OPTICAL TOMOGRAPHY IN TWO DIMENSIONS

PLAMEN STEFANOV AND GUNTHER UHLMANN

Abstract. We consider in two dimensions, the inverse boundary problem of reconstructing the absorption and scattering coefficient of an inhomogeneous medium by probing it with diffuse light. The problem is modeled as an inverse boundary problem for the stationary linear Boltzmann equation. The information is encoded in the albedo operator. We show that we can recover the absorption and the scattering kernel from this information provided that the latter is small in an appropriate topology. We also give stability estimates.

Optical Tomography concerns with the determination of spatially-varying optical absorption and scattering properties of a medium by measuring the response of the medium to transmitted near-infrared light. This has been proposed recently as a possible diagnostic tool in medicine (see [A] for a review). A probabilistic approach to optical tomography is the so-called diffused tomography [SGKZ]. Other areas of applications are atmospheric remote sensing [Bi] , nuclear physics (see [MC] for a review), etc. We describe in more detail the mathematical problem below.

Let X ⊂ R^{2} be an open convex set with smooth boundary and consider the 2D stationary transport equation

(1)  −v(θ) · ∇_{x}f − σ(x, θ)f + ∫_{S^{1}} k(x, θ′, θ)f(x, θ′) dθ′ = 0,

where x ∈ X, θ, θ′ ∈ S^{1} and throughout this paper we will use the notation

v(θ) = (cos θ, sin θ).

Here f(x, θ) represents the density of a particle flow, σ ≥ 0 is the absorption coefficient, while k(x, θ′, θ) ≥ 0 is the collision kernel. We assume that σ and k are in L^{∞}. Denote Γ_{±} = {(x, v) ∈ ∂X × S^{1}; ±n(x) · v > 0}, where n(x) is the outer normal to ∂X at x ∈ ∂X. Let us denote by f the solution (if exists) to (1) satisfying the boundary condition

(2)  f|_{Γ_{−}} = f_{−},

where f_{−} ∈ C_{0}^{∞}(Γ_{−}). The albedo operator A mapping the incoming flux f_{−} on Γ_{−} to the outgoing one on Γ_{+} is defined as follows

A : f_{−} ⟼ f|_{Γ_{+}}.

In this paper we work with k sufficiently small and under this assumption (1), (2) is always uniquely solvable with f ∈ L^{∞}(X × S^{1}) and Af ∈ L^{∞}(Γ_{+}).

Research at MSRI is supported in part by NSF grants DMS-0196440 (P.S.) and DMS-0070488 and a John Simon Guggenheim fellowship (G.U.).

The inverse problem we are interested in, is the problem of recovering (σ, k) from knowledge of the albedo operator A. First, if k = 0, it is obvious that A does not determine uniquely σ since in this case all we get from the knowledge of A is ∫ σ(x + t(cos θ, sin θ), θ) dt for any θ ∈ S^{1} and any x and this is not enough to recover σ. For this reason in our main theorems we assume that σ depends on x only. In dimensions n ≥ 3 this problem is formally overdetermined for the recovery of k and uniqueness under the assumption above is proved in [CS1], [CS2] by studying the singularities of the distributional kernel of A (see also [Bo]). Some of the earlier works where this problem is treated under more restrictive assumptions are [AB], [CZ], [L1], [L2], [MC], [PV]. Stability estimates under various assumptions are proved in [R2],

[R3] (see also the references therein), [W].

In two dimensions, this inverse problem is formally determined for the recovery of k. Uniqueness and stability are proved for small k in the case when k = k(x, cos(θ − θ′)) in [R1], and in the case k = k(θ, θ′) uniqueness for small k is proved in [T]. Note that in those cases the inverse problem is still formally overdetermined. The purpose of this paper is to prove a uniqueness results for general (small) k(x, θ′, θ) and a stability estimate.

The first main result in this paper is the following.

Theorem 1. Define the class

(3)  U_{Σ,ε} = {(σ(x), k(x, θ′, θ)) ; ∥σ∥_{L^{∞}} ≤ Σ, ∥k∥_{L^{∞}} ≤ ε} .

