1 Introduction
Since the work of Levinson [Reference Levinson4], it has been known that one can obtain lower bounds for the proportion of zeros of the Riemann zeta-function on the critical line by computing upper bounds for the mollified second moment
$$\begin{eqnarray}\displaystyle I_{N}(T_{1},T_{2}):=\int _{T_{1}}^{T_{2}}|M_{N}({\textstyle \frac{1}{2}}+\text{i}t)|^{2}|\unicode[STIX]{x1D701}({\textstyle \frac{1}{2}}+\text{i}t)|^{2}\,dt, & & \displaystyle\end{eqnarray}$$
where
$M_{N}(s)$
is a mollifier roughly of the form
$$\begin{eqnarray}M_{N}(s):=\mathop{\sum }_{n\leqslant N}\frac{\unicode[STIX]{x1D707}(n)}{n^{s}}\biggl(1-\frac{\log n}{\log N}\biggr),\end{eqnarray}$$
with
$N\geqslant 2$
an integer. Levinson [Reference Levinson4] computed the asymptotic formula
$$\begin{eqnarray}\displaystyle \lim _{T\rightarrow \infty }\frac{I_{T^{\unicode[STIX]{x1D703}}}(0,T)}{T}=1+\frac{1}{\unicode[STIX]{x1D703}} & & \displaystyle\end{eqnarray}$$
for
$0<\unicode[STIX]{x1D703}<1/2$
, and used this result to deduce that
$\unicode[STIX]{x1D705}>1/3$
, where
$$\begin{eqnarray}\unicode[STIX]{x1D705}:=\frac{\#\{\unicode[STIX]{x1D70C}\mid \unicode[STIX]{x1D701}(\unicode[STIX]{x1D70C})=0,~0<\Im \,\unicode[STIX]{x1D70C}<T,~\Re \,\unicode[STIX]{x1D70C}=\frac{1}{2}\}}{\#\{\unicode[STIX]{x1D70C}\mid \unicode[STIX]{x1D701}(\unicode[STIX]{x1D70C})=0,~0<\Im \,\unicode[STIX]{x1D70C}<T\}}\end{eqnarray}$$
is the proportion of the non-trivial zeros of
$\unicode[STIX]{x1D701}(s)$
that lie on the critical line. Conrey [Reference Conrey1] later proved that (1.2) (with a slightly different mollifier) remains valid for
$\unicode[STIX]{x1D703}<4/7$
, and thereby deduced that
$\unicode[STIX]{x1D705}>2/5$
.
Initially, it was believed (see [Reference Farmer2]) that (1.2) does not hold when
$\unicode[STIX]{x1D703}>1$
. However, Farmer [Reference Farmer2] produced a heuristic argument suggesting that it holds for every
$\unicode[STIX]{x1D703}>0$
, and called this the “
$\unicode[STIX]{x1D703}=\infty$
conjecture”. Moreover, he proved that this conjecture implies that
$\unicode[STIX]{x1D705}=1$
, in other words, that
$100\%$
of the non-trivial zeros of
$\unicode[STIX]{x1D701}(s)$
lie on the critical line. He also argued that a slightly stronger form of the conjecture implies Montgomery’s pair correlation conjecture. More recently, Radziwiłł [Reference Radziwiłł6] showed that, as
$\unicode[STIX]{x1D703}\rightarrow \infty$
,
$M_{T^{\unicode[STIX]{x1D703}}}(t)$
is essentially the best possible mollifier of length
$T^{\unicode[STIX]{x1D703}}$
for
$\unicode[STIX]{x1D701}(s)$
. In particular, his work implies that Levinson’s method can give
$\unicode[STIX]{x1D705}=1$
only if it is used with mollifiers of length
$T^{\unicode[STIX]{x1D703}}$
, where
$\unicode[STIX]{x1D703}$
is arbitrarily large.
