I would like to know if the proof of the following is correct : consider $u(x,y)$ a real valued bounded function defined on $X\times Y$ , then we have
$$ \sup_{x\in X}\inf_{y\in Y}u(x,y)\leq\inf_{y\in Y}\sup_{x\in X}u(x,y) $$
My proof is : first we notice that infinimum and supremum are all well defined since $u$ is bounded. Now for all $(x,y)$ we have $\inf_{y\in Y}u(x,y)\leq u(x,y)\leq\sup_{x\in X}u(x,y)$. Then we notice that $u(x,y)$ is a majorant (a minorant) of $q(x) = \inf_{y\in Y}u(x,y)$ ( respectively $w(y) = \sup_{x\in X}u(x,y)$). So there exists $x_0$ and $y_0$ such that $\sup_{x\in X}q(x)\leq u(x_0,y)$ and $\inf_{y\in Y}w(y)\geq u(x,y_0)$. Thus we get for $(x_0,y_0)$
$$ \sup_{x\in X}\inf_{y\in Y}u(x,y)\leq u(x_0,y_0)\leq\inf_{y\in Y}\sup_{x\in X}u(x,y) $$
Is this seems correct please ? Thank you a lot
The proof is clearly incorrect, because it implies the existence of a saddle point $(x_0, y_0) \in X\times Y$. Consider the following real-valued bounded function $\mathbb{R}^2\ni (x,y)\mapsto u(x,y) = \arctan(x)$. There is no $(x_0,y_0) \in \mathbb{R}^2$ such that $$\pi/2 = \sup_{x\in\mathbb{R}} \inf_{y \in\mathbb{R}}\arctan(x) \leq \arctan(x_0) \leq \inf_{y \in\mathbb{R}}\sup_{x\in\mathbb{R}}\arctan(x) = \pi/2$$.
Basically you start from the right place, by writing that $$\inf_{\tilde{y}\in Y} u(x,\tilde{y}) \leq u(x,y) \leq \sup_{\tilde{x}\in X} u(\tilde{x},y), \qquad\forall(x,y)\in X\times Y.$$ Get rid of the middle term, i.e., $$\inf_{\tilde{y}\in Y} u(x,\tilde{y}) \leq \sup_{\tilde{x}\in X} u(\tilde{x},y), \qquad\forall(x,y)\in X\times Y.$$ Then simply note that LHS is independent from $y$, and take an infimum over $y$ on the right side. Similarly, RHS is independent from $x$, so we can take supremum over $x$ on the left side.