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A compactly supported contrast with volume-operator norm greater than two

Example

Assume Countable Choice. At wavenumber $k = 1$ , let $D = [0, 1 / 4]^{3}$ and let the contrast be the constant $q (y) = 512 π$ on $D$ , extended by zero outside. In the outgoing volume model, the associated operator satisfies $∥ K_{q, 1} ∥_{L^{2} (D) \to L^{2} (D)} > 2$ . Thus this contrast lies outside the small-norm domain used for the Born linearization. This example makes no claim that the exact scattering equation has no solution or is nonunique.

Example references

The small-norm domain is $ρ_{k} (q) = ∥ K_{q, k} ∥ < 1$ .

The outgoing operator uses $G_{k} (z) = e^{i k ∣ z ∣} / (4 π ∣ z ∣)$ .

The L2 operator is well-defined and bounded for bounded contrasts on finite-volume sets.

A vector-valued function with derivative norm at most $L$ is $L$ -Lipschitz.

The operator norm is the supremum over the unit ball.

The constant $π$ is positive.

Facts & Assumptions

Given: Countable Choice, the cube $D = [0, 1 / 4]^{3}$ , $k = 1$ , $q = 512 π 1_{D}$ , and the volume operator $K_{q, 1}$ .

[F1]

The volume model defines $K_{q, 1}$ by the kernel $e^{i ∣ x - y ∣} / (4 π ∣ x - y ∣)$ off the diagonal and uses the zero representative on the diagonal; the operator lemma identifies its L2 value with this a.e. kernel integral (The three-dimensional volume integral scattering model, The volume integral operator is well defined and bounded on L2).

[F2]

The closed box $D$ is Lebesgue measurable and has volume $(1 / 4)^{3} = 1 / 64$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F3]

The Euclidean norm on $R^{3}$ is $\sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}$ , so every $x, y \in D$ satisfy $∣ x - y ∣ \leq \sqrt{3} / 4 < 1 / 2$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F4]

For real $t$ , $∣ \exp (i t) ∣ = 1$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[F5]

For real $r \geq 0$ , the curve $s \mapsto \exp (i r s)$ has derivative $i r \exp (i r s)$ (The complex exponential is entire and its complex derivative is itself); a differentiable curve with derivative norm bounded by $r$ is $r$ -Lipschitz (If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz).

[F6]

The constant function $1_{D}$ is a nonzero vector of $L^{2} (D)$ with norm $\sqrt{∣ D ∣}$ , and the complex $L^{2}$ pairing obeys Cauchy–Schwarz (The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz).

[F7]

The operator norm is $∥ T ∥ = \sup_{∥ x ∥ \leq 1} ∥ T x ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F8]

The number $π$ is positive (Pi as twice the smallest positive zero of cosine).

[F9]

The small-norm domain is $U_{D, k} = {q : ∥ K_{q, k} ∥ < 1}$ (Fixed-frequency far-field data on the small-norm volume branch).

[A1]

Countable Choice is assumed in the volume model and bounded-operator supplier (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Verification

Given: The data above.

1.1givenF2F3

By [F2], $M = ∣ D ∣ = 1 / 64$ . By [F3], every distance in $D$ is at most $\sqrt{3} / 4$ , and opposite corners attain this value, so $d : = diam (D) = \sqrt{3} / 4 < 1 / 2$ . The indicator $1_{D}$ is a nonzero $L^{2}$ class.

2.1F4F5step 1.1algebra

For $0 \leq r \leq d$ , the curve $γ (s) = e^{i r s}$ on $[0, 1]$ has derivative $γ^{'} (s) = i r e^{i r s}$ , whose modulus is $r$ by [F4]. Fact [F5] gives $∣ e^{i r} - 1 ∣ \leq r$ , and hence $Re (e^{i r}) \geq 1 - r \geq 1 - d > 1 / 2$ .

3.1A1F1F8step 1.1step 2.1algebra

For almost every $x \in D$ , the kernel integral in [F1] represents $K_{q, 1} 1_{D} (x)$ . For such an $x$ , almost every $y \in D$ has $r = ∣ x - y ∣$ in $(0, d]$ by step 1.1, and [F1] gives $Re G_{1} (x - y) = \frac{Re (e^{i r})}{4 π r} > \frac{1}{8 π d} .$ The measurable function $y \mapsto Re G_{1} (x - y) - 1 / (8 π d)$ is positive almost everywhere on the positive-measure set $D$ , so some set where it exceeds $1 / n$ has positive measure and its integral is strictly positive. Integrating the pointwise lower bound against $q = 512 π > 0$ therefore yields $Re (K_{q, 1} 1_{D}) (x) > q M / (8 π d) = 1 / d = 4 / \sqrt{3} > 2$ for almost every $x \in D$ ; the last inequality follows by squaring the positive sides of $\sqrt{3} < 2$ .

4.1A1F1F6F7F9step 1.1step 3.1algebra∎

By step 3.1, $g : = ∣ K_{q, 1} 1_{D} ∣^{2} - 4$ is positive almost everywhere on $D$ ; [F1] and [A1] put $K_{q, 1} 1_{D}$ in $L^{2} (D)$ , so $g$ is integrable. Since $M > 0$ by step 1.1, the sets ${x \in D : g (x) > 1 / n}$ cannot all be null, as their countable union contains the positive-measure set where $g > 0$ . Thus for some $n$ , the set ${x \in D : g (x) > 1 / n}$ has positive measure, so $\int_{D} g \geq ∣ {g > 1 / n} ∣ / n > 0$ ; hence $∥ K_{q, 1} 1_{D} ∥_{2}^{2} > 4 M = 4 ∥ 1_{D} ∥_{2}^{2}$ . The unit-ball supremum in [F7] gives $∥ K_{q, 1} ∥ > 2$ . By [F9] the contrast lies outside $U_{D, 1}$ , so the small-norm model branch does not apply; no nonexistence or nonuniqueness conclusion follows.

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Sources

Fioralba Cakoni, David Colton, and Houssem Haddar, Inverse Scattering Theory and Transmission Eigenvalues, CBMS 88 (standard reference, not scraped)
Otmar Scherzer and Leonidas Mindrinos, Inverse Problems: Lecture Notes, University of Vienna, Sommersemester 2015 (standard reference, not scraped)