The purpose of this note is to show that the
$\unicode[STIX]{x1D703}=\infty$
conjecture actually implies the Riemann hypothesis. Indeed, we show that even an upper bound of the form
$I_{N}(0,T)\ll T^{1+\unicode[STIX]{x1D700}}$
, for some
$\unicode[STIX]{x1D703}>1$
and all
$N$
in the range
$2\leqslant N\leqslant T^{\unicode[STIX]{x1D703}}$
, implies a zero-free region for the zeta-function of the form
$\Re \,s>1-\unicode[STIX]{x1D6FF}$
, for some
$\unicode[STIX]{x1D6FF}>0$
depending on
$\unicode[STIX]{x1D703}$
: in other words, a quasi-Riemann hypothesis.
Theorem 1. Let
$\unicode[STIX]{x1D703}>0$
and assume that, for every
$\unicode[STIX]{x1D700}>0$
,
$I_{N}(0,T)\ll _{\unicode[STIX]{x1D700}}T^{1+\unicode[STIX]{x1D700}}$
for
$N$
in the range
$2\leqslant N\leqslant T^{\unicode[STIX]{x1D703}}$
. Then
$\unicode[STIX]{x1D701}(s)$
has no zero in the half-plane
$\Re \,s>1/2+1/2\unicode[STIX]{x1D703}$
. In particular, if
$I_{N}(0,T)\ll _{\unicode[STIX]{x1D700}}T^{1+\unicode[STIX]{x1D700}}$
for
$2\leqslant N\leqslant T^{\unicode[STIX]{x1D703}}$
with
$\unicode[STIX]{x1D703}$
arbitrarily large, then the Riemann hypothesis is true.
In a number of recent works on mean values of
$L$
-functions in the
$t$
-aspect, the integral is taken over
$[T,2T]$
rather than over
$[0,T]$
. Thus, it is natural to ask whether one can obtain a version of Theorem 1 for the interval
$[T,2T]$
. Usually, there is no difficulty in passing from one interval to the other. In our case, however, the problem for
$[T,2T]$
is more subtle because one needs an
$\unicode[STIX]{x1D6FA}$
-result for
$M_{N}(t)$
that is uniform in
$t$
. Using ideas from [Reference Pintz5] and [Reference Gonek, Graham and Lee3], we prove the following.
Theorem 2. Let
$\unicode[STIX]{x1D703}>0$
and assume that, for every
$\unicode[STIX]{x1D700}>0$
,
$I_{N}(T,2T)\ll _{\unicode[STIX]{x1D700}}T^{1+\unicode[STIX]{x1D700}}$
for
$N$
in the range
$2\leqslant N\leqslant T^{\unicode[STIX]{x1D703}}$
. Then
$\unicode[STIX]{x1D701}(s)$
has no zero in the half-plane
$\Re \,s>1/2+2/\unicode[STIX]{x1D703}$
. In particular, if
$I_{N}(T,2T)\ll _{\unicode[STIX]{x1D700}}T^{1+\unicode[STIX]{x1D700}}$
for
$2\leqslant N\leqslant T^{\unicode[STIX]{x1D703}}$
with
$\unicode[STIX]{x1D703}$
arbitrarily large, then the Riemann hypothesis is true.
