Leland Morrow

2022-06-24

Define the function
$f\left(a,b,c,\alpha ,\beta ,\gamma ,x\right)=max\phantom{\rule{negativethinmathspace}{0ex}}\left(0,\phantom{\rule{thinmathspace}{0ex}}max\phantom{\rule{negativethinmathspace}{0ex}}\left(\left(a+x\right)\alpha ,\left(b+x\right)\beta \right)-\left(c+x\right)\gamma \right),$
where
$a,b,c,\alpha ,\beta ,\gamma ,x\in \left[0,M\right].$
Is it true that for any
$\xi =\left(a,b,c,\alpha ,\beta ,\gamma \right),$
the maximum of $f\left(\xi ,\cdot \right)$ occurs either when $x=0$ or $x=M$?

I think the answer is yes, but I have trouble prooving it.
My argument is as follows:
Given any $\xi$, $f\left(\xi ,c\right)$ will be
$f\left(\xi ,x\right)=\left\{\begin{array}{ll}0& \text{case A}\\ \left(a+x\right)\alpha -\left(c+x\right)\gamma & \text{case B}\\ \left(b+x\right)\beta -\left(c+x\right)\gamma & \text{case C}\end{array}$
Hence, $f\left(\xi ,x\right)$ is a linear function of $x$ in all three cases, and the result follows.

I think this argument works only if each case is independent of $x$, but this is not the case.

As, when $\xi$ is given, judiciously choosing $x$ may put $f$ in another case.

What would be a right way to prove this result?

Leland Ochoa

Expert

It appears that $f$ is a concatenation of convex functions (namely "max" and linear operations), and therefore $f$ is convex. It follows that one of the end points will always be maximal.