1
$\begingroup$

For any $r\equiv(r_1,...,r_M)\in \mathbb{R}$, consider the function $$a\equiv(a_1,...,a_M) \in \mathbb{R}^M\mapsto G(a)\equiv \max_{i\in \{1,...,M\}} (a_i+ r_i)$$

Is $G$ strictly convex or convex? Could you help me to show it?

$\endgroup$
2
  • $\begingroup$ it's definitely convex, I don't think it's strictly convex though. $\endgroup$ – TSF May 25 '18 at 11:43
  • $\begingroup$ Any proof for non strict convexity @TonyS.F.? $\endgroup$ – TEX May 25 '18 at 11:53
2
$\begingroup$

The maximum of convex functions is convex.

See here.

Since $\phi_i(a)=a_i+r_i$ is convex for every $i$, then $G(a)=\max \{\phi_i(a)\} $ is convex.

Consider $M=1$ and $r=0$, since $G(a)=a$, $G$ is not strictly convex.

$\endgroup$
2
  • $\begingroup$ Thank you very much. Two clarifications: 1) $\phi_i$ is linear and hence convex and concave, correct? 2) $G(a)=a$ implies $G$ not strictly convex because it is a linear function (hence convex and concave), correct? $\endgroup$ – TEX May 25 '18 at 12:00
  • $\begingroup$ $\phi_i$ is affine, but that does not change the conclusion. You are correct. $\endgroup$ – nicomezi May 25 '18 at 12:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.