arXiv:1801.05914v5 [math.NT] 3 Jul 202
                                         Forum of Mathematics, Pi (2021), vol. ???, e1, 61 pages
                                                                                                                                                                        1
                                         doi:???




                                                             THE DE BRUIJN-NEWMAN CONSTANT IS
                                                                       NON-NEGATIVE

                                                                      BRAD RODGERS1 and TERENCE TAO2
                                                    1
                                                    Dept. of Math. and Stat., Queen’s University, Kingston ON K7L 3N6, Canada;
                                                                              brad.rodgers@queensu.ca
                                               2
                                                 Department of Mathematics, UCLA, Los Angeles CA 90095, USA; tao@math.ucla.edu


                                                                                                Received ???


                                                                                                     Abstract

                                              For each t ∈ R, define the entire function
                                                                                    Z ∞
                                                                                            2
                                                                           Ht (z) ≔      etu Φ(u) cos(zu) du
                                                                                                 0

                                         where Φ is the super-exponentially decaying function
                                                                                     ∞
                                                                                     X
                                                                          Φ(u) ≔           (2π2 n4 e9u − 3πn2 e5u ) exp(−πn2 e4u ).
                                                                                     n=1

                                         Newman showed that there exists a finite constant Λ (the de Bruijn-Newman constant) such that
                                         the zeroes of Ht are all real precisely when t ≥ Λ. The Riemann hypothesis is the equivalent to the
                                         assertion Λ ≤ 0, and Newman conjectured the complementary bound Λ ≥ 0.
                                              In this paper we establish Newman’s conjecture. The argument proceeds by assuming for
                                         contradiction that Λ < 0, and then analyzing the dynamics of zeroes of Ht (building on the work
                                         of Csordas, Smith, and Varga) to obtain increasingly strong control on the zeroes of Ht in the
                                         range Λ < t ≤ 0, until one establishes that the zeroes of H0 are in local equilibrium, in the sense
                                         that locally behave (on average) as if they were equally spaced in an arithmetic progression, with
                                         gaps staying close to the global average gap size. But this latter claim is inconsistent with the
                                         known results about the local distribution of zeroes of the Riemann zeta function, such as the pair
                                         correlation estimates of Montgomery.

                                         2010 Mathematics Subject Classification: 11M06 (primary)




                                         © The Author(s) 2021. This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence
                                         (<http://creativecommons.org/licenses/by/3.0/>), which permits unrestricted re-use, distribution, and reproduction in any medium,
                                         provided the original work is properly cited.
                      Brad Rodgers and Terence Tao                                2

                                   1. Introduction
Let H0 : C → C denote the function
                                               !
                                        1 1 iz
                                H0 (z) ≔ ξ +     ,                              (1)
                                        8 2 2
where ξ denotes the Riemann xi function
                                    s(s − 1) −s/2  s 
                          ξ(s) ≔            π Γ         ζ(s)                    (2)
                                        2           2
and ζ is the Riemann zeta function. Then H0 is an entire even function with
functional equation H0 (z) = H0 (z), and the Riemann hypothesis is equivalent to
the assertion that all the zeroes of H0 are real.
    It is a classical fact (see [30, p. 255]) that H0 has the Fourier representation
                                       Z ∞
                              H0 (z) =     Φ(u) cos(zu) du
                                          0

where Φ is the super-exponentially decaying function
                          ∞
                          X
                 Φ(u) ≔         (2π2 n4 e9u − 3πn2e5u ) exp(−πn2 e4u ).         (3)
                          n=1

The sum defining Φ(u) converges absolutely for negative u also. From Poisson
summation one can verify that Φ satisfies the functional equation Φ(u) = Φ(−u)
(i.e., Φ is even).
    De Bruijn [4] introduced the more general family of functions Ht : C → C
for t ∈ R by the formula
                                 Z ∞
                                        2
                        Ht (z) ≔     etu Φ(u) cos(zu) du.                  (4)
                                      0

As noted in [11, p.114], one can view Ht as the evolution of H0 under the
backwards heat equation ∂t Ht (z) = −∂zz Ht (z). As with H0 , each of the Ht are
entire even functions with functional equation Ht (z) = Ht (z). From results of
Pólya [22] it is known that Ht has purely real zeroes for some t then Ht′ has
purely real zeroes for all t′ > t. De Bruijn showed that the zeroes of Ht are purely
real for t ≥ 1/2. Strengthening these results, Newman [18] showed that there is
an absolute constant −∞ < Λ ≤ 1/2, now known as the De Bruijn-Newman
constant, with the property that Ht has purely real zeroes if and only if t ≥ Λ.
The Riemann hypothesis is then clearly equivalent to the upper bound Λ ≤ 0.
Newman conjectured the complementary lower bound Λ ≥ 0, and noted that this
conjecture asserts that if the Riemann hypothesis is true, it is only “barely so”.
                   The De Bruijn-Newman constant is non-negative                                 3


Table 1. Previous lower bounds on Λ. Dates listed are publication dates. The final four results use
                         the method of Csordas, Smith, and Varga [11].
    Lower bound on Λ         Reference
    −∞                       Newman 1976 [18]
    −50                      Csordas-Norfolk-Varga 1988 [8]
    −5                       te Riele 1991 [29]
    −0.385                   Norfolk-Ruttan-Varga 1992 [19]
    −0.0991                  Csordas-Ruttan-Varga 1991 [10]
    −4.379 × 10−6            Csordas-Smith-Varga 1994 [11]
    −5.895 × 10−9            Csordas-Odlyzko-Smith-Varga 1993 [9]
    −2.63 × 10−9             Odlyzko 2000 [20]
    −1.15 × 10−11            Saouter-Gourdon-Demichel 2011 [23]


As progress towards this conjecture, several lower bounds on Λ were established:
see Table 1.
    We also mention that the upper bound Λ ≤ 1/2 of de Bruijn [4] was
sharpened slightly1 by Ki, Kim, and Lee [14] to Λ < 1/2. See also [25], [5]
on work on variants of Newman’s conjecture, and [3, Chapter 5] for a survey.
    The main result of this paper is to affirmatively settle Newman’s conjecture:

Theorem 1. One has Λ ≥ 0.

    We now discuss the methods of proof. Starting from the work of Csordas-
Smith-Varga [11], the best lower bounds on Λ were obtained by exploiting the
following repulsion phenomenon: if Λ was significantly less than zero, then
adjacent zeroes of H0 (or of the Riemann ξ function) cannot be too close to
each other (as compared with the other nearby zeroes). See [11, Theorem 1]
for a precise statement. In particular, a negative value of Λ gives limitations
on the quality of “Lehmer pairs” [15], which roughly speaking refer to pairs of
adjacent zeroes of the Riemann zeta function that are significantly closer to each
other than the average spacing of zeroes at that level. The lower bounds on Λ
in [11], [9], [20], [23] then follow from numerically locating Lehmer pairs of
increasingly high quality. (See also [26] for a refinement of the Lehmer pair
concept used in the above papers.)
    In principle, one could settle Newman’s conjecture by producing an infinite
sequence of Lehmer pairs of arbitrarily high quality. As suggested in [20],
we were able to achieve this under the Gaussian Unitary Ensemble (GUE)

1
    Added in press: this bound has recently been improved to Λ ≤ 0.22 in [21].
                         Brad Rodgers and Terence Tao                                            4

hypothesis on the asymptotic distribution of zeroes of the Riemann zeta function;
we do not detail this computation here2 as it is superseded by our main result.
However, without the GUE hypothesis, the known upper bounds on narrow gaps
between zeroes (e.g. [6]) do not appear to be sufficient to make this strategy
work, even if one assumes the Riemann Hypothesis (which one can do for
Theorem 1 without loss of generality). Instead, we return to the analysis in [11]
and strengthen the repulsion phenomenon to a relaxation to local equilibrium
phenomenon: if Λ is negative, then the zeroes of H0 are not only repelled
from each other, but will nearly always be arranged locally as an approximate
arithmetic3 progression, with the gaps between zero mostly staying very close to
the global average gap that is given by the Riemann-von Mangoldt formula.
    To obtain the local relaxation to equilibrium under the hypothesis that Λ < 0
requires a sequence of steps in which we obtain increasingly strong control on the
distribution of zeroes of Ht for Λ < t ≤ 0 (actually for technical reasons we will
need to move t away from Λ as the argument progresses, restricting instead to
ranges such as Λ/2 ≤ t ≤ 0 or Λ/4 ≤ t ≤ 0). The first step is to obtain Riemann-
von Mangoldt type formulae for the number of zeroes of Ht in an interval such
as [0, T ] or [T, T + α] where T ≥ 2 and 0 < α ≤ o(T ). When t = 0, we can
                         T       T     T                         α
obtain asymptotics of 4π   log 4π   − 4π  + O(log T ) and 4π       log T + o(log T ) by the
classical Riemann-von Mangoldt formula and a result of Littlewood respectively;
this gives good control on the zeroes down to length scales α ≍ 1. For Λ < t < 0,
we were only able to obtain the weaker bounds of 4π        T
                                                              log 4πT   T
                                                                      − 4π + O(log2 T ) and
 α                2
4π log T + o(log T ) respectively down to length scales α ≍ log T , but it turns
out that these bounds still (barely) suffice for our arguments; see Section 3. A
key input in the proof of the Riemann-von Mangoldt type formula will be some
upper and lower bounds for Ht (x − iy) when y is comparable to log x; see Lemma
4 for a precise statement. The main tool used to prove these bounds is the saddle
point method, in which various contour integrals are shifted until they resemble
the integral for the Gamma function, to which the Stirling approximation may be
applied.
    It was shown in [11] that in the region Λ < t ≤ 0, the zeroes x j (t) of Ht are
simple, and furthermore evolve according to the system of ordinary differential
equations
                                         X            1
                           ∂t xk (t) = 2                        ;                        (5)
                                               x
                                         j: j,k k
                                                  (t) − x j (t)


2
  A sketch of the argument may be found at terrytao.wordpress.com/2018/01/20.
3
  To illustrate the equilibrium nature of arithmetic progressions under backwards heat flow,
                                          2
consider the entire functions Ft (z) ≔ etu cos(zu) for some fixed real u > 0. These functions
                                                        1
                                                  2π(k+ 2 )
all have zeroes on the arithmetic progression {       u       : k ∈ Z} and solve the backwards heat
equation ∂t Ft = −∂zz F.
                  The De Bruijn-Newman constant is non-negative                                       5

see Theorem 11 for a more precise statement. One can view this equation as
describing the dynamics of a system of “particles” x j , in which every pair of
particles x j , xk experiences a repulsion4 that is inversely proportional to their
separation. By refining the analysis in [11], we can obtain a more quantitative
lower bound on the gap x j+1 (t) − x j (t) between adjacent “particles” (zeroes), in
particular establishing a bound of the form
                                           1
                            log                       ≪ log2 j log log j
                                  x j+1 (t) − x j (t)
for all large j in the range Λ/2 ≤ t ≤ 0; see Proposition 13 for a more precise
statement. While far from optimal, this bound almost allows one to define the
Hamiltonian                        X                   1
                          H(t) ≔          log                    ,
                                 j,k: j,k
                                              |x j (t) − xk (t)|

although in practice we will have to apply some spatial cutoffs in j, k to make this
series absolutely convergent. For the sake of this informal overview we ignore
this cutoff issue for now. The significance of this quantity is that the system (5)
can (formally, at least) be viewed as the gradient flow for the Hamiltonian H(t).
In particular, there is a formal monotonicity formula
                                         ∂t H(t) = −4E(t)                                           (6)
where the energy E(t) is defined as
                                   X                         1
                          E(t) ≔                                        .
                                           j,k: j,k
                                                    |x j (t) − xk (t)|2

Again, in practice one needs to apply spatial cutoffs to j, k to make this quantity
finite, and one then has to treat various error terms arising from this cutoff, which
                                                             1
among other things “renormalizes” the summands |x j (t)−x      k (t)|
                                                                     2 so that the renor-

malized energy vanishes when the zeroes are arranged in the equilibrium state of
an arithmetic progression; we ignore these issues for the current discussion. A
further formal calculation indicates that E(t) is monotone non-increasing in time
(so that H(t) is formally convex in time, as one would expect for the gradient
flow of a convex Hamiltonian). Exploiting (a variant of) the equation (6), we are
                                                                        R0
able to control integrated energies that resemble the quantities Λ/2 E(t) dt; see


 4
   We caution however that the dynamics here are not Newtonian in nature, since (5) prescribes the
velocity ∂t xk of each particle rather than the acceleration ∂2t xk . Nevertheless we found the physical
analogy to be helpful in locating the arguments used in this paper.
                        Brad Rodgers and Terence Tao                                   6

first the weak preliminary integrated energy bound in Proposition 15, and then
the final integrated energy bound in Theorem 17. By exploiting local monotonic-
ity properties of the energy (and using a pigeonholing argument of Bourgain [2]),
we can then obtain good control (a truncated version of) the energy E(t) at time
t = 0, which intuitively reflects the assertion that the “particles” x j (t) are close to
local equilibrium at time t = 0. This implies that the zeroes of the Riemann zeta
function behave locally like an arithmetic progression on the average. However,
this can be ruled out by the existing results on the local distribution of zeroes,
such as pair correlation estimates of Montgomery [16]. As it turns out, it will be
convenient to make use of a closely related estimate of Conrey, Ghosh, Goldston,
Gonek, and Heath-Brown [7].
     It may be possible to use the methods of this paper to also address the
generalized Newman conjecture introduced in [25], but we do not pursue this
direction here5 .

Remark 2. It is interesting to compare this with the results in [14, Theorem
1.14], which show that regardless of the value of Λ, the zeroes of Ht will be
spaced like an arithmetic progression on average for any positive t.

Remark 3. Added in press: we note that in forthcoming work, Alex Dobner has
found a proof that Λ ≥ 0 which avoids the heat equation approach we have used
here. Dobner’s approach instead relies on a Riemann-Siegel type approximation
for Ht in order to demonstrate the existence of zeros off the critical line. There is
also some very intriguing numerical work of Rudolph Dwars (see the comments
to terrytao.wordpress.com/2018/12/28) that suggest that many of the
zeroes of Ht , t < 0 away from the critical line organize around deterministic
curves.

1.1. Acknowledgments. The first author received partial support from the
NSF grant DMS-1701577 and an NSERC grant. The second author is supported
by NSF grant DMS-1266164 and by a Simons Investigator Award. We thank
anonymous referees for useful suggestions, and likewise we thank Charles
Newman for helpful comments and Alex Dobner for corrections.

5
  Note added in proof: the generalized Newman conjecture has now been established, with a
significantly simpler proof than the one given here: see [12].
               The De Bruijn-Newman constant is non-negative                        7

1.2. Notation. Throughout the rest of the paper, we will assume for sake of
contradiction that Newman’s conjecture fails:
                                        Λ < 0.
In particular this implies the Riemann hypothesis (which, as mentioned previ-
ously, is equivalent to the assertion Λ ≤ 0).
    We will have a number of logarithmic factors appearing in our upper bounds.
To avoid the minor issue of the logarithm occasionally being negative, we will
use the modified logarithm
                               log+ (x) ≔ log(2 + |x|)
for several of these bounds. We also use the standard branch of the complex
logarithm, with imaginary part in the interval (−π, π], and the standard branch
z1/2 ≔ exp( 21 log z) of the square root, defined using the standard branch of the
complex logarithm.
      Let Λ < t ≤ 0, then the zeroes of Ht are all real, and symmetric around the
origin. It is a result of Csordas, Smith, and Varga [11, Corollary 1] that the zeroes
are also distinct and avoid the origin. Thus we can express the zeroes of Ht as
(x j (t)) j∈Z∗ , where Z∗ ≔ Z\{0} are the non-zero integers,
                              0 < x1 (t) < x2 (t) < . . . ,
and x− j (t) = −x j (t) for all j ≥ 1.
   For any real numbers j− ≤ j+ , we use [ j− , j+ ]Z∗ to denote the discrete interval
                        [ j− , j+ ]Z∗ ≔ { j ∈ Z∗ : j− ≤ j ≤ j+ }.
    We use the usual asymptotic notation X ≪ Y, Y ≫ X, or X = O(Y) to denote
a bound of the form |X| ≤ CY for some absolute constant C, and write X ≍ Y for
X ≪ Y ≪ X. Note that as Λ is also an absolute constant, C can certainly depend
on Λ; thus for instance |Λ| ≍ 1. If we need the implied constant C to depend on
other parameters, we will indicate this by subscripts, thus for instance X = Oκ (Y)
denotes the estimate |X| ≤ Cκ Y for some C depending on κ. If the quantities X, Y
depend on an asymptotic parameter such as T , we write X = oT →∞ (Y) to denote
a bound of the form |X| ≤ c(T )Y, where c(T ) is a quantity that goes to zero as
T → ∞.
    For X and Y depending on an asymptotic parameter T , we will also use the
notation X / Y or X = Õ(Y) for X ≪ Y logO(1) T in the last two sections of this
paper.
    Furthermore, in sums that will appear which depend on a parameter T , we
say that indices j, k are nearby, and write j ∼T k, if one has 0 < | j − k| <
(T 2 + | j| + |k|)0.1 .
                           Brad Rodgers and Terence Tao                                           8

    We will use a marked sum to indicate principle value summation:
                                    ′
                                    X                   X
                                          · · · = lim            ··· .
                                                J→∞
                                      j                 | j|≤J

In cases where there is any chance of confusion for the range of summation we
record the index being summed and use a colon to indicate its range; e.g. we
       P
write j: j,k to indicate that the summation is over j, and j is to not equal k
(where k is fixed outside the sum). Semicolons are used to separate additional
conditions.
    We use the phrase for almost every t throughout this paper to denote that a
relation holds for all t except a set of null Lebesgue measure.

