3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $(x_{0},y_{0})$"... $\endgroup$ – Paata Ivanishvili Jul 14 '17 at 16:24
  • 1
    $\begingroup$ Ekeland -- missing d at the end $\endgroup$ – Aryeh Kontorovich Jul 15 '17 at 18:27
2
$\begingroup$

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 $$ 1=\inf_x\sup_y f(x,y)>\sup_y\inf_x f(x,y)=0.$$

$\endgroup$
  • 2
    $\begingroup$ or $f(x,y)=x+y$ on $\mathbb{R}\times \mathbb{R}$. $\endgroup$ – Paata Ivanishvili Jul 14 '17 at 16:40
  • $\begingroup$ Even better -- and simpler! $\endgroup$ – Aryeh Kontorovich Jul 15 '17 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.