Strictly decreasing function with a horizontal asymptote is convex?

Question

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?

Answer 1

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.