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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.

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1 Answer 1

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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.

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I don't know, but if this is the case, then the $n$-restriction on $\Delta$ in this example would be in $(x,4x]$, which seems quite unusual. –  ccire Jul 1 '13 at 9:59
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The point is, in the Intro they were dealing with a function $f$ that was not finitely supported, so $\Delta$ had the $x$-cutoff in it, to make it easier to talk about. When they introduce the convolution of Motohashi, then the sequence $\alpha\star\beta$ is already finitely supported (though in a way that depends on $x$), so the prior convention is not any longer in value. –  v08ltu Jul 1 '13 at 11:52

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