A Proof of the Generalized Riemann Hypothesis

A Proof of the Generalized Riemann Hypothesis
Charaf Ech-Chatbi
Quantitative Analyst
charaf@melix.net
August 14, 2019
Charaf Ech-Chatbi Generalized Riemann Hypothesis August 14, 2019 1 / 32

Overview
1 Dirichlet L−functions
Dirichlet series
Dirichlet character
Dirichlet L−functions
2 The Generalized Riemann Hypothesis
3 Road to the proof of GRH
Proof Strategy
Asymptotic Expansion
The Case of Dirichlet character modulo an odd prime q
The Case of Riemann Hypothesis
4 Proof of the GRH
Case One: 1
2 < (s) ≤ 1 and χ non-trivial
Case Two: 1
2 < (s) ≤ 1 and χ trivial
Conclusion

Dirichlet L−functions Dirichlet series
Dirichlet series
Let’s (zn)n≥1 be a sequence of complex numbers. A Dirichlet series is a
series of the form
∞
n=1
zn
ns
, where s is complex.
If (zn)n≥1 is a bounded, then the corresponding Dirichlet series
converges absolutely on the open half-plane where (s) > 1.
If the set of sums zn + zn+1 + ... + zn+k for each n and k ≥ 0 is
bounded, then the corresponding Dirichlet series converges on the
open half-plane where (s) > 0.
In general, if zn = O(nk), the corresponding Dirichlet series converges
absolutely in the half plane where (s) > k + 1.
The function L(s) =
∞
n=1
zn
ns
is analytic on the corresponding open half
plane.

Dirichlet L−functions Dirichlet character
Dirichlet character
A function χ : Z −→ C is a Dirichlet character modulo q if it satisﬁes the
following criteria:
(i) χ(n) = 0 if (n, q) = 1.
(ii) χ(n) = 0 if (n, q) > 1.
(iii) χ is periodic with period q :χ(n + q) = χ(n) for all n.
(iv) χ is multiplicative :χ(mn) = χ(m)χ(n) for all integers m and n.
The trivial character is the one with χ0(n) = 1 whenever (n, q) = 1. For
any integer n we have χ(1) = 1. Also if (n, q) = 1, we have
(χ(n))φ(q) = 1 with φ is Euler’s totient function. χ(n) is a φ(q)−th root
of unity. Therefore, |χ(n)| = 1 if (n, q) = 1, and |χ(n)| = 0 if (n, q) > 1.
The cancellation property for Dirichlet characters modulo q: For any n
integer
q
i=1
χ(i + n) =
φ(q), if χ = χ0 the trivial character
0, if otherwise, χ = χ0
(1)

Dirichlet L−functions Dirichlet L−functions
Dirichlet L−functions
The Dirichlet L−functions are simply the sum of the Dirichlet series. Let’s
χ be a Dirichlet character modulo q, The Dirichlet L−function L(s, χ) is
deﬁned for (s) > 1 as the following:
L(s, χ) =
+∞
n=1
χ(n)
ns
(2)
Where the series n≥1
χ(n)
ns is convergent when (s) > 0 and L(s, χ) is
analytic in (s) > 0. In the particular case of the trivial character χ0,
L(s, χ0) extends to a meromorphic function in (s) > 0 with the only pole
at s = 1.
Also, like the Riemann zeta function, the Dirichlet L−functions have their
Euler product. For (s) > 1:
L(s, χ) =
p Prime
1 −
χ(p)
ps
(3)

Dirichlet L−functions Dirichlet L−functions
Functional equation
Let’s q be the smallest divisor (prime) of q. Let’s χ be the Dirichlet
character χ mod q . For any integer n such that (n, q) = 1 we have also
(n, q ) = 1 and χ(n) = χ (n). χ is called primitive and L(s, χ) and
L(s, χ ) are related analytically such that:
L(s, χ) = L(s, χ )
p/q
1 −
χ (p)
ps
(4)
L(s, χ) and L(s, χ ) have the same zeros in the critical strip 0 ≤ (s) ≤ 1.
Also, for a primitive character χ, (i.e.χ = χ ) L(s, χ) has the following
functional equation:
τ(χ)Γ(
1 − s + a
2
) L(1 − s, χ) =
√
π(
q
π
)s
ia
q
1
2 Γ(
s + a
2
) L(s, χ) (5)
Where Γ is the Gamma function and a = 0 if χ(−1) = 1 and a = 1 if
χ(−1) = −1, and τ(χ) = q
k=1 χ(k) exp(2πki
q ).

