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Consider two functions $f: [0,1) \rightarrow \mathbb{R}$ and $g: [0,1) \rightarrow \mathbb{R}$. Suppose $f(x) > g(x)$ for all $x \in [0,1)$. Suppose further that $f$ and $g$ are infinitely differentiable and all derivatives of both $f$ and $g$ are strictly positive on the interior $(0,1)$ of the domain and non-negative at $x=0$: $f^{(n)}(x) > 0$ and $g^{(n)}(x) > 0$ for all $x \in (0,1)$, $f^{(n)}(0) \geq 0$, $g^{(n)}(0) \geq 0$, for all $n$. (For lack of a better term, I call $f$ and $g$ 'fully monotonic', since 'totally' or 'completely' monotonic requires the signs to alternative. See here.) Let $x_0 \in (0,1)$ be a fixed interior point.

Is it true that $$ \frac{f(x)}{f(x_0)} = \frac{g(x)}{g(x_0)} $$ only holds at $x = x_0$? That is, do the functions $f(x)/f(x_0)$ and $g(x)/g(x_0)$ cross only once?

A standard sufficient condition for uniqueness is $f' > g'$, or the slightly weaker $f'(x)/f(x_0) > g'(x)/g(x_0)$, but this is not assumed here. Intuitively it seems that full monotonicity and $f > g$ imply any rescaled functions $f(x)/C_1$ and $g(x)/C_2$ where $C_1 > 1$ and $0 < C_2 < 1$ can cross at most once, but I have not been able to prove this, nor can I construct a counterexample.

Also, in case these added properties matter: I know that $f(0) = 1$, $g(0) = 0$ and both $f(x)$ and $g(x)$ diverge to infinity as $x \rightarrow 1$. $f$ and $g$ are both power series.

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2 Answers 2

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Let $g\colon[0,1)\to\mathbb{R}$ be a $C^\infty$ function such that $g(0)=0$, $g^{(n)}(x)>0$ for all $x\in(0,1)$ and all $n\ge0$, $g^{(n)}(0)\ge0$ for all $n\ge0$ and $\lim_{x\to1^-}g(x)=\infty$. One example is $g(x)=\tan(\pi\, x/2)$. Let $f(x)=1+(g(x))^2$. Then $f(x)>g(x)$ for all $x\in(0,1)$, $f$ satisfies the required conditions about the positivity of its derivatives and $\lim_{x\to1^-}f(x)/g(x)=\infty$. Take $x_0\in(0,1)$ such that $g(x_0)<1$ (this is possible since $g(0)=0$). Then the equation $$ \frac{f(x)}{f(x_0)}=\frac{g(x)}{g(x_0)} $$ has at least two solutions. One is $x=x_0$. A simple calculation shows that $$ \Bigl(\frac{f(x)}{f(x_0)}\Bigr)'\Bigr|_{x=x_0}<\Bigl(\frac{g(x)}{g(x_0)}\Bigr)'\Bigr|_{x=x_0}. $$ Then $g(x)/g(x_0)>f(x)/f(x_0)$ on $(x_0,x_0+\epsilon)$ for some $\epsilon>0$. But $\lim_{x\to1^-}f(x)/g(x)=\infty$, so that the grapg of $f(x)/f(x_0)$ must cut the graph of $g(x)/g(x_0)$ on some point $x>x_0$.

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Nice counterexample---thanks! –  Aelmore Jun 26 '12 at 16:46

You can have fully monotonic functions that cross each other infinitely many times :

Let $f(x) = \sum_{k=0}^\infty (1-x)^{-(4k+1)}/(4k+1)!$ and $g(x) = \sum_{k=0}^\infty (1-x)^{-(4k+3)}/(4k+3)!$. $f$ and $g$ are fully monotonic (because $x \mapsto (1-x)^{-1}$ is, and the set of fully monotonic functions is closed by rescaling, addition, multiplication, and good enough convergence), but $f(x) - g(x) = \sin((1-x)^{-1})$, which changes sign infinitely many times in $[0;1)$

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+1: Very nice! I considered a similar approach, but initially missed that the domain of definition was $[0,1)$, so was working to adapt my answer. –  Cameron Buie Jun 25 '12 at 16:35

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