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A Lipschitz domain is an open connected subset of Euclidean space whose boundary is locally the graph of a Lipschitz continuous function after rotating and translating coordinates. Locally the domain lies on one side of that graph. This permits corners but excludes boundary cusps that cannot be represented with a finite Lipschitz constant. Bounded Lipschitz domains are a standard setting for Sobolev spaces and the Poincaré inequality for total variation.

On a bounded connected Lipschitz domain, let $u\ge 0$ belong to the BV space, and let $T$ be a positivity-preserving operator with $T1\ne 0$. Writing $u=c1+(u-c1)$, $c=u_{\Omega }\ge 0$, the triangle inequality and Poincaré inequality for total variation give $c\Vert T1{\Vert }_1\le \Vert Tu{\Vert }_1+\Vert T\Vert \Vert u-c{\Vert }_1\le \Vert Tu{\Vert }_1+C_{\Omega }\Vert T\Vert \mathrm{TV}(u)$. Thus forward-image and variation bounds control the missing constant mode, and hence the full $BV$ norm.

On a bounded domain of positive measure, the constant functions $u_k\equiv k$ satisfy $\mathrm{TV}(u_k)=0$ while $\Vert u_k{\Vert }_{BV}=k|\Omega |\to \infty$. Thus the total variation seminorm on a domain is neither coercive in the $BV$ norm nor strictly convex. Additional control of the mean can remove this obstruction; the Poincaré inequality for total variation supplies the needed bound on suitable connected domains.

Positivity means $u>0$ almost everywhere, since the real logarithm must belong to $L^1$. A positivity-preserving operator gives $s=Tu\ge 0$. For fixed $x$, the scalar shifted Poisson data fidelity is $f_x(s)=s-g(x)\log (1+s)$, withComposition with the linear operator $T$ proves convexity in $u$, and strict convexity holds along pairs whose forward images differ on a set of positive measure. The admissible class itself is convex: the concavity of the logarithm gives $\log (\theta u+(1-\theta )v)\ge \theta \log u+(1-\theta )\log v$, controlling its negative part, while ${\log }^{+}w\le w$ controls its positive part.

Because $|\Omega |=1$ and $\log (1+s)\ge 0$, the Jensen inequality gives the requested bound:The final equality uses nonnegativity and the unit area. Since $t-G\log (1+t)\to \infty$, this controls the forward-image norm on energy sublevels.

The printed strictly positive problem has no minimizer. This is nonattainment under strict positivity for shifted Poisson fidelity, rather than a failure of the coercivity calculation. Indeed, with $0<g<1$ and $s>0$, $\log (1+s)<s$ implies $f_x(s)>0$. Also $Tu$ cannot vanish identically for a strictly positive $u$. To see this, let $E_n=\{u\ge 1/n\}$. If $Tu=0$, positivity and $0\le {\chi }_{E_n}\le nu$ give $T{\chi }_{E_n}=0$. But ${\chi }_{E_n}\to {\chi }_{\Omega }$ in $L^1$ and continuity of $T$ would imply $T{\chi }_{\Omega }=0$, contradicting the hypothesis. Thus every admissible $u$ has positive fidelity and hence positive total energy.

Conversely, the constants $u_{\epsilon }=\epsilon {\chi }_{\Omega }$ are admissible, have zero total variation, and satisfyTheir logarithms are integrable for each $\epsilon >0$, but the limit is excluded. ThereforeA bounded minimizing sequence and bounded-variation compactness do not repair a nonclosed positivity/logarithm constraint. In particular, a literal existence or uniqueness proof for that domain is impossible.

The natural correction is to minimize over $BV(\Omega )$ with $u\ge 0$, omitting the unnecessary $\log u\in L^1$ condition: the fidelity only contains $\log (1+Tu)$, which is already integrable. Here is the full existence for nonnegative shifted Poisson regularization argument, also valid for any bounded nonnegative data $g$. Let $G=\Vert g{\Vert }_{\infty }$, $m_G={\inf }_{t\ge 0}\{t-G\log (1+t)\}>-\infty$, and take a minimizing sequence of energy at most $C$. The displayed bound gives $\Vert T{u}_n{\Vert }_1\le C_1$ and $\alpha \mathrm{TV}(u_n)\le C-m_G$. Write $c_n={\int }_{\Omega }{u}_n\ge 0$. The Poincaré inequality for total variation and mean control for positive imaging operators yieldThe denominator is nonzero, so the full $BV$ norm is bounded. Bounded-variation compactness gives $u_n\to u$ in $L^1$ along a subsequence, with $u\ge 0$. Continuity gives $T{u}_n\to Tu$ in $L^1$. On $s\ge 0$, $|f_x^{\prime }(s)|\le 1+G$, so the fidelity converges in the integral; lower semicontinuity of variation completes the direct method in the calculus of variations.

For the actual printed data $0<g<1$, the corrected problem has the unique minimizer $u=0$, even if $T$ is not injective. Zero attains energy zero. Any other zero-energy candidate would have both $Tu=0$ and $\mathrm{TV}(u)=0$; on the connected square, zero variation makes $u$ a nonnegative constant, and $T{\chi }_{\Omega }\ne 0$ forces that constant to vanish. For general positive bounded data, an injective $T$ is a sufficient uniqueness condition, because its fidelity is strictly convex; injectivity is not a necessary condition in every instance.

In the finite-dimensional interpretation, let ${\lambda }_{ij}=1+(Tu{)}_{ij}$. Independent Poisson observations with these intensities have negative log-likelihood ${\sum }_{ij}[{\lambda }_{ij}-g_{ij}\log {\lambda }_{ij}]+C(g)$. Removing the constant $\sum 1$ gives precisely the stated fidelity. Thus the model is Poisson counting noise with a unit background intensity, or an approximate version of it for rescaled/continuous grey values. Literal Poisson counts are integers; the constraint $0<g<1$ is a grey-value normalization, not a literal unscaled count sample.

For $u\in L^1(\Omega )$, its total variation seminorm on a domain isThe corresponding bounded-variation space and its zero-mean subspace areIf $u_{\Omega }=|\Omega {|}^{-1}{\int }_{\Omega }udx$, the Poincaré inequality for total variation isIn particular, $\Vert u{\Vert }_{L^1}\le C_{\Omega }\mathrm{TV}(u)$ for $u\in B{V}_0(\Omega )$.

The zero-mean bounded-variation space isOn a bounded connected Lipschitz domain, removing the constant nullspace makes total variation coercive on this subspace by the Poincaré inequality for total variation.