NEAR OPTIMAL BOUNDS IN FREIMAN'S THEOREM

CALDERÓN INVERSE PROBLEM ON RIEMANN SURFACES 113
for some smooth differential operator P (x,θ; x∂x,∂θ) in the vector fields x∂x,∂θ
down to x = 0. Let us define Vb := x −2 V , which is compactly supported, and
H 2m
b := u ∈ L 2 (X, dvolgb); m
gb u ∈ L2 (X, dvolgb) , m ∈ N0.
We also define the following spaces for α ∈ R:
Fα := ker x α H 2 b (gb + Vb).
Since the eigenvalues of S 1 are {j 2 ; j ∈ N0}, the relative index theorem of Melrose
[27, Section 6.2] shows that gb + Vb is Fredholm from xαH 2 b to xαH 0 b if α /∈ Z.
Moreover, from [27, Section 2.2.4], we have that any solution of (gb + Vb)u = 0 in
xαH 2 b has an asymptotic expansion of the form
u ∼
ℓj
j>α,j∈Z ℓ=0
x j (log x) ℓ uj,ℓ(θ) as x → 0
for some sequence (ℓj)j of nonnegative integers and some smooth function uj,ℓ on
S 1 . In particular, it is easy to check that kerL 2 (X,dvolg)(g + V ) = F1+ɛ for ɛ ∈ (0, 1).
THEOREM 6.2
Let (X, g) be an asymptotically Euclidean surface, let V1,V2 be two compactly supported
smooth potentials, and let x be a boundary-defining function. Let ɛ ∈ (0, 1),
and assume that, for any j ∈ Z and any function ψ ∈ ker x j−ɛ H 2 b (g + V1), thereisa
ϕ ∈ ker x j−ɛ H 2 b (g + V2) such that ψ − ϕ = O(x ∞ ), and conversely. Then V1 = V2.
Proof
The idea is to reduce the problem to the compact case. First, we notice that by unique
continuation, ψ = ϕ where V1 = V2 = 0. Now it remains to prove that, if Rη denotes
the restriction of smooth functions on X to {x ≥ η} and if V is a smooth compactly
supported potential in {x ≥ η}, then the set ∞ j=0 Rη(F−j−ɛ) is dense in the set NV
of H 2 ({x ≥ η}) solutions of (g + V )u = 0. The proof is well known for positive
frequency scattering (see, e.g., [28, Lemma 3.2]). Here it is very similar, so we do not
give details. The main argument is to show that it converges in the L2 sense and then
uses elliptic regularity; the L2 convergence can be shown as follows. Let f ∈ NV
such that
∞
fψdvolg = 0, ∀ ψ ∈ F−j−ɛ.
x≥η
Then we want to show that f = 0. By[27, Proposition 5.64], there exists k ∈ N
and a generalized right inverse Gb for Pb = gb + Vb (here, as before, x 2 Vb = V )
j=0