Sions minimax theorem ([wiki](http://en.wikipedia.org/wiki/Sion%27s_minimax_theorem), [paper](http://projecteuclid.org/download/pdf_1/euclid.pjm/1103040253)) 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](http://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_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.



On compactness in Sion's minimax theorem
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
$$ 1=\inf\_x\sup\_y f(x,y)>\sup\_y\inf\_x f(x,y)=0.$$


