Let $T \in \mathbb{R}$ be large and $\rho$ be a non-trivial zero of the Riemann zeta function. Assume that $|\rho|=|\rho\_T| \approx T$ and let $\varepsilon\_T \approx \frac{\log \log T}{\log T}$. Is it true that


$$|\zeta(1-\rho\_T + \varepsilon\_T)|=o(1)$$


uniformly for $T \geq T\_0$?



On the upper bound for $|\zeta(s)|$ near the zeta zeros
If $\rho\_T$ is a zero of $\zeta(s)$, so is $1-\rho\_T$ by the functional equation, and $|1-\rho\_T|\approx T$ (depending on what you mean by $\approx$), so the $1-\ldots$ is superfluous.


A Taylor expansion of $\zeta(s)$ at $\rho\_T$ gives
$$
\zeta(\rho\_T+\epsilon)=\zeta^\prime(\rho\_T)\cdot \epsilon+O(\epsilon)^2.
$$
Even assuming the Riemann Hypothesis, all that is known is that
$$
\zeta^\prime(\rho\_T)=O(T^{\epsilon^\prime})
$$
for any $\epsilon^\prime$ (via Lindelof and the Cauchy Integral Formula for derivatives), so there's not much hope for a uniform bound.




---


Edit: I'm expanding on DavidFarmer's comment below. Selberg proved that for $E\subset \mathbb R$, we have that the limit as $T\to\infty$ of
$$
\frac{1}{T}\,\mu(T\le t\le 2T\,|\,\log|\zeta(1/2+i t)|/(1/2\log\log t)^{1/2}\in E)
$$
where $\mu$ is Lebesgue measure, is equal to the integral over $E$ of a Gaussian Random Variable with mean $0$ and standard deviation $1$:
$$
\frac{1}{\sqrt{2\pi}}\int\_E \exp(-x^2/2)\, dx.
$$
which is the basis for David's comment that $|\zeta(1/2+it)|$ is equal $1$ "on average". Since the count of zeros up to height $T$ is $\sim T/(2\pi)\log(T)$, at height $T$ the average spacing between zeros is $2\pi/\log(T)$. So the Hardy function $Z(t)$ is like $\sin(t \log(T)/2)$ near height $T$, and it's derivative has a multiplier of $\log(T)$.


