I'm reading the proof of Theorem 1.38. from section 1.6.2 $c$-cyclical monotonicity and duality of Santambrogio's Optimal transport for applied mathematicians.
My understanding: It seems for the inequalities
- $\color{blue}{c(x, y)>c\left(x_i, y_i\right)-\varepsilon}$ to make sense, it has to be $c\left(x_i, y_i\right) \neq +\infty$.
- $\color{blue}{\int c \mathrm{~d} \gamma-\int c~d \widetilde{\gamma}} = \varepsilon_0 \sum_{i=1}^k \int c \mathrm{~d} \gamma_i-\varepsilon_0 \sum_{i=1}^k \int c~d \widetilde{\gamma}_i$ to make sense, it has to be $\int c~\mathrm d \gamma \neq \pm\infty$.
Could you confirm if my understanding is correct? Thank you so much for your elaboration!
We start from the translation to the $c$-concave case of the definition of cyclical monotonicity.
Definition 1.36. Once a function $c: \Omega \times \Omega \rightarrow \mathbb{R} \cup\{\color{blue}{+\infty}\}$ is given, we say that a set $\Gamma \subset$ $\Omega \times \Omega$ is $c$-cyclically monotone (briefly $c$-CM) if, for every $k \in \mathbb{N}$, every permutation $\sigma$ and every finite family of points $\left(x_1, y_1\right), \ldots,\left(x_k, y_k\right) \in \Gamma$ we have $$ \sum_{i=1}^k c\left(x_i, y_i\right) \leq \sum_{i=1}^k c\left(x_i, y_{\sigma(i)}\right) . $$
Theorem 1.38. If $\gamma$ is an optimal transport plan for the cost $c$ and $c$ is continuous, then $\operatorname{spt}(\gamma)$ is a c-CM set.
Proof. Suppose by contradiction that there exist $k, \sigma$ and $\left(x_1, y_1\right), \ldots,\left(x_k, y_k\right) \in \operatorname{spt}(\gamma)$ such that $$ \sum_{i=1}^k c\left(x_i, y_i\right)>\sum_{i=1}^k c\left(x_i, y_{\sigma(i)}\right) $$ Take now $\varepsilon<\frac{1}{2 k}\left(\sum_{i=1}^k c\left(x_i, y_i\right)-\sum_{i=1}^k c\left(x_i, y_{\sigma(i)}\right)\right)$. By continuity of $c$, there exists $r$ such that for all $i=1, \ldots, k$ and for all $(x, y) \in B\left(x_i, r\right) \times B\left(y_i, r\right)$ we have $\color{blue}{c(x, y)>c\left(x_i, y_i\right)-\varepsilon}$ and for all $(x, y) \in B\left(x_i, r\right) \times B\left(y_{\sigma(i)}, r\right)$ we have $c(x, y)<c\left(x_i, y_{\sigma(i)}\right)+\varepsilon$. Now consider $V_i:=B\left(x_i, r\right) \times B\left(y_i, r\right)$ and note that $\gamma\left(V_i\right)>0$ for every $i$, because $\left(x_i, y_i\right) \in \operatorname{spt}(\gamma)$. Define the measures $\gamma_i:=\gamma\left\llcorner V_i / \gamma\left(V_i\right)\right.$ and $\mu_i:=\left(\pi_x\right)_{\#} \gamma_i, v_i:=\left(\pi_y\right)_{\#} \gamma_i$. Take $\varepsilon_0<\frac{1}{k} \min _i \gamma\left(V_i\right)$.
For every $i$, build a measure $\widetilde{\gamma}_i \in \Pi\left(\mu_i, v_{\sigma(i)}\right)$ at will (for instance take $\widetilde{\gamma}_i=\mu_i \otimes$ $\left.v_{\sigma(i)}\right)$. Now define $$ \widetilde{\gamma}:=\gamma-\varepsilon_0 \sum_{i=1}^k \gamma_i+\varepsilon_0 \sum_{i=1}^k \widetilde{\gamma}_i $$
We want to find a contradiction by proving that $\widetilde{\gamma}$ is a better competitor than $\gamma$ in the transport problem, i.e. $\widetilde{\gamma} \in \Pi(\mu, v)$ and $\int c d \widetilde{\gamma}<\int c \mathrm{~d} \gamma$
First we check that $\widetilde{\gamma}$ is a positive measure. It is sufficient to check that $\gamma-\varepsilon_0 \sum_{i=1}^k \gamma_i$ is positive, and, for that, the condition $\varepsilon_0 \gamma_i<\frac{1}{k} \gamma$ will be enough. This condition is satisfied since $\varepsilon_0 \gamma_i=\left(\varepsilon_0 / \gamma\left(V_i\right)\right) \gamma\left\llcorner V_i\right.$ and $\varepsilon_0 / \gamma\left(V_i\right) \leq \frac{1}{k}$ Now, let us check the condition on the marginals of $\widetilde{\gamma}$. We have $$ \begin{gathered} \left(\pi_x\right)_{\#} \widetilde{\gamma}=\mu-\varepsilon_0 \sum_{i=1}^k\left(\pi_x\right)_{\#} \gamma_i+\varepsilon_0 \sum_{i=1}^k\left(\pi_x\right)_{\#} \widetilde{\gamma}_i=\mu-\varepsilon_0 \sum_{i=1}^k \mu_i+\varepsilon_0 \sum_{i=1}^k \mu_i=\mu \\ \left(\pi_y\right)_{\#} \widetilde{\gamma}=v-\varepsilon_0 \sum_{i=1}^k\left(\pi_y\right)_{\#} \gamma_i+\varepsilon_0 \sum_{i=1}^k\left(\pi_y\right)_{\#} \widetilde{\gamma}_i=v-\varepsilon_0 \sum_{i=1}^k v_i+\varepsilon_0 \sum_{i=1}^k v_{\sigma(i)}=v . \end{gathered} $$ Finally, let us estimate $\int c \mathrm{~d} \gamma-\int c d \widetilde{\gamma}$ and prove that it is positive, thus concluding the proof. We have $$ \begin{aligned} \color{blue}{\int c \mathrm{~d} \gamma-\int c d \widetilde{\gamma}} &=\varepsilon_0 \sum_{i=1}^k \int c \mathrm{~d} \gamma_i-\varepsilon_0 \sum_{i=1}^k \int c d \widetilde{\gamma}_i \\ & \geq \varepsilon_0 \sum_{i=1}^k\left(c\left(x_i, y_i\right)-\varepsilon\right)-\varepsilon_0 \sum_{i=1}^k\left(c\left(x_i, y_{\sigma(i)}\right)+\varepsilon\right) \\ &=\varepsilon_0\left(\sum_{i=1}^k c\left(x_i, y_i\right)-\sum_{i=1}^k c\left(x_i, y_{\sigma(i)}\right)-2 k \varepsilon\right)>0 \end{aligned} $$ where we used the fact that $\gamma_i$ is concentrated on $B\left(x_i, r\right) \times B\left(y_i, r\right), \widetilde{\gamma}_i$ on $B\left(x_i, r\right) \times$ $B\left(y_{\sigma(i)}, r\right)$, and that they have unit mass (due to the rescaling by $\gamma\left(V_i\right)$ ).
