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1. Suggestion

We are interested in a probability distribution :<math>\rho:X \rightarrow \mathbb R_+</math>, such that

To assure positivity we set $\rho = e^\varphi$ and obtain $\rho \log \rho = \varphi e^\varphi$ for the entropy. For simplicity we assume for $\mathbb P_i$ only discrete probabilities $\mathbb P_i = \sum_k p_{i,k} \delta_{x_{i,k}}$

The Gâteaux differential $D^{Gâteaux}$ in direction $\psi$ is given by $\partial_h F(\varphi + h\psi)\mid_{h=0}$

$$ 0 = D^{Gâteaux} \left( \int -\varphi e^\varphi + \sum_i \lambda_i \sum_k p_{i,k} \delta_{x_{i,k}}\cdot 1_{d_i^{-1} (p_{i,k})} e^\varphi d\mathcal L \right)(\varphi; \psi) \\ = \partial_h \int -(\varphi + h\psi) e^{\varphi + h\psi} + \sum_{i,k} \lambda_{i,k}' e^{\varphi + h\psi} \cdot 1_{d_i^{-1} (p_{i,k})}d\mathcal L \\ =-\int \psi e^\varphi \left(1 + \varphi - \sum_{i,k} \lambda_{i,k}' \cdot 1_{d_i^{-1} (p_{i,k})} \right) d\mathcal L \\$$

The last bracket has to be zero (a.e.) because $\psi$ is of arbitrary choice and $e^\varphi \ne 0$.

Absorbing '+1' in the $\lambda$'s leads to the condition

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