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Colecţia

STEF Academic

LUCILIUS VASILIU

EIGHT MAJOR THEORETICAL BREAKTHROUGHS IN THE SUPERIOR MATHEMATICS

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ISBN 978-973-1809-00-7 © 2007, Lucilius Vasiliu, Iaşi. All Rights Reserved.

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LUCILIUS VASILIU

EIGHT MAJOR THEORETICAL

BREAKTHROUGHS

IN THE SUPERIOR MATHEMATICS

STEF, IAŞI, 2007

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PREFACE

Motto “A Hilbert space is a giant computer”

Vasiliu Lucilius

This book represents a sequence of the book “Five Major Breakthroughs in Superior Mathematics”. This happened after all nowadays’ challenges, when Mathematics is so far behind all other Sciences, life Informatics and Genetics, which are evolving with huge steps every day.

There are a lot of entire Mathematical domains that have remained unchanged since the year of 1900.

As a Mathematician, with the burning desire to developing also Mathematics Sciences, I have been trying over the years, using daring and original ideas, to give another impulse to this science.

In 2007, I have managed to make another three huge and extraordinary theoretical breakthroughs which brought up to the light some extremely difficult domains like: The Algebraic Theory of Numbers and The Algebraic Geometry.

I would like to pay special regards this way to my dear professor Eugen Campu from the Mathematical Faculty of Bucharest, to whom I have collaborated over the hard times I have been through trying to solve Riemann Hypothesis. He has helped me solve it after a terrible and tremendous battle I had with all the obstacles I have been through in my way to solve this Hypothesis. I have to mention also that professor Eugen Campu has also been a great moral support for me during all this time.

I would also like to thank to the prestigious Mathematician and professor Daniel Bump from Stanford University who was so willing to read and to comment my work, the one that contained Riemann Hypothesis.

When it comes to the mathematical section, some intricacies may be

found in the rationations. If the ones who will read the book will find these intricacies, the author

will be receptive at a civilized and constructive criticism and will apply himself to correct them in a later edition.

Vasiliu Lucilius, May 2007

Internet address: vasiliulucilius@yahoo.com Str. Ion Creangă nr. 106, bl. C1, et. 5, ap. 3 Iaşi

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7

Chapter 1

Fractional powers with rational exponent of the monotonous operators of class C¹ on .

The study of functional equation f(f(x))=g(x), g:[a, b]→ g∈C¹ [a, b] have started after the year of 1700, great mathematicians like Leonard Euler or K. F. Gauss have thought about it.

During periods of time, an entire theory has been developed around this equation, mathematicians all around the world trying to solve more complicated functional equation, like P(f(x))=g(x), where P(x) is a polynomial of n degree with real or complex coefficients. Along with the evolution of mathematics, had been replaced at the definition of functions g with more complicated spaces, like ⁿ, n , or at the beginning of the 20th century with Hilbert H infinite dimensional spaces or even with Banach spaces.

Moving on, we will solve the functional equation in the most basic case, the one researched ever since 1700.

Theorem 1 Considering g:[a, b]→ , g of class C¹ on [a, b], g monotonically

increasing. Then it exists and is unique a monotonically increasing f function and continuous, with the property that f(f(x))=g(x).

Demonstration It is being considered a division Δ 0 of the interval [a, b] comprised

of n 0 equidistant points a = x o , x 1 , …, x0n = b and ׀x 1j+ –x j ε=׀

∀ j 1, on∈ . On each interval [x j , x 1j+ ] it is approximated the function g(x) with the segment determined by the points (x j , g(x j )) and (x 1j+ , g(x 1j+ )) from the plane xOy. This approximation is possible because the function g is continuous.

Thus it is obtained an affine approximant g1 (x) continuous of the function g. Applying a process of infinite dichotomy to the 0Δ division we obtain a sequence of g n (x) of continuous function, afine on every interval type [x i , x 1i+ ] of the division and monotonically increasing.

So,

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0ε∀ > n∃ ∈ , n>1, therefore ׀g n (x)–g(x)׀<ε x∀ ∈[a, b]. On every interval [x i , x 1i+ ] of a division type nΔ , n>1, the function

g n (x) is g n (x)=Ax+B, where A>0, because g n (x) is monotonically increasing.

But it is known that for every function f1 (x)=Ax+B with A>0 we have the function f 2 (x)=a1 x+b1 with the property that f 2 (f 2 (x))=a1 (a1 x+b1 )+b1 = 2

1a 1x a+ 1b + 1b = 2 ( )f x Ax B= + .

Out of the identification of the coefficients we obtain 1a A= and

1 1Bb

A=

+, so:

f 2 (x)= Ax +1

BA +

So, for every function ( ) :[ , ]ng x a b → n ∗∈ , ( )nf x∃ , 1n > , in order to have:

( ( )) ( )n n nf f x g x= . Let us demonstrate now that every function ( )nf x is continuous in

x , [ , ]x a b∀ ∈ . It may be observed that the function ( )nf x is continuous on the

intervals 1[ , ]i ix x + , being affine. The only thing left to analyze is the continuity in the points ix of the division nΔ . In order to do that we will do a translation of coordinates so the point ( , ( ))i ix g x to become the new origin of the orthonormal system of axes.

So 1 1

2 1

, [ , ]( ) , 0

, [ , ]i i

n ii i

A x if x x xg x x

A x if x x x−

+

∈ ⎧= =⎨ ∈ ⎩

But in this case we have

1 1

2 1

, [ , ]( ) , 0

, [ , ]i i

n i

i i

A x x x xf x x

A x x x x−

+

⎧ ∈ ⎪= =⎨ ∈ ⎪⎩

It is very obvious that ( )nf x is continuous in 0.ix = We repeat the procedure in every point of the nΔ division and we obtain that ( )nf x is a continuous function on [a, b].

Further more, on every interval 1[ , ]i ix x + of the division nΔ , n>1, if ( )ng x Ax B= + with 0A > , we will have:

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( )1n

Bf x AxA

= ++

So ( ) ( )1n n

Bg x A f xA

= ++

on every interval 1[ , ]i ix x + of the

division. This equality is obtained trough the direct calculus. But the sequence ( )ng x is a Cauchy sequence of affine continuous

functions on the interval [ , ]a b . This is to be known from the construction of ( )ng x and from the fact that g(x) is continuous on [ , ]a b .

So, 0 Mnε ∗∀ > ∃ ∈ , so that ׀ ( ) ( )n mg x g x− ׀ Mε< , , [ , ]Mn m n x a b∀ > ∀ ∈ and M>0.

In this case, M is considered as the maximum of the A slope which is bigger than 0 and finite, as a direct consequence of Lagrange’s theorem applied to the g function of class C¹ on [a, b].

