Resonances for steplike potentials: forward and inverse results

We consider resonances associated to the operator $-\frac{d^2}{dx^2}+V(x)$, where $V(x)=V_+$ if $x>x_M$ and $V(x)=V_-$ if $x<-x_M$, with $V_+\not = V_-$. We obtain asymptotics of the resonance-counting function in several regions. Moreover, we show that in several situations, the resonances, $V_+$, and $V_-$ determine $V$ uniquely up to translation.


Introduction
Suppose V ∈ L ∞ (R; R) is "steplike" in the sense that there exist constants x M , V + , V − , V + = V − so that V (x) = V ± when ±x > x M . We will study the resonances associated with the operator − d 2 dx 2 + V . The resolvent R(z) = (− d 2 dx 2 + V − z) −1 is bounded on L 2 (R) for all but a finite number of z ∈ C \ R + . Considered as an operator from L 2 comp (R) to H 2 loc (R), it has a meromorphic continuation to the Riemann surfaceẐ which is the minimal Riemann surface on which (z − V + ) 1/2 and (z − V − ) 1/2 are single-valued holomorphic functions. This follows from the same techniques as used to study manifolds with cylindrical ends [7,11], although this case is much simpler (see Section 3 for an explanation). In fact, one motivation for studying steplike potentials is the similarity their scattering theory has to the scattering theory for manifolds with cylindrical ends, and a hope that an understanding of the steplike case will shed some light on what can happen in the cylindrical ends case.
Choose a χ ∈ C ∞ c (R) so that χ = 1 on the support of V ′ . Then the poles of χR(z)χ are independent of the choice of χ with this property and we set R = {z j ∈Ẑ : χR(z)χ has a pole at z j }.
We list the elements with their multiplicities.
Let Π :Ẑ → C be the natural projection operator and let r ± (z) = (z − V ± ) 1/2 . If V + = V − ,Ẑ is a four-fold cover of the plane. We identify the part ofẐ with Im r + (z) > 0 and Im r − (z) > 0 as the "physical sheet," that is, the sheet ofẐ on which R(z) is a bounded operator on L 2 (R).
Poles of R(z) are called resonances. They are in some sense analogous to eigenvalues and can, in many settings, be viewed as corresponding to decaying waves. See [16,20] for an introduction to resonances and a survey of some results on their distribution, or [21] for a light-hearted introduction.
Although resonances have been studied in many settings, there are relatively few cases where the asymptotics of the resonance counting function are known. Many of these are in some sense one-dimensional or degenerate, cf. [5,12,13,14,15,18]. Our first theorem is Theorem 1.1. Suppose V + = V − , and the convex hull of the support of V ′ is [a, b]. Then #{z j ∈ R : |Π(z j )| ≤ r 2 , Im r + (z j ) < 0, and Im r − (z j ) < 0} = 2 π (b − a)r + o(r).
In case V + = V − , this theorem is due to [13,18]. Our proof has more in common with the proof of [5], though we have to make adaptations due to the nature ofẐ. In addition, rather than study the zeros of a function holomorphic on all ofẐ, we study the zeros of functions meromorphic onẐ, in a region where they are holomorphic except at a finite number of points. The next theorem follows by similar techniques, although we must make stronger assumptions on the perturbation.
Here and elsewhere H is the Heaviside function. If β ∈ (−b 1 , b 1 ), then, we see that the distribution of poles on these sheets depends rather heavily on the value of β. It therefore seems unlikely that on one of these sheets (say, Im r + > 0, Im r − < 0) an asymptotic formula holds in general. However, we can prove the weaker Zworski [22] and Korotyaev [8] showed that resonances determine a compactly supported, even potential in one dimension if 0 is not a pole of the resolvent, and that there are two such potentials with the same resonances if 0 is a pole of the resolvent. The papers [1] and [9] have studied a similar question for compactly supported potentials on a half-line. Here we give an example of our inverse results for steplike potentials. Further results can be found in Section 5. Proposition 1.1. Suppose we know all the poles of R(z), V + , V − (with V + = V − ), and know that our potential V has V ′ compactly supported. If, in addition, Π −1 (V + ) ∩ R = ∅, then V is uniquely determined up to translation.
It is clear that the qualifier "up to translation" cannot be removed, as V (x) and V (x − c), c ∈ R, have the same resonances. The techniques of the proof are similar to those Zworski [22] used in recovering an even, compactly supported potential on R. This problem is more difficult because onẐ we do not have the simple Weierstrass factorization that we have on C. On the other hand, because V + = V − , some "symmetry" is broken and we have more identities at our disposal than when V + = V − . A related issue is thatẐ is "bigger" than C, and thus there are "more" resonances and they can carry more information.
