on the riemann hypothesis - gerasimos pergaris
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On the Riemann Hypothesis GERASIMOS PERGARIS Department of Informatics University of Piraeus 80 Karaoli and Dimitriou Street, 18534, Piraeus GREECE e-mail: [email protected] Abstract: A connection between a strictly monotonic function and the Riemann zeta function has been established. The extended use of a computer mathematical tool together along with human interaction led to the observation that the function: 1 1 1 1 ζ x iy ζ x iy ζ x iyTRANSCRIPT
On the Riemann Hypothesis
GERASIMOS PERGARIS Department of Informatics
University of Piraeus 80 Karaoli and Dimitriou Street, 18534, Piraeus
GREECE e-mail: [email protected]
Abstract: A connection between a strictly monotonic function and the Riemann zeta function has been established. The extended use of a computer mathematical tool together along with human interaction led to the observation that the function:
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2i i i i
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2
′ ′ ′ ′− − − + + + + − − + − + − − − + + + + −
, x , y∈ℝ
can be connected to the Riemann hypothesis and is strictly monotonic. The motive behind this brute force examination of a variety of functions related to the Riemann’s zeta was the belief that the difficulty in solving the famous hypothesis is the non explicit character of the zeta function. The main idea was to examine a variety of functions related to zeta, that can be translated to known functions, i.e. expressions of the gamma function, so that their monotonicity could easily be proved. This is possible for a variety of functions through the functional equation of the Riemann zeta function. After the choice of the function has been made, a mathematical setting in the sequel is proposed that leads to a proof of the statement. Keywords: Riemann Hypothesis, Riemann Zeta Zeros, Riemann Zeta Function
1. Introduction The Riemann zeta function is the function of the complex variable s, defined in the half-plane ( )Re s 1> by the absolutely convergent series:
( ) sn 1
1ζ s
n
∞
=
=∑ , ( )1
and in the whole complex plane ℂ by analytic continuation. As shown by Riemann,
( )ζ s extends to ℂ as a meromorphic function with only a simple pole at s 1= , with
residue 1, and satisfies the functional equation, [1], [7]:
( ) ( ) ( )1 s /2s/2 s 1 sπ Γ ζ s π Γ ζ 1 s
2 2− −− − = −
, ( )2
or stated as , [3], [6]:
( ) ( ) ( )s s 1 πsζ s 2 π sin Γ 1 s ζ 1 s
2− = − −
( )3
where ( )Γ s is the gamma function of the complex variable s σ it= + , defined for
σ 0> by the integral:
( ) t σ 1
0Γ σ e t dt
∞ − −= ∫ , ( )4
and in the whole complex plane by analytic continuation. It is well known, [4], that
( )Γ s extends to ℂ as a meromorphic function with simple poles at the non-positive
integers 0, 1, 2,− − … , with residue at s n= − , ( )n1 / n!− , and satisfies the functional
equation , [4]: ( ) ( ) ( )Γ s s 1Γ s 1= − − . ( )5
We recall that for all s∈ℂ with s 0, 1, 2,≠ − − … , we have, [6]:
( ) ( )Γ s Γ s= ∈ℂ ( )6
which follows easily from Gauss’s limit definition:
( )( ) ( )
s
n
n!nΓ s lim
s s 1 s n→∞=
+ +⋯, ( )7
also that, [4]:
( )
( )1 sinπsΓ 1 s
Γ s π= − ( )8
holds, with the function ( )1
Γ s being entire with simple zeros at s 0, 1, 2,= − − … .
We recall that for all s∈ℂ with s 1≠ , we have, [7]:
( ) ( )ζ s ζ s= ∈ℂ , ( )9
which follows easily from the analytic continuation to the whole complex plane (Knopp, Hasse 1930):
( ) ( ) ( )n
k s
1 s n 1n 0 k 0
n1 1ζ s 1 k 1
k1 2 2
∞−
− += =
= − + −
∑ ∑ , ( )10
with a simple pole at s 1= . We recall that the trivial zeros of the function ζ are simple and occur at the negative even integers, also that the non-trivial zeros of the function ζ lie on the critical strip:
( )0 Re s 1< < , ( )11
and are purely complex, initially proved, [3], by Riemann for the case ( )0 Re s 1≤ ≤
and then by Hadamard and de la Vallee Poussin for the case ( )Re s 0,1= .
We recall that the function ξ : →ℂ ℂ :
( ) ( ) ( )s
21 s
ξ s s s 1π Γ ζ s2 2
− = −
, ( )12
is entire, has the property: ( ) ( )ξ s ξ 1 s= − for all s∈ℂ , and can be expanded as the
infinite product:
( ) ( )ρ
sξ s ξ 0 1
ρ
= −
∏ , ( )13
where ρ ranges over all solutions of the equation ( )ξ s 0= , and ( )logξ 0 log 2= − ,
[3]. Combining equations ( )12 and ( )13 , canceling all poles of the Γ function at the
negative integers by the trivial zeros of the ζ function and canceling the pole of the
Γ function at the point s 0= by s, leads us to the observation that the zeros of the ξ function are exactly the non-trivial zeros of the ζ function, [3], [4]. We recall a formula for the logarithmic derivative of the function ζ , used for the derivation of the well known von Mangoldt’s Formula, [3], that is:
( )( ) ( ) ( )
( )( )ρ n 1
ζ s ζ 0s s s
ζ s s 1 ρ s ρ 2n s 2n ζ 0
∞
=
′ ′= − + − +
− − +∑ ∑ , ( )14
where the first summation is over all roots ρ of the equation ( )ξ s 0= , and
( )( )
( )ζ 0
log 2πζ 0
′= . The series
ρ
1
s ρ−∑ converges uniformly in any disc s K≤ when
the terms ρ , 1 ρ− are paired, so when K is fixed the series
( )ρ ρ
1 1 s
s ρ ρ ρ s ρ
+ = − −
∑ ∑ converges by an elementary theorem
[ ( )n n n nα β α β+ = +∑ ∑ ∑ when nα∑ , nβ∑ both converge], [3]. The terms of
the sum over ρ could be taken in the order of increasing Imρ , [3].
We recall the formula, [2]:
( )( ) n 1
Γ s 1 1 1γ
Γ s s s n n
∞
=
′ = − − − − + ∑ , ( )15
for the logarithmic derivative of the gamma function, for { }s 0, 1, 2,∈ − − −ℂ … ,
where γ 0.577216≃ is the Euler's or Euler-Mascheroni constant. Remark 1: We shall refer to the functions Γ and ζ meaning their analytic continuations to the whole complex plane as stated before, rather than their definitions.
Proposition 1: If one of ( )0ζ s , ( )0ζ s , ( )0ζ 1 s− , ( )0ζ 1 s− is a non-trivial zero then
all are.
Proof: We have 0s 1≠ . From ( )9 we have ( )0ζ s 0= ⇔ ⇔ ( )0ζ s 0= ,
( )0ζ 1 s 0− = ( )0ζ 1 s 0− = ⇔ ( )0ζ 1 s 0− = ⇔ ( )0ζ 1 s 0− = .
By ( )2 : ( )2⇒ ( )
01 s02
0
1 sπ Γ ζ 1 s 0
2
−− − − =
⇒ ( )0
0
1 sΓ ζ 1 s 0
2
− − =
,
and by: ( )8 ⇒ 01 sΓ 0
2
− ≠
we have: ⇒ ( )0ζ 1 s 0− = .
