$\sup\limits_{x\in X}\left(\sup\limits_{y\in y}|f(x,y)|\right) = \sup\limits_{y\in Y}\left(\sup\limits_{x\in X}|f(x,y)|\right)$ Let $X,Y$ be non empty sets and $f: X \times Y \to \mathbb R$ be a bounded function. Since it's bounded we can define $s_1 : X\to \mathbb R$ by $$s_1(x_0)=\displaystyle \sup_{y\in Y}|f(x_0,y)|$$ and similarly $s_2 : Y\to \mathbb R$ by $$s_2(y_0)=\displaystyle \sup_{x\in X}|f(x,y_0)|$$ 
prove that $\displaystyle \sup_{x\in X} s_1(x)=\displaystyle \sup_{y\in Y} s_2(y)$ in other words
$$\displaystyle \sup_{x\in X}(\displaystyle \sup_{y\in y}|f(x,y)|) = \displaystyle \sup_{y\in Y}(\displaystyle \sup_{x\in X}|f(x,y)|)$$
What if we consider $\displaystyle \sup_{(x,y)\in X \times Y}|f(x,y)|$ it's the same result? If not there is a reasonable case in when it is?
I don't know how to prove this but I'll be very useful to know.

Since $f$ is bounded, let $u=\sup\{|f(x,y)|:\langle x,y\rangle\in X\times Y\}$. For any $\epsilon>0$ there is an $\langle x,y\rangle\in X\times Y$ such that $|f(x,y)|>u-\epsilon$; clearly $s_1(x)>u-\epsilon$ and $s_2(y)>u-\epsilon$. Thus,
$$u-\epsilon<\sup_{x\in X}s_1(x)\le u$$
and
$$u-\epsilon<\sup_{y\in Y}s_2(y)\le u$$
for all $\epsilon>0$, and it follows that
$$\sup_{x\in X}s_1(x)=\sup_{y\in Y}s_2(y)=u\;.$$
