TV source certificates and level-set convergence

Every $L^2$ function is integrable on compact subsets by Cauchy–Schwarz, so its distributional derivatives are defined. For $i=1,2$, set $L_i(\psi ):=-{\int }_{\mathrm{\Omega }}u{\mathrm{\partial }}_i\psi dx$ on $C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{R})$. If $\psi \mathrm{\ne }0$, the vector field $\phi =\psi e_i\mathrm{/}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}$ and its negative are admissible in [F1]; hence $\mathrm{\mid }{L}_i(\psi )\mathrm{\mid }\le M\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}$. The same bound is immediate for $\psi =0$. By [F3], $L_i$ extends uniquely to a bounded real linear functional on $C_0(\mathrm{\Omega })$.

Countable Choice and [F4] represent each extension by a finite signed Radon measure ${\mu }_i$. Formula [F2] identifies ${\mu }_i$ with $D_{i}u$, so $\mu =({\mu }_1,{\mu }_2)$ has finite vector Radon variation by [F5]. For every continuous compactly supported $g$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, approximate uniformly by a smooth compactly supported $h$ with $\mathrm{\parallel }h-g{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$, and replace $h$ by $h\mathrm{/}\max \{1,\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}\}$. The replacement has norm at most one and differs from $g$ by at most $2\epsilon$. Since ${\sum }_i\mathrm{\mid }{\mu }_i\mathrm{\mid }(\mathrm{\Omega })<\mathrm{\infty }$, its integral against $\mu$ approaches that of $g$ as $\epsilon \downarrow 0$. Thus the supremum over smooth unit fields equals the continuous-field supremum in [F5]. By [F2], $\int u\mathrm{div}\phi =-\int \phi \cdot d\mu$; symmetry of the unit test class removes this minus sign. Consequently $\mathrm{\mid }Du\mathrm{\mid }(\mathrm{\Omega })=M=T{V}_2(u;\mathrm{\Omega })$, proving the assertion.

The zero function has $T{V}_2(0;\mathrm{\Omega })=0$, so the functional is proper; its defining test class is symmetric under $\phi \mapsto -\phi$, hence its supremum is nonnegative. For $\lambda >0$, linearity of every test integral gives $T{V}_2(\lambda u;\mathrm{\Omega })=\lambda T{V}_2(u;\mathrm{\Omega })$, and the equality also holds for $\lambda =0$. For $\theta \in [0,1]$, every admissible field gives $\int (\theta u+(1-\theta )v)\mathrm{div}\phi \le \theta T{V}_2(u;\mathrm{\Omega })+(1-\theta )T{V}_2(v;\mathrm{\Omega })$; taking the supremum proves convexity, with the extended-real cases automatic.

By [F3], every fixed test integral is a continuous linear functional on $L^2$. For each real $a$, the sublevel set $\{u:T{V}_2(u;\mathrm{\Omega })\le a\}$ is the intersection, over all admissible $\phi$, of the closed half-spaces $\{u:\int u\mathrm{div}\phi \le a\}$; hence it is closed and $T{V}_2$ is lower semicontinuous. If $u\in BV(\mathrm{\Omega })\cap L^2(\mathrm{\Omega })$, [F1] and [F2] give $T{V}_2(u;\mathrm{\Omega })=TV(u;\mathrm{\Omega })$.

First suppose $G\subset {\mathbb{R}}^2$ is a bounded open set with compact smooth boundary $\mathrm{\Gamma }$. The boundary is a compact smooth one-dimensional manifold without boundary. Connected coordinate arcs cover $\mathrm{\Gamma }$; compactness gives a finite subcover, so $\mathrm{\Gamma }$ has finitely many connected components. By [F5] each component is a smooth embedded circle. Parametrize component $i$ by an orientation-compatible unit-speed smooth periodic curve ${\gamma }_i:[0,L_i]\to {\mathbb{R}}^2$, where $L_i$ is its length. The divergence formula in [F3] gives $D{\chi }_G=-\nu {\mathcal{H}}^1{\mathrm{\mid }}_{\mathrm{\Gamma }}$ with $\nu$ the outward normal, and [F4] therefore gives $P(G;{\mathbb{R}}^2)={\mathcal{H}}^1(\mathrm{\Gamma })={\sum }_i{L}_i$. Applying Stokes in [F3] to $\omega =(xdy-ydx)\mathrm{/}2$, whose exterior derivative is $dx\wedge dy$, gives $\mathrm{\mid }G\mathrm{\mid }=\frac{1}{2}{\sum }_i{\int }_0^{L_i}({\gamma }_{i,1}{\gamma }_{i,2}^{\mathrm{\prime }}-{\gamma }_{i,2}{\gamma }_{i,1}^{\mathrm{\prime }})ds\mathrm{.}$ For each $i$, translating ${\gamma }_i$ by its arclength mean ${\bar{\gamma }}_i=L_i^{-1}{\int }_0^{L_i}{\gamma }_i(s)ds$ does not change this integral, since each coordinate derivative integrates to zero over a period. Write $z_i={\gamma }_i-{\bar{\gamma }}_i$.

