I am reading about Dirichlet polynomials in the book Analytic Number Theory by the said authors. Can anyone justify the following inequality? Assume that $a(n),b(m)$ are sequences of non-negative numbers such that $a(n)=0$ for all large enough $n$ and $b(m)=0$ for all large enough $m$. How does one prove $$ \sum_{\substack{ n,m , n_1, m_1 \in \mathbb N \\ n m =n_1 m_1 }}a(n)b(m)a(n_1) b(m_1) \leq \sum_n \sum_m a(n)^2 b(m)^2 \tau(nm )?$$ Here $\tau$ is the function that counts the number of positive integer divisors. This is after equation (9.7) in page 231.
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$\begingroup$ There are infinitely many divisor functions. Would you give more detail about $\tau$ ? By the way, $\tau$ is often used for Ramanujan's function, which is not a divisor function. $\endgroup$– Denis SerreCommented Mar 20, 2020 at 15:54
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2$\begingroup$ Don't feel offended. Wikipedia page en.wikipedia.org/wiki/Divisor_function speaks of functions $\sigma_x(n)$ parametrized by a real $x\ge0$. Apparently, yours is $\sigma_0$. I am confined at home and don't have the book next to me. $\endgroup$– Denis SerreCommented Mar 20, 2020 at 16:07
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13$\begingroup$ Just use $a(n)b(m)a(n_1)b(m_1) \le (a(n)^2b(m)^2 + a(n_1)^2 b(m_1)^2)/2$, and note that given $n$, $m$, there are at most $\tau(nm)$ choices for $n_1$, $m_1$. $\endgroup$– LuciaCommented Mar 20, 2020 at 16:16
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$\begingroup$ @Denis: in case you or your confined colleagues need references in this book, I have it at home, so feel free to ask. It would be nice to create some shared file among members of this site to know who owns which book, so that we can help one another. $\endgroup$– Sylvain JULIENCommented Mar 20, 2020 at 18:13
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Write the first sum as $$ \sum_{m,n }\sum_{d|mn}a(m)b(n)a(d)b(mn/d) $$ So we are considerung triples $(m,n,d)$ of natural numbers with $d|mn$. Let $S$ be the set of all triples $(m,n,d)$ with $m=d$. The triples not in $S$ come in pairs $$ (m,n,d)\quad\text{and}\quad (d,mn/d,n). $$ Pick one of each pair and let $T$ be the set of all the triples you picked in this way. Each $(m,n,m)\in S$ yields a summand $a(m)^2b(n)^2$. The two elements of a pair of triples yield the same summand, so for each $(m,n,d)\in T$ get a contribution $$ 2a(m)b(n)a(d)b(mn/d). $$ Let $A=a(m)b(n)$ and $B=a(d)b(mn/d)$. The estimate $2AB\le A^2+B^2$ now yields $$ \sum_{m,n }\sum_{d|mn}a(m)b(n)a(d)b(mn/d) =\sum_S+\sum_T\le \sum_{(m,n,m)\in S}a(m)^2b(n)^2\\ +\sum_{(m,n,d)\in T}a(m)^2b(n)^2+a(d)^2b(mn/d)^2\\ =\sum_{m,n}\sum_{d|mn}a(m)^2b(n)^2=\sum_{m,n}a(m)^2b(n)^2\tau(mn). $$