Let $\mu$ be the Möbius function. Say we have a bound on $\check{M}(x) = \sum_{n\leq x} \mu(n) \log \frac{x}{n}$ of the form $|\check{M}(x)|\leq \epsilon x$ for all $x\geq x_0$.
It is then easy to prove a bound $|M(x)|\leq 2\sqrt{\epsilon} x$ (plus a tiny error term) by a very simple procedure that is valid for any bounded $a_n$ instead of $\mu(n)$; see (Explicit) Tauberian theorems: removing $(\log x/n)$
It is also easy to see that we can do a little better for $\mu(n)$: since $\mu$ is supported on square-free numbers, we can aim at a bound of the form $|M(x)|\leq 2\frac{\sqrt{6 \epsilon}}{\pi} x$.
The question is really: can one hope to do better, using the fact that $\mu(n)$ is, well, $\mu(n)$? There are all sorts of nice identities relating different sums of $\mu$ to each other, sometimes in non-obvious ways, and it is possible I have missed one.
Or is there a real reason why a bound $|\check{M}(x)|\leq \epsilon x$ can only lead to a bound $|M(x)|\leq c \sqrt{\epsilon} x$, for some constant $c$? It makes sense that $\check{M}(x)$ should be easier to estimate -- the Mellin transform of $t\mapsto \max(1/t,0)$ is $1/s^2$, whereas that of $1_{[0,1]}$ is $1/s$. But can one give at least a plausible argument for why $|\check{M}(x)|\leq \epsilon x$ can only lead to a bound $|M(x)|\leq c \sqrt{\epsilon} x$, and not any better?
