[proofplan] The proof is a direct unpacking of admissibility and the definitions of the two $c$-transforms. Fixing one variable turns the admissibility inequality into a family of upper bounds for the corresponding potential. Taking the infimum over the remaining variable gives the desired comparison with the appropriate $c$-transform. [/proofplan] [step:Fix $y$ and take the infimum over $X$ to bound $\psi$ by $\varphi^c$] Fix $y \in Y$. Since $(\varphi,\psi)$ is admissible, for every $x \in X$ we have \begin{align*} \varphi(x) + \psi(y) \leq c(x,y). \end{align*} Subtracting the real number $\varphi(x)$ from both sides gives \begin{align*} \psi(y) \leq c(x,y) - \varphi(x) \end{align*} for every $x \in X$. Since $X$ is nonempty and $\psi(y)$ is a lower bound for the set $\{c(x,y)-\varphi(x): x \in X\}$, it is bounded above by the infimum of that set: \begin{align*} \psi(y) \leq \inf_{x \in X}\{c(x,y)-\varphi(x)\}. \end{align*} By the definition of $\varphi^c(y)$, this is exactly \begin{align*} \psi(y) \leq \varphi^c(y). \end{align*} [guided] Fix a point $y \in Y$. The goal is to compare the single number $\psi(y)$ with the infimum defining $\varphi^c(y)$. By admissibility, the inequality \begin{align*} \varphi(x) + \psi(y) \leq c(x,y) \end{align*} holds for every $x \in X$. Since $\varphi(x)$ is a real number, we may subtract it from both sides and obtain \begin{align*} \psi(y) \leq c(x,y) - \varphi(x). \end{align*} This inequality is valid for every $x \in X$, so $\psi(y)$ is a lower bound for the entire family of [real numbers](/page/Real%20Numbers) $\{c(x,y)-\varphi(x):x\in X\}$. Because $X$ is nonempty, the infimum of this family is the greatest lower bound, and therefore every lower bound is less than or equal to it. Hence \begin{align*} \psi(y) \leq \inf_{x \in X}\{c(x,y)-\varphi(x)\}. \end{align*} The right-hand side is precisely the definition of the $c$-transform $\varphi^c(y)$, so \begin{align*} \psi(y) \leq \varphi^c(y). \end{align*} Since the point $y \in Y$ was arbitrary, this proves $\psi \leq \varphi^c$ pointwise on $Y$. [/guided] [/step] [step:Fix $x$ and take the infimum over $Y$ to bound $\varphi$ by $\psi^c$] Fix $x \in X$. By admissibility, for every $y \in Y$, \begin{align*} \varphi(x) + \psi(y) \leq c(x,y). \end{align*} Subtracting the real number $\psi(y)$ from both sides gives \begin{align*} \varphi(x) \leq c(x,y) - \psi(y) \end{align*} for every $y \in Y$. Since $Y$ is nonempty, taking the infimum over $y \in Y$ yields \begin{align*} \varphi(x) \leq \inf_{y \in Y}\{c(x,y)-\psi(y)\}. \end{align*} By the definition of $\psi^c(x)$, this becomes \begin{align*} \varphi(x) \leq \psi^c(x). \end{align*} Since $x \in X$ was arbitrary, the two pointwise inequalities hold for all $y \in Y$ and all $x \in X$, respectively. [/step]