I am working through the proof of the Bombieri-Vinogradov theorem in Analytic Number Theory (Iwaniec, Kowalski). My problem is that on page 424, it is said that $\mu(m)$ satisfies $D_f(x;q,a)\ll (\sum \limits_{n \le x} |f(n)|^2)^{1/2} \, \, x^{1/2}(\log x)^{-A}$ (17.12) by the Siegel-Walfisz theorem (so that we can apply Theorem 17.4). How does the Siegel-Walfisz theorem apply to the Möbius function?


The Siegel-Walfisz principle for a function $f(n)$ states that for all $A>0$ fixed then whenever $a$ modulo $q$ is a residue class with $a$ and $q$ coprime then one has $$\sum_{\substack{n\leq x \\ n \equiv a\hspace{-0,3cm}\mod{\hspace{-0,1cm}q}}}f(n)=\frac{1}{\phi(q)}\sum_{\substack{n\leq x\\ \gcd(n,q)=1}}f(n)+O_{A,f}\left(\sqrt{q}\frac{x}{(\log x)^A}\right).$$ Bear in mind that this is non--trivial only in the case that $q\leq (\log x)^B$ for some $B>0$ fixed. It has been proved for a lot of multiplicative functions, one example being $f(n)=\mu(n)$. The proof for this case is similar to the proof for primes, one simple exposition is Corollary 5.29, Equation 5.80, of the book you are reading. It gives a good upper bound for the quantities $$\sum_{n \leq x}\chi(n) \mu(n)$$ for all primitive Dirichlet characters and in order to deduce Siegel-Walfisz for $\mu(n)$ you use orthogonality of characters. Orthogonality means $$\sum_{\substack{n\leq x \\ n \equiv a\hspace{-0,3cm}\mod{\hspace{-0,1cm}q}}}\hspace{-0,5cm}\mu(n)=\frac{1}{\phi(q)}\sum_{\chi\hspace{-0,3cm}\mod{\hspace{-0,1cm}q}}\left(\sum_{n \leq x}\chi(n) \mu(n)\right).$$

Finally to answer your question, the bound $$D_f(x;q,a)\ll \left(\sum \limits_{n \le x} |f(n)|^2\right)^{1/2} \, \, x^{1/2}(\log x)^{-A}$$ for $q\leq (\log x)^B$ and $f=\mu$ is equivalent to $$D_\mu(x;q,a)\ll x (\log x)^{-A}$$ owing to $$\sum_{n\leq x}|\mu(n)|^2=\frac{6}{\pi^2}x+O(x^{\frac{1}{2}})\gg x.$$ Let me know if something is not clear.

  • $\begingroup$ In your first sentence, may one add (n,q)=1 to the right side? So instead of summing over all n less than x, only those coprime less than x? $\endgroup$ – Mats Hansen May 21 '15 at 13:04
  • $\begingroup$ I am afraid you are right, I edited it accordingly. $\endgroup$ – Captain Darling May 21 '15 at 17:38

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