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Dec 14, 2012 at 0:41 comment added George Lowther @Greg: So, you have $L(x)=\sum_d M(x/d^2)$ giving $\limsup_{x\to\infty}\lvert L(x)/x\rvert\le\limsup_{x\to\infty}\lvert M(x)/x\rvert\sum_d d^{-2}$ (=0). Similarly, $M(x)=\sum_d \mu(d)L(x/d^2)$, so you can go in either direction.
Dec 14, 2012 at 0:18 comment added Greg Martin For example, problem 11(b) in section 6.2 of Montgomery and Vaughan's book "Multiplicative Number Theory I" is to use the method of proof Peter described to prove that $|L(x)| \le Ax \exp({-}c\sqrt{\log x})$ for some positive constants $A$ and $c$. Interestingly, problem 11(c) suggests that it is easy to derive this bound from the corresponding bound for $M(x) = \sum_{n\le x} \mu(n)$, which might be easier to find in the literature. The link between the two functions is given by $\lambda(n) = \sum_{d\colon d^2\mid n} \mu(n/d^2)$.
Dec 14, 2012 at 0:00 comment added Peter Humphries Also, you may be interested in the contents of this paper: staff.science.uu.nl/~dahme104/DistributionOmega.pdf
Dec 13, 2012 at 23:59 comment added Peter Humphries I don't know of a direct reference of such a proof, but this theorem is folklore and it's pretty easy to see why it's true: one can show (say, via comparing Euler products) that $\sum_{n=1}^{\infty}\lambda(n)n^{-s}=\zeta(2s)/\zeta(s)$ for $\Re(s)>1$, then use the fact that this extends meromorphically to the entire complex plane and has no poles in the region $\Re(s)>1-c/\log(|\Im(s)|+2)$. So basically the usual way of proving the prime number theorem, just with a different Dirichlet series.
Dec 13, 2012 at 23:37 vote accept Joel Moreira
Dec 13, 2012 at 23:37 comment added Joel Moreira Thanks! So this seems to be well known among number theorists. Can you give a reference where I can find a proof that $L(x)=o(x)$?
Dec 13, 2012 at 22:59 history answered Peter Humphries CC BY-SA 3.0