By ( )2 : ( )0ζ 1 s 0− = ( )2⇒ ( )0s /2 0
0
sπ Γ ζ s
2−
⇒ ( )00
sΓ ζ s 0
2 =
,
and by: ( )8 ⇒ 0sΓ 0
2 ≠
we have: ( )0ζ 1 s 0− = ⇒ ,
so: ( )0ζ s 0= ⇔ ( )0ζ 1 s 0− = and the proof follows.
□
( )0ζ s 0=
⇔
( )0ζ s 0=
( )0ζ s 0=
( )0ζ s 0=
The sequel is divided into four parts: Part A: Basic observations, Part B: Statement of a conjecture, Part C: Monotonicity proofs, Part D: Proof of the conjectured.
2. Main Part
Part A We restrict our attention to the fraction:
1 1ζ x iy ζ x iy
2 21 1
ζ x iy ζ x iy2 2
− − − + + + + −
.
If 1
x iy s2− − = , then:
1x iy s
2− + = ,
1x iy 1 s
2+ + = − and
1x iy 1 s
2+ − = − ,
so the four ζ’s of the fraction vanish simultaneously. Remark 3: Our area of interest is an arbitrary small punctured disc centered at the point 0s , with ( )0ζ s 0= being a non-trivial zero.
Proposition 2: Let 0 0x , y ∈ℝ , 0
1 1x
2 2− < < , 0y 0≠ and 0 0 0
1s x iy
2= − − with
( )0ζ s 0= . Let 2h : →ℝ ℂ be defined in an arbitrary small open disc U with radius
r 0> , centered at the point ( ) 20 0x , y ∈ℝ , not defined at the point ( )0 0x , y , with:
( )
1 1ζ x iy ζ x iy
2 2h x, y
1 1ζ x iy ζ x iy
2 2
− − − + = + + + −
.
Then:
( )h x, y 0> for every ( ) ( ){ }0 0x, y U x , y∈ − .
Proof: It is clear that 0s gives a random non-trivial zero of the function ζ.
Let hD ( ){ } 20 0U x , y= − ⊆ ℝ , be the definition field of h .
We have: 1 1
x iy x iy2 2− − = + + and
1 1x iy x iy
2 2+ + = + −
and by the property: ( ) ( )ζ s ζ s= , we have: 2
1 1 1ζ x iy ζ x iy ζ x iy
2 2 2 − − − + = − −
,
21 1 1
ζ x iy ζ x iy ζ x iy2 2 2
+ − + + = + +
.
By ( )3 one can see that the function:
( )( )
( )( )
( )
2 2
2
1 1ζ x iy ζ x iy
ζ s ζ s2 2h x, y
1 1 ζ 1 sζ 1 sζ x iy ζ x iy2 2
− − − + = = =
− −+ + + −
, for 1
s x iy2
= − − ,
is the square of the modulus of the holomorphic function:
( )s s 1 πs2 π sin Γ 1 s
2− −
.
Since this function has no zeros for ( )Im s 0≠ , the function ( )h x, y is real analytic
in U as a function of two real variables. Also ( )h x, y 0> as the square of the
modulus of a non-vanishing function. □ Proposition 3: Let g : →ℝ ℝ be defined in ( )0 0y δ, y δ− + ⊆ ℝ , with δ 0> and
arbitrary small, not defined at the point 0y , with:
( )0 0
0 0
1 1ζ x iy ζ x iy
2 2g y
1 1ζ x iy ζ x iy
2 2
− − − + = + + + −
,
with 0x , 0y being as in function h .
Then: ( )g y 0> for every gy D∈ .
Proof: It is clear that:
( ) ( )0g y h x , y 0= > , for all ( )0 hx , y D∈ ,
and with gD being ( )0 0y δ, y δ− + ⊆ ℝ , except the point 0y , with δ 0> and arbitrary
small, we choose δ smaller than r and the proposition is proved. □ Proposition 4: Let g : →ℝ ℝ be defined at ( ) ( )g 0 0 0 0D y δ, y y , y δ≡ − ∪ + , δ , 0x ,
0y as in proposition 3, with:
( )0 0
0 0
1 1ζ x iy ζ x iy
2 2g y
1 1ζ x iy ζ x iy
2 2
− − − + = + + + −
.
Then:
( )0y y
lim g y→
=
0 0
0 00 0
y y del 'Hospital y y
0 0 0 0
d 1 11 1ζ x iy ζ x iyζ x iy ζ x iy
dy 2 22 2lim lim
1 1 d 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 dy 2 2
→ →
− − − +− − − + = = + + + − + + + −
with ( )0y y
lim g y k→
= ∈ℝ , also k 0> .
Proof: The functions:
0 0
1 1ζ x iy ζ x iy
2 2 − − − +
, 0 0
1 1ζ x iy ζ x iy
2 2 + + + −
are differentiable in the real variable y everywhere in ( )0 0y δ, y δ− + as products of
differentiable functions. In addition, 0
0 0y y
1 1lim ζ x iy ζ x iy 0
2 2→
− − − + =
and
00 0y y
1 1lim ζ x iy ζ x iy 0
2 2→
+ − + + =
. Therefore, we can use de l’ Hospital rule, if
the limit exists. From the functional equation of the function ζ in the form:
( ) ( ) ( )s s 1 πsζ s 2 π sin Γ 1 s ζ 1 s
2− = − −
for 1
s x iy2
= − − , we have:
1 1
x iy x iy2 2
1ζ x iy
π 1 122 π sin x iy Γ x iy
1 2 2 2ζ x iy
2
− − − − −
− − = − − + + + +
( )16
for 1
s x iy2
= − + , we have:
1 1
x iy x iy2 2
1ζ x iy
π 1 122 π sin x iy Γ x iy
1 2 2 2ζ x iy
2
− + − − +
− + = − + + − + −
( )17
, by multiplication of ( )16 , ( )17 , we have:
1 1ζ x iy ζ x iy
2 21 1
ζ x iy ζ x iy2 2
− − − + = + + + −
1 2x 1 2x π 1 π 1 1 12 π sin x iy sin x iy Γ x iy Γ x iy
2 2 2 2 2 2− − − = − − − + + + + −
( )18
and by the observation, that: π 1 π 1
sin x iy sin x iy2 2 2 2
− − − + =
π 1 π 1 π 1 π 1cos x iy x iy cos x iy x iy
2 2 2 2 2 2 2 2
2
− + − − − − − + − − − = =
( )
( ) ( )( )π
cos iπy cos πx12
cosh πy sin πx2 2
− − = = − ( )19
equation ( )18 becomes:
( )20
Setting 0x x= and letting gy D∈ , the right-hand side of ( )20 is a product of
differentiable functions of the real variable y for 1 1
x ,2 2
∈ −
, { }y 0∈ −ℝ ,
therefore continuous, hence the limit ( )0y y
lim g y→
exists by equality. Differentiability of
the functions 1
Γ x iy2
+ +
, 1
Γ x iy2
+ −
in the real variable y comes from the
analyticity of these functions in the area 1 1
x ,2 2
∈ −
, y∈ℝ .
