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Let $u : \mathbb{R} \rightarrow \mathbb{R}$ be such that $\forall x : u'(x) > 0, u''(x) < 0$.

Furthermore let $p \in (0, 1)$.

Is it then true that the function

$x \mapsto \frac{u^{-1} (p \cdot u (x))}{x}$ is decreasing?

You may assume that $u(0) = 0$ if necessary.

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The answer is no. Here is my counter example:

Consider any strictly increasing, strictly concave, and $C^2$ function $u:\mathbb{R}\mapsto\mathbb{R}$ satisfying $$u(1)>4/3,\;u(2)=2,\;u(4)=3.$$ Because of $$\frac{2-4/3}{2-1}>\frac{3-2}{4-2}$$ such a function exists. Note that $u^{-1}(2)=2$. Furthermore set $p=2/3$.

I claim that $U(2)<U(4)$, where $$U(x)=\frac{u^{-1}(p\cdot u(x))}x.$$ We have $$U(2)=\frac{u^{-1}((2/3)\cdot u(2))}2=\frac{u^{-1}(4/3)}2$$ and $$U(4)=\frac{u^{-1}((2/3)\cdot u(4))}4=\frac{u^{-1}(2)}4=\frac12.$$ The condition $U(2)<U(4)$ is therefore equivalent to $4/3<u(1)$, which holds by assumption.

Construction of $u$. For example, you can set $$u(x)=\left\{\begin{array}{cl}f(x)&\mbox{ for }x\leq 1\\g(x)&\mbox{ if }x\geq1\end{array}\right.$$ where $f(x)=p+qe^{-rx}$ and $g(x)=ax+\sqrt{bx+c}$. Now you determine the values $a$, $b$, and $c$ from the three equations $$g(1)=17/12,\;g(2)=2,\;g(4)=3.$$ Then you compute $g_1=g'(1)$ and $g_2=g''(1)$ and determine $p$, $q$, and $r$ from the equations $$f(1)=17/12,\;f'(1)=g_1,\;f''(1)=g_2.$$

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  • $\begingroup$ If my answer was helpful you may want to accept it. $\endgroup$ – Gerhard S. Sep 26 '17 at 6:48
  • $\begingroup$ Thank you for the answer. I see that this holds for such a $u$. Is there an easy way to prove that such a $u$ exists? I wrote down ln(p(x)) such that this function goes through the desired points, but it's not entirely clear to me that it's strictly increasing and concave. $\endgroup$ – Christian Sep 26 '17 at 9:31
  • $\begingroup$ I have edited my answer to explain one possible construction of $u$. There are many others. $\endgroup$ – Gerhard S. Sep 26 '17 at 10:18
  • $\begingroup$ I checked it and you can indeed solve for a, b, c. I suppose the function won't be smooth in 1 though, just C^2 ? $\endgroup$ – Christian Sep 26 '17 at 12:26
  • $\begingroup$ You are right. It is only $C^2$. But you did not assume anything else. $\endgroup$ – Gerhard S. Sep 26 '17 at 12:29

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