The techniques developed here for the forward problem will be applied in a future paper to obtain results for the distribution of resonances for potential scattering on cylinders.
We shall assume throughout that V + < V − .

Preliminaries from Complex Analysis
In this section we recall some results and language of complex analysis, e.g. [10], and prove a theorem we shall need on the distribution of zeros of functions which are "good" in a half-plane.
We shall often work with functions that are holomorphic not in the whole plane but are holomorphic within an angle (θ 1 , θ 2 ). A function F holomorphic in an angle (θ 1 , θ 2 ) is of order ρ there if lim r→∞ ln ln(sup θ∈θ1,θ2 |F (re −θ )|) r = ρ.
A function of order 1 and type τ < ∞ (in an angle (θ 1 , θ 2 )) is said to be of exponential type there. Of course, ρ and τ can depend on (θ 1 , θ 2 ). The indicator of a function F holomorphic in an angle θ 1 ≤ arg z ≤ θ 2 and of order ρ is A function F is of completely regular growth within the angle ( where the set E ⊂ R + is of zero relative measure and the convergence is uniform for θ ∈ (θ 1 , θ 2 ). We shall abuse notation slightly and also use the language above for a function that is holomorphic for θ 1 ≤ arg z ≤ θ 2 and z outside of a compact set.
For a function f defined in the upper half plane, define n f + (r) = #{z j : Im z j > 0 and f (z j ) = 0}.
Here, as elsewhere, all zeros are counted with multiplicities. Then A related result is Lemma 3.2 of [6].
Proof. We first adopt some arguments of [10, Chapter III, Section 2]; a similar argument is used to prove Lemma 6.1 of [6]. By the principle of the argument, We use the Cauchy-Riemann equations to rewrite this as Dividing both sides of the equation by 2πr, and integrating from 0 to r, results in ln |f (re iϕ )|dϕ.
so that we can, given ǫ > 0, choose r ǫ so that for all r > r ǫ . Now we roughly follow the proof of [10, Theorem 3, Chapter III, Section 3]. By [10, Theorem 6, Chapter V, Section 4], (1) and (2) imply that f (z) is a function of class A and completely regular growth for 0 < arg z < π, and h f (θ) = k sin θ. Because f is of completely regular growth, if r does not belong to the exceptional set E for the function f , then for a sufficiently large r ǫ and all r > r ǫ we have Then we have, for r > r ǫ not in the exceptional set, In the same manner as [10], proof of Theorem 3, Chapter III, we can show that the limit holds without the restriction that r ∈ E. Again following [10], we have by the monotonicity of n f + (r) that for k > 1, giving us lim sup for any k > 1. Taking the limit as k ↓ 1 we obtain lim sup By a similar argument (taking 0 < k < 1), we obtain This proves the theorem.
We shall use another theorem whose proof is very similar to the previous one. If f (z) is holomorphic in {z ∈ C : |z| ≥ R 1 }, let n f,R1 (r) = #{z j : f (z j ) = 0, R 1 < |z j | ≤ r}. Proof. As in the proof of the previous theorem, the principle of the argument followed by an application of the Cauchy-Riemann equations gives us, for r > R 1 , Dividing both sides of the equation by 2πr and integrating from R 1 to r, results in Then the proof follows the proof of the previous theorem.

Elementary scattering theory of
To a point z ∈Ẑ we may associate the two roots r ± (z), and (if we are consistent) we can determine a mappingẐ →Ẑ by its action on r + (z) and r − (z). We define three maps ω + , ω − , and ω +− :Ẑ →Ẑ, by and r ± (ω +− (z)) = −r ± (z). We shall return to these mappings shortly.
We define two functions φ ± (x, z) with the following properties: Recall that we assume that V + < V − . The following identities are well known when π(z) ∈ R (e.g. [4, Theorem 2.1]) and can be extended toẐ using the meromorphy of R ± , T ± . We have translated them to statements onẐ using the mappings ω ± , ω +− defined above to obtain We shall also want a family of reference operators, which we define below. For β ∈ R, define the potential V β to be We define the corresponding set of generalized eigenfunctions, φ ±,β (x, z), which in this case can be given explicitly: and We shall denote From the explicit representation of φ ±,β and the corresponding expression for the Schwartz kernel of R β (z) (cf. (19)), it is relatively easy to see that R β (z) : also has a meromorphic continuation toẐ.