                                   2. Asymptotics of H t
    In this section we establish some upper and lower bounds on Ht (z) and its
                        H′
logarithmic derivative Htt (z). We will be able to obtain reasonable upper bounds
in the regime where z = x − iy with y = O(log+ x), and obtain more precise
asymptotics when y ≍ log+ x (as long as the ratio y/ log+ x is large enough); this
will be the key input for the Riemann-von Mangoldt type asymptotics in the next
section. More precisely, we show

Lemma 4. Let z = x − iκ log+ x for some x ≥ 0 and 0 ≤ κ ≤ C, and let Λ < t ≤ 0.
Then one has6                            πx           
                          Ht (z) ≪ exp − + OC (log2+ x) .                   (7)
                                           8
Furthermore, there is an absolute constant C ′ > 0 (not depending on C) such
that if κ ≥ C ′ , then one has the refinement
                                         πx           
                                                   2
                           Ht (z) = exp − + OC (log+ x) ,                   (8)
                                          8
as well as the additional estimate
                            Ht′
                                                             !
                                     i     iz       log+ x
                                (z) = log        + OC          ,                                (9)
                            Ht       4     4π           x
using the standard branch of the complex logarithm.

6
                                                                 √
  The reader is advised not to take the numerous factors of π, 2, etc. appearing in this section
too seriously, as the exact numerical values of these constants are not of major significance in the
rest of the arguments.
               The De Bruijn-Newman constant is non-negative                        9

Remark 5. With a little more effort one could replace the hypothesis Λ < t here
by −C < t; in particular (in contrast to the remaining arguments in this paper)
these results are non-vacuous when Λ ≥ 0. However, we will need to assume
Λ < t in the application of these estimates in the next section, particularly with
regards to the proof of (49). Our proof methods also allow for a more precise
version of the asymptotic (8) (as one might expect given the level of precision
in (9)), but such improvements do not seem to be helpful for the rest of the
arguments in this paper. In the t = 0 case, one can essentially obtain Dirichlet
                                  H′
series expansions for H01(z) or H00 (z) which allow one to also obtain bounds such
as (8) or (9) when the imaginary part of z is much smaller than log+ x. However,
in the t < 0 case there does not appear to be any usable series expansions for
 1         Ht′
 Ht (z) or Ht (z) that could be used to prove (8) or (9). Instead, we will prove these
estimates by computing Ht (z) to a high degree of accuracy, which we can only
do when y is greater than or equal to a large multiple of log+ x in order to ensure
that the series expansions we have for Ht (z) converge rapidly.
    We begin by treating the easy case t = 0, in which we can exploit the identity
(1). We have the very crude bound
                              ζ(σ + iτ) ≪ (1 + |τ|)O(1)                          (10)
whenever σ ≥ 1/2 and τ ∈ R (this follows for instance from [30, Theorem
4.11]). In the region σ ≥ 1/4, we also have the Stirling approximation (see e.g.
[1, 6.1.41])
                                                           √
                             !                                                !!
                           1                                            1
Γ(σ + iτ) = exp σ + iτ −       log(σ + iτ) − (σ + iτ) + log 2π + O               ,
                           2                                         |σ + iτ|
                                                                            (11)
where we use the standard branch of the logarithm; in particular
                                                                 !
                                   1                        τ
             Γ(σ + iτ) ≪ exp (σ − ) log |σ + iτ| − τ arctan − σ .           (12)
                                   2                        σ
As arctan στ = π2 sgn(τ) + O( σ+|τ|
                               σ
                                    ), we have in particular that
                                      π                            
                  Γ(σ + iτ) ≪ exp − |τ| + O(σ log+ (|σ| + |τ|)) .
                                        2
Inserting these bounds into (1), (2), we obtain the crude upper bound
                                                                       !
                                       π|x|
               H0 (x − iy) ≪ exp −          + O((1 + y) log+ (|x| + y))          (13)
                                        8
for x ∈ R and y ≥ 0. This gives the s = 0 case of (7). As is well known, when
σ ≥ 2 (say) we can improve (10) to
                                   |ζ(σ + iτ)| ≍ 1
                      Brad Rodgers and Terence Tao                                10

and so we obtain the improvement
                                                                     !
                                     π|x|
               H0 (x − iy) = exp −        + O((1 + y) log+ (|x| + y))
                                      8
when y ≥ C ′ log+ x (in fact in this case it would suffice to have y ≥ 4, say). This
gives the s = 0 case of (8). Finally, from taking logarithmic derivatives of (1),
(2) one has
              H0′                                   1 Γ′  s  ζ ′
                                                                     !
                        i 1        1      1
                  (z) =       +        − log π +              + (s)
              H0        2 s s−1 2                   2Γ 2        ζ

where s ≔ 12 + iz2 . From taking log-derivatives of (11) using the Cauchy integral
formula, one has the well known asymptotic
                             Γ′  s 
                                                       !
                                           s        1
                                      = log + O
                              Γ 2          2       |s|
                              ′                                               ′
for the digamma function ΓΓ , and from the Dirichlet series expansion ζζ (s) =
  P Λ(n)      P∞ log n
− ∞ n=1 n s ≪  n=2 nRe s one can easily establish the bound

                                    ζ′        1
                                       (s) ≪
                                    ζ        |s|
in the regime C ′ log+ x ≤ y ≤ C log x. Putting all this together, one obtains (9) in
this case.
    Henceforth we address the t < 0 case. We begin with the proof of the upper
bound (7). Here it will be convenient to exploit the fundamental solution for the
(backwards) heat equation to relate Ht with H0 . Indeed, for any t < 0, we have
the classical heat equation (or Gaussian) identity
                                    Z
                  2             1          2
                                                                  
               etu exp(izu) = √         e−r /4 exp i(z + r|t|1/2 )u dr          (14)
                                 4π R
for any complex numbers z, u; replacing z, r by −z, −r and averaging we conclude
that                              Z
               tu2             1         2
                                                               
              e cos(zu) = √           e−r /4 cos (z + r|t|1/2 )u dr.
                               4π R
Multiplying by Φ(u), integrating u from 0 to infinity, and using Fubini’s theorem,
we conclude that
                                  Z
                              1          2
                   Ht (z) = √         e−r /4 H0 (z + r|t|1/2 ) dr.            (15)
                               4π R
              The De Bruijn-Newman constant is non-negative                       11

Applying (13), the triangle inequality, and the hypothesis Λ < t ≤ 0, we conclude
that
                                                        !
                        π|x|
  Ht (x − iy) ≪ exp −        + O((1 + y) log+ (|x| + y))
                         8
                                                  r2
                                      Z                                    !
                                    ×     exp − + O((1 + y + |r|) log+ r) dr.
                                        R          4
Using (1 + |r|) log+ r ≤ εr2 + Oε (1) and y log+ r ≪ εr2 + Oε (y2 ) for any absolute
constant ε > 0, we have
           r2                                            r2
        −     + O(|r|) + O((1 + y + |r|)(1 + log+ r)) ≤ − + O((1 + y)2 ),
           4                                              8
thus arriving at the bound
                                                                       !
                                 π|x|
            Ht (x − iy) ≪ exp −       + O((1 + y) log+ |x| + (1 + y)2 ) .
                                  8
Since y = OC (log+ x), this gives (7).
    To prove the remaining two bounds (8), (9), it is convenient to cancel off the
t = 0 case that has already been established, and reduce to showing that
                             Ht                         
                                 (z) = exp OC (log2+ x) ,                        (16)
                             H0
and
                            Ht′         H′          log+ x
                                 (z) − 0 (z) ≪C            ,                     (17)
                             Ht         H0            x
when κ ≥ C ′ . To prove these estimates, the heat equation approach is less
effective due to the significant oscillation present in H0 . Instead we will use
the method of steepest descent (also known as the saddle point method) to shift
contours to where the phase is stationary rather than oscillating. We allow all
implied constants to depend on C. We may assume that x is larger than any
specified constant C ′′ (depending on C), as the case x = OC (1) follows trivially
from compactness, since the zeroes of Ht for t ≥ Λ are all real, so that Ht (z) is
bounded away from zero in this region of interest.
    Now suppose that z = x − iy where y = κ log+ x for some C ′ ≤ κ ≤ C; in
particular C is large since C ′ is. As Φ is even, we may write (4) as
                                        Z
                                      1       2
                           Ht (z) =        etu Φ(u)eizu du.
                                      2 R
From (3) and Fubini’s theorem (which can be justified when t < 0) we conclude
that
                      ∞
                   1X 2 4
         Ht (z) =        2π n It (πn2 , 9 + y + ix) − 3πn2 It (πn2 , 5 + y + ix) (18)
                   2 n=1
                          Brad Rodgers and Terence Tao                                             12

where It (b, ζ) denotes the oscillatory integral
                                  Z
                      It (b, ζ) ≔    exp(tw2 − be4w + ζw) dw,                                   (19)
                                        R

which is an absolutely convergent integral for t < 0 whenever Re b > 0.
    We therefore need to obtain good asymptotics on It (b, ζ) for b ≥ 1 and ζ in
the region
               Ω ≔ {y + ix : x ≥ C ′′ ; C ′ log+ x ≤ y ≤ 2C log+ x}.        (20)
Observe that the phase tw2 − be4w + ζw has a stationary point at the origin when
4b = ζ. In general, 4b will not equal ζ; however, for any complex number w0 in
the strip                                          π
                            w0 ∈ C : 0 ≤ Im (w0 ) <    ;                     (21)
                                                    8
we see from shifting the contour in (19) to the horizontal line {w + w0 : w ∈ R}
that we have the identity
                       It (b, ζ) = exp(tw20 + ζw0 )It (be4w0 , ζ + 2tw0 )                       (22)
whenever b > 0 (so that be4w0 has positive real part). We will thus be able to
reduce to the stationary phase case 4b = ζ if we can solve the equation
                                       4be4w0 = ζ + 2tw0                                        (23)
in the strip (21). This we do in the following lemma7 :

Lemma 6. If b ≥ 1 and ζ ∈ Ω, then there exists a unique w0 = w0 (b, ζ) in the
strip (21) such that (23) holds. Furthermore we have the following estimates:
  (i) Re (4be4w0 ) ≥ 1.
  (ii) (Precise asymptotic for small and medium b) If ζ = y + ix and b ≤
                  1/2
       x exp(100 x|t| ), then

                                                  t log 4bx       log2+ x
                                     !                                    !!
                     1      x      1      π   y
              w0 =     log    + OR     +i   −   −           + OCR
                     4     4b      x      8 4x       8x            x3/2
       where the superscript in the O() notation indicates that these quantities are
       real-valued.

 7
   One could also write w0 explicitly in terms of the Lambert W-function as w0 = − 2tζ +
1     8b      2ζ
4 W(− t exp(− t )), but we will not use this expression in this paper, and in fact will not explicitly
invoke any properties of the W-function in our arguments.
              The De Bruijn-Newman constant is non-negative                        13

                                                                        1/2
 (iii) (Crude bound for huge b) If ζ = y + ix and b > x exp( x|t| ), then Re w0 is
       negative; in fact we have
                                               1
                                    −Re w0 ≥     log+ b.
                                               8
Proof. The function w0 7→ 4be4w0 −2tw0 traverses the graph {a+i( π|t|
                                                                    4 +4be
                                                                            2a/|t|
                                                                                   ):
                              a     π
a ∈ R} on the upper edge { 2|t| + i 8 : a ∈ R} of the strip (21), while the lower
edge of the strip is of course mapped to the real axis. Since |t| ≤ Λ and C, C ′
are large, the region Ω lies between these two curves, and so from the argument
principle (and observing that the map w0 7→ 4be4w0 −2tw0 sends the line segments
{−R + iβ : 0 < β < π/8} and {R + iβ : 0 < β < π/8} well to the left and right of ζ
respectively for R large enough), for every ζ ∈ Ω there exists exactly one w0 in
the strip (21) such that 4be4w0 − 2tw0 = ζ, which is of course equivalent to (23).
The uniqueness implies that the holomorphic function w0 7→ 4be4w0 − 2tw0 has
non-zero derivative at this value of w0 .
    Now write ζ = y + ix as per (20), and write w0 = α + iβ for some α ∈ R
and 0 < β < π/8. Taking real and imaginary parts in (23) we have the system of
equations
                              4be4α cos 4β = y + 2tα                          (24)
and
                              4be4α sin 4β = x + 2tβ.                             (25)
                                                              4w0
   To prove (i), suppose for contradiction that Re (4be             ) < 1, thus
                                     4α
                                 4be cos 4β ≤ 1.                                  (26)
                                                   4α
Since t, β = O(1), we see from (25) that 4be sin 4β ≪ x, and hence from
sin2 4β + cos2 4β = 1 we have
                                     4be4α ≪ x
and hence (since b ≥ 1) α ≤ 14 log+ x+O(1). In particular −2tα ≤ |t|2 log+ x+O(1).
Inserting this into (24) and using (26) one then has
                                   |t|
                               y≤      log+ x + O(1),
                                    2
which contradicts (20) since |t| ≤ Λ and C ′ is large.
   Now we show (ii). From (25) and sin 4β ≤ 1, t, β = O(1) one has
                                 4be4α ≥ x − O(1)
and hence on taking logarithms (and using the fact that b ≥ 1 and x is large)
                                                 !
                                1      x       1
                            α ≥ log      −O        .                          (27)
                                4     4b       x
                          Brad Rodgers and Terence Tao                           14

On the other hand, from squaring (24), (25) and summing we have
                           (4be4α )2 = (y + 2tα)2 + (x + 2tβ)2 .                (28)
Crudely bounding x + 2tβ = O(x), y = O(x), b ≥ 1, and t = O(1) we conclude
that
                             e8α ≪ x2 + α2
                                                                          1/2
which implies that α ≤ O(log+ x). From the hypothesis b ≤ x exp(100 x|t| ) and
(27) we also have α ≥ −O(x1/2 /t), thus tα ≪ x1/2 . Returning to (28) and using
2tβ = O(1) and y ≪ x1/2 we conclude that
                                   (4be4α )2 = x2 + O(x)
so on taking square roots
                                     4be4α = x + O(1)                           (29)
and hence on taking logarithms we have the matching upper bound
                                                !
                                1     x       1
                            α ≤ log      +O
                                4    4b       x
to (27). In particular,
                                              t log 4bx
                                                             !
                                                           1
                             y + 2tα = y +              +O     .
                                                  2        x
Inserting this and (29) into (24), we have
                                                   x
                                    y t log 4b
                                                               !
                                                     1
                            cos 4β = +         + O 3/2
                                    x    2x        x
and hence (by Taylor expansion of the arc cosine function)
                                         x
                         π y t log 4b             log2+ x
                                                          !
                    4β = − −               + OC             ,
                         2 x         2x            x3/2
giving (ii).
    Finally, we prove (iii). From the identity (28) and crudely bounding y, tβ =
O(x) we have
                              (4be4α )2 ≪ x2 + t2 |α|2
and hence either
                                                   b
                                         e−4α ≫
                                                   x
or
                                                    b
                                       e−4α ≫           .
                                                 |t||α|
                                        1/2
Under the hypothesis b > x exp( x|t| ), so that 1/|t| and x are O(b1/10) (say), so
both options force −α ≥ 18 log b as claimed.
                The De Bruijn-Newman constant is non-negative                            15

    We combine the above lemma with the following asymptotic.

Lemma 7. Let b be a complex number with Re b ≥ 1. Then
                              r                           !!
                                π         1          1
                 It (b, 4b) =     exp(−b) √ + O                                        (30)
                                8          b       |b|3/2
using the standard branch of the square root.

Proof. One could establish this from Laplace’s method, but we will instead use
the Stirling approximation8 (11). Writing
                                   Z
                               tw2
                              e =       e4iξw dµ(ξ)
                                               R

where µ is the Gaussian probability measure
                                                2    2
                                       dµ(ξ) ≔ √ e−4ξ /|t|
                                                π|t|
of mean zero and variance |t|/8, and applying Fubini’s theorem, we obtain
                         Z Z                               !
                                        4w
            It (b, 4b) =       exp(−be + 4(b + iξ)w) dw dµ(ξ).
                               R       R

Making the change of variables r = Rbe4w (and contour shifting or analytic
                                         ∞
continuation) and the definition Γ(s) = 0 e−r r s−1 dr of the Γ function, we see
that
       Z
                                         1
           exp(−be4w + 4(b + iξ)w) dw = exp(−(b + iξ) log b)Γ(b + iξ)
         R                               4
and hence
                                   Z
                               1
                It (b, 4b) =            exp(−(b + iξ) log b)Γ(b + iξ) dµ(ξ).
                               4   R

We divide integral into regions |ξ| ≤ 10|t|1/2 |b|1/2 and |ξ| > 10|t|1/2 |b|1/2 . By
applying Stirling’s approximation, the integral over the first region becomes
  Z                                !! √
1                            1          2π                        iξ
                      1+O                    exp((b + iξ) log(1 + ) − b − iξ) dµ(ξ).
4 |ξ|≤10|t|1/2 |b|1/2     |b + iξ|
                                     p
                                      b + iξ                      b


8
  We thank Alex Dobner for pointing out some issues in the original proof of this lemma, and
suggesting a repaired proof which is reproduced here.
                               Brad Rodgers and Terence Tao                                      16

Note we may assume |b| is sufficiently large so that |ξ| < |b|/2 in this region (the
small |b| case of the lemma follows trivially from compactness),
                                                              2     and so we have
                                                              |ξ|
                 |ξ|                         iξ
 √ 1                      1
       = 1 + O |b|        √ and (b + iξ) log(1 + b ) = iξ + O |b| . Substituting these
                                b
  b+iξ
expressions into the integrand we get
              r
                                                       1 + |ξ|2
                             Z                                  !!
                  π
                    exp(−b)                        1+O             dµ(ξ),
                 8b            |ξ|≤10|t|1/2 |b|1/2       |b|
and now the integral evaluates to 1 + O( |b|1 ). Thus it will suffice to establish the
tail bound
      Z
                 exp(−(b + iξ) log b)Γ(b + iξ) dµ(ξ) ≪ exp(−Re (b))|b|−3/2 .
         |ξ|>10|t|1/2 |b|1/2

By applying the triangle inequality and bounding the integrand with
                                                           π
           | exp(−(b + iξ) log b)| ≤ exp(−Re (b) log |b| + (|b| + |ξ|))
                                                           2
and
              |Γ(b + iξ)| ≤ Γ(Re (b)) ≤ exp(Re (b) log |b| − Re (b))
we get the following upper bound:
                                                                                  4ξ 2
                                                 Z
                                π      2                                   π
                   exp(−Re (b) + |b|) √                                exp( |ξ| −      ).
                                2       π|t|       |ξ|>10|t|1/2 |b|1/2     2       |t|
                                                                                      2     2
Now again we assume |b| is large enough so that we have π2 |ξ| − 4ξ|t| ≤ − ξ|t| for all
ξ in the given region, and hence the integral is bounded above by
                                                         ξ2
                               Z
                    π      2
    exp(−Re (b) + |b|) √                            exp(− ) ≪ exp(−Re (b) − 10|b|)
                    2      π|t| |ξ|>10|t|1/2 |b|1/2      |t|
(say), and the claim follows.

