c-cyclical monotonicity

Remarks

Example: let $X = Y = \R$, $c(x, y) = (x - y)^2$ and $\Gamma = \set{(x, x) : x \in \R}$, the diagonal. The left side of the inequality is $0$ and the right side is a sum of squares, so the inequality holds and $\Gamma$ is $c$-cyclically monotone.

Non-example: with the same cost, $\Gamma = \set{(0, 1), (1, 0)}$ fails for $N = 2$: the left side is $1 + 1 = 2$ and the right side is $c(0, 0) + c(1, 1) = 0$.

Equivalently, the inequality may be required for every permutation $\sigma$ of $\set{1, \dots, N}$, with $y_{\sigma(i)}$ in place of $y_{i+1}$, since every permutation is a product of disjoint cycles. For $X = Y = \R^n$ and $c(x, y) = \frac12 \abs{x - y}^2$, a set is $c$-cyclically monotone exactly when it is cyclically monotone in the sense of Rockafellar, that is, $\sum_i \ip{y_i}{x_{i+1} - x_i} \le 0$ with $x_{N+1} = x_1$.