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Given two strictly concave, strictly increasing and everywhere derivable functions $f,g: \mathbb{R}^+_0 \to [0,1]$ where $f(0)=g(0)=0$ and $$\lim_{x\to\infty} f(x)=\lim_{x\to\infty} g(x)=1$$ Excluding $x=0$, what is the maximum number of the other intersections between them?

I think $2$ points, but I did not find a good proof of that.

Fili7.5
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  • Why do you think so? – ViktorStein Jan 20 '20 at 16:48
  • It seems intuitive. But I don't exclude that a counterexample exists. – Fili7.5 Jan 20 '20 at 16:55
  • There is always the case $f=g$, which satisfies your assumptions as they are currently written. Beyond this, suppose there is a point $x$ other than $0$ such that $f(x)=g(x)$. Can you construct an example, or derive a contradiction? Perhaps you can consider the problem with $f,g:[0,1]\rightarrow[0,1]$ instead; try drawing $f$ and $g$ concave and intersecting somewhere and see what happens. – R_B Jan 20 '20 at 17:00
  • Do you want the functions to be continuous at $0$ or it doesn't matter? The current conditions only imply continuity in $(0,\infty)$. – bjorn93 Jan 21 '20 at 01:04
  • Yes. But River Li has provided a good counterexample, so my conjecture is wrong. Maybe we have to add some other constraints to obtain an upper bound to the number of intersections. – Fili7.5 Jan 21 '20 at 13:23

1 Answers1

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Let \begin{align} f(x) &= 1 - \mathrm{e}^{-x}, \\ g(x) &= 1 - \mathrm{e}^{-x} + \frac{1}{4}\mathrm{e}^{-x}\sin x. \end{align} We have $f(0) = g(0) = 0$ and $\lim_{x\to \infty} f(x) = \lim_{x\to \infty} g(x) = 1$. Also, we have, on $[0, \infty)$,
\begin{align} f'(x) &= \mathrm{e}^{-x} > 0, \\ f''(x) &= -\mathrm{e}^{-x} < 0, \\ g'(x) &= \mathrm{e}^{-x} - \frac{1}{4}\mathrm{e}^{-x}\sin x + \frac{1}{4}\mathrm{e}^{-x}\cos x > 0\\ g''(x) &= -\mathrm{e}^{-x} - \frac{1}{2}\mathrm{e}^{-x}\cos x < 0. \end{align} Thus, $f(x)$ and $g(x)$ are both strictly concave, strictly increasing.

However, $f(x) = g(x)$ for $x = \pi, 2\pi, 3\pi, \cdots$.

River Li
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