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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.

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1 Answer 1

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Just use the definition of the Fenchel transform:

Define $G(\nu) = \sum_{j=0}^N F^\ast_j(\nu)$ and calculate $G^\ast(0)$.

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  • $\begingroup$ Can you explain more about the definition? Thanks. $\endgroup$
    – Hermi
    Dec 14, 2020 at 0:50
  • $\begingroup$ $G^\ast(\zeta) = \sup_\nu\{\langle\nu,\zeta\rangle - G(\nu)\}$ $\endgroup$
    – masi
    Dec 15, 2020 at 10:23

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