N(T)\left[ {\widehat}g(0)+2\int_{0}^{1}{{\widehat}g}(x)x{\,\text{\rm d}}x + o(1)\right]  = N(T)\left[\frac{{\mathcal{Z}}(f)}{r(f)} + o(1)\right].\end{aligned}$$ On the other hand, clearly we have $$\begin{aligned}
\sum_{0<\gamma, \gamma'\leq T} g\bigg((\gamma-\gamma')\dfrac{\log T}{2\pi}\bigg)w(\gamma-\gamma') \geq g(0)\sum_{0<\gamma\leq T} m_\rho = \frac{N^*(T)}{r(f)}.\end{aligned}$$ Combining these results we show the first inequality in the theorem.

Assume now GRH. It is then shown in [@GGOS] that for any fixed and sufficiently small $\delta >0$ we have $$\label{Fasymptoticmore}
F(x,T) \geq \frac{3}{2}-|x| - o(1),$$ uniformly for $1\leq |x|\leq \frac{3}{2}-\delta$ as $T\to \infty$. Using this estimate and the fact that ${\widehat}g(x)\leq 0$ for $|x|\geq 1$ we obtain $$\begin{aligned}
\sum_{ 0<\gamma, \gamma'\leq T}{\widehat}g\bigg((\gamma-\gamma')\dfrac{\log T}{2\pi}\bigg)w(\gamma-\gamma')& \leq N(T)\left[{{\widehat}g}(0) + 2\int_{1}^{\frac{3}{2}-\delta}{{\widehat}g}(x)\left(\frac{3}{2}-x\right){\,\text{\rm d}}x + o(1)
\right] \\
& = N(T)\left[\frac{{\widetilde}{\mathcal{Z}}(f)}{r(f)}+o(1) +O(\delta)\right].\end{aligned}$$ Arguing as before we finish the proof.

To analyze $N(x,T)$ we define the [function]{} $${\mathcal{P}}(f) = \inf \big\{{\lambda}>0: p_f({\lambda}) > 0 \big\},$$ where $$p_f({\lambda})=-1+\frac{{\lambda}}{r(f)}+ \frac{2r(f)}{{\lambda}}\int_0^{\frac{{\lambda}}{r(f)}}{\widehat}f(x)x{\,\text{\rm d}}x,$$ and the [function]{} $${\widetilde}{\mathcal{P}}(f) = \inf\big\{{\lambda}>0: {\widetilde}p_f({\lambda}) > 0 \big\},$$ where $${\widetilde}p_f({\lambda})=-1+\frac{{\lambda}}{r(f)}+ \frac{2r(f)}{{\lambda}}\int_0^{\frac{{\lambda}}{r(f)}}{\widehat}f(x)x{\,\text{\rm d}}x +3\int_{\frac{{\lambda}}{r(f)}}^{\frac{3{\lambda}}{2r(f)}}{\widehat}f(x){\,\text{\rm d}}x -  \frac{2r(f)}{{\lambda}}\int_{\frac{{\lambda}}{r(f)}}^{\frac{3{\lambda}}{2r(f)}}{\widehat}f(x)x{\,\text{\rm d}}x.$$ Note that these functions are well defined since $p_f$ and ${\widetilde}p_f$ are $C^1$ functions that assume $-1$ at ${\lambda}=0$, and using the fact that $\widehat{f}\in L^1(\mathbb{R})$ one can show $$\lim_{{\lambda}\to\infty} \frac{p_f({\lambda})}{{\lambda}} = \lim_{{\lambda}\to\infty} \frac{{\widetilde}p_f({\lambda})}{{\lambda}} =\frac{1}{r(f)} > 0.$$

\[thmsmallgapszerosofxi\] Let $f\in {\mathcal{A}}_{LP}$ and ${\varepsilon}>0$. Assuming RH and we have $$N({\mathcal{P}}(f)+\varepsilon,T)\gg 