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 horizontal distribution of zeros of $\zeta^\prime(s)$
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 $\gg \log T$. 