So de l’ Hospital’s rule is applicable, and for δ small enough, so that ( )0 0y δ, y δ− +
has a distance from zero, we have:
( )
( )( )
( ) ( ) ( )
2x 1 2x
2
0 0
0 0
2 π 0 , for x
1 1 1Γ x iy Γ x iy Γ x iy 0 , for 0 y y δ, y δ
2 2 2
sin πx 1 , for xcosh πy sin πx 0 , for 0 y y δ, y δ
cosh πy 1 , for y 0
− − −
> ∈ + + + − = + + > ≠ ∈ − +
≤ ∈ ⇒ − > ≠ ∈ − + > ≠
ℝ
ℝ
⇒
( ) ( )( )2x 1 2x 1 12 π Γ x iy Γ x iy cosh πy sin πx ε 0
2 2− − − + + + − − > >
( )0 0, for 0 y y δ, y δ≠ ∈ − +
Setting 0x x= and applying limits as 0y y→ on both sides of equation ( )20 ,
we conclude that k 0> and the proof follows. □ Proposition 5: Let 0x , 0y be as in proposition 4. Then:
0
0 0
0 0
y y
0 0
0 0
1 1ζ x iy ζ x iy
2 2i i1 1
ζ x iy ζ x iy2 2
lim 11 1
ζ x iy ζ x iy2 2i i1 1
ζ x iy ζ x iy2 2
→
′ ′− − − + − + − − − + = ′ ′+ + + − − + + + −
.
( ) ( )( )2x 1 2x
1 1ζ x iy ζ x iy
1 12 22 π Γ x iy Γ x iy cosh πy sin πx
1 1 2 2ζ x iy ζ x iy
2 2
− − −
− − − + = + + + − − + + + −
Proof: We have:
0
0 0
0 0
y y
0 0
0 0
1 1ζ x iy ζ x iy
2 2i i1 1
ζ x iy ζ x iy2 2
lim1 1
ζ x iy ζ x iy2 2i i1 1
ζ x iy ζ x iy2 2
→
′ ′− − − + − + − − − + = ′ ′+ + + − − + + + −
0
0 0 0 0
0 0
y y
0 0 0 0
0 0
1 1 1 1iζ x iy ζ x iy i ζ x iy ζ x iy
2 2 2 21 1
ζ x iy ζ x iy2 2
lim1 1 1 1
iζ x iy ζ x iy iζ x iy ζ x iy2 2 2 2
1 1ζ x iy ζ x iy
2 2
→
′ ′− − − − + + − − − +
− − − + =
′ ′+ + + − − + + + −
+ + + −
0
0 0
0 0
y y
0 0
0 0
d 1 1ζ x iy ζ x iy
dy 2 21 1
ζ x iy ζ x iy2 2
limd 1 1ζ x iy ζ x iy
dy 2 21 1
ζ x iy ζ x iy2 2
→
− − − + − − − + = + + + −
+ + + −
0
0 0
0 0
y y
0 0
0 0
d 1 1ζ x iy ζ x iy
dy 2 2
d 1 1ζ x iy ζ x iy
dy 2 2lim
1 1ζ x iy ζ x iy
2 21 1
ζ x iy ζ x iy2 2
→
− − − +
+ + + − =
− − − + + + + −
0
0
0 0
y y
0 0
0 0
y y
0 0
d 1 1ζ x iy ζ x iy
dy 2 2lim
d 1 1ζ x iy ζ x iy
dy 2 2 k1
1 1 kζ x iy ζ x iy
2 2lim1 1
ζ x iy ζ x iy2 2
→
→
− − − +
+ + + − = = =
− − − + + + + −
.
□
Part B We state the following conjecture: Conjecture 1: Let 0x , 0y be as in proposition 5. Then:
0
0 0 0 0
x x
0 0 0
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2lim i i i i 0
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2
→
′ ′ ′ ′− − − + + + + − − + − + = − − − + + + + −
□ We have: From the functional equation of the ζ function we have:
( ) ( ) ( )s s 1 πsζ s 2 π sin Γ 1 s ζ 1 s
2− = − −
,
setting 1
s x iy2
= − − , we have:
1 1x iy x iy 1
2 2
1π x iy
1 1 12ζ x iy 2 π sin Γ 1 x iy ζ x iy
2 2 2 2
− − − − −
− − − − = − − − + +
,
setting 1
s x iy2
= − + , we have:
1 1x iy x iy 1
2 2
1π x iy
1 1 12ζ x iy 2 π sin Γ 1 x iy ζ x iy
2 2 2 2
− + − + −
− + − + = − − + + −
,
letting ( )1 1
x iy x iy 12 2
1π x iy
12a x, y 2 π sin Γ 1 x iy
2 2
− − − − −
− − = − − −
and ( )1 1
x iy x iy 12 2
1π x iy
12b x, y 2 π sin Γ 1 x iy
2 2
− + − + −
− + = − − +
,
we have:
( )1 1ζ x iy a x, y ζ x iy
2 2 − − = + +
, ( )21
( )1 1ζ x iy b x, y ζ x iy
2 2 − + = + −
. ( )22
Differentiating equation ( )21 on the variable y and reformulating, we have:
( ) ( )1 d 1 1iζ x iy a x, y ζ x iy ia x, y ζ x iy
2 dy 2 2 ′ ′− − − = ⋅ + + + + +
⇒
( )( )
d 11 1a x, y ζ x iyζ x iy ζ x iy
dy 22 2i ia x, y
1 1 1ζ x iy ζ x iy ζ x iy
2 2 2
′ ′⋅ + +− − + + − = +
− − − − − −
⇒
( )
( )( )
( )
d 11 1a x, y ζ x iyζ x iy ζ x iy
dy 22 2i ia x, y
1 1 1ζ x iy a x, y ζ x iy a x, y ζ x iy
2 2 2
′ ′⋅ + +− − + + − = +
− − + + + +
( )21⇒
( )
( )
d1 1a x, yζ x iy ζ x iy
dy2 2i i
1 1a x, yζ x iy ζ x iy
2 2
′ ′− − + + − = + − − + +
⇒
( )
( )
d1 1a x, yζ x iy ζ x iy
dy2 2i i
1 1 a x, yζ x iy ζ x iy
2 2
′ ′− − + + − − = − − + +
. ( )23
Differentiating equation ( )22 on the variable y and reformulating, we have:
( ) ( )1 d 1 1iζ x iy b x, y ζ x iy ib x, y ζ x iy
2 dy 2 2 ′ ′− + = ⋅ + − − + −
⇒
( )( )
d 11 1b x, y ζ x iyζ x iy ζ x iy
dy 22 2i ib x, y
1 1 1ζ x iy ζ x iy ζ x iy
2 2 2
′ ′⋅ + −− + + − = −
− + − + − +
( )22⇒
( )
( )( )
( )
d 11 1b x, y ζ x iyζ x iy ζ x iy
dy 22 2i ib x, y
1 1 1ζ x iy b x, y ζ x iy b x, y ζ x iy
2 2 2
′ ′⋅ + −− + + − = −
− + + − + −
⇒
( )
( )
d1 1b x, yζ x iy ζ x iy
dy2 2i i
1 1b x, yζ x iy ζ x iy
2 2
′ ′− + + − = − − + + −
⇒
( )
( )
d1 1b x, yζ x iy ζ x iy
dy2 2i i
1 1 b x, yζ x iy ζ x iy
2 2
′ ′− + + − + = − + + −
. ( )24
Adding equations ( )23 , ( )24 and reformulating, we have:
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2i i i i
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2
′ ′ ′ ′− − − + + + + − − + − + = − − − + + + + −
( )
( )
( )
