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Suppose $f$ is a strictly decreasing function with a horizontal asymptote at $t \rightarrow + \infty$. Hence, there exists a $t_{0}$ such that $\forall t>t_{0}, ~f(t)$ is a convex fuction. Is this the case?

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    $\begingroup$ What happens if you take a function whose graph is a smooth staircase? I mean, think of a step, then a smooth "jump" to a lower step, and so on. I guess it is not a convex function. You can also pretend it is strictly decreasing, by letting the width of each step tend to zero. $\endgroup$ – Siminore Apr 2 '14 at 11:07
  • $\begingroup$ I understand, thank you $\endgroup$ – jacie Apr 2 '14 at 11:41
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For completeness, a concrete example: $$f(t) = \int_t^\infty \frac{\sin^2 x}{x^2}\,dx$$ The function is strictly decreasing, since the integrand is nonnegative and vanishes only at isolated points. It tends to $0$ as $t\to\infty$. On the other hand, $f'(\pi n) =0$ for every $n$, so $f'$ is definitely not monotone.

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