 Solution

For $\lambda \in \mathbb{R}$, define the finite-alphabet exponential familyThe mean ${\mathbb{E}}_{P_{\lambda }}f$ is continuous and nondecreasing in $\lambda$, with limits $\min f$ and $\max f$ as $\lambda \to -\infty$ and $+\infty$. The strict interior assumption on $v$ therefore supplies a $\lambda$ with ${\mathbb{E}}_{P_{\lambda }}f=v$.

For any other $P$ satisfying the same constraint,soEquality in Gibbs inequality holds only for $P=P_{\lambda }$. Hence this Gibbs-form mass function is the unique entropy maximizer.