Asymptotics for the number of resonances
In this section we use knowledge about the relationship between poles of the resolvent and poles of R ± , T ± to obtain results on the asymptotics for the number of poles. Many or all of these results could be obtained by more closely following the methods of [5]. However, we shall want some of the intermediate results we obtain here for the inverse results of Section 5.
To bound the number of resonances, we shall use the following proposition. It is essentially a restating of Proposition 3.3 of [3] for this special case. We remark that while the proofs of [3] are given for the Laplacian, or Laplacian plus compactly supported potential, on a manifold with cylindrical ends, they work almost without change for our setting.
Proposition 4.1. Let z ∈Ẑ, and suppose Π(z) is not in the point spectrum of −d 2 dx 2 +V . Then, if Im r ± (z) < 0 but Im r ∓ (z) > 0, then z is a pole of χRχ if and only if z is a pole of R ∓ , and the multiplicities coincide. If Im r − (z) < 0 and Im r + (z) < 0, then z is a pole of χRχ if and only if it is a pole of R − R + − T − T + , and the multiplicities coincide.
In fact, in this simple case we can say more. Looking at the expansions for φ + at infinity, if R − has a pole of order k at z 0 , then so does T − . Then using (5), unless π(z 0 ) is V + or V − , T + has a pole of the same order, and thus so does and since T ± , R ± are regular at ω +− (z), the multiplicity of z as a pole of R − R + − T − T + is equal to its multiplicity as a pole of R + .
We recall some results and notation of [3] that will be helpful in extending Proposition 4.1. Although the following lemma may be well known, one reference is Lemma 3.1 of [3].
Lemma. Suppose A(z) is a d × d-dimensional meromorphic matrix, invertible for some value of z. Then, near z 0 , it can be put into the form where E(z), F (z), and their inverses are holomorphic near z 0 , and P i P j = δ ij P i , tr P 0 = d − p ′ , tr P i = 1, i = 1, ..., p ′ . The k j and l j are, up to rearrangement, uniquely determined.
Using the notation of the lemma, the "maximal multiplicity" µm of A at z 0 is The following proposition improves on Proposition 4.1.
Proof. By Proposition 4.1 and the subsequent remarks, we need only consider what happens when Π(z) lies in the spectrum of − d 2 Thus R ± , T ± cannot have a pole of multiplicity greater than the multiplicity of the pole of R. First, suppose z corresponds to an eigenvalue; that is, z lies in the physical sheet and Π(z) lies in the point spectrum of − d 2 dx 2 + V . Then χRχ, T ± , R ± all have poles of order one at z. Then T ± , R ± are regular at ω + (z), ω − (z), and ω +− (z). By [3, Theorem 3.1], then, χRχ is regular at ω + (z), ω − (z), and ω +− (z).
By [11,Proposition 6.7] and the fact that there are no eigenvalues embedded in the continuous spectrum, if z ∈ R lies on the boundary of the physical sheet and Π(z) ∈ [V + , ∞), then Π(z) = V + or Π(z) = V − . Moreover, again by Proposition 6.27 of [11], if Π(z) = V − , z is not a pole of the resolvent, so the entries in the scattering matrix must be regular there. Because , and hence T ± (z), R + (z) have no poles when Π(z) ∈ (V + , V − ). Then, by [11,Proposition 6.7] and [3, Theorem 3.1], χRχ has no poles in this region.
Next suppose that Π(z) ∈ (V − , ∞) but that z does not lie on the boundary of the physical sheet. Let By Theorem 3.1 of [3], the multiplicity of z as a pole of R is equal to its "maximal multiplicity" as a pole of S 2 . Since, for the z we are considering, (S 2 (z)) −1 = S 2 (ω +− (z)) = S 2 (z), the determinant of S 2 is regular for these values of z. Therefore, since S 2 is a 2 × 2 matrix, the maximal multiplicity of z as a pole of S 2 is equal to its multiplicity as a pole of R + (or T ± or R − ). Finally, we consider what happens for z 0 such that Π(z 0 ) = V + . If z 0 lies on the boundary of the physical sheet, by [11,Proposition 6.7], z 0 is a pole of order one of χRχ if and only if T − (z 0 ) = 0. Notice in this case T + and R + have a pole at z 0 but are regular at ω − (z 0 ). If T + is regular at z 0 , χRχ is regular at z 0 . Now suppose z 0 does not lie on the boundary of the physical sheet and Π(z 0 ) = V + . Then we will use Here we use the notation Note that φ + is holomorphic at z 0 and ω +− (z 0 ).
