Osgood condition Let $h$ and $g$ be continuous, non-decreasing and concave functions in the interval $[0,\infty)$ with $h(0)=g(0)=0$ and $h(x)>0$ and $g(x)>0$ for $x>0$ such that both satisfy the Osgood condition $$\int_{0+}\frac{dx}{f(x)}=\infty.$$ 
Does there exist a concave function $F$ such that $F(x)\geq h(x)$ and $F(x)\geq g(x)$ for all $x$, and satisfies the Osgood condition? 
 A: I think that there are plenty of redundant informations in this question. If $f(x)$ is positive and concave on $\mathbb{R}^+$ then it must be continuous (continuity is a consequence of concavity) and non-decreasing, since otherwise, assuming $a<b$ and $f(b)<f(a)$, the graphics of $f$ over $(b,+\infty)$, lying under the line through $(a,f(a))$ and $(b,f(b))$, must intersect the $x$-axis at some point $c>b$. A concave function is also almost-everywhere differentiable, so for almost every $z\in\mathbb{R}^+$ we have:
$$\forall x>z,\quad f(x)< f'(z)(x-z)+f(z),\qquad f(z),f'(z)>0$$
for the same reasons as above. It follows that:
$$\int_{z}^{M+z}\frac{dx}{f(x)}\geq\int_{0}^{M}\frac{dx}{f'(z)\,x+f(z)}=\frac{1}{f'(z)}\log\left(1+\frac{f'(z)}{f(z)}M\right)$$
so the Osgood condition is fulfilled without further assumptions. It follows that if $g(x)$ and $h(x)$ are positive concave functions on $\mathbb{R}^+$, then $F(x)=g(x)+h(x)$ is greater than both, positive and concave, so it satisfies the Osgood condition.
