To find $h=\max_{a,b}(\min\{a, f(a,b)\}), a\geq 0, b\geq 0$

Question

How to find $$h=\max_{a,b}(\min\{a, f(a,b)\}), a\geq 0, b\geq 0$$ where $f$ is some fixed non negative continuous function. My teacher told me put $a=f(a,b)$ and then solve $b$ in terms of $a$. But I am not getting the point. Usually this type of $h$ is needed to find interval of existence of unique solution by Existence and uniqueness theorem of Picard for ordinary differential equation. I have confusion when $a<f(a,b)$ or $a>f(a,b)$, then how can we put $a=f(a,b)$? Please clear how to find this $h$ in easy way and how by putting $a=f(a,b)$ is working?

Answer 1

we are given $$h= \max_{a,b} \min(a,f(a,b)); a \ge 0, b \ge 0$$ The $\min$ function either selects its first or second argument (or they are equal in which case it doesn't matter which argument is selected.)

case 1 (arg 1 selected): $$ h= \max_{a,b} a; f(a,b) \ge a \ge 0, b \ge 0$$

case 2 (arg 2 is selected): $$ h= \max_{a,b} f(a,b); a \ge f(a,b), a \ge 0, b \ge 0$$

In the first case, we can increase $a$ until $a=f(a,b)$, but no further, and so $h=a$ when $a=f(a,b)$ In the second case, $f(a,b)$ might be able to be increased until $f(a,b)=a$, but no further, in which case $h=f(a,b)$ when $f(a,b)=a$. However, we have to consider the possibility that $\forall a,b. f(a,b) \lt a$. There are again two cases. In the first case, $f$ may take on any value and still have $f(a,b)<a$. For example, if $f$ were defined such that $f(a,b)=a-1$. In that case, the "maximum" value, if it is said to exist, would be $h=f(a,b)=\infty$, and since $f(a,b)<a$ then $a=\infty$. We might write this as $f(\infty,b)=\infty$

The final case is that $f(a,b)$ reaches some finite maximal value $k$ for any $a,b$ while $\forall a,b. a\gt f(a,b) \ge 0$. An example of such a function would be $$f(a,b)=\min(k,\max(0,a-1)$$

It is clear that in that case $h=k$ and the solution is not related to setting $f(a,b)=a$