Minimum of maximum function

Question

I'm having a bit of trouble understanding what the minimum of the maximum function means.

For example, let's consider fixed $a, b$, and we are looking to find: $$\min_{0\leq r\leq 1} \max\left(\frac{f(a)}{r},\frac{f(b)}{1-r}\right)$$

In this case, $f(a)$ and $f(b)$ are just values since we have fixed $a, b$. So we first want to find the maximum first, but how can we do that without knowing the value of $r$? What exactly does the above expression mean?

Answer 1

Consider a simpler problem: $$\min_{0 \le x \le 1} \max(x,1-x).$$ You want to find the value of $x\in[0,1]$ that minimizes $g(x)=\max(x,1-x)$. Notice that the arguments of the $\max$ are equal when $x=1/2$. Also, $g(x)=x$ if $x \ge 1/2$ and $g(x)=1-x$ if $x \le 1/2$. In both cases, $g(x) \ge 1/2$, so the minimum occurs when $x=1/2$, yielding $$\min_{0 \le x \le 1} \max(x,1-x)=\max(1/2,1-1/2)=1/2.$$

https://www.desmos.com/calculator/hcqyaxchgp

Answer 2

Actually in those case, it is important to be very precise on the notation :

For instance here one can have several cased

First

Find

$$ \min_{0\leq r \leq 1}\max_{a,b}(\frac{f(a)}{r},\frac{f(b)}{r-1})$$

Second

Let's $a$ fixed (resp. b)

Find

$$ \min_{0\leq r \leq 1}\max_{b}(\frac{f(a)}{r},\frac{f(b)}{r-1})$$

Conclusion and comments

You see here that the two problems are different, in the first case $(\frac{f(a)}{r})$ and $(\frac{f(b)}{r-1})$ can 'move' for $r$ fixed, to find the highest values depending on $r$ for $a$ and $b$ moving. Then $r$ is fixed by the seek of a minimum.

In the second one, $a$ (for instance) is fixed. So only $(\frac{f(b)}{r-1})$ can move in the maximum evaluation. Then $r$ can move to find the minimum.

I hope it is clear. Don't hesitate for further explainations.