TV source certificates and level-set convergence — Examples and Counterexamples

The function $u^{†}=\mathrm{\mid }{x}_1\mathrm{\mid }$ is bounded on the bounded Lipschitz domain $\mathrm{\Omega }$, hence belongs to $L^2(\mathrm{\Omega })$. Fubini and one-dimensional integration by parts show that its distributional gradient is $D{u}^{†}=\mathrm{sgn}(x_1)e_{1}dx$: for every compactly supported smooth vector field $\phi$, the $x_1$ derivative pairing is $-\int \mathrm{\mid }{x}_1\mathrm{\mid }{\mathrm{\partial }}_1\phi =\int \mathrm{sgn}(x_1){\phi }_1$, and the $x_2$ derivative pairing vanishes. This bounded density defines a finite vector Radon measure by [F6], so $u^{†}\in BV(\mathrm{\Omega })$. Since $A=I$, its Hilbert adjoint is also $I$. The equation $Au=f$ forces $u=f=u^{†}$, so $u^{†}$ is the unique exact solution and therefore minimizes $R$ among exact solutions.

For $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{R})$ and $t\in \mathbb{R}$, the distributional gradient of $u^{†}+t\phi$ has density $g_t=\mathrm{sgn}(x_1)e_1+t\mathrm{\nabla }\phi$. For any integrable vector density $g$, its partition variation mass is $\int \mathrm{\mid }g\mathrm{\mid }$: every partition gives ${\sum }_j\mathrm{\mid }{\int }_{B_j}g\mathrm{\mid }\le \int \mathrm{\mid }g\mathrm{\mid }$; conversely, for $\epsilon >0$ choose a finite-valued measurable simple function $s={\sum }_{k=1}^N{c}_k{\mathbf{1}}_{B_k}$ with $\mathrm{\parallel }g-s{\mathrm{\parallel }}_1<\epsilon$, where the disjoint $B_k$ partition the set and $s=0$ off their union. Then this partition gives ${\sum }_k\mathrm{\mid }{\int }_{B_k}g\mathrm{\mid }\ge {\sum }_k\mathrm{\mid }{c}_k\mathrm{\mid }\mathrm{\mid }{B}_k\mathrm{\mid }-\mathrm{\parallel }g-s{\mathrm{\parallel }}_1=\mathrm{\parallel }s{\mathrm{\parallel }}_1-\mathrm{\parallel }g-s{\mathrm{\parallel }}_1\ge \mathrm{\parallel }g{\mathrm{\parallel }}_1-2\epsilon$. Taking arbitrary $\epsilon$ proves the formula. Consequently $R(u^{†}+t\phi )={\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{sgn}(x_1)e_1+t\mathrm{\nabla }\phi \mathrm{\mid }dx\mathrm{.}$ For almost every $x$, the vector $q=\mathrm{sgn}(x_1)e_1$ has norm one and ${\lim }_{t\downarrow 0}(\mathrm{\mid }q+th\mathrm{\mid }-\mathrm{\mid }q\mathrm{\mid })\mathrm{/}t=q\cdot h$. The quotient is bounded in absolute value by $\mathrm{\mid }h\mathrm{\mid }$, by the reverse triangle inequality. Dominated convergence with $h=\mathrm{\nabla }\phi$ therefore yields the one-sided derivative $L(\phi ):={\lim }_{t\downarrow 0}\frac{R(u^{†}+t\phi )-R(u^{†})}{t}={\int }_{\mathrm{\Omega }}\mathrm{sgn}(x_1){\mathrm{\partial }}_1\phi dx=-2{\int }_{-1}^1\phi (0,x_2)d{x}_2,$ where the last equality is integration by parts on $(-1,0)$ and $(0,1)$; compact support removes the endpoint terms. Replacing $\phi$ by $-\phi$ gives directional derivative $-L(\phi )$.

Suppose a certificate $p\in L^2(\mathrm{\Omega })$ existed. The subgradient inequality [F3] applied to $u^{†}+t\phi$ for $t>0$, divided by $t$ and passed to the limit in step 1.2, gives ${\int }_{\mathrm{\Omega }}p\phi \le L(\phi )$. Applying the same inequality to $u^{†}-t\phi$ gives $-{\int }_{\mathrm{\Omega }}p\phi \le -L(\phi )$. Thus ${\int }_{\mathrm{\Omega }}p\phi =L(\phi )$ for every compactly supported smooth $\phi$.