So, for every division, max sup '( )A M g x≤ = [ , ]x a b∈

But ( ) ( )1n n

Bg x A f xA

= ++

on each interval 1[ , ]i ix x + of the

division. So, ׀ ( ) ( )n mg x g x− ׀ A= ׀ ( ) ( )n mf x f x−

or ׀ ( ) ( )n mf x f x− =׀ ( ) ( ) ( ) ( )n m n mg x g x g x g xA M

− −≤

where max sup g'(x)A M≤ = )0( [ , ]x a b∈

The inequality (0) shows us that also ( )nf x is a Cauchy sequence

of continuous functions. So, if we shift to limit in the equality ( ( )) ( )n n nf f x g x= , we obtain that it exists and that is unique a function

f continuous on [ , ]a b , ( ) lim ( )nnf x f x

→∞= so that

( ( )) ( )f f x g x= . The uniqueness of f is directly obtained out of the construc-

tion procedure for ( )ng x and ( )nf x Q.E.D. After this theoretical breaktrough we are now able to demonstrate

Theorem 2.

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Theorem 2 Considering :[ , ]g a b → , g of class C¹ on [a, b], g strictly

increasing. Thus it exists and is unique a function f strictly increasing and continuous with the property that [ ] ( ) ( )kf x g x= , where through [ ] ( )kf x we have noted the function ...f f fο ο ο , the composition taking place for k times.

Demonstration It is being considered a division 0Δ of the interval [a, b] made by

0n equidistant points.

01, , ...,o na x x x b= = and ׀ 1j jx x+ − ׀ ε= 01,j n∀ ∈ . On each interval 1[ , ]j jx x + the function g(x) is being approximated with the segment determined by the points ( , ( ))j jx g x and 1 1( , ( ))j jx g x+ + from the plane xOy. This approximation is possible because the g function is continuous.

Thus it is obtained an affine approximant 1( )g x continuous of the function g. Applying a process of infinite dichotomy to the division 0Δ , we obtain a sequence of continuous ( )ng x affine function on each interval type 1[ , ]i ix x + of the division and strictly increasing.

So, 0ε∀ > , 1n n∗∃ ∈ > , so that ׀ ( ) ( )ng x g x− ׀ [ , ]x a bε< ∀ ∈ . On each interval 1[ , ]i ix x + of a , 1n nΔ > division, the function

( )ng x has the form ( )ng x Ax B= + , where A>0, because ( )ng x is strictly increasing.

But it is known that for every function 1( )f x Ax B= + with A>0, we have the function 2 1 1( )f x a x b= + with the property that:

[ ] [ 1]2 1 1 1 1

1 21 1 1 1 1 1 1 1

1 21 1 1 1 1

1

( ) ( ( ) )

....

( ... 1)( )

k k

k k k

k k k

f x a a f x b b

a x a b a b a b b

a x b a a af x Ax B

−

− −

− −

= + + =

= + + + + + =

= + + + + + == = +

From the identification of the coefficients we obtain

1ka A= and 1 1 2( ) ( ) ... ( ) 1

k kkk k

BbA A A

− −=

+ + + +.

So, for every function

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( ) :[ , ]ng x a b → , n ∗∈ ( ), 1nf x n∃ > , in order to have: [ ] ( ) ( )k

n nf x g x= . Let’s demonstrate now that every function ( )nf x is continuous in x [ , ]x a b∀ ∈ . It is to be observed the fact that the function ( )nf x is continuous on

the intervals 1[ , ]i ix x + , being affine. What it remains to be seen the continuity in the points ix of division nΔ . For this we will do a translation of coordinates so that the point ( , ( ))i ix g x to become the new origin for the orthonormal system of axes.

So, 1 1

2 1

, [ , ]( ) , 0

, [ , ]i i

n ii i

A x if x x xg x x

A x if x x x−

+

∈⎧= =⎨ ∈⎩

But in this case we have:

1 1

2 1

, [ , ]( ) , 0

, [ , ]

ki i

n ik

i i

A x x x xf x x

A x x x x−

+

⎧ ∈⎪= =⎨ ∈⎪⎩

It is obvious that ( )nf x is continuous in 0ix = . We repeat the procedure in every point of the division nΔ and we obtain that ( )nf x is a continuous function on [a, b].

More over, on each interval 1[ , ]i ix x + of the division nΔ , 1n > , if ( )ng x Ax B= + with A>0, we will have:

1 2( )

( ) ( ) ... 1k

n k kk k k

Bf x AxA A A− −

= ++ + + +

Making basic calculations it is shown that:

( ) ( )1

11 2

1( ) ( )... 1

k kk k

n n k kk k k

Ag x A f x BA A A

−−

− −

⎡ ⎤−⎢ ⎥= ⋅ + ⎢ ⎥

+ + + +⎢ ⎥⎣ ⎦

on

each interval 1[ , ]i ix x + of the division. But the series ( )ng x is a Cauchy sequence of continuous affine functions on the interval [a, b]. This is known from the construction of ( )ng x and from the fact that g(x) is continuous on [a, b].

So, 0 Mnε ∗∀ > ∃ ∈ , so that 1( ) ( ) ,k k

n m Mg x g x M n m nε −− < ∀ > [ , ]x a b∀ ∈

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In this case, M is taken as the maximum of the A slop which is bigger than 0 and finite as a direct consequence of Lagrange’s theorem applied to g function of class 1C on [a,b].

So, for every division, max sup '( )A M g x≤ = [ , ]x a b∈ But

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1 2

1( ) ( ) [( ) ( ) ... 1

k kk k

n n k kk k k

Ag x A f x BA A A

−−

− −

−= ⋅ +

+ + + +] on every

interval 1[ , ]i ix x + of the division. So,

1( ) ( ) ( ) ( )k kn m n mg x g x A f x f x−− = − or

1 1

( ) ( ) ( ) ( )( ) ( ) n m n m

n m k kk k

g x g x g x g xf x f x

A M− −

− −− = ≤ (00)

where max sup '( )A M g x≤ = [ , ]x a b∈ The (00) inequality shows us that also ( )nf x is a Cauchy series of

continuous functions. So, if we shift to limit in the equality [ ] ( ) ( )k

n nf x g x= , we obtain that it exists and that is unique a f continuous function on [a,b]

( ) lim ( )nnf x f x

→∞= so that [ ]( ) ( )kf x g x= .

Its uniqueness is directly obtained out of the construction procedure of ( )ng x and ( )nf x .

Q.E.D. So, we also defined the k class root of g(x). If we want to define fractional powers with rational exponent for

g(x), we have the following possibility:

Considering res

= ,r s ∈ and be it ( ) :[ , ]g x a b → 1[ , ]g C a b∈ ,

g strictly increasing. We build the function f(x) with the property that [ ]( ) ( )kf x g x=

from Theorem 2 and then we compose f(x) for r times. This way we managed to define fractional powers with rational

exponent for :[ , ]g a b → 1[ , ]g C a b∈ , g strictly increasing. The interested readers may expand these results when instead of

it is taken ,n n or we have g operators defined on Hilbert spaces. The method is similar.

Also more complicated equations can be studied, such as P(f)=g, where P is a polynominal of n degree with real or complex coefficient.

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Chapter 2

Using the symmetry group in demonstrating the radiality of several semilinear biharmonic equations’ solutions.

