Primal compactness and selection of a minimum-TV limit

If $\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }=0$, the adjoint identity implies $A=0$ and $f=0$. Since $0$ is exact, minimality gives $R(u^{†})=0$. The data term of $J_n$ is then constant, so every global minimizer has $R(u_n)=0$. In the full-space case [F6] makes $u_n$ constant and $L^2({\mathbb{R}}^2)$ forces it to be zero; in the Dirichlet case apply [F6] to its zero extension, which vanishes outside bounded $\mathrm{\Omega }$, and again obtain zero; in the Neumann case [F6] makes it constant and the prescribed zero-mean choice makes it zero. Thus $u_0=0$ is an exact minimum-TV solution. Since $R\ge 0$ and $R(0)=0$, $0\in \mathrm{\partial }R(0)=\mathrm{\partial }R(u_0)$, so $A^{\ast }{p}^{†}=0\in \mathrm{\partial }R(u_0)$ and the assertion holds. For the remainder assume $\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }>0$.

Put $p_n=(f+w_n-A{u}_n)\mathrm{/}{\alpha }_n$. By [F2], $p_n=p_{{\alpha }_n,w_n}$ and $p_{{\alpha }_n,0}$ is bounded. The noise hypothesis gives $\mathrm{\parallel }{w}_n{\mathrm{\parallel }}_Y\mathrm{/}{\alpha }_n\le \eta \mathrm{/}\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }$, hence [F2] makes $(p_n)$ bounded and $\mathrm{\parallel }{w}_n{\mathrm{\parallel }}_Y\to 0$. The identity $A{u}_n=f+w_n-{\alpha }_n{p}_n$ therefore gives $A{u}_n\to f$ strongly in $Y$. Comparing the objective at $u_n$ and $u^{†}$ gives $\frac{1}{2}\mathrm{\parallel }A{u}_n-(f+w_n){\mathrm{\parallel }}_Y^2+{\alpha }_{n}R(u_n)\le \frac{1}{2}\mathrm{\parallel }{w}_n{\mathrm{\parallel }}_Y^2+{\alpha }_{n}R(u^{†}),$ so $R(u_n)\le R(u^{†})+\mathrm{\parallel }{w}_n{\mathrm{\parallel }}_Y^2\mathrm{/}(2{\alpha }_n)\le R(u^{†})+{\eta }^2{\alpha }_n\mathrm{/}(2\mathrm{\parallel }{A}^{\ast }{\mathrm{\parallel }}^2)$. In particular the variations are bounded and ${\mathrm{lim\hspace{0.17em}sup}}_{n}R(u_n)\le R(u^{†})$.

In the Neumann branch on bounded connected $\mathrm{\Omega }$, write $m_n=\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{-1}{\int }_{\mathrm{\Omega }}{u}_n$ and $v_n=u_n-m_n$. The Poincaré inequality [F14] gives $\mathrm{\parallel }{v}_n{\mathrm{\parallel }}_2\le C_{\mathrm{\Omega }}TV(v_n;\mathrm{\Omega })=C_{\mathrm{\Omega }}R(u_n)$, which is uniformly bounded by step 1.2.

If $A{\mathbf{1}}_{\mathrm{\Omega }}=0$, subtracting $m_n$ leaves both $A{u}_n$ and Neumann TV unchanged, so we use the zero-mean minimizer specified in the Statement, which is uniformly bounded in $L^2$ by step 2.1. If $A{\mathbf{1}}_{\mathrm{\Omega }}\mathrm{\ne }0$, then [F1] and step 1.2 give bounded $\mathrm{\parallel }A{u}_n{\mathrm{\parallel }}_Y$ and $\mathrm{\mid }{m}_n\mathrm{\mid }\mathrm{\parallel }A{\mathbf{1}}_{\mathrm{\Omega }}{\mathrm{\parallel }}_Y=\mathrm{\parallel }A{u}_n-A{v}_n{\mathrm{\parallel }}_Y\le \mathrm{\parallel }A{u}_n{\mathrm{\parallel }}_Y+\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }{v}_n{\mathrm{\parallel }}_2,$ so the means, and therefore $\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_2$, are bounded. In the full-space case [F5] gives $\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_2\le R(u_n)$. In the Dirichlet case [F4] applied to ${\tilde{u}}_n$ gives $\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_2\le R(u_n)$. Thus in every branch the selected sequence is bounded in $L^2$.

