= Solution
Positivity means $u>0$ almost everywhere, since the real <logarithm> must belong to $L^1$. A <positivity-preserving operator> gives $s=Tu\ge0$. For fixed $x$, the scalar <shifted Poisson data fidelity> is $f_x(s)=s-g(x)\log(1+s)$, with
$$
f_x'(s)=1-\frac{g(x)}{1+s},\qquad f_x''(s)=\frac{g(x)}{(1+s)^2}>0.
$$
Composition 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 $\log(1+s)\ge0$, the <Jensen inequality> gives the requested bound:
$$
\begin{aligned}
\int_\Omega[s-g\log(1+s)]\,dx
&\ge\|s\|_1-\|g\|_\infty\int_\Omega\log(1+s)\,dx\\
&\ge\|s\|_1-\|g\|_\infty\log(1+\|s\|_1)\\
&=\|Tu\|_1-\|g\|_\infty\log\|Tu+1\|_1.
\end{aligned}
$$
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.
\b[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 $E_n=\{u\ge1/n\}$. 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 $u_\varepsilon=\varepsilon\chi_\Omega$ are admissible, have zero <total variation>, and satisfy
$$
0<E(u_\varepsilon)\le\varepsilon\|T\chi_\Omega\|_1\longrightarrow0.
$$
Their logarithms are integrable for each $\varepsilon>0$, but the <limit> is excluded. Therefore
$$
\boxed{\inf E=0,\qquad\operatorname{argmin}E=\varnothing\quad\text{in the printed domain}.}
$$
A 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 $u\ge0$, 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 $G=\|g\|_\infty$, $m_G=\inf_{t\ge0}\{t-G\log(1+t)\}>-\infty$, and take a minimizing sequence of energy at most $C$. The displayed bound gives $\|Tu_n\|_1\le C_1$ and $\alpha\operatorname{TV}(u_n)\le C-m_G$. Write $c_n=\int_\Omega u_n\ge0$. The <Poincaré inequality for total variation> and <mean control for positive imaging operators> yield
$$
c_n\|T\chi_\Omega\|_1\le\|Tu_n\|_1+\|T\|\|u_n-c_n\|_1
\le C_1+C_P\|T\|\operatorname{TV}(u_n).
$$
The 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 $u\ge0$. <Continuity> gives $Tu_n\to Tu$ in $L^1$. On $s\ge0$, $|f_x'(s)|\le1+G$, 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 \b[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 $\operatorname{TV}(u)=0$; on the connected square, zero variation makes $u$ a nonnegative constant, and $T\chi_\Omega\ne0$ 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 $\sum1$ gives precisely the stated fidelity. Thus the model is \b[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.
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