# On compactness in Sion's minimax theorem

Sions minimax theorem (wiki, paper) can be stated as follows:

Let $X$ be a compact convex subset of a linear topological space and $Y$ a convex subset of a linear topological space. Let $f$ be a real-valued function on $X \times Y$ such that

$1.$ $f(x, \cdot)$ is upper semicontinuous and quasi-concave on $Y$ for each $x \in X$.

$2.$ $f( \cdot, y)$is lower semicontinuous and quasi-convex on $X$ for each $y \in Y$.

Then: $$\min_{x \in X}\sup_{y \in Y}f(x,y) = \sup_{y \in Y}\min_{x \in X}f(x,y)$$

The result of Sion's minimax theorem can be restated as

$$\min_{h_l \in X}\sup_{h_k \in Y}f(h_l,h_k) = \sup_{h_k \in Y}\min_{h_l \in X}f(h_l,h_k)$$

where $h_l$ and $h_k$ are continuous functions on $X$ and $Y$, respectively, and $f$, is then a functional.

Let

$$h_l^\kappa(x) = \begin{cases} 0, &{l}(x)<\rho \\ \kappa(x), & {l}(x)=\rho\\ 1, & {l}(x)> \rho \end{cases}$$

where $l$ is a positive continuous function on a closed interval of real numbers $I$, $\kappa\in [0,1]$ is an increasing function on an interval $A=\{x:l(x)=\rho\}$, $\lambda(A)>0$, where $\lambda$ is the Lebesgue measure, and $\min_x l(x)<\rho<\max_x l(x)$ is a positive number.

Let $$X=\{h_l^\kappa:\forall l,\forall \kappa\}$$

$Q_1:$ Do we need $X$ to be a compact space w.r.t. say supremum norm such that Sion's theorem holds? If yes what is a counterexample sucht that $X$ is not compact and Sion's minimax theorem fails?

My work on this problem is based on the Arzelà–Ascoli's theorem:

$1.$ It is true that $|h_l^k(x)|\leq M=1$ for all $x$ and $l$, which implies uniform boundedness

$2.$ But $\kappa$ can be chosen arbitrarily on $A$ as long as it is increasing. This means one can choose a sequence of $\kappa_n$ on $A$ such that $\kappa_n$, $n=1,2..$ is not an equicontinuous family. Therefore, $X$ is not compact.

When I look back at Sion's theorem, It seems that I need the compactness argument such that the minimum exists. Namely, there must be an $h_l^k\in X$, which minimizes $f$. If the domain of $h_l^k$ would be $\mathbb{R}$, It would be easy to take a sequence of functions $h_{l_n}^k$, $n=1,2..$ each belonging to $X$ and the minimizing $h_l^k$ could be obtained for $\lim_{n\to\infty}h_{l_n}^k$, clearly implying that Sion's theorem would fail due to lack of compactness. But the domain is given as a closed interval; $I\subset \mathbb{R}$ and I have difficulties to understand why I must expect that for $I$, Sion's minimax theorem can fail? All functions are on $I$ and all are bounded by $1$ and at least one of them will be a minimizer and I can determine it.

Thanks for reading and for any help.

• If you want minmax theorem to hold for noncompact case $X$ and $Y$ you need an extra assumption on $f(x,y)$, say the following one: there exists $(x_{0}, y_{0})\in X\times Y$ such that $\lim_{|x|\to ]\infty}f(x,y_{0})=-\infty$ and $\lim_{|y|\to \infty} f(x_{0},y)=+\infty$. See Proposition 2.2, page 173 in I. Ekelan, R. Temam, Convex analysis and variational problems, North-Holland publishing Company, Amsterdam, 1979. There are also weaker assumptions: for example "sup-inf compactness at some $(x_{0},y_{0})$"... – Paata Ivanishvili Jul 14 '17 at 16:24
• Ekeland -- missing d at the end – Aryeh Kontorovich Jul 15 '17 at 18:27

Here is a counterexample where neither $X$ nor $Y$ are compact. Consider $f(x,y)=y/(x+y)$ on $X\times Y$, where $X=Y=[1,\infty)$. Then $$1=\inf_x\sup_y f(x,y)>\sup_y\inf_x f(x,y)=0.$$
• or $f(x,y)=x+y$ on $\mathbb{R}\times \mathbb{R}$. – Paata Ivanishvili Jul 14 '17 at 16:40