Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been trying to prove this statement the whole weekend...

Let $g_1,\dots,g_m$ be concave functions on $\mathbb{R}^n$. Prove that the set $S=\{x:g_i(x)\geq{0},\ i=1,\dots,m\}$ is convex.

share|cite|improve this question
up vote 6 down vote accepted

Take $x,y\in S$ and $a\in [0,1]$ and $i\in\{1,\dots,m\}$. Then by concavity of $g_i$, $$g_i(ax+(1-a)y)\geq ag_i(x)+(1-a)g_i(y).$$ This quantity is non-negative, because so are $a$, $1-a$, $g_i(x)$ and $g_i(y)$. We conclude that $ax+(1-a)y\in S$, hence $S$ is convex.

share|cite|improve this answer
    
Oh! Ok I see it now. I just somehow couldn't see the logical leap from g's concavity to x and y's convexity. Thank you so much! – Jeff Jul 2 '12 at 9:54
    
You are welcome. – Davide Giraudo Jul 2 '12 at 10:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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