THE INGHAM-HUXLEY ZERO-DENSITY THEOREM
Notes by Tim Jameson
Introduction
We show that for σ ≥ 12 , T ≥ 2, the number of zeros (counting multiplicity) ρ = β + iγ
of ζ(s) with σ ≤ β ≤ 1 and 0 < γ ≤ T is
N (σ, T ) T A(1−σ) logc T
where
A=
12
.
5
As far as I am aware, this is still the smallest known (constant) allowable value of A. The
‘density hypothesis’ is that one may take A = 2. This would follow from the Lindelöf
hypothesis (see for example Titchmarsh) or from Montgomery’s large values conjecture (see
for example Ben Green’s notes [Grn]).
A particular consequence of the zero-density theorem is
1
π(x + x
So we have this with 1 −
1
A
=
7
.
12
1
1− A
+
x1− A +
) − π(x) ∼
.
log x
(It is conjectured to hold with A = 1 but even on RH we
can’t get any better than A = 2. The first result like this was Hoheisel’s A = 33000.) The
proof of this uses the Landau formula (truncated explicit formula for ψ(x)) and also requires
a zero-free region slightly stronger than Weyl’s (which in turn is slightly stronger than de
la Vallée-Poussin’s). As far as I can tell all known proofs of a sufficiently strong zero-free
region are at least as hard as Vinogradov-Korobov.
The original zero-density theorem was N (σ, T ) σ T (σ > 21 ), which was proved by
Bohr and Landau (see Edwards or Titchmarsh). I think for certain bogus ‘zeta functions’
(not possesing Euler products and not satisfying RH) it has been shown that you can’t do
much (any?) better than this.
By following the same methods as used here one can make use of results like
2
Z
T
T +T 3
2
|ζ( 12 + it)|4 dt T 3 +
(Iwaniec – deep; previously Heath-Brown had this with
7
8
in place of
2
)
3
to bound
N (σ, T + ∆) − N (σ, T ). Backlund showed that the Lindelöf hypothesis is equivalent to
this quantity (with ∆ = 1) being o(log T ) for σ ≥ const > 1 (see Edwards or Titchmarsh).
1
Ingham published his result (giving A = 3) in 1940 [Ing2], Huxley in 1972. Huxley
obtained his result (actually generalised to L-functions) whilst preparing his book [Hux1].
His corresponding paper [Hux2] gave a stronger result than the book when σ > 34 .
The basic prerequisites for the present exposition are N (T + 1) − N (T ) log T , the
simplest van der Corput type exponential sum estimates and the approximate functional
equation. Some boring details are omitted, including the smallness of various quantities
using crude bounds for the gamma function. We shan’t concern ourselves with trying to
make c as small as possible.
Large values theorems for Dirichlet polynomials
First we derive a bound of a type due to Davenport using a couple of lemmas of
Gallagher [Gall] (mean values and mean-to-max). Better constants could be obtained by the
methods of Montgomery and Vaughan or Selberg. Suppose m 6= n ⇒ |λm − λn | ≥ δ. For
x ≤ 21 , | sin πx| ≥ 2|x| so sinc x ≥ 2/π. Write Λ(x) = max(0, 1 − |x|).
2
N
2
Z U +T /2 X
N
2 Z ∞ X
π
t
an e(λn (U + t)) sinc2 dt
an e(λn t) dt ≤
4 −∞ n=1
T
U −T /2 n=1
Z
∞
π2 X
t
=
am ān e((λm − λn )U )
e((λm − λn )t)sinc2 dt
4 m,n
T
−∞
π2 X
am ān e((λm − λn )U )T Λ((λm − λn )T ).
4 m,n
P
If δT ≥ 1 then the RHS is (π 2 /4)T n |an |2 . But if T < 1/δ then we can extend the interval
=
of integration to one of length 1/δ. Thus in any case we have
2
N
Z U +T /2 X
N
π2
1 X
an e(λn t) dt ≤
max T,
|an |2 .
4
δ
U −T /2
n=1
n=1
In the case λn = (log n)/(2π), e(λn t) = nit we have (for N ≥ 2) 2πδ = log(N/(N −1)) ≥
1/N so for
S(t) =
N
X
an nit
n=1
we get
Z
T2
|S(t)|2 dt (T2 − T1 + N )
X
T1
|an |2 .
n
We now deduce a discrete version of this.
