Ref: <http://link.springer.com/article/10.1007%2FBF02399204>


My question is about the proof of Theorem 0(b). On p.213, we see the expression


$\Delta\_{\alpha\ast\beta}(x;q,a)=\frac{1}{\varphi(q)}\sum\_{\chi\neq\chi\_0}\bar{\chi}(a)\left(\sum\_m\alpha\_m\chi(m)\right)\left(\sum\_n\beta\_n\chi(n)\right).$


I suppose I know how to get this:
\begin{eqnarray\*}
\Delta\_{\alpha\ast\beta}(x;q,a)
&=&
\sum\_{n\leq x, n\equiv a(q)}\alpha\ast\beta(n)-\frac{1}{\varphi(q)}\sum\_{n\leq x,(n,q)=1}\alpha\ast\beta(n)\\
&=&
\sum\_n\alpha\ast\beta(n)\frac{1}{\varphi(q)}\sum\_{\chi}\chi(n)\bar{\chi}(a)-\frac{1}{\varphi(q)}\sum\_{n\leq x,(n,q)=1}\alpha\ast\beta(n)\\
&=&
\frac{1}{\varphi(q)}\sum\_{\chi\neq\chi\_0}\bar{\chi}(a)\sum\_n\alpha\ast\beta(n)\chi(n),
\end{eqnarray\*}


and then we split the innermost sum into a product of two sum, by the definition of convolution $\alpha\ast\beta$.


My question is that, the supports of $\alpha$ and $\beta$ is $(M,2M]$ and $(N,2N]$ respectively, where $NM=x$. But the range $\Delta\_{\alpha\ast\beta}$ considers is just $n\leq x$. Even using the usual practice $x<n\leq 2x$ when we define $\Delta\_{\alpha\ast\beta}$, it cannot cover $(x,4x]=(NM,4NM]$. Therefore the aforementioned innermost sum may not be split. 


I then try to assume the supports of $\alpha$ and $\beta$ is $(0,M]$ and $(0,N]$. This makes everything fine, but when I try to use this result obtained to recover Theorem 0(b) I just fail.


Can someone help me? Thanks a lot.



Question about BFI's "Primes in arithmetic progressions to large moduli"
I think you just want to forget the $n\le x$ conditions in the $\Delta$ definition of (1.1). Once they start at the bottom of p206 to switch to Motohashi's formulation, then the $n$-restriction on any $\Delta$-sum is better seen as coming from the support of the $\alpha$ and $\beta$. Prior to that, they needed to introduce the $x$-parameter to be able to cogently discuss the Introduction, but the later rewriting overrides this. At least that is what I think is going on.


