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Let $X$ be a compact subset of $\mathbb{R}$ and $c(x_1,x_2,x_3,x_4)$ be a fixed bounded continuous functions on $X^4$. Assume $\mu,\nu$ are probability measures on $X^2$, and $\mu\otimes\nu$ is the product measure on $X^4$ (i.e. $\mu\otimes\nu(A\times B)=\mu(A)\nu(B)$). Now consider the problem \begin{equation} K(c):=\sup_{f,g,h}\left\{\int f(x_1,x_2)d\mu+\int g(x_3,x_4)d\nu+\int \hat{h}(x_1,x_2,x_3,x_4)d(\mu\otimes\nu) \right\}, \end{equation} where $f,g,h$ satisfying:

(1)$f,g,h$ are all bounded continuous functions on $X^2$;

(2)$\hat{h}(x_1,x_2,x_3,x_4):=h(x_2,x_3)$;

(3)$f(x_1,x_2)+g(x_3,x_4)+h(x_2,x_3)\leq c(x_1,x_2,x_3,x_4)$.

Now I want to show there exists a triple $(f_0,g_0,h_0)$ such that the supreme $K(c)$ can be obtained at it.

I was thinking about constructing a maximizing sequence $\{(f_i,g_i,h_i)\}_{i=1}^{\infty}$, then use Arzel`a-Ascoli theorem to obtain a uniformly convergent subsequence. But I stuck at showing equicontinuity of this function family.

Indeed, if there is no such $h$ term, this problem can be solved by some standard conjugate argument, for example see Proposition 1.11 of this book which heavily relies on the fact that $f$ and $g$ are separated.

So my question is, are there any possible methods to show the equicontinuity of the maximizing sequence, or directly prove the existence of $(f_0,g_0,h_0)$?

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  • $\begingroup$ Is there anything in your assumptions which implies the uniform boundedness of the sequence $\{(f_i,g_i,h_i)\}$? (3) would imply this if $f,g,h \geq 0$. $\endgroup$
    – Dieter Kadelka
    Commented Dec 7, 2019 at 11:32
  • $\begingroup$ @DieterKadelka No, once we have equicontinuity, then we can assume $f_n, g_n$ equals to 0 at some fixed point, then uniform boundedness can be obtained since $X$ is compact $\endgroup$
    – aurora_borealis
    Commented Dec 7, 2019 at 12:07
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    $\begingroup$ Just a remark: as the problem is stated, you can't prove the equicontinuity of all maximizing sequences: for instance if the support of $\mu$ is smaller than $X^2$, the value of the integral remains unchanged changing $f$ in the open set $X^2\setminus \text{supp} (\mu)$. Therefore you may always modify any maximizing sequence making it not equicontinuous. $\endgroup$
    – Pietro Majer
    Commented Dec 7, 2019 at 13:14
  • $\begingroup$ @PietroMajer Yes you are right, so actually I want to show there EXISTS a equicontinuous maximizing sequence so that I can apply Arzela-Ascoli theorem $\endgroup$
    – aurora_borealis
    Commented Dec 7, 2019 at 13:24

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