Z 1
Z
2
12
0
1
1
(x − 2 )f (x)dx = (x − 2 )f (x) 0 −
0
0
2
1
2
f=
1
f (0)
2
Z
−
1
2
f
0
Z
0
(x +
− 21
1
)f 0 (x)dx
2
1
)f 0 (x)dx
2
(x +
− 12
− 21
1
2
Z
|f (0)| ≤
1
2
|f | +
− 12
Suppose q 6= r ⇒ 1 ≤ |tq − tr | ≤ T , say T1 +
X
− 12
0
Z
f+
f (0) =
f=
2
1
2
Z
Z
1
2
1
2
Z
+
0
1
2
Z
−T1
Z
0
−
f
− 12
|f 0 |
− 12
T2
|f | +
1
f (0)
2
(x − 12 )f 0 (x)dx
1
2
≤ tr ≤ T2 −
|f (tr )| ≤
r
0
Z
0
= (x + 12 )f (x) − 1 −
1
2
T2
Z
with T2 − T1 = T + 1. Then
|f 0 |
−T1
With f = S 2 ,
X
2
|S(tr )|
Z
T2
Z
2
T2
|SS 0 |
|S| +
≤
−T1
−T1
r
Z
T2
Z
21 Z
T2
T2
|S|2
|S 0 |2
−T1
−T1
−T1 1
X
1
2
1 + 1 2 (log N ) 2 (T + N )
|an |2
|S|2 +
≤
21
n
= (T + N )(log eN )
X
2
|an |
n
Results like this are called ‘large values’ theorem. The idea is that we can use such a result
to give a bound for R when we assume S(tr ) ≥ V , i.e. we can count the number of times
S(t) is ‘large’. We could have carried on to the discrete case retaining the generality of λn .
The case of λn = n, T = 1 and the tr -spaced with small is the ‘large sieve’, and sometimes
people still call it ‘large sieve’ when it is a more general statement.
Now suppose the tr are as above and sr = σr +tr with 0 ≤ σr ≤ 12 . For 1 ≤ U ≤ x ≤ 2U ,
write
A(x, t) =
X
an n−it
U <n≤x
Then
X
U <n≤2U
an n
−σ−it
Z
2U +
x−σ dA(x, t)
=
U+
−σ
= (2U )
Z
2U
A(2U, t) +
σx−σ−1 A(x, t)dx
U
Z 2U
X
1
−σ−it |A(x, t)|dx
an n
≤ |A(2U, t)| +
2U U
U <n≤2U
3
Z 2U X
X X
X
1
−sr an n ≤
|A(2U, tr )| +
|A(x, tr )|dx
2U U
r
r
r
U <n≤2U
X
|A(x, tr )|
max
U <x≤2U
r
! 21
≤
R
max
U <x≤2U
X
|A(x, tr )|2
r
! 21
R(T + U )(log 2U )
X
|an |2
U <n≤2U
The situation has been simplified by assuming an upper bound on the σr and by the range
being u ∼ U instead of u ≤ U .
Now we derive a bound of a different shape (depending in an essential way on R)
which we will call the Halász-Montgomery-Huxley large values theorem (HMH). This proof
is essentially due to Graham [Gra] (for the bound on the sum over r) and Huxley (for the
bound on R). By duality,
2
X
X X
|an |2
an n−sr ≤ ∆
r
n
n
is equivalent to
2
X
X X
yr n−sr ≤ ∆
|yr |2 ,
r
r
n
the LHS of which is
X
yq ȳr
q,r
X
n−σq −σr −itq +itr
n
X
2
2
−σq −σr −itq +itr 1
≤
(|y
|
+
|y
|
)
n
q
r
2
n
q,r
X
X X
−σq −σr −itq +itr |yr |2
n
=
r
q
n
!
X
X X
2
−σq −σr −itq +itr ≤
|yr | max
n
.
r
n
r
q
X
Now for t ≥ 1, 1 ≤ U ≤ x ≤ 2U we have
U
X
(for U ≥ Ct where C > 1/(2π), by Kusmin-Landau)
it
t
n 1
t 2 (for U t, by the exponent pair ( 21 , 21 )).
