We consider $\alpha = (\alpha_i)_{i \in I} \in (\mathbb{R}_+^*)^I$ and $\beta = (\beta_j)_{j \in J} \in (\mathbb{R}_+^*)^J$ such that $\sum_{i \in I} \alpha_i = \sum_{j \in J} \beta_j = 1$. We denote by $\boldsymbol{p}$ the element of $F(\alpha, \beta)$ defined by $p_{ij} = \alpha_i \beta_j > 0$ for all $(i,j) \in I \times J$. Let $C = (C_{ij})_{(i,j) \in I \times J} \in \mathbb{R}_+^{I \times J}$ and $\epsilon > 0$. We consider $J_\epsilon : Q \rightarrow \mathbb{R}$ defined by $$J_\epsilon(\boldsymbol{q}) = \sum_{ij} q_{ij} C_{ij} + \epsilon \operatorname{KL}(\boldsymbol{q}, \boldsymbol{p})$$ and $\boldsymbol{q}(\epsilon)$ the unique minimizer of $J_\epsilon$ on $F(\alpha, \beta)$. (a) Verify that $q(\epsilon)_{ij} > 0$ for all $(i,j) \in I \times J$ (Hint: One may reason by contradiction and consider for all $t \in ]0,1[$ $\boldsymbol{q}(\epsilon, t) = (1-t)\boldsymbol{q}(\epsilon) + t\boldsymbol{p}$ then observe the behavior of $\varphi(x)$ near $x = 0$). (b) Show that this is no longer true if we assume that $\epsilon = 0$.
We consider $\alpha = (\alpha_i)_{i \in I} \in (\mathbb{R}_+^*)^I$ and $\beta = (\beta_j)_{j \in J} \in (\mathbb{R}_+^*)^J$ such that $\sum_{i \in I} \alpha_i = \sum_{j \in J} \beta_j = 1$. We denote by $\boldsymbol{p}$ the element of $F(\alpha, \beta)$ defined by $p_{ij} = \alpha_i \beta_j > 0$ for all $(i,j) \in I \times J$. Let $C = (C_{ij})_{(i,j) \in I \times J} \in \mathbb{R}_+^{I \times J}$ and $\epsilon > 0$. We consider $J_\epsilon : Q \rightarrow \mathbb{R}$ defined by
$$J_\epsilon(\boldsymbol{q}) = \sum_{ij} q_{ij} C_{ij} + \epsilon \operatorname{KL}(\boldsymbol{q}, \boldsymbol{p})$$
and $\boldsymbol{q}(\epsilon)$ the unique minimizer of $J_\epsilon$ on $F(\alpha, \beta)$.\\
(a) Verify that $q(\epsilon)_{ij} > 0$ for all $(i,j) \in I \times J$ (Hint: One may reason by contradiction and consider for all $t \in ]0,1[$ $\boldsymbol{q}(\epsilon, t) = (1-t)\boldsymbol{q}(\epsilon) + t\boldsymbol{p}$ then observe the behavior of $\varphi(x)$ near $x = 0$).\\
(b) Show that this is no longer true if we assume that $\epsilon = 0$.