A unit contrast on a small cube lies in the Born regime

Given: Countable Choice, $D=[0,\mathrm{\ell }{]}^3$ with $\mathrm{\ell }>0$ and ${\mathrm{\ell }}^4=\pi \mathrm{/}(2\sqrt{3})$, $k=1$, and $q=1_D$.

The small-norm domain is ${\mathcal{U}}_{D,1}=\{q:{\rho }_1(q)<1\}$ (Fixed-frequency far-field data on the small-norm volume branch).

The volume operator satisfies ${\rho }_1(q)\le \mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\sqrt{\mathrm{\mid }D\mathrm{\mid }\mathrm{diam}(D)\mathrm{/}(4\pi )}$ (The volume integral operator is well defined and bounded on L2).

For $q\in {\mathcal{U}}_{D,1}$, the uniform Born remainder is at most $\frac{M}{4\pi }\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\frac{{\rho }_1(q)}{1-{\rho }_1(q)}$ and, when $C_D\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}<1$, at most $\frac{M{C}_D}{4\pi }\frac{\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}^2}{1-C_D\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}}$ (The first Born data are the Fréchet linearization at zero contrast).

The Lebesgue measure of the closed cube $[0,\mathrm{\ell }{]}^3$ is ${\mathrm{\ell }}^3$ (A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included).

The Euclidean norm in ${\mathbb{R}}^3$ is $\sqrt{x_1^2+x_2^2+x_3^2}$ (The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$).

The unique nonnegative real $\mathrm{\ell }$ satisfying ${\mathrm{\ell }}^4=\pi \mathrm{/}(2\sqrt{3})$ exists, and it is positive because the right-hand side is positive (Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$, Pi as twice the smallest positive zero of cosine).

Countable Choice is assumed in the L2 operator and Neumann-series suppliers (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)).