( )( ) ( )( )
d da x, y b x, y
ddy dyln a x, y b x, y
a x, y b x, y dy= + = , ( )25
and by replacing a and b in ( )25 , or by the observation that:
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2i i i i
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2
′ ′ ′ ′− − − + + + + − − + − + = − − − + + + + −
1 1ζ x iy ζ x iy
d 2 2ln
1 1dyζ x iy ζ x iy
2 2
− − − + = + + + −
and by equation ( )20 , we have:
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2i i i i
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2
′ ′ ′ ′− − − + + + + − − + − + = − − − + + + + −
( ) ( )( )2x 1 2xd 1 1ln 2 π Γ x iy Γ x iy cosh πy sin πx
dy 2 2− − − = + + + − −
( ) ( ) ( )( )2x 1 2xd d 1 1 dln 2 π ln Γ x iy Γ x iy ln cosh πy sin πx
dy dy 2 2 dy− − − = + + + + − + −
( )( ) ( )
1 1Γ x iy Γ x iy
π sinh πy2 2i i
1 1 cosh πy sin πxΓ x iy Γ x iy
2 2
′ ′+ + + − ⋅ = − +− + + + −
⇒
( )
( ) ( )
1 1Γ x iy Γ x iy
π sinh πy2 2i i
1 1 cosh πy sin πxΓ x iy Γ x iy
2 2
′ ′+ + + − ⋅ = − +− + + + −
. ( )26
We set 0y y= to equation ( )26 , and it suffices to show that the function:
( ) ( )( ) ( )
0 00
00 0
1 1Γ x iy Γ x iy
π sinh πy2 2J x i i
1 1 cosh πy sin πxΓ x iy Γ x iy
2 2
′ ′+ + + − ⋅ = − +− + + + −
,
defined for 1 1
x ,2 2
∈ −
, is:
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2i i i i
1 1 1 1ζ x iy ζ x iy ζ x iy ζ x iy
2 2 2 2
′ ′ ′ ′− − − + + + + − − + − + = − − − + + + + −
• Strictly decreasing for 0y 0<
• Identically equal to zero for 0y 0=
• Strictly increasing for 0y 0>
, also has a zero at the point 0x 0= for every 0y ∈ℝ .
Part C
At this part, we present the monotonia of the function J . By formula ( )15 we have:
( )( ) n 1
Γ s 1 1 1γ
Γ s s s n n
∞
=
′ = − − − − + ∑ , for { }s 0, 1, 2,∈ − − −ℂ …
We have:
0 0
0 0
1 1Γ x iy Γ x iy
2 2i i
1 1Γ x iy Γ x iy
2 2
′ ′+ + + − − = + + + −
n 10 0
1 1 1i γ
1 1 nx iy x iy n2 2
∞
=
= − − − − − + + + + +
∑
n 1
0 0
1 1 1i γ
1 1 nx iy x iy n2 2
∞
=
− − − − − + − + − +
∑
n 1 n 10 0 0 0
i i 1 1 1 1i i
1 1 1 1n nx iy x iy x iy n x iy n2 2 2 2
∞ ∞
= =
= − + − − + − + + + − + + + + − +
∑ ∑
( )0
2 2n 0 0
8y
1 2x 2n 4y
∞
=
= −+ + +
∑ .
Remark 4: The last equality above is valid by an elementary theorem [3]: [ ( )n n n nα β α β+ = +∑ ∑ ∑ when nα∑ , nβ∑ both converge].
Differentiating in the variable x , we have:
( ) ( )0 0
2 22 2n 0 n 00 0
8y 8yd d
dx dx1 2x 2n 4y 1 2x 2n 4y
∞ ∞
= =
− = − =
+ + + + + + ∑ ∑
( )
( )( )( )
( )( )0 0
2 22 22 2n 0 n 00 0
32 1 2x 2n y 32 1 2x 2n y
1 2x 2n 4y 1 2x 2n 4y
∞ ∞
= =
+ + + + = − − =
+ + + + + +
∑ ∑
Remark 5: Termwise differentiation above is valid due to the uniform convergence of the last sum. It is well known , [ ]3 , that a series can be differentiated termwise if
the result is uniformly convergent. For 1 1
x ,2 2
∈ −
, n 0> , we have:
( )
( )( )( )
( )( )
2
002 22 22 2
0 0
32 1 2x 2n y 1 2x 2n32 y
1 2x 2n 4y 1 2x 2n 4y
+ + + +≤
+ + + + + +
( )
( )( ) ( )
2 20
0 0 0 02 2 2 22 20
1 2x 2n 4y 1 1 132 y 32 y 32 y 8 y
4n n1 2x 2n1 2x 2n 4y
+ + +≤ ≤ ≤ =
+ ++ + +.
The series 0 2n 1
18 y
n
∞
=∑ converges to
2
0
π8 y
6 independently of x , therefore the series
( )
( )( )0
22 2n 10
32 1 2x 2n y
1 2x 2n 4y
∞
=
+ +
+ + +∑ passes the Weierstrass M-test and hence converges
uniformly. The first term can be differentiated as a rational function of x . We have:
( )( ) ( )
( ) ( )( ) ( )( )
20 0
20 0
π sinh πy π cos πx sinh πyd
dx cosh πy sin πx cosh πy sin πx
⋅=
− −.
The sign of the last fraction follows the sign of its numerator and by the fact that
( )cos πx 0≥ for 1 1
x2 2
− ≤ ≤ , it follows the sign of ( )0 0πy πy
0
e esinh πy
2
−−= , which is:
• negative for 0y 0<
• equal to zero for 0y 0=
• positive for 0y 0>
The same are true for ( )
( )( )0
22 2n 00
32 1 2x 2n y
1 2x 2n 4y
∞
=
+ +
+ + +∑ . For
1 1x ,
2 2 ∈ −
we have:
1 2x 2n 0+ + > , so every term of the above sum follows the sign of 0y .
Conclusively:
( ) ( )
( )( )( ) ( )
( ) ( )( )
20 0
2 22 2n 0 00
32 1 2x 2n y π cos πx sinh πydJ x
dx cosh πy sin πx1 2x 2n 4y
∞
=
+ += +
−+ + +∑ is:
• negative for 0y 0<
• equal to zero for 0y 0=
• positive for 0y 0>
so the function ( )J x is:
• strictly decreasing for 0y 0<
• constant for 0y 0=
• strictly increasing for 0y 0>
for 0y 0= we have:
( ) ( )( ) ( )
1 1Γ x Γ x
πsinh π 02 2J x i i 0
1 1 cosh π 0 sin πxΓ x Γ x
2 2
′ ′+ + ⋅ = − + =⋅ − + +
so the function ( )J x is identically equal to zero for 0y 0= .
