a-points of the Riemann zeta function
Youness Lamzouri (York University)
Joint work with Steve Lester (Tel Aviv University)
and Maksym Radziwill (Institute for Advanced Study)
Number Theory at Illinois
A Conference in Memory of the Batemans and Heini Halberstam
June 5, 2014
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
1 / 18
History of the subject and previous results
Definition
Let a be a nonzero complex number. The roots of ζ(s) = a are called
a-points of ζ(s). They are denoted by ρa = βa + iγa .
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
2 / 18
History of the subject and previous results
Definition
Let a be a nonzero complex number. The roots of ζ(s) = a are called
a-points of ζ(s). They are denoted by ρa = βa + iγa .
For every a 6= 0 there is n0 (a) such that for all n ≥ n0 (a) there is an
a-point of ζ(s) quite close to s = −2n, and there are at most finitely
many other a-points in Re(s) ≤ 0. These are called the trivial
a-points of ζ(s).
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
2 / 18
History of the subject and previous results
Definition
Let a be a nonzero complex number. The roots of ζ(s) = a are called
a-points of ζ(s). They are denoted by ρa = βa + iγa .
For every a 6= 0 there is n0 (a) such that for all n ≥ n0 (a) there is an
a-point of ζ(s) quite close to s = −2n, and there are at most finitely
many other a-points in Re(s) ≤ 0. These are called the trivial
a-points of ζ(s).
The remaining a-points all lie in a strip 0 < Re(s) < A, where A
depends on a, and we call these nontrivial a-points.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
2 / 18
History of the subject and previous results
Definition
Let a be a nonzero complex number. The roots of ζ(s) = a are called
a-points of ζ(s). They are denoted by ρa = βa + iγa .
For every a 6= 0 there is n0 (a) such that for all n ≥ n0 (a) there is an
a-point of ζ(s) quite close to s = −2n, and there are at most finitely
many other a-points in Re(s) ≤ 0. These are called the trivial
a-points of ζ(s).
The remaining a-points all lie in a strip 0 < Re(s) < A, where A
depends on a, and we call these nontrivial a-points.
Let
Na (T ) =
X
1.
1<γa <T
βa >0
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
2 / 18
Riemann-von Mangoldt type formula
T
Na (T ) =
log
2π
T
2π
−
T
log 2
− δ1 (a)
T + Oa (log T ),
2π
2π
where δ1 (a) = 1 if a = 1 and equals 0 otherwise.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
3 / 18
Riemann-von Mangoldt type formula
T
Na (T ) =
log
2π
T
2π
−
T
log 2
− δ1 (a)
T + Oa (log T ),
2π
2π
where δ1 (a) = 1 if a = 1 and equals 0 otherwise.
The number a = 1 is special, since ζ(σ + it) = 1 + O(2−σ ) for σ large.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
3 / 18
Riemann-von Mangoldt type formula
T
Na (T ) =
log
2π
T
2π
−
T
log 2
− δ1 (a)
T + Oa (log T ),
2π
2π
where δ1 (a) = 1 if a = 1 and equals 0 otherwise.
The number a = 1 is special, since ζ(σ + it) = 1 + O(2−σ ) for σ large.
Theorem (Levinson, 1975)
Almost all of the a-points are close to the critical line Re(s) = 1/2. More
precisely, for any δ > 0
X
1
1 = Na (T ) 1 + Oδ
.
log T
1/2−δ<βa <1/2+δ
1<γa <T
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
3 / 18
The distribution of a-points in Re(s) ≤ 1/2
Conjecture (Selberg)
For any a 6= 0, three quarters of the a-points of ζ(s) are to the left
of the critical line Re(s) = 1/2. More precisely,
X
3
+ o(1) Na (T ).
1=
4
0<βa <1/2
1<γa <T
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
4 / 18
The distribution of a-points in Re(s) ≤ 1/2
Conjecture (Selberg)
For any a 6= 0, three quarters of the a-points of ζ(s) are to the left
of the critical line Re(s) = 1/2. More precisely,
X
3
+ o(1) Na (T ).
