If $\mu(x)$ is Mobius function can it be true that $$ \sum_{a\leq x}\mu(a)O(\frac{x}{a})=O(x\log x), $$ where $a$ is natural number and $O$ is used as big-O notation?
I came to this in one other problem I was solving, and it didnt came as part of those sets being equal. So I proved that if we take function in $O(\frac{x}{a})$ than left side belongs to $O(x\log x)$, which was only thing I needed and is fairly easy. But I was thinking can the other part be true as well, like if I take function $f$ from $O(x\log x)$ and lets say that $g$ is function such that
$\sum_{a\leq x}\mu(a)g(\frac{x}{a})=f(x)$,
I know that
$g(x)= \sum_{a\leq x}f(\frac{x}{a})$.
Is $g$ neccesseraly function from $O(\frac{x}{a})$, that is $O(x)$?
The notation $f(x) = O(g(x))$ is defined to mean that $\left|f(x)\right|\leq C g(x)$ for some suitable constant $C$ which is independent of $x$. As TravorLZH said in his comment, the equation you have written is completely trivial, since $\left|\mu(a)\right| \leq 1$ for all integers $a$. By the definition of the notation, it then follows that $$ \mu(a) O\left(\frac{x}{a} \right) = O\left(\frac{x}{a} \right), $$ and so we have $$ \sum_{a\leq x} \mu(a) O\left( \frac{x}{a}\right) = \sum_{a\leq x} O\left( \frac{x}{a}\right) = O\Big( \sum_{a\leq x}\frac{x}{a} \Big) = O\Big( x\sum_{a\leq x}\frac{1}{a} \Big) = O(x\log x). $$ It is fruitful to work through each step of this equation. Using only the definition of big $O$ notation and the inequality $\left|\mu(a)\right| \leq a$, you should be able to verify each step rigorously.
Edit: It should be noted that the usage of this notation in analytic number theory differs from its usage in a discipline such a computer science. In computer science, the notation $O(g(x))$ typically describes a set, whereas in analytic number theory, it is used in the sense stated above (personally I prefer the latter definition of the notation, but I'm happy to politely disagree with my CS brethren, as I've been told that the former definition is more natural in that discipline).