Choose $\eta ,\psi \in C_c^{\mathrm{\infty }}((-1,1))$ with $\eta (0)=1$ and ${\int }_{-1}^1\psi \mathrm{\ne }0$, and for $0<\epsilon <1$ set ${\phi }_{\epsilon }(x_1,x_2)=\eta (x_1\mathrm{/}\epsilon )\psi (x_2)$. Then ${\phi }_{\epsilon }\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and $\mathrm{\parallel }{\phi }_{\epsilon }{\mathrm{\parallel }}_2=\sqrt{\epsilon }\mathrm{\parallel }\eta {\mathrm{\parallel }}_2\mathrm{\parallel }\psi {\mathrm{\parallel }}_2\to 0$. But step 2.1 and the formula in step 1.2 give ${\int }_{\mathrm{\Omega }}p{\phi }_{\epsilon }=-2{\int }_{-1}^1\psi (x_2)d{x}_2$, a fixed nonzero number, whereas Cauchy–Schwarz [F4] forces these integrals to tend to zero. This contradiction proves that no $L^2$ source certificate exists.

Put $S=\sigma (S^1)<\mathrm{\infty }$. For each fixed $z\in S^1$, polar coordinates [F5] give ${\int }_0^2{\int }_{S^1}\mathrm{\mid }u(z+t\omega ){\mathrm{\mid }}^{2}td\sigma (\omega )dt={\int }_{B(z,2)}\mathrm{\mid }u(x){\mathrm{\mid }}^{2}dx<\mathrm{\infty },$ so for almost every $t>0$ the circle section is square-integrable. Cauchy–Schwarz [F4] then gives $\mathrm{\mid }(A_{0}u)(z,t){\mathrm{\mid }}^2\le S{t}^2{\int }_{S^1}\mathrm{\mid }u(z+t\omega ){\mathrm{\mid }}^{2}d\sigma (\omega )\mathrm{.}$ Integrating in $z,t$, using polar coordinates [F5] around each center and Tonelli [F6], yields $\mathrm{\parallel }{A}_{0}u{\mathrm{\parallel }}_{L^2(\mathrm{\Sigma })}^2\le S{\int }_{S^1}{\int }_{\mathrm{\mid }x-z\mathrm{\mid }<2}\mathrm{\mid }x-z\mathrm{\mid }\mathrm{\mid }u(x){\mathrm{\mid }}^{2}dxd\sigma (z)\le 2S^2\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2({\mathbb{R}}^2)}^2,$ where $\mathrm{\mid }x-z\mathrm{\mid }<2$ and $\sigma (S^1)=S$. Thus $A_0$ is real-linear and bounded on real Schwartz functions with norm at most $\sqrt{2}S$.

The zero vector $u^{†}=0$ satisfies $A{u}^{†}=f=0$ and $T{V}_2(u^{†})=0$. Since $T{V}_2\ge 0$, it minimizes TV over the exact solution set, regardless of the kernel of $A$. Its adjoint certificate $p^{†}=0$ satisfies $A^{\ast }{p}^{†}=0\in \mathrm{\partial }T{V}_2(0)$ because $T{V}_2(v)\ge 0=T{V}_2(0)+⟨0,v⟩$ for every $v\in L^2$.

By [F7], real Schwartz functions are dense in real $L^2$: approximate a real function by complex Schwartz classes and take real parts, which cannot increase the $L^2$ error. By [F8], the target real $L^2(\mathrm{\Sigma })$ is complete. For any real $u\in L^2$, choose real Schwartz $u_k\to u$; step 1.1 makes $(A_0{u}_k)$ Cauchy, and its limit is independent of the approximating sequence by the same bound. This defines a bounded real-linear extension $A$ with $\mathrm{\parallel }A\mathrm{\parallel }\le \sqrt{2}S$. If another bounded extension agrees with $A_0$ on the dense Schwartz subspace, continuity forces equality on all of $L^2$.

For $w_n=0$, the regularized objective is $J_n(u)=\frac{1}{2}\mathrm{\parallel }Au{\mathrm{\parallel }}_2^2+{\alpha }_{n}T{V}_2(u)\ge 0$, and $J_n(0)=0$. Hence $u_n=0$ is a global minimizer for every $n$. If $t>0$, $\{u_n>t\}=\mathrm{\varnothing }$; if $t<0$, $\{u_n<t\}=\mathrm{\varnothing }$. The same holds for $u^{†}$, so all canonical density-one sets and ambient boundaries agree as empty sets; symmetric-difference measures and the extended Hausdorff distances are zero. The noise ratio is $\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }\mathrm{\parallel }{w}_n{\mathrm{\parallel }}_2\mathrm{/}{\alpha }_n=0<2\sqrt{\pi }$, so the full-space threshold is met.