1=
4
0<βa <1/2
1<γa <T
For any a 6= 0, there are at most only finitely many a-points on the
critical line Re(s) = 1/2.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
4 / 18
The distribution of a-points in Re(s) ≤ 1/2
Conjecture (Selberg)
For any a 6= 0, three quarters of the a-points of ζ(s) are to the left
of the critical line Re(s) = 1/2. More precisely,
X
3
+ o(1) Na (T ).
1=
4
0<βa <1/2
1<γa <T
For any a 6= 0, there are at most only finitely many a-points on the
critical line Re(s) = 1/2.
Theorem (Selberg, unpublished)
Assume the Riemann Hypothesis. Then at least (1/2 + o(1))Na (T ) of
the a-points of ζ(s) with ordinates in (1, T ) lie to the left of the critical
line.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
4 / 18
The distribution of a-points in Re(s) > 1/2
Let a be a nonzero complex number and 1/2 < σ1 < σ2 . Define
X
Na (σ1 , σ2 ; T ) :=
1.
σ1 <βa <σ2
1<γa <T
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
5 / 18
The distribution of a-points in Re(s) > 1/2
Let a be a nonzero complex number and 1/2 < σ1 < σ2 . Define
X
Na (σ1 , σ2 ; T ) :=
1.
σ1 <βa <σ2
1<γa <T
Theorem (Borchsenius and Jessen, 1948)
For any 1/2 < σ1 < σ2 , we have
Na (σ1 , σ2 ; T ) ∼ ca (σ1 , σ2 )T .
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
5 / 18
The distribution of a-points in Re(s) > 1/2
Let a be a nonzero complex number and 1/2 < σ1 < σ2 . Define
X
Na (σ1 , σ2 ; T ) :=
1.
σ1 <βa <σ2
1<γa <T
Theorem (Borchsenius and Jessen, 1948)
For any 1/2 < σ1 < σ2 , we have
Na (σ1 , σ2 ; T ) ∼ ca (σ1 , σ2 )T .
Theorem (Matsumoto, 1988)
For any 1 ≤ σ1 < σ2 , we have
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + OA
Youness Lamzouri (York)
T
(log log T )A
a-points of the Riemann zeta function
.
June 5, 2014
5 / 18
Our results
Na (σ1 , σ2 ; T ) =
X
1.
σ1 <βa <σ2
1<γa <T
Theorem 1 (L, Lester and Radziwill, 2014)
Let 1/2 < σ1 < σ2 < 1. We have
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + O
Youness Lamzouri (York)
T log log T
(log T )σ1 /2
a-points of the Riemann zeta function
.
June 5, 2014
6 / 18
Our results
Na (σ1 , σ2 ; T ) =
X
1.
σ1 <βa <σ2
1<γa <T
Theorem 1 (L, Lester and Radziwill, 2014)
Let 1/2 < σ1 < σ2 < 1. We have
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + O
T log log T
(log T )σ1 /2
.
Remark
Our method also gives an improved version of Matsumoto’s result for
σ1 ≥ 1. Namely, for 1 ≤ σ1 < σ2 we have
T log log T
√
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + O
.
log T
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
6 / 18
The work of Borchsenius and Jessen
Na (σ1 , σ2 ; T ) :=
X
1.
σ1 <βa <σ2
1<γa <T
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
7 / 18
The work of Borchsenius and Jessen
Na (σ1 , σ2 ; T ) :=
X
1.
σ1 <βa <σ2
1<γa <T
Borchsenius and Jessen studied the zeros of general almost periodic
functions and applied their results to the function ζ(s) − a.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
7 / 18
The work of Borchsenius and Jessen
Na (σ1 , σ2 ; T ) :=
X
1.
σ1 <βa <σ2
1<γa <T
Borchsenius and Jessen studied the zeros of general almost periodic
functions and applied their results to the function ζ(s) − a.
They showed that the Jessen function
Z
1 T
fa (σ) = lim
log |ζ(σ + it) − a|dt
T →∞ T 1
exists and is twice differentiable in σ, for 1/2 < σ < 1.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
7 / 18
The work of Borchsenius and Jessen
Na (σ1 , σ2 ; T ) :=
X
1.
σ1 <βa <σ2
1<γa <T
Borchsenius and Jessen studied the zeros of general almost periodic
functions and applied their results to the function ζ(s) − a.