Notice that Theorem 2 only implies a quasi-Riemann hypothesis when
$\unicode[STIX]{x1D703}>4$
, so, in this respect, it is weaker than Theorem 1. However, Theorem 2, whose proof is more difficult than that of Theorem 1, is, in a certain sense, best possible. If, for example, one assumes that
$\unicode[STIX]{x1D701}(s)$
has a unique simple zero
$\unicode[STIX]{x1D70C}_{0}=\unicode[STIX]{x1D6FD}_{0}+\text{i}\unicode[STIX]{x1D6FE}_{0}$
such that
$\unicode[STIX]{x1D6FE}_{0}>0$
and
$\unicode[STIX]{x1D6FD}_{0}>1/2$
, one can show that
$$\begin{eqnarray}\displaystyle I_{N}(T,2T) & = & \displaystyle c_{1}\frac{N^{2\unicode[STIX]{x1D6FD}_{0}-1}}{T^{3}}\frac{\log T}{\log ^{2}N}\biggl(1+\Re \biggl(N^{2\text{i}\unicode[STIX]{x1D6FE}_{0}}\frac{|\unicode[STIX]{x1D701}^{\prime }(\unicode[STIX]{x1D70C}_{0})|^{2}}{\unicode[STIX]{x1D701}^{\prime }(\unicode[STIX]{x1D70C}_{0})^{2}}\biggr)+o(1)\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,O\biggl(T^{1+\unicode[STIX]{x1D700}}+\frac{N^{\unicode[STIX]{x1D6FD}_{0}-1/2+\unicode[STIX]{x1D700}}}{T}\biggr)\nonumber\end{eqnarray}$$
for some constant
$c_{1}>0$
as
$T\rightarrow \infty$
, and this is consistent with the assumption that
$I_{T^{\unicode[STIX]{x1D703}}}(T,2T)\ll T^{1+\unicode[STIX]{x1D700}}$
if
$\unicode[STIX]{x1D703}<4$
. For the sake of comparison, we note that, with the same zero configuration,
$$\begin{eqnarray}\displaystyle I_{N}(0,T)=\frac{N^{2\unicode[STIX]{x1D6FD}_{0}-1}}{\log ^{2}N}(C(N)+o(1))+O(T^{1+\unicode[STIX]{x1D700}}+N^{\unicode[STIX]{x1D6FD}_{0}-1/2+\unicode[STIX]{x1D700}}T^{\unicode[STIX]{x1D700}}) & & \displaystyle \nonumber\end{eqnarray}$$
for some positive function
$C(N)$
bounded away from zero, so that
$I_{T^{\unicode[STIX]{x1D703}}}(0,T)\ll T^{1+\unicode[STIX]{x1D700}}$
implies that
$\unicode[STIX]{x1D6FD}_{0}\leqslant 1/2+1/2\unicode[STIX]{x1D703}$
, which is consistent with Theorem 1.
2 Proof of the theorems
We will prove Theorems 1 and 2 at the same time. It should be pointed out, however, that an easier argument would suffice for the former.
We begin by extending our earlier definition of
$M_{N}(s)$
slightly by writing
$$\begin{eqnarray}\displaystyle M_{x}(s)\log x=\mathop{\sum }_{n\leqslant x}\frac{\unicode[STIX]{x1D707}(n)}{n^{s}}\log (x/n) & & \displaystyle\end{eqnarray}$$
for
$x>0$
(with
$M_{1}(s):=0$
). Notice that the right-hand side is zero when
$0<x\leqslant 1$
and that this also allows us to extend the definition of
$I_{N}(T_{1},T_{2})$
in (1.1) to
$I_{x}(T_{1},T_{2})$
. Now, for
$t\in \mathbb{R}$
,
$$\begin{eqnarray}\displaystyle M_{x}({\textstyle \frac{1}{2}}+\text{i}t)\log x=\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{1-\text{i}\infty }^{1+\text{i}\infty }\frac{x^{z}}{\unicode[STIX]{x1D701}(\frac{1}{2}+\text{i}t+z)}\,\frac{dz}{z^{2}}. & & \displaystyle \nonumber\end{eqnarray}$$
Thus, by Mellin inversion, we see that
