Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $f,g$ be two convex function on $D\in\mathbb{R}$ (what about $\mathbb{R}^n$?) satisfying that there is no point $x$ in $D$ such that $f(x)<0$ and $g(x)<0$ at the same time.

I want to prove that there exists $p\in(0,1)$ such that $pf+(1-p)g\geq0$ on $D$.

I'm not sure whether we need to add some constraints on $D$, e.g. $D$ is compact.

Any idea?

share|cite|improve this question
up vote 1 down vote accepted

There is some $a$ with $f(a) < 0$ and some $b$ with $g(b) < 0$, else you could take $p=1$ or $0$. By assumption, $g(a) \ge 0$ and $f(b) \ge 0$. Assume without loss of generality $a < b$. By the Intermediate Value theorem there is $c$ with $a < c < b$ and $f(c) = g(c)$, and by assumption this common value $\ge 0$. Now by convexity $f(x) \ge f(c) + f'(c) (x-c)$ and $g(x) \ge f(c) + g'(c) (x-c)$ for all $x \in D$, and so $p f(x) + (1-p) g(x) \ge f(c) + (p f'(c) + (1-p) g'(c)) (x-c)$. Moreover, $f'(c) > 0$ and $g'(c) < 0$. Take $p = \dfrac{-g'(c)}{f'(c) - g'(c)}$. Then $0 < p < 1$ and $p f'(c) + (1-p) g'(c) = 0$, so that $p f(x) + (1-p) g(x) \ge f(c) \ge 0$.

EDIT: If $f$ or $g$ is not differentiable at $c$, take the left or right one-sided derivative.

share|cite|improve this answer
Beautiful proof, thank you! – hxhxhx88 Nov 15 '12 at 6:29
Do you have any idea about high dimension case? – hxhxhx88 Nov 15 '12 at 6:38

Your Answer


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.