The symmetry problem in nature had fascinated the minds of

people since antiquity. Great thinkers of humanity were trying to understand why the waves of the lake are radially symmetrical at the surface of the water, when you throw a little stone into the lake. The understanding of these problems began in the Middle Ages along with the inventing of integral and differential calculus by I. Newton and G.F. Leibnitz. So, the processes which were governing the world of physics had been introduced to differential equations. However, solving the symmetry problem seemed at that time a very complicated matter. And it was complicated, because the mathematical apparatus which was used until 1900 used to be very rudimentary.

Late development of mathematics in the 20th century permitted the apparition of a very sophisticated method called the symmetrical rearrangement which partially solves this problem. It is to be mentioned here the famous article of Gidas-Ni-Nirenberg who uses the method in demonstrating the radiality of positive solutions of a semilinear equation of order 2. the method uses the principle of maximum which is valid only for elliptical operators of order 2.

This result has been obtained in 1979 and it can be found on the Internet and is considered a famous result and the core of the development of a reach Internet literature.

The article ingeniously uses different variants of the principle of maximum and an interesting procedure of domain reflection.

The article [1] establishes that the positive solutions for the following elliptical semilinear equations.

( ),0,

u f u inBu on B−Δ = ⎧

⎨ = ∂⎩

where B is a ball in n and f is a continuous Lipschitz, are radial symmetrical.

But the principle of the maximum is no longer valid for the operators type ( )P Δ in spite of various attempts which are made now,

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especially after 2000, in order to obtain different forms of the maximum principle for bi-Laplacian.

Also the property of solution’s positiveness in article [1] plays a very important role. But in nature we also have solutions which have negative parts.

Next, I will present my ideas regarding these types of equations. The key to the reasoning is represented by the following theorem:

Theorem 1 [2] Considering ( )ijA a= and ( )ijU u= two matrices type n n× with

real numbers as elements. We consider the following operators:

2

, 1

( ) : ( , ) ( , )

( )

n n

n

iji j l j

P D C C

P D ax x

∞ ∞

=

→

∂=

∂ ∂∑and

: ( , ) ( , ), ( )n nU C C U f f Uο∞ ∞→ = ,for any ( , )nf C∞∈ .

Then, ( ) ( )P D U U P Dο ο= if and only if tA UAU= .

Demonstration If { }1 2, ,..., ne e e represents the canonic base from n , then

1( )

n

i ij jj

U e U e=

= ∑ for 1, 2,...,i n= . For any { }1, 2,...,k n∈ we consider the

sequence functional : nkU →

1 21

( ) ( , ,..., )n

nk r rk n

rU x x u x x x x

=

= ∀ = ∈∑

If 1

n

i ii

x x e=

= ∑ , then

1 1 1 1

( ) ( ) ( ( ))n n n n

i i i ij j j ji i j j

U x xU e xU e U x e= = = =

= = =∑ ∑∑ ∑

Because

1 1

[ ( )]n n

ik ik ik ij jk

i ij j

xU x u u ux x

δ= =

∂∂= = =

∂ ∂∑ ∑ it results that

2

, 1 , 1 1[ ( ( ))] ( ) ( )

n n n

ij ij ki j i j ki j i k j

fa f U x a Ux U xx x x x x= = =

⎡ ⎤∂ ∂ ∂ ∂= ⎢ ⎥

∂ ∂ ∂ ∂ ∂⎢ ⎥⎣ ⎦∑ ∑ ∑ =

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=, , 1

( )n

ij jki j k i k

fa u Uxx x=

⎡ ⎤∂ ∂⎢ ⎥∂ ∂⎣ ⎦

∑ =

=2 2

, , , 1 , , , 1

( ) ( ) ( )n n

ij jk r ij jk iri j k r i j k rr k i r k

f fa u Ux U x a u u Uxx x x x x= =

∂ ∂ ∂⋅ =

∂ ∂ ∂ ∂ ∂∑ ∑ .

Out of the equalities

2

, , , 1

( ) ( ) ( )n

ij jk iri j k r r k

fP D U f x a u u Uxx x=

∂=

∂ ∂∑

2

, 1

( ( ) ) ( )n

rkr k r k

fU P D f x a Uxx x=

∂=

∂ ∂∑ we obtain that ( ) ( )P D U UP D=

if and only if , 1

n

ir ij jk rki j

u a u a=

=∑ for any , 1,2,...,k r n= , which is equivalent

to tUAU A= . A remarkable particular case it represents the Laplace’s differential

operator 2 2 2

2 2 21 2

...nx x x

∂ ∂ ∂Δ = + + +

∂ ∂ ∂ which, as to the previous theorem, is

invariant to the group of orthogonal transformations of n U U Δ = Δ for any orthogonal matrices U.

In what follows next I will work with a classical example taken after G. Dinca [3] “Variational methods and applications”.

Considering { },nB x x R= ∈ ≤ . The function 4 ( )v C B∈ is being searched for, which satisfies the biharmonic semilinear problem with homogeneous conditions at the frontier.

2 2 2 20 0 10 20 0

1

0

( ) ( , ), ...

( ) 0

monotonically increasing in v and continuous in (r,v)

r is fixed

n

BB

Av v v f r v r x x x

vP vn

f

∂∂

⎧ = Δ = Δ Δ = = + + +⎪⎪ ∂

= =⎨ ∂⎪⎪−⎩

These types of problems have been obtained ever since the 19th century by Sophie Germain in the study of the homogeneous and isotropic embedded plates bending at the frontier.

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Theorem [Vasiliu Lucilius 1988], Theorem ∗ The biharmonic semilinear problem (P1 ) admits a unique solution

which is radial.

Demonstration In [3] it is being demonstrated that A operator is a positively

defined symmetrical operator. Because – 0( , )f r v is monotonically increasing in v, it results that the operator 0( , )Lv Av f r v= − is monotonically maximal. So, the equation 0Lv = admits a unique solution.

Now we will demonstrate the radiality of the solution. According to Theorem 1 from [2], making x Ux→ , we have that if

( )v x is solution, then ( )v Ux is also solution U∀ orthogonal matrices. Here is essential also the sphere’s invariation to the orthogonal changes.

I will proceed now to the last state of the demonstration. In [2], we have the following Theorems:

Theorem 2[2] Considering ,x y ∈ with the property that x y= . Then it exists

an orthogonal matrices with the property Ax y= . The norm considered on n is the Euclidian one.

Theorem 3[2] Considering : nf → an invariant function to the orthogonal

transformations, meaning it has the property ( ) ( )f Ax f x= for every A orthogonal matrices type n n× and for every nx ∈ . Then f is a radial function, meaning that it exists :g + → so that ( ) ( )f x g x= for

every nx ∈ . Applying the Theorems 2 and 3 to the unique solution we obtain that ( )v x is a radial function. Q.E.D.

To make it simpler to understand, we will say that the operators’ symmetry group overlaps the symmetry group of the domain, resulting the radiality, or the solution spins around but remains the same, giving the radiality.

The interested readers may search the Internet for similar problems debating existential problems for biharmonic or polyharmonic equations and their radiality can be easily obtained by applying the group of symmetry to the solution.