They showed that the Jessen function
Z
1 T
fa (σ) = lim
log |ζ(σ + it) − a|dt
T →∞ T 1
exists and is twice differentiable in σ, for 1/2 < σ < 1.
They proved
Na (σ1 , σ2 ; T ) ∼ ca (σ1 , σ2 )T ,
where
ca (σ1 , σ2 ) =
Youness Lamzouri (York)
fa0 (σ2 ) − fa0 (σ1 )
.
2π
a-points of the Riemann zeta function
June 5, 2014
7 / 18
Strategy of the proof of Theorem 1
Theorem 1 (L, Lester and Radziwill, 2014)
Let 1/2 < σ1 < σ2 < 1. We have
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + O
Youness Lamzouri (York)
T log log T
(log T )σ1 /2
a-points of the Riemann zeta function
.
June 5, 2014
8 / 18
Strategy of the proof of Theorem 1
Theorem 1 (L, Lester and Radziwill, 2014)
Let 1/2 < σ1 < σ2 < 1. We have
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + O
T log log T
(log T )σ1 /2
.
Littlewood’s Lemma
Let R be the rectangle with vertices σ1 + i, σ2 + i, σ2 + iT , σ1 + iT .
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
8 / 18
Strategy of the proof of Theorem 1
Theorem 1 (L, Lester and Radziwill, 2014)
Let 1/2 < σ1 < σ2 < 1. We have
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + O
T log log T
(log T )σ1 /2
.
Littlewood’s Lemma
Let R be the rectangle with vertices σ1 + i, σ2 + i, σ2 + iT , σ1 + iT .
Let φ(s) be holomorphic inside and on the boundary of R.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
8 / 18
Strategy of the proof of Theorem 1
Theorem 1 (L, Lester and Radziwill, 2014)
Let 1/2 < σ1 < σ2 < 1. We have
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + O
T log log T
(log T )σ1 /2
.
Littlewood’s Lemma
Let R be the rectangle with vertices σ1 + i, σ2 + i, σ2 + iT , σ1 + iT .
Let φ(s) be holomorphic inside and on the boundary of R.
Let ν(σ, T ) be the number of zeros of φ(s) inside the rectangle R
and such that Re(s) > σ.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
8 / 18
Strategy of the proof of Theorem 1
Theorem 1 (L, Lester and Radziwill, 2014)
Let 1/2 < σ1 < σ2 < 1. We have
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + O
T log log T
(log T )σ1 /2
.
Littlewood’s Lemma
Let R be the rectangle with vertices σ1 + i, σ2 + i, σ2 + iT , σ1 + iT .
Let φ(s) be holomorphic inside and on the boundary of R.
Let ν(σ, T ) be the number of zeros of φ(s) inside the rectangle R
and such that Re(s) > σ.
Z σ2
Z
1
ν(σ, T )dσ = −
log φ(s)ds.
2πi ∂R
σ1
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
8 / 18
Na (σ, T ) :=
X
1.
βa >σ
1<γa <T
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
9 / 18
Na (σ, T ) :=
X
1.
βa >σ
1<γa <T
Littlewood’s Lemma
σ2
Z
Na (σ, T )dσ
σ1
Z T
Z T
1
1
log |ζ(σ1 + it) − a|dt −
log |ζ(σ2 + it) − a|dt
2π 1
2π 1
Z σ2
Z σ2
1
1
−
arg(ζ(σ + i) − a)dσ +
arg(ζ(σ + iT ) − a)dσ
2π σ1
2π σ1
=
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
9 / 18
Na (σ, T ) :=
X
1.
βa >σ
1<γa <T
Littlewood’s Lemma
Z
σ2
Na (σ, T )dσ
σ1
Z T
Z T
1
1
log |ζ(σ1 + it) − a|dt −
log |ζ(σ2 + it) − a|dt
2π 1
2π 1
Z σ2
Z σ2
1
1
−
arg(ζ(σ + i) − a)dσ +
arg(ζ(σ + iT ) − a)dσ
2π σ1
2π σ1
Z T
Z T
1
1
log |ζ(σ1 + it) − a|dt −
log |ζ(σ2 + it) − a|dt
=
2π 1
2π 1
+Oa (log T ).