   From the above two lemmas and (22), we have the asymptotic
                 r                                                 !!
                   π       2     4w0        1              1
     It (b, ζ) =     exp(tw0 − be + ζw0 ) √      +O                                             (31)
                   8                       be4w0       |be4w0 |3/2
for any b ≥ 1 and ζ ∈ Ω, where w0 = w0 (b, ζ) is the quantity in Lemma 6.
    Now we can control the sum (18). As before we assume that z = x − iy where
y = κ log+ x for some C ′ ≤ κ ≤ C. From (18) one has
                                                         ∞
                                                     1X
                                       Ht (x − iy) =       Qt,n                                 (32)
                                                     2 n=1
               The De Bruijn-Newman constant is non-negative                        17

where Qt,n is the quantity
             Qt,n ≔ 2π2 n4 It (πn2 , 9 + y + ix) − 3πn2 It (πn2 , 5 + y + ix).
We first consider the estimation of Qn in the main case when n is not too huge,
in the sense that
                                             x1/2
                                                  !
                              n ≤ x exp 100         .                      (33)
                                              |t|
In this case, if we apply Lemma 6(ii) with ζ = 9 + y + ix and b = πn2 we have
that the quantity w0 = w0,t,n arising in that lemma obeys the asymptotics
                                                        x           2 !!
                                     π 9 + y t log 4πn2
                               !
         1        x      R 1                                   R log+ x
   w0 = log          + O         + i   −        −          + O C          , (34)
         4      4πn2         x       8     4x        8x           x3/2
which when combined with (23), gives
                               4be4w0 = ix + OC (x1/2 ).
                                                
In particular, the factor √ 14w0 + O |be4w10 |3/2 in (31) can be expressed as
                              be

                                     1             
                                   √     1 + OC x−1/2 ,
                                    ix/4
and thus by (31)
                              r
                                   π             ζ tw0                
         2
   |It (πn , 9 + y + ix)| =           exp Re tw20 − −    + ζw0 + OC x−1/2
                                   2x              4  2
where we have again used (23). From (34) (and using t = O(1) and y =
OC (log
     + x)ζ to bound some
                        small error terms), we can calculate the quantity
       2       tw0
Re tw0 − 4 − 2 + ζw0 to be
                                                                                 x
 t        x      tπ2 9 + y t            x      9+y         x    πx 9 + y t log 4πn2     
                                                                                            −1/2
                                                                                                 
   log2       −     −         −   log       +       log       −    +      +         +OC   x
16      4πn2 64           4     8     4πn2      4        4πn2 8      4        8
and thus on cancelling and gathering terms we obtain
                                        x  9+y4                    
             |It (πn2 , 9 + y + ix)| =       2
                                                  Jt Kt,n exp OC x−1/2
                                        4πn
where Jt = Jt (x) and Kt,n = Kt,n (x) are the positive quantities
                             r
                                                         tπ2 πx
                                                                  !
                                 π        t        x
                      Jt ≔          exp      log2     −      −                   (35)
                                2x       16       4π 64         8
and                            t    x        t      
                    Kt,n ≔ exp − log     log n + log2 n .
                                4    4π         4
                            Brad Rodgers and Terence Tao                                18

A similar computation gives
                                   x  5+y4                    
                   |It (πn2 , 5 + y + ix)| =
                                       2
                                             Jt Kt,n exp OC x−1/2
                                   4πn
In particular we have the upper bound
                                      x  9+y
                                             4
                           Qt,n ≪ n4      2
                                                Jt Kt,n
                                      4πn
                               1/2
for 1 ≤ n ≤ x exp(100 x|t| ), and for n = 1 we have the refinement
                                                 x  9+y
                                                            4
                                  2        −1/2
                    |Qt,1 | = 2π + OC x                         Jt .                   (36)
                                                      4π
Using the crude bound
                                 t      x                 t
                   Kt,b ≤ exp − log            log n ≤ n− 4 log x
                                   4     4π
we conclude that                      1+y t
                             Qt,n ≪ n− 2 − 4 log x |Qt,1 |.
                                                          1/2
Since y ≥ C ′ log+ x, the 2 ≤ n ≤ x exp(100 x|t| ) terms sum to O(|Qt,1 |/x), thus
                                                         !!
                            X                          1
                                     Qt,n = 1 + OC          Qt,1
                                x1/2
                                                       x
                             n≤x exp(100 |t| )

Also, from (35) we have
                                      x  9+y
                                             πx
                                            4
                                                               
                       |Qt,1 | ≍    Jt = exp − + OC (log2+ x) .
                             4π                8
Thus,   to finish the proof of (8) (or (16)), one just needs to show that the tail
                      is negligible compared with the main term Qt,1 = exp(− πx
P
               1/2 Q                                                            +
  n>x exp(100 x ) t,n
             |t|
                                                                             8
                                                                1/2
OC (log2+ x)). Suppose now that n > x exp(100 x|t| ). If we now apply Lemma
6(iii) with ζ = 9 + y + ix and b = πn2 , and write w0 = α + iβ with 0 < β < π/8,
we have that α is negative with
                                     1
                                       log n,    −α ≥
                                     8
while from (31) and (23) (and Lemma 6(i)) we have
                                                              ζ tw0
              It (πn2 , 9 + y + ix) ≪ exp(Re (tw20 −             −     + ζw0 ))
                                                              4      2
                                                             |t||α|
                                            ≪ exp(−|t||α|2 −        + OC (log2+ x)).
                                                                2
               The De Bruijn-Newman constant is non-negative                          19

                                                             1/2               1/2
Similarly for It (πn2 , 5 + y + ix). Since log n ≥ 100 x|t| , we have |α| ≥ 10 x|t| and
thus
                                 |t||α|2 ≥ 10x1/2 log n.
In particular, n4 exp(−|t||α|2 ) ≪ exp(−9x1/2 log n) and thus
                      Qt,n ≪ exp(−8x1/2 log n + OC (log2+ x))
(say). Summing, we conclude that
                         X
                                              Qt,n ≪ exp(−100x/|t|)
                                     1/2
                        n>x exp(100 x |t| )

(say), which is certainly O(|Q1 |/x). Inserting these bounds into (18), we conclude
that
                                                 log2+ x
                                                         !!
                                       1
                       Ht (x − iy) =     + OC               Qt,1 ,
                                       2           x
which already gives (7). Sending t to 0, taking absolute values, and then dividing
using (36) and (35), we obtain after cancelling all the t-independent terms that
                                     log2+ x                      tπ2
                                             !!                       !
            Ht                                       t     2 x
                (x − iy) = 1 + OC               exp    log     −        .
            H0                         x            16       4π 64
Since the ratio HH0t is holomorphic in the region of interest, we can thus find a
holomorphic branch of log HH0t for which

                                                tπ2        log2+ x
                                                                   !
                        Ht        t       z
               Re log      (z) −    log2     =−     + OC
                        H0       16      4πi    64             x
for all z = x − iy in this region. Varying x, y by O(log+ x) (adjusting the constants
C, C ′ , C ′′ slightly as necessary) and using the Borel-Carathéodory theorem and
the Cauchy integral formula, we conclude that
                                                  !              !
                       d      Ht        t    2 z          log+ x
                          log    (z) −    log       = OC           ,
                      dz      H0       16     4πi             x
which gives (17) after a brief calculation.

                   3. Riemann-von Mangoldt type formulae
   For any Λ < t ≤ 0, the zeroes of Ht are all real and simple [11, Corollary
1]. For any interval I ⊂ R, let Nt (I) denote the number of zeroes of Ht in
I. The classical Riemann-von Mangoldt formula (see e.g. [30, Theorem 9.4]),
combined with (1), gives the asymptotic
                           N0 ([0, T ]) = Ψ(T ) + O(log+ T )                         (37)
                            Brad Rodgers and Terence Tao                                               20

for all T ≥ 0, where we use Ψ : R+ → R to denote9 the function
                                                T      T  T
                                    Ψ(T ) ≔        log   − .                                        (38)
                                                4π     4π 4π
For future reference, we record the derivative of Ψ as
                                                     1    T
                                       Ψ′ (T ) =       log ,                                        (39)
                                                    4π    4π
in particular Ψ is increasing for T > 4π. Applying (37) with T replaced by T + α
and subtracting, we conclude from the mean value theorem that
                                                  α log+ T
                           N0 ([T, T + α]) =               + O(log+ T )                             (40)
                                                     4π
for all T ≥ 0 and 0 ≤ α ≤ C for any fixed C, where the implied constants in the
asymptotic notation are allowed to depend on C. Because we are assuming the
Riemann hypothesis (and hence the Lindelöf hypothesis), one can improve this
latter bound10 to
                                               α log+ T
                        N0 ([T, T + α]) =               + oT →∞ (log+ T ),                          (41)
                                                  4π
a result of Littlewood (see [30, Theorem 13.6]). A key input in these bounds is a
lower bound on |ζ(s)| when Re(s) is somewhat large, e.g. between 2 and 3; this is
                                                            P µ(n)
                                                       1
easily obtained through the Dirichlet series identity ζ(s) = ∞n=1 n s that is valid
in this region.
    Define the classical location ξ j of the jth zero for j ≥ 1 to be the unique
quantity in (1, +∞) solving11 the equation
                                              Ψ(ξ j ) = j,                                          (42)
and extend this to negative j by setting ξ− j ≔ −ξ j . Clearly the ξ j are increasing
in j. For future reference we record the following bounds on the ξ j :

Lemma 8 (Spacing of the classical locations).

9
  It is traditional to also insert the lower order term − 78 here, but this term will not be of use in our
analysis and will therefore be discarded. The factors of 4π are not of particular significance and
may be ignored by the reader on a first reading.                                     log T 
 10
  Indeed, on the Riemann hypothesis one can improve the error term to O log log           +
                                                                                                ; see [30,
                                                                                        +   +T
Theorem 14.13]. However, we will not need this further refinement in this paper.
 11
  As with the quantity w0 introduced in Lemma 6, one could express ξ j explicitly in terms of the
Lambert W function if desired as ξ j = 4πe exp(W( j/e)), but we will not use this relation in this
paper.
               The De Bruijn-Newman constant is non-negative                     21

  (i) For any j ≥ 1, one has
                                                               4π j
                                 ξ j = (1 + o| j|→∞ (1))                       (43)
                                                              log+ j

      In particular, ξ j ≍ logj j and log+ ξ j ≍ log+ j.
                             +

                       ∗
 (ii) For any j, k ∈ Z , one has
                                                      |k − j|
                                 |ξk − ξ j | ≍                         .       (44)
                                                 log+ (|ξ j | + |ξk |)

 (iii) If 1 ≤ j ≍ k, then one has the more precise approximation
                                                    |k − j|2 
                                                                 
                                      4π(k − j)
                           ξk − ξ j =           +O               .         (45)
                                       log ξ j         j log2 ξ j
      Of course, the implied constant in the error term in (45) can depend on the
      implied constants in the hypothesis j ≍ k.

Proof. If j ≥ 1, then from (38) one has

                           ξ j log+ ξ j = (1 + o j→∞ (1))4π j                  (46)

which implies that j1/2 ≪ ξ j ≪ j (say), which implies that log+ ξ j ≍ log+ j;
substituting this back into (46) yields
                                                j
                                      ξj ≍          .
                                             log+ j
This in turn implies that log+ ξ j = (1 + o j→∞ (1)) log+ j, and using (46) one last
time gives (42).
    Now we obtain (ii). If j, k have opposing sign, then (44) follows from (43), so
by symmetry we may assume that j, k are both positive. If j is much larger than
k or vice versa, then the bound (44) follows from (43) and the triangle inequality,
so we may now restrict attention to the case 1 ≤ j ≍ k. The estimates (44) and
(45) are trivial for j = O(1), so we may assume j to be large.
    From (42) we have
                                Ψ(ξk ) − Ψ(ξ j ) = k − j
and hence by the mean value theorem and (39) we have
                              1    T
                                log (ξk − ξ j ) = k − j
                             4π    4π
                         Brad Rodgers and Terence Tao                                         22

for some T between ξk and ξ j . From (43) we see that T ≍ ξ j , and so (44) follows.
Furthermore, we can conclude that
                                                              !
                                                      |k − j|
                   T = ξ j + O(|ξk − ξ j |) = ξ j + O
                                                      log ξ j
and hence
                                                     !                       !
                                          |k − j|                    |k − j|
                 log T = log ξ j + O                   = log ξ j + O
                                         ξ j log ξ j                    j
and                                                             
                                1       1           |k − j| 
                                    =         + O              
                              log T   log ξ j         j log2 ξ j
giving (45).

     Applying (37) to T = x j (0) for some j ≥ 1, we conclude in particular that
                            Ψ(x j (0)) − Ψ(ξ j ) = O(log+ x j (0)).
From (39) and the mean value theorem12 we conclude that
                                      x j (0) = ξ j + O(1)                                 (47)
for all j ≥ 1, and hence for all j ∈ Z∗ by symmetry. In particular, from (43) and
the fact that x1 (0) > 0 we conclude that
                                                j              j
                                 x j (0) ≍              ≍
                                             log+ ξ j       log+ j
for all j ≥ 1.
    In a similar vein, if 1 ≤ j < k ≤ j + log+ j, then from applying (41) with
T = x j (0) and α equal to (or slightly less than) xk (0) − x j (0), we have
                               xk (0) − x j (0)
                     k− j=                      log+ ξ j + o j→∞ (log+ ξ j )
                                     4π
and hence
                                               4π(k − j)
                          xk (0) − x j (0) =             + o j→∞ (1).
                                                log+ ξ j
Informally, this asserts that the zeroes x j (0) behave like an arithmetic progression
of spacing log4π ξ j at spatial scales between o(1) and 1. (In fact, when combined
                +




12
  One may wish to treat the bounded case j = O(1) separately, to avoid the minor issue that Ψ(T )
becomes decreasing for T < 1.
               The De Bruijn-Newman constant is non-negative                                 23

with (47) and (45), we see that this behavior persists for all scales between o(1)
and o(ξ j ).)
     In this section we use the asymptotics on Ht obtained in the previous section
to establish analogous, but weaker, bounds for the zeroes x j (t) of the functions
Ht , in which we lose an additional logarithm factor in the error estimates.

Theorem 9 (Riemann-von Mangoldt type formulae). Let Λ < t ≤ 0, T > 0, and
let 0 ≤ α ≤ C for some C > 0. Then one has
                            Nt ([0, T ]) = Ψ(T ) + O(log2+ T )                              (48)
and
                                      α log2+ T
                Nt ([T, T + α log+ T ]) =       + oT →∞ (log2+ T ).    (49)
                                         4π
The decay rate in the oT →∞ () error term is permitted to depend on C but is
otherwise uniform in α.

     Repeating the previous analysis, we conclude

Corollary 10 (Macroscopic structure of zeroes). Let Λ < t ≤ 0. Then one has
                                  x j (t) = ξ j + O(log+ ξ j )                              (50)
for all j ∈ Z∗ ; in particular
                                                j              j
                                 x j (t) ≍              ≍                                   (51)
                                             log+ ξ j       log+ j
for all j ≥ 1. We also have
                                           4π(k − j)
                      xk (t) − x j (t) =             + o j→∞ (log+ ξ j )                    (52)
                                            log+ ξ j
whenever 1 ≤ j < k ≤ j + log2+ ξ j .

    Informally, this corollary asserts that the zeroes x j (t) behave like an arith-
metic progression of spacing log4π ξ j at spatial scales between o(log+ ξ j ) and o(ξ j ).
                                   +
This level of spatial resolution is worse by a factor of log+ ξ j than what one
can achieve for x j (0), but will still (barely) be enough for our applications. We
remark that a significantly sharper estimate (with an error term of just O(1) in
the analog of (48)) is available for any fixed t > 0; see13 [14, Theorem 1.4].
    We now turn to the proof of the two bounds in Theorem 9.