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Bibliography to Chapter 2 [1] Gidas, B.; Ni;Wei Ming; Nirenberg, L., Symmetry and related

properties via the maximum principle. Comm. Math. Phys. 68 (1979), 209-243.

[2] Sorin Radulescu, Marius Radulescu, Theorems and Problems of Mathematical Analysis, Didactical and Pedagogical Publishing House, Bucharest, 1982.

[3] George, Dinca, Variational Methods and Applications, The Technical Publishing House, Bucharest, 1980.

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Chapter 3

A study on zeroes of the function ( )zζ of Riemann

24th of August, 2000 by Lucilius Vasiliu

Riemann’s hypothesis is considered at this moment the most important mathematical problem due to its numerous connections which it realizes to different domains, which apparently seem separated to those of mathematics.

In this study, the author manages to transform the problem from a Complex Analysis problem into a Convex Functional Analysis problem, thus obtaining a geometrical algorithm for generating the zeroes to Riemann’s function ( )zζ .

So, the problem is geometrized and algorithmized. In 2004 it has

been considered, even by prestigious specialists from U.S.A., as the best writing in this extremely fascinating and difficult domain at the same time, the one of analytical theory of numbers.

The reader who is interested of these aspects, may find in this study essential ideas to approaching some problems which are situated beyond Riemann’s Hypothesis, such it would be Riemann’s Generalized Hypothesis or The Great Hypothesis of Riemann.

The First Stage

Problem putting

We have 1

1( ) , , for >1.zn

z z itn

ζ σ σ σ∞

≥

= = + ∈∑

And also, 1 1 1 1 12 ( ) 2 ... ... for Rez>1

2 4 6 (2 )z

z z z zzn

ζ− ⎛ ⎞⋅ = + + + + + ⎜ ⎟

⎝ ⎠

So we have 1 1

1

1(1 2 ) ( ) ( 1) for Rez>1 (1)z nz

n

zn

ζ∞

− −

≥

− ⋅ = − ⋅∑

Then we use the following integral representation of the sum of a Dirichlet series, whose coefficients satisfy a certain asymptotic condition.

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Lemma 1 Let us consider 1( )n na ≥ a numerical succession ( ) n

n x

z aψ≤

= ∑ and z

a complex number for which ( )lim 0.zx

xx

ψ→∞

= Then 110

( )nz z

n

a xz dxn x

ψ∞ ∞

+=

= ⋅∑ ∫

(on the condition that at least one of the two members be defined).

Demonstration As ( )xψ is constant for each interval [ ), 1n n + for any complex

number 0z ≠ we shall have 1 1

111 1

( ) 1 1 1 1 ( 1)( )( 1)

k kk nz z z z z

n n

ax kdx nx z n n z n k

ψ ψψ− −

+= =

⎡ ⎤ −⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥+ ⎣ ⎦⎣ ⎦∑ ∑∫ where

for the last equality I used the relation ( ) ( 1) , 2nn n a nψ ψ− − = ≥ .

Thus, 111

( ) ( )k knz z

n

a k xz dxn k x

ψ ψ+

=

= +∑ ∫ and lemma is obtained doing

0k → . Taking in particular 1( 1)nna −= − we have 1( ) ( 1)n

n xxψ −

≤

= −∑ .

It can be noticed that [ )( ) 2 1,x xψ < ∀ ∈ +∞ and ( )lim 0zn

xx

ψ→∞

= for

Rez>0 , and lemma gives us 111

1

1 ( )( 1) , for Rez>1.nz z

n

xz dxn x

ψ∞ ∞−+

=

− = ⋅∑ ∫

So, we have the identity: 1

11

( )(1 2 ) ( ) , Rez>1 (2)zz

xz z dxx

ψζ∞−

+− ⋅ = ⋅ ∫

But the function 1(1 2 ) ( )z zζ−− ⋅ exists also for Rez>0, because

[ )( ) 2 1,x xψ < ∀ ∈ +∞ the integral 11

( )z

x dxx

ψ∞

+∫ exists also for Rez>0.

So, we have the identity: 1

11

( )(1 2 ) ( ) , Rez>0. (3)zz

xz z dxx

ψζ∞−

+− ⋅ = ⋅ ∫

Now we work in band 0<Rez<1. We have 1 ln 2

-it 1 1 1

2 2 2 ,but 2 so

2 1 and 2 2 2 2 1 for 0 Re 1

it i it it it

z it

e

z

σ σ

σ σ

− − − − − −

− − − −

= ⋅ =

= = ⋅ = ≠ < <

So, it results that 11 2 0 for 0 Re 1.z z−− ≠ < < We know from The Dictionary of Mathematical Analysis,

Scientific and Encyclopedic Publishing House, Bucharest,1989, on page

20

136, that the zeroes of ( ) with z z itζ σ= + are in the critical band 0 1σ< < (from Hadamard and de la Valée Poussin’s theorem). Using the identity (3) and the fact that 11 2 0 for 0<Rez<1z−− ≠ , it results that in the critical band 0 1σ< < the zeroes of the function ( )zζ coincide with the

zeroes of the function 11

( )( ) .z

xf z z dxx

ψ∞

+= ⋅ ∫ So, we must show that the

function 1 11

( )( ) z

xf z dxx

ψ∞

+= ∫ doesn’t admit symmetrical zeroes given to

the critical axis 12

σ = . Our problem has been reduced to this.

Lemma 1 is taken from the book Complex Analysis. Classical and Modern Aspects, coordinator Cabiria Andreian Cazacu, Scientific and Encyclopedic Publishing House, Bucharest,1989, 405 pages.

We note with 1S the critical band 0 Re 1z< < . Afterwards, we shall work only in the critical band 1S . We define the function ( , )g z v .

[ ]1

1

( , ): 0,10

( , ) ( ) v

g z v S C

g z v f z e

× →

= ⋅

It can be noticed that [ ]0 0,10ve v≠ ∀ ∈ . Thus the zeroes of the function ( , )g z v are the zeroes of the

function 1( )f z , which, in its turn, are the zeroes of the function ( )zζ from the critical band 0 Re 1z< < .

So, 11

( )( , ) vzxg z v e dx

xψ∞

+= ⋅∫ .

Observation. Instead of [ ]0,10v ∈ , I can choose [ ]0, ,v ∈ Τ Τ being

however much along our study. The factor ve is a factor of homothety.

Stage 2 We define the space Hilbert complex [ )2 1,L ∞ . I do not write the problems connected to defining these spaces,

considering them as a classical known theory. We have:

[ ) [ ){2 1, : 1,L f C f∞ = ∞ → is measurable Lebesgue and 2f is integrable Lebesgue }

21

[ )21 11, 1

, ( ) ( )L

u v u x v x dx∞

∞= ⋅∫

It is evident that [ )21

( ) 1,z

x Lx

ψ+ ∈ ∞ according to the criterion in α

for 0 Re 1z< < because ( ) 2xψ < .

Stage 3 We consider two certain zeroes symmetrical 1z and 2z symmetri-

cal to the axis 12

σ = .

