# The distribution of Hardy's function $Z(t)$ and the argument function $S(t)$

Posted in
Speaker:
Aleksandar Ivic
Affiliation:
Serbian Academy of Sciences and Arts
Date:
Mon, 2018-09-03 16:30 - 17:00
Location:
MPIM Lecture Hall

Hardy's function is ($t \in \mathbb R$)
$$Z(t) := \zeta(1/2+it)\bigl(\chi(1/2+it)\bigr)^{-1/2},$$
where $\zeta(s) \;=\; \chi(s)\zeta(1-s)$ is the functional equation for the Riemann zeta-function $\zeta(s)$.
The argument function is
$$S(t) \;:=\; \frac{1}{\pi}\arg \zeta(1/2 + it)\qquad(t>0, \;t \ne \gamma),$$
where $\rho = \beta+i\gamma$ denotes generic complex zeros of $\zeta(s)$. If $t=\gamma,$
$S(t) \;=\; S(t+0)$. These important functions are real-valued, so one may naturally ask
about the distribution of their positive and negative values.

S.M. Gonek and the speaker (2017) proved that ($\mu(\cdot)$ denotes measure)
$$\mu\bigl(T\in [T,2T]\;:\; Z(t)>0\bigr) \;\gg\; T, \quad \mu\bigl(T\in [T,2T]\;:\; Z(t)<0\bigr)\; \gg\; T,\quad(1)$$
and explicit bounds if one assumes the RH and the pair correlation conjecture.

The speaker (2018) derived the identity
$$Z(t) = (-1)^{N(t)+1}|\zeta(1/2+it)|\qquad(t>0),$$
where, as usual, $N(T) = \sum\limits_{0<\gamma\le T}1$.

Recently M.A. Korolev and the speaker (2018) proved the analogue of (1) for $S(t)$:
Suppose that $0<\varepsilon<10^{-3}$ is an arbitrary small fixed constant,
$T\ge T_{0}(\varepsilon)$, $T^{c+\varepsilon} \le H \le T$,
where $c = \tfrac{27}{82}$. Then
$$\mu\Bigl\{t\in [T, T+H]\;:\; S(t)>0\Bigr\} = \frac{H}{2} + O\left(H\frac{\log\log\log T}{\varepsilon\sqrt{\log\log T}}\right),\quad(2)$$
where the $O$-constant is absolute. A formula analogous to (2) holds for the values $S(t) <0$.
This result is derived from an asymptotic formula
for the distribution of values of $S(t)$ over short intervals.
This is uniform in the relevant parameters,
which  is of crucial importance. This in fact depends on the distribution of values of the
Dirichlet polynomial which approximates $S(t)$, namely ($p$ denotes primes)
$$V_{y}(t)\,=\,\sum\limits_{p\le y}\frac{\sin{(t\log{p})}}{\sqrt{p}}.$$

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