What is $O(o(x))$ as $x\to\infty $? I think it is $o (x)$ or $O (x^\delta) $ where $\delta <1$

Question

Let $M (x)=\sum_{n\le x}\mu (n) $. We know that

$$M (x)\log x+\sum_{n\le x}M (\frac xn)\Lambda (n)=O (x) $$

I want to prove $$M (x)\log x+\sum_{p\le x}M (\frac xp)\log p=O (x) $$

We have

$$M (x)\log x+\sum_{p\le x}M (\frac xp)\log p=M (x)\log x+\sum_{n\le x}M (\frac xn)\Lambda (n)-\sum_{m=2}\sum_{p\le x^{1/m}}M (\frac x {p^m})\log p $$

But

$$\sum_{m=2}\sum_{p\le x^{1/m}}M (\frac x {p^m})\log p=O (o (x \sum_{p\le\sqrt x}\frac {\log p}{p^2}))=O (o (x))$$

Now I need to know $O (o (x)) $.

Answer 1

If $f\in o(x)$ and $g\in O(f(x))$, then $\lim_{x\to \infty}\frac{f(x)}x=0$ and $|g(x)|\le M f(x)$ for some $M$ and all $x\gg 0$. It follows that $\lim_{x\to \infty}\frac{g(x)}x=0$, i.e., $g\in o(x)$.

(But $f\in o(x)$ and $g\in o(x)$ does not imply $g\in O(f(x))$)