I am reading the paper "OPTIMAL INEQUALITIES IN PROBABILITY THEORY: A CONVEX OPTIMIZATION APPROACH" by BERTSIMAS and POPESCU. In the paper, the authors derived a duality problem for an infinite-dimensional optimization problem. I am not sure how to derive the following:


The primal is:


$\max\_\mu \quad \int\_S \textbf{1}d\mu$


subject to $\int\_\Omega \bar z^kd\mu=\sigma\_\kappa$, $\forall \kappa\in J\_k$.


The dual is:


$\min\_{y\in \mathcal{R}^{|J\_k|}} \quad \sum\_{\kappa\in J\_k}y\_\kappa \sigma\_\kappa$


subject to $g(\bar z)=\sum\_{\kappa\in J\_k}y\_\kappa\bar z^\kappa\geq 1$, $\forall \bar z \in S$, and $g(\bar z)=\sum\_{\kappa\in J\_k}y\_{\kappa}\bar z^\kappa\geq 0$, $\forall \bar z\in\Omega$.


In the above, $\mu$ is a probability measure and $S\subseteq \Omega\subseteq\mathcal{R}^n$.
Moreover, $\bar z=(z\_1,\ldots,z\_n)'$, $\kappa=(k\_1,\ldots,k\_n)'$, $\bar z^\kappa=z\_1^{k\_1}\cdots z\_n^{k\_n}$, and
$$
J\_k=\{ \kappa=(k\_1,\ldots,k\_n)'|k\_1+\cdots+k\_n\leq k,~k\_j\in\mathcal{Z}\_+,~j=1,\ldots,n \}.
$$
Could anyone tell me how to deal with this kind of infinite-dimensional optimization problem?



Duality problem of an infinite dimensional optimization problem
This is a special case (with $f=1\_S$) of the duality
$$s=i,\tag{1}$$
where
$$s:=\sup\Big\{\int f\,d\mu\colon\mu\text{ is a measure, }\int g\_j\,d\mu=c\_j\ \;\forall j\in J\Big\},$$
$$i:=\inf\Big\{\sum b\_j c\_j\colon f\le\sum b\_jg\_j\Big\},$$
$\int:=\int\_\Omega$, $\sum:=\sum\_{j\in J}$, $f$ and the $g\_j$'s are given measurable functions, the $c\_j$'s are given real numbers, and $J$ is a finite set such that (say) $0\in J$, $g\_0=1$, and $c\_0=1$, so that the restriction $\int g\_0\,d\mu=c\_0$ means that $\mu$ is a probability measure.


In turn, (1) is a special case of the von Neumann-type minimax duality
$$IS=SI,\tag{2}$$
where
$$IS:=\inf\_b\sup\_\mu L(\mu,b),\quad SI:=\sup\_\mu\inf\_b L(\mu,b),$$
$\inf\_b$ is the infimum over all $b=(b\_j)\_{j\in J}\in\mathbb R^J$, $\sup\_\mu$ is the supremum over all probability measures $\mu$ over $\Omega$, and $L$ is the Lagrangian given by the formula
$$L(\mu,b):=\int f\,d\mu-\sum b\_j\Big(\int g\_j\,d\mu-c\_j\Big)
=\int \Big(f-\sum b\_j g\_j\Big)\,d\mu+\sum b\_j c\_j.$$


Indeed, $\inf\_b L(\mu,b)=\int f\,d\mu$ if $\int g\_j\,d\mu=c\_j$ for all $j$, and $\inf\_b L(\mu,b)=-\infty$ otherwise. So,
$$SI=s.\tag{3}$$


On the other hand,
$$IS=i.\tag{4}$$
Indeed,
\begin{align}
IS&=\inf\_b\Big\{\Big[\sup\_\mu \int \Big(f-\sum b\_j g\_j\Big)\,d\mu\Big]+\sum b\_j c\_j\Big\} \\ 
&=\inf\_b\Big\{\Big[\sup\Big(f-\sum b\_j g\_j\Big)\Big]+\sum b\_j c\_j\Big\}, 
\end{align}
which is clearly no greater than $i$. On the other hand, if for some $b$ we have $s\_b:=\sup\big(f-\sum b\_j g\_j\big)\in\mathbb R$, then $f\le\sum \tilde b\_jg\_j$ and
$$\sum\tilde b\_jc\_j=s\_b+\sum b\_jc\_j=\Big[\sup\Big(f-\sum b\_j g\_j\Big)\Big]+\sum b\_j c\_j,$$
where $\tilde b\_j:=b\_j+s\_b\,1\_{j=0}$. So, $i$ is no greater than $\inf\_b\big\{\big[\sup\big(f-\sum b\_j g\_j\big)\big]+\sum b\_j c\_j\big\}=IS$. Thus, (4) is verified as well.


So, by (3) and (4), (1) indeed follows from (2).


In turn, the von Neumann-type minimax duality (2) follows under [general conditions](https://en.wikipedia.org/wiki/Minimax_theorem) when $L(\mu,b)$ is affine in $\mu$ and in $b$ (as it is in our case). A necessary and sufficient condition for the minimax duality
$$\inf\_y\sup\_x F(x,y)=\sup\_x \inf\_y F(x,y)$$
whenever $F(x,y)$ is concave in $x$ and convex in $y$ was given in [this paper](https://link.springer.com/content/pdf/10.1007/BF01158303.pdf).


