Vasy [23]. • (M 0 , g 0 is a simple surface. In this case, there is a Fredholm type inversion formula which was derived by Pestov and Uhlmann [17]; see also the article of Krishnan [11].
The problem of constructive inversion of local or global geodesic ray transforms is of independent interest in integral geometry.
The structure of the paper is as follows. In Section 2 we give some preliminaries about trace operators and Green's identity for the space H ∆g (M ). We also consider the well-posedness of the Dirichlet problem for the Schrödinger equation (−∆ g + q)u = 0 with boundary condition in H g (∂M ). Section 3, following the arguments of [15] and modifying them, is devoted to the construction of the new Green's operators for the Laplace-Beltrami operator, and in Section 4 the corresponding single layer potentials are constructed. The solvability of the required boundary integral equation is given in Section 5. Then we construct complex geometrical optics solutions in Section 6, and we use these solutions to give reconstruction procedures in Section 7.

Trace operators and the Dirichlet-to-Neumann map
Let (M, g) be a compact Riemannian manifold with boundary. We use the notation d Vol g for the volume form of (M, g) and d(∂M ) g for the induced volume form on the boundary ∂M . For any two functions u, v on M , define an inner product and the corresponding norm will be denoted by · L 2 (M) . For any two functions f, h on Γ ⊂ ∂M , define an inner product and by · Γ will be denoted the corresponding norm. We also write for short . Following Bukhgeim and Uhlmann [1], we work with the following Hilbert space which is the largest domain of the Laplace-Beltrami operator ∆ g : The norm on H ∆g (M ) is The proof of the following result is essentially the same as in [1] (see also, for example [12]). We include it here for the completeness and accuracy of the exposition. Proof. First, we show that the trace map tr has an extension to a bounded operator H ∆g (M ) → H −1/2 (∂M ). Let u ∈ C ∞ (M ) and w ∈ H 1/2 (∂M ). By the surjectivity of the trace map on H 2 (M ), there is v ∈ H 2 (M ) such that Using Green's formula, we get Therefore, This proves that the map tr : Next, we show that the trace map tr ν has an extension to a bounded operator H ∆g (M ) → H −3/2 (∂M ). Let u ∈ C ∞ (M ) and w ∈ H 3/2 (∂M ). By the surjectivity of the trace map on H 2 (M ), there is v ∈ H 2 (M ) such that Using Green's formula, we get Therefore, This proves that the map tr ν : C ∞ (M ) → H −3/2 (∂M ) is bounded and controlled by the H ∆g (M )-norm. Since C ∞ (M ) is dense in H ∆g (M ), we can extend tr ν to a bounded linear map H ∆g (M ) → H −3/2 (∂M ). Now, we give the proof of the last statement. First, we consider the case when tr(u) = 0. Let u ∈ C ∞ (M ) with tr(u) = 0. Using, Green's identity, we have for some constant C > 0. By [21,Theorem 1.3] in Chapter 5, we have , for some another constant C > 0. Combining this with (2.1), we obtain , tr(u) = 0. By density arguments, we obtain , tr(u) = 0. This proves the last statement for the case when tr(u) = 0.
Suppose now that u ∈ H ∆g (M ) with tr(u) ∈ H 3/2 (∂M ). By the surjectivity of the trace operator, there is v ∈ H 2 (M ) such that tr(v) = tr(u). Set w := u − v, then w ∈ H ∆g (M ) with tr(w) = 0. By what we have proved above, w ∈ H 2 (M ), and hence u ∈ H 2 (M ).
The proof of Proposition 2.1 gives the following.
where ·, · H −s,s (∂M) is the duality between H −s (∂M ) and H s (∂M ). Now we introduce the following space on the boundary ∂M : Assume that q ∈ L ∞ (M ) and let us introduce the Bergman space b q (M ) as follows The topology on this space is a subspace topology in L 2 (M ). It is not difficult to check that b q (M ) is a closed subspace of L 2 (M ).
We need the following result to define a topology on H g (∂M ): is one-to-one and onto.
. We define the norm on H g (∂M ) as f Hg(∂M) = P 0 f L 2 (M) . In particular, by Proposition 2.3, this implies that tr : b 0 → H g (∂M ) as well as P 0 : H g (∂M ) → b 0 are bounded. Next, we give the following solvability result of the Dirichlet problem with boundary data in H g (∂M ): is bounded and since tr(w) = tr(u), we can conclude that the map u → tr(u) is bounded H ∆g (M ) → H g (∂M ).
Since the inclusion b q ֒→ H ∆g (M ) is bounded, by the first part of the proposition, the map tr : b q → H g (M ) is bounded. Bijectivity of the latter map, which follows from Proposition 2.3, together with Open Mapping Theorem, implies the last statement.
We also extend the domain of the Dirichlet-to-Neumann map to H g (∂M ): Moreover, the following integral identity holds Proof. Suppose that f, h ∈ H g (∂M ). Let u ∈ H ∆g (M ) be the unique solution to the boundary value problem and let u 0 be the unique solution to the boundary value problem By the last statement of Proposition 2.1, we can conclude that w ∈ H 2 (M ). Note that by h. Now, we can apply Corollary 2.2 and get 3) The right-hand side depends continuously on f, h ∈ H g (∂M ). Hence, so does the left hand-side and this together with (2.3) implies that the result.