The Generalized Riemann Hypothesis
Generalized Riemann Hypothesis (GRH)
The GRH states that the Dirichlet L−functions have all their non-trivial
zeros on the critical line (s) = 1
2.
Non trivial in this case means L(s, χ) = 0 for s ∈ C and 0 < (s) < 1. So
for any primitive character χ modulo q, all non-trivial zeros of L(s, χ) lies
in the critical strip {s ∈ C : 0 < (s) < 1}. From the functional equation
above we have that if:
s0 is a non-trivial zero of L(s, χ), then 1 − s0 is a zero of L(s, χ).
s0 is a non-trivial zero of L(s, χ), then 1 − s0 is a zero of L(s, χ).
Therefore, we just need to prove that for all primitive character χ modulo
q, there is no non-trivial zeros of L(s, χ) in the right hand side of the
critical strip {s ∈ C : 1
2 < (s) < 1}.

Road to the proof of GRH Proof Strategy
Proof Strategy
Let’s denote for each n ≥ 1: zn = xn + iyn = χ(n).
We are going to develop the sequence ZN(s) = N
n=1
zn
ns as follows: For
N ≥ 1
ZN =
N
n=1
xn + iyn
na0+ib0
=
N
n=1
Un
na0
+ i
N
n=1
Vn
na0
= AN + iBN (6)
Where
Un =
xn cos(b0 ln (n)) + yn sin(b0 ln (n))
na0
(7)
Vn =
yn cos(b0 ln (n)) − xn sin(b0 ln (n))
na0
(8)
It is always insightful to work with the norm when working with complex:
|ZN|2
= A2
N + B2
N =
N
n=1
Un
2
+
N
n=1
Vn
2
(9)

Proof Strategy cont
A2
N =
N
n=1
U2
n + 2
N
n=1
Un
n−1
k=1
Uk = −
N
n=1
U2
n + 2
N
n=1
UnAn (10)
B2
N = −
N
n=1
V 2
n + 2
N
n=1
VnBn (11)
Let’s now deﬁne Fn and Gn as follows:
Fn = UnAn, Gn = VnBn (12)
Therefore
A2
N = 2
N
n=1
Fn −
N
n=1
U2
n , B2
N = 2
N
n=1
Gn −
N
n=1
V 2
n (13)

Proof Strategy cont
s is a zero for L(s, χ) = 0, if and only if
lim
N→∞
AN = 0 and lim
N→∞
BN = 0 (14)
Equally, s is a L(s, χ) zero, L(s, χ) = 0, if and only if
lim
N→∞
A2
N = 0 and lim
N→∞
B2
N = 0 (15)
The idea is to prove that in the case of a complex s that is in the right
hand side of the critical strip 1
2 < a0 ≤ 1 and that is a L(s, χ) zero, that
the limit limn→∞ A2
n = +/ − ∞ OR the limit limn→∞ B2
n = +/ − ∞. This
will create a contradiction. Because if s is a L(s, χ) zero then the
limn→∞ A2
n should be 0 and the limn→∞ B2
n should be 0. And therefore the
sequences ( N
n=1 Fn)N≥1 and ( N
n=1 Gn)N≥1 should converge and their
limits should be: limn→∞
N
n=1 Fn = 1
2 limN→∞
N
n=1 U2
n < +∞ and
limn→∞
N
n=1 Gn = 1
2 limN→∞
N
n=1 V 2
n < +∞.

Road to the proof of GRH Asymptotic Expansion
Asymptotic Expansion
If the set of the partial sums zn + zn+1 + ... + zn+k for n and k ≥ 0 is
bounded, then we can write An = λn
na0 where (λn) is a bounded sequence.
So the asymptotic expansion of Fn+1 − Fn is as follows:
Fn+1 − Fn =
γ2
n
n2a0
+
αnAn
na0
+
βnAn
na0+1
+ 2
γnβn
n2a0+1
+ O(
1
na0+2
) (16)
We have An = λn
na0 where λn is bounded. Therefore
Fn+1 − Fn =
γ2
n + αnλn
n2a0
+
λn + 2γn βn
n2a0+1
+ O(
1
na0+2
) (17)