Then, for any Σ > 0 there exists ε > 0 such that a pair (σ, k) ∈ U_{Σ,ε} is uniquely determined by its albedo operator A in the class U_{Σ,ε}.

The proof of Theorem 1 follows by a careful analysis of the singularities of the distribution kernel α(x, θ, x′, θ′) of A (see also [CS1], [CS2] for similar analysis in dimensions n ≥ 3). In the two dimensional case we show that A admits the singular decomposition (see Lemma 1 and Proposition 1)

(4)  α = \frac{δ(θ − θ′)δ_{\{x′+τ_{+}(x′,θ′)v(θ′)\}}(x)}{n(x) · v(θ)}e^{−a(x′,θ′)} + b(x, θ, x′, θ′),

where n(x) is the exterior normal to ∂X, the function τ_{+}(x′, θ′) is the “exit time” defined in next section, and

(5)  a(x′, θ′) = ∫_{0}^{τ_{+}(x′,θ′)} σ(x′ + tv(θ′), θ′) dt, sin(θ − θ′)b ∈ L^{∞}.

In particular, knowing A, we can uniquely determine a and b. Let we have two pairs (σ, k) and (σ̃, k̃) with albedo operators A and Ã, respectively. Set

(6)  δ_{1} = ∥a − ã∥_{H^{1}(Γ_{−})}, δ_{2} = ∥(b − b̃) sin(θ − θ′)∥_{L^{∞}(Γ_{−}×Γ_{+})}.

Our second result is the following stability estimate.

Theorem 2. Let

V_{Σ,ε}^{s} = {(σ(x), k(x, θ′, θ)) ∈ H^{s}(X) × C(X × S^{1} × S^{1}); ∥σ∥_{H^{s}} ≤ Σ, ∥k∥_{L^{∞}} ≤ ε} .

Then, for any s > 1, Σ > 0, there exists ε > 0 such that for any (σ, k) ∈ V_{Σ,ε}^{s} and (σ̃, k̃) ∈ V_{Σ,ε}^{s} and 0 < μ < 1 − 1/s, there exists C > 0 such that

∥σ − σ̃∥_{L^{∞}} ≤ Cδ_{1}^{1−1/s−μ},
∥k − k̃∥_{L^{∞}} ≤ C (δ_{1}^{1−1/s−μ} + δ_{2}) .

1. The uniqueness, proof of Theorem 1

Recall that σ and k are L^{∞} in all variables. It is convenient to think later that σ and k are extended as 0 for x ∉ X. Denote the left-hand side of (1) by T f. Denote

T_{1}f = −v(θ) · ∇_{x}f − σf, Kf = ∫_{S^{1}} k(x, θ′, θ)f(x, θ′)dθ′,

so that T = T_{1} + K. First, we reduce the boundary value problem

(7)  \{ \begin{matrix}T f & = & 0 in X × S^{1}, \\ f|_{Γ_{−}} & = & f_{−} ∈ C_{0}^{∞}(Γ_{+}) \end{matrix}

to an integral equation by a standard argument. For (x, θ) ∈ X̄×S^{1}, define τ_{±}(x, θ) ≥ 0 by the condition (x ± τ_{±}(x, θ)v(θ), θ) ∈ Γ_{±}. Introduce the operator M by

(8)  Mf = ∫_{0}^{∞} e^{− ∫_{0}^{t} σ(x−sv(θ),θ)ds}(Kf)(x − tv(θ), θ) dt.

If we denote the solution to T_{1}u = f, u|_{Γ_{−}} = 0 by

T_{1}^{−1}f = − ∫_{0}^{τ_{−}(x,θ)} e^{− ∫_{0}^{t} σ(x−sv(θ),θ)ds}f(x − tv(θ), θ) dt,

then M = −T_{1}^{−1}K. Problem (7) can be written as (T_{1} + K)f = 0, f|_{Γ_{−}} = f_{−}. Applying T_{1}^{−1} to the first equation, we get that f solves

(9)  (I − M)f = Jf_{−},

where

Jf_{−} = e^{− ∫_{0}^{τ_{−}(x,v)} σ(x−sv,v)ds}f_{−}(x − τ_{−}(x, v)v, v).