13
 Added in press: even sharper estimates have recently been obtained in [21, Theorem 1.5].
                      Brad Rodgers and Terence Tao                                 24

Proof (Proof of (48)). We make use of the argument principle in exactly the
same manner as in the classical proof of the Riemann-von Mangoldt formula.
By perturbing T slightly if necessary, we may assume that T is not a zero of Ht .
Let κ > 0 be a sufficiently large absolute constant. Then the argument principle
yields
                                                   Ht′
                                               Z
                                            1
                            Nt ([0, T ]) =             (z) dz,
                                           2πi Γ Ht
where Γ is the counterclockwise contour carved out by a straight line from
iκ log+ 0 = iκ log 2 to −iκ log+ 0 = −iκ log 2, then along the curve ΓI parame-
terized by x − iκ log+ x for x ∈ [0, T ], then along the line ΓII from T − iκ log+ T
to T , then along the vertical line conjugate to ΓII and the curve conjugate to ΓI ,
leading back to i log 2. As the integrand is odd, the, the integral along the line
from iκ log+ 0 to −iκ log+ 0 vanishes. Using the symmetry Ht (z) = Ht (z), we thus
have                                        Z      Z ! ′
                                      1                  Ht
                      Nt ([0, T ]) = Im          +           (z) dz.
                                      π       ΓI    ΓII Ht
From (9), (39) one sees that
                         1 Ht′
                                                               !
                                    d            log+ x
                               (z) = (Ψ(iz)) + O
                         π Ht       dz             x
for z = x − iκ log+ x on ΓI (extending Ψ to the right half-plane using the standard
branch of the logarithm), and hence by the fundamental theorem of calculus
                    Ht′
              Z
        1
          Im            (z) dz = Im Ψ(iT + κ log+ T ) − Ψ(log+ 0) + O(log2+ T )
        π       Γ I Ht
                           = Ψ(T ) + O(log2+ T ).
On the other hand, if we let θ be a phase so that eiθ Ht (T − iκ log+ T ) is real and
positive, then
                                   Ht′
                              Z
                         Im            (z) dz ≤ π(m + 1),
                               ΓII Ht

where m is the number of zeroes of Re eiθ Ht (z) along the contour ΓII , since the
left hand side is the change in arg eiθ Ht (z) as z varies over this contour, and for
each increment of π in the value of arg eiθ Ht (z), we must have that Re eiθ Ht (z) is
zero for some z. Note that the number of zeroes of Re Ht (z) along this contour is
the same as the number of zeroes of
                    g(s) ≔ 12 (e−iθ Ht (is + T ) + eiθ Ht (−is + T ))
as s ranges along the line from 0 to κ log+ T . Hence m is no more than the number
of zeroes m′ of g(s) in the disc of radius κ log+ T centered at κ log+ T .
               The De Bruijn-Newman constant is non-negative                              25

    The count m′ we can estimate with Jensen’s formula as follows. Let M be
the maximum of g(s) in a disc centered at κ log+ T of radius 2κ log+ T . Using (7)
and the conjugate symmetry of Ht (z), we have
                                               π        2
                                     M ≪ e− 8 T +O(log+ T ) .
                                                                                 π   2
Since from (8) we have g(κ log+ T ) = eiθ Ht (T − iκ log+ T ) = e− 8 T +O(log+ T ) , it
therefore follows from Jensen’s formula (see e.g. [17, Lemma 6.1]) that
                                         m′ ≪ log2+ T.
                                                                   H′
This induces a corresponding bound on the integral of Htt over ΓII and therefore
establishes the claimed estimate for Nt ([0, T ]).
Proof (Proof of (49)). We will use a “limiting profile argument” (also known
as a “compactness argument” or “normal families argument”), in which one
extracts and then studies a limit of suitably rescaled versions of a family of
analytic functions to conclude asymptotic information about these functions.
We remark that this sort of argument can also be used in a similar fashion to
deduce the Lindelöf hypothesis from the Riemann hypothesis: see Theorem 1 of
terrytao.wordpress.com/2015/03/01.
    Suppose for contradiction that this claim failed, then there exists a sequence
T n → ∞, and bounded sequences Λ < tn ≤ 0 and 0 ≤ αn ≤ C, as well as an
ε > 0, such that
                                                    αn log2+ T n
               Ntn ([T n , T n + αn log+ T n ]) −                > ε log2+ T n           (53)
                                                        4π
for all n. By perturbing T n slightly we may assume that Htn does not vanish at
T n or T n + αn .
    Let κ > 0 be a sufficiently large absolute constant. By the hypothesis Λ < tn ,
the function Htn has no zeroes in the lower half-plane. Thus we can define
holomorphic functions F n on the lower half-plane by the formula
                                    1        Ht (T n + z log+ T n )
                     F n (z) ≔      2
                                          log n
                                 log+ T n    Htn (T n − iκ log+ T n )
with the branch of the logarithm chosen so that F n (−iκ) = 0. From (7) we see
that the F n are uniformly bounded on any compact subset of the lower half-plane.
Thus, by Montel’s theorem (see [24, Sec 3.2]), we may pass to a subsequence
and assume that the F n converge locally uniformly to a holomorphic function F
on the lower half-plane; since the F n all vanish on −iκ, F does also. Then by the
Cauchy integral formula, the derivatives
                                         1 Ht′n
                        F n′ (z) =                 (T n + z log+ T n )
                                      log+ T n Htn
                          Brad Rodgers and Terence Tao                                         26

converge locally uniformly to F ′ . Comparing this with (9), we conclude that
                                                     1
                                             F ′ (z) =
                                                     4
whenever the imaginary part of z is sufficiently large and negative. By unique
continuation, we thus have F ′ (z) = 14 for all z in the lower half-plane; as F
vanishes on −iκ, we thus have
                                                z + iκ
                                        F(z) =
                                                    4
on the lower half-plane. Since F n converges locally uniformly to F, we conclude
that
                                                                                   !
                                                         z + iκ + on→∞ (1)    2
  Htn (T n + z log+ T n ) = Htn (T n − iκ log+ T n ) exp                   log+ T n (54)
                                                                 4
uniformly for z in a compact subset of the lower half-plane. Similarly, since F n′
converges locally to F, we have
                       Ht′n                           1 + on→∞ (1)
                              (T n + z log+ T n ) =                log+ T n                 (55)
                       H tn                                4
uniformly for z in a compact subset of the lower half-plane.
     Let δ > 0 be a small constant. As in the proof of (48), we can use the
argument principle (and a rescaling) to write
                                                                 Z ! ′
                                                                         H tn
                                              Z        Z
                                  log+ T n
Ntn ([T n , T n +αn log+ T n ]) =          Im        +         +              (T n +z log+ T n ) dz,
                                     π          ΓI,n     ΓII,n    ΓIII,n Htn

where ΓI,n , ΓII,n , ΓIII,n trace the line segments from 0 to −iδ, from −iδ to αn − iδ,
and from αn − iδ to αn respectively. By (55), the contribution of the ΓII,n integral
is α+on→∞4π(1)+O(δ) log2+ T n (we allow the decay rate in the on→∞ (1) errors to depend
on δ). Using the Jensen formula argument used to prove (48), we see that the
contribution of the ΓI,n integral is bounded in magnitude by
                           Z 1
                       ≪        log |gn (δ + 2δe2πiα)| − log |gn (δ)| dα
                               0
where
                                                                                    
           gn (s) ≔ 21 e−iθn Htn (T n + is log+ T n ) + eiθn Htn (T n − is log+ T n )
and the phase θn is chosen so that eiθn Htn (T n − iδ log+ T n ) is real and positive.
                                                                                   √
Applying (54) (and the functional equation Htn (z) = Htn (z)) when |Im(z)| ≥ δ
(say), and (7) (and the
                     functional equation)    otherwise, we conclude that the ΓI,n
                                    √  2
integral is equal to on→∞ (1) + O( δ) log+ T n . Similarly for the ΓIII,n integral.
Taking δ to be sufficiently small and n sufficiently large, we contradict (53).
              The De Bruijn-Newman constant is non-negative                       27

                             4. Dynamics of zeroes
   As remarked in the introduction, the functions Ht solve a backwards heat
equation. As worked out in [11], this induces a corresponding dynamics on the
zeroes x j of Ht :

Theorem 11 (Dynamics of zeroes). For Λ < t ≤ 0, the zeroes x j (t) depend in a
continuously differentiable fashion on t for each j, with the equations of motion
                                           ′
                                           X           1
                           ∂t xk (t) = 2                                        (56)
                                                 x (t) − x j (t)
                                           j: j,k k

for k ∈ Z∗ and Λ < t ≤ 0, where recall the tick denotes principal value
summation over j ∈ Z∗ (which will converge thanks to (50), (43)).

Proof. This follows from [11, Lemma 2.4] (the continuity of the derivative
following for instance from [11, Lemma 2.1]).

    Informally, the ODE (56) indicates that the zeroes xk (t) will repel each other
as one goes forward in time. On the other hand, if the xk (t) are arranged
(locally, at least) in an arithmetic progression, then the ODE (56) suggests that
the zeroes will be in equilibrium. If the xk are not arranged in an arithmetic
progression, and instead have some fluctuation in the spacing between zeroes,
then heuristically the ODE (56) suggests that the zeroes would move away from
the more densely spaced regions and towards more sparsely spaced regions,
thus converging towards the equilibrium of an arithmetic progression. This
is the intuition behind the convergence to local equilibrium mentioned in the
introduction.
    One can estimate the speed of this local convergence to equilibrium by the
following heuristic calculation. Consider the zeroes in a region [T, T + α] of
space, where T > 0 is large and α is reasonably small (e.g. α = O(log+ T )).
                                                                      α
From Theorem 9 (or (50), (43)), we see that we expect about 4π           log T zeroes
                                                 4π
in this interval, with an average spacing of log T . Suppose for sake of informal
                                                  +
discussion that there is some moderate fluctuation in this spacing, for instance
                                                               α
suppose that the left half of the interval contains about 1.5 8π log T zeroes and the
                                     α
right half contains only about 0.5 8π log+ T zeroes. Then a back of the envelope
calculation suggests that for xk (t) near the middle of this interval, the right-hand
                                                         α log T
side of (56) would be positive and have magnitude ≍ α + = log+ T . Since the
length of the interval is α, one may then predict that the time needed to relax to
equilibrium is about α/ log+ T . Since we can flow for time |Λ| ≍ 1, one would
expect to attain equilibrium at the final time t = 0 if the initial length scale α of
the fluctuation obeys the bound α = oT →∞ (log+ T ). Happily, this upper bound is
                         Brad Rodgers and Terence Tao                                         28

precisely what the asymptotic (52) gives, so we heuristically expect to (barely)
be able to establish local equilibrium at time t = 0.
    Of course, one has to make this intuition more precise. Our strategy for
doing so involves exploiting14 the formal gradient flow structure of the ODE
(56). Indeed, one may formally write (56) as the gradient flow

                               ∂t xk (t) = −∂ xk H((x j (t)) j∈Z∗ ),

where H is the formal “Hamiltonian”
                                                  X                     1
                           H((x j ) j∈Z∗ ) ≔                  log
                                               j,k∈Z∗ : j,k
                                                                    |xk − x j |

where we ignore for this non-rigorous discussion the fact that the series defining
H is not absolutely convergent. The Hamiltonian is convex, so one expects the
quantity                                        X
                    H(t) ≔ H((x j (t)) j∈Z∗ ) =       H jk (t)
                                                              j,k∈Z∗ : j,k

to be decreasing and convex in time, and for the state (x j (t)) j∈Z∗ to converge to a
critical point of the Hamiltonian, where
                                                            1
                                H jk (t) ≔ log                                             (57)
                                                   |x j (t) − xk (t)|
denotes the Hamiltonian interaction between x j (t) and xk (t). Indeed, a formal
calculation using (56) yields the identity

                                      ∂t H(t) = −4E(t)

where E is the “energy”
                                                 X
                                  E(t) ≔                      E kk′ (t)
                                           k,k′ ∈Z∗ : k,k′

and
                                                        1
                                 E kk′ (t) ≔                                               (58)
                                               |xk (t) − xk′ (t)|2

 14
  This strategy was loosely inspired by the work of Erdős, Schlein, and Yau [13] exploiting the
Hamiltonian structure of Dyson Brownian motion to obtain local convergence to equilibrium, since
the equations for Dyson Brownian motion resemble that in (56) (but with an additional Brownian
motion term). Indeed, Dyson Brownian motion is the diffusion related to the Gibbs measure Z1 e−βH
for the Hamiltonian studied here.
                    The De Bruijn-Newman constant is non-negative                                                      29

denotes the “interaction energy” betwen xk (t) and xk′ (t), and we once again
ignore the issue that the series is not absolutely convergent. A further formal
calculation using (56) again eventually yields
                                                                                                                        2
                   X                    2                     X                                  1                    
∂t E(t) = −2                 
                                                      −                                                                 
                                 |x   (t) −  x   ′ (t)|2                       (x   ′′ (t) − xk (t))(xk′′ (t) − xk′ (t))
                                    k          k                                  k
                                                                                                                             
             k,k′ ∈Z∗ : k,k′                               k′′ ∈Z∗ : k′′ ,k,k′

suggesting that H(t) and E(t) are decreasing and that H(t) is convex, as claimed.
    In order to deal with the divergence of the infinite series appearing above,
we will need to truncate the Hamiltonian and energy before differentiating them.
The following lemma records some of the identities that arise when doing such
truncations:

Lemma 12 (Identities). For brevity, we suppress explicit dependence on the time
parameter t ∈ (Λ, 0]. Let K ⊂ Z∗ be a finite set of some cardinality |K|. All
summation indices such as i, j, k are assumed to lie in Z∗ .

   (i) (Dynamics of a gap, cf. [11, Lemma 2.4]) If j, k ∈ Z∗ are distinct, then
                                 4                      X                  1
             ∂t (xk − x j ) =          − 2(xk − x j )                                   .
                              xk − x j                i: i,k, j
                                                                (x i − x k )(x i − x j)


  (ii) (Cross-energy inequality, cf. [11, Lemma 2.5]) One has
                              X                 X          8
                         ∂t          E jk ≥ −
                            k∈K; j<K          k∈K; j<K
                                                       (xk − x j )4

        in the weak sense that
                                                    Z t2 X
                   X                                                                  8
                        E jk (t2 ) − E jk (t1 ) ≥ −                                            (t) dt
                       k∈K; j<K                                     t1   k∈K; j<K
                                                                                  (xk − x j )4

        whenever Λ < t1 < t2 ≤ 0.

 (iii) (Energy identity) One has
           X                X                                            4
       ∂t         E kk′ =
            k,k′ ∈K: k,k′                 j<K
                                                    (xk − xk′ )2 (xk − x j )(xk′ − x j )
                                    k,k′ ∈K: k,k′
                                                                                                                            2
                                              X                   2                X                        1             
                                    −2                         (xk − xk′ )2 −
                                                                                                                             .
                                          k,k′ ∈K: k,k′                          k′′ ∈K: k′′ ,k,k′
                                                                                                   (xk′′ − xk )(xk′′ − xk′ ) 
                             Brad Rodgers and Terence Tao                                                        30

 (iv) (Virial15 identity) One has
           X                                                     X                         X               4
      ∂t          (xk −xk′ )2 = 4|K|2 (|K|−1)−                               (xk −xk′ )2
         k,k′ ∈K: k,k′                                       k,k′ ∈K: k,k′                 j<K
                                                                                               (x k − x j )(x k′ − x j )


  (v) (Hamiltonian identity) One has
             X                   X                                       X                      1
        ∂t          Hkk′ = −4                            E kk′ + 2                                           .
             k,k′ ∈K: k,k′               k,k′ ∈K: k,k′                     j<K
                                                                                     (x j − xk )(x j − xk′ )
                                                                     k,k′ ∈K: k,k′

    A key point in the identities (iii), (iv), (v) is that if one ignores the “cross
terms” involving interactions between indices in K (representing some “local
subsystem” of particles) and indices outside of K (representing the “environ-
ment” that that subsystem interacts with), the right-hand side has a definite sign
(negative in the case of (iii) and (v), and positive in the case of (iv)). This gives a
number of useful “monotonicity formulae” as long as cross terms are under con-
trol. As discussed above, many of these various monotonicity formulae reflect
the formal convexity properties of the Hamiltonian H. With more effort one can
obtain a precise formula for the defect in the inequality in (ii); see [11, Lemma
2.5].

Proof. From (56) one has
                                     2         2      X          2       2
             ∂t xk − ∂t x j =             −        +                 −
                                  xk − x j x j − xk i: i,k, j xk − xi xk − x j

which gives (i). Note that the series is now absolutely convergent thanks to (50),
(43).
   Now we prove (ii). By monotone convergence, it suffices to show that
                                                Z t2
            X                                        X        8
                    E jk (t2 ) − E jk (t1 ) ≥ −                        (t) dt
            k∈K                                  t1  k∈K
                                                          (xk − x j )4
          j∈[−R,R]Z∗ \K                                         j∈[−R,R]Z∗ \K

for all Λ < t1 ≤ t2 ≤ 0 and all sufficiently large R. By the fundamental theorem
of calculus, it suffices to show that
                           X                  X           8
                     ∂t           E jk ≥ −                         .
                           k∈K                k∈K
                                                      (xk − x j )4
                             j∈[−R,R]Z∗ \K               j∈[−R,R]Z∗ \K



15
  The terminology here is in analogy with the virial identity in N-body classical gravitational
physics; see e.g., [27, Exercise 1.48].
               The De Bruijn-Newman constant is non-negative                                         31

we can expand the left-hand side as
                                                X          ∂t (xk − x j )
                                    −2
                                              k∈K
                                                           (xk − x j )3
                                          j∈[−R,R]Z∗ \K

which by (i) becomes
             X          8                              X                         1
       −                         +4
             k∈K
                   (x k − x j )4                     k∈K
                                                                (xk − x j )2 (xi − xk )(xi − x j )
          j∈[−R,R]Z∗ \K                         j∈[−R,R]Z∗ \K
                                                   i: i, j,k

and so it will suffice to show that
                         X                                 1
                                                                              ≥ 0.
                          k∈K
                                         (xk − x j )2 (xi − xk )(xi − x j )
                     j∈[−R,R]Z∗ \K
                        i: i, j,k

If R is large enough that [−R, R]Z∗ contains k, we can split this sum into three
parts, depneding on whether i ∈ K, i ∈ [−R, R]Z∗ \K, or i < [−R, R]Z∗ . The
contribution of the case i ∈ K can be rewritten as
                        X               4(xk′ − x j )
                               j<K
                                         (xk − x j )2 (xk′ − x j )2 (xk − xk′ )
                          k,k ∈K: k,k′
                            ′



which equals
                                     X                   2
                                            (x   − x   )2 (x ′ − x )2
                                 j∈[−R,R] ∗    k     j      k     j
                                           Z
                                k,k′ ∈K: k,k′

after symmetrising in k and k′ , which is clearly non-negative. Similarly the
contribution of the case i ∈ [−R, R]Z∗ \K is
                             X                       2
                                                    2 (x − x ′ )2
                                                                  ,
                             k∈K
                                        (x k − x j )    k   j
                           j, j′ ∈[−R,R]Z∗ \K: j, j′

which is also clearly non-negative. Finally, for i < [−R, R]Z∗ , all summands are
already non-negative. This gives (ii).
    For (iii), we can similarly expand the left-hand side as
                                               X        ∂t (xk − xk′ )
                                     −2
                                          k,k′ ∈K: k,k′
                                                        (xk − xk′ )3
                                  Brad Rodgers and Terence Tao                                                                 32

which by (i) becomes
              X             8                X                       1
       −                          4
                                    + 4                         2
                                                                                       .
         k,k′ ∈K: k,k′
                       (xk − xk′ )      k,k′ ∈K:k,k′
                                                     (xk − xk′ ) (xi − xk )(xi − xk′ )
                                                           i: i,k,k′

