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For example, consider $y$ such that $$ y=\cases{c, & if $c\le0$,\\0, & if $c>0$.} $$ Then, it can be easily expressed as $$y=\min\{c, 0\}.$$

Using the above trick, can I express the following $y$ using $\max$ or $\min$ ? $$ y = \cases{c, & if $c\le d$,\\ 0, & if $c>d$.}$$

Danny_Kim
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    $\max$ and $\min$ are continuous functions, whereas your function is not continuous. So there is no obvious solution that I can see. – TonyK May 26 '20 at 08:51
  • @TonyK Your brief answer gave me a lot of knowledge. Thank you. – Danny_Kim May 26 '20 at 08:53

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The second function is discontinuous, you will need an extra "step" term, like

$$y=\min\{c-d,0\}+d\cdot(c\le d).$$

If you don't mind having an undefined value at $c=d$, you can "emulate" with

$$y=\min\{c-d,0\}\frac c{c-d}.$$

  • Actually, I wanted to make it using max or min or both, but I knew that it is impossible. Thank you for proposing another way to make it. – Danny_Kim May 26 '20 at 08:56