Lastly, for x 0= , we have:
( )( )
( )( ) ( )
0002 22
n 0 n 000 20
sinh πy8y 1J 0 π π 2πy
cosh πy1 2n 4y 2n 1 πy
2
∞ ∞
= =
= − + = − ⋅ ⋅ + + + + +
∑ ∑
( )0π tanh πy+ ⋅
and by the series of ( )tanh z we conclude that:
( ) ( ) ( )0 0J 0 π tanh πy π tanh πy 0= − ⋅ + ⋅ = ,
that is, there is a zero for the function J at the point 0x 0= , for every 0y ∈ℝ .
Due to the fact that ( )J x is a strictly monotonic function of x for 0y 0≠ , we
conclude that ( )J 0 is the only zero for the function in 1 1
x ,2 2
∈ −
with 0y 0≠ .
Part D
Recall formula ( )14 :
( )( ) ( ) ( )
( )( )ρ n 1
ζ s ζ 0s s s
ζ s s 1 ρ s ρ 2n s 2n ζ 0
∞
=
′ ′= − + − +
− − +∑ ∑ .
Setting 1
s x iy2
= − − , we have:
( )( )ρ n 1
1 1 1 1ζ x iy x iy x iy x iy ζ 02 2 2 211 1 1 ζ 0x iy 1ζ x iy ρ x iy ρ 2n x iy 2n22 2 2
∞
=
′ − − − − − − − − ′ = − + − + − − −− − − − − − − +
∑ ∑
( )27
Setting 1
s x iy2
= − + , we have:
( )( )ρ n 1
1 1 1 1ζ x iy x iy x iy x iy ζ 02 2 2 211 1 1 ζ 0x iy 1ζ x iy ρ x iy ρ 2n x iy 2n22 2 2
∞
=
′ − + − + − + − + ′ = − + − + − + −− + − + − − + +
∑ ∑
. ( )28
Multiplying ( )27 by i− , ( )28 by i and adding the results, we have:
0 0
0 0
1 1ζ x iy ζ x iy
2 2i i
1 1ζ x iy ζ x iy
2 2
′ ′− − − + − + = − − − +
0 0 0 0
ρ ρ0 0 0 0
1 1 1 1x iy x iy x iy x iy
2 2 2 2i i1 1 1 1x iy x iy ρ x iy ρ ρ x iy ρ2 2 2 2
− − − + − − − + = − − − +
− − − − − + − − − − + −
∑ ∑
0 0
n 1 n 10 0
1 1x iy x iy
2 2i1 1
2n x iy 2n 2n x iy 2n2 2
∞ ∞
= =
− − − +
+ − − − + − − +
∑ ∑ ( )29
For the first summand of ( )29 , we have:
( )
0 0
2 2
0 0
1 1x iy x iy 8y2 2i
1 1 1 2x 4yx iy x iy2 2
− − − + − = −
+ + − − − − − +
.
For the first sum of the third summand of ( )29 , we have:
00
n 1 n 1 n 100 0
11 x iy 2n 2nx iy 1 12211 1 2n x iy 2n2n x iy 2n 2n x iy 2n22 2
∞ ∞ ∞
= = =
− − + −− − = = − = − − +− − + − − +
∑ ∑ ∑
n 10
1 1 11 12 nx iy n2 2
∞
=
= − − =
− − +
∑
n 10 0
1 1 1 11 1 1 12 n1 x iy 1 x iy n2 2 2 2
∞
=
= − − −
+ − − + − − +
∑
and by equation ( )15 it follows that:
00
n 10 0
1 11 Γ 1 x iyx iy 2 212 γ1 2 1 12n x iy 2n Γ 1 x iy2 2 2
∞
=
′ + − − − − = +
− − + + − −
∑ .
Analogously for the second sum, we have:
00
n 10 0
1 11 Γ 1 x iyx iy 2 212 γ1 2 1 12n x iy 2n Γ 1 x iy2 2 2
∞
=
′ + − + − + = +
− + + + − +
∑ .
So, rewriting equation ( )29 , we have:
( )
0 002 2
00 0
1 1ζ x iy ζ x iy
8y2 2i i
1 1 1 2x 4yζ x iy ζ x iy2 2
′ ′− − − + − + = − + +− − − +
0 0
0 0
i 1 1 i 1 1Γ 1 x iy Γ 1 x iy
2 2 2 2 2 2
1 1 1 1Γ 1 x iy Γ 1 x iy
2 2 2 2
′ ′− ⋅ + − − ⋅ + − + − +
+ − − + − +
0 0
ρ ρ0 0
1 1x iy x iy
2 2i1 1
ρ x iy ρ ρ x iy ρ2 2
− − − +
− − − − − − + −
∑ ∑ ⇒
( )
0 002 2
00 0
1 1ζ x iy ζ x iy
8y2 2i i
1 1 1 2x 4yζ x iy ζ x iy2 2
′ ′− − − + − + = − − + +− − − +
0
20 0
ρ ρy y0 0
1 1x iy x iyd 1 1 2 2ln Γ 1 x iy i
1 1dy 2 2ρ x iy ρ ρ x iy ρ
2 2=
− − − + − + − − − − − − − − + −
∑ ∑
( )30
In order to apply the limit function 0x x
lim→
we first observe that the first two summands
of equation ( )30 converge to a point, say 1k ∈ℝ . This is due to the continuity of
these functions at a neighborhood of the point ( )0 0x , y , where 0 0
1ζ x iy 0
2 − − =
.
For the third summand of ( )30 , roots ρ , 1 ρ− is understood to be paired, so for the
first sum we have: 0
ρ0
1x iy
21
ρ x iy ρ2
− −=
− − −
∑
0
j 0j j 0 j j
1x iy
2(1 1 1
x iy x iy x iy2 2 2
∞
=
− −= +
− − − − − − −
∑
0
j j 0 j j
1x iy
2 )1 1 1
x iy x iy x iy2 2 2
− −+
+ + − − − + +
( )31
where the pair j j
1x , y
2 − −
ranges over the real and imaginary parts respectively of
all non-trivial roots of the ζ function, except of those that coincide with j j
1x , y
2 +
.
This can easily be done by the constraint jy 0> for all j 0= , 1, 2 , … .
Remark 6: The order of the summation over j in the last equation is just optical. The
fact that the variables jx , jy coincide with 0x , 0y at j 0= is just for simplicity
reasons. The actual order of the series j 0
∞
=∑ follows the order of the series
ρ
∑ for all
values of j sufficiently large, that is of increasing Imρ .
Reformulating ( )31 , we have:
( )( )
( ) ( )( ) ( )( )
2
0 0 j j0
2 22ρ j 0
0 j j j j0
11 x iy x iy 2 x iyx iy22
1 x iy x iy 1 4 x iyρ x iy ρ2
∞
=
− − + + +− − =
− + + + − +− − −
∑ ∑
( )32
Analogously we have:
0
ρ0
1x iy
21
ρ x iy ρ2
− +=
− + −
∑
0
j 0j j 0 j j
1x iy
2(1 1 1
x iy x iy x iy2 2 2
∞
=
− += +
− + − + − − +
∑
0
j j 0 j j
1x iy
2 )1 1 1
x iy x iy x iy2 2 2
− ++
+ − − + − + −
( )33
where the pair j j
1x , y
2 −
ranges over the real and imaginary parts respectively of
all non-trivial roots of the ζ function except of those that coincide with j j
1x , y
2 + −
.