The Green's operators
Let (M, g) be an admissible manifold and let q ∈ L ∞ (M ). Let us introduce certain notations which will be used thoughout the paper. For τ ∈ R, we consider the Constructions of Green's operators and the corresponding single layer potentials, as well as construction of complex geometrical optics solutions are based on the following Carleman estimates with boundary terms for the conjugated operator e τ x1 (−∆ g + q)e −τ x1 . Proposition 3.1. Let (M, g) be an admissible manifold and let q ∈ L ∞ (M ). There are constants C 0 , τ 0 > such that for all τ ∈ R with |τ | ≥ τ 0 and δ > 0, we have Proof. This estimate was proven by Kenig and Salo; see [8,Proposition 4.2].
. The aim of this section is to prove the following result.
Theorem 3.2. Let (M, g) be an admissible manifold. There is a constant τ 0 > 0 such that for all τ ∈ R with |τ | ≥ τ 0 , there is a linear operator Proposition 3.4. Let (M, g) be an admissible manifold. There is τ 0 > 0 such that for all τ ∈ R with |τ | ≥ τ 0 and for a given v ∈ L 2 (M ), there is a unique solution u ∈ L 2 (M ) of the equation where in the last step we have used the Carleman estimate (3.1). By the Hahn-Banach theorem, we may extend L to a linear continuous functional L on L τ . On the orthogonal complement of L τ in L 2 (M ) we define L to be zero. By the Riesz representation theorem, there exists u ∈ L 2 (M ) such that In particular, Since L ≡ 0 on the orthogonal complement of L τ in L 2 (M ), we have that u ∈ L τ and hence π τ u = u.
Let H τ : L 2 (M ) → L 2 (M ) be the solution operator obtained in the previous result. In other words, the operator H τ is defined by H τ v = u, where u and v are as in Proposition 3.4. The following is an immediate corollary of the preceeding result.
Corollary 3.5. Let (M, g) be an admissible manifold. There is τ 0 > 0 such that for all τ ∈ R with |τ | ≥ τ 0 , there is a linear operator and π τ H τ = H τ . This operator satisfies Thus, the operator H τ satisfies Theorem 3.2 except (3.2). We shall accodingly modify H τ to obtain (3.2). We need the technical result.
where in the first step we have used the fact that π * −τ = π −τ (since π −τ is projection) and in the last step we have used that H * τ π τ = H * τ (this follows from π τ H τ = H τ which is true by Corollary 3.5). Also note that T −τ π ⊥ τ = H −τ π τ π ⊥ τ = 0. Thus, T * τ π ⊥ τ = T −τ π ⊥ τ = 0, and hence, to prove the lemma it is sufficient to show that Since by Corollary 3.5 we know that tr(H τ e τ x1 (−∆ g )e −τ x1 w) is supported in S + τ , we can use Green's identity and the fact that v ∈ D + τ to get Since e τ x1 (−∆ g )e −τ x1 H τ = Id by Corollary 3.5, we obtain Here, in the last step we used the Green's identity and that w| ∂M = v| ∂M = 0.
Using that tr(H −τ e −τ x1 (−∆ g )e τ x1 v) is supported in S + −τ , w ∈ D + −τ and the Green's identity, we obtain In the last step we used the fact that e −τ x1 (−∆ g )e τ x1 H −τ = Id (by Corollary 3.5). The proof of the lemma is thus complete.
Proof of Theorem 3.2. Define G τ = H τ +π ⊥ τ H * −τ . By Corollary 3.5 and Lemma 3.3, it follows that It is left to prove (3.2). For this, we need first to show that G * τ = G −τ . Using Lemma 3.6, we can show We have shown that tr(G −τ f ) is supported in S + −τ . This fact together with u ∈ D + −τ allows us to use the generalized Green's identity from Corollary 2.2 and get Here, in the last step we used the already proven fact that e −τ x1 (−∆ g )e τ x1 G −τ = Id. This finishes the proof.