U <n≤x
The same bound holds for
P
n
n−s (where 0 ≤ σ ≤ 1) by partial summation (an = 1 in an
earlier identity). Then by summing over U = N/2, N/4, N/8, . . . we see that
N
X
n
−s
1
2
t log 2N +
n=1
4
N
if N t .
t
Thus we may take
1
X
∆ N (from q = r) + RT 2 log 2N + max
r
q
1≤|tq −tr |N
1
N
(N + RT 2 )(log 2N ).
|tq − tr |
Now suppose that
X
an n−sr ≥ V
n
P
for each r. Write L = log 2N and G = n |an |2 . Then (for some constant C)
1
RV 2 ≤ C(N + RT 2 )LG,
1
R(V 2 − CT 2 GL) ≤ CGN L.
1
1
If V 2 ≤ 2CT 2 GL then the LHS is ≥ V 2 /2 so R GN V −2 L. If however V 2 > 2CT 2 GL,
1
then define T0 by V 2 = 2CT02 GL and split the interval for the tr into ≤ (1+T /T0 ) subintervals
of length ≤ T0 . Applying the above to each subinterval and adding up then gives (in all
cases)
T
R 1+
GN V −2 L
T0
G 2 L2
GN V −2 L
1+T
V4
= GN V −2 L + G3 N T V −6 L3 .
If the range for n is U < n ≤ 2U instead of n ≤ N then the above is simplified and we get
P P
1
−sr 2
| (U L + RT 2 )G and R GN V −2 L + G3 N T V −6 L, where L now means
n an n
r|
log 2U . (In the application this has not been taken into account yet.)
Montgomery’s conjecture (modified to cope with Bourgain’s counterexample to the
original version) is
2
X X
an n−sr (N + R)N T max |an |2 .
n
r
n
Discrete fourth power moment
We prove this by an averaging technique from [Hux1]. See also my notes [Jam]. Let
√
C = 1/(2π). Then for t T , xy = Ct, x y T , the approximate functional equation
says
1
ζ( 12 + it) = S(t, x) + χ( 12 + it)S(t, y) + O(T − 4 )
5
where
S(t, x) =
X
1
n− 2 −it .
n≤x
5
1
(Actually the very crude error term O((log T ) 4 ) in place of O(T − 4 ) would suffice.) So
ζ( 12 + it)4 |S(t, x)|4 + |S(t, y)|4 + T −1 ,
ζ( 21 + it)4
1
√
T
Z
1
√
T
Z
√
2 T
(|S(t, x)|4 + |S(t, Ct/x)|4 ) dx + T −1
√
T
√
B T
√
A T
|S(t, x)|4 dx + T −1
for suitable B > A > 0, since
Z
√
2 T
Z
4
√
√
Ct/ T
|S(t, Ct/x)| dx =
|S(t, y)|4
√
Ct/(2 T )
√
2C T
T
Z
Ct
dy
y2
|S(t, y)|4 dy.
√
C T /2
So for a 1-spaced set of values t1 , . . . , tR T (necessarily with R T ) we have
X
r
1
|ζ( 12 + itr )| √
T
4
Z
√
B T
√
A T
X
|S(tr , x)|4 dx + 1
r
Now
S(t, x)2 =
X
1
a(n, x)n− 2 −it
n≤x2
where
a(n, x) =
X
1 ≤ d(n)
de=n
d,e≤x
so
X
|S(tr , x)|4 (T + x2 )(log T )
r
X d(n)2
n
2
n≤x
5
T log T
and hence
X
|ζ( 12 + itr )|4 T log5 T
r
By summing over T, T /2, T /4, . . . the same result holds for 0 < t1 , . . . , tR ≤ T .
6
Zero counting mechanism
ζ(ρ) = 0, ρ = β + iγ with
1
2
< σ ≤ β < 1, 0 < γ ≤ T . Using N (T + 1) − N (T ) log T ,
pick a representative set of such ρ with the γ something like log2 T -spaced. X, Y (to be
chosen) are between fixed powers of T .