Reformulating ( )33 , we have:
( )( )
( ) ( )( ) ( )( )
2
0 0 j j0
2 22ρ j 0
j j 0 j j0
11 x iy x iy 2 x iyx iy22
1 x iy x iy 1 4 x iyρ x iy ρ2
∞
=
− + − + −− + =
− − − − −− + −
∑ ∑
( )34
In order to prove the convergence of the third summand of ( )30 , we first note that
the critical term for both of the sums ( )32 and ( )34 occurs at j 0= where the
denominator of the corresponding term vanishes for 0x x= . However, when we apply
limit where 0x x→ to their subtraction it turns out that this is a removable
singularity! We have:
0
0 0
x xρ ρ
0 0
1 1x iy x iy
2 2lim1 1
ρ x iy ρ ρ x iy ρ2 2
→
− − − +
− = − − − − + −
∑ ∑
( )( )( ) ( )( ) ( )( )0
2
0 0 j j
2 22x xj 0
0 j j j j
1x iy x iy 2 x iy
2lim (
x iy x iy 1 4 x iy
∞
→=
− − + + + = −
− + + + − +∑
( )( )
( ) ( )( ) ( )( )
2
0 0 j j
2 22j 0
j j 0 j j
1x iy x iy 2 x iy
2)
x iy x iy 1 4 x iy
∞
=
− + − + − −− − − − −
∑
( )( )( ) ( )( ) ( )( )0
2
0 0 j j
2 22x xj 0
0 j j j j
1x iy x iy 2 x iy
2lim (
x iy x iy 1 4 x iy
∞
→=
− − + + + = −
− + + + − +∑
( )( )
( ) ( )( ) ( )( )
2
0 0 j j
2 22
j j 0 j j
1x iy x iy 2 x iy
2)
x iy x iy 1 4 x iy
− + − + − −− − − − −
where the critical term occurs when j 0= . For j 0= we have:
( )( )( ) ( )( ) ( )( )0
2
0 0 0 0
2 2 2x x0 0 0 0 0
1x iy x iy 2 x iy
2lim (
x iy x iy 1 4 x iy→
− − + + + −
− + + + − +
( )( )( ) ( )( ) ( )( )
2
0 0 0 0
2 2 2
0 0 0 0 0
1x iy x iy 2 x iy
2)
x iy x iy 1 4 x iy
− + − + − − =− − − − −
( )( )( ) ( )( )( )( )( ) ( ) ( )( )0
2 2 2 40 0 0 0 0 0 0 0 0 0 0
22 2x x 2 4 2 20 0 0 0 0 0
4iy 1 8y 8 x 2x 4xx x 1 2x 1 x 4x 2 x y 2ylim
x x 4y 16x 8x 1 4y 1 4y→
+ + + + − + + + + +=
+ + + − + + +
and it is clear that after multiplication by i− the result is a number, say 2k ∈ℝ . So,
all of the terms of the last sum are finite and the proof of the convergence is
analogous to the proof of the convergence of the series ρ
1
s ρ−∑ for all s ρ≠ .
As in [3], the infinite series ρ
1
s ρ−∑ converges uniformly in any discs K≤ when the
terms ρ and 1 ρ− are paired
( ) ( )( ) 11s ρ s 1 ρ
−−− + − −
1 11 1 1 1
s ρ s ρ2 2 2 2
− − = − − − + − + −
2
2 2
12 s
12constρ
21 1s ρ
2 2
− − = ≤ −
− − −
,
for all sufficiently large ρ once Κ is fixed and because 2
ρ
1ρ
2
−
−∑ converges.
Fixing the imaginary part of s as well as setting ( ) 1Re s x
2= − ,
1 1x
2 2− < < ,
won’t affect the uniform convergence, therefore 0
ρ0
1x iy
21
ρ x iy ρ2
− −
− − −
∑ ,
0
ρ0
1x iy
21
ρ x iy ρ2
− +
− + −
∑ both converge uniformly for 0x x≠ ,
hence 0 0
ρ0 0
1 1x iy x iy
2 21 1
ρ x iy ρ ρ x iy ρ2 2
− − − +
− − − − − + −
∑ converges uniformly for every
0x x≠ . We have:
0 0
ρ0 0
1 1x iy x iy
2 21 1
ρ x iy ρ ρ x iy ρ2 2
− − − +
− = − − − − + −
∑ …pairing roots ρ , 1 ρ− …
( )( ) ( )( ) ( ) ( )( ) ( ) ( )( )
2
0 j
22 2 2 2 4 2 2 2j 0j 0 j j 0 j j j j j
iy x x A
x x y y x x y y 16x 8x 1 4y 1 4y
∞
=
−=
− + − + + + + − + + +∑
, where A is a polynomial in the variable x .
From the last expression we observe that the partial sums N
j 0=∑ of the series are
continuous functions of the variable x as rational functions whose denominator never equals zero. Therefore by the uniform convergence the result of the infinite summation is a continuous function of x . Analogously we have:
0
0 0
x x
0 0
1 1ζ x iy ζ x iy
2 2lim i i
1 1ζ x iy ζ x iy
2 2
→
′ ′+ − + + − + ∈ + − + +
ℝ .
Summarizing results so far from part D we have:
0
0 0
x x
0 0
1 1ζ x iy ζ x iy
2 2lim i i
1 1ζ x iy ζ x iy
2 2
→
′ ′− − − + − + ∈ − − − +
ℝ ,
0
0 0
x x
0 0
1 1ζ x iy ζ x iy
2 2lim i i
1 1ζ x iy ζ x iy
2 2
→
′ ′+ − + + − + ∈ + − + +
ℝ ,
also we have from part A that:
0
0 0
0 0
y y
0 0
0 0
1 1ζ x iy ζ x iy
2 2i i1 1
ζ x iy ζ x iy2 2
lim 11 1
ζ x iy ζ x iy2 2i i1 1
ζ x iy ζ x iy2 2
→
′ ′− − − + − + − − − + = ′ ′+ + + − − + + + −
.
In order to prove the Conjecture 1 stated at Part B, it suffices to prove that:
0
0 0
0 0
x x
0 0
0 0
1 1ζ x iy ζ x iy
2 2i i1 1
ζ x iy ζ x iy2 2
lim 11 1
ζ x iy ζ x iy2 2i i1 1
ζ x iy ζ x iy2 2
→
′ ′− − − + − + − − − + = ′ ′+ − + + − + + − + +
.
