In this paper: Matching for teams by G. Carlier and I. Ekeland (2010), <https://www.jstor.org/stable/25619994?seq=1#metadata_info_tab_contents>: they claim that follows a standard argument in Ekeland and Teman (1999), we rewrite


$$\inf\_{\nu\in\mathcal{M}(Z)} \sum\_{j=0}^N F\_j^\*(\nu)=-(\sum\_{j=0}^N F\_j^\*)^\*(0) \qquad (\*)$$
where $F^\*$ is the Fenchel transform on page 407.


Here for $\varphi\in C^0(Z,\mathbb{R})$, define
$$F\_j(\varphi):=-\int\_{X\_j} \varphi\_j^{c\_j}d\mu\_j$$
where $\varphi\_{j^{c\_j}}=inf\_{z\in Z}\{c(x,z)-\varphi(z)\}$.


I am confused about using which argument to get this identity $(\*)$. Why can we have


$$\inf F\_j^\*(\nu) =-(F\_j^\*)^\*(0) ?$$


I guess we need some arguments in convex analysis.



One question about how to get $\inf_{\nu\in\mathcal{M}(Z)} \sum_{j=0}^N F_j^*(\nu)=-(\sum_{j=0}^N F_j^*)^*(0)$?
Just use the definition of the Fenchel transform:


Define $G(\nu) = \sum\_{j=0}^N F^\ast\_j(\nu)$ and calculate $G^\ast(0)$.


