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In a paper by Saffari and Vaughan there appears a complicated-looking double sum

$$\Sigma_1=\sum_{\rho_1}\sum_{\rho_2}\frac{(1+\theta)^{\rho_1}-1}{\rho_1}\cdot \frac{(1+\theta)^{\bar{\rho_2}}-1}{\bar{\rho_2}} \cdot \frac{2^{1+\rho_1+\bar{\rho_2}}-2^{-1-\rho_1-\bar{\rho_2}}}{1+\rho_1+\bar{\rho_2}} \cdot \frac{2^{2+\rho_1+\bar{\rho_2}}-1}{2+\rho_1+\bar{\rho_2}} \cdot x^{1+\rho_1+\bar{\rho_2}},$$

where both sums are over all non-trivial zeros $\rho_k=\beta_k+i\gamma_k$ of the Riemann zeta-function, and $\theta \in (0,1]$ is a parameter. Using the inequality $|z_1z_2|\leq\frac{1}{2}(|z_1|^2+|z_2|^2)$, they state that one can bound $\Sigma_1$ by

$$\Sigma_1\ll \sum_{\rho_1}\sum_{\rho_2} x^{1+2\beta_1} \min(\theta^2, \gamma_1^{-2}) (1+|\gamma_1-\gamma_2|)^{-2}.$$

I can see where the final factor $(1+|\gamma_1-\gamma_2|)^{-2}$ comes from, however the other two factors confuse me. The Riemann hypothesis is assumed in the final result, but does not appear to have been used here. Even if it was, I can only see how this implies the bound $x^{1+\rho_1+\bar{\rho_2}} \ll x^{1+2\beta_1}.$ What I mainly can't understand is how they bound

$$\frac{(1+\theta)^{\rho_1}-1}{\rho_1}\cdot \frac{(1+\theta)^{\bar{\rho_2}}-1}{\bar{\rho_2}} \ll \min(\theta^2, \gamma_1^{-2}).$$

Unconditionally, I can show that $$\frac{(1+\theta)^{\rho_1}-1}{\rho_1}\cdot \frac{(1+\theta)^{\bar{\rho_2}}-1}{\bar{\rho_2}} \ll \theta^2 \left(\frac{1}{\gamma_1^2} + \frac{1}{\gamma_2^2} \right).$$

But, to get the original bound, it seems like you would need to bound $\gamma_2$ with $\gamma_1$, which sounds unlikely at best. Am I missing any magic tricks to do with double-sums/convergence/Riemann zeros?

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    $\begingroup$ Is $\sum_{ρ1}\sum_{ρ2} \min(\theta^2, \gamma_1^{-2})$ even a convergent sum if the sum runs over all $\rho_2$ while the summand not depending on $\rho_2$? $\endgroup$ Dec 17, 2021 at 5:50
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    $\begingroup$ Thanks for pointing that out - the notation I used was a bit fishy. I've updated the question to be more clear. $\endgroup$ Dec 17, 2021 at 6:02

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Using $\theta\in[0,1]$ and $\mathrm{Re}(\rho)\leq 1$, we see that $$\frac{(1+\theta)^\rho-1}{\rho}=\int_1^{1+\theta}t^{\rho-1}\,dt$$ has absolute value at most $\min(\theta,3|\rho|^{-1})$. Therefore, $$\Sigma_1\ll x\sum_{\rho_1}\sum_{\rho_2}x^{\beta_1+\beta_2}\min(\theta,|\gamma_1|^{-1})\min(\theta,|\gamma_2|^{-1})(1+|\gamma_1-\gamma_2|)^{-2}.$$ Now we apply $$x^{\beta_1+\beta_2}\min(\theta,|\gamma_1|^{-1})\min(\theta,|\gamma_2|^{-1})\ll x^{2\beta_1}\min(\theta^2,\gamma_1^{-2})+x^{2\beta_2}\min(\theta^2,\gamma_2^{-2})$$ and the symmetry $\rho_1\leftrightarrow\rho_2$ to arrive at $$\Sigma_1\ll x\sum_{\rho_1}\sum_{\rho_2}x^{2\beta_1}\min(\theta^2,\gamma_1^{-2})(1+|\gamma_1-\gamma_2|)^{-2}.$$

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    $\begingroup$ Thanks so much! This was precisely the magic I was looking for. $\endgroup$ Dec 17, 2021 at 9:45
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    $\begingroup$ @DanielJohnston Thanks for the feedback. I should add that this kind of trick occurs frequently in analytic number theory. In particular, the baby Cauchy-Schwarz inequality $2|z_1z_2|\leq |z_1|^2+|z_2|^2$ is more useful than one would think. It can save a lot of trouble or simplify an argument considerably. $\endgroup$
    – GH from MO
    Dec 17, 2021 at 10:48

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