Considering the expansion at infinity of the residue of the kernel of R(z), we see that R has a pole at z 0 only if φ − , and thus T + , has a pole at z 0 . Thus, using (17), z 0 ∈ R if and only if T + has a pole of order 1 at z 0 .
We shall now study the behaviour of T ± (z), R ± (z), when z lies on the closure of the physical sheet (Im r + (z) > 0, Im r − (z) > 0). It will be more convenient to introduce the variable k = r + (z) on its closure. We write z = z(k) = (k 2 + V + ) 1/2 in this region. When k 0 ∈ R, we understand z(k 0 ) to be the element ofẐ obtained by taking the limit of z(k) as k → k 0 , Im k > 0.
The following proposition is related to Lemma 3.3 and Corollary 3.4 of [5].
when z is on the closure of the physical sheet and |z| is sufficiently large. We shall apply this identity and (17) with β = 0. The operator e ±ikx R β (z(k))e ∓ikx has Schwartz kernel given by where is the Wronskian. Letχ ∈ C ∞ c (R) and note that when z is in the closure of the physical sheet, r + (z) = r − (z) + O(1/|z| 1/2 ). Examining (14), (15), (19), and (20), we see that when k is in the closed upper half plane and z is in the closure of the physical sheet. (The constant C depends onχ, of course.) Therefore, in this same region and, when |k| is sufficiently large, Let Then g is supported in [−b 1 , b 1 ], and using (14) and (21) we have e ik· g(·, k) ≤ C.
Using (17), (18), and the explicit expression for the Schwartz kernel of R β (z), we obtain Thus |T − (z(k)) − 1| ≤ C |k| for k in the closed upper half plane, and z in the closure of the physical space. The result for T + follows from the relation (5).
We recall a slight modification of Lemma 4.1 of [5]. Then e ±ikx h(x)(1 − f (x, k))dx has exponential type at least 1 for k in the upper half plane.
We will use this lemma to prove is the convex hull of the support of V ′ . Then R ± (z(k)) are, for Im k > 0, functions of exponential type and completely regular growth. Moreover, R ± (z(k)) is of type 2b 1 in this region. For t ∈ R and fixed α ≥ 0, and z(t + iα) in the closure of the physical sheet, R ± (z(t + iα)) = O(|t| −1 ).
Proof. The proof of this Proposition resembles that of the previous one, though we must use Lemma 4.1.
We give the proof for R − , as the proof for R + is similar. Using the functions g and W 0 defined in (22) and (20), respectively, we have where Using (23), |I 1 (k)| ≤ C, |I 3 (k)| ≤ C when Im k ≥ 0 and |k| is sufficiently large. We rewrite where The second integral in (27) is clearly bounded. Using (21), the first integral is, for large |k|, bounded by Ce 2b1 Im k . This shows that |R − (z(t + iα))| = O(|t| −1 ). The function f 1 (x, k) may have poles a finite number of points with Im k ≥ 0 (and Im r − (z(k)) > 0). These poles correspond to eigenvalues of − d 2 dx 2 + V . Using (21), for t real, and for all but a finite number of Rdt). Now, by a shift of variable (using k − iα, for α chosen greater than the largest imaginary part of a pole of R(z(k))) followed by a rescaling, we may apply Lemma 4.1 to see that I 2 (k), and thus R − (z(k)), is of type 2b 1 .
Proof of Theorem 1.1. We continue to use the variable k = r + (z), working in the closed upper half plane (with Im r − (z(k)) ≥ 0).
Suppose k 1 , k 2 , ..., k n are the poles, listed with multiplicity, of φ(k) = (T − T + − R − R + )(z(k)) for k in the closed upper half plane. Let p(k) = n j=1 (1 − k/k j ). Then φ 1 (k) = p(k)φ(k)/p(−k) is a holomorphic function in Im k > 0, with a continuous extension to the closed half plane. Now let using Proposition 4.4 and the fact that R − (z(k)), R + (z(k)) are of completely regular growth in the upper half plane.
is not holomorphic in the closed upper half plane, it is holomorphic in the closed upper half plane except at the points k = ±(V − − V + ) 1/2 . It is continuous at these points (in fact, a stronger statement can be made), and that is enough to guarantee that we may apply Theorem 2.1 to obtain Theorem 1.1.
