Large Sieve Inequality – Large Sieve Type Inequality in Number Theory

Question

Let $S_x(t)=\sum_{n\le x} a_n e(nt)$, where $e(x)=e^{2\pi i x}$. Then, the large sieve inequality tells us that
$$
\sum_{q\le Q} \sum_{\substack{0\lt a \lt q \\ (a,q)=1}}|S_x(a/q)|^2 \le (Q^2+4\pi x)\sum_{n\le x} |a_n|^2.
$$
Is it possible to obtain a similar inequality for $\sum_{q\le Q} \frac{1}{q}\sum_{\substack{0\lt a \lt q \\ (a,q)=1}}|S_x(a/q)|$? I tried using Cauchy Schwartz but then we have
$$
\sum_{q\le Q} \frac{1}{q}\sum_{\substack{0\lt a \lt q \\ (a,q)=1}}|S_x(a/q)| \le \frac{\pi}{6}\bigg(\sum_{q\le Q}\Big(\sum_{\substack{0\lt a \lt q \\ (a,q)=1}}|S_x(a/q)| \Big)^{2}\bigg)^{\frac{1}{2}},
$$
and from here I don't know the most optimal way of applying the large sieve inequality. I was thinking of using for instance an inequality of the form $$
(x_1+\ldots+x_m)^2\le m(x_1^2+\ldots+x_m^2)
$$
for positive $x_1,\ldots x_m$, but I think I lose too much information using this inequality and don't obtain a very optimal bound. Is there any way to do this easier?

Answer 1

You can use integration by parts, but since I am lazy I am going to just use dyadic dissection and positivity. For any nonnegative function $f$, $$\sum_{q \le Q} \frac{f(q)}{q} \le \sum_{k\ge 0:\, 2^k \le Q} \frac{2^{k+1}}{Q}\sum_{Q/2^{k+1}<q \le Q/2^k}f(q)\le \sum_{k\ge 0:\, 2^k \le Q} \frac{2^{k+1}}{Q}\sum_{q \le Q/2^k}f(q).$$ Take $f(q)=\sum_{(a,q)=1}|S_x(a/q)|^2$ to obtain $$\begin{align}(\star)\, \sum_{q \le Q} \frac{1}{q}\sum_{(a,q)=1}|S_x(a/q)|^2&\le \sum_{k \ge 0:\, 2^k \le Q} \frac{2^{k+1}}{Q}\left( \left( \frac{Q}{2^k}\right)^2 + 4 \pi x\right)\sum_{n \le x} |a_n|^2 \\&\le 4(Q+4 \pi x)\sum_{n \le x} |a_n|^2 .\end{align}$$

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If you want a reference for $(\star)$ see equation (5.13) in Brüdern's book "Einführung in die analytische Zahlentheorie" where it is shown more generally that $$\sum_{R<q \le Q} \frac{1}{q}\sum_{(a,q)=1} |S_{x}(a/q)|^2 \ll \left( Q R^{-1}+x\right) \sum_{n \le x}|a_n|^2$$ for any $Q \ge R \ge 1$. See equations (5.12) and (5.14) in the same book for other variants of the Large Sieve, where $1/q$ is replaced by $(N+qQ)^{-1}$ and $\log(3Q/q)/q$, respectively.