=
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
9 / 18
An asymptotic formula for
RT
1
log |ζ(σ + it) − a|dt
Borchsenius and Jessen
Z
T
log |ζ(σ + it) − a|dt ∼ T · fa (σ).
1
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
10 / 18
An asymptotic formula for
RT
1
log |ζ(σ + it) − a|dt
Borchsenius and Jessen
Z
T
log |ζ(σ + it) − a|dt ∼ T · fa (σ).
1
How can we obtain a quantitative asymptotic formula?
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
10 / 18
An asymptotic formula for
RT
1
log |ζ(σ + it) − a|dt
Borchsenius and Jessen
Z
T
log |ζ(σ + it) − a|dt ∼ T · fa (σ).
1
How can we obtain a quantitative asymptotic formula?
Unlike most classical mean value problems where Dirichlet polynomial
approximations are used, there is no simple expression of
log |ζ(σ + it) − a| as a Dirichlet series for σ > A.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
10 / 18
An asymptotic formula for
RT
1
log |ζ(σ + it) − a|dt
Borchsenius and Jessen
Z
T
log |ζ(σ + it) − a|dt ∼ T · fa (σ).
1
How can we obtain a quantitative asymptotic formula?
Unlike most classical mean value problems where Dirichlet polynomial
approximations are used, there is no simple expression of
log |ζ(σ + it) − a| as a Dirichlet series for σ > A.
Instead, we exploit the fact that ζ(s) can be very well modeled by a
random Euler product for Re(s) > 1/2.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
10 / 18
A random model for the values log ζ(σ + it)
Let {X (p)}p prime be a sequence of independent random variables
uniformly distributed on the Unit circle U = {z ∈ C : |z| = 1}.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
11 / 18
A random model for the values log ζ(σ + it)
Let {X (p)}p prime be a sequence of independent random variables
uniformly distributed on the Unit circle U = {z ∈ C : |z| = 1}. Define
Y
X (p) −1
ζ(σ, X ) :=
1− σ
.
p
p
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
11 / 18
A random model for the values log ζ(σ + it)
Let {X (p)}p prime be a sequence of independent random variables
uniformly distributed on the Unit circle U = {z ∈ C : |z| = 1}. Define
Y
X (p) −1
ζ(σ, X ) :=
1− σ
.
p
p
This random product converges almost surely for any σ > 1/2.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
11 / 18
A random model for the values log ζ(σ + it)
Let {X (p)}p prime be a sequence of independent random variables
uniformly distributed on the Unit circle U = {z ∈ C : |z| = 1}. Define
Y
X (p) −1
ζ(σ, X ) :=
1− σ
.
p
p
This random product converges almost surely for any σ > 1/2.
Bohr and Jessen (1930)
For any σ > 1/2, the distribution of log ζ(σ + it) converges, as
t → ∞, to the distribution of the random variable log ζ(σ, X ).
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
11 / 18
A random model for the values log ζ(σ + it)
Let {X (p)}p prime be a sequence of independent random variables
uniformly distributed on the Unit circle U = {z ∈ C : |z| = 1}. Define
Y
X (p) −1
ζ(σ, X ) :=
1− σ
.
p
p
This random product converges almost surely for any σ > 1/2.
Bohr and Jessen (1930)
For any σ > 1/2, the distribution of log ζ(σ + it) converges, as
t → ∞, to the distribution of the random variable log ζ(σ, X ).
More precisely, for any rectangle R in C with sides parallel to the
coordinate axes we have
lim
T →∞
1
meas{t ∈ [1, T ] : log ζ(σ + it) ∈ R} = P (log ζ(σ, X ) ∈ R) .
T
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
11 / 18
A discrepancy bound for the distribution of log ζ(σ + it)
Let
Dσ (T ) = sup PT log ζ(σ + it) ∈ R − P log ζ(σ, X ) ∈ R R
where the supremum is taken aver all rectangles R in C with sides parallel
to the coordinate axes and where
PT (f (t) ∈ R) :=
Youness Lamzouri (York)
1
meas{t ∈ [1, T ] : f (t) ∈ R}.
