# $\displaystyle\lim_{p \rightarrow \infty} p h'(p) = 0$ for $h$ positive, decreasing, and log-convex

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.