$$\begin{eqnarray}\displaystyle H_{t}(w):=\int _{1}^{\infty }M_{x}({\textstyle \frac{1}{2}}+\text{i}t)(\log x)x^{-w}\,dx=\frac{1}{(w-1)^{2}\unicode[STIX]{x1D701}(w-\frac{1}{2}+\text{i}t)} & & \displaystyle \nonumber\end{eqnarray}$$
for
$\Re \,w>3/2$
. Next, assuming that
$\unicode[STIX]{x1D70C}_{0}=\unicode[STIX]{x1D6FD}_{0}+\text{i}\unicode[STIX]{x1D6FE}_{0}$
is a fixed zero of
$\unicode[STIX]{x1D701}(w)$
with
$\unicode[STIX]{x1D6FD}_{0}\geqslant 1/2$
, we define
$$\begin{eqnarray}\displaystyle G_{t}(w):=\frac{(w-1)^{2}(w-\frac{3}{2}+\text{i}t)\unicode[STIX]{x1D701}(w-\frac{1}{2}+\text{i}t)}{(w+1)^{2}(w-\frac{1}{2}+\text{i}t-\unicode[STIX]{x1D70C}_{0})(w+\text{i}t+1)^{4}}. & & \displaystyle \nonumber\end{eqnarray}$$
In the half-plane
$\Re \,w\geqslant 0$
,
$G_{t}(w)$
is holomorphic and satisfies
$G_{t}(w)\ll (1+|w+\text{i}t|)^{-5/2}$
. Thus, setting
$$\begin{eqnarray}\displaystyle g_{t}(u)=\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{3-\text{i}\infty }^{3+\text{i}\infty }G_{t}(w)u^{-w}\,dw & & \displaystyle \nonumber\end{eqnarray}$$
for
$u>0$
,
$$\begin{eqnarray}\displaystyle g_{t}(u)=\left\{\begin{array}{@{}ll@{}}0\quad & \text{if }u>1,\\ O(1)\quad & \text{if }0\leqslant u\leqslant 1,\end{array}\right. & & \displaystyle\end{eqnarray}$$
as can be seen by moving the line of integration to
$\Re \,w=+\infty$
when
$u>1$
, and to
$\Re \,w=0$
when
$0\leqslant u\leqslant 1$
.
Now consider the integral
$$\begin{eqnarray}\displaystyle \hspace{-10.00002pt}J_{t}(x) & := & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{3-\text{i}\infty }^{3+\text{i}\infty }G_{t}(w)H_{t}(w)x^{w}\,dw\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{3-\text{i}\infty }^{3+\text{i}\infty }\frac{(w-\frac{3}{2}+\text{i}t)x^{w}}{(w+1)^{2}(w-\frac{1}{2}+\text{i}t-\unicode[STIX]{x1D70C}_{0})(w+\text{i}t+1)^{4}}\,dw,\end{eqnarray}$$
where, from this point on, we assume that
$x\geqslant 2$
. On the one hand, by the convolution formula for products of Mellin transforms, and since
$M_{y}(1/2+it)\log y=0$
when
$0<y\leqslant 1$
,
$$\begin{eqnarray}\displaystyle J_{t}(x)=\int _{1}^{\infty }M_{y}({\textstyle \frac{1}{2}}+\text{i}t)(\log y)g_{t}(y/x)\,dy. & & \displaystyle \nonumber\end{eqnarray}$$
Thus, by (2.2),
$$\begin{eqnarray}\displaystyle J_{t}(x)\ll \int _{1}^{x}|M_{y}({\textstyle \frac{1}{2}}+\text{i}t)|\log y\,dy & & \displaystyle\end{eqnarray}$$
for
$x\geqslant 2$
. On the other hand, moving the line of integration in (2.3) to
$\Re \,w=0$
, we see that
$$\begin{eqnarray}\displaystyle \hspace{-10.00002pt}J_{t}(x)=\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{-\text{i}\infty }^{+\text{i}\infty }G_{t}(w)H_{t}(w)x^{w}\,dw+\frac{x^{\unicode[STIX]{x1D70C}_{0}+1/2-\text{i}t}(\unicode[STIX]{x1D70C}_{0}-1)}{(\frac{3}{2}+\unicode[STIX]{x1D70C}_{0}-\text{i}t)^{2}(\unicode[STIX]{x1D70C}_{0}+\frac{3}{2})^{4}}. & & \displaystyle\end{eqnarray}$$
