# 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?

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Strictly concave, or is falt okay? Because the one that suggests itself is to take the line that is the max of the derivatives at 0 of $h$ and $g$. – Ray Yang Feb 4 '13 at 12:43
@RayYang: It is possible that $h'(0)=\infty$, e.g. if $h$ behaves like $-x\ln x$ near $0$. – Jonas Meyer Sep 11 '13 at 6:42

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.

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What if $f(x)=\sqrt{x}$, isn't it positive concave on $\mathbb{R}^+$ that does not satisfy Osgood condition? – Omran Kouba Jul 1 '14 at 10:47
$$\int_{0}^{M}\sqrt{x}\,dx = \frac{2}{3}M^{3/2}$$ goes to infinity as $M$ goes to infinity. – Jack D'Aurizio Jul 1 '14 at 11:21
No, Osgood condition has to do the behavior of the function near $0^+$. That is $\int_0^\alpha\frac{dx}{f(x)} =+\infty$ for every $\alpha>0$. – Omran Kouba Jul 1 '14 at 11:34
Oh, god. I believed that $\int_{0^+}$ stood for $\int_{\mathbb{R}^+}$. My fault. Anyway, taking the convex envelope of the two graphics should do the trick, in this case. – Jack D'Aurizio Jul 1 '14 at 13:52