Note that Jf_{−} solves the problem T_{1}Jf_{−} = 0, Jf_{−}|_{Γ_{−}} = f_{−}. Then f is given by

(10)  f = (I − M)^{−1}Jf_{−},

provided that I − M is invertible in a suitable space. The definition (8) of M implies immediately that

∥Mf∥_{L^{∞}(X×S^{1})} ≤ C∥f∥_{L^{∞}(X×S^{1})},

where C = diam(X)∥k∥_{L^{∞}}. Therefore, if (σ, k) ∈ U_{Σ,ε} and ε < 1/diam(X), then I − M is invertible in L^{∞}(X × S^{1}), and then the solution f to (7) is given by (10). By using Neumann series, it is not hard to see that the trace (I − M)^{−1}f|_{Γ_{+}} is well defined in L^{∞}(Γ_{+}) for any f ∈ L^{∞}(X × S^{1}). This proves in particular that A maps C_{0}^{∞}(Γ_{−}) into L^{∞}(Γ_{+}) under the smallness assumption on k above. The same arguments also show that Af_{−} can be defined for any f_{−} ∈ L^{∞}(Γ_{−}) as well but we will not need this since we work with the distribution kernel of A. The mapping properties of A in L^{1} spaces are discussed in [CS2].

We proceed with the study of the distribution kernel of A. Define first the fundamental solution ϕ(x, θ, x′, θ′) of the boundary value problem (7) as follows. For (x′, v′) ∈ Γ_{−}, let ϕ(x, v, x′, v′) solve

(11)  \{ \begin{matrix}T ϕ & = & 0 in X × S^{1}, \\ ϕ|_{Γ_{−}} & = & |n(x′) · v(θ′)|^{−1}δ_{x′}(x)δ(θ − θ′), \end{matrix}

where δ_{x′}(x) ∈ D′(∂X) is defined by ∫_{∂X} δ_{x′}(x)φ(x) dμ(x) = φ(x′), with dμ(x) the induced measure on ∂X. Then the albedo operator A has distribution kernel α(x, θ, x′, θ′) = ϕ(x, θ, x′, θ′)|_{(x,θ)∈Γ_{+}}, (x′, θ′) ∈ Γ_{−}, (x, θ) ∈ Γ_{+}, i.e.,

(Af_{−})(x, v) = ∫_{Γ_{−}} α(x, θ, x′, θ′)f_{−}(x′, θ′) dξ(x′, θ′), ∀f_{−} ∈ C_{0}^{∞}(Γ_{+}),

where dξ(x′, θ′) = |n(x′)·v(θ′)|dξ(x′, θ′) dμ(x′) dθ′ and n(x) is the unit exterior normal to ∂X.

As in [CS2], we construct a singular expansion of ϕ as follows. Note that for any f_{−} ∈ C_{0}^{∞}(Γ_{−}), the outgoing flux Af_{−} is given by

Af_{−} = Jf_{−}|_{Γ_{+}} + MJf_{−}|_{Γ_{+}} + (I − M)^{−1}M^{2}Jf_{−}|_{Γ_{+}}.

Let

E(x, θ, t) = e^{∓ ∫_{0}^{t} σ(x+sv(θ),θ)ds}, ±t ≥ 0

be the total absorption along the path [x, x + tv(θ)]. Then

ϕ = ϕ_{0} + ϕ_{1} + ϕ_{2},

where

(12)  ϕ_{0} = Jϕ_{−}, ϕ_{1} = MJϕ_{−}, ϕ_{2} = (I − M)^{−1}M^{2}Jϕ_{−},

and ϕ_{−} = |n(x′) · v(θ′)|^{−1}δ_{x′}(x)δ(θ − θ′) as in (11). As in [CS2],

ϕ_{0} = E(x, θ, −∞)δ(θ − θ′) ∫_{0}^{τ_{+}(x′,v(θ′))} δ(x − x′ − tv(θ′))dt.