The integral on the right is
$O(1)$
since
$H_{t}(w)G_{t}(w)\ll (1+|w|)^{-2}$
for
$\Re \,w=0$
. Thus, from (2.4) and (2.5), we deduce that
$$\begin{eqnarray}\displaystyle \frac{x^{\unicode[STIX]{x1D6FD}_{0}+1/2}}{(1+|t|)^{2}}+1\ll \int _{1}^{x}|M_{y}({\textstyle \frac{1}{2}}+\text{i}t)|\log y\,dy. & & \displaystyle \nonumber\end{eqnarray}$$
It follows from the Cauchy–Schwarz inequality that
$$\begin{eqnarray}\displaystyle \frac{x^{2\unicode[STIX]{x1D6FD}_{0}}}{(1+|t|)^{4}}+\frac{1}{x}\ll \int _{1}^{x}|M_{y}({\textstyle \frac{1}{2}}+\text{i}t)|^{2}\log ^{2}y\,dy & & \displaystyle \nonumber\end{eqnarray}$$
for
$x\geqslant 2$
. Multiplying both sides by
$|\unicode[STIX]{x1D701}(1/2+\text{i}t)|^{2}$
and integrating with respect to
$t$
over the interval
$[T_{1},T_{2}]$
, where
$0\leqslant T_{1}\leqslant T_{2}/2$
, we obtain
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{T_{1}}^{T_{2}}|\unicode[STIX]{x1D701}({\textstyle \frac{1}{2}}+\text{i}t)|^{2}\bigg(\frac{x^{2\unicode[STIX]{x1D6FD}_{0}}}{(1+t)^{4}}+\frac{1}{x}\bigg)\,dt\nonumber\\ \displaystyle & & \displaystyle \quad \ll \int _{1}^{x}\log ^{2}y\int _{T_{1}}^{T_{2}}|M_{y}({\textstyle \frac{1}{2}}+\text{i}t)\unicode[STIX]{x1D701}({\textstyle \frac{1}{2}}+\text{i}t)|^{2}\,dt\,dy\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \log ^{2}x\int _{1}^{x}I_{y}(T_{1},T_{2})\,dy.\nonumber\end{eqnarray}$$
Now
$\int _{T_{1}}^{T_{2}}|\unicode[STIX]{x1D701}(1/2+\text{i}t)|^{2}\,dt\gg T_{2}\log (T_{2}+2)$
for
$0\leqslant T_{1}\leqslant T_{2}/2$
, so
$$\begin{eqnarray}\displaystyle \frac{x^{2\unicode[STIX]{x1D6FD}_{0}}\log (T_{1}+2)}{|1+T_{1}|^{3}}+\frac{T_{2}\log (T_{2}+2)}{x}\ll \log ^{2}x\int _{1}^{x}I_{y}(T_{1},T_{2})\,dy. & & \displaystyle \nonumber\end{eqnarray}$$
Thus, if
$I_{N}(0,T)\ll _{\unicode[STIX]{x1D700}}T^{1+\unicode[STIX]{x1D700}}$
holds for
$2\leqslant N\leqslant T^{\unicode[STIX]{x1D703}}$
and for every
$\unicode[STIX]{x1D700}>0$
, then, taking
$T_{1}=0$
,
$T_{2}=T$
and
$x=T^{\unicode[STIX]{x1D703}}$
, we obtain
$$\begin{eqnarray}\displaystyle T^{2\unicode[STIX]{x1D6FD}_{0}\unicode[STIX]{x1D703}}\ll _{\unicode[STIX]{x1D700}}T^{1+\unicode[STIX]{x1D700}+\unicode[STIX]{x1D703}}. & & \displaystyle \nonumber\end{eqnarray}$$
Letting
$T\rightarrow \infty$
and letting
$\unicode[STIX]{x1D700}>0$
be sufficiently small, we obtain
$\unicode[STIX]{x1D6FD}_{0}\leqslant 1/2+1/2\unicode[STIX]{x1D703}$
, as claimed in Theorem 1. Theorem 2 follows in the same way on taking
$T_{1}=T$
and
$T_{2}=2T$
.
Acknowledgements
The first author would like to thank Brian Conrey and Jon Keating for bringing this problem to his attention. Both authors wish to thank the organizers of the workshop “Computational Aspects of L-functions” and ICERM for providing an excellent environment for collaboration.
