# Prove for monotone $L \colon (0, \infty) \to (0,\infty)$, $L$ is slowly-varying if and only if there exists $x$ such that $\frac{L(tx)}{L(t)} \to 1$

$$L \colon (0,\infty) \to (0,\infty)$$ is slowly varying if for all $$x>0$$, $$x \neq 1$$ $$\lim_{t \to \infty} \frac{L(tx)}{L(t)} = 1$$.

I want to prove that for if $$L$$ is monotone, $$L$$ is slowly varying if and only if there exists $$x > 0$$ such that $$\frac{L(tx)}{L(t)} = 1$$.

I have tried writing for $$1>x>0$$, $$\frac{L(ty)}{L(t)} \leq \frac{L(ty)}{L(tx)}$$, but I can make no further progress from this.

Also, what would be an example of $$L$$ such that for some $$x \neq 1$$, $$x>0$$, $$\frac{L(tx)}{L(t)} \to 1$$, but $$L$$ is not slowly varying?

Thank you.

• What is $t$ in the statement you want to prove?
– Will
Commented May 28, 2023 at 7:02
• It probably should be $\frac{L(tx)}{L(t)} \to 1$, as in the title. But that always holds for $x=1$, so something is missing here. Commented May 28, 2023 at 7:18
• Commented May 28, 2023 at 7:54
• @MartinR I forgot to type that $x \neq 1$
– Phil
Commented May 28, 2023 at 14:53

If $$x>0, x \neq 1$$ and $$\frac {L(tx)} {L(t)} \to 1$$ as $$t \to \infty$$ then $$\frac {L(ty)} {L(t)} \to 1$$ as $$t \to \infty$$ for every $$y$$.
Proof: Since $$\frac {L(tx)} {L(t)} \to 1$$ is equivalent to $$\frac {L(t\frac 1 x)} {L(t)} \to 1$$ we may sppose $$x>1$$. By iteration we get $$\frac {L(tx^{n})} {L(t)} \to 1$$ for $$n =0,1,2...$$ and any number $$y \in (1,\infty)$$ lies between $$x^{n}$$ and $$x^{n+1}$$ for some, $$n \geq 0$$. By monotonicity we get $$\frac {L(ty)} {L(t)} \to 1$$. Once again taking reciprocal yields $$\frac {L(ty)} {L(t)} \to 1$$ for $$y <1$$.