# Strictly decreasing function with a horizontal asymptote is convex?

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?

• 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. – Siminore Apr 2 '14 at 11:07
• I understand, thank you – jacie Apr 2 '14 at 11:41

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.