Next,

(13)  ϕ_{1} = χE(y, θ′, −∞)\frac{k(y, θ′, θ)}{| sin(θ − θ′)|}E(y, θ, ∞),

where y = y(x′, θ′, x, v) is the point of intersection of the rays (0, ∞) ∋ s ↦ x′+sv(θ′) and (−∞, 0) ∋ t ↦ x+tv(θ) and χ = χ(x, θ, x′, θ′) equals 1, if those two rays intersect in X̄, otherwise χ = 0. Recall that X is convex.

To estimate ϕ_{2} we need the following.

Lemma 1. Let (σ, k) and (σ, k̃) be in L^{∞}. Let K, M and K̃, M̃ be related to k and k̃ (not necessarily non-negative), respectively. Then there exists C > 0 depending on diam(X) only such that

|(M̃Mϕ_{0})(x, θ, x′, θ′)| ≤ C∥k̃∥_{L^{∞}}∥k∥_{L^{∞}} (1 + log \frac{1}{sin |θ − θ′|})

almost everywhere on X × S^{1} × Γ_{−}, and also almost everywhere on Γ_{+} × Γ_{−}.

Proof. Note first that for any f,

|Mf| ≤ ∫_{0}^{∞} ∫_{S^{1}} |k(x − tv(θ), θ′, θ)f(x − tv(θ), θ′)| dθ′ dt,

therefore

|(K̃Mf)(x, θ)| ≤ ∫_{S^{1}} ∫_{0}^{∞}∫_{S^{1}} |k̃(x, θ′′, θ)k(x − tv(θ′′), θ′θ′′)f(x − tv(θ′′), θ′)| dθ′ dt dθ′′
= ∫ ∫_{S^{1}} \frac{|k̃(x, arg(x − y), θ)k(y, θ′, arg(x − y))|}{|x − y|}|f(y, θ′)| dθ′ dy.

Here arg(z) is the polar angle of z ∈ R^{2}. Setting f = ϕ_{0}, we get

|(K̃Mϕ_{0})(x, θ, x′, θ′)| ≤ ∫_{y∈l(x′,θ′)} \frac{|k̃(x, arg(x − y), θ)k(y, θ′, arg(x − y))|}{|x − y|}dl(y),

where l(x′, θ′) is the line through x′ parallel to θ′ and dl is the Euclidean measure on it. The following estimate

∫_{−A}^{A} \frac{ds}{\sqrt{ν^{2} + s^{2}}} ≤ 2 (1 + log \frac{A}{ν}) , 0 < ν ≤ A.

is easy to prove by splitting the interval of integration into two parts: |s| ≤ 1 and 1 ≤ |s| ≤ A. Therefore,

(14)  |(K̃Mϕ_{0})(x, θ, x′, θ′)| ≤ 2∥k̃∥_{L^{∞}}∥k∥_{L^{∞}} (1 + log \frac{d}{dist\{x, l(x′, θ′)\}}) ,

where d = diam(X). Applying T_{1}^{−1}, we get

|(M̃Mϕ_{0})(x, θ, x′, θ′)| ≤ 2∥k̃∥_{L^{∞}}∥k∥_{L^{∞}} ∫_{0}^{d} (1 + log \frac{d}{dist\{x − tv(θ), l(x′, θ′)\}}) dt
≤ C∥k̃∥_{L^{∞}}∥k∥_{L^{∞}} (1 + log \frac{1}{| sin(θ − θ′)|}) ,

with C = C(d). This proves the lemma. □

For ϕ_{2} we have therefore ϕ_{2} = (I − M)^{−1}ϕ_{2}^{#} with ϕ_{2}^{#} = M^{2}ϕ_{0} and by Lemma 1,

0 ≤ ϕ_{2}^{#}(x′, θ′, x, θ) ≤ C∥k∥_{L^{∞}}^{2} (1 + log \frac{1}{| sin(θ − θ′)|}) .

To estimate ϕ_{2}, write ϕ_{2} = ϕ_{2}^{#} + (I − K)^{−1}Kϕ_{2}^{#}. Since the singularity in the r.h.s. of (15) is integrable, we get that Kϕ_{2}^{#} ∈ L^{∞}, therefore ∥ϕ_{2} − ϕ_{2}^{#}∥_{L^{∞}} ≤ C∥k∥_{L^{∞}}^{3}, with some constant C > 0, and in particular, ϕ_{2} − ϕ_{2}^{#} ∈ L^{∞}. From now on, we will denote by C different positive constants. Therefore, ϕ_{2} also satisfies (15). Observe that 0 ≤ ϕ_{2} because (I − K)^{−1} preserves the non-negative functions.