Now let $E$ be as in the Statement and put $u={\chi }_E$. By [F1], choose $u_j\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$ with $e_j:=\mathrm{\parallel }{u}_j-{\chi }_E{\mathrm{\parallel }}_1\to 0$ and $a_j:=\int \mathrm{\mid }\mathrm{\nabla }{u}_j\mathrm{\mid }\to P(E;{\mathbb{R}}^2)$. If $\mathrm{\mid }E\mathrm{\mid }=0$, the result is immediate. If some $e_j=0$, continuity forces $u_j$ to take only the values $0$ and $1$ everywhere; connectedness of ${\mathbb{R}}^2$ and compact support then force $u_j=0$, contradicting $\mathrm{\mid }E\mathrm{\mid }>0$. Thus $e_j>0$. Let ${\tau }_j=\sqrt{e_j}$ and discard finitely many terms so $0<{\tau }_j<\min \{1\mathrm{/}2,\mathrm{\mid }E\mathrm{\mid }\mathrm{/}2\}$. For every $t\in I_j:=[{\tau }_j,1-{\tau }_j]$, a point where ${\chi }_E$ and ${\chi }_{\{u_j>t\}}$ differ contributes at least ${\tau }_j$ to $\mathrm{\mid }{u}_j-{\chi }_E\mathrm{\mid }$; hence $\mathrm{\mid }\{u_j>t\}\mathrm{△}E\mathrm{\mid }\le e_j\mathrm{/}{\tau }_j={\tau }_j$.

For a smooth $L$-periodic scalar function $z$ of mean zero, set $v(\theta )=z(L\theta \mathrm{/}(2\pi ))$. Its zeroth Fourier coefficient vanishes. Integration by parts gives $\widehat{v^{\mathrm{\prime }}}(k)=ik\widehat{v}(k)$ for every integer $k$. Parseval [F6] and $k^2\ge 1$ for $k\mathrm{\ne }0$ imply ${\int }_0^{2\pi }\mathrm{\mid }v{\mathrm{\mid }}^{2}d\theta \le {\int }_0^{2\pi }\mathrm{\mid }{v}^{\mathrm{\prime }}{\mathrm{\mid }}^{2}d\theta$. Changing variables yields ${\int }_0^L\mathrm{\mid }z(s){\mathrm{\mid }}^{2}ds\le {\left(\frac{L}{2\pi }\right)}^2{\int }_0^L\mathrm{\mid }{z}^{\mathrm{\prime }}(s){\mathrm{\mid }}^{2}ds\mathrm{.}$ Apply this to both coordinates of $z_i$ and add. Since $\mathrm{\mid }{z}_i^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }{\gamma }_i^{\mathrm{\prime }}\mathrm{\mid }=1$, the Cauchy–Schwarz inequality gives $\mid \frac{1}{2}{\int }_0^{L_i}(z_{i,1}{z}_{i,2}^{\mathrm{\prime }}-z_{i,2}{z}_{i,1}^{\mathrm{\prime }})ds\mid \le \frac{1}{2}\mathrm{\parallel }{z}_i{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }{z}_i^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}\le \frac{L_i^2}{4\pi }\mathrm{.}$ Summing over the finitely many boundary components and using ${\sum }_i{L}_i^2\le ({\sum }_i{L}_i{)}^2$ proves $4\pi \mathrm{\mid }G\mathrm{\mid }\le P(G;{\mathbb{R}}^2{)}^2$.