MX (s) =
X
µ(d)d−s
d≤X
an =
X
µ(d)
d|n
d≤X
1 < n ≤ X ⇒ an = 0
|an | ≤ d(n)
By the theory of Mellin transforms
−1/Y
e
+
X
−ρ −n/Y
an n e
= MX (1)Γ(1−ρ)Y
n>X
1−ρ
1
+
2πi
Z
1
−β+i∞
2
ζ(w+ρ)MX (w+ρ)Γ(w)Y w dw,
1
−β−i∞
2
where a residue term has vanished since ζ(ρ) = 0. Since e−1/Y is about 1, all ρ (except
possibly a few stupid ones) are of class (i) or class (ii) where class (i) zeros satisfy
X
an n−ρ e−n/Y 1,
X<n≤Y log Y
and class (ii) zeros satisfy
Z 1
2 −β+i log2 T
ζ(w + ρ)MX (w + ρ)Γ(w)Y w dw 1.
1 −β−i log2 T
2
Put w =
1
2
− β + iτ , w + ρ = 21 + iγ + iτ . The integral above has modulus
Z ∞
1
1
≤
|Γ( 2 − β + iτ )|dτ Y 2 −β |ζ( 21 + itρ )MX ( 12 + itρ )|
−∞
1
1
Y 2 −σ |ζ( 21 + itρ )MX ( 21 + itρ )|
log
1
σ−2
where tρ ∈ [γ − log2 T, γ + log2 T ] is chosen to maximise |ζ( 21 + itρ )MX ( 12 + itρ )|, since for
− 12 < α =
1
2
− β < 0,
Z
∞
Z
|Γ(α + iτ )|dτ −∞
0
1
1
1
dτ
= log 1 +
log
.
|α| + τ
|α|
|α|
7
Class (i) – Ingham
Split the range X < n ≤ Y log Y into subranges n ∼ U with U = 2r X. Write
L = log T . The number of class (i) zeros (not counting multiplicity) is
X X
−ρ −n/Y W1 an n e
ρ
X<n≤Y log Y
X X X
an n−ρ e−n/Y ≤
ρ
n∼U
U
! 21
X
X
W1 (T + U )L
d(n)2 n−2σ
U
n∼U
X
=
X
=
X
1
(W1 (T + U )L · U L3 · U −2σ ) 2
U
1
(W1 (T U 1−2σ + U 2−2σ )L4 ) 2
U
1
(W1 (T X 1−2σ + (Y L)2−2σ )L4 ) 2
U
Hence
X
W1 (
1)2 (T X 1−2σ + Y 2−2σ )L5
U
(T X 1−2σ + Y 2−2σ )L7
Class (i) – Huxley (σ ≥ 34 )
Applying HMH with
G
X
d(n)2 n−2σ U 1−2σ L3 ,
n∼U
N = 2U , V
−1
L says that the number of ρ with
X
1
−ρ −n/Y a
n
e
n
L
n∼U
is
U 1−2σ L3 U L2 L + (U 1−2σ L3 )3 U T L6 L3
= U 2−2σ L6 + T U 4−6σ L18
13
Y 2−2σ L 2 + T X 4−6σ L18
W1 is L×this so
W1 (Y 2−2σ + T X 4−6σ )L19
8
Class (ii) – Ingham
Fix p, a, b > 0 (to be chosen) with (1/a) + (1/b) = 1. Then W2 inequality gives
1
log
σ−
−p
1
2
1
Y −p( 2 −σ) W2 X
1p and Hölder’s
P
ρ
|ζ( 12 + itρ )MX ( 12 + itρ )|p
ρ
! a1
X
≤
! 1b
X
|ζ|pa
ρ
|MX |pb
ρ
We choose the exponents so that pa = 4, pb = 2. Thus
p p
+ = 1,
4 2
4
p= ,
3
1
p
1
= = ,
a
4
3
1
p
2
= =
b
2
3
and the above is
! 23
! 13
X
|ζ|4
X
|MX |2
(T L5 )
X µ2 (d)
(T + X)L
d
d≤X
1
3
ρ
ρ
1
2
(T + T 3 X 3 )L3
Hence (not counting multiplicity)
W2 1
log
σ−
43
1
2
Y
2
(1−2σ)
3
1
2
(T + T 3 X 3 )L3
Class (ii) – Huxley (σ ≥ 34 )
1
|ζ( 12 + itρ )MX ( 12 + itρ )| Y σ− 2 ⇒ |ζ| ≥ U or |MX | ≥ V
where
1
U V Y σ− 2
The number of ρ with |ζ| ≥ U is
T U −4 L5 T Y 2−4σ V 4 L5
Applying HMH with
G
X µ2 (d)
L,
d
d≤X
N = X, says that the number of ρ with |MX | ≥ V is
LXV −2 L + L3 XT V −6 L3 = XV −2 L2 + T XV −6 L6
9
! 23
Suppose we choose V so that V 4 T . Then the second term dominates the first term. That
is, we ignore the first term and choose V to minimise
T Y 2−4σ V 4 + T XV −6 ,
that is
V 10 = XY 4σ−2
giving
3
2
W2 T X(XY 4σ−2 )− 5 Lc = T X 5 Y
3
(2−4σ)
5
Lc
provided V 4 T .