Thus 1 1 0z itσ= + and 2 2 0z itσ= + where 2 11σ σ= − . We define the vectors ( , )B vσ and ( )A t and it is shown that

( )A t [ )2 1,L t R∈ +∞ ∀ ∈ and

[ )2( , ) 1,B v Lσ ∈ +∞ [ ]1 1( , ) ,1 0,102 2

v σ σσ ⎛ ⎞∀ ∈ − ×⎜ ⎟⎝ ⎠

.

Demonstration Let us consider the function [ )2: 1,A L→ +∞

[ )1

12

( )( )( ) 1,it

xA t x xx x

α

ψ+

= ∀ ∈ +∞

Where 1α is chosen so that 11 3

σα < .

We notice that A(t)(x) is measurable in α on [ )0,∞ . lnit it xx e=

We have 1

1 1

22

1 1 2ln ln2 2

1 ( ) 1 ( ) ( )( )( ) it x iy x

x x xA t x t Re e x x

x xαα α

ψ ψ ψ+ +

= ⋅ ⋅ ⋅ = ∀ ∈⋅

.

Thus 2( )( )A t x is integrable Lebesgue on [ )1,∞ , according to the

criterion in α . So, [ )2( ) 1,A t L t R∈ ∞ ∀ ∈ .

The function [ ] [ )1 12: ,1 0,10 1,

2 2B Lσ σ⎛ ⎞− × → ∞⎜ ⎟

⎝ ⎠ is defined

through:

22

1

111

2

( , )( )3

veB v x wherex

σ α

σσ α+ −

= <

We can notice that ( , )( )B v xσ is continuous in x on [ )1,∞ , thus it

is measurable Lebesgue on [ )1,∞ and 1

22 1

12( )( , )( ) ,3

veB v xx x σ α

σσ α−= <⋅

is integrable Lebesgue on [ )1,∞ according to the criterion in α .

Thus, [ ) [ ]1 12( , )( ) 1, ( , ) ,1 0,10

2 2B v x L v σ σσ σ ⎛ ⎞∈ ∞ ∀ ∈ − ×⎜ ⎟

⎝ ⎠.

We have

[ )

1 1

2 1,

1 111 12 2

( ) ( , )

( ) ( )( , )

( ), ( , )

vv

zit

A t B v

L

x x eg z v e dx dxx x x x

A t B t

α σ α

σ

ψ ψ

σ∞

∞ ∞

+ + + −= = ⋅ =

⋅

=

∫ ∫

It is seen that the function B is a real function, so

[ ]1 1( , ) ( , ) ( , ) ( , ) ,1 0,102 2

B v B v B v v σ σσ σ σ σ ⎛ ⎞= = ∀ ∈ − ×⎜ ⎟⎝ ⎠

.

Stage 4 We show that the family of functions ( ),A t t ∈ is a continuous

set in the space [ )2 1,L ∞ . Let us consider 0t ∈ . We know that lnit it xx e= . We have:

1 10

2

20 1 11 1

2 2

( ) ( )( ) ( )itit

x xA t A t dx dxx x x x

α α

ψ ψ∞ ∞

+ +− = − =

⋅ ⋅∫ ∫

0 011

22 2

1 ln2 ln1 12

( ) 1 1 ( ) 1 1 2it it xit it x

x xdx dxx x x x e e

xαα

ψ ψ∞ ∞

+

⎡ ⎤⎢ ⎥= ⋅ − = ⋅ − =

⋅⎢ ⎥⎣ ⎦

∫ ∫

0

1

2ln ln

21

( ) 2it x it xx e e dxx x α

ψ∞= ⋅ −

⋅∫ that 2ln 1,it xe t= ∀ ∈ and

[ )1,x∀ ∈ ∞ and [ )02it ln

0e 1 and 1,x t x= ∀ ∈ ∀ ∈ ∞ . Thus

23

0

1

1

22ln ln2

0 21 1

22 2

0ln 0 1 021

( )( ) ( )

( ) (cos ln cos ) (sin ln sin ln ) ( , )

it x it xxA t A t dx e e dxx x

x t x t t x t x dx I t tx x

α

α

ψ

ψ

∞ ∞

∞

− = ⋅ − =⋅

⎡ ⎤− + − =⎣ ⎦⋅

∫ ∫

∫

We have: 0

2lnln 2 20 0(cos ln cos ln ) (sin ln sin ln )it xit xe e t x t x t x t x⎡ ⎤− = − + − ≤⎣ ⎦

2 22 2 8≤ + = . As 1 0 0( , ) it is , it results that >0 M 1I t t t t εε∀ ∈ ∀ ∃ > so that:

0

1 1

22 2ln ln

12 2

( ) ( ) 8 (D )4

it x it x

M M

x xe e dx dxx x x xε ε

α α

ψ ψ ε∞ ∞− < ⋅ <

⋅ ⋅∫ ∫

On the other hand,

[ ]0 0

1 1

2 22ln lnln ln22 21 1,

( ) ( )( 1) max (D )M it x it xit x it x

x M

x xe e dx M e ex x x x

ε

εεα α

ψ ψ∈

− < − ⋅ −⋅ ⋅∫

Applying a consequence of the theorem of Langrange of the finite

increases, we have: 2 2

0 0(cos t ln x-cos t ln ) (sin t ln x-sin t ln )x x+ ≤2 2

0 0(t ln x-t ln x) (t ln x-t ln )x≤ + < 2 2 2 20 0 0ln ( ) ( ) 2( ) lnx t t x t t t t x− − = − .

Thus the inequality 2( )D becomes

0

1 1

2 22ln ln 20 32 21 [1, ]

( ) ( )( 1) max 2( ) ln (D )M it x it x

x M

x xe e dx M t t xx x x x

ε

εεα α

ψ ψ∈

− < − ⋅ −⋅ ⋅∫

3M being fixed from the stage 1( )D , from 3( )D , it results that

1 0εδ∃ > so that t∀ ∈ with 0 1t-t εδ< to have:

[ ] 1

22 2

0 421,

( )2( ) ( 1) max ln (D )4x M

xt t M xx xε

εψ ε

∈− − ⋅ <

⋅

From 4( )D and 3( )D it results that 1 0εδ∃ > so that t∀ ∈ with

0 1t-t εδ< to have:

0

1

22ln ln

521

( ) (D )4

M it x it xx e e dxx x

ε

α

ψ ε− <

⋅∫

And from 5( )D and 1( )D it results that 10 0εε δ∀ > ∃ > so that

0 1t with t-t εδ∀ ∈ < to have:

24

0

1

22ln ln

621

( ) (D )2

it x it xx e e dxx x α

ψ ε∞− <

⋅∫

Thus 1 0 10 so that t with t-t we have:ε εδ δ∃ > ∀ ∈ < 2

0 71( ) ( ) (D )

2A t A t dx ε∞

− <∫

From 7( )D it results that 10 0εε δ∀ > ∃ > so that t∀ ∈ with

0 1t-t εδ< we have:

2

220 0[1, ] 1

( ) ( ) ( ) ( )L

A t A t A t A t dx ε∞

∞− = − <∫

0t being taken arbitrarily, it results that A(t) is a continuous set in

[ )2 1,L +∞ .