T
a-points of the Riemann zeta function
June 5, 2014
12 / 18
A discrepancy bound for the distribution of log ζ(σ + it)
Let
Dσ (T ) = sup PT log ζ(σ + it) ∈ R − P log ζ(σ, X ) ∈ R R
where the supremum is taken aver all rectangles R in C with sides parallel
to the coordinate axes and where
PT (f (t) ∈ R) :=
1
meas{t ∈ [1, T ] : f (t) ∈ R}.
T
Theorem 2 (L, Lester and Radziwill, 2014)
Let 1/2 < σ < 1 be fixed. Then
Dσ (T ) σ
Youness Lamzouri (York)
1
.
(log T )σ
a-points of the Riemann zeta function
June 5, 2014
12 / 18
We use a“two-dimensional
integration by part” technique to relate
RT
the integral 1 log |ζ(σ + it) − a|dt to the discrepancy Dσ (T ).
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
13 / 18
We use a“two-dimensional
integration by part” technique to relate
RT
the integral 1 log |ζ(σ + it) − a|dt to the discrepancy Dσ (T ).
However, in order for our method to work
the
we need to control
contribution of the points t for which log |ζ(σ + it) − a| is large.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
13 / 18
We use a“two-dimensional
integration by part” technique to relate
RT
the integral 1 log |ζ(σ + it) − a|dt to the discrepancy Dσ (T ).
However, in order for our method to work
the
we need to control
contribution of the points t for which log |ζ(σ + it) − a| is large.
To bound the contribution of those t with |ζ(σ + it)| large, we use
bounds for the distribution of extreme values of log |ζ(σ + it)|
(L, 2011).
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
13 / 18
We use a“two-dimensional
integration by part” technique to relate
RT
the integral 1 log |ζ(σ + it) − a|dt to the discrepancy Dσ (T ).
However, in order for our method to work
the
we need to control
contribution of the points t for which log |ζ(σ + it) − a| is large.
To bound the contribution of those t with |ζ(σ + it)| large, we use
bounds for the distribution of extreme values of log |ζ(σ + it)|
(L, 2011).
To bound the contribution of those t such that ζ(σ + it) is very close
to a we use the following L2k bound.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
13 / 18
We use a“two-dimensional
integration by part” technique to relate
RT
the integral 1 log |ζ(σ + it) − a|dt to the discrepancy Dσ (T ).
However, in order for our method to work
the
we need to control
contribution of the points t for which log |ζ(σ + it) − a| is large.
To bound the contribution of those t with |ζ(σ + it)| large, we use
bounds for the distribution of extreme values of log |ζ(σ + it)|
(L, 2011).
To bound the contribution of those t such that ζ(σ + it) is very close
to a we use the following L2k bound.
Proposition (L, Lester and Radziwill)
Let 1/2 < σ < 1. Then, there exists a constant C = C (a, σ), such that for
any integer k ≥ 1
Z
T
(log |ζ(σ + it) − a|)2k dt T (Ck)4k .
1
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
13 / 18
An asymptotic formula for
quantitative error term
RT
1
log |ζ(σ + it) − a|dt with a
Theorem 3 (L, Lester and Radziwill, 2014)
Let 1/2 < σ < 1 be fixed. Then
Z
T
log |ζ(σ + it) − a|dt = T · E (log |ζ(σ, X ) − a|) + O
1
T (log log T )2
(log T )σ
.
In particular,
fa (σ) = E (log |ζ(σ, X ) − a|).
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
14 / 18
An asymptotic formula for
quantitative error term
RT
1
log |ζ(σ + it) − a|dt with a
Theorem 3 (L, Lester and Radziwill, 2014)
Let 1/2 < σ < 1 be fixed. Then
Z
T
log |ζ(σ + it) − a|dt = T · E (log |ζ(σ, X ) − a|) + O
1
T (log log T )2
(log T )σ
.
In particular,
fa (σ) = E (log |ζ(σ, X ) − a|).
Remark
A part from the power of log log T , the error term is the best that we can
obtain using our discrepancy bound.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
14 / 18
Completing the proof of Theorem 1
Recall that Na (σ, T ) = |{βa > σ, 1 < γa < T }|.
Littlewood’s Lemma
Z
σ2
σ1
1
Na (σ, T )dσ =
2π
−
Youness Lamzouri (York)
1
2π
Z
T
log |ζ(σ1 + it) − a|dt
1
Z T
log |ζ(σ2 + it) − a|dt + Oa (log T ) .