This is the case because the limits of the numerator and denominator of the above exist in ℝ and therefore by an elementary theorem:
[If ( )( )
f xlim 1
g x= and ( )limf x , ( )limg x exist in { }\ 0ℝ then ( ) ( )( )lim f x g x 0− = ]
The conditions become even easier after the following theorem, called “The theorem on interchange of two limits” [8], which in terms of generalized sequences it is called Moore’s Theorem, [9]. Theorem 1: (The theorem on interchange of two limits) Let f : A B M× → be a mapping into a complete metric space, where A and B are subsets of metric spaces
1M and 2M , respectively, and let 0x A \ A′∈ , 0y B \ B′∈ . If
( )i ( ) ( )0x x
lim f x, y ψ y→
= exists for each y B∈ ;
( )ii ( ) ( )0y y
lim f x, y φ x→
= exists uniformly in x A∈ ,
then the three limits
( )0 0x x y y
lim lim f x, y→ →
, ( )0 0y y x x
lim lim f x, y→ →
and ( ) ( )
( )0 0x,y x ,y
lim f x, y→
All exist and are equal. □ Let:
( )
2
2
1 1ζ x iy ζ x iy
2 2i id 11 1 ln ζ x iyζ x iy ζ x iydy 22 2
q x, y1 1 d 1ζ x iy ζ x iy ln ζ x iy2 2 dy 2i i1 1
ζ x iy ζ x iy2 2
′ ′− − − + − +
− −− − − + = = = ′ ′+ − + + + + − + + − + +
2
2
2
d 1ζ x iy
dy 2
d 1ζ x iy
dy 2
1ζ x iy
21
ζ x iy2
− −
+ + =
− − + +
It can be seen from equation ( )20 that the denominator of the last expression is a real
analytic non-vanishing function, therefore the denominator does not affect the existence of the double limit. We state the following line of equalities:
( ) ( ) ( ) ( )0 0 0 0 0 0
0y y y y x x x x y y x x1 lim q x , y lim lim q x, y lim lim q x, y lim q x, y α
→ → → → → →= = = = = ∈ℝ .
The first and the last have been proved. Due to the continuity of the function everywhere near the critical point, it is an easy task to prove the second and the fourth. In order to prove the third one has to prove that either:
( ) ( )0x x
lim q x, y ψ y→
= or ( ) ( )0y y
lim q x, y φ x→
=
exist uniformly. This should mean that the double limit at the critical point exists and then the hypothesis would follow. Remark 7: Another elegant way of proving that the double limit exists is by observing that the first factor of:
( )
2
2 2
d 1ζ x iy
dy 21q x, y
1 d 1ζ x iy ζ x iy
2 dy 21
ζ x iy2
− − = ⋅
− − + + + +
is a real analytic function as the reciprocal of a non-vanishing real analytic function therefore our attention goes to the second factor. The second factor is the quotient of two real analytic functions which vanish at the point ( )0 0x , y .
The functions 1
ζ x iy2
− −
, 1
ζ x iy2
+ +
are complex analytic in the open disc
s r< with s x iy= + , so their real and imaginary parts are harmonic functions in that
disc. Then harmonic functions are real analytic and by the properties that the sum or the product of real analytic functions is real analytic we conclude that the functions:
21
ζ x iy2
− −
, 2
1ζ x iy
2 + +
are real analytic in an open disc ( ) ( )2 2
0 0x x y y r− + − < as the sum of the squares
of two harmonic functions. Therefore their partial derivatives are also real analytic.
Furthermore the quotient
2
2
d 1ζ x iy
dy 2
d 1ζ x iy
dy 2
− −
+ +
is a separate real analytic function in the
open disc. Fixing one of the variables enables a factorization of both numerator and denominator. For example, fixing 0x x= gives:
( ) ( )( ) ( )
2
m00 0
2 n
0 00
d 1ζ x iy
y y f x , ydy 2
y y g x , yd 1ζ x iy
dy 2
− − − =− + +
where f , g are real analytic non-vanishing functions in the open disc and then m n= follows from the fact that:
0 0
2 2
0 0
2 2y y y y
0 0
d 1 1ζ x iy ζ x iy
dy 2 2lim lim k
d 1 1ζ x iy ζ x iy
dy 2 2
→ →
− − − − = = ∈ + + + +
ℝ , k 0≠
Fixing 0y y= gives:
( )0
0x xlim q x, y→
=
0
0
0 00 0
x x
00 0
x x
0 0
0 0
1 11 1ζ x iy ζ x iyζ x iy ζ x iy
2 22 2 lim i ii i11 1
ζ x iy ζζ x iy ζ x iy22 2
lim1 1
ζ x iy ζ x iy2 2i i1 1
ζ x iy ζ x iy2 2
→
→
′ ′− − − +′ ′− − − + − +− + − −− − − + = =
′ ′+ − + + − + + − + +
0
0
0 0
x x
0 0
1x iy
2
1 1ζ x iy ζ x iy
2 2lim i i1 1
ζ x iy ζ x iy2 2
→
− + ′ ′+ − + +
− + + − + +
, where both numerator and denominator of the last fraction are real numbers other than zero. Then, after reformulating we have:
( )0 0 0
2
0
0 2 2x x x x x x
0 0
0
d 1ζ x iy
dy 21lim q x, y lim lim
1 d 1ζ x iy ζ x iy
2 dy 21
ζ x iy2
→ → →
− − = ⋅
− − + + + +
and by division, the last limit is a real number other than zero. Then, further techniques could be applied in order to prove that the separate real analytic function ( )q x, y is in fact real analytic in the joint sense.
□ The proof here follows a different way, it will be proved that:
0 0
2 2
0 0
2 2x x y y
0 0
d 1 d 1ζ x iy ζ x iy
dy 2 dy 2lim lim
d 1 d 1ζ x iy ζ x iy
dy 2 dy 2
→ →
− − − − = − + − +
, where 0x , 0y are as usual.
This is sufficient for the equality:
0
0 0
0 0
x x
0 0
0 0
1 1ζ x iy ζ x iy
2 2i i1 1
ζ x iy ζ x iy2 2
lim 11 1
ζ x iy ζ x iy2 2i i1 1
ζ x iy ζ x iy2 2
→
′ ′− − − + − + − − − + = ′ ′+ − + + − + + − + +
to hold, and therefore for the proof of the Conjecture 1. We recall that the function:
2
2
11 1ζ x iyζ x iy ζ x iy
22 21 1 1ζ x iy ζ x iy ζ x iy2 2 2
− +− − − + = + + + − + +
is real analytic in the variables x , y . Theorem 1: Let f , g : 2 →ℝ ℝ be of squared modulus holomorphic type, [13], i.e.
( ) ( )2
f x, y F x iy= + and ( ) ( )2
g x, y G x iy= + where F, G are holomorphic in a disc
s K≤ and let ( ) ( )F s F s= , ( ) ( )G s G s= , also ( ) ( )F 0 G 0 0= = . If the quotient
function ( )( )
f x, y
g x, y is analytic and non-vanishing at the origin, then the following
equality holds:
( )
( )
( )
( )
( )( )x 0 y 0 y 0
d df x,0 f 0, y
f 0, ydy dylim lim lim
d d g 0, yg x,0 g 0, ydy dy
→ → →= = .
Proof: Let ( ) ( )
( )( )
{ }x,y 0,0
f x, ylim k \ 0
g x, y→= ∈ℝ .
There exist real-analytic functions ( )1f x , ( )1g x, y , ( )1h y and ( )2f x , ( )2g x, y ,
( )2h y and non-negative numbers α , β , γ , δ not all zero, such that:
( ) ( ) ( ) ( )α β γ δ
1 1 1f x, y x f x x y g x, y y h y= + + ,
( ) ( ) ( ) ( )α β γ δ
2 2 2g x, y x f x x y g x, y y h y= + +
and ( )1f 0 , ( )1g 0,0 , ( )1h 0 , ( )2f 0 , ( )2g 0,0 , ( )2h 0 exist in { }\ 0ℝ . Indeed,
from the convergence of ( )
α
f x,0
x and
( )α
g x,0
x we conclude that the series that
correspond to the functions 1f , 2f converge at the origin.