Stage 5 We show that the family of functions ( , )B vσ is a continuous set in

the space [ )2 1,L +∞ in function of depending on

[ ]1 1( , ) ,1 0,102 2

v σ σσ ⎛ ⎞∈ − ×⎜ ⎟⎝ ⎠

.

Demonstration

Let us consider [ ]1 10 0( , ) ,1 0,10

2 2v σ σσ ⎛ ⎞∈ − ×⎜ ⎟

⎝ ⎠. We have:

2

0

1 0 1

220 0 0 0[1, ] 1

22

0 0 2 0 01 11 12 2

( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , , , )

L

vv

B v B v B v B v dx

e eB v B v dx dx I v vx x

σ α σ α

σ σ σ σ

σ σ σ σ

∞

∞

∞ ∞

+ − + −

− = −

− = − =

∫

∫ ∫

Because 2 0( , , , )oi v vσ σ there is

1 10 0( ,v), ( , ) ,1 [1,10]

2 2v σ σσ σ ⎛ ⎞∀ ∈ − ×⎜ ⎟

⎝ ⎠, it results that

>0 M 1εε∀ ∃ > so that

0

1 0 1

2

11 12 2

(F )4

vv

M

e e dxx x

ε σ α σ α

ε∞

+ − + −− <∫

25

We notice 1

1 12

1 11

( , , )

: ,1 [1,10] [1, ]2 2

veg v xx

g M

σ

ε

σ

σ σ

+ −∝=

⎛ ⎞− × × →⎜ ⎟⎝ ⎠

It can be noticed that derivatives 1g σ and 1vg are limited on

1 1,1 [0,10] [1, ]2 2

Mεσ σ⎛ ⎞− × ×⎜ ⎟

⎝ ⎠ as well.

Thus 2M ε∃ so that 1 2g ( , , )v x Mσ εσ < and 1v 2g ( , , )v x M εσ < . Applying a consequence the theorem of Langrange of two

variables, where x is considered a parameter, we obtain: 22 2

1 1 0 0 2 0 0( , , ) ( , , )g v x g v x M v vεσ σ σ σ⎡ ⎤− < − + −⎣ ⎦ . Thus

2 221 1 0 0 2 0 0 21( , , ) ( , , ) ( 1) (F )

Mg v x g v x dx M M v vε

ε εσ σ σ σ⎡ ⎤− < − ⋅ − + −⎣ ⎦∫

From 2( )F it results that 2 0εδ∃ > so that:

1 1( ,v) ,1 [0,10]2 2

σ σσ ⎛ ⎞∀ ∈ − ×⎜ ⎟⎝ ⎠

with 20 0 2( ,v)-( , )v εσ σ δ< to have:

222 0 0 3( 1) (F )

4M M v vε ε

εσ σ⎡ ⎤− ⋅ − + − <⎣ ⎦

From 2 3( ) and (F )F it results that 2 0εδ∃ > so that:

2

1 10 0 2( , ) ,1 [0,10] with ( ,v)-( , to have

2 2v v ε

σ σσ σ σ δ⎛ ⎞∀ ∈ − × <⎜ ⎟⎝ ⎠

:

0

1 0 1

2

41 112 2

(F )4

vvM e e dxx x

ε

σ α σ α

ε+ − + −

− <∫

From 1( )F and 4( )F it results that 20 0 εε δ∀ > ∃ > so that

21 1

0 0 2( , ) ,1 [0,10] with ( ,v)-( , )2 2

v v εσ σσ σ σ δ⎛ ⎞∀ ∈ − × <⎜ ⎟

⎝ ⎠to have:

0

1 0 1

2

51 112 2

(F )4 4 2

vve e dxx x

σ α σ ε

ε ε ε∞

+ − + −− < + =∫

26

From 5( )F it results that 20 0εε δ∀ > ∃ > so that

1 1( ,v) ,1 [0,10]2 2

σ σσ ⎛ ⎞∀ ∈ − ×⎜ ⎟⎝ ⎠

with 20 0 2( , ) ( , )v v εσ σ δ− < we have:

[ )2

220 0 0 01, 1

( , ) ( , ( , ) ( , )2L

B v B v B v B v dx εσ σ σ σ∞

∞− = − <∫ , 0 0( , )vσ

being taken arbitrarily, it results that ( , )B vσ is a continuous set.

Stage 6 We show that ( ) 0 tA t ≠ ∀ ∈ . We show that when the vector A(t)

runs continuously the set in the space [ )2 1,L ∞ , then the hyperplan tH perpendicular on A(t) also varies continuously as geometric position.

We have 1 1

21 11 12 2

( ) ( )( ) ( ) ( )it it

x xA t A t A t dx dxx x x x

α α

ψ ψ∞ ∞

+ += ⋅ = ⋅ =

⋅ ⋅∫ ∫

1 1

2 2

2 21 1

( ) 1 1 ( ) 0 tit it

x xdx dxx x x x x xα α

ψ ψ∞ ∞= ⋅ ⋅ = > ∀ ∈

⋅ ⋅∫ ∫ . Thus ( ) 0A t t≠ ∀ ∈ .

We know that Hilbert space has a very similar geometry with Euclidian spaces.

More demonstrations can be given here with reasoning of type ,ε δ considering the generators of the hyperplan tH which at a continuous movement of A(t) they also move continuously.

But I think that the simplest demonstration is this one. Considering the vector system A(t) and the tH hyperplan as a rigid

in Hilbert space through the continuous movement of A(t) we have evidently the continuous movement too of tH in Hilbert space.

Stage 7

Putting Riemann’s problem in terms of functional analysis From the Stage 1 we know that we have reduced the study of

zeroes of ( )zζ from the critical band 0 Re 1z< < to the study of the zeroes of function g(z,v) also in the critical band 0 Re 1z< < .

11

( )( , ) 0<Rez<1vz

xg z v e dxx

ψ∞

+= ∫

We have showed in Stage 3 that 2[1, )

( , ) ( ), ( , )L

g z v A t B tσ+∞

= .

We know, according to the Dictionary of Mathematical Analysis, Scientific and Encyclopedic Publishing house, Bucharest, 1989, on page

27

136, that the zeroes of ( )zζ are in the critical band 0 1σ< < and are

placed symmetrically reported to the critical axis 12

σ = .

Also, we know from The Dictionary as well that it has been

demonstrated that there is an infinity of zeroes on the critical axis 12

σ =

(Hardy, 1914).

If the function ( )zζ has two zeroes 1 1 0 2 2 0 and zz it itσ σ= + = + is

symmetrical to the axis 12

σ = and g(z,v) has these zeroes as well.

Thus the hyperplan 0t

H will be in contact with the set ( , )B vσ in

points with the form of and 1 2( , ) and B( , )B v vσ σ .

Also, let us consider 3 112

z it= + one zero on the critical axis, in this

case, the hyperplan 1t

H will have in common with the surface ( , )B vσ the

points of form 1( , )2

B v .

Thus the study of the symmetrical zeroes of the function ( , )g z v and of what happens in this case reduces to the study of aspect, form of intersection between the set ( , )B vσ and the hyperplan tH .

