Consider a minimax problem $$\inf_{x\in X} \sup_{y\in Y} L(x,y)$$ where $L$ is convex in $x$ and concave in $y$. Assume that $$\inf_x \sup_y L(x,y) = \sup_y \inf_x L(x,y)$$ and all the inf's and sup's are attainable(if they are not infinity). I also want to add that $X$, $Y$ are typically closed convex set in $\mathbb{R}^n$(not necessarily compact).
The background of the problem is that such convex-concave saddle point problems often occur in the primal-dual formulation of convex optimization. For example, in a standard linear programming, the primal-dual problem is $\min_{x\ge 0}\max_{y\in \mathbb{R}^m} L(x,y)= c^T x+y^T b-y^T A x$.
Now let $z=(x,y)$ and $Z^*$ denote the solution set of $z$ for the minimax problem $L(z)=L(x,y)$.
Let $X^*$ denote the set of $x$ which minimizes $\sup_y L(x,y)$. Similarly, let $Y^*$ denote the set of $y$ which maximizes $\inf_x L(x,y)$.
I wonder if there is some relation beween $Z^*$ and $X^* \times Y^*$. I want to show that $X^*\times Y^* = Z^*$. I have shown that $X^*\times Y^* \subset Z^*$, but I cannot prove the other way. Can somebody give me a hint on whether the statement is true or not?
Remark 1 This is a "concurrent" question with this. That question is still open and I tried to answer it but finally found that the two questions are equivalent, i.e., if I can show that $Z^*=X^*\times Y^*$, then I can also show that the solution of a minimax problem is a saddle point and vice versa.
Remark 2 To show $X^*\times Y^* \subset Z^*$, we only need to notice that for any $(x^*,y^*)\in (X^*,Y^*)$,
$$\sup_{y} \inf_{x} L(x,y)= \inf_{x} L(x,y^*)\le L(x^*,y^*)\le \sup_y L(x^*,y)= \inf_x \sup_{y} L(x,y) $$ Since we also assume minimax equality holds, $\sup_{y} \inf_{x} L(x,y)= \inf_x \sup_{y} L(x,y)$, the equalities holds throughout above.
