Suppose that I have 2 increasing, convex functions $f_1$ and $f_2$ such that:

  1. $f_k:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ for $k = 1, 2$,

  2. $f_k(0) = 0$ for $k = 1,2$ and

  3. $f'_1(x) < f'_2(x)$ for all $x \in \mathbb{R}^+$.

Pick $y,z\in \mathbb{R}^+ $ such that:

$f'_1(y) = f'_2(z)$.

Is it true that $f_1(y) < f_2(z)$?

Every example that I try suggests that the claim is true, but I'm unsure how to prove it.


This is not true. For example, let $f_2(x)=2x+x^2$ and $$ f_1(x) = \begin{cases}x,\quad &x\le 1000 \\x + (x-1000)^2, \quad & x\ge 1000\end{cases} $$ Then $f_2'(1)=4$, but at the point $y$ where $f_1'(y)=4$ we have $y> 1000$, hence $f_1(y)> 1000$.

  • $\begingroup$ Thanks @Woodface! What if the functions are strictly convex? $\endgroup$ – Jason Mar 12 '15 at 6:12
  • $\begingroup$ Add $10^{-6}e^x$. $\endgroup$ – user147263 Mar 12 '15 at 6:14

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.