Integrable slowly varying function

Question

We say a function $L$ is slowly varying if $$\lim_{t\to\infty} \frac{L(tx)}{L(t)} = 1$$ for every $x > 0$.

Are there such $L$ that are integrable? Say $L$ is defined on $[0,\infty)$ and is continuous with $\lim\limits_{t\to\infty}L(t) = 0?$ I'm thinking of $L$ as the tail distribution of some non-negative random variable.

Answer 1

Unfortunately I can't think of an example with the extra requirements, but there certainly exists such $L$ that are integrable.

Take $L(x) =\log(x)^b$, where $b$ is a real number.

and let's show using L'Hopital that this function indeed is slowly varying.

$$ \lim_{x \to \infty} \frac{\log(tx)^b}{\log(x)^b} = \frac{\frac{b\log(xt)^{b-1}}{x}}{\frac{b\log(x)^{b-1}}{x}} = \frac{\log(tx)^{b-1}}{log(x)^{b-1}} $$ Repeated application will yield the desired result.

Now, $\log(x)^b$ is certainly integrable, although with a hideous expression resorting to special functions. Wolfram Alpha

Answer 2

No slowly varying function $L$ can be integrable on $[0,\infty)$.

It follows from the Potter bounds (e.g., see Theorem 1.5.6 of the book Regular Variation by Bingham, Goldie, and Teugel), that for any $\delta>0$, one has $$ L(x) \ge C_\delta x^{-\delta} $$ for $x$ sufficiently large, where $C_\delta>0$.

This bound is also not difficult to prove directly from the uniform convergence theorem, which states that the limit $L(tx)/L(t)\to1$ is attained uniformly for $x$ in any compact subset of $(0,\infty)$: One shows that for all large enough $t$, $$ L(te^k) \ge L(t)e^{-\delta k}$$ by induction for $k=1,2,\ldots$, then writes $x$ in the form $te^k$ with $t$ in a fixed compact set $[T,Te^1]$. (Note $e^{-k}\ge x/(Te)$.)