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H1 logarithmic stability from centered three-dimensional NtD data

Statement

Assume the Axiom of Choice, Countable Choice and Dependent Choice. Fix 5/2<s<7/2, 0<ρ<1, 0<m<1<M and A>0. Let σj=1+aj be real conductivities on R3, with aj supported in Bρ‾, ∥aj∥Hs≤A, and m≤σj≤M, for j=1,2. Let Nj be their boundary-mean-zero NtD maps on S2=∂B1, and let δ=(∑ℓ≥1∑k=12ℓ+1∥(N1−N2)Yℓk∥L2(S2)2)1/2. Then there is C>0 depending only on s,ρ,m,M,A such that ∥σ1−σ2∥H1(R3)≤C[log⁡(3+δ−1)]−(s−3/2)(δ>0). If δ=0, the conductivities are equal. The zero-noise modulus is therefore defined to be zero. Restricting the left side to B1 gives the same H1 norm because the difference is supported in Bρ‾.

Facts & Assumptions

Given: The choice assumptions and fixed collared Sobolev conductivity class in the Statement.

[F1]

With qj=(Δσj)/σj and Q=q1−q2, one has ∥Q∥Hs−2≤C, and ∥σ1−σ2∥H1≤C∥Q∥H−1. Collared Sobolev conductivities have regular Liouville potentials and H1 recovery

[F2]

The exact centered NtD/CGO identity has boundary term at most CδeCt. For each ∣ξ∣≤t, its remaining Fourier term is the bilinear pairing of e−iξ⋅xQ with a cutoff remainder of H1/2 norm at most C/t, uniformly in phase direction. Centered NtD data control the exact CGO identity with its modulation factor

[F3]

For every b≥1, ∫∣ξ∣≤b⟨ξ⟩−1∥e−iξ⋅xQ∥H−1/22dξ≤Cb2∥Q∥H−1/22. Sharp weighted absorption of modulated CGO remainders

Proof

1.1A1F1givenalgebra

Put N=∥Q∥H−1/2 and κ=s−3/2>0. For b≥1, the high-frequency contribution satisfies ∫∣ξ∣>b⟨ξ⟩−1∣Q^(ξ)∣2dξ≤Cb−2κ∥Q∥Hs−22≤Cb−2κ. Indeed the ratio of the two Fourier weights is ⟨ξ⟩−1−2(s−2)=⟨ξ⟩−2κ.

2.1F2F3step 1.1algebra

For t≥max⁡{T,b}, [F2] gives the pointwise inequality ∣Q^(ξ)∣≤CδeCt+(C/t)∥e−iξ⋅xQ∥H−1/2 for ∣ξ∣≤b. The right side is measurable by the nonnegative Fourier integral in [F3]; thus this inequality can be integrated without selecting a measurable family of CGO phases. Square it using (u+v)2≤2u2+2v2, multiply by ⟨ξ⟩−1 and integrate. Since ∫∣ξ∣≤b⟨ξ⟩−1dξ≤Cb2, [F3] yields a low-frequency bound Cb2δ2eCt+Cb2t−2N2, after increasing the exponential constant. Combining with step 1.1 gives N2≤Cb−2κ+Cb2δ2eCt+Cb2t−2N2. This uses the integrated modulated estimate, not an unmodulated pointwise Sobolev pairing.

3.1step 2.1choosealgebra

Choose a fixed K≥max⁡{T,1} large enough that C/K2≤1/2 for the last constant in step 2.1, and set t=Kb. Absorb that term into the left side and take square roots. Increasing fixed constants and using b≤eb for b≥1 gives N≤Cb−κ+CδeDb for some D>0 depending only on the fixed class. If δ=0, let b→∞ to obtain N=0. If 0<δ<e−2D, take b=log⁡(1/δ)/(2D)≥1. Then δeDb=δ, which is bounded by a constant times [log⁡(1/δ)]−κ because e−x/2xκ is bounded for x≥1. Hence N≤C[log⁡(1/δ)]−κ for all sufficiently small positive δ.

4.1F1step 3.1algebra∎

By [F1] and the Fourier weight inequality, ∥σ1−σ2∥H1≤C∥Q∥H−1≤CN. This proves the claimed small-noise rate and, if δ=0, equality of the conductivities. For larger positive δ, the uniform bound ∥a1−a2∥H1≤2A and the positive lower bound for [log⁡(3+δ−1)]−κ on that region prove the same displayed estimate after increasing C. On a sufficiently small noise interval, log⁡(3+δ−1) is comparable to log⁡(1/δ), so the two small-noise formulations agree. All constants came from the fixed class.

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