Prove that there exists $s > 0$ such that $E[X^s] < 1$ given $E[X^r] < ∞$ for some $r$ and $E[\log X] < 0$

Question

Let $X$ be a random variable with $X ≥ 0$ a.s. and such that $E[X^r] < ∞$ for some $r > 0$ and $E[\log X] < 0$. Prove that there exists $s > 0$ such that $E[X^s] < 1$.

I know when $s$ goes to 0+ $E[X^s]$ should be very close to 1. But I am not able to prove the inequality. I have tried to used Taylor expansion and construct contradiction but it does not look straightforward for me.

Answer 1

Replacing $X$ by $X^r$, we can assume that $r=1$. We know that $$\lim_{s\to 0}\left(\mathbb E\left[X^s\right]\right)^{1/s}=\exp\left(\mathbb E\left[\log X\right]\right)$$ and the right hand side is strictly smaller than $1$.