28

Stage 8

The form of the set ( , )B vσ ( , )B vσ= . It is evident that the function

12 1

20

1( )fx

σ ασ

+ −= is convex in [ )0 x 1,σ ∀ ∈ +∞ .

[ ] [ ] [ )

1

3 4 3 4

1 13 4

v

12

So B( , ) (1 ) ( , ) ( (1 ) , )

0,10 , 0,1 , , ,1 and 1,+ ( )2 2

ewhere B( ,v)=x

v B v B v

v

σ α

τ σ τ σ τσ τ σσ στ σ σ

σ+ −

+ − > + −

⎛ ⎞∀ ∈ ∀ ∈ ∀ ∈ − ∞ ∗⎜ ⎟⎝ ⎠

So the set ( , )B vσ ( , )B vσ= is convex in σ .

Stage 9

Final considerations on the zeroes of ( )zζ

We show that the family of function ( , )B vσ is a differentiable set

in the space [ )2 1,L +∞ depending on 1 1,12 2

σ σσ ⎛ ⎞∈ −⎜ ⎟⎝ ⎠

Demonstration

Let it be 1 10 0( , ) ,1 [0,10]

2 2v σ σσ ⎛ ⎞∈ − ×⎜ ⎟

⎝ ⎠

We have:

[ )1,2

0 0

1 0 1

2 2

0 0 0 0 0 0

0 01

2

1 12 2

3 0 001

( , ) ( , ) ( , ) ( , )

( , , )

L

vve e

B v B v B v B v dx

x x dx I vσ α σ α

σ σ σ σσ σ σ σ

σ σσ σ

∞

∞

−+ − + −∞

− −= =

− −

= =−

∫

∫

We have to show that 0

3 0 0lim ( , , )I vσ σ

σ σ→

exists and is finite.

Passing to the limit we obtain:

29

03 0 0lim ( , , )I v

σ σσ σ

→0

1

2

11 2

1lnve x dxx

σ α

∞

+ −= −∫

Let us show now that 1

2

11 2

1ln x dxx

σ α

∞

+ −−∫ exists and is finite where

1 1 11,1 and

2 2 3σ σ σσ α⎛ ⎞∈ − <⎜ ⎟

⎝ ⎠.

Let us show that ln rx x< , where r>0 starting from one 4x>M .

Applying the exponential, we have to show actually that rxx e< where

0r > starting from one 4x M> .

Let it be 3( )rxel x

x= . We have to show that 3 4( ) 1,l x x M> ∀ > .

Let it be

( )'1

3 2 2 2

( 1)( ) , 0

r rr r r rx xx r x x x re x e xe rx e x e e rxl x r

x x x x

−−⎛ ⎞ − −= = = = >⎜ ⎟⎜ ⎟

⎝ ⎠

It is obvious that starting from one 4

x M> , we have '

3 3l ( ) 0 so l ( )x x> is strictly increasing.

We calculate limrx

x

ex→∞

We are in one of the cases where the rule of 1’Hopital applies. We have

4lim lim , (x)' 0 x>M

rx

x xe x

→∞ →∞= = +∞ ≠ ∀ and

' 1'( )lim lim lim , 0

' 1

r r rx r x r x

x x x

e rx e rx e rx x

−

→∞ →∞ →∞∃ = = >

We calculate now limrr x

x

x ex→∞

. We are still in one of these cases

where the rule of 1’Hopital applies:

30

' '

1 1 1

2

4

( ) ( ) ' ( )lim lim lim( ) ' 1

( )lim lim1 1

( )lim lim , 0, 0

r r r r

r r r

r r

r x r x r x r x

x x x

r x r r x r x r

x x

r r x r x

x x

x e x e x e x ex x

rx e x rx e x e r x r

x r x r e rx e x M rx x

→∞ →∞ →∞

− − −

→∞ →∞

→∞ →∞

+= = =

+ += = =

+= ≥ > > >

We try to calculate 2

limrr x

x

x ex→∞

We still are in one of the cases where a variant of the theorem of 1’Hopital applies.

2 2 ' 2 ' 2

2 1 1 2 2 3

3

( ) ( ) ' ( )lim lim lim( ) ' ( ) '

2 (2 )lim lim1

lim

r r r r

r r r

r

r x r x r x x r

x x x

r x r x r x r r

x x

r x

x

x e x e x e e xx x x

rx e rx e x e rx rxx

rx ex

→∞ →∞ →∞

− −

→∞ →∞

→∞

+= = =

+ += = ≥

≥

Going recursively, we apply the rule of 1’Hopital and obtain: 2 2 3 3

4

lim lim lim lim ...

lim ... and x>M 0, 0

r r r r

r

x r x r x r

x x x x

n nr x

x

e rx e r x e r x ex x x x

r x e n rx

→∞ →∞ →∞ →∞

→∞

= ≥ ≥ ≥

≥ ≥ ∀ ∈ > >

As 0

0r n∗

> ⇒ ∃ ∈ so that 0

1n r > .

Therefore 0

n∃ so that 0

01

lim limr

rn r x

n r x

x x

x e x ex

−

→∞ →∞= = +∞ because

1 0o

n r − > .

Therefore, lim ,rx

x

e r ox→∞

= +∞ > as well.

So we have the function 3( )

rxel xx

= .

31

We know that 4

0M∃ > thus that 3( )l x is strictly increasing for

4 3 and lim ( )

xx M l x

→∞> = +∞ .

Therefore 5

0M∃ > thus that 3 5( ) 1 x>Ml x > ∀ .

So 5

1rxe x M

x> ∀ > therefore

50, 0

rxx e x M r< ∀ > > > ⇒

5ln , 0, 0 ( )rx x x M r⇒ < ∀ > > > ∗∗

Therefore [ )1

1 22

1ln 1,x Lx

σ α+ −− ∈ +∞ where 1 1,1

2 2

σ σσ

⎛ ⎞∈ −⎜ ⎟⎜ ⎟

⎝ ⎠ and

11 3

αα < therefore

10σ σ− > according to the inequality ( )∗∗ and to

the criteria in α . Therefore the family of function ( , )B vσ is a differentiable set in

the space [ )2

1,L +∞ depending on 1 1,12 2

σ σσ

⎛ ⎞∈ −⎜ ⎟⎜ ⎟

⎝ ⎠ and

[ )0

1

110 2 12

1( , ) ln 1, ,3

vB v e x L

xσ σ α

σσ α

+ −= − ∈ +∞ < .

Therefore, the tangent vector to 0

( , )B vσ is

[ )0

110 22

1B ( , ) ( ln ) 1,v

v e x Lx

σ σ ασ

+ −= − ∈ +∞ .

The vector determines one tangent plan to the set ( , )B vσ , after we apply the homothety factor [ ], 0,10ve v ∈ .

We consider 3 1

12

z it= + a zero on the critical axis 12

σ = .

In this case, the hyperplan t

H will have in common with the set

( , )B vσ the points of the form of 1 ,2

B v⎛ ⎞⎜ ⎟⎝ ⎠

.