By [F2], the measurable level-perimeter function has integral at most $a_j$ on $I_j$, whose length is $1-2{\tau }_j$. For any $\epsilon >0$, the set of $t\in I_j$ with $P(\{u_j>t\};{\mathbb{R}}^2)\le a_j\mathrm{/}(1-2{\tau }_j)+\epsilon$ has positive measure; otherwise its integral over $I_j$ would exceed $a_j$. The critical values of $u_j$ form a null set by [F2], so take $\epsilon =1\mathrm{/}(j+1)$ and choose such a regular value $t_j$. Since $t_j>0$ and $u_j$ is compactly supported, $G_j:=\{u_j>t_j\}$ is bounded and its closure is a compact smooth domain with boundary $\{u_j=t_j\}$. Thus step 2.1 applies to $G_j$, and $\mathrm{\mid }{G}_j\mathrm{△}E\mathrm{\mid }\le {\tau }_j,P(G_j;{\mathbb{R}}^2)\le \frac{a_j}{1-2{\tau }_j}+\frac{1}{j+1}\mathrm{.}$ The right side has limit $P(E;{\mathbb{R}}^2)$, while $\mathrm{\mid }{G}_j\mathrm{\mid }\to \mathrm{\mid }E\mathrm{\mid }$.

Step 2.1 gives $4\pi \mathrm{\mid }{G}_j\mathrm{\mid }\le P(G_j;{\mathbb{R}}^2{)}^2$ for every selected $j$. Taking upper limits, the convergence $\mathrm{\mid }{G}_j\mathrm{\mid }\to \mathrm{\mid }E\mathrm{\mid }$ and the perimeter upper bound from step 3.1 give $4\pi \mathrm{\mid }E\mathrm{\mid }\le P(E;{\mathbb{R}}^2{)}^2$. This proves the claimed inequality for every finite-measure finite-perimeter set.

By [F2], choose $\chi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$ with $0\le \chi \le 1$, $\chi =1$ on $B(0,1)$, $\mathrm{supp}\chi \subset B(0,2)$, and let ${\chi }_R(x)=\chi (x\mathrm{/}R)$ for integers $R\ge 1$. This proves existence of a family with ${\sup }_{R}R\mathrm{\parallel }\mathrm{\nabla }{\chi }_R{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$. Now fix any family satisfying all four conditions in the statement, write $C={\sup }_{R}R\mathrm{\parallel }\mathrm{\nabla }{\chi }_R{\mathrm{\parallel }}_{\mathrm{\infty }}$, and put $v_R={\chi }_{R}u$. Then $v_R\in L^2$ has compact support, hence is in $L^1$, and $v_R\to u$ in $L^2$ because ${\chi }_R=1$ on $B(0,R)$ and $u\in L^2$.

For every scalar $\psi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$, local integrability of $u$ and the definition of distributional derivative give $-\int {\chi }_{R}u{\mathrm{\partial }}_i\psi =-\int u{\mathrm{\partial }}_i({\chi }_R\psi )+\int u\psi {\mathrm{\partial }}_i{\chi }_R=\int {\chi }_R\psi d{D}_{i}u+\int u\psi {\mathrm{\partial }}_i{\chi }_R\mathrm{.}$ Thus $D{v}_R={\chi }_{R}Du+u\mathrm{\nabla }{\chi }_{R}dx$. The variation of the first term is at most $\mathrm{\mid }Du\mathrm{\mid }({\mathbb{R}}^2)$, and the second has variation at most $\frac{C}{R}{\int }_{B(0,2R)\setminus B(0,R)}\mathrm{\mid }u\mathrm{\mid }dx\le C^{\mathrm{\prime }}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2({\mathbb{R}}^2\setminus B(0,R))},$ by Cauchy–Schwarz and the annulus area bound $\mathrm{\mid }B(0,2R)\setminus B(0,R){\mathrm{\mid }}^{1\mathrm{/}2}\le C^{\mathrm{\prime }\mathrm{\prime }}R$. This error tends to zero. Since $v_R\in L^1$ and its derivative is a finite vector Radon measure, $v_R\in BV({\mathbb{R}}^2)$, and [F1] yields ${\mathrm{lim\hspace{0.17em}sup}}_{R}TV(v_R;{\mathbb{R}}^2)\le \mathrm{\mid }Du\mathrm{\mid }({\mathbb{R}}^2)=T{V}_2(u;{\mathbb{R}}^2)$. The estimate applies to the arbitrary controlled family fixed in step 1.1 and proves the statement.

Put $q^{†}=A^{\ast }{p}^{†}$. Applying [F2] at $u^{†}$ with $z=0$ and $z=2u^{†}$, and using $R(0)=0$ and $R(2u^{†})=2R(u^{†})$, gives $⟨q^{†},u^{†}{⟩}_X=R(u^{†})$. For arbitrary $z\in X$, [F2] then yields $R(z)\ge ⟨q^{†},z{⟩}_X$, so $q^{†}\in \mathrm{\partial }R(0)$ and $p^{†}\in K$. Every $p\in K$ satisfies $⟨f,p{⟩}_Y=⟨u^{†},A^{\ast }p{⟩}_X\le R(u^{†})$; equality holds for $p^{†}$. Thus $P_0$ is nonempty, closed, and convex.