We have therefore proved the following.

Proposition 1. For ε > 0 small enough, the fundamental solution ϕ of (7) defined by (11) admits the representation ϕ = ϕ_{0} + ϕ_{1} + ϕ_{2} with

ϕ_{0} = E(x, θ, −∞)δ(θ − θ′) ∫_{0}^{τ_{+}(x′,v(θ′))} δ(x − x′ − tv(θ′))dt.
ϕ_{1} = χE(y, θ′, −∞)\frac{k(y, θ′, θ)}{| sin(θ − θ′)|}E(y, θ, ∞),
0 ≤ ϕ_{2} ≤ C∥k∥_{L^{∞}}^{2} (1 + log \frac{1}{| sin(θ − θ′)|}) ,

where y, χ are as in (13) and C = C(diam(X)).

Fix Σ > 0 and assume that we have two pairs (σ, k), (σ̃, k̃) in U_{Σ,ϵ} with the same albedo operator and σ, σ̃ depending on x only. Denote by ϕ_{j} and ϕ̃_{j}, j = 0, 1, 2, the corresponding components of the fundamental solutions ϕ and ϕ̃. Then

(15)  ϕ_{0} + ϕ_{1} + ϕ_{2} = ϕ̃_{0} + ϕ̃_{1} + ϕ̃_{2} for (x, θ) ∈ Γ_{+}.

As in [CS1], the most singular terms must agree, therefore,

(16)  ϕ_{0}(x′, θ′, x, θ) = ϕ̃_{0}(x′, θ′, x, θ) for (x′, θ′) ∈ Γ_{−}, (x, θ) ∈ Γ_{+}.

Thus the X-ray transform of σ(x) and σ̃(x) coincide, therefore,

σ(x) = σ̃(x).

Next,

χE(y, θ′, −∞)\frac{k(y, θ′, θ) − k̃(y, θ′, θ)}{| sin(θ − θ′)|}E(y, θ, ∞) = ϕ̃_{2} − ϕ_{2} on Γ_{−} × Γ_{+},

where y, χ are as in (13). This, together with (15), (16), leads to the inequality (17) χ|k(y, θ′, θ)−k̃(y, θ′, θ)| ≤ C| sin(θ−θ′)||ϕ̃_{1}−ϕ_{1}| = C| sin(θ−θ′)||ϕ̃_{2}−ϕ_{2}| on Γ_{−} × Γ_{+},

where C = e^{2dΣ}, d = diam(D). Our goal is to show that

(18)  ess sup_{Γ_{−}×Γ_{+}} | sin(θ − θ′)||ϕ̃_{2} − ϕ_{2}| ≤ Cε∥k − k̃∥_{L^{∞}(X×S^{1}×S^{1})}

with C > 0 depending on diam(D) only. Then by (17), (18),

∥k − k̃∥_{L^{∞}(X×S^{1}×S^{1})} ≤ Cε∥k − k̃∥_{L^{∞}(X×S^{1}×S^{1})},

and for ε > 0 small enough this implies k = k̃.

It remains to prove (18). By (12),

ϕ_{2} − ϕ̃_{2} = (I − M)^{−1}ϕ_{2}^{#} − (I − M̃)^{−1}ϕ̃_{2}^{#}
= (I − M)^{−1}(M^{2}ϕ_{0} − M̃^{2}ϕ_{0}) + (I − M̃)^{−1}(M − M̃)(I − M)^{−1}ϕ̃_{2}^{#}
(19)  = I_{1} + I_{2}.

Writing M^{2} − M̃^{2} = M(M − M̃) + (M − M̃)M̃, the first term I_{1} can be estimated by Lemma 1 and the paragraph after (15), thus

(20)  |I_{1}| ≤ Cε∥k − k̃∥_{L^{∞}} (1 + log \frac{1}{| sin(θ − θ′)|}) .

