# How to prove or disprove this proposition: convex combination of two convex function is non-negative (under some assumption)

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?

-

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.