A minimum-TV truth can lack an L² source certificate

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.