The set $K$ is nonempty because $0\in \mathrm{\partial }R(0)$, convex by the subgradient inequality, and closed because $\mathrm{\partial }R(0)=\{q:R(z)\ge ⟨q,z{⟩}_X\text{ for all }z\}$ is an intersection of closed half-spaces and $A^{\ast }$ is continuous. For any nonempty closed convex subset $C$ of a Hilbert space, let $d={\inf }_{c\in C}\mathrm{\parallel }x-c\mathrm{\parallel }$ and use Countable Choice to select $c_n\in C$ with $\mathrm{\parallel }x-c_n{\mathrm{\parallel }}^2\le d^2+1\mathrm{/}n$. The midpoint belongs to $C$, so the parallelogram identity gives $\mathrm{\parallel }{c}_n-c_m{\mathrm{\parallel }}^2\le 2\mathrm{/}n+2\mathrm{/}m$; thus $(c_n)$ is Cauchy. Completeness and closedness yield a nearest point, and strict convexity of squared norm makes it unique. Applying this to $K$ and $P_0$ defines their projections.

Completing the square gives $D_{\alpha ,w}(p)=\frac{\mathrm{\parallel }f+w{\mathrm{\parallel }}_Y^2}{2\alpha }-\frac{\alpha }{2}\mathrm{\parallel }p-(f+w)\mathrm{/}\alpha {\mathrm{\parallel }}_Y^2$. Hence its unique maximizer on $K$ is $p_{\alpha ,w}={\mathrm{Proj}}_K((f+w)\mathrm{/}\alpha )$.

Let $u_{\alpha ,w}$ be a supplied global minimizer and put $p=(f+w-A{u}_{\alpha ,w})\mathrm{/}\alpha$. For any $h\in X$ and $t>0$, objective minimality and expansion of the square give $[R(u_{\alpha ,w}+th)-R(u_{\alpha ,w})]\mathrm{/}t\ge ⟨A^{\ast }p,h{⟩}_X-t\mathrm{\parallel }Ah{\mathrm{\parallel }}_Y^2\mathrm{/}(2\alpha )$. As $t\downarrow 0$, the lower limit of these secants is at least $⟨A^{\ast }p,h{⟩}_X$; convexity makes the unit secant at least every smaller positive secant, so $R(u_{\alpha ,w}+h)-R(u_{\alpha ,w})\ge ⟨A^{\ast }p,h{⟩}_X$. This for every $h$ proves $A^{\ast }p\in \mathrm{\partial }R(u_{\alpha ,w})$. Applying [F2] with $z=0$ and $z=2u_{\alpha ,w}$ gives $⟨A^{\ast }p,u_{\alpha ,w}{⟩}_X=R(u_{\alpha ,w})$, and then [F2] for arbitrary $z$ gives $A^{\ast }p\in \mathrm{\partial }R(0)$, so $p\in K$. By [F4] at $0$, $R^{\ast }(A^{\ast }q)=0$ for every $q\in K$; Fenchel–Young in [F4] therefore gives $⟨A^{\ast }q,u_{\alpha ,w}{⟩}_X\le R(u_{\alpha ,w})$. Thus $D_{\alpha ,w}(q)\le R(u_{\alpha ,w})+\alpha ⟨p,q{⟩}_Y-\alpha \mathrm{\parallel }q{\mathrm{\parallel }}_Y^2\mathrm{/}2=D_{\alpha ,w}(p)-\alpha \mathrm{\parallel }q-p{\mathrm{\parallel }}_Y^2\mathrm{/}2$. Hence $p=p_{\alpha ,w}$. Finally $A{u}_{\alpha ,w}=f+w-\alpha p$, so $\frac{1}{2}\mathrm{\parallel }A{u}_{\alpha ,w}-(f+w){\mathrm{\parallel }}_Y^2+\alpha R(u_{\alpha ,w})=\alpha D_{\alpha ,w}(p)$, establishing the normalized value identity in the Statement.

