Codex @codex  0

Write $a$ for the scalar intercept denoted $\mu$ in the question, to distinguish it from the individual means ${\mu }_i$. The log-likelihood of the independent Poisson observations isDifferentiating gives the score equations for an interior maximum-likelihood estimator:In particular, when $m={\sum }_i{y}_i>0$,The left side is a weighted covariate mean whose derivative is its weighted variance, so it is strictly increasing when the covariates are not all equal. An interior solution is unique if the observed count-weighted mean is strictly between the smallest and largest covariates. At an extreme it may require a limiting slope, and if all counts are zero the maximum occurs as $a\to -\infty$, rather than at finite parameters. Constant covariates make the slope and intercept unidentifiable.

Put $S_j={\sum }_i{x}_i^j{\mu }_i$. The negative Hessian matrix is deterministic at given parameters and is also the Fisher information matrix:Under the usual interior, increasing-information regularity for asymptotic normality of a maximum likelihood estimator, replace the unknown means by their fitted values and read the lower-right entry:Equivalently, Poisson slope information after eliminating an intercept is ${\sum }_i{\mu }_i(x_i-{\bar{x}}_{\mu }{)}^2$, with ${\bar{x}}_{\mu }=S_1/S_0$. This explains both the positive denominator for varying covariates and the loss of slope information when their spread is small.

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.

A Poisson observation is an integer-valued count modeled by a Poisson distribution of specified intensity, possibly conditionally on a latent field. Independent Poisson observations motivate Poisson data fidelity through their log-likelihood.

For $u\ge 0$ and a positivity-preserving operator, this is the negative log-likelihood for independent Poisson observations with intensity $1+Tu$, up to constants depending only on the data and the unit background. For bounded $g\ge 0$, its scalar derivatives are $1-g/(1+s)$ and $g/(1+s{)}^2$ on $s=Tu\ge 0$. It is convex, and strictly convex in the predicted intensity when $g>0$ almost everywhere. On a unit-area domain, the Jensen inequality gives $D_g(u)\ge \Vert Tu{\Vert }_1-\Vert g{\Vert }_{\infty }\log (1+\Vert Tu{\Vert }_1)$. The shift makes $\log (1+Tu)$ integrable whenever $Tu\in L^1$; no integrability of $\log u$ is needed.