Suppose there is a random variable, $X$, with finite variance, and c.d.f. $F(x)$. Does this imply that the upper Matuszewska index of $\bar F(x)$ exists and is strictly smaller than $-2$?

The upper Matuszewska index of $f(x)$ is defined as the infimum of $\alpha$ such that there exists a $C$ such that for each $\Lambda > 1$,

$$ f(\lambda x)/f(x) \le C(1 + o(1))\lambda^\alpha \,\,\,(x \rightarrow \infty)\text{ uniformly in }\lambda \in [1, \Lambda]$$

I ask this because I was reading a paper where a result was proven for distributions with upper Matuszewska index less than $-2$, but in the conclusion the authors stated that the result held for finite variance distributions, seemingly without justification.