To prove (iii), it thus suffices to establish the identity
                                                                                     2
     X              2                   X                            1             
                 (xk − xk′ )2 −
                                                                                     
k,k′ ∈K: k,k′                        k′′ ∈K: k′′ ,k,k′
                                                        (xk′′ − xk )(xk′′ − xk′ ) 
            X               4                              X                                            1
     =                               4
                                         −2                                                     2 (x ′′ − x )(x ′′ − x ′ )
                                                                                                                           .
       k,k′ ∈K: k,k′
                      (x k −  x k ′ )         k,k′ ,k′′ ∈K: k,k′ ,k′′ distinct
                                                                               (x k − x    k ′ )    k      k   k      k

The left-hand side expands as
       X              4                    X                     4                 X                          1
                               4
                                  −                                       2
  k,k′ ∈K: k,k′
                (x k −  x k ′ )       k,k′ ∈K: k,k′
                                                         (x k  −   x k ′ )  k′′ ∈K: k′′ ,k,k′
                                                                                               (x k ′′ − x k )(x k′′ − xk′ )
              X               X                                1
       +                                                      2 (x ′′ − x ′ )2
         k,k′ ∈K: k,k′ k′′ ∈K: k′′ ,k,k′
                                           (x  k ′′ −    x k )     k          k
              X                  X                                                    1
       +                                                                                                              .
         k,k′ ∈K: k,k′ k′′ ,k′′′ ∈K: k′′ ,k,k′
                                                (x  k ′′  −   x k )(x  k ′′ −  x k ′ )(x k′′′ − xk )(xk′′′ − xk′ )

The final sum can be rewritten as
           X                                                          (xk − xk′ )(xk′′ − xk′′′ )
                                                                                                                            .
k,k′ ,k′′ ,k′′′ ∈K: k,k′ ,k′′ ,k′′′ distinct
                                             (xk′′ − xk )(xk′′ − xk′ )(xk′′′ − xk )(xk′′′ − xk′ )(xk − xk′ )(xk′′ − xk′′′ )

The denominator is a Vandermonde determinant and is totally antisymmetric
in k, k′ , k′′ , k′′′ . All the monomials appearing in the numerator disappear upon
antisymmetrization, so the final sum vanishes. To conclude the proof of (iii), it
suffices to show that
       X                X               1
    k,k′ ∈K: k,k′ k′′ ∈K: k′′ ,k,k′
                                      (xk′′ − xk )2 (xk′′ − xk′ )2
                                             X             2                X                          1
                                      =                             2
                                                                                                                      .
                                        k,k′ ∈K:k,k′
                                                     (x k −  x k ′ )  k′′ ∈K: k′′ ,k,k′
                                                                                        (x k ′′ − x k )(x k′′ − xk′ )

The difference between the LHS and RHS can be written as
                    X         (x − x ′ )2 − 2(x ′′ − x )(x ′′ − x )
                                                       k        k             k        k     k         k
                                                                                                           .
                     k,k′ ,k′′ ∈K: k,k′ ,k′′ distinct
                                                      (xk′′ − xk )2 (xk′′ − xk′ )2 (xk − xk′ )2
                   The De Bruijn-Newman constant is non-negative                                            33

The denominator is totally symmetric in k, k′ , k′′ , while the numerator sym-
metrizes to zero, giving the claim.
   Now we prove (iv). The left-hand side expands as
                              X
                         2          (xk − xk′ )∂t (xk − xk′ )
                                        k,k′ ∈K: k,k′

which by (i) becomes
                                                X                          X                1
             8|K|(|K| − 1) − 4                              (xk − xk′ )2                                .
                                            k,k′ ∈K: k,k′                  i,k,k′
                                                                                  (xi − xk )(xi − xk′ )
It will thus suffice to show that
                 X                (xk − xk′ )2
                                                                            = −|K|(|K| − 1)(|K| − 2).
         k,k′ ,k′′ ∈K: k,k′ ,k′′ distinct
                                            (xk′′ − xk )(xk′′ − xk′ )
But the left-hand side can be written as
               X                               (xk − xk′ )3
                                                                        = −|K|(|K| − 1)(|K| − 2).
  k,k′ ,k′′ ∈K: k,k′ ,k′′ distinct
                                   (xk′′ − xk )(xk′′ − xk′ )(xk − xk′ )
The denominator is totally antisymmetric in k, k′ , k′′ . The numerator antisym-
metrizes to −(xk′′ − xk )(xk′′ − xk′ )(xk − xk′ ), giving the claim.
   Finally we prove (v). The left-hand side expands as
                                       X ∂t (xk − xk′ )
                                −
                                  k,k′ ∈K: k,k′
                                                xk − xk ′
which by (i) becomes
                 X             4               X          X                 1
         −                           2
                                       +2                                               .
            k,k′ ∈K: k,k′
                          (xk − xk′ )     k,k′ ∈K: k,k′ i: i,k,k′
                                                                  (xi − xk )(xi − xk′ )
It thus suffices to show that the expression
                              X                                              1
                              k,k′ ,k′′ ∈K:k,k′ ,k′′ distinct
                                                              (x k ′′ − x k )(x k′′ − xk′ )

vanishes. But the summand antisymmetrizes to zero, giving the claim.

                                        5. A weak bound on gaps
                                                                            P
   In order to analyze (truncated versions) of the Hamiltonian H(t) = j,k H jk (t),
we will need some upper bounds on the individual terms H jk (t). It was shown in
[11, Corollary 1] that these quantities are finite (i.e., the zeroes are simple) when
Λ < t ≤ 0. It turns out that by refining the analysis in [11] (and by narrowing the
range of times t to the region Λ/2 ≤ t ≤ 0), one can establish a more quantitative
lower bound:
                      Brad Rodgers and Terence Tao                            34

Proposition 13 (Lower bound on gaps). For any j ∈ Z∗ and any Λ/2 ≤ t ≤ 0,
one has
                      max
                       ∗
                           H jk (t) ≪ (log2+ j) log+ log+ j          (59)
                       k∈Z : k, j

    The bound in (59) is probably not optimal, but for our application any bound
that grows more slowly than (say) | j|0.1 as j → ∞ would suffice.
    To prove this proposition, we first need the following variant of a result in
[11]:
Lemma 14. Let K be a finite subset of Z∗ of cardinality |K| ≥ 2, and let
Λ/2 ≤ t ≤ 0. Then
                      X                                  |K|3
                               (xk (t) − xk′ (t))2 ≫    P             .
                 k,k′ ∈K: k,k′
                                                     1 + k∈K E jk (t)
                                                                  j<K

   Informally, this lemma asserts that the gaps within K cannot be too small,
unless there is also a small gap between an element of K and an element outside
of K. The strategy will be to iterate this observation to show that a very small
gap will therefore propagate until it contradicts (52).
Proof. Let A = A(t) and B = B(t) denote the functions
                                 X
                       A(t) ≔          (xk (t) − xk′ (t))2
                                     k,k′ ∈K: k,k′
                                     X
                          B(t) ≔               E k j (t).
                                         k∈K
                                         j<K

The function A(t) is continuously differentiable. The corresponding claim for
B(t) is not obvious; however, the sum defining B(t) is uniformly convergent
(thanks to (51)) and hence B(t) is at least continuous. From Lemma 12(ii) we
have the lower bound
                              ∂t′ B(t′ ) ≥ −8B(t′ )2
(cf. [11, Lemma 2.5]) in the weak sense for Λ < t′ ≤ 0. In particular, if there
exists a time Λ < t− < t such that
                              sup B(t′ ) = B(t− ) = 2B(t)
                             t− ≤t′ ≤t

then we have
                           B(t) − B(t− ) ≥ −8B(t)2 (t − t− )
which rearranges as
                                                         1
                                         t − t− ≥             .
                                                        8B(t)
              The De Bruijn-Newman constant is non-negative                      35

By continuity, we conclude that B(t′ ) cannot attain or exceed the value 2B(t)
                                         1
anywhere in the interval (−Λ, t] ∩ (t − 8B(t) , t), that is to say that

                                         B′ (t) < 2B(t)
whenever
                                          1
                                t−             , Λ < t′ ≤ t.
                                         8B(t)
by hypothesis, this is a range of size at least
                               Λ   1          1
                           min( ,       )≫          .
                               2 16B(t)    1 + B(t)
On the other hand, for t′ in the above range, we see from Lemma 12(iv) that
                     ∂t′ A(t′ ) = 4|K|2 (|K| − 1) + O(B(t′ )A(t′ ))
                              = 4|K|2 (|K| − 1) + O(B(t)A(t′ ))
and hence by Gronwall’s inequality one has
                                            4|K|2 (|K| − 1)
                               A(t) ≫                       ,
                                               1 + B(t)
giving the claim.

    Now we fix a time Λ/2 ≤ t ≤ 0, and drop the dependence on t. For any finite
set K ⊂ Z∗ with |K| ≥ 2, set δ(K) := maxk,k′ ∈K |xk − xk′ | to be the largest gap in
K. Then                   X
                                (xk − xk′ )2 ≤ |K|2 δ(K)2
                         k,k′ ∈K: k,k′

and so from the above lemma we have
                            X
                         1+     E k j ≥ |K|−5 δ(K)−2 .
                                  k∈K
                                  j<K

In particular, if δ(K) ≤ c|K|−5/2 for a sufficiently small absolute constant c > 0,
then we have                  X
                                  E k j ≥ |K|−5 δ(K)−2 ,
                               k∈K
                               j<K

and hence by the pigeonhole principle there exists k ∈ K such that
                           X
                               E k j ≫ |K|−6 δ(K)−2 .
                               j<K
                          Brad Rodgers and Terence Tao                                36

From (52), (58) we have
                    X
                                E k j ≪ 1 + (log2+ ξk ) min |xk − x j |−2 .
                                                         j<K
                          j<K

We conclude that if δ(K) ≤ c|K|−3 for a sufficiently small c > 0, then there exists
k ∈ K such that
                     (log2+ ξk ) min |xk − x j |−2 ≫ |K|−6 δ(K)−2
                                         j<K

or equivalently
                                min |xk − x j | ≪ |K|3 δ(K) log+ ξk .
                                j<K

Now suppose that K is a discrete interval [k− , k+ ]Z∗ for some 1 < k− < k+ . Then

                   min |xk − x j | ≥ min(|xk− − xk− −1 |, |xk+ − xk+ +1 |)
                    j<K

and thus (assuming that δ(K) ≤ c|K|−3 ) we have

                min(|xk− − xk− −1 |, |xk+ − xk+ +1 |) ≪ |K|3 δ(K) log+ k+

which implies that
                                      δ(K ′ ) ≪ |K|3 log(k+ )δ(K)                   (60)
whenever δ(K) ≤ c|K|−3 , where K ′ is either the interval K ′ = [k− − 1, k+ ]Z∗ or
K ′ = [k− , k+ + 1]Z∗ . In either case, we call K ′ an enlargement of K.
    Now we can prove Proposition 13. By symmetry we may assume j is
positive. We can also assume j is large, as the claim follows from compactness
for bounded j. As before, we suppress the dependence on t. It thus suffices to
show that
                                     1
                           log                ≪ (log2 j) log log j
                               |x j+1 − x j |
for large positive j.
    By iterating (60) at most log j times starting from the interval K1 ≔ [ j, j +
1]Z∗ , we can find a sequence

                            [ j, j + 1]Z∗ = K1 ⊂ K2 ⊂ · · · ⊂ Kr

of discrete intervals Ki = [k−,i , k+,i ]Z∗ for some 1 ≤ r ≤ log2+ ξ j with the following
properties:

  (i) For each 1 ≤ i < r, Ki+1 an enlargement of Ki with δ(Ki+1 ) ≪ |Ki |3 δ(Ki ) log+ k+,i .

 (ii) Either δ(Kr ) > c|Kr |−3 , or r + 1 > log2+ ξ j .
              The De Bruijn-Newman constant is non-negative                      37

Since |Ki | ≤ r +1 ≪ log2+ ξ j ≪ log2 j and k+,i ≤ j+r ≪ j, we have from property
(i) that
                               δ(Ki+1 ) ≪ j log2 jδ(Ki )
for all 1 ≤ i < r, and hence
                     δ(Kr ) ≪ exp(O(log2 j log log j))δ(K1 ).
On the other hand, from property (ii), using the bound |Kr | ≤ r + 1 ≪ log2 ξ j in
the first case and (52) and the pigeonhole principle in the second case, we have
                           δ(Kr ) ≫ log−6 ξ j ≫ log−6 j.
Combining the two estimates, we obtain the claim.

                    6. A weak bound on integrated energy
    In addition to truncations of the Hamiltonian, we will also need to control
                               P
truncations of the energy j,k E jk (t). While Proposition 13 provides some
control on the summands here, it is too weak for our purposes (being of worse
than polynomial growth in j, k), and we will need the following integrated bound
that, while still weak, is at least of polynomial growth:

Proposition 15 (Weak bound on integrated energy). Let J > 0. Then
                    Z 0 X
                                E jk (t) dt ≪ J 2 logO(1)
                                                     +    J.
                        Λ/2 J≤ j<k≤2J


    We will use this bound to justify an interchange of a derivative and an infinite
series summation in the next section.

Proof. We may take J to be large, as the claim is trivial for J in the compact
region J = O(1). For any discrete interval I, let QI denote the quantity
                                Z 0 X
                          QI ≔                E jk (t) dt.
                                    Λ/2 j,k∈I: j,k

From (52) we have a crude lower bound
                               Q[J,2J]Z∗ ≫ J log−O(1) J
while from Proposition 13 we have an extremely crude upper bound
                      Q[0.5J,3J]Z∗ ≪ exp(O(log2 J log log J)).
                         Brad Rodgers and Terence Tao                                           38

                                                                                         0.1
The ratio between Q[0.5J,3J]Z∗ and Q[J,2J]Z∗ is thus less than (1 + J −0.1 )0.5J/J . By
the pigeonhole principle, we can then therefore find an interval K ≔ [J− , J+ ]Z∗
containing [J, 2J]Z∗ and contained in [0.5J + J 0.1 , 3J − J 0.1 ]Z∗ , such that
                                   QK ′ ≤ (1 + J −0.1 )QK ,                                    (61)
where K ′ ≔ [J− − J 0.1 , J+ + J 0.1 ]Z∗ is a slight enlargement of K. Next, we apply
Lemma 12(v) and use the fundamental theorem of calculus to obtain the identity
                                             Z 0
     X                                                X                        1
              Hkk′ (Λ/2)−Hkk′ (0) = 4QK −2                                                         dt.
k,k′ ∈K: k,k′                                  Λ/2    j<K
                                                            (x j (t) − x k (t))(x j (t) − xk′ (t))
                                                           k,k′ ∈K: k,k′

From Proposition 13, the left-hand side is O(J 2 logO(1) J), thus
                        Z 0        X                             1
   QK ≪ J 2 logO(1) J +                                                              dt.
                          Λ/2       j<K
                                             (x  j (t) − x k (t))(x j (t) − xk′ (t))
                                    k,k′ ∈[J− ,J+ ]Z∗ : k,k′

Using ab ≪ a2 + b2 , we thus have
                                               Z 0 X
                                                                   1
                   QK ≪ J 2 logO(1) J +                                       dt.
                                                  Λ/2 j<K (x j (t) − xk (t))2
                                                        k∈K

Using (52), the contribution to the integral of those j outside of K ′ may be
crudely bounded by O(J 2 logO(1) J) (in fact one can improve this bound to
O(J logO(1) J) if desired, although this will not help us significantly here). The
contribution of those j inside K ′ may be bounded by
                                   QK ′ − QK ≤ J −0.1 QK ,
thanks to (61). We conclude that
                                     QK ≪ J 2 logO(1) J
and the claim follows.

