Exercise 15, § 4

Let $I$ be an interval contained in $]0, +\infty[$; now prove that, if $f(1/x)$ is convex on $I$, the same is true of $x f(x)$, and conversely.

¹ That is to say that $\lim_{x \to +\infty} (f(x) - (\alpha x + \beta)) = 0$.