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

Question

$L \colon (0,\infty) \to (0,\infty)$ 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.

Answer 1

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 $y <1$.