                       7. Strong control on integrated energy
    As discussed previously, the strategy to establish convergence to local equi-
librium is to study (a suitable variant of) the formal Hamiltonian
                                          X
                               H(t) =            H jk (t)
                                               j,k∈Z∗ : j,k
                  The De Bruijn-Newman constant is non-negative                                   39

and its derivatives, with the intention of controlling (suitable variants of) inte-
grated energies such as
                              Z 0 X
                                            E jk (t) dt.
                                          Λ/4 j,k∈Z∗ : j,k

Unfortunately, even with the bound just obtained in Proposition 13, the above
expression is far from being absolutely convergent. To address this issue we
need to mollify and renormalize the Hamiltonian and the energy in a number of
ways. We renormalize the inverse square function x 7→ |x|12 for x , 0 that appears
in the definition of the energy interactions E jk (t) by introducing the modified
potential
                                    1
                          V(x) ≔ 2 − 1 + 2(|x| − 1),
                                  |x|
which (for positive x) is x12 minus the linearization 1 − 2(x − 1) of that function
at x = 1. As x12 is convex, V is non-negative, and one can verify the asymptotics
                                      1
                             V(x) ≍         for |x| ≤ 1/2
                                    |x|2
                                                                                                 (62)
                             V(x) ≍ (|x| − 1)2 for 1/2 < |x| ≤ 2
                             V(x) ≍ |x| for |x| > 2.
For any distinct j, k and any Λ/2 ≤ t ≤ 0, we define the renormalization
                                                                    !
                                        1          xk (t) − x j (t)
                        Ẽ jk (t) ≔              V
                                    |ξk − ξ j |2      ξk − ξ j
of the interaction energy E jk (t); we observe that
                                             1           (xk (t) − ξk ) − (x j (t) − ξ j )
                Ẽ jk (t) = E jk (t) −              2
                                                      +2                                   .     (63)
                                         |ξk − ξ j |              (ξk − ξ j )3
    For any discrete interval I ⊂ Z∗ , we define the renormalized energy
                                          X
                               Ẽ I (t) ≔      Ẽ jk (t);
                                                    j,k∈I: j,k

this is clearly a non-negative quantity that is non-decreasing in I. It can also be
simplified up to negligible error as follows:

Lemma 16. If I = [I− , I+ ]Z∗ is a discrete interval and Λ/2 ≤ t ≤ 0, then
                                                          
           I
                     X                         1        
                                                              + O(logO(1) (|I− | + |I+ |)).
         Ẽ (t) =              E jk (t) −            2              +
                        j,k∈I: j,k
                                              |ξk − ξ j |      
                          Brad Rodgers and Terence Tao                                        40

Proof. By symmetry and the triangle inequality we may assume without loss of
generality that 0 ≤ I− ≤ I+ ; we may then assume that I+ is large, as the claim is
trivial for I+ in the compact region I+ = O(1). By (63), it suffices to show that
                      X        (xk (t) − ξk ) − (x j (t) − ξ j )
                                                    3
                                                                 ≪ logO(1) I+ .
                    j,k∈I: j,k
                                        (ξk  − ξ j )

We may desymmetrize the left-hand side as
                                X                 X            1
                            2    (x j (t) − ξ j )                       .
                                j∈I
                                                            (ξ − ξ j )3
                                                   k∈I: k, j k

By (50), it thus suffices to show that

                           X X                1
                                                     3
                                                       ≪ logO(1) I+ .                       (64)
                            j∈I k∈I: k, j
                                          (ξk − ξ j )
                            P                 1
Consider the inner sum          k∈I: k, j (ξk −ξ j )3 . From (44) we see that the contribution to
                                                         O(1) 
this inner sum of those k with |k − j| ≥ 12 j (say) is O log j2 I+ . For the remaining
range |k − j| < 21 j, we can use (45) to estimate

                                    log3 ξ j            logO(1) I+
                                                                             !
                         1                    1
                                  =                 + O
                     (ξk − ξ j )3    (4π)3 (k − j)3      j(k − j)2
and so on summing we obtain

                                log3 ξ j                                 logO(1) I+
                                                                                    !
          X          1                           X             1
                              =                                      + O              .
                  (ξ − ξ j )3
         k∈I: k, j k
                                 (4π)3                    1
                                                            (k − j)3         j
                                           k∈I: 0<|k− j|< 2 j


As k 7→ k−1 j is odd around j, the sum on the right-hand side can be estimated as
O( max(| j−I1− |,|I+ − j|)2 ). Using this bound we obtain (64).

   In this section we will establish the following significant improvement to
Proposition 15:

Theorem 17. For any T > 0, one has
              Z 0
                   Ẽ [0.5T log T,3T log T ]Z∗ (t) dt = oT →∞ (T log3+ T ).                 (65)
                    Λ/4
                  The De Bruijn-Newman constant is non-negative                               41

     The remainder of this section is devoted to a proof of Theorem 17. The claim
is trivial for T in any compact region T = O(1), so we may assume without
loss of generality that T is large. Recall the notation X / Y or X = Õ(Y) for
X ≪ Y logO(1) T introduced in the notation section of the paper; this will be
convenient to use in the argument that follows. (Typically, when we use this
notation, we will also have some sort of power gain T −c that will safely absorb
all the logO(1) T factors.) Let ψT : Z∗ → R+ be the weight function
                                                    !−100
                                               | j|
                             ψT ( j) := 1 +               .                  (66)
                                            T log T
This is a smooth positive weight that is mostly localised to the region j =
O(T log T ) and fairly rapidly decaying away from this region.
   We introduce the smoothly truncated renormalized energy
                                   X
                        Ẽ T (t) ≔       ψT ( j)ψT (k)Ẽ jk (t)         (67)
                                          j,k∈Z∗ : j,k

for Λ/2 ≤ t ≤ 0. This is clearly non-negative, and from Proposition 15, (66),
(62), and Fubini’s theorem we see that Ẽ T is absolutely integrable in time (in
particular, it is finite for almost every Λ/2 ≤ t ≤ 0). Since the E jk are non-
negative, we see from (66) that to prove (65) it will suffice to show that
                           Z 0
                                Ẽ T (t) dt = oT →∞ (T log3+ T ).           (68)
                                 Λ/4

    We have an analogue of Lemma 16:

Lemma 18. For almost every Λ/2 ≤ t ≤ 0, one has
                                                                           !
                        X                                        1         
         Ẽ T (t) =               ψT ( j)ψT (k) E jk (t) −            2 
                                                                                 + Õ(1).
                                ∗
                           j,k∈Z : j,k
                                                                |ξk − ξ j |      

Proof. For almost every t, one sees from Proposition 15, (66), and Fubini’s
theorem (and (51)) that the series
                                                                                  !
              X                                   1          |x j (t)| + |xk (t)|
                     ψT ( j)ψT (k) E jk (t) +              +
            j,k: j,k
                                              |ξk − ξ j |2        |ξk − ξ j |3

is absolutely convergent. Thus, by Fubini’s theorem and (63), it will suffice to
show that
                   X                      (xk (t) − ξk ) − (x j (t) − ξ j )
                            ψT ( j)ψT (k)                                   / 1.
               j,k∈Z∗ : j,k
                                                   (ξk − ξ j )3
                       Brad Rodgers and Terence Tao                                 42

We may desymmetrize the left-hand side (again using Fubini’s theorem) as
                  X                              X           ψT (k)
                2      ψT ( j)(x j (t) − ξ j )                          ,
                  j∈Z∗                         k∈Z∗ : k, j
                                                           (ξk − ξ j )3

and so it will suffice to establish the bound
                              X        ψ (k)
                                           T                1    1
                                                       /       +
                            k∈Z∗ : k, j
                                        (ξk − ξ j )3       | j| T

for all j ∈ Z∗ .
    As in the proof of Lemma 16, we see from (44) that the contribution of those
k with |k − j| ≥ 12 j is acceptable. For the remaining range |k − j| < 21 j, we again
use (45) to estimate

                                    log3 ξ j
                                                                   !
                          1                   1              1
                                  =                 + Õ
                     (ξk − ξ j )3    (4π)3 (k − j)3      j(k − j)2
and similarly                                            !
                                                 |k − j|
                           ψT (k) = ψT ( j) + Õ           ,
                                                    T
and the claim follows by direct computation using the fact that k 7→ k−1 j is odd
around j.

    Recall that two indices j, k ∈ Z∗ are said to be nearby, and we write j ∼T k,
if one has
                         0 < | j − k| < (T 2 + | j| + |k|)0.1 .
This is clearly a symmetric relation.
   Next, for Λ/2 ≤ t ≤ 0, we define the smoothly truncated renormalized
Hamiltonian
                                                                               !
                            X                                           1
              H̃T (t) ≔               ψT ( j)ψT (k) H jk (t) − log               . (69)
                        j,k∈Z∗ : j∼ k
                                                                   |ξ j − ξk |
                                 T


From (66), Proposition 13, and (45) we see that the sum here is absolutely
convergent for every Λ/2 ≤ t ≤ 0. We can also express it in terms of non-
negative quantities plus a small error, in a manner similar to Lemma 18, as
follows. We first introduce the renormalization
                                                 1
                               L(x) := log          + |x| − 1
                                                |x|
                  The De Bruijn-Newman constant is non-negative                                       43

of the logarithm function x 7→ log |x|1 ; this is a convex nonnegative function on
R\{0} that vanishes precisely when |x| = 1, and obeys the asymptotics
                                             1
                               L(x) ≍ log+       for 0 < |x| ≤ 1/2
                                            |x|
                                                                                                     (70)
                               L(x) ≍ (|x| − 1)2 for 1/2 < |x| ≤ 2
                               L(x) ≍ |x| for |x| > 2.

For any Λ/2 ≤ t ≤ 0 and distinct j, k ∈ Z∗ , we define the normalization
                                                         !
                                        x j (t) − xk (t)
                          H̃ jk (t) ≔ L                                                              (71)
                                             ξ j − ξk
of the Hamiltonian interaction H jk (t); this is symmetric in j, k and non-negative,
vanishing precisely when xk (t) − x j (t) = ξk − ξ j .

Lemma 19. For every Λ/2 ≤ t ≤ 0, one has
                      X
           H̃T (t) =        ψT ( j)ψT (k)H̃ jk (t) + oT →∞ (T log3+ T ).
                              j,k∈Z∗ : j∼T k


Proof. From (71) one has
                                                        1        (x j (t) − ξ j ) − (xk (t) − ξk )
              H̃ jk (t) = H jk (t) − log                       −
                                                   |ξ j − ξk |               ξ j − ξk
so by (69) it suffices to show that
            X                            (x j (t) − ξ j ) − (xk (t) − ξk )
                         ψT ( j)ψT (k)                                     = oT →∞ (T log3+ T ).
        j,k∈Z∗ : j∼T k
                                                     ξ j − ξk

Note from (43), (51), (45), (66) that the sum here is absolutely convergent.
Desymmetrizing, it suffices to show that

               X                                     X         ψT (k)
                      ψT ( j)|x j (t) − ξ j |                           = oT →∞ (T log3 T ).
               j∈Z∗                                k∈Z∗ : j∼ k
                                                               ξ j − ξk
                                                          T


The inner sum can be crudely bounded by Õ(1) for all j thanks to (66), (45). By
(66), (50), (43), it thus suffices to show that
                              X ψT (k)
                                             = oT →∞ (log T )               (72)
                                      ξ − ξk
                           k∈Z∗ : j∼ k j       T
                            Brad Rodgers and Terence Tao                              44

whenever T 0.5 ≤ | j| ≤ T 1.5 (say). For j ∼T k, one has ψT (k) = ψT ( j) + Õ(T −0.8 ),
and the contribution of the error term is acceptable by (45), so it suffices to show
that                         X          1
                                             = oT →∞ (log T )                     (73)
                                     ξ
                          k∈Z∗ : j∼ k j
                                        − ξk
                                        T

                    0.5
whenever | j| ≥ T         . But from (45) we have
                                                                             
                                        4π                   | j − k|2 
                           ξ j − ξk =         ( j − k) + O                  
                                      log ξ j                  | j| log2+ ξ j
and hence                                                    !
                                   1      log ξ j 1      1
                                        =            +O        .                    (74)
                               ξ j − ξk     4π j − k    | j|
         log ξ  1
As k 7→ 4π j j−k  is odd around j, and the set {k : j ∼T k} is very nearly symmetric
around j, it is then easy to establish (73) as required.

   In contrast to the non-normalized interaction H jk (t), the quantity H̃ jk (t) is
well controlled when k and j are far apart:

Lemma 20 (Long-range decay of H̃ jk ). Let j, k be distinct elements of Z∗ , and let
t be in the range Λ/2 ≤ t ≤ 0. There exists a quantity ε( j) that goes to zero as
| j| → ∞, such that if |k − j| ≥ ε( j)−1 log2+ ξ j , then

                                                   log4+ (| j| + |k|)
                                     H̃ jk (t) ≪                      ,
                                                      |k − j|2
and if ε( j) log2+ ξ j ≤ |k − j| ≤ ε( j)−1 log2+ ξ j , one has the refinement

                                                            log4+ j
                                      H̃ jk (t) ≪ ε( j)2            .
                                                           |k − j|2
Finally, in the remaining region |k − j| < ε( j) log2+ ξ j , one has the crude bound
                                  H̃ jk (t) ≪ (log2+ j) log+ log+ j.

Proof. First suppose that |k − j| ≥ 12 | j| (so in particular |k − j| ≍ | j| + |k|). From
(50) one has
                   xk (t) − x j (t) = ξk − ξ j + O(log+ (| j| + |k|))
while from (44) one has
                                                       | j| + |k|
                                    |ξk − ξ j | ≫
                                                    log+ (| j| + |k|)
               The De Bruijn-Newman constant is non-negative                           45

and thus
                          xk (t) − x j (t)     log2+ (| j| + |k|)
                                           −1≪                    ,
                             ξk − ξ j             | j| + |k|
and the claim then follows from (70) (noting that the case | j| + |k| = O(1) can be
treated by compactness).
    Now suppose that ε( j)−1 log2+ ξ j ≤ |k − j| < 12 | j|. By symmetry we can take
 j positive; we may also assume j large, as the bounded case j = O(1) may be
treated by compactness. From (50) one then has

                           xk (t) − x j (t) = ξk − ξ j + O(log j)

and from (44) and one has
                                                    |k − j|
                                    |ξk − ξ j | ≍                                    (75)
                                                     log j
and hence
                         xk (t) − x j (t)     log2 ( j)
                                          −1≪           ≤ ε( j).
                            ξk − ξ j           |k − j|
The claim then follows from (70).
   Next, suppose that ε( j) log2+ ξ j ≤ |k − j| ≤ ε( j)−1 log2+ ξ j . In this case, from
(52) (iterated O(ε( j)−1 ) times) and (45) we have

                     xk (t) − x j (t) = ξk − ξ j + o j→∞ (ε( j)−1 log j)

while from (44) we continue to have (75), and hence
                      xk (t) − x j (t)                         2    !
                                                        −1 log ( j)
                                       − 1 = o j→∞ ε( j)              ,
                         ξk − ξ j                           |k − j|
with the decay rate in the o j→∞ notation independent of the choice of function
ε(). For ε( j) going to zero sufficiently slowly, the claim once again follows from
(70).
      Finally, for the remaining case |k − j| < ε( j) log2+ ξ j (which implies xk (t) −
x j (t) ≪ log2+ j thanks to (52)) the claim follows from Proposition 13 and (70).

   We call a (time-dependent) quantity moderately sized if it is of the form
O(T log3+ T + Ẽ T (t)), and negligible if it is of the form oT →∞ (T log3+ T + Ẽ T (t)).
The following lemma gives some examples of moderately sized and negligible
quantities:

Lemma 21. Let t be in the range Λ/2 ≤ t ≤ 0.
                       Brad Rodgers and Terence Tao                              46

  (i) The quantity
                                           X           ψT ( j)ψT (k)
                                        j,k∈Z∗ : j,k
                                                     |x j (t) − xk (t)|2
      is moderately sized.
 (ii) The quantity
                                                   X         ψT ( j)ψT (k)
                                   (log+ T )
                                               j,k∈Z∗ : j,k
                                                            |x j (t) − xk (t)|

      is moderately sized.
 (iii) For any absolute constants C, c > 0, the expression
                                         X         ψT ( j)ψT (k)
                         (logC+ T )
                                       ∗      1−c
                                                  |x j (t) − xk (t)|
                                           j,k∈Z : | j|,|k|≤T

      is negligible.
 (iv) For any absolute constants C, c > 0, the expression
                                        X         ψT ( j)ψT (k)
                        (logC+ T )
                                      ∗      1+c
                                                 |x j (t) − xk (t)|
                                           j,k∈Z : | j|,|k|≥T

      is negligible.
Similarly if the xi (t) are replaced by ξi throughout.
Proof. For brevity we omit the explicit dependence on the time t. Also, all
summation indices i, j, k are understood to range in Z∗ .
   From (44) we see that
                              X        1
                                             2
                                               ≪ log2+ j
                                    |ξ
                             k: k, j j
                                       − ξk |
for all j ∈ Z∗ , and hence
                       X                               1
                                  ψT ( j)ψT (k)                ≪ T log3+ T.
                       j,k: j,k
                                                  |ξ j − ξk |2
From this and Lemma 16 we conclude (i). Using
                        log+ T                1
                                    ≤                    + log2+ T
                  |x j (t) − xk (t)| |x j (t) − xk (t)|2
we then obtain (ii). If instead we use
                       logC+ T         1            1
                                   ≤                           + log2C+1
                                                                    +    T
                 |x j (t) − xk (t)| log+ T |x j (t) − xk (t)|2
we obtain (iii) and (iv). Similarly if the xi are replaced by ξi throughout.
                  The De Bruijn-Newman constant is non-negative                                               47

    We now have the following crucial derivative computation:

Proposition 22. In the range Λ/2 ≤ t ≤ 0, the function HT is absolutely
continuous, and the derivative ∂t H̃T (t) is equal to −4Ẽ T (t) plus negligible terms
for almost all t. In other words, one has
                                                                     
                   ∂t H̃T (t) = −4Ẽ T (t) + oT →∞ T log3 T + Ẽ T (t)             (76)
for almost every t.

Remark 23. This may be compared with Lemma 12(v) or indeed the formal
identity (57). That the right hand side is approximated in terms the renormalized
energy, rather than just the energy, may be thought of heuristically as being a
result of ∂t H vanishing when the zeros x j settle on an equilibrium, being spaced
like the points ξ j .