Road to the proof of GRH The Case of Dirichlet character modulo an odd prime q
Let’s q be an odd prime integer. Let’s χ be a non-trivial Dirichlet
character modulo q. Let’s denote that for each k, zk = xk + iyk = χ(k).
Therefore we have the following asymptotic expansion:
q
i=1
γ2
qn+i + αqn+i λqn+i
(nq + i)2a0
+
q
i=1
(λnq+i + 2γqn+i )βnq+i
(nq + i)2a0 + 1
(18)
=
Ta + Tb cos(2b0 ln (qn)) + Tc sin(2b0 ln (qn)
(nq)2a0 + 1
+ O(
1
(qn)2a0+2
) (19)
Where m ∈ [1, q − 2] is deﬁned such that χm(g) = e
2πmi
q−1 where g is the
generator of the cyclic group Z/qZ∗.
Ta =



−a0
2
5q−1
2 −
(q−1) sin( 2πm
q−1
2−2 cos( 2πm
q−1
+ b0
2
q−1
2 +
(q−1) sin( 2πm
q−1
1−cos( 2πm
q−1
, m = q−1
2
−a0
4 5q − 1 + b0
4 (q − 1), m = q−1
2
(20)Charaf Ech-Chatbi Generalized Riemann Hypothesis August 14, 2019 12 / 32

Road to the proof of GRH The Case of Dirichlet character modulo an odd prime q
Tb =



a0
2 q + 1 −
(q−1) sin(2πm
q−1
2−2 cos( 2πm
q−1
− b0
2
q−1
2 +
(q−1) sin( 2πm
q−1
1−cos( 2πm
q−1
, m = q−1
2
a0
4 4q2 + 2q − 2 − b0
4 (q − 1), m = q−1
2
(21)
Tc =



a0
2
q−1
2 +
(q−1) sin( 2πm
q−1
1−cos( 2πm
q−1
− b0
2 q + 1 +
(q−1) sin( 2πm
q−1
2−2 cos( 2πm
q−1
, m = q−1
2
a0
4 (q − 1) − b0q, m = q−1
2
(22)

Road to the proof of GRH The Case of Riemann Hypothesis
The Case of Riemann Hypothesis
The Riemann zeta function is extended to the part of the complex plane
where (s) > 0 by the Dirichlet η function: η(s) = +∞
n=1
χ(n)
ns . Where
q = 2 and χ(kq + 1) = 1, χ(kq + 2) = −1. In this case, the function χ is
not a Dirichlet character but it is not a stop to apply our method as our
method is not Dirichlet character speciﬁc. Therefore we have the same
asymptotic expansion as in the previous slide:
q
i=1
γ2
(nq + i)2a0
+
q
i=1
(nq + i)2a0 + 1
(23)
=
Ta + Tb cos(2b0 ln (qn)) + Tc sin(2b0 ln (qn)
(nq)2a0 + 1
+ O(
1
(qn)2a0+2
) (24)
with
Ta = −a0, Tb = −a0, c = −b0 (25)

Proof of the GRH Case One: 1
2
< (s) ≤ 1 and χ non-trivial
Divergence of n≥1 Fn
For large N ≥ N0 we have the following expression of FN+1:
FN+1 = FN0 +
N
n=N0
γ2
n + αnλn
n2a0
+
λn + 2γn βn
n2a0+1
+
n
na0+2
(26)
= FN0 +
N
n=N0
Cn
n2a0
+
Dn
n2a0+1
+
n
na0+2
(27)
Where
Cn = γ2
n + αnλn (28)
Dn = λn + 2γn βn (29)

2
N
n=N0
Fn+1 = (N + 1)
(N − N0 + 1)
N + 1
FN0 (30)
+
N
n=N0
Cn
n2a0
+
Dn
n2a0+1
+
εn
na0+2
(31)
−
1
N + 1
N
n=N0
Cn
n2a0−1
+
Dn
n2a0
+
εn
na0+1
(32)
As we have 2a0 > 1, So limn→∞
Cn
n2a0−1 + Dn
n2a0
+ εn
na0+1 = 0
Following C´esaro theorem: limN→∞
1
N+1
N
n=N0
Cn
n2a0−1 + Dn
n2a0
+ εn
na0+1 = 0

2
We are going to prove that:
lim
N→∞
N
n=N0
Fn+1 = +∞ (33)
As we have 2a0 > 1, the series n≥1
Cn
n2a0
+ Dn
n2a0+1 + εn
na0+2 is converging
absolutely. Let’s denote:
RN0 = lim
N→+∞
Nq
n=qN0+1
Cn
n2a0
+
Dn
n2a0+1
+
εn
na0+2
(34)
If we can ﬁnd N0 such that RN0 + FN0 > 0 then we can conclude that
limN→∞
N
n=N0
Fn+1 = +∞.