The multiplier $m$ is real, even, and satisfies $0<m\le 1$, so [F1] makes $A$ a contraction on complex $L^2$. For real Schwartz $g$, the integral formula [F2] gives $\widehat{g}(-\xi )=\overline{\widehat{g}(\xi )}$, and multiplication by $m$ preserves this symmetry. Approximate $m\widehat{g}$ in $L^2$ by Schwartz functions using [F3], then replace each approximant $s$ by $\frac{1}{2}(s(\xi )+\overline{s(-\xi )})$. These symmetrized approximants remain Schwartz, converge to $m\widehat{g}$, and have real inverse Fourier transforms by [F2]; [F1] and the closedness observation above pass this property to $Ag$. Density of real Schwartz functions and boundedness of $A$ extend real-valuedness to real $L^2$. By [F7], the complex Plancherel pairing restricts to the real Hilbert pairing. Multiplication by the real function $m$ is self-adjoint, so $⟨Au,v⟩=⟨u,Av⟩$ for real $u,v$, and $A^{\ast }=A$. If $Au=0$, then $m\widehat{u}=0$ almost everywhere; strict positivity of $m$ and injectivity of the unitary Fourier transform give $u=0$. Thus $A$ is bounded, self-adjoint, and injective.

Because $\varphi$ is supported away from zero and from $b$, $u^{†}$ is constant near the origin, smooth across radii $a,b$, and zero for $\mathrm{\mid }x\mathrm{\mid }\ge b$; hence it is a compactly supported smooth real function. Its gradient is $\mathrm{\nabla }{u}^{†}(x)=-\varphi (\mathrm{\mid }x\mathrm{\mid })x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ wherever $\varphi (\mathrm{\mid }x\mathrm{\mid })\mathrm{\ne }0$ and is zero elsewhere. The cutoff in the Statement exists by [F6] and gives $z\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2;{\mathbb{R}}^2)$ with $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$. Integration by parts gives ${\int }_{{\mathbb{R}}^2}{u}^{†}\mathrm{div}zdx=-{\int }_{{\mathbb{R}}^2}z\cdot \mathrm{\nabla }{u}^{†}dx={\int }_{{\mathbb{R}}^2}\varphi (\mathrm{\mid }x\mathrm{\mid })dx={\int }_{{\mathbb{R}}^2}\mathrm{\mid }\mathrm{\nabla }{u}^{†}\mathrm{\mid }dx\mathrm{.}$ For every admissible test field $y$, the same integration by parts gives $\int u^{†}\mathrm{div}y\le \int \mathrm{\mid }\mathrm{\nabla }{u}^{†}\mathrm{\mid }$, while the field $z$ attains equality. Thus $T{V}_2(u^{†})=\int u^{†}\mathrm{div}z$. Setting $q=\mathrm{div}z$, [F4] gives $T{V}_2(v)\ge \int vq=T{V}_2(u^{†})+⟨q,v-u^{†}⟩$ for every $v$, so $q\in \mathrm{\partial }T{V}_2(u^{†})$.

The vector $q$ is smooth and compactly supported, so $q,{\mathrm{\partial }}_{1}q,{\mathrm{\partial }}_{2}q\in L^2$. Integration by parts using [F2] gives $\widehat{{\mathrm{\partial }}_{j}q}=2\pi i{\xi }_j\widehat{q}$, and Plancherel [F1] yields ${\int }_{{\mathbb{R}}^2}(1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2)\mathrm{\mid }\widehat{q}(\xi ){\mathrm{\mid }}^{2}d\xi =\mathrm{\parallel }q{\mathrm{\parallel }}_2^2+\mathrm{\parallel }\mathrm{\nabla }q{\mathrm{\parallel }}_2^2<\mathrm{\infty }\mathrm{.}$ Therefore $\widehat{p}(\xi ):=(1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\widehat{q}(\xi )$ is in $L^2$, defining $p={\mathcal{F}}_2^{-1}\widehat{p}\in L^2$. The integral formula [F2] gives Hermitian symmetry of $\widehat{q}$, and the real even multiplier preserves it. As in step 1.1, approximate $\widehat{p}$ by Schwartz functions, symmetrize them, apply inverse Fourier transformation, and use Plancherel continuity and closedness of the real subspace to conclude that $p$ is real. By construction $A^{\ast }p=Ap=q$, so [F5] gives the required source certificate. Injectivity from step 1.1 makes $u^{†}$ the unique exact solution, hence the unique minimum-TV exact solution.

Step 2.1 provides a certificate at the unique minimum-TV exact truth. For every positive ${\alpha }_n$ and zero noise $w_n=0$, the parameter ratio is $\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }\mathrm{\parallel }{w}_n{\mathrm{\parallel }}_2\mathrm{/}{\alpha }_n=0$. Thus this example verifies the source condition and the zero-noise parameter side of the full-space deblurring setup without asserting existence of a regularized minimizer.