Proof. As before, we omit the explicit dependence on t, and all summation
indices are understood to lie in Z∗ . By (56) we have
                                         ′           ′
                                                                 
                                  2  X 1          X     1 
               ∂t H jk (t) = −                    −             . (77)
                               xk − x j    xk − xi     x j − xi 
                                                     i: i,k                i: i, j

If we formally insert this into (69), and desymmetrize in j and k, we would obtain
the identity
                                                                               ′
                                       X                             1 X 1
                    ∂t H̃T = −4                   ψT ( j)ψT (k)                           .             (78)
                                     j,k: j∼T k
                                                                  xk − x j i: i,k xk − xi

However, we need to justify the interchange of the derivative and the infinite
summation. First, we use the fundamental theorem of calculus to rewrite (77) in
integral form as
                            Z 0                  ′                          ′
                                                                                                    
                                       1         X          1          X               1         
 H jk (0) − H jk (t0 ) = −2                                          −                              dt
                             t0 xk (t) − x j (t) i: i,k xk (t) − xi (t)           x   (t) − x   (t)
                                                                                   j         i
                                                                                                        
                                                                          i: i, j

for any Λ/2 ≤ t0 ≤ 0. Multiplying by ψT ( j)ψT (k), we conclude that
                                                                Z 0
                                  X                                          1
   HT (0) − HT (t0 ) = −2                     ψT ( j)ψT (k)
                                 j,k: j∼T k                      t0   xk (t) − x j (t)
                                                        ′                     ′
                                                                                                  
                                                        X        1          X         1        
                                                   ×                     −                       dt.
                                                       i: i,k
                                                             xk (t) − xi (t)     x j (t) − xi (t) 
                                                                                     i: i, j
                                 Brad Rodgers and Terence Tao                                                      48

By the dominated convergence theorem, we can interchange the outer sum and
the integral as soon as we can show that the expression
                                                                                                            
  X                           Z 0
                                      1           ′  X              1           X   ′
                                                                                                   1         
           ψT ( j)ψT (k)                                                      +                              dt
 j,k: j∼ k               t 0
                             |x k (t) − x j (t)|     
                                                        i: i,k
                                                               x k (t) − x i (t)   i: i, j
                                                                                           x j (t) − x i (t)     
      T



is finite. By symmetry in j and k, it suffices to show that
                                              Z 0                        ′
                  X                                       1          X             1
                              ψT ( j)ψT (k)                                                 dt                  (79)
                 j,k: j∼T k                    t0 |xk (t) − x j (t)| i: i,k xk (t) − xi (t)


is finite. But using (52), (50) we can crudely bound
′                                                                                                                       !
X           1                1                              1                   1
                    ,                   ≪ logO(1)
                                             + (k)                    +
      x (t) − xi (t) |xk (t) − x j (t)|
i: i,k k
                                                   |xk (t) − xk−1 (t)| |xk (t) − xk+1 (t)|

(using the convention x0 (t) = 0), so the expression (79) may in turn be crudely
bounded by
                                                    Z 0
    X                                                              1                        1
          ψ2T (k)(T + |k|)0.1 logO(1)
                                 + (k)                                       2
                                                                               +                       dt,
      k                                              t0 |x k (t) − x k−1 (t)|    |x k (t) − xk+1 (t)|2

and this will be finite thanks to Proposition 15 and (66). We conclude (after
desymmetrizing in j and k) that
                                    Z 0 X                                 ′
                                                            1         X            1
  HT (0) − HT (t0 ) = −4              ψT ( j)ψT (k)                                         dt.
                         t0 j,k: j∼ k               x k (t) − x j (t)       x
                                                                      i: i,k k
                                                                               (t) − xi (t)
                                                T



The above analysis also shows that the integrand is absolutely integrable in time.
From the Lebesgue differentiation theorem, we conclude that H̃T is absolutely
continuous and that (78) holds at almost every time t.
    To conclude the proof of the proposition, it will thus suffice to show that
                                                                    ′
                                    X                        1 X 1
                                            ψT ( j)ψT (k)                                                       (80)
                                  j,k: j∼ k
                                                          xk − x j i,k xk − xi
                                      T



is equal to Ẽ T plus negligible terms. We can split this expression as X1 + X2 +
              The De Bruijn-Newman constant is non-negative                                49

X3 + X4 , where
                          X                              1
              X1 ≔                 ψT ( j)ψT (k)
                      j,k: j∼T k
                                                    (xk − x j )2
                          X                            1       X          1
              X2 ≔                 ψT ( j)ψT (k)
                      j,k: j∼T k
                                                    xk − x j i: i∼ j,k xk − xi
                                                                  T
                          X                            1            X               1
              X3 ≔                 ψT ( j)ψT (k)
                      j,k: j∼T k
                                                    xk − x j i: i∼ k; i≁ j; i, j xk − xi
                                                                 T     T
                                                                 ′
                          X                            1         X           1
              X4 ≔                 ψT ( j)ψT (k)                                  .
                      j,k: j∼T k
                                                    xk − x j i: i≁ k; i,k xk − xi
                                                                  T

   We first claim that X4 is negligible. From (50) we have
                           xk − xi = ξk − ξi + O(log+ (|i| + |k|))
and hence (by (44))
                                             log2+ (|i| + |k|)
                                                               !
                             1       1
                                 =        +O                     ,
                          xk − xi ξk − ξi       |k − i|2
which implies that
                     ′                           ′
                     X             1           X           1
                                         =                      + Õ(T −0.1 ).
                  i: i≁ k; i,k
                               x k − x i   i: i≁ k; i,k
                                                        ξk − ξi
                      T                             T


From (44) we may crudely bound this sum by Õ(1). By Lemma 21(iii), this
shows that the contribution to X4 of those k for which |k| ≤ T 0.9 or |k| ≥ T 1.1
(say) is negligible, so we may assume T 0.9 ≤ |k| ≤ T 1.1 . Let A ≥ 2 be a large
constant. Using (44) we may write
            ′                                                             !
          X          1         X        1        X          1       log T
                         =                  +                    +O         .
                  ξ − ξi
      i: i≁ k; i,k k       0.2
                                     ξk − ξi i: |i|≥A|k| ξk − ξi      A
         T                   i: T    ≤|k−i|≤A|k|

For the first sum on the right-hand side, we use (45) (as in the proof of (74)) as
well as (43) to conclude that
                                                        !
                           1      log ξk 1           1
                                =             + OA        ,
                        ξk − ξi     4π k − i        |k|
where the subscript in the OA notation means that the implied constant can
depend on A. As i 7→ log4πξk k−i
                              1
                                   is odd around k, we conclude that
                                   X           1
                                                    = OA (1).
                               0.2
                                            ξk − ξi
                              i: T    ≤|k−i|≤A|k|
                             Brad Rodgers and Terence Tao                                           50

Meanwhile, combining the i and −i terms and using (44), (43) we have
                                                                   !
                 X         1            X          1         log T
                               = −2ξk                    = O         .
                        ξ − ξi
             i: |i|≥A|k| k
                                               ξ 2 − ξk2
                                      i: i≥A|k| i
                                                               A
Sending A slowly to infinity, we conclude that
                              ′
                            X          1
                                           = oT →∞ (log T )
                                    x − xi
                        i: i≁ k; i,k k      T

and the negligibility of X4 then follows from Lemma 21(ii).
    Now we claim that X2 is negligible. Thanks to the restrictions on i, j, k, we
see that                                                
                    ψT (i), ψT ( j) = 1 + Õ (T + |k|)−0.8 ψT (k)
and hence
        ψT ( j)ψT (k) = ψT (i)2/3 ψT ( j)2/3 ψT (k)2/3 + Õ((T + |k|)−0.8 ψT ( j)ψT (k)).
The sum
                                        X         ψT (i)2/3 ψT ( j)2/3 ψT (k)2/3
                             i, j,k: j∼ k; i∼ j,k
                                                      (xk − x j )(xk − xi )
                                        T       T

symmetrises to zero, and hence
                          X                                                  ψT ( j)ψT (k)
                X2 /                                      (T + |k|)−0.8                         .
                                 i, j,k: j∼T k; i∼T j,k
                                                                          |xk − x j ||xk − xi |
                                           1
Estimating |xk −x j1||xk −xi |1
                       ≪ |xk −x  j|
                                   2 + |x −x |2 and performing the i or j summation
                                         k   i
respectively, we conclude that
                                X                    ψT ( j)ψT (k)
                      X2 /             (T + |k|)−0.6
                             j,k: j∼ k
                                                      |xk − x j |2
                                                    T

and so X2 is negligible thanks to Lemma 21(i).
   We have shown that the expression (80) is equal to X1 + X3 plus negligible
terms. A similar argument (replacing xi with ξi throughout) shows that the
expression
                                                           ′
                         X                       1 X 1
                                ψT ( j)ψT (k)                            (81)
                      j,k: j∼ k
                                              ξk − ξ j i: i,k ξk − ξi
                                        T

is equal to X1′ + X3′ plus negligible terms, where
                          X                          1
                X1′ ≔             ψT ( j)ψT (k)
                       j,k: j∼T k
                                                (ξk − ξ j )2
                          X                        1            X               1
                X3′ ≔             ψT ( j)ψT (k)                                      .
                       j,k: j∼ k
                                                ξk − ξ j i: i∼ k; i≁ j; i, j ξk − ξi
                                    T                                       T     T
                       The De Bruijn-Newman constant is non-negative                                               51

From Lemma 18, we see that Ẽ T is equal to
                                                                                                     !
                                             X                               1            1
                  X1 − X1′ +                             ψT ( j)ψT (k)               −                        (82)
                                       j,k: j≁T k; j,k
                                                                         (xk − x j )2 (ξk − ξ j )2

up to negligible terms. From (43), (44) we have

                                      1              1          logO(1)
                                                                   + (| j| + |k|)
                                             2
                                               −            2
                                                              /
                                  (xk − x j )    (ξk − ξ j )        |k − j|3
when j , k and j ≁T k, so the final term in (82) is negligible. Thus, to complete
the proof of the proposition, it will suffice to show that the expression (81) and
the difference X3 − X3′ are both negligible.
    The expression (81) may be rearranged as
                                                                           
                     X              X ψT ( j)   ′     X 1 
                         ψT (k)                                      .
                       k              j: j∼ k
                                              ξk − ξ j
                                                         
                                                               i: i,k
                                                                      ξk − ξi
                                                                                
                                                           T


By (44), both inner sums are Õ(1), so the contribution of those |k| ≤ T 0.5 or
|k| ≥ T 1.5 (say) are negligible. For T 0.5 < |k| ≤ T 1.5 , we see from (72) that
the factor j: j∼T k ξψkT−ξ
                         ( j)
            P
                            j
                              is oT →∞ (log T ), and from (73), (44), and the triangle
inequality we also see that ′i: i,k ξk 1−ξi = O(log T ). Thus (81) is negligible as
                                  P
required.
    Finally, we show that X3 − X3′ is negligible. This quantity may be written as
                     X                                                      1                     1
                                                 ψT ( j)ψT (k)(                        −                      ).
                                                                  (xk − x j )(xk − xi ) (ξk − ξ j )(ξk − ξi )
   i, j,k: i, j∼T k; |i− j|>(T 2 +|i|+| j|)0.1

Observe that if |k − j| and |k − i| are both larger than or equal to T 0.1 , then from
(50), (44) one has

          1                     1            logO(1) (|i| + | j| + |k|)   logO(1)
                                                                             + (|i| + | j| + |k|)
                     −                      ≪ 0.1+                      ≪                         ,
(xk − x j )(xk − xi ) (ξk − ξ j )(ξk − ξi )  T |ξk − ξ j ||ξk − ξi |        T 0.1 |k − j||k − i|
and so the contribution of this case is negligible. From the triangle inequality,
we see that it is not possible for |k − j| and |k − i| to both be less than T 0.1 , so it
remains to treat the components
                                                                                       !
             X                                       1                     1
                            ψT ( j)ψT (k)                       −
                  2     0.1
                                           (xk − x j )(xk − xi ) (ξk − ξ j )(ξk − ξi )
    i, j,k: 0<| j−k|<(T +| j|+|k|)
  0<|i−k|<T 0.1 ; |i− j|>(T 2 +|i|+| j|)0.1
                                                                                                              (83)
                                     Brad Rodgers and Terence Tao                                              52

and
                                                                                                                  !
                       X                                                        1                     1
                                                        ψT ( j)ψT (k)                      −                        .
                                                                      (xk − x j )(xk − xi ) (ξk − ξ j )(ξk − ξi )
             i, j,k: 0<| j−k|<T 0.1
0<|i−k|<(T 2 +|i|+|k|)0.1 ; |i− j|>(T 2 +|i|+| j|)0.1
                                                                               (84)
   Consider first (83). From the triangle inequality we have | j − k| ≫ T 0.2 , and
hence by (50)
                           1                            1
                                 = (1 + Õ(T −0.2 ))          .
                        xk − x j                     ξk − ξ j
                                                 1          1
By Lemma 21(ii) and (44) we may thus replace xk −x j
                                                      by ξk −ξ j
                                                                 at negligible cost
in (83), leaving us with
                                                                 !
                    X                          1          1          1
                              ψT ( j)ψT (k)          −
                       2  0.1
                                            xk − xi ξk − ξi ξk − ξ j
                 i, j,k: 0<| j−k|<(T +| j|+|k|)
               0<|i−k|<T 0.1 ; |i− j|>(T 2 +|i|+| j|)0.1

up to negligible errors. But by (45) and the hypothesis |i − k| ≤ T 0.1 , one may
bound                      X             ψT ( j)
                                                    / T −0.1 ψT (k)
                              2     0.1
                                        |ξk − ξ j |
                                  j: 0<| j−k|<(T +| j|+|k|)
                                      |i− j|>(T 2 +|i|+| j|)0.1

when T 0.9 ≤ |k| ≤ T 1.1 , and use the weaker bound
                                 X            ψT ( j)
                                                        / ψT (k)
                                    2    0.1
                                             |ξk − ξ j|
                                       j: 0<| j−k|<(T +| j|+|k|)
                                           |i− j|>(T 2 +|i|+| j|)0.1

for all other k, so this expression is also negligible by Lemma 21(ii), (iii), (iv)
(noting that ψT (k) and ψT (i) are comparable). A similar argument also handles
(84).
    To use Proposition 22, we need estimates that ensure Ẽ T is large when H̃T is
large. To this end we have
Lemma 24. Let m be a natural number, and let Λ/2 ≤ t ≤ 0. Let T > 0, and let
δ = δ(T ) go to zero as T → ∞ sufficiently slowly. If H̃T (t) ≥ δmT log3+ T , then
Ẽ T (t) ≫ δ22m T log3+ T where the implied constant is absolute.
Proof. As before, we suppress explicit dependence on t, and we may assume T
to be large as the claim is trivial from compactness for T = O(1). From Lemma
19 we have (for δ decaying sufficiently slowly) that
                      X                                  99
                                ψT ( j)ψT (k)H̃ jk (t) ≥     δmT log3 T.
                  j,k∈Z∗ : j∼ k
                                                         100
                                          T
                 The De Bruijn-Newman constant is non-negative                                       53

From Lemma 20 we see that
                        X
                                                                 H̃ jk (t) ≪ ε( j) log2+ j
                             k: j∼T k; |k− j|≥ε( j) log2+ ξ j

for any j ∈ Z∗ , which implies that
                       X                                                              1
                                                       ψT ( j)ψT (k)H̃ jk (t) ≤         δT log3 T
                                                                                      2
                  j,k: j∼T k; |k− j|≥ε( j) log2+ ξ j

if δ(T ) goes to zero slowly enough. By (69), we conclude that
                        X
                                  ψT ( j)ψT (k)H̃ jk (t) ≫ δmT log3+ T                              (85)
                 j,k: j∼T k; |k− j|<ε( j) log2+ ξ j

   We now claim that
                   X
                                                      ψT ( j)ψT (k)H̃ jk (t) ≪ δ2 mT log3 T         (86)
                 j,k: j∼T k; |k− j|<ε( j) log2+ ξ j
                      |x j −xk |≥2−m |ξ j −ξk |

(say). To see this, we use (70) and (45) to bound
                                              |x j − xk |     |x j − xk |
                          L jk ≪ m +                      ≪m+             log T
                                              |ξ j − ξk |      | j − k|
and also ψT ( j) ≍ ψT (k) for j, k in the sum. Thus we may bound (86) by
                 X                           X                 |x j − xk |
       m                   ψT ( j)2 +                 ψT ( j)2             log T.
                        2                           2
                                                                | j − k|
          j,k: |k− j|<ε( j) log+ ξ j                   j,k: 0<|k− j|<ε( j) log+ ξ j

We may directly compute
                                          X
                                                                 ψT ( j)2 ≪ δ2 T log3 T
                            j,k: j∼T k; |k− j|<ε( j) log2+ ξ j

if δ = δ(T ) goes to zero slowly enough. Thus it will suffice to show that
                         X                  |x j − xk |
                                   ψT ( j)2             ≪ δ2 T log2 T.                              (87)
                                2
                                             | j − k|
                      j,k: 0<|k− j|<ε( j) log+ ξ j

But for any natural number n, we see from telescoping series and (50) that
                             X                           2n
                                    |x j − x j+h | ≪ |h|
                           n    n+1
                                                         n
                                       j: 2 ≤| j|<2
                          Brad Rodgers and Terence Tao                                   54

whenever |h| ≪ 2n ; summing over |h| < ε( j) log2+ ξ j , we conclude that
                          X          |x j − xk |
                                                 ≪ ε(2n )2n n
                          n    n+1
                                      | j − k|
                              j,k: 2 ≤| j|<2
                            0<|k− j|<ε( j) log2+ ξ j

which gives (87) if δ goes to zero slowly enough.
   From (85) and (86) we have
                      X
                                  ψT ( j)ψT (k)H̃ jk (t) ≫ δmT log3 T.
              j,k: j∼T k; |k− j|<ε( j) log2+ ξ j
                    |x j −xk |≤2−m |ξ j −ξk |

But for j, k in this sum, we see from (70), (62) that
                                   |ξ j − ξk |   m2−2m |ξ j − ξk |2
             H̃ jk (t) ≪ log                   ≪                    ≪ m2−2m Ẽ jk
                                   |x j − xk |     |x j − xk |2
and the claim follows.

   We can now shrink H̃T down to a reasonable size in finite time:

Corollary 25. One has H̃T (t) = O(δT log3+ T ) for Λ/4 ≤ t ≤ 0.

Proof. We may take T to be large. From Proposition 22 and Lemma 24, we see
that for any natural number m, and for almost every time t for which one has
                                          H̃T (t) ≥ δmT log3 T,
one has
                                    ∂t H̃T (t) ≤ −cδ22m T log3 T
for some absolute constant c > 0. In particular, if m is larger than some large
absolute constant m0 , and Λ/2 ≤ t ≤ Λ/4 is such that
                        δmT log3 T ≤ H̃T (t) ≤ δ(m + 1)T log3 T,                        (88)
                                                                     ′              3
then it is not possible (for m0 large enough) to have H̃T (t ) ≥ δmT log T for all
t ≤ t′ ≤ t + c−1 2−2m , as this would violate the fundamental theorem of calculus
for absolutely continuous functions. Thus, by the intermediate value theorem,
there exists t ≤ t′ ≤ t + c−1 2−2m such that
                       δ(m − 1)T log3 T ≤ H̃T (t′ ) ≤ δmT log3 T,
and on iterating this we conclude (for m0 large enough) that there exists t ≤ t′′ ≤
t + 2c−1 2−2m0 such that
                             H̃T (t′′ ) ≤ δm0 T log3 T.                       (89)
               The De Bruijn-Newman constant is non-negative                                55

We run this argument with t set equal to Λ/2, and m the unique integer obeying
(88), to conclude (for m0 large enough) that there exists Λ/2 ≤ t′′ ≤ Λ/4
obeying (89). (Note that this conclusion is immediate if the initial value of m
was already less than m0 .) On the other hand, from Proposition 22 we have
∂t H̃T (t) ≤ O(δT log3 T ) for almost every t′′ ≤ t ≤ 0, if δ decays sufficiently
slowly. The claim now follows from the fundamental theorem of calculus
(absorbing m0 into the implied constants), recalling that H̃T is non-negative.