2
Let’s now study the term Nq
n=qN0+1
Cn
n2a0
+ Dn
n2a0+1 .
We have from above results that:
Nq
n=qN0+1
Cn
n2a0
+
Dn
n2a0+1
(35)
=
N−1
n=N0
q
i=1
γ2
(nq + i)2a0
+
q
i=1
(nq + i)2a0 + 1
(36)
=
N−1
n=N0
α + β cos(2b0 ln (qn)) + γ sin(2b0 ln (qn)
(qn)2a0 + 1
+
N−1
n=N0
ξn
(qn)2a0+2
(37)

2
RN,N0 =
Nq
n=qN0+1
Cn
n2a0
+
Dn
n2a0+1
+
εn
na0+2
(38)
=
N−1
n=N0
α + β cos(2b0 ln (qn)) + γ sin(2b0 ln (qn)
(qn)2a0+1
(39)
+
N−1
n=N0
ξn
(qn)2a0+2
+
Nq
n=qN0+1
εn
na0+2
(40)
=
N−1
n=N0
g(n) +
ξn
(qn)2a0+2
+
Nq
n=qN0+1
εn
na0+2
(41)

2
Let’s denote the function g as: g(x) =
α+β cos(2b0 ln (qx))+γ sin(2b0 ln (qx)
(qx)2a0+1
We have the following inequality:
+∞
n=N0+1
g(n) −
+∞
N0+1
g(x) ≤
K
(N0)2a0+1
(42)
And the primitive function G of the function x → 1
q g(x
q ):
G(x) = 1
q − α
(2a0)x2a0
+
2b0β−(2a0)γ sin(2b0 ln (x))+ −2a0β−2b0γ cos(2b0 ln (x))
(2b0)2+(2a0)2 x2a0
Therefore
+∞
n=N0+1
g(n) + G(q(N0 + 1)) ≤
K
(N0)2a0+1
(43)

2
Concerning the term FN0 . We have
FN0 =
uN0
(N0)a0
AN0 =
uN0 λN0
(N0)2a0
(44)
Where the sequence (λn) is bounded. If we choose N0 = n0q as a multiple
of q we have from the deﬁnition of un0q = 0 because χ(n0q) = 0.
Therefore:
FN0 = Fn0q = 0 (45)
The remaining terms in the expression of RN0 are all of the order of
1
(N0)2a0+1 and above. So the dominant term in the expression RN0 is the
term −G(N0 + 1). The function G is a nonzero function. Therefore this
term is the dominant term in the expression of 1
N
N−1
n=N0
Fn+1. Hence
based on the sign of G(N0 + 1) we can show that the limit of N−1
n=1 Fn+1
can be both +∞ and −∞.

2
Let’s define the function f :
f (x) = − α
2a0
+
2b0β−(2a0)γ sin(2b0x)+ −2a0β−2b0γ cos(2b0x)
(2b0)2+(2a0)2 . We simplify
further the notations: f (x) = α2 + β2 cos(2b0x) + γ2 sin(2b0x). Without
loss of generality, let’s define β0 = − f (0)
q2a0+1 > 0. See[2] for more detail.
Let’s fix > 0 to be very small such that 0 < < min 1, β0
10000 .
Let’s define γ0 = |β2|+|γ2|
q2a0+1 . Without loss of generality, let’s N1 be such that
|cos(2b0 ln q(N1 + 1)) − 1| ≤ /γ0 and |sin(2b0 ln q(N1 + 1)) − 0| ≤ /γ0.
Therefore:
G q(N1 + 1) −
f (0)
q q(N1 + 1)
2a0
≤
N1 + 1
2a0
(46)
Because...

2
G q(N1 + 1) −
f (0)
q q(N1 + 1)
2a0
≤ (47)
(γ2) sin(2b0 ln q(N1 + 1)) + (β2) cos(2b0 ln q(N1 + 1))
q q(N1 + 1)
2a0
(48)
−
(γ2) sin(2b00) + (β2) cos(2b00)
q q(N1 + 1)
2a0
(49)
≤
|γ2| |sin(2b0 ln q(N1 + 1))| + |β2| |cos(2b0 ln q(N1 + 1)) − 1|
q q(N1 + 1)
2a0
(50)