If $P_{K}x=p$ and $P_{K}y=q$, minimizing squared distance along the segment from $p$ to any $z\in K$ gives $⟨x-p,z-p⟩\le 0$. Apply this with $z=q$, and the corresponding inequality for $P_{K}y$ with $z=p$. Adding yields $\mathrm{\parallel }p-q{\mathrm{\parallel }}_Y^2\le ⟨x-y,p-q{⟩}_Y\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_Y\mathrm{\parallel }p-q{\mathrm{\parallel }}_Y$ by [F7], so $\mathrm{\parallel }p-q{\mathrm{\parallel }}_Y\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_Y$. Taking $x=(f+w)\mathrm{/}\alpha$ and $y=f\mathrm{/}\alpha$ proves the first noise-stability bound; applying $A^{\ast }$ proves the second.

Write $p_n=p_{{\alpha }_n,0}$. Maximality against $p_0\in P_0$ and the support bound $⟨f,p_n{⟩}_Y\le R(u^{†})$ give $\mathrm{\parallel }{p}_n{\mathrm{\parallel }}_Y\le \mathrm{\parallel }{p}_0{\mathrm{\parallel }}_Y$ and $⟨f,p_n{⟩}_Y\to R(u^{†})$. The closed span of any subsequence is separable; Gram–Schmidt on rational finite linear combinations of its terms gives a finite or countable orthonormal family with dense span. Repeated use of [F8] and [A1] selects a further subsequence whose coordinates all converge. For each finite initial segment, expanding the nonnegative squared norm after orthogonal projection shows that the sum of squared limiting coordinates is at most $\mathrm{\parallel }{p}_0{\mathrm{\parallel }}_Y^2$; hence the coordinates define a vector $p$ in that closed span. Approximation by finite linear combinations proves weak convergence there, and [F9] represents each functional $x\mapsto ⟨x,y{⟩}_Y$ restricted to the span, so convergence is weak in $Y$. The limit lies in $K$: by [F2], $K={\bigcap }_{z\in X}\{q:⟨q,Az{⟩}_Y\le R(z)\}$, an intersection of weakly closed half-spaces. It also attains $⟨f,p{⟩}_Y=R(u^{†})$, so $p\in P_0$. Weak lower semicontinuity of the norm and the bound force $\mathrm{\parallel }p{\mathrm{\parallel }}_Y=\mathrm{\parallel }{p}_0{\mathrm{\parallel }}_Y$, hence uniqueness gives $p=p_0$. Every subsequence therefore has a further subsequence weakly converging to $p_0$; testing against any fixed $y\in Y$ shows that every scalar subsequence has a further subsequence converging to $⟨p_0,y{⟩}_Y$, so $⟨p_n,y{⟩}_Y\to ⟨p_0,y{⟩}_Y$. Thus $p_n\rightharpoonup p_0$. Weak lower semicontinuity and the uniform upper bound then give $\mathrm{\parallel }{p}_n{\mathrm{\parallel }}_Y\to \mathrm{\parallel }{p}_0{\mathrm{\parallel }}_Y$. The norm identity [F10] yields strong convergence, and boundedness of $A^{\ast }$ yields strong convergence of $A^{\ast }{p}_n$.

For $u\in \mathrm{dom}R$, [F1] and [F7] give $-⟨A^{\ast }{p}^{†},u-u^{†}{⟩}_X=⟨p^{†},f-Au{⟩}_Y\le \mathrm{\parallel }{p}^{†}{\mathrm{\parallel }}_Y\mathrm{\parallel }Au-f{\mathrm{\parallel }}_Y$. Therefore the VSC [F5] holds with $\beta =0$, $p=2$, and $\psi (s)=C\sqrt{s}$, $C=1+\mathrm{\parallel }{p}^{†}{\mathrm{\parallel }}_Y$, on every regularizer sublevel; [F3] identifies its Bregman term with the distance in the Statement. Its conjugate is $H(q)=C^2\mathrm{/}(4q)$ for $q>0$. For an exact minimizer with $\mathrm{\parallel }w{\mathrm{\parallel }}_Y\le \delta$, comparison to $u^{†}$ gives $R(u_{\alpha ,w})\le R(u^{†})+{\delta }^2\mathrm{/}(2\alpha )$; if $\alpha \asymp \delta$, this is below any fixed $R_0>R(u^{†})$ for small $\delta$. With $c_p=1\mathrm{/}4$, ${\rho }_{\delta }=H(c_p\mathrm{/}\alpha )=C^2\alpha \asymp \delta$ and $H(c_p\mathrm{/}(2\alpha ))=2{\rho }_{\delta }$, so all parameter-coupling hypotheses in [F6] hold and the stated rates follow.