From the convergence of ( )δ
f 0, y
y and
( )δ
g 0, y
y we conclude that the series that
correspond to the functions 1h , 2h converge at the origin.
Then, from the convergence of:
( ) ( ) ( )α δ
1 1β γ
f x, y x f x y h y
x y
− − and
( ) ( ) ( )α δ
2 2β γ
g x, y x f x y h y
x y
− −
we conclude that the series that correspond to the functions 1g , 2g converge at the
origin. We have:
( )( )
( ) ( ) ( )( ) ( ) ( )
α β γ δ
1 1 1α β γ δ
2 2 2
f x, y x f x x y g x, y y h y
g x, y x f x x y g x, y y h y
+ +=
+ +
and
( )
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
β γ 1 β γ δ 1 δ
1 1 1 1
β γ 1 β γ δ 1 δ
2 2 2 2
d d df x, y γx y g x, y x y g x, y δy h y y h y
dy dy dyd d d
g x, y γx y g x, y x y g x, y δy h y y h ydy dy dy
− −
− −
+ + +=
+ + +
We have:
( )( )
( )( )x 0 y 0
f x,0 f 0, ylim lim
g x,0 g 0, y→ →= so
( )( )
( )( )
1 1
1 2
f 0 h 0k
g 0 h 0= = .
We have:
( )
( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( )( )
δ 1 δ
1 1 1 11
y 0 y 0 y 0δ 1 δ 22 2 2 2
d d df 0, y δy h y y h y δh y y h y
h 0dy dy dylim lim lim k
d d d h 0g 0, y δy h y y h y δh y y h ydy dy dy
−
→ → →−
+ += = = =
+ +.
If γ δ> , we have:
( )
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
β γ δ β γ δ 11 1 1 1
β γ δ β γ δ 12 2 2 2
d d df x, y γx y g x, y x y g x, y δh y y h y
dy dy dyd d d
g x, y γx y g x, y x y g x, y δh y y h ydy dy dy
− − +
− − +
+ + +=
+ + +,
and ( )
( )
( )( )
( )( )
1 1
x 02 2
df x,0
δh 0 h 0dylim k
d δh 0 h 0g x,0dy
→= = = .
If γ δ= , we have:
( )
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
β β
1 1 1 1
β β
2 2 2 2
d d df x, y δx g x, y x y g x, y δh y y h y
dy dy dyd d d
g x, y δx g x, y x y g x, y δh y y h ydy dy dy
+ + +=
+ + +,
and ( )
( )
( ) ( )( ) ( )
( )( )
( )( )
β
1 1 1 1βx 0 x 0
2 2 2 2
df x,0
δx g x,0 δh 0 δh 0 h 0dylim lim k
d δx g x,0 δh 0 δh 0 h 0g x,0dy
→ →
+= = = =
+
If γ δ< , we have:
( )
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
β β δ γ δ γ 11 1 1 1
β β δ γ δ γ 12 2 2 2
d d df x, y γx g x, y x y g x, y δy h y y h y
dy dy dyd d d
g x, y γx g x, y x y g x, y δy h y y h ydy dy dy
− − +
− − +
+ + +=
+ + +,
and ( )
( )
( )( )
( )( )
β
1 1βx 0 x 0
2 2
df x,0
γx g x,0 g 0,0dylim lim
d γx g x,0 g 0,0g x,0dy
→ →= = .
It suffices to show that ( )( )
1
2
g 0,0k
g 0,0= .
Set κx t≡ and λy t≡ for real positive κ , λ , then:
( )( )
( ) ( ) ( )( ) ( ) ( )
κ λ κα κ κβ λγ κ λ λδ λ
1 1 1
κ λ κα κ κβ λγ κ λ λδ λ
2 2 2
f t , t t f t t g t , t t h t
g t , t t f t t g t , t t h t
+
+
+ +=
+ +
If β α< and γ δ< then fix λ small enough, so that: κβ λγ κα+ <
and κ small enough, so that: κβ λγ λδ+ < ,
then ( )( )
( ) ( ) ( )( ) ( ) ( )
κ λ κα κβ λγ κ κ λ λδ κβ λγ λ
1 1 1
κ λ κα κβ λγ κ κ λ λδ κβ λγ λ
2 2 2
f t , t t f t g t , t t h t
g t , t t f t g t , t t h t
− − − −
− − − −
+ +=
+ + , with κα κβ λγ 0− − >
and λδ κβ λγ 0− − > , therefore ( )( )
( )( )
κ λ
1
κ λt 02
f t , t g 0,0lim
g 0,0g t , t→= and from continuity of the
function ( )( )
f x, y
g x, y at the origin, we conclude that:
( )( )
1
2
g 0,0k
g 0,0= .
It suffices to show that β α< . We assume multiplicity m 0> for the zero of the holomorphic functions ( )F s , ( )G s
at the point s 0= , so ( ) ( )m1F s s F s= and ( ) ( )m
1G s s G s= with 1F , 1G holomorphic
and non-vanishing at the point s 0= . Set ( ) j
1 jj 0
F s a s≥
=∑ and ( ) j1 j
j 0
G s b s≥
=∑ with 0a 0≠ and 0b 0≠ .
Then, after the translation s x iy= + , a power series expansion about the point ( )0,0
can be used to derive:
( ) ( ) ( ) ( ) ( )mj j 2 2 2j j 0
j 0 j 0
f x, y a x iy a x iy x y a≥ ≥
= + − = + +∑ ∑ … and
( ) ( ) ( ) ( ) ( )mj j 2 2 2j j 0
j 0 j 0
g x, y b x iy b x iy x y b≥ ≥
= + − = + +∑ ∑ …
It can easily be verified that odd powers of y are absent from the expansions of the functions f , g so for the multiplicity m , m 1= is not the case, because then γ δ≥ would occur. For m 1> we have α δ 2m= = and β γ 1= = , therefore β α< . □
Setting ( ) 0 0
1F s ζ x iy s
2 = − − −
, ( ) 0 0
1G s ζ x iy s
2 = + + +
gives the desired result.
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Fundamental Research, Bombay, 1953. [3] Edwards H. M.: Riemann’s Zeta Function, Dover Publications Inc., Mineola,
New York, 1974. [4] Stein, E. M., Shakarchi, R.: Princeton Lectures in Analysis II: Complex Analysis,
Princeton University Press, Princeton and Oxford, 2003. [5] Spiegel M. R., Lipschutz S., Schiller J., Spellman D.: Complex Variables(2nd
edition), Schaum’s outlines, The McGraw-Hill Companies, Inc. 2009 [6] Wikipedia, the free encyclopedia, web site, Riemann zeta function, Gamma
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mathematics, 2005, vol. XIII, 1, pp. 15-29 [9] Moore E. H., H. L. Smith, A general theory of limits, American Journal of
Mathematics, Vol. 44, No. 2, (Apr. 1922) pp. 102 -121 Further References: [10] Steinlage R. C.: Nearly uniform convergence and interchange of limits,
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[14] Aleksandar Ivic: The Riemann Zeta function: Theory and Applications, Courier Dover Publications, book
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[16] Krantz S. G., Parks H. R.: A primer of real analytic functions, Birkhauser Advanced Texts, book [17] Titchmarsh E. C., Heath-Brown D.R.: The theory of the Riemann-Zeta function,
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