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I'm trying to prove a conjecture and need some help. Consider a continuous, twice differentiable function $p(a)$ such that $p(0) = 0$ and $\forall a$, $p'(a) > 0$ and $p''(a) < 0$ and $p$ is bounded above by 1. I believe it's the case that for $\gamma > 1$, $\exists a' | p'(a) = \gamma p'(\gamma a)$. Another way of stating that is that if you have two functions, $p(a)$ and $p(\gamma a)$, there is some point $a'$ where the slopes of the two curves are the same.

I'm trying to prove this conjecture but I'm not making much progress. I feel like I'm making this harder than it is, but in in terms of a proof sketch / intuition, I think one approach would be to:

  • Show that $\gamma p'(0) - p'(0) > 0$

  • Show that $p(a)$ and $p(\gamma a)$ both have to converge some the same value, say $B$.

  • Argue that for the $p(a)$ curve to "catch up" to $p(\gamma a)$---because they converge to $B$ and the $p(a)$ curve is everywhere below the $p(\gamma a)$ curve---there has to be some point $\hat{a}$ where $p'(\hat{a}) > \gamma p(\gamma \hat{a})$ and hence $\gamma p'(\gamma \hat{a}) - p'(\hat{a}) < 0$ and so by the IVT, there has to be some $a'$ such that $\gamma p'(\gamma a') - p'(a') = 0$, which is what I'm looking for.

While I think this (might) work, the "catch up" notion is really informal and I suspect there's some much simpler, more rigorous way of showing that it's true. Thanks.

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The differentiable function $x\mapsto p(\gamma x)-p(x)$ vanishes at zero and at infinity, and it is non-negative because $p$ is increasing, so it has a maximum point $a$, which is what you want. –  Pietro Majer Aug 13 '12 at 6:42
    
Of course - thank you! If you migrate this to an answer, I can approve it. –  John Horton Aug 13 '12 at 14:46
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The differentiable function $x\mapsto p(\gamma x)−p(x)$ vanishes at zero and at infinity, and it is non-negative because p is increasing, so it has a maximum point a, which is what you want.

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