Find $$\min_{y \in \mathbb{R}} \max_{0 \le x \le 1} |x^2 - xy|.$$
I haven't really tried much yet, and I'm not sure how to start this. I tried setting specific values for $x$ in the interval, then solving the problem, but it didn't quite work out. I would greatly appreciate it if anyone could be me any hints!
Thanks in advance!!
Imagin to fix a certain value of $y\in \mathbb{R}$, because $f(x)=x^2-xy$ is a parable, its maximum in the interval $[0,1]$ can only be realized in $0$ or in $1$, so the maximum of $|f(x)|$ in $[0,1]$ can only be realized in $0$, in $1$ or in the vertice of the parable, in other words it can only be $|f(0)|=0$ or $|f(1)|=|1-y|$ or $|f(y/2)|=y^2/4$.
In particular when $y^2/4<|1-y|$ the maximum of $|f|$ is $|1-y|$, in the other cases the maximum is $y^2/4$.
It's easy to compute that $y^2/4<|1-y|$ iff $y\in (-2\sqrt{2}-2,2\sqrt{2}-2)$, so we obtain that if $y\in (-2\sqrt{2}-2,2\sqrt{2}-2)$ then $g(y):=\max\limits_{x\in[0,1]}|x^2-xy|=|1-y|=1-y$, otherwise $g(y):=\max\limits_{x\in[0,1]}|x^2-xy|=y^2/4$.
Now unfix $y \in \mathbb{R}$ and look for the minimum $\min\limits_{y\in\mathbb{R}}g(y)$, it can be founded studying separately the case $y\in (-2\sqrt{2}-2,2\sqrt{2}-2)$ and $y\in\mathbb{R}\setminus (-2\sqrt{2}-2,2\sqrt{2}-2)$ and we see that it is $3-2\sqrt{2}$.