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I have the following question: Suppose $f$ and $g$ $: [0, l) \to \mathbb{R}$ are continuous, concave and increasing where $l < \infty$. Can we claim that they intersect at most finitely many points? What if we replace $l$ with $\infty$?

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No.

Start with some continuous, concave and increasing $f$ and find the points $\left( \frac{l}{2} ,f\left(\frac{l}{2}\right) \right)$, $\left( \frac{l}{4} ,f\left(\frac{l}{4}\right) \right)$, $\left( \frac{l}{8} ,f\left(\frac{l}{8}\right) \right)$, ...

Then draw a continuous, concave and increasing function $g$ through these points which is not identical everywhere to $f$; this will almost always be possible with some exceptions when $f$ is only "weakly concave" and "weakly increasing".

For replacing $l$ with $\infty$, chose an arbitrary positive $l$ and repeat.

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    I think I couldn't state well my prev question, I was gonna ask about boundedness. If $f,g:[0, \infty) \to R$ are both concave , increasing, bounded and smooth can we claim that they intersect at most finitely many points in [0,1]? [0,1)? Does the inclusion of the endpoint 1 matter here as f&g are bounded on $[0, \infty)?2012-09-01