   From Proposition 22 and the fundamental theorem of calculus for absolutely
continuous functions, one has
                                            Z 0
      H̃T (Λ/4) − H̃T (0) = (4 + oT →∞ (1))     Ẽ T (t) dt + oT →∞ (T log3+ T )
                                                     Λ/4

and the claim (68) now follows from Corollary 25. This concludes the proof of
Theorem 17.

                        8. Controlling the energy at time 0
    In the previous section we controlled a time average of the energy. Now,
using monotonicity properties of the energy, we can in fact control energy at
time zero:

Proposition 26 (Energy bound at time zero). Let T be large. Then
                        Ẽ [T log T,2T log T ]Z∗ (0) = oT →∞ (T log3 T ).

    Proposition 26 will be proven16 as follows. First we locate a good initial
interval I 0 = [I−0 , I+0 ]:

Proposition 27 (Locating a good interval). Let T be large. Then there exists an
interval I 0 = [I−0 , I+0 ]Z∗ containing [0.9T log T, 2.1T log T ]Z∗ and contained in
[0.8T log T, 2.2T log T ]Z∗ such that
               Z 0 X             X                0   n 0    n
                                         2−n Ẽ [I± −2 ,I± +2 ]Z∗ (t) dt = Õ(1)
                  Λ/4 ± 1≤2n ≤0.1T log T

where ± ranges over both choices of sign +, − and n ranges over natural numbers
with 1 ≤ 2n ≤ 10T log T .

     Recall that Õ(1) is any quantity which is O(logO(1) T ).

16
 We thank Ofer Zeitouni for pointing out an error in a previous version of this argument.
                                Brad Rodgers and Terence Tao                                        56

Proof. By the pigeonhole principle, it suffices to show that
 Z 0.9T log T Z 2.2T log T Z 0 X  X                  0   n 0    n
                                            2−n Ẽ [I± −2 ,I± +2 ]Z∗ (t) dtdI+0 dI−0 / T 2 .
   0.8T log T      2.1T log T      Λ/4 ± 1≤2n ≤0.1T log T


By the triangle inequality (and the definition of Õ) it suffices to show that
          Z 0.9T log T Z 2.2T log T Z 0
                                             0   n 0    n
                                        Ẽ [I± −2 ,I± +2 ]Z∗ (t) dtdI+0 dI−0 / 2n T 2
                0.8T log T      2.1T log T       Λ/4

for either choice of sign ± and any 1 ≤ 2n ≤ 0.1T log T . But from the Fubini–
Tonelli theorem and the definition of modified energies Ẽ I we see that
    Z 0.9T log T Z 2.2T log T
                                   0   n 0    n
                              Ẽ [I± −2 ,I± +2 ]Z∗ (t) dI+0 dI−0 / 2n T Ẽ [0.5T log T,3T log T ]Z∗ (t)
      0.8T log T      2.1T log T

and the claim now follows from Theorem 17.

    Once the interval I 0 is located, the main step is to iterating the following
claim:

Proposition 28 (Energy propagation inequality). Let T be large, let I = [I− , I+ ]Z∗
be an interval containing [T log T, 2T log T ]Z∗ and contained in [0.5T log T, 3T log T ]Z∗ ,
                                                                                 1
let I 0 be as in Lemma 27, and let Λ/4 ≤ t1 ≤ t2 ≤ 0 be such that t2 ≤ t1 + 100 log2 .
                                                                                    T
Then
                        ′
                    Ẽ I (t2 ) ≤ Ẽ I (t1 ) + Õ(1 + |I− − I−0 | + |I+ − I+0 |),
where I ′ ≔ [I− + log3 T, I+ − log3 T ]Z∗ is a slightly shrunken version of I.

    Let us assume Proposition 28 for the moment and finish the proof of Propo-
sition 26. From Theorem 17 we have
                Z 0
                     Ẽ [0.5T log T,3T log T ]Z∗ (t) dt = oT →∞ (T log3+ T )
                         Λ/4

and so by the pigeonhole principle, we may find Λ/4 ≤ t0 ≤ 0 such that
                             Ẽ [0.5T log T,3T log T ] (t0 ) = oT →∞ (T log3+ T ).
In particular
                                             0
                                      Ẽ I (t0 ) = oT →∞ (T log3+ T ).
Applying Proposition 28 O(log2 T ) times to get from t0 to 0 (starting from the
interval I 0 and shrinking it by at most O(log5 T ) during the entire process), we
conclude that
                           Ẽ [I− ,I+ ] (0) ≤ oT →∞ (T log3+ T )
                The De Bruijn-Newman constant is non-negative                             57

for some interval [I− , I+ ]Z∗ containing [T log T, 2T log T ]Z∗ and contained in
[0.5T log T, 3T log T ]Z∗ . Since Ẽ I (0) is monotone in I, Proposition 26 follows.
    It remains to establish Proposition 28. We use an argument due to Bourgain
[2, §4] that combines local conservation laws (or, in this case, local monotonicity
formulae) with the pigeonhole principle.
    The first step is to locate a good subset of particles indexed by an interval
close to [I− , I+ ] which does not gain too much energy due to interactions its
environment, due to separation between these particles and the environment.
From (50) and the pigeonhole principle, one can find natural numbers
        I− ≤ j− − 1 < j− ≤ I− + log3 T ≤ I+ − log3 T ≤ j+ < j+ + 1 ≤ I+
such that
                                                                    1
                                 x j− (t2 ) − x j− −1 (t2 ) ≥                            (90)
                                                                  log T
(say) and similarly
                                                1
                                 x j+ +1 (t2 ) − x j+ (t2 ) ≥
                                                     .
                                             log T
    From Lemma 12(iv) applied to K = { j− − 1, j− } we have
                                  ∂t (x j− (t) − x j− −1 (t))2 ≤ 8
                                           1
for all t1 ≤ t ≤ t2 . Since t2 − t1 ≤ 100 log2 , we conclude from the fundamental
                                              T
theorem of calculus and (90) that
                                                                   1
                                  x j− (t) − x j− −1 (t) ≫                               (91)
                                                                 log T
for all t1 ≤ t ≤ t2 . Similarly
                                                                   1
                                 x j+ +1 (t) − x j+ (t) ≫              .                 (92)
                                                                 log T
The basic point is that because the particles x j− , . . . , x j+ never get too close to the
remaining particles x j , j < j− and x j , j > j+ in the system, the total energy of
former set of particles will remain approximately conserved over short periods of
time thanks to Lemma 12. More precisely, let K now denote the discrete interval
K ≔ [ j− , j+ ]Z∗ , and define the un-normalized energy
                                           X
                                E K (t) ≔        E kk′ (t).
                                                 k,k′ ∈K: k,k′

From Lemma 12 we have
                                    X                           4
                 ∂t E K (t) ≤                                                        .
                                      j<K
                                                (xk − xk′ )2 (xk − x j )(xk′ − x j )
                                k,k′ ∈K: k,k′
                                   Brad Rodgers and Terence Tao                                                             58

for t1 ≤ t ≤ t2 . But from (91), (92), (50) we have
         X                       4
                             2
                   (xk − xk ) (xk − x j )(xk′ − x j )
                           ′
         j<K
     k,k′ ∈K: |xk −xk′ |≥1
                                                                                                                        !
                                          X              1       X     1              1
                             /                                                 +
                               k,k′ ∈K: |x −x ′ |≥1
                                                    (xk − xk′ ) ± 1 + |k − j± | 1 + |k′ − j± |
                                                               2
                                            k       k

                             /1
and
       X                                4
        j<K
                        (xk − xk′ )2 (xk − x j )(xk′ − x j )
k,k′ ∈K: |xk −xk′ |<1
                                   X                   1      X      1
                        /                                   2
                             k,k′ ∈K: |x −x ′ |<1
                                                  (xk − xk ) ±
                                                          ′     1 + |k − j± |
                                      k     k

                        / (1 + |I− − I−0 | + |I+ − I+0 |)
                                     X                  1       X     1
                            ×
                              k,k′ ∈K: |x −x ′ |<1
                                                   (xk − xk′ ) ± 1 + |k − I±0 |
                                                              2
                                                k       k

                        / (1 + |I− − I−0 | + |I+ − I+0 |)
                                                                                                                                   
                                     X               X                    n      n
                                                                                                                                    
                               ×                          2−n Ẽ [I± −2 ,I± +2 ] (t) + T −1 Ẽ [0.5T log T,3T log T ]Z∗ (t) .
                                     ± 1≤2n ≤0.1T log T

Inserting these bounds and integrating using the fundamental theorem of calcu-
lus, Proposition 27, and Theorem 17, , we conclude that
                      E K (t2 ) ≤ E K (t1 ) + Õ(1 + |I− − I−0 | + |I+ − I+0 |)
which by monotonicity of E K in K implies that
                           ′
                      E I (t2 ) ≤ E I (t1 ) + Õ(1 + |I− − I−0 | + |I+ − I+0 |).
Applying Lemma 16, we conclude that
            ′
                                                                         X            1
        Ẽ I (t2 ) ≤ Ẽ I (t1 ) + Õ(1 + |I− − I−0 | + |I+ − I+0 |) + 2
                                                                        j∈I\I ′
                                                                                 (ξ j − ξk )2
                                                                                          k∈I: j,k
                                                                                                ′
(the factor of two coming because if j, k ∈ I are not both in I then at least one of
the cases j ∈ I\I ′ , k ∈ I and k ∈ I\I ′ , j ∈ I occurs). But from (44) one has
                                    X           1
                                                        /1
                                   j∈I\I ′
                                           (ξ j − ξk )2
                                                    k∈I: j,k
and Proposition 28 follows.
                 The De Bruijn-Newman constant is non-negative                                   59

                           9. Contradicting pair correlation
   It remains to see that Proposition 26 is in contradiction with results that are
known to be the case for the points x j (0). Note in particular that
                                                                     !
                X                     1          x j+1 (0) − x j (0)
                                             2
                                               V                       ≤ Ẽ [T log T,2T log T ] (0).
     T log T ≤ j, j+1≤2T log T
                               |ξ j+1 − ξ j |        ξ j+1 − ξ j

In this range using (43) and (45) we have ξ j+1 − ξ j ∼ 4π/ log+ T , and so
Proposition 26 implies that
                                                               !
            2
                          X                x j+1 (0) − x j (0)
         log T                           V                       = oT →∞ (T log3 T )
               T log T ≤ j, j+1≤2T log T
                                               ξ j+1 − ξ j

By Markov’s inequality (see [28, Ch. 1]), this implies that
                                               !
                           x j+1 (0) − x j (0)
                       V                         = oT →∞ (1)
                               ξ j+1 − ξ j
for a fraction 1 − oT →∞ (1) of j ∈ [T log T, 2T log T ]. But using the properties
(62) of the function V, this implies that
                               x j+1 (0) − x j (0)
                                                   = 1 + oT →∞ (1)
                                   ξ j+1 − ξ j
or
                                                     4π + oT →∞ (1)
                             x j+1 (0) − x j (0) =                  ,                         (93)
                                                         log T
for a fraction 1 − oT →∞ (1) of j ∈ [T log T, 2T log T ].
    In particular since the points x j (0) are twice the imaginary ordinates of
nontrivial zeroes of the Riemann zeta function, this implies that the gaps between
the zeroes of the zeta function are rarely much larger or smaller than the mean
spacing. But this contradicts perhaps most strikingly results of Montgomery
[16] who determined on the Riemann Hypothesis the pair correlation measure
for the zeroes, measured against a class of band-limited functions. As noted by
Montgomery his result implies that a positive proportion of zeroes have a spacing
between them strictly smaller than mean spacing. The proof of this claim is not
written down in [16], but Conrey, et. al. prove as their main result of [7] (using
different ideas) that for any λ > .77 there exists a constant c(λ) > 0 such that at
least a proportion c(λ) of j ≤ T log T satisfy
                                                              4π
                                  x j+1 (0) − x j (0) ≤ λ         .
                                                            log T
This contradicts (93) and therefore the assumption that Λ < 0.
                         Brad Rodgers and Terence Tao                                           60

                                         References
 [1]   M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions, (Dover, New York,
       1965).
 [2]   J. Bourgain, Global wellposedness of defocusing critical nonlinear Schrödinger equation
       in the radial case, J. Amer. Math. Soc. 12 (1999), no. 1, 145–171.
 [3]   K. Broughan, Equivalents of the Riemann hypothesis. Vol. 2. Analytic equivalents. En-
       cyclopedia of Mathematics and its Applications, 165. Cambridge University Press, Cam-
       bridge, 2017.
 [4]   N. C. de Bruijn, The roots of trigonometric integrals, Duke J. Math. 17 (1950), 197–226.
 [5]   A. Chang, D. Mehrle, S. J. Miller, T. Reiter, J. Stahl, D. Yott, Newman’s conjecture in
       function fields, J. Numb. Thy. 157 (2015), 154–169.
 [6]   J. B. Conrey, A. Ghosh, S. M. Gonek, A note on gaps between zeros of the zeta function,
       Bull. London Math. Soc. 16 (1984), no. 4, 421–424.
 [7]   J. B. Conrey, A. Ghosh, D. Goldston, S. M. Gonek, D. R. Heath-Brown, On the distribution
       of gaps between zeros of the zeta-function, Q.J. Math, 36 (1985), 43 –51.
 [8]   G. Csordas, T. S. Norfolk, R. S. Varga, A lower bound for the de Bruijn-Newman constant
       Λ, Numer. Math. 52 (1988), 483–497.
 [9]   G. Csordas, A. M. Odlyzko, W. Smith, R. S. Varga, A new Lehmer pair of zeros and a
       new lower bound for the De Bruijn-Newman constant Lambda, Electronic Transactions on
       Numerical Analysis. 1 (1993), 104–111.
[10]   G. Csordas, A. Ruttan, R.S. Varga, The Laguerre inequalities with applications to a
       problem associated with the Riemann hypothesis, Numer. Algorithms, 1 (1991), 305–329.
[11]   G. Csordas, W. Smith, R. S. Varga, Lehmer pairs of zeros, the de Bruijn-Newman constant
       Λ, and the Riemann hypothesis, Constr. Approx. 10 (1994), no. 1, 107–129.
[12]   A. Dobner, A New Proof of Newman’s Conjecture and a Generalization, preprint.
[13]   L. Erdős, B. Schlein, H.-T. Yau, Universality of random matrices and local relaxation flow,
       Invent. Math. 185 (2011), no.1, 75–119.
[14]   H. Ki, Y. O. Kim, and J. Lee, On the de Bruijn-Newman constant, Advances in Mathemat-
       ics, 22 (2009), 281–306.
[15]   D. H. Lehmer, On the roots of the Riemann zeta-function, Acta Math. 95 (1956) 291–298.
[16]   H. L. Montgomery, The pair correlation of zeros of the zeta function, Analytic number
       theory (Proc. Sympos. Pure Math., Vol. XXIV, St. Louis Univ., St. Louis, Mo., 1972), pp.
       181–193. Amer. Math. Soc., Providence, R.I., 1973.
[17]   H. L. Montgomery, R. C. Vaughan, Multiplicative number theory. I. Classical theory. Cam-
       bridge Studies in Advanced Mathematics, 97. Cambridge University Press, Cambridge,
       2007.
[18]   C. M. Newman, Fourier transforms with only real zeroes, Proc. Amer. Math. Soc. 61
       (1976), 246–251.
[19]   T. S. Norfolk, A. Ruttan, R. S. Varga, A lower bound for the de Bruijn-Newman constant Λ
       II., in A. A. Gonchar and E. B. Saff, editors, Progress in Approximation Theory, 403–418.
       Springer-Verlag, 1992.
[20]   A. M. Odlyzko, An improved bound for the de Bruijn-Newman constant, Numerical
       Algorithms 25 (2000), 293–303.
[21]   D. H. J. Polymath, Effective approximation of heat flow evolution of the Riemann ξ
       function, and a new upper bound for the de Bruijn-Newman constant, Res. Math. Sci.
       6 (2019), no. 3, Paper No. 31, 67 pp.
[22]   G. Pólya, Über trigonometrische Integrale mit nur reelen Nullstellen, J. Reine Angew.
       Math. 58 (1927), 6–18.
                The De Bruijn-Newman constant is non-negative                               61

[23]   Y. Saouter, X. Gourdon, P. Demichel, An improved lower bound for the de Bruijn-Newman
       constant, Mathematics of Computation. 80 (2011), 2281–2287.
[24]   E.M. Stein, R. Shakarchi. Complex analysis (Vol. 2). Princeton University Press, 2010.
[25]   J. Stopple, Notes on Low discriminants and the generalized Newman conjecture, Funct.
       Approx. Comment. Math., vol. 51, no. 1 (2014), pp. 23–41.
[26]   J. Stopple, Lehmer pairs revisited, Exp. Math. 26 (2017), no. 1, 45–53.
[27]   T. Tao, Nonlinear dispersive equations. Local and global analysis. CBMS Regional Confer-
       ence Series in Mathematics, 106. Published for the Conference Board of the Mathematical
       Sciences, Washington, DC; by the American Mathematical Society, Providence, RI, 2006.
[28]   T. Tao, V.H. Vu. Additive Combinatorics (Vol. 105). Cambridge University Press, 2006.
[29]   H. J. J. te Riele, A new lower bound for the de Bruijn-Newman constant, Numer. Math., 58
       (1991), 661–667.
[30]   E. C. Titchmarsh, The Theory of the Riemann Zeta-function, Second ed. (revised by D. R.
       Heath-Brown), Oxford University Press, Oxford, 1986.
