Suppose you have two continuous, positive convex functions $F(x)$ and $G(x)$, $x\in\mathbb{R}$ such that: $$\lim_{x\rightarrow\pm\infty}F'(x)=\lim_{x\rightarrow\pm\infty}G'(x)=\pm 1$$ and that $-1\leq F'(x)\leq 1$ and $-1\leq G'(x)\leq 1$. I'm trying to assert that such functions can cross at most once. Is this true in general?
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2I don't think so. Try to imagine two "V" one being deeper than the second one. Both having the same vertical symmetry. – mathcounterexamples.net Jun 08 '15 at 11:21
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Start with the curve $f(x)=\sqrt{x^2+1} $ and let $F$ be the piecewise linear interpolation between the points $(2n,f(2n))$, $n\in\mathbb Z$, and $G$ the inear interpolation between the points $(2n+1,f(2n+1))$. Then the graphs of $F,G$ intersect infinitely often.
Hagen von Eitzen
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No -- in fact you can get infinitely many crossings.
If you choose a starting point you can start by having $F'(x)=0$, $G'(x)=\frac12$ until $G$ has crossed above $F$. Then increase $F'(X)$ (as smoothly as you want) to $\frac23$ (but keep $G'(x)=\frac12$) until $F$ has crossed above $G$, then increase $G'(x)$ to $\frac34$ until $G$ has crossed above $F$ and so forth.
Computing the precise points where you increase the slope of one of the functions is left as an exercise for the reader.
hmakholm left over Monica
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1@user42397: Not really -- you can use the same recipe with the slopes converging to $\frac 12$ instead of $1$, and then add $\frac12\sqrt{1+x^2}$ to both functions in order to make sure their slopes increase monotonically. – hmakholm left over Monica Jun 08 '15 at 11:24
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