An example of a function which satisfies $\sup_{y}\inf_{x}f(x,y)<\inf_{x}\sup_{y}f(x,y)$

Question

Let $f:X\times Y\to \mathbb{R}$. I could not come up with an example that satisfies $$\sup_{y\in Y}\inf_{x\in X}f(x,y)<\inf_{x\in X}\sup_{y\in Y}f(x,y).$$ Any help would be appreciated. Thanks!

Answer 1

I am not sure, but this seems work $$ f(x,y)=\begin{cases} 0 & x<y\\ 1 & x\geq y \end{cases} $$

Answer 2

We can put $X=Y=(0,1)$, and construct a function $f$ on $X\times Y$ such that for any fixed $y,$ $\inf_{x\in X}f(x,y)=0$, and for any fixed $x,$ $\sup_{y\in Y}f(x,y)=1$.

Let's start with a "framework function" $g$ defined on the set $$\bigl(\{0\}\times[0,1)\bigr)\cup\bigl([0,1)\times\{0\}\bigr)\cup\bigl(\{1\}\times(0,1]\bigr)\cup\bigl((0,1]\times \{1\}\bigr)$$ (this is just the boundary of $X\times Y$, with the points $\langle 0,1\rangle$ and $\langle 1,0\rangle$ removed), given by $$g(x,y)=\begin{cases}0 & x=0\text{ or }y=0\\1 & x=1\text{ or }y=1.\end{cases}$$ We'd like to extend $g$ to a continuous function on the original domain of $g$, together with $X\times Y$, so that the restriction of this extension to $X\times Y$ will have the desired property. It turns out to be simplest to use linear interpolation, here. I choose to consistently interpolate along line segments with ($xy$-plane) slope $1.$ (I'd be glad to include the details, but right now I'm typing on my phone, so it's rather inconvenient to do a bunch of LaTeX.) The function I develop in this way is $$f(x,y)=\frac{x+y-|x-y|}{2-2|x-y|},$$ or equivalently, $$f(x,y)=\frac{\min(x,y)}{1-|x-y|}.$$ This agrees with $g$ on the domain of $g$, and has the desired properties on $X\times Y$.

There are even simpler examples if you don't require $f$ to be continuous. Try $f(x,y)=0$ if $y<1-x$ and $f(x,y)=1$ otherwise.