I am going though a proof, one of the steps of which reads like this:

Let $h$ be a $C^3$, strictly decreasing, and log-convex function from $R_{++}$ to $R_{++}$. Then:

$\lim_{p \rightarrow \infty} p h'(p) = 0$

The proof goes like this:

By Fundamental Theorem of Calculus: $$h(p) = h(1) + \int_1^p h'(x)dx = h(1) + ph'(p) - h'(1) - \int_1^p x h''(x)dx$$

where the second equality is obtained integrating by parts. Then:

$$p h'(p) = h(p) - h(1) + h'(1) + \int_1^p x h''(x)dx$$

Since $h$ is positive and decreasing it has a finite limit when $p \rightarrow \infty$.

Since $h$ is log-convex $(\log(h))'' = \frac{h''h - {h'}^2}{h^2} \geq 0$, which implies $h'' \geq 0$.

Everything is good so far. Now, the proof follows:

Therefore, the function $p \rightarrow \int_1^p x h''(x)dx$ is non-decreasing, and that function has a limit at infinity.

I get that if $h'' \geq 0$ the integral over a positive interval would be non-decreasing. But how does it follow from that (or anything else) that the limit of $\int_1^p x h''(x)dx$ exists?


Since $h(p)\ge 0$ and $h'(p)\le 0$ we have $$ \int_1^pxh''(x)\,dx=h(1)-h'(1)-h(p)+ph'(p)\le h(1)-h'(1), $$ hence, bounded above. Plus increasing.

  • $\begingroup$ Thanks. That makes a lot more sense. It feels as though they should have made that explicit. $\endgroup$ – Francisco Andrés Dec 14 '18 at 23:15

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