# $\max_{i\in \{1,…,M\}} (a_i+ r_i)$ strictly convex?

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?

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

## 1 Answer

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.

• 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? – TEX May 25 '18 at 12:00
• $\phi_i$ is affine, but that does not change the conclusion. You are correct. – nicomezi May 25 '18 at 12:04