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.


1 Answer 1


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 $\gg \log T$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.