To estimate I_{2}, write

(21)  |I_{2}| ≤ C∥k − k̃∥_{L^{∞}} ∫_{S^{1}} ∥(I − M)^{−1}ϕ̃_{2}∥_{L^{∞}(X)} dθ ≤ Cε^{2}∥k − k̃∥_{L^{∞}}.

By (19), (20), (21),

|ϕ_{2} − ϕ̃_{2}| ≤ Cε∥k − k̃∥_{L^{∞}} (1 + log \frac{1}{| sin(θ − θ′)|}) .

for a.e. (x, θ, x′, θ′) ∈ X × S^{1} × Γ_{−} and for a.e. (x, θ, x′, θ′) ∈ Γ_{+} × Γ_{−}. The last statement follows from the fact that M = T_{1}^{−1}K and from the smoothing properties of T_{1}^{−1}. Since sin |θ − θ′| log(1/ sin |θ − θ′|) is bounded, this proves (18).

This completes the proof of Theorem 1. □

Remark. It follows from (17) and (18) that we can choose ε = C(d)e^{−2dΣ}, where d = diam(D).

2. The stability estimate, proof of Theorem 2

According to Proposition 1, for the distribution kernel α(x, θ, x′, θ′) of A we have the representation (4), where a is as in (5), while b = (ϕ_{1} + ϕ_{2})|_{(x,θ)∈Γ_{+}}. Then b is a function and an elementary calculation show that b ∈ L^{∞}(Γ_{+}, L^{1}(Γ_{−})). Proposition 1 shows that b sin(θ−θ′) ∈ L^{∞}. Assume that we have two pairs of continuous functions (σ, k) and (σ̃, k̃) in V_{Σ,ε}^{s} with albedo operators A and Ã, respectively. In what follows we will denote the quantities a, b, α, etc., related to the second pair by putting a tilde sign over it. Also, we will use the notation ΔA to denote the difference ΔA = A−Ã, and similarly Δa = a − ã, etc. In this paper, Δ never stands for the Laplacian.

Let δ_{1} and δ_{2} be as in (6). According to Proposition 1, δ_{2} < ∞. Our goal is to estimate Δσ and Δk in terms of δ_{1} and δ_{2}. Observe that, as in the uniqueness proof, Δe^{−a} and Δ(α_{1} +α_{2}) can be recovered from Δα by separating the most singular part of Δα from the rest. Therefore, δ_{1} measures the magnitude of the singular part of Δα, while δ_{2} measures the magnitude of the regular part.

We start with estimating Δσ. By a result of Mukhometov [M],

(22)  ∥Δσ∥_{L^{2}} ≤ Cδ_{1}.

Next we estimate Δk in terms of δ_{1} and δ_{2}. Set E_{1}(y, θ, θ′) = E(y, θ′, −∞)E(y, θ, ∞). Then by (13), sin |θ − θ′|α_{1}(x, θ, x′, θ′) = χ(E_{1}k)(y, θ′, θ) with y, χ as in (13). Our starting point is the relation Δ(E_{1}k) = kΔE_{1} + Ẽ_{1}Δk. By the inequality |e^{−s_{1}} − e^{−s_{2}}| ≤ |s_{1} − s_{2}| for 0 ≤ s_{i}, i = 1, 2, we get that |ΔE_{1}| ≤ 2d|Δσ| ≤ 2dδ′_{1}, where

δ′_{1} = ∥Δσ∥_{L^{∞}}.

Hence, on supp χ,

Ẽ_{1}|Δk|(y, θ, θ′) ≤ |Δ(E_{1}k)| + k|ΔE_{1}|
≤ |Δ(α_{1} + α_{2}) sin |θ − θ′| − Δα_{2} sin |θ − θ′|| + Cδ′_{1}
(23)  ≤ δ_{2} + |Δα_{2}| sin |θ − θ′| + Cδ′_{1}.