1
a-points of the Riemann zeta function
June 5, 2014
15 / 18
Completing the proof of Theorem 1
Recall that Na (σ, T ) = |{βa > σ, 1 < γa < T }|.
Littlewood’s Lemma
Z
σ2
σ1
1
Na (σ, T )dσ =
2π
−
1
2π
Z
T
log |ζ(σ1 + it) − a|dt
1
Z T
log |ζ(σ2 + it) − a|dt + Oa (log T ) .
1
Theorem 3 (L, Lester and Radziwill)
Z
T
log |ζ(σ + it) − a|dt = T · fa (σ) + O
1
Youness Lamzouri (York)
a-points of the Riemann zeta function
T (log log T )2
(log T )σ
.
June 5, 2014
15 / 18
Completing the proof of Theorem 1
Recall that Na (σ, T ) = |{βa > σ, 1 < γa < T }|.
Littlewood’s Lemma
Z
σ2
σ1
1
Na (σ, T )dσ =
2π
−
1
2π
Z
T
log |ζ(σ1 + it) − a|dt
1
Z T
log |ζ(σ2 + it) − a|dt + Oa (log T ) .
1
Theorem 3 (L, Lester and Radziwill)
Z
T
log |ζ(σ + it) − a|dt = T · fa (σ) + O
1
T (log log T )2
(log T )σ
.
Borchsenius and Jessen
fa (σ) = E (log |ζ(σ, X ) − a|) is twice differentiable in σ, for 1/2 < σ < 1.
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
15 / 18
Recall that Na (σ, T ) = |{βa > σ, 1 < γa < T }|.
1
h
Z
σ
σ+h
T
Na (σ, T )dσ =
2π
Youness Lamzouri (York)
fa (σ) − fa (σ + h)
h
a-points of the Riemann zeta function
+O
T (log log T )2
h(log T )σ
June 5, 2014
16 / 18
Recall that Na (σ, T ) = |{βa > σ, 1 < γa < T }|.
1
h
Z
σ
σ+h
T fa (σ) − fa (σ + h)
T (log log T )2
Na (σ, T )dσ =
+O
2π
h
h(log T )σ
T 0
T (log log T )2
= − fa (σ) + O hT +
.
2π
h(log T )σ
Since fa (σ + h) = fa (σ) + hfa0 (σ) + Oa (h2 ).
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
16 / 18
Recall that Na (σ, T ) = |{βa > σ, 1 < γa < T }|.
1
h
Z
σ
σ+h
T fa (σ) − fa (σ + h)
T (log log T )2
Na (σ, T )dσ =
+O
2π
h
h(log T )σ
T 0
T (log log T )2
= − fa (σ) + O hT +
.
2π
h(log T )σ
Since fa (σ + h) = fa (σ) + hfa0 (σ) + Oa (h2 ).
Finally using that Na (σ, T ) is decreasing in σ, and choosing h optimally,
leads to
Na (σ1 , σ2 ; T ) =
X
σ1 <βa <σ2
1<γa <T
Youness Lamzouri (York)
f 0 (σ2 ) − fa0 (σ1 )
1= a
T +O
2π
a-points of the Riemann zeta function
T log log T
(log T )σ1 /2
June 5, 2014
.
16 / 18
Last remarks
Our result
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + O
T log log T
(log T )σ1 /2
,
Relies on the discrepancy bound
1
sup meas{t ∈ [1, T ] : log ζ(σ + it) ∈ R} − P (log ζ(σ, X ) ∈ R)
R T
1
σ
,
(log T )σ
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
17 / 18
Last remarks
Our result
Na (σ1 , σ2 ; T ) = ca (σ1 , σ2 )T + O
T log log T
(log T )σ1 /2
,
Relies on the discrepancy bound
1
sup meas{t ∈ [1, T ] : log ζ(σ + it) ∈ R} − P (log ζ(σ, X ) ∈ R)
R T
1
σ
,
(log T )σ
Any improvement in the discrepancy bound for the distribution of
log ζ(σ + it) will lead to an improvement in the error term of the number
of a-points of ζ(s).
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
17 / 18
Thank you for your attention !
Youness Lamzouri (York)
a-points of the Riemann zeta function
June 5, 2014
18 / 18