2
Therefore
+∞
n=N1+1
g(n) −
β0
(N1 + 1)2a0
≤
(N1 + 1)2a0
+
K
(N1)2a0+1
(51)
And
+∞
n=N1+1
g(n) ≥
β0 −
(N1 + 1)2a0
−
K
(N1)2a0+1
(52)

2
Therefore, without loss of generality let’s suppose N1 = qN2.
RN1 =
+∞
k=qN2+1 g(k) + FqN2
=0
+ limN→+∞
N−1
n=N2
ξn
(qn)2a0+2 + Nq
n=qN2+1
εn
na0+2
≥ β0−
(qN2+1)2a0
− K
(qN2)2a0+1 + limN→+∞
N−1
n=N2
ξn
(qn)2a0+2 + Nq
n=qN2+1
εn
na0+2
We have the sequences (ξk) and (εk) are bounded. Plus we have
a0 + 2 ≥ 2a0 + 1 > 2a0 > 1, therefore the series k≥1
ξk
k2a0+2 and
k≥1
εk
ka0+2 are converging absolutely.

2
Let the positive constant M such that:
For each k ≥ N2 and N ≥ N2 :
|ξk| ≤ M (53)
|εk| ≤ M (54)
Therefore:
+∞
n=N2
ξn
(qn)2a0+2
+
+∞
n=qN2+1
εn
na0+2
≤ (55)
M
(2a0 + 1)q2a0+2(N2 − 1)2a0+1
+
M
(a0 + 1)(qN2)a0+1
(56)

2
Therefore
RN1 ≥ β0−
(qN2+1)2a0
− K
(qN2)2a0+1 − M
(2a0+1)q2a0+2(N2−1)2a0+1 − M
(a0+1)(qN2)a0+1
≥ 1
(qN2+1)2a0
β0 − + δqN2
Where the sequence (δn) is deﬁned as the following:
δqn = −K(qn+1)2a0
(qn)2a0+1 − M(qn+1)2a0
(2a0+1)q2a0+2(n−1)2a0+1 − M(qn+1)2a0
(a0+1)(qn)a0 + 1
As we have 1
2 < a0 < 1, therefore
0 < 1 − a0 < 1
2 < a0 < 2a0 < a0 + 1 < 3a0 < 2a0 + 1.
Therefore the limn→+∞ δqn = 0.
So we can choose N1 such that |δN1 | < , i.e δN1 ≥ − . Therefore
RN1 ≥
1
(N1 + 1)2a0
β0 − 2 > 0 (57)
And limN→+∞
N
n=N1
Fn = +∞.

2
Divergence of n≥1 Fn: Conclusion
We have the series n≥1 U2
n is converging absolutely thanks to 2a0 > 1.
We have from the previous slide that:
lim
N→∞
N
n=1
Fn = +∞ (58)
Therefore
lim
N→∞
A2
N = +∞ (59)
This result is in contradiction with the fact that s is a L(s, χ) zero and
therefore with the fact that the limit limN→∞ AN = 0 should be zero.
Therefore s with 1
2 < a0 = (s) < 1 cannot be a zero for L(s, χ).

Proof of the GRH Case Two: 1
2
< (s) ≤ 1 and χ trivial
Case Two: 1
2 < (s) ≤ 1 and χ trivial
For the trivial character χ0 modulus q we have:
ζ(s) = L(s, χ0)
p Prime,p/q
1 −
1
ps
−1
(60)
Where ζ is the Riemann Zeta function.
The product p Prime,p/q 1 − 1
ps
−1
is ﬁnite, bounded and nonzero as
1
2 < (s) = a0 ≤ 1. From the equation above we have: if s is a zero for
L(s, χ0), then it is also a zero for the Riemann zeta(s). We saw in the
previous case that if 1
2 < (s) = a0 ≤ 1, it is not possible for the Riemann
zeta function to have such a zero. Therefore L(s, χ0) cannot have a zero
where 1
2 < (s) = a0 ≤ 1.

Proof of the GRH Conclusion
Conclusion
We saw that if s is a L(s, χ) zero, then real part (s) can only be 1
2 as all
other possibilities can be discarded using the functional equation like in
[1]. Therefore the Generalized Riemann hypothesis is true: The non-trivial
zeros of L(s, χ) have real part equal to 1
2.

Proof of the GRH References
References
Charaf Ech-Chatbi (2019)
A Proof of the Riemann’s Hypothesis
Charaf Ech-Chatbi (2019)
A Proof of the Generalized Riemann Hypothesis

Proof of the GRH QED
The End

A Proof of the Generalized Riemann Hypothesis

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