Single layer operators
The aim of this section is to construct the single layer operators S τ corresponding to the Green's operators G τ constructed in the previous section.
In other words, (tr • G τ ) * defined for h ∈ e −τ x1 (H g (∂M )) * by , and that the support of h is in ∂M \ B.
Let us now show that the support of tr((tr • G τ ) * h) is in S + −τ . For arbitrary f ∈ D + −τ , using the generalized Green's identity from Corollary 2.2, we get where in the last step we have used (4.1).

Now, we prove the last statement of the proposition. If h is supported in
. This is because by the last statement of Theorem 3.2, (tr • G τ )f is supported in S + τ . The proof of the proposition is thus complete.

Boundary integral equation
In the present section, we prove the solvability of the following boundary integral equation: for τ ∈ R with |τ | ≥ τ 0 To prove the solvability of (5.1), we need the following result on basic properties of the operator S τ (Λ g,q − Λ g,0 ).
, and for f ∈ H g (∂M ), S τ (Λ g,q − Λ g,0 )f is supported in B and can be computed from the knowledge of Λ q f | ∂M − sgn(τ ) . Moreover, the following factorization identity holds Proof. First part of the proposition is a consequence of Proposition 2.5 and Proposition 4.2. To prove the last statement, consider h ∈ (H g (∂M )) * and f ∈ H g (∂M ). Then The proof is thus finished.
The following result shows that the boundary integral equation is equivalent to the certain integral equation; compare with [9, Proposition 3.2].
Proof. Suppose that f, h ∈ H g (∂M ) satisfies (Id +S τ (Λ g,q − Λ g,0 )) h = f . Note that by Theorem 3.2, we can show that Therefore, it is enough to prove that tr Id +e −τ x1 G τ e τ x1 q P q (h) = f, or equivalently h + tr e −τ x1 G τ e τ x1 qP q (h) = f. Using the factorization identity in Proposition 5.1, we can see that the left hand-side is (Id +S τ (Λ g,q − Λ g,0 )) h, which is equal to f by assumption.
The converse direction can be shown by applying tr to the both sides of the identity Id +e −τ x1 G τ e τ x1 q P q (h) = P 0 (f ) and using the factorization identity in Proposition 5.1. Corollary 5.3. Suppose that q ∈ L ∞ (M ) and 0 is not a Dirichlet eigenvalue of −∆ g + q in M . There is τ 0 > 0 such that for all τ ∈ R with |τ | ≥ τ 0 , the operator Id +S τ (Λ g,q − Λ g,0 ) : H g (∂M ) → H g (∂M ) is an isomorphism if and only if so is the operator The following proposition combined together with the above two results implies the solvability of the boundary integral equation (5.1).