Parameters – Ingham
For σ close to
1
2
we can just use N (T ) T L, so we may assume σ ≥const>
ignore the (− log(σ −
1 34
))
2
1
2
and can
factor. Thus
N (σ, T ) (T X 1−2σ + Y 2−2σ + T Y
2
(1−2σ)
3
1
2
+ T 3X3Y
2
(1−2σ)
3
)Lc .
First choose Y to minimise the terms not involving X. Thus
2
Y 2−2σ = T Y − 3 (2σ−1) ,
2
Y = T 1/(2−2σ+ 3 (2σ−1)) = T 3/(2−σ) ,
giving
Y 2−2σ = T 3(1−σ)/(2−σ) .
Now we show that we can choose X so that the other terms are no worse than this. We have
2
T X −(2σ−1) ≤ T Y − 3 (2σ−1) ⇔ Y
⇔ Y
2
(2σ−1)
3
2
3
≤ X 2σ−1
≤X
⇔ T 1/(2−σ) ≤ X
and
1
2
T 3X3Y
2
(1−2σ)
3
≤ TY
2
(1−2σ)
3
⇔ X ≤ T,
so we are OK provided 1/(2 − σ) ≤ 1, which is fine because σ ≤ 1. We have now proved
Ingham’s theorem:
N (σ, T ) T 3(1−σ)/(2−σ) logc T.
Ingham had c = 5. Note that 3/(2 − σ) is increasing but 3(1 − σ)/(2 − σ) is decreasing.
10
Parameters – Huxley (σ ≥ 43 )
Provided V 4 T (see above), we have
2
N (σ, T ) (Y 2−2σ + T X 4−6σ + T X 5 Y
3
(2−4σ)
5
)Lc
Given Y , we choose X to minimise the last two terms, so
2
X 5 +6σ−4 = Y
3
(4σ−2)
5
⇔ X 30σ−18 = Y 12σ−6
⇔ X 5σ−3 = Y 2σ−1 ,
giving
T X 4−6σ = T Y −(6σ−4)(2σ−1)/(5σ−3) ,
so that we choose
Y
= T 1/(2−2σ+(2σ−1)(6σ−4)/(5σ−3))
= T (5σ−3)/((2−2σ)(5σ−3)+(2σ−1)(6σ−4))
= T (5σ−3)/(10σ−6−10σ
= T (5σ−3)/(2σ
2 +2σ−2)
2 +6σ+12σ 2 −8σ−6σ+4)
,
giving
Y 2−2σ = T (5σ−3)(1−σ)/(σ
2 +σ−1)
.
