I would like to compute $\sup_{x}\inf_{y} 4yx-x-2y$ and $\inf_{y}\sup_{x} 4yx-x-2y$ where $x,y\in[0,1]$.
Since I had no idea to how to tackle the problem, I tried to take the derivative (1) with respect to $x$ when looking for the infimum over $y$ and (2) the derivative with respect to $y$ when looking for the supremum over $x$.
I get for (1)
$$ 4x-2 =0 \implies x=\frac{1}{2} $$
Thus plug this in the expression I found $2y -\frac{1}{2} - 2y = -\frac{1}{2}$ thus the supremum over $x$ of $4yx-x-2y$ seems to be equal to $-\frac{1}{2}$ and thus $\inf_{y}\sup_{x} 4yx-x-2y= -\frac{1}{2}$.
Similarly when I take the derivative with respect to $y$ for (2) I get
$$ 4y-1 =0\implies y=\frac{1}{4} $$
Plug in this un the expression I get $x-x-\frac{1}{2}$ so $\sup_{x}\inf_{y} 4yx-x-2y=-\frac{1}{2}$
I think the result is good because I should find that there are equal which is the case. However I do not understand why it works by taking the derivative with respect to the variable $x$ when I want to find the infimum over $y$ and vice versa.
Someone has an explanation please ?
It worked because the critical point $(\frac{1}{2},\frac{1}{4})$ of $f(x,y)=4yx-x-2y$ is a saddle point, but I do not recommend the use of derivative here. It suffices to apply the definition of $\sup$ and $\inf$ in the given order.
As regards the first one we have that, if $x\in [0,1]$ then $$\inf_{y\in [0,1]} (4yx-x-2y)=\inf_{y\in [0,1]} (2y(2x-1))-x= \begin{cases} 0(2x-1)-x=-x &\text{if $x\in [\frac{1}{2},1]$,}\\ 2(2x-1)-x=3x-2 &\text{if $x\in [0,\frac{1}{2}]$.}\\ \end{cases}$$ Hence $$\sup_{x\in [0,1]}(\inf_{y\in [0,1]} (4yx-x-2y))=-\frac{1}{2}.$$ Can you verify that $\inf_{y\in [0,1]}(\sup_{x\in [0,1]} (4yx-x-2y))=-\frac{1}{2}$ by using a similar approach?