Complex geometrical optics solutions
Let q ∈ L ∞ (M ) be such that 0 is not a Dirichlet eigenvalue of −∆ g + q in M , and let τ ∈ R with |τ | ≥ τ 0 . In this section we construct the complex geometrical optics solutions for the Schrödinger equation (−∆ g + q)u = 0 in M whose trace is supported in Γ sgn(τ ) .
is in the closure of M τ . However, by Lemma 6.1, (u − u ′ , u + τ − u ′+ τ ) is orthogonal to the closure of M τ . Thus, we obtain u − u ′ = 0 which finishes the proof.
be the solution operator obtained in the previous result. In other words, the operator R τ is defined by where u, f, f − τ are as in Proposition 6.2. 6.2. Construction of complex geometrical optics solutions. Now, we are ready to construct complex geometrical optics solutions whose traces are supported in Γ sgn(τ ) . These are the solutions of the form u = e −τ x1 (a + r 0 ), (6.1) where r 0 is a correction term and a is an amplitude.
We want to ensure that tr(u 0 ) is supported in Γ sgn(τ ) where Γ sgn(τ ) ⊃ ∂M sgn(τ ) ∪Γ a . To achieve this, following [8], we take a small parameter δ > 0 to be chosen later, and define the following sets For the boundary condition, we set Defining f − τ,δ in such a way, we have f − τ,δ | Γ τ,δ a ∩∂Mtan = 0. Recall that Γ i ⊂ R × (∂M 0 \ E) and a was chosen in a way to satisfy a = 0 near ∂M 0 \ E. Therefore, f − τ,δ | V τ,δ ∩∂Mtan = 0, and hence we have 1, we obtain the following estimates If we set r 0 = R τ (f, f − τ,δ ), then, by Proposition 6.2, r 0 solves (6.2) with tr(r 0 )| S − τ = f − τ,δ and satisfies Thus, there is constant C 0 > 0 such that We choose δ such that C 0 o δ→0 (1) ≤ ε/2. Then we take |τ | ≥ τ 0 large enough so that Therefore, we get r 0 L 2 (M) → 0 as τ → ∞. This will give the complex geometrical optics solution u 0 = e −τ x1 (a + r 0 ) to (−∆ g )u 0 = 0 whose trace is supported in Γ sgn(τ ) . Thus, we have proved the following proposition. Remark 6.4. Modifying the above arguments in appropriate places, one can construct complex geometrical optics solutions whose traces are supported in ∂M sgn(τ ) if ∂M tan has zero measure in ∂M . Let us indicate these modifications. Up to (6.3) everything is same except that we do not put any restrictions on b, so that we do not require a to vanish on any part of the boundary. In order to ensure that supp(tr(u)) ⊂ ∂M sgn(τ ) , for fixed δ > 0, we set 1 and σ ∂M (∂M tan ) = 0, we obtain the following estimates We use Proposition 6.2 to solve (6.2) for r 0 with tr(r 0 )| S − τ = f − τ,δ and to show that r 0 satisfies the same estimate as before for some C 0 > 0 constant: Thus, we have constructed the complex geometrical optics solution u 0 ∈ H ∆g (M ) to (−∆ g )u 0 = 0 of the form u 0 = e −τ x1 (a + r 0 ) whose trace is supported in ∂M sgn(τ ) and r 0 L 2 (M) → 0 as τ → ∞.

Proofs of the main results
Proof of Theorem 1.1. Suppose that q ∈ C(M ) such that 0 is not a Dirichlet eigenvalue of −∆ g + q. Assume the knowledge of (M, g) and Λ g,q f on Γ − for all f ∈ H(∂M ) supported in Γ + . Then by Proposition 2.5, the following integral identity holds where u 1 ∈ H ∆g (M ) is a solution of (−∆ g + q)u 1 = 0 in M with tr(u 1 ) supported in Γ + , and u 2 ∈ H ∆g (M ) is a solution of (−∆ g )u 2 = 0 in M with tr(u 2 ) supported in Γ − .
Now, we show that tr(u 1 ) can be reconstructed from the above mentioned partial knowledge of Λ g,q . By Proposition 6.5 and Proposition 5.2, one can check that tr(u 1 ) satisfies the following boundary integral equation (Id +S τ (Λ g,q − Λ g,0 ))tr(u 1 ) = tr(u ′ 1 ). Then Corollary 5.3 and Proposition 5.4 imply solvability of the above boundary integral equation for sufficiently large τ . Substituting this solution tr(u 1 ) into the left hand-side of (7.1), we can determine (u 2 |qu 1 ) L 2 (M) for all complex geometrical optics solutions u 1 , u 2 of the above form.
Varying b ∈ C ∞ (S n−2 ) so that the support of b is sufficiently close to θ 0 and noting that the term in the brackets is the one-dimensional Fourier transform of q with respect to the x 1 -variable, which we denote by q, we determine ∞ 0 e −2λr (cq)(2λ, r, θ 0 ) dr.
Since q is compactly supported in x 1 -variable, its Fourier transform (cq)(λ, ·) is analytic with respect to λ. Therefore, we have reconstructed the Taylor series expansion of (cq)(λ, ·) in O. Then we determine q in M ∩ (R × O) by inverting the one-dimensional Fourier transform of cq with respect to the x 1 -variable.
Proof of Theorem 1.3. Let (M, g) be a known admissible manifold such that ∂M tan is of measure zero in ∂M . Suppose that q ∈ C(M ) such that 0 is not a Dirichlet eigenvalue of −∆ g + q. Assume the knowledge of Λ g,q f on ∂M − for all f ∈ H(∂M ) supported in ∂M + .