Now we check
5
V 4 ≤ T ⇔ V 10 ≤ T 2
5
⇔ XY 4σ−2 ≤ T 2
(2σ − 1) + (4σ − 2)(5σ − 3)
5
⇔
≤
2σ 2 + 2σ − 2
2
⇔ 2σ − 1 + (4σ − 2)(5σ − 3) ≤ 5(σ 2 + σ − 1)
⇔ 2σ − 1 + 20σ 2 − 12σ − 10σ + 6 ≤ 5(σ 2 + σ − 1)
⇔ 20σ 2 − 20σ + 5 ≤ 5(σ 2 + σ − 1)
⇔ 4σ 2 − 4σ + 1 ≤ σ 2 + σ − 1
⇔ 3σ 2 − 5σ + 2 ≤ 0
2
5
25 2
1
⇔
σ−
≤
− =
6
36 3
36
5 1
⇔ σ − ≤
6
6
2
⇔
≤ σ ≤ 1,
3
11
which is OK. So we have proved (as in Huxley’s book [Hux1])
N (σ, T ) T (5σ−3)(1−σ)/(σ
2 +σ−1)
logc T
( 34 ≤ σ ≤ 1).
We leave showing that (5σ − 3)/(σ 2 + σ − 1) is decreasing on at least the interval in question
as another tedious exercise. [Note added by Graham Jameson: this is easy. Write σ 2 +
√
σ − 1 = (σ − σ1 )(σ − σ2 ), where σ1 = 12 ( 5 − 1) > 53 and σ2 < 0. Then the expression is
5/(σ − σ2 ) + δ/(σ 2 + σ − 1), where δ > 0.] We’ve said all this for σ ≥ 34 , but in fact it would
all work for σ ≥
2
3
or something. However we see that Ingham’s result is better for σ <
since 3/(2 − σ) = (5σ − 3)/(σ 2 + σ − 1) =
12
5
3
4
when σ = 34 , which gives our main statement.
By an elaboration of the argument, Huxley’s paper [Hux2] shows that
N (σ, T ) T 3(1−σ)/(3σ−1) logc T
which is superior for σ >
3
4
( 34 ≤ σ ≤ 1),
and gives (better than) the density hypothesis for σ ≥ 56 . although
in fact one obtains a larger value of c here so that for our final result we would want to use
the book one.
Further improvements (but only for σ > 43 ) by people like Jutila are demonstrated in
Ivić’s book. These use things like Heath-Brown’s result for the 12th power moment
Z T
|ζ( 21 + it)|12 dt T 2 log17 T.
0
There are also delicate results (Selberg) for σ = σ(T ) close to 21 , whereas for σ near 1 one
can use the known zero-free region.
References (added by Graham Jameson)
[BL]
H. Bohr and E. Landau, Sur les zéros de la fonction ζ(s) de Riemann, Comptes
Rendus Acad. Sci. Paris 158 (1914), 106–110.
[Edw]
H. M. Edwards, Riemann’s Zeta Function, Academic Press (1974).
[Gall]
P. X. Gallagher, The large sieve, Mathematika 14 (1967), 14–20.
[Gra]
S. W. Graham, An algorithm for computing optimal exponent pairs, J. London Math. Soc.
(2) 33 (1986), 203–218.
[Grn]
B. J. Green, The Kakeya problem, www.maths.bris.ac.uk/˜mabjg/papers.
[Hoh]
G. Hoheisel, Primzahlprobleme in der Analysis, Sitz. Preuss. Akad. Wiss. 33
(1930), 580–588.
[Hux1]
M. N. Huxley, The Distribution of Prime Numbers, Oxford Univ. Press (1972).
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[Hux2]
M. N. Huxley, On the difference between consecutive primes, Invent. Math. 15
(1972), 164–170.
[Ing1]
A. E. Ingham, Mean-value theorems in the theory of the Riemann zeta function,
Proc. London Math. Soc. 27 (1928), 273–300.
[Ing2]
A. E. Ingham, On the estimation of N (σ, T ), Quart. J. Math. 11 (1940),
291–292.
[Iv]
A. Ivić, The Riemann Zeta Function, Wiley (1985).
[Jam]
Tim Jameson, The fourth power moment of ζ( 21 + it), at
www.maths.lancs.ac.uk/~jameson
[MV]
H. L. Montgomery and R. C. Vaughan, Hilbert’s inequality, J. London Math.
Soc. (2) 8 (1974), 73-82.
[Ti]
E. C. Titchmarsh, The Theory of the Riemann Zeta Function, 2nd ed., Oxford
Univ. Press (1986).
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