Proof. We give the proof for R − . Using (17) and (18), for large |k|, k ∈ R we can write Therefore, Since g 3 (·, z(k)) = O(|k| −2 ) for k ∈ R, we need only concern ourselves with the contributions of g 1 and g 2 .
We have We will use the fact that r − (z(k)) = k + O(|k| −1 ). Using integration by parts in the first integral and in the first term of the second one, we obtain The first term is O(|k| −2 ) using the fact that r − (z(k)) = k + O(|k| −1 ). If h ∈ L 1 (R), then the Fourier transformĥ satisfiesĥ(k) = o(1). Therefore, the two integrals in (28) contribute terms which are o(|k| −1 ), Note that g 2 (·, z(k)) = O(|k| −1 ) and g 2 has compact support, so that Similarly, g 2 (x, k)−(pR β pe −ik· )(x, z(k)) L 2 (Rx) = O(|k| −2 ) so that we may further simplify our calculations. Moreover, using the explicit formula for the resolvent in terms of φ ±,β , To show that (29) is o(|k| −1 ), we can integrate by parts in x ′′ when x ′′ < x ′ and in x ′ when x ′′ > x ′ , thus finishing the proof of the lemma.
We shall use the previous lemma, Theorem 2.1, Proposition 4.1, and Proposition 4.4 to prove Theorem 1.2.
Proof of Theorem 1.2. If R − (k) has no poles in Im k > 0 we shall apply Theorem 2.1 to the function Moreover, by Proposition 4.4, h F (ϕ) = 2(b 1 − β) sin ϕ. Then we may apply Theorem 2.1 to find that the number of zeros of F (k) in the upper half plane with norm less than r is given by 2(b 1 − β)(π) −1 r + o(r). Then the first part of the theorem follows by applying Proposition 4.1.
If R − has any poles in the upper half-plane, it has only finitely many, and these may be handled as in the proof of Theorem 1.1. The second part of the theorem follows in an analogous way, using the estimates on R + .
We now give the proof of the last of our principle forward results, Theorem 1.3.
Proof of Theorem 1.3. By translation, we may assume that the convex hull of the support of V ′ is [−b 1 , b 1 ].
As before, use the coordinate k = r + (z) to describe points in the closure of the physical sheet ofẐ.
Note that F (k) depends on the values of R ± , T − only on the closure of the physical sheet, although F is We remark that T − (z(k)) is nonzero in this region, and that if T − has a pole, then R ± has a pole of the same order at the same place. The function R − (z(k))/T − (z(k)) is holomorphic on {k : Im k ≥ 0 and k ∈ [−ǫ − V − − V + , V − − V + + ǫ]}, any ǫ > 0. A similar statement holds for R + (z(−k))/T − (z(−k)) for k in the closed lower half-plane. The relationships (5) and (8) for any ǫ > 0. From Propositions 4.3 and 4.4, we know that h F (ϕ) = 2b 1 sin(|ϕ|). Thus applying Theorem 2.2, we obtain, using the notation of that theorem, It now remains to relate the zeros of F to the poles of the resolvent in the desired region. The function T − is nonzero, except, perhaps, at points which project to V + . By Proposition 4.1 and (6), the zeros of F in the upper half plane correspond, with multiplicity, to poles of the resolvent on the sheet {z ∈Ẑ : Im r + < 0, Im r − > 0}. Similarly, using (11), the zeros of F in the lower half-plane correspond, with multiplicity, to poles of the resolvent on the sheet {z ∈Ẑ : Im r + > 0, Im r − < 0}. By Proposition 4.2, the zeros of F (k) with k ∈ R coincide with poles at z ∈Ẑ with Π(z) ∈ R, Π(z) > V − , and r + (z) and r − (z) having opposite signs.
Finally, we include a proposition which we shall need for our inverse results. Proposition 4.5. We have Proof. The proof follows from a application of Carleman's Theorem, using the variable k = r + (z), to, in turn, R − , R + , and R − R + − T − T + in the physical plane. We also use Propositions 4.2, 4.3, and 4.4.

Inverse Results
In this section we prove our inverse results. We must recover the reflection and transmission coefficients, R ± , T ± on the boundary of the physical space. Because V ′ is compactly supported, this information is enough to recover the eigenvalues and the norming constants. Then, using results of [4], we recover the potential.
We will work with functions meromorphic in the plane whose zeros and poles are determined by R. We can recover these functions, up to a finite number of unknown constants, by applying the Weierstrass factorization theorem. The difficulty is to find enough such functions to be able to recover R ± and T ± .
