Uniform density and Hausdorff convergence of TV level-set boundaries

The TV functional is proper, convex, lower semicontinuous and positively one-homogeneous by [F12]. On a bounded domain $L^2\subset L^1$, so the relative functional agrees with that test-field functional by [F8]; the Dirichlet version is its composition with the linear isometric zero-extension map into full-space $L^2$. Thus the hypotheses of [F1] hold. Comparison of each objective with its finite value at zero also gives finite TV at a supplied minimizer. Put $z_n=v_{{\alpha }_n,0}=A^{\ast }{p}_{{\alpha }_n,0}$. By [F1], $z_n\to v_0=A^{\ast }{p}_0$ strongly in $L^2(\mathrm{\Omega })$ and $v_n=z_n+e_n$ with $\mathrm{\parallel }{e}_n{\mathrm{\parallel }}_2\le \eta$. Thus the exact-field sequence, its limit and the noisy fields have precisely the coefficient decomposition required by [F10]. In particular their uniform small-measure moduli and, in full space, exact-field tail moduli follow from strong convergence, as proved there, without an assumed uniform integrability of the noisy perturbations.

For each supplied minimizer, [F1] and positive one-homogeneity imply $R(u_n)=⟨v_n,u_n⟩$ and $v_n\in \mathrm{\partial }R(0)$. The selected TV is even, since its test-field class is symmetric and zero extension is linear. Applying its inequality to both ${\chi }_F$ and $-{\chi }_F$ gives $P_{\ast }(F)\ge \sigma {\int }_F{v}_n$ for either sign $\sigma$, for every finite-area finite-selected-perimeter competitor; $P_{\ast }$ denotes the selected branch perimeter throughout this proof. Indeed these indicators belong to $L^2$ and their selected TV is exactly $P_{\ast }$ by [F8] and the boundary conventions. In full space every nonzero level has finite area by $\mathrm{\mid }\{\mathrm{\mid }{u}_n\mathrm{\mid }>\mathrm{\mid }t\mathrm{\mid }\}\mathrm{\mid }\le \mathrm{\parallel }{u}_n{\mathrm{\parallel }}_2^2\mathrm{/}{t}^2$. Bounded-domain levels have finite area automatically. Coarea [F4] gives finite selected perimeter for almost every nonzero level. Relative-perimeter Neumann levels also have finite ambient perimeter by [F11]. Signed layer cake and Cauchy–Schwarz give an absolutely integrable level pairing: the integral over levels of ${\int }_{E_n(t)}\mathrm{\mid }{v}_n\mathrm{\mid }$ is $\int \mathrm{\mid }{u}_n\mathrm{\mid }\mathrm{\mid }{v}_n\mathrm{\mid }\le \mathrm{\parallel }{u}_n{\mathrm{\parallel }}_2\mathrm{\parallel }{v}_n{\mathrm{\parallel }}_2$. Hence [F11] and coarea show that the integral of the nonnegative gaps $P_{\ast }(E_n(t))-\mathrm{sgn}(t){\int }_{E_n(t)}{v}_n$ equals $R(u_n)-⟨v_n,u_n⟩=0$. Each gap therefore vanishes almost everywhere. Their value zero is the global minimum of $P_{\ast }(F)-\mathrm{sgn}(t){\int }_F{v}_n$, since every competitor has nonnegative value. This proves the stated equality and even global branch minimization, which in particular implies compact-perturbation variational curvature [F5]. Taking the countable union of exceptional level sets makes this true simultaneously for all $n$.

The threshold and actual obstacle assumptions are exactly those of [F10]. Apply it to every calibrated set from step 1.2, with either level sign. It supplies one $c,r_0>0$ for both ambient density inequalities at every point of each canonical boundary, independently of the index and the nonexceptional level. In the Dirichlet branch the actual obstacle extension uses perimeter submodularity from [F10]; in the Neumann branch its proof uses relative cuts and the specified $C_{\mathrm{\Omega }}$, and at points on $\mathrm{\partial }\mathrm{\Omega }$ it uses the uniform Lipschitz exterior area for the ambient complement. Thus no false relative-complement estimate is imposed on the valid case $E=\mathrm{\Omega }$. In full space [F10] compares $E$ with actual radial cuts and proves finite-radius extinction of $\mathrm{\mid }E\setminus B(0,r)\mathrm{\mid }$, giving one common ball containing the density-one sets and their topological boundaries. Bounded-domain representatives and boundaries lie in $\bar{\mathrm{\Omega }}$. These are the full density and support conclusions, rather than assumptions for the transfer below.

The limiting field calibrates every minimum-TV exact solution $u_0$ under consideration. More explicitly, the dual supplier [F1] defines $p_0\in \{p:A^{\ast }p\in \mathrm{\partial }R(0),⟨f,p⟩=R(u^{†})\}$. For any such exact minimum $A{u}_0=f$ and $R(u_0)=R(u^{†})$, one has $⟨v_0,u_0⟩=⟨p_0,f⟩=R(u_0)$, while $v_0\in \mathrm{\partial }R(0)$. Its subgradient inequality is therefore also the inequality at $u_0$. Repeat step 1.2 with $u_0,v_0$; its almost-everywhere level sets are calibrated global minimizers. Adjoin the limiting exact field with perturbation zero in [F10]. The very same coefficient moduli and noise threshold used in step 2.1 then give the same uniform density and support conclusions for this limit. This includes any selected minimum-TV exact limit carrying the inherited certificate; it does not identify that solution with $u^{†}$ without uniqueness.

Fix a nonzero nonexceptional level for which the corresponding sets converge in measure to such a canonical limit $E_0$. If $E_0=\mathrm{\varnothing }$ and infinitely many $E_n$ are nonempty, their bounded common support makes each boundary nonempty: a boundaryless subset of connected ${\mathbb{R}}^2$ is clopen, and the only bounded clopen set is empty. Choose a boundary point and one fixed $0<r<r_0$; the density estimate gives $\mathrm{\mid }{E}_n\mathrm{△}{E}_0\mathrm{\mid }\ge c\pi r^2$, a contradiction. If $E_0$ is nonempty, it has positive measure by its density-one definition, so convergence rules out infinitely many empty $E_n$. The empty status therefore agrees eventually; two empty boundaries have Hausdorff distance zero. In the nonempty case, if $x_n\in \mathrm{\partial }{E}_n$ stays at distance at least $\delta >0$ from $\mathrm{\partial }{E}_0$, common compact support and [F9] give a subsequence $x_n\to x$. For $r<\min (r_0,\delta \mathrm{/}4)$ the connected ball $B(x_n,r)$ misses $\mathrm{\partial }{E}_0$ and is entirely in $E_0$ or entirely outside it. The two-sided density of $E_n$ forces symmetric difference at least $c\pi r^2$, again a contradiction. Conversely, if $y_n\in \mathrm{\partial }{E}_0$ stays at distance at least $\delta$ from $\mathrm{\partial }{E}_n$, take a convergent subsequence with limit $y\in \mathrm{\partial }{E}_0$. A fixed smaller connected ball about $y$ misses every corresponding $\mathrm{\partial }{E}_n$, so along a further subsequence its status is always full or always empty. Measure convergence makes $E_0$ full or empty almost everywhere on that ball. By the density-one definition it is then actually full or empty in a smaller ball, contradicting $y\in \mathrm{\partial }{E}_0$. Both directed Hausdorff distances tend to zero, with all point and subsequence selections allowed by [A1].