BV translation modulus for the extended conductivity

Set $u:=\gamma -1$ on $\mathrm{\Omega }$. Since constants have zero distributional derivative, [F1] gives $u\in BV(\mathrm{\Omega })$, $TV(u;\mathrm{\Omega })=TV(\gamma ;\mathrm{\Omega })\le R$, and $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^1(\mathrm{\Omega })}\le \mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }(M+1)$. The smooth bounded domain $\mathrm{\Omega }$ is Lipschitz, so [F2] applies.

By the zero-extension part of [F2], $\tilde{u}\in BV({\mathbb{R}}^2)$ and $TV(\tilde{u};{\mathbb{R}}^2)\le C_{\mathrm{\Omega }}(\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }(M+1)+R)=:V_0\mathrm{.}$ If two representatives differ only on a null set $N$, their values at $x+h$ differ only on $N-h$, which is null by [F3]. Translated classes are therefore well-defined. Choosing ${\gamma }_e=1+\tilde{u}$ pointwise, constants cancel in the difference; the resulting class identity holds for every choice of representatives. Thus $\mathrm{\parallel }{\gamma }_e(\cdot +h)-{\gamma }_e{\mathrm{\parallel }}_1=\mathrm{\parallel }\tilde{u}(\cdot +h)-\tilde{u}{\mathrm{\parallel }}_1\le V_0\mathrm{\mid }h\mathrm{\mid }$. Thus the first inequality holds with $C_1=V_0$.

The bound $m\le \gamma \le M$ inside $\mathrm{\Omega }$ and the background value $1$ outside show $\mathrm{\mid }{\gamma }_e\mathrm{\mid }\le B$ almost everywhere; hence $\mathrm{\mid }{\gamma }_e(\cdot +h)-{\gamma }_e\mathrm{\mid }\le 2B$ almost everywhere. Applying [F3] and step 2.1 yields $\mathrm{\parallel }{\gamma }_e(\cdot +h)-{\gamma }_e{\mathrm{\parallel }}_p^p\le (2B{)}^{p-1}{V}_0\mathrm{\mid }h\mathrm{\mid }\mathrm{.}$ Taking $p$-th roots and then the supremum over $\mathrm{\mid }h\mathrm{\mid }\le t$ gives the second inequality with $C_p=((2B{)}^{p-1}{V}_0{)}^{1\mathrm{/}p}$. The bound is uniform in $\gamma$, so taking the outer supremum proves the claim when $X_2(R)$ is nonempty; when it is empty, the stated convention gives zero. At $t=0$ both sides vanish.