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Is there any proof that two "completely monotonic" functions ($f,g: (0, \infty) \rightarrow \mathbb{R}$) would intersect at most at one point?

Completely monotonic means: The $n$'th derivative of each function satisfies $(−1)^ n f^{(n)}(x) \geq 0$, $(−1)^ n g^{(n)}(x) \geq 0$, $x \in (0, \infty)$.

Mister Benjamin Dover
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Tareq Ahmed
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1 Answers1

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This is incorrect. Two completely monotonic functions can intersect at more than $1$ point, in fact they can intersect at infinitely many points.

Consider $$\color{red}{f(x) = \left\lfloor \dfrac{2x}{\pi} \right\rfloor+\sin\left(x-\left\lfloor \dfrac{2x}{\pi} \right\rfloor\cdot \dfrac{\pi}2\right) - 0.125}$$ and $$\color{blue}{g(x) = 1.125+ \left\lfloor \dfrac{2x}{\pi} \right\rfloor-\cos\left(x-\left\lfloor \dfrac{2x}{\pi} \right\rfloor\cdot \dfrac{\pi}2\right)}$$

The plot of these functions look as below.

enter image description here

Adhvaitha
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  • But are $f$ and $g$ satisfying the condition for "complete monotonicity":
    The $n$'th derivatives of each function satisfies $(−1)^ n f^{(n)}(x) \geq 0$, $(−1)^ n g^{(n)}(x) \geq 0$, $x \in (0, \infty)$?     – Tareq Ahmed May 06 '15 at 18:22