# The horizontal distribution of zeros of $\zeta^\prime(s)$

I have a question about a detail in the proof of Proposition 1.6 in "The horizontal distribution of zeros of $\zeta^\prime(s)$", K. Soundararajan, Duke J. Math. vol. 91 1998. Throughout I will simplify by assuming the Riemann Hypothesis. (His results are more general, merely assuming no counterexamples 'nearby')

Lemma 2.1 (p. 40) states that if $\rho_1=\beta_1+i\gamma_1$ is a zero of $\zeta^\prime(s)$ with $T<\gamma_1<2T$, and $\rho$ is a zero of $\zeta(s)$, then $$\left|\rho-\rho_1\right|^2\ge \frac{2(\beta_1-1/2)}{\log T}.$$ The proof requires the Hadamard factorization formula evaluated at $s=\rho_1$, taking real parts, and Stirling's formula.

Proposition 1.6 (stated on p. 39) states that if $1/2+i\gamma^\prime$ and $1/2+i\gamma^{\prime\prime}$ are two consecutive zeros of $\zeta(s)$, $T<\gamma^\prime,\gamma^{\prime\prime}<2T$, then the rectangle $$\left\{s=\sigma+i t\,:\, 1/2<\sigma<1/2+1/\log T,\quad \gamma^\prime<t<\gamma^{\prime\prime}\right\}$$ contains at most one zero of $\zeta^\prime(s)$. The proof (on p. 53) proceeds by contradiction: suppose $s_1=\sigma_1+it_1$ and $s_2=\sigma_2+it_2$ are two zeros of $\zeta^\prime$ in the rectangle. Evaluating $\zeta^\prime/\zeta(s_1)=0=\zeta^\prime/\zeta(s_2)$ in the Hadamard factorization and taking the difference, one obtains $$0=\sum_{\rho=1/2+i\gamma}\frac{s_1-s_2}{(s_1-\rho)(s_2-\rho)}+O\left(\frac{|s_1-s_2|}{T}\right).$$ Dividing by $s_1-s_2$ and taking real parts one obtains $$\sum_{\rho=1/2+i\gamma}\frac{(\sigma_1-1/2)(\sigma_2-1/2)-(t_1-\gamma)(t_2-\gamma)}{|s_1-\rho|^2|s_2-\rho|^2}=O\left(\frac1T\right)$$

"We will show that for $s_1$, $s_2$ in the rectangle, the numerator of the left side is always negative. This is clearly untenable and would yield the proposition."

I follow the proof that the numerator is negative (sketched below for completeness). The actual sign is irrelevant because of the Big O (or the arbitrary ordering of $s_1$ and $s_2$.) Rather, the point must be that there is no cancellation in the sum, and so it is too large to be $O(1/T)$ But this is where I'm stuck.

Why isn't this sum $O(1/T)$?

The proof of non-negativity proceeds as follows: Since $\gamma^\prime<t_1<t_2<\gamma^{\prime\prime}$ are consecutive zeros, $(t_1-\gamma)(t_2-\gamma)$ is non-negative, and we need only show that $$|t_1-\gamma||t_2-\gamma|\ge(\sigma_1-1/2)(\sigma_2-1/2).$$ But (for $j=1,2$) $$2(\sigma_j-1/2)^2\le \frac{2(\sigma_j-1/2)}{\log T}$$ by hypothesis, and $$\frac{2(\sigma_j-1/2)}{\log T}\le (\sigma_j-1/2)^2+(t_j-\gamma)^2$$ by Lemma 2.1. This gives the desired inequality.

The zeta function will have $\gg \log T$ zeros with ordinates in the interval $[t_2+1,t_2+2]$. The contribution of these zeros to the sum is of absolute value $\gg \log T$.