Proof. Note that R − (z)R − (ω − (z)) is a meromorphic function of r + (z). The poles of R − (z) are the same as the elements of R, except that R − (z) is always regular at Π −1 (V + ). The zeros of R − (z) are those z ′ ∈Ẑ such that ω + (z ′ ) ∈ R, Π(z ′ ) = V + . Therefore, using Proposition 4.4 and the Weierstrass Factorization Theorem, where γ 1 and δ 1 are constants to be determined. The product converges by Proposition 4.5. But using (5), (6), (8) and (9). When z lies on the physical sheet, using Propositions 4.3 and 4.4, this is a function of completely regular growth of type determined by the length of the convex hull of the support of V ′ . By Theorem 1.1, this is determined by the resonances, so that the resonances determine δ 1 .
A similar argument determines R + (z)R + (ω + (z)) as a meromorphic function of r − (z). Next we consider T − (z)T − (ω − (z)) and T − (z)T − (ω + (z)), which are meromorphic functions of r + (z) and r − (z), respectively. Note that T − (z) is nonzero, except, perhaps, if z ∈ Π −1 (V + ). If z ′ ∈ Π −1 (V + ), then T − has a zero of order 1 at z ′ if z ′ ∈ R and is nonzero at (11) ensures that at most one of the points which projects to V + can be a resonance.) Again, the product converges by Proposition 4.5.
To see that δ 2 is determined by R, note that Taking the reciprocal, we obtain a function which is, when z is in the physical space, a function of completely regular growth as a function of r + (z). Just as for R − (z)R − (ω − (z)), the type is determined by R, and thus δ 2 is determined. Since T − (ω + (z))T − (ω +− (z)) is the complex conjugate of T − (z)T − (ω − (z)) when V + < Π(z) < V − , γ 2 is determined up to a real multiple.
A similar argument shows that T − (z)T − (ω + (z)) is determined by R, except for a real multiple. Now is therefore determined up to a constant, real, multiple. But, by Proposition 4.3, T − (z)/T − (ω +− (z)) → 1 when z lies on the boundary of the physical sheet and |z| → ∞. This, then, completely determines The next lemma makes an assumption, that T − (z)T − (ω − (z)) is determined, that we have yet to prove. Proving this assumption is the sticking point in proving the inverse result in general. We are able to prove it in some special cases, and this is what enables us to prove the specialized inverse results.
Proof. Using (7), For the remainder of this proof, we assume that z lies on the boundary of the physical space, with Π(z) > V − . Then T − (z) = T − (ω +− (z)), R − (z) = R − (ω +− (z)). Then, away from the zeros of R − (ω +− (z)), T − (z)T − (ω − (z)) determines the argument of R − (ω +− (z)), modulo 2π. Using (9) and (5), where f (z) ≥ 0 is known. Thus , it is clear that we must take the "+" sign since R − (z) → 0 as |z| → ∞ (recalling that z lies on the boundary of the physical sheet, and using (9) and Proposition 4.3). On the other hand, if R − (z) does have such a zero, it is easy to see that we must take the "+" sign in the choice of ρ to get ρ = 0 at a pole of f . Thus, for z on the boundary of the physical space with Π(z) ≥ V − , we know the argument (modulo 2π) and norm of R − , and thus know R − there. Knowing R − on an interval uniquely determines it on all of Z.
Lemma 5.5. Suppose V is a real-valued steplike function, V + and V − are known, and it is known a priori that V = p + V β for some (unknown) β ∈ R and (unknown) p ∈ C 0 comp (R) with p ′ ∈ L 1 (R). Then R determines T − (z)T − (ω + (z)).
We remark that by Lemma 5.1 and (5), T + (z 0 )/T + (ω − (z 0 )) is determined by R. This ratio must be real valued because for z with V + ≤ Π(z) ≤ V − , T + (z) = T + (ω + (z)) and ω + (z 0 ) = z 0 . Checking this ratio in the simple case of a potential which takes on only three (distinct) values shows that this ratio can have either sign (see e.g. [17, Appendix 1]).
We summarize our inverse results in the following theorem.
Theorem 5.1. Knowledge of V + , V − , and R uniquely determines, up to translation, a real steplike potential V with V ′ compactly supported provided at least one of the following criteria is met: • Π −1 (V + ) = ∅ • It is known a priori that V = V β + p for some (unknown) β ∈ R and (unknown) p ∈ C 0 comp (R) with p ′ ∈ L 1 (R).