0
$\begingroup$

Let $f:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$, and $g:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ be two functions. Suppose that $f(0) > g(0) = 0$, and $f$ is strictly increasing and concave, while $g$ is strictly increasing and strictly convex (so $f'(x)>0,\; f''(x) \leq 0,\; g'(x)>0$, and $g''(x) >0)$. Suppose that for some $x>0$ we have that $f(x)- g(x)>0$. Is it true that then for every $y \in [0,x]$ it must be that $f(y) - g(y) \geq 0$? Seems to be correct, at least for some special cases but, but I'm unable to prove it for the above outlined general case.

$\endgroup$
0
$\begingroup$

$h(x) = f(x) - g(x)$ is strictly concave. If $h(0)\ge 0$ and $h(x) \ge 0$ for some $x > 0$ then $$ h(y) > \frac{x-y}{x} \, h(0) + \frac yx \, h(x) \ge 0 $$ for $y \in (0, x)$.

The monotony of $f$ and $g$ is not needed for this conclusion.

$\endgroup$
1
  • $\begingroup$ Very nice. Thanks! $\endgroup$
    – Nidjsi
    Oct 17 '18 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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