If $\mathrm{\Omega }$ is bounded, the $L^2$ bound gives an $L^1$ bound by [F11]; the interior variation is bounded by $R(u_n)$ in both boundary conventions [F7]. [F3] therefore supplies a subsequence converging strongly in $L^1(\mathrm{\Omega })$. If $\mathrm{\Omega }={\mathbb{R}}^2$, restrict to each ball $B(0,m)$: the $L^1$ norms are bounded by $\mathrm{\mid }B(0,m){\mathrm{\mid }}^{1\mathrm{/}2}{\sup }_n\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_2$, and [F9] bounds the variation there. Applying [F3] on each ball and taking a countable diagonal subsequence using [A1] gives convergence in $L^1$ on every bounded ball.

The resulting sequence is still bounded in the Hilbert space $L^2(\mathrm{\Omega })$. Its closed linear span is separable, since rational finite linear combinations of its terms are dense. Gram–Schmidt gives a finite or countable orthonormal family with dense span. Each coordinate sequence is bounded by [F11]; repeated use of [F10] and [A1] yields a subsubsequence on which every coordinate converges. Pythagoras applied to each finite orthogonal projection bounds the sums of squared limiting coordinates by the common squared norm bound, so these coordinates define a vector $u_0$ in the closed span. Approximation by finite linear combinations proves convergence of inner products against every vector in the span. For any $z\in L^2(\mathrm{\Omega })$, [F12] represents $x\mapsto ⟨x,z⟩$ restricted to that span by a vector in the span, so the convergence is weak in all of $L^2$. Testing against compactly supported smooth functions shows that this weak limit and the strong $L^1$ or local-$L^1$ limit from step 4.1 define the same distribution; injectivity [F13] makes them equal almost everywhere.

Weak convergence and boundedness of $A$ imply $A{u}_n\rightharpoonup A{u}_0$, while step 1.2 gives $A{u}_n\to f$ strongly; hence $A{u}_0=f$. In the full-space case, for every admissible test field $\phi$, local $L^1$ convergence gives $\int u_0\mathrm{div}\phi ={\lim }_n\int u_n\mathrm{div}\phi \le {\mathrm{lim\hspace{0.17em}inf}}_{n}R(u_n)$; taking the supremum gives $R(u_0)\le {\mathrm{lim\hspace{0.17em}inf}}_{n}R(u_n)$. In the Neumann case the same lower semicontinuity follows from [F8] and strong $L^1(\mathrm{\Omega })$ convergence. In the Dirichlet case the zero extensions converge strongly in $L^1({\mathbb{R}}^2)$, so [F8] gives the same inequality for their full-space TV. Thus $R(u_0)\le {\mathrm{lim\hspace{0.17em}inf}}_{n}R(u_n)\le {\mathrm{lim\hspace{0.17em}sup}}_{n}R(u_n)\le R(u^{†})$. Exactness and the minimum property of $u^{†}$ force equality, so $u_0$ is a minimum-TV exact solution.

Let $\xi =A^{\ast }{p}^{†}$. Since $A{u}_0=A{u}^{†}=f$, the adjoint identity in [F1] gives $⟨\xi ,u_0-u^{†}{⟩}_X=⟨p^{†},A(u_0-u^{†}){⟩}_Y=0$. For every $v\in X$, the subgradient inequality at $u^{†}$ therefore gives $R(v)\ge R(u^{†})+⟨\xi ,v-u^{†}{⟩}_X=R(u_0)+⟨\xi ,v-u_0{⟩}_X\mathrm{.}$ This is exactly $A^{\ast }{p}^{†}\in \mathrm{\partial }R(u_0)$, as required.