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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.

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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 $f_1(y)> 1000$.

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  • $\begingroup$ 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

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