# Specific application of Cauchy-Schwarz and Large Sieve

Im reading a paper by Matomaki here, and the following is stated (I'm paraphrasing):

"By the Cauchy-Schwarz inequality and the large sieve, we have $$\sum_{q \leq Q}\frac{q}{\phi(q)}\sum_{\substack{\chi\text{(mod q)}\\\text{primtiive}}} \big{|}\sum_{a \in \mathcal{A}}\chi(a)\sum_{b \in \mathcal{B}}\chi(b) \big{|} \leq (Q^2 + N)(AB)^{1/2}$$ where $$Q,N$$ are positive integers and $$\mathcal{A}, \mathcal{B}\subseteq \{1,...,N\}$$ and $$|\mathcal{A}| = A$$ and $$|\mathcal{B}| = B$$."

Now I am not so concerned with the application of the large sieve, but I am a little confused about how she applied Cauchy-Schwarz. Of course the large sieve she is referring to states $$\sum_{q \leq Q}\frac{q}{\phi(q)}\sum_{\substack{\chi\text{(mod q)}\\\text{primtiive}}} \big{|}\sum_{n \leq N }a_n\chi(n)\big{|}^2 \leq (Q^2 + N)\sum_{n \leq N}|a_n|^2.$$

But I am unsure of how she used Cauchy-Schwarz, especially with multiplicative characters. Does anyone have any thoughts?

• See my response for a more direct (or more formal) argument. May 9, 2022 at 3:08

As she writes, first apply Cauchy-Schwarz, and only then apply the large sieve (twice). The relevant instance of Cauchy-Schwarz is $$|x_1 x_2| \le \frac{|x_1|^2+|x_2|^2}{2},$$ which, by replacing $$x_1$$ and $$x_2$$ by $$x_1\sqrt{C}$$ and $$x_2/\sqrt{C}$$ ($$C>0$$) becomes $$|x_1 x_2| \le \frac{C |x_1|^2 + C^{-1} |x_2|^2}{2}.$$ We apply it with $$x_1=\sum_{a \in \mathcal{A}} \chi(a)$$ and $$x_2 = \sum_{b \in \mathcal{B}}\chi(b)$$ and with $$C$$ to be determined later (but independent of $$\chi$$). We obtain that the relevant sum is $$\le \frac{1}{2}\left( C\sum_{q \le Q} \frac{q}{\phi(q)}\sum_{\substack{\chi \bmod q\\ \text{primitive}}}\left|\sum_{a \in \mathcal{A}} \chi(a)\right|^2 + C^{-1}\sum_{q \le Q} \frac{q}{\phi(q)}\sum_{\substack{\chi \bmod q\\ \text{primitive}}} \left|\sum_{b \in \mathcal{B}} \chi(b)\right|^2 \right),$$ which, by two applications of the large sieve, is $$\le \frac{1}{2}(Q^2 + N)\left( CA+ C^{-1} B\right).$$ Now take $$C=\sqrt{B/A}$$.
The answer of Ofir Gorodetsky is perfectly fine, but one can also apply the Cauchy-Schwarz inequality for $$L^2$$ spaces directly.
Indeed, let us consider the $$L^2$$ space of functions on the set of pairs $$(q,\chi)$$, where the measure of the pair $$(q,\chi)$$ is $$q/\phi(q)$$. Then, for any complex-valued functions $$f$$ and $$g$$ on the set of these pairs, the inequality says that $$\sum_{\substack{q \leq Q\\\text{\chi mod q}\\\text{primitive}}}\frac{q}{\phi(q)}|f(q,\chi)g(q,\chi)|\leq\Biggl(\sum_{\substack{q \leq Q\\\text{\chi mod q}\\\text{primitive}}}\frac{q}{\phi(q)}|f(q,\chi)|^2\Biggr)^{1/2}\Biggl(\sum_{\substack{q \leq Q\\\text{\chi mod q}\\\text{primitive}}}\frac{q}{\phi(q)}|g(q,\chi)|^2\Biggr)^{1/2}.$$ Applying this for $$f(q,\chi):=\sum_{a\in\mathcal{A}}\chi(a) \qquad\text{and}\qquad g(q,\chi):=\sum_{b\in\mathcal{B}}\chi(b)$$ yields the claimed result (using the large sieve inequality).
• You don't have to use the language of measure as the inequality becomes evident when you split $q/\varphi(q)$ into two square roots. Aug 9, 2022 at 18:23