Therefore,

(24)  ∥Δk∥_{L^{∞}} ≤ C (δ′_{1} + δ_{2} + ∥Δα_{2} sin |θ − θ′|∥_{L^{∞}})

We want to prove an estimate similar to (18) on the last term in the r.h.s. above. Recall that in (18), we have σ = σ̃, while here we only know how to estimate the difference Δσ = σ − σ̃. Nevertheless, we can proceed along similar lines. Denote by γ the trace on Γ_{+}. Clearly, γ : C(X̄ × S^{1}) → C(Γ_{+}) is bounded with norm 1. Note also that since we assume here that σ and k are continuous, (I − M)^{−1} exists in C(X × S^{1}) for ε ≪ 1. Similarly to (18),

Δα_{2} = γ(I − M)^{−1}ϕ_{2}^{#} − γ(I − M̃)^{−1}ϕ̃_{2}^{#}
= γ(I − M)^{−1}(M^{2}ϕ_{0} − M̃^{2}ϕ̃_{0}) + γ(I − M̃)^{−1}(M − M̃)(I − M)^{−1}ϕ̃_{2}^{#}
(25)  = I_{1} + I_{2}.

To estimate I_{1}, we write M^{2}ϕ_{0}−M̃^{2}ϕ̃_{0} = (M^{2}−M̃^{2})ϕ_{0}+M̃^{2}Δϕ_{0}. In view of (19), (20), the contribution to I_{1} of the first term in the r.h.s above can be estimated similarly to (20) with ∥Δk∥_{L^{∞}} there replaced by ∥Δk∥_{L^{∞}}+∥Δσ∥_{L^{∞}}. Next, about the contribution of M̃^{2}Δϕ_{0} to I_{1}, the first formula in Proposition 1 easily implies that |M̃^{2}Δϕ_{0}| = O(δ′_{1})/(1 − log sin |θ − θ′|), hence |γ(I − M)^{−1}M̃^{2}Δϕ_{0}| = O(δ′_{1})/(1 − log sin |θ − θ′|). This shows that we have the following analogue of (20)

(26)  |I_{1}| ≤ Cε (∥Δk∥_{L^{∞}} + δ′_{1}) (1 + log \frac{1}{| sin(θ − θ′)|}) .

In a similar way, we obtain the version of (21) in this case

(27)  |I_{2}| ≤ Cε^{2} (∥Δk∥_{L^{∞}} + δ′_{1}) ,

Combining (25), (26) and (27), we get, similarly to (18),

(28)  |Δα_{2} sin |θ − θ′|∥_{L^{∞}} ≤ Cε (∥Δk∥_{L^{∞}} + δ′_{1})

Therefore, for ε > 0 small enough, (24) implies the estimate

(29)  ∥Δk∥_{L^{∞}} ≤ C (δ′_{1} + δ_{2}) .

Estimate (29) is the base of our stability estimate. We are going now to use the interpolation inequality

∥f∥_{α_{1}s_{1}+α_{2}s_{2}} ≤ ∥f∥_{s_{1}}^{α_{1}}∥f∥_{s_{2}}^{α_{2}}, α_{1} + α_{2} = 1, α_{j} ≥ 0,

where, in our case, ∥ · ∥_{s} will be the norm in the Sobolev space H^{s}(X). Let σ and σ̃ belong to V_{Σ,ε}^{s}. Then

∥Δσ∥_{1+sμ} ≤ C(Σ)∥Δσ∥_{L^{2}}^{1−1/s−μ}

for any fixed 0 ≤ μ ≤ 1 − 1/s. By a standard Sobolev embedding theorem and (22),

(30)  δ′_{1} = ∥Δσ∥_{L^{∞}} ≤ C(Σ)∥Δσ∥_{L^{2}}^{1/s+μ} ≤ C′(Σ)δ_{1}^{1−1/s−μ}.

Therefore, (29) yields

(31)  ∥Δk∥_{L^{∞}} ≤ C (δ_{1}^{1−1/s−μ} + δ_{2}) .

Estimates (30) and (31) complete the proof of Theorem 2. □

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Plamen Stefanov, Department of Mathematics, Purdue University, West Lafayette, IN 47906, USA

  E-mail address: stefanov@math.purdue.edu

Gunther Uhlmann, Department of Mathematics, University of Washington, Seattle, WA 98195

  E-mail address: gunther@math.washington.edu