Two-dimensional TV level-set convergence under a source condition

Apply [F1] to the supplied minimizers, retaining its subsequence, and call its limit $u_0$. Then $A{u}_0=f$, $R(u_0)$ is minimal among exact solutions, and $A^{\ast }{p}^{†}\in \mathrm{\partial }R(u_0)$. In bounded domains the subsequence already converges strongly in $L^1(\mathrm{\Omega })$. In full space, [F5] supplies a common ball containing the regularized and limit level sets for almost every nonzero level. By [F2] and the positive and negative layer-cake identities, for each fixed $n$ the integral of $\mathrm{\mid }{u}_n\mathrm{\mid }$ outside that ball is the integral over levels of the measures of those level sets outside it; Tonelli [F3] makes this integral zero. The same argument applies to $u_0$, so local $L^1$ convergence from [F1] is global $L^1$ convergence.

By [F2], for each real-valued $u$ the positive and negative level sets use $\{u>t\}$ for $t>0$ and $\{u<t\}$ for $t<0$. They satisfy the pointwise layer-cake identity $\mathrm{\mid }u(x)-u_0(x)\mathrm{\mid }={\int }_0^{\mathrm{\infty }}\mathrm{\mid }{\mathbf{1}}_{\{u>t\}}-{\mathbf{1}}_{\{u_0>t\}}\mathrm{\mid }dt+{\int }_{-\mathrm{\infty }}^0\mathrm{\mid }{\mathbf{1}}_{\{u<t\}}-{\mathbf{1}}_{\{u_0<t\}}\mathrm{\mid }dt\mathrm{.}$ Integrating and applying [F3] gives $\mathrm{\parallel }{u}_n-u_0{\mathrm{\parallel }}_1={\int }_0^{\mathrm{\infty }}\mathrm{\mid }{E}_n^{+}(t)\mathrm{△}{E}_0^{+}(t)\mathrm{\mid }dt+{\int }_{-\mathrm{\infty }}^0\mathrm{\mid }{E}_n^{-}(t)\mathrm{△}{E}_0^{-}(t)\mathrm{\mid }dt\mathrm{.}$ The right side tends to zero by step 1.1.

Let $g_n(t)$ denote the symmetric-difference measure of the positive level sets for $t>0$ and of the negative level sets for $t<0$. Step 2.1 says $g_n\to 0$ in $L^1(\mathbb{R}\setminus \{0\})$. Choose a further subsequence with ${\sum }_k\mathrm{\parallel }{g}_{n_k}{\mathrm{\parallel }}_1<\mathrm{\infty }$; Tonelli implies ${\sum }_k{g}_{n_k}(t)<\mathrm{\infty }$ for almost every nonzero $t$, hence the level sets converge in measure there.

Fix a nonzero level for which step 3.1 gives measure convergence. If the limit level set is empty and infinitely many regularized level sets are nonempty, [F5] gives compact common support, so each such set has a boundary point; choose one for each index using [A1]. Its two-sided density estimate at a fixed radius $r<r_0$ forces symmetric-difference measure at least $c\pi r^2$, contradicting convergence to the empty limit. If the limit set is nonempty, it has positive measure because it is a nonempty density-one representative; measure convergence therefore rules out infinitely many empty regularized sets. Hence empty status agrees eventually, and the empty boundaries have Hausdorff distance zero by [F2]. For the remaining nonempty sets, [F5] supplies the two-sided density and support bounds. If boundary points of $E_{n_k}(t)$ stayed a fixed distance from $\mathrm{\partial }{E}_0(t)$, choose them using [A1]; bounded support and [F4] give a convergent subsequence. A ball around its limit lies on one side of $E_0(t)$, while the two-sided density bound forces a fixed positive symmetric-difference measure, contradicting step 3.1. Conversely, if a boundary point of $E_0(t)$ stayed a fixed distance from every $\mathrm{\partial }{E}_{n_k}(t)$, choose these points using [A1]; compact support and [F4] give a subsequence converging to $y\in \mathrm{\partial }{E}_0(t)$. A connected ball around $y$ misses those boundaries for all large $k$, so each $E_{n_k}(t)$ is either full or empty there. A subsequence with one fixed status and measure convergence would make $E_0(t)$ full or empty almost everywhere on a smaller ball, contradicting that $y$ lies on the boundary of its canonical density-one representative. Thus the canonical boundaries converge in Hausdorff distance. The limit is the selected solution from [F1]; if the exact minimum-TV solution is unique it equals $u^{†}$, and otherwise no such identification is asserted.