I am reading Maynard's paper "Small gaps between primes" and I don't understand how to deduce the second line from the first line of below's inequality. How does $\dfrac{1}{(p-1)^2}$ appear?
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3$\begingroup$ Here's my guess. I think it's in general $\frac{1}{(p-1)^k}$, but since $k \ge 2$, you can replace $k$ with $2$. Because of how products work, all you need to show is $$\sum_{\substack{u < R \\ p \mid u \\ (u,W) = 1}} \frac{\mu(u)^2}{\varphi(u)} \le \frac{1}{p-1}\sum_{\substack{u < R \\ (u,W) = 1}} \frac{\mu(u)^2}{\varphi(u)},$$ which follows from writing $u = p^mu'$ (where $p^m \mid \mid u$), using the multiplicativity of $\varphi$, and then trivially upper bounding the sum over $u' < R/p$ by the sum over $u' < R$. $\endgroup$– mathworker21Commented Jun 17, 2022 at 15:33
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$\begingroup$ @mathworker21 the same $u'$ corresponds to several $u$'s, that's why I would instead introduce $\tilde{u}=u/p$ $\endgroup$– Fedor PetrovCommented Jun 17, 2022 at 15:55
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1$\begingroup$ @mathworker21 thank you for your explanation $\endgroup$– Laurence PWCommented Jun 17, 2022 at 16:00
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$\begingroup$ @FedorPetrov But $\varphi$ is not completely multiplicative, so how do you plan on getting the $\frac{1}{p-1}$ out? But you bring up a good point -- I guess my comment needs some more details. $\endgroup$– mathworker21Commented Jun 17, 2022 at 16:58
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$\begingroup$ @mathworker21 It is not completely multiplicative, but enjoys the inequality $\varphi(px)\geqslant (p-1)\varphi(x)$ $\endgroup$– Fedor PetrovCommented Jun 17, 2022 at 19:43
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1 Answer
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For your questions it suffices to study the inner sum $$\sum_{\substack{u_1,....,u_k < R \\ p\mid u_i,u_j \\ (u_i,W)=1 \forall i}}\prod_{i=1}^k \frac{\mu(u_i)}{\varphi(u_i)}.$$ As was pointed out in the comments, we have the inequality $\varphi(pu_i')\geq (p-1)\varphi(u_i')$. Thus writing $u_i = pu_i', u_j = pu_j'$, we may bound the sum by $$\frac{1}{(p-1)^2}\sum_{\substack{u_1,....,u_k < R \\ (u_i,W)=1 \forall i}}\prod_{i=1}^k \frac{\mu(u_i)}{\varphi(u_i)}.$$ (Note that I did not change the variables' names, this does not matter of course). The reason why the exponent is $2$, and not say $k$ is because the condition in the original sum is just that $p$ divides (at least) $2$ of the variables.
Now, the variables have been completely decoupled from each other and we may separate the sum into a product of $k$ sums and thus we find the above equals $$\frac{1}{(p-1)^2}\left(\sum_{\substack{u < R \\ (u,W)=1}}\frac{\mu(u)}{\varphi(u)}\right)^k.$$
