[proofplan] We argue by contradiction. If two distinct probability measures both minimize the entropy, then strict displacement convexity along a Wasserstein geodesic between them forces every interior point of that geodesic to have entropy strictly below the common minimum value. This contradicts the definition of a minimizer, so two distinct minimizers cannot exist. [/proofplan]

[step:Assume two distinct minimizers and name the minimum value] Suppose, toward a contradiction, that there exist two distinct measures $\mu_0,\mu_1\in\mathcal P_2(\mathbb R^n)$ which both minimize $\operatorname{Ent}_\gamma$ on $\mathcal P_2(\mathbb R^n)$. Define the minimum value $m\in[0,+\infty]$ by \begin{align*} m:=\inf_{\nu\in\mathcal P_2(\mathbb R^n)}\operatorname{Ent}_\gamma(\nu). \end{align*} Since $\operatorname{Ent}_\gamma(\nu)\ge 0$ for every $\nu\in\mathcal P_2(\mathbb R^n)$ and $\operatorname{Ent}_\gamma(\gamma)=0$, we have $m=0$. Since $\mu_0$ and $\mu_1$ are minimizers, we have \begin{align*} \operatorname{Ent}_\gamma(\mu_0)=m \end{align*} and \begin{align*} \operatorname{Ent}_\gamma(\mu_1)=m. \end{align*} Because $\operatorname{Ent}_\gamma(\mu_0)=m=0$ and $\operatorname{Ent}_\gamma(\mu_1)=m=0$, both minimizers have finite entropy. Hence $\mu_0,\mu_1\in\mathcal D(\operatorname{Ent}_\gamma)$. [/step]

[step:Apply strict displacement convexity along a Wasserstein geodesic]By the assumed geodesic existence property of $(\mathcal P_2(\mathbb R^n),W_2)$, choose a constant-speed $W_2$-geodesic \begin{align*} \mu_\bullet:[0,1]\to\mathcal P_2(\mathbb R^n) \end{align*} joining $\mu_0$ to $\mu_1$, so that $\mu_\bullet(0)=\mu_0$ and $\mu_\bullet(1)=\mu_1$. Since $\mu_0\ne\mu_1$ and both endpoints lie in $\mathcal D(\operatorname{Ent}_\gamma)$, strict displacement convexity gives, for every $t\in(0,1)$, \begin{align*} \operatorname{Ent}_\gamma(\mu_t)<(1-t)\operatorname{Ent}_\gamma(\mu_0)+t\operatorname{Ent}_\gamma(\mu_1). \end{align*} Substituting the equalities $\operatorname{Ent}_\gamma(\mu_0)=m$ and $\operatorname{Ent}_\gamma(\mu_1)=m$ gives \begin{align*} \operatorname{Ent}_\gamma(\mu_t)<(1-t)m+tm=m. \end{align*}[/step]

[guided]The purpose of choosing the geodesic is that displacement convexity is not ordinary linear convexity in the [vector space](/page/Vector%20Space) of signed measures; it is convexity along $W_2$-geodesics in $\mathcal P_2(\mathbb R^n)$. By the assumed geodesic existence property of $(\mathcal P_2(\mathbb R^n),W_2)$, we therefore take a map \begin{align*} \mu_\bullet:[0,1]\to\mathcal P_2(\mathbb R^n) \end{align*} which is a constant-speed $W_2$-geodesic with endpoints $\mu_\bullet(0)=\mu_0$ and $\mu_\bullet(1)=\mu_1$. Now we verify the hypotheses needed to use strict displacement convexity. The endpoints are distinct by the contradiction assumption, so $\mu_0\ne\mu_1$. Define \begin{align*} m:=\inf_{\nu\in\mathcal P_2(\mathbb R^n)}\operatorname{Ent}_\gamma(\nu). \end{align*} They belong to the effective domain because the entropy is nonnegative and normalized by $\operatorname{Ent}_\gamma(\gamma)=0$. Indeed, this infimum satisfies $0\le m\le\operatorname{Ent}_\gamma(\gamma)=0$, so $m=0$. Since both endpoints are minimizers, \begin{align*} \operatorname{Ent}_\gamma(\mu_0)=\operatorname{Ent}_\gamma(\mu_1)=0<+\infty. \end{align*} Therefore \begin{align*} \mu_0,\mu_1\in\mathcal D(\operatorname{Ent}_\gamma). \end{align*} Thus the strict displacement convexity assumption applies to this geodesic and yields, for each $t\in(0,1)$, \begin{align*} \operatorname{Ent}_\gamma(\mu_t)<(1-t)\operatorname{Ent}_\gamma(\mu_0)+t\operatorname{Ent}_\gamma(\mu_1). \end{align*} The two endpoint entropies are equal because both endpoints are minimizers. Therefore \begin{align*} \operatorname{Ent}_\gamma(\mu_0)=m \end{align*} and \begin{align*} \operatorname{Ent}_\gamma(\mu_1)=m. \end{align*} Substituting these two identities into the strict convexity inequality gives \begin{align*} \operatorname{Ent}_\gamma(\mu_t)<(1-t)m+tm. \end{align*} Since $(1-t)+t=1$, the right-hand side is exactly $m$, and hence \begin{align*} \operatorname{Ent}_\gamma(\mu_t)<m. \end{align*} This is the key contradiction: strict convexity forces an interior geodesic point to lie strictly below the common endpoint value.[/guided]

[step:Contradict minimality and conclude uniqueness] Choose any $t_0\in(0,1)$. The previous step gives \begin{align*} \operatorname{Ent}_\gamma(\mu_{t_0})<m. \end{align*} But $\mu_{t_0}\in\mathcal P_2(\mathbb R^n)$, while $m$ was defined as the infimum of $\operatorname{Ent}_\gamma$ over all of $\mathcal P_2(\mathbb R^n)$. Hence no element of $\mathcal P_2(\mathbb R^n)$ can have entropy strictly smaller than $m$. This contradiction shows that two distinct minimizers cannot exist. Therefore $\operatorname{Ent}_\gamma$ has at most one minimizer in $\mathcal P_2(\mathbb R^n)$. [/step]