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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 $\rho =e^{\phi }$ and obtain $\rho \log \rho =\phi e^{\phi }$ 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 ${\partial }_{h}F(\phi +h\psi ){\mid }_{h=0}$

$$0=D^{Gâteaux}(\int -\phi e^{\phi }+\sum _i{\lambda }_i\sum _k{p}_{i,k}{\delta }_{x_{i,k}}\cdot 1_{d_i^{-1}(p_{i,k})}{e}^{\phi }d\mathcal{L})(\phi ;\psi )={\partial }_h\int -(\phi +h\psi )e^{\phi +h\psi }+\sum _{i,k}{\lambda }_{i,k}^{\prime }{e}^{\phi +h\psi }\cdot 1_{d_i^{-1}(p_{i,k})}d\mathcal{L}=-\int \psi e^{\phi }(1+\phi -\sum _{i,k}{\lambda }_{i,k}^{\prime }\cdot 1_{d_i^{-1}(p_{i,k})})d\mathcal{L}\$\$$$

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

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

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