What happens if g(z,v) will have the other two zeroes 1 1 0

z itσ= +

and 2 2 0

z itσ= + ?

In this case, the hyperplan t

H moves continuously by [ )2

1,L +∞

reaching the situation showed in Figure 1.

32

What happens at points 1 ,2

B v⎛ ⎞⎜ ⎟⎝ ⎠

, respectively ( , )B vσ and

2( , )B vσ ?

At these points, the hyperplans 1

tH and

0t

H are tangent to the set

( , )B vσ and this is the only possible situation, because if the hyperplan stings the set ( , )B vσ the continuous variation of

tH combined with the

continuity of the set ( , )B vσ will render us a continuous set of zeroes for the function g(z,v) and therefore for the analytical function ( )zζ what is ABSURD. Therefore the set ( , )B vσ is caught between the hyperplans

1t

H and 0

tH .

Therefore, at points 1

( , )B vσ and 2

( , )B vσ we have a bidimensio-

nal plan determined by the vectors tangent to the set ( , )B vσ , plan tangent to ( , )B vσ .

These vectors are:

[ )0

1 111 0 22

1( , ) ln 1,v

B v e x Lx

σ σ ασ

+ −= ∈ +∞ and

[ )0

2 112 0 22

1( , ) ln 1,v

B v e x Lx

σ σ ασ

+ −= − ∈ +∞

1 1,12 2

σ σσ

⎛ ⎞∈ −⎜ ⎟⎜ ⎟

⎝ ⎠ and 1

1 3

σα < .

33

This bidimensional plan is tangent to the set ( , )B vσ in 1

( , )B vσ

and 2( , )B vσ , and we will mark it with T. Let us show now that the vectors

1 0( , )B v

σσ and

2 0( , )B v

σσ are

independently linear vectors in T so they make up a basis for the plan T. We have to show that:

1 1 0 2 2 0( , ) ( , ) 0B v B v

σ σβ σ β σ+ = implies

1 20β β= = .

We have:

[ )0 0

1 1 2 11 11 22 2

1 1( ) ln ( ) ln 0, 1,v v

e x e x xx x

σ α σ αβ β

+ − + −− + − = ∀ ∈ +∞

So [ )1 1 2 1

1 11 22 2

1 1 0, 1,xx x

σ α σ αβ β

+ − + −+ = ∀ ∈ +∞

So [ )2 11 2

1 0, 1,xxσ σβ β −+ = ∀ ∈ +∞

So 2 12 1

1xσ σβ β− = − = constant, [ )1,x∀ ∈ +∞

So 2 12 1

1xσ σβ β− = − = constant, [ )

2 11, ,x σ σ∀ ∈ +∞ > and 1

1 3

σα < ,

1

12

σ < . So 2 1 1

1 02

σ σ α+ − − > .

We see that the last equality between the continuous unconstant

function 2 12

1xσ σβ − and –

1β on the interval [ )

11, , β+∞ − = constant leads

us to only one conclusion: 1 2

0 and 0β β= = .

So the vectors 1 0

( , )B vσ

σ and 2 0

( , )B vσ

σ are independently linear

vectors in T so they make up a basis for the T plan, an bidimensional plan. But the vector

1 0( , )B v Tσ ∈ .

If 0

1 111 0 1 02

( , ) , ( , )v

eB v T B v Tx

σ ασ σ

+ −∈ = ∈ then

1 2,l l∃ ∉ so that

1 0 1 1 0 2 2 0( , ) ( , ) ( , )B v l B v l B v

σ σσ σ σ= + so that

[ )0

0 0

1 1 1 1 2 11 1 11 22 2 2

1 1( ) ln ( ) ln , 1,v

v ve l e x l e x xx x x

σ α σ α σ α+ − + − + −= − ⋅ ⋅ + − ⋅ ⋅ ∀ ∈ +∞ .

34

That is [ )2 11 2

11 ln ln , 1,l x l x xxσ σ−= − − ∀ ∈ +∞

So [ )2 11 2

11 ln , 1,x l l xxσ σ−

⎡ ⎤= − + ∀ ∈ +∞⎢ ⎥

⎣ ⎦

So a constant function on [ )1, +∞ is equal to a continuous function

on [ )1, +∞ .

If we mark 2 11 2

1( ) lnU x x l lxσ σ−

⎡ ⎤= − +⎢ ⎥

⎣ ⎦.

We have 12 1 1 2 1 1

1 and so 03 2

σσ σ α σ σ α> < + − − > .

We calculate lim ( )x

U x→∞

.

Because 2 1 1

1 02

σ σ α+ − − > . We have the limit 2 1

2ln

lim 0x

l x

xσ σ−→∞= .

This happens because on the same Stage 9 we determined the inequality ( )∗∗ .

5ln , 0, 0 ( )rx x x M r< ∀ > > > ∗∗

So 2 1

1

1 2 1 1

1

0, if 0

1lim ( ) ln lim ln , if 0

, if 0

x x

l

U x x l l xl lx

l

σ σ−→∞ →∞

⎧ =⎪

⎡ ⎤ ⎪= − + = − = +∞ <⎨⎢ ⎥⎣ ⎦ ⎪

⎪−∞ >⎩

So the equality:

[ )2 11 2

11 ln , 1,x l l xxσ σ−

⎡ ⎤= − + ∀ ∈ +∞⎢ ⎥

⎣ ⎦ is impossible.

So there is no 1 2,l l ∈ so that

1 0 1 1 0 2 2 0( , ) ( , ) ( , )B v l B v l B v

σ σσ σ σ= + .

This means that the bidimensional plan 1

Tσ that is tangent to

the set ( , )B vσ in 1

( , )B vσ is not the same with the bidimensional

plan 2

Tσ that is tangent to the set ( , )B vσ in 2

( , )B vσ

Let us see now in what relation is the tangent bidimensional plan

1Tσ with the set ( , )B vσ . If this plan would be sting the set ( , )B vσ then

as we know that 10( )A t Tσ⊥ because

1 0tT Hσ ⊂ the continuous variation

35

of A(t) combined with the continuity of the set ( , )B vσ would lead us to a continuous set of zeroes, for the function g(z,v) so also for the analytical function ( )zζ . ABSURD.

If there would be any other tangential contact points like 3

( , )B vσ

with the set then we would have 1 3

T Tσ σ= but this is impossible because of the reasoning previously presented at Stage 9 that showed that

1 2T Tσ σ≠ .

So the plan 1

Tσ does not have in common with the set 1

( , )B vσ

only the points like 1

( , )B vσ the set ( , )B vσ being located completely

on one part of this plan. We put the set ( , )B vσ in a convex prismatic wrapper made by

bidimensional planes. Two faces of the convex prismatic wrapper will be included in the planes

1Tσ and

2Tσ .

This also means that the hyperplan 0t

H that is tangent to the set

( , )B vσ in 1

( , )B vσ does not have in common with the set ( , )B vσ only

points of form 1

( , )B vσ the set ( , )B vσ being located completely on

one side of this hyperplan 0t

H that divides the Hilbert space

[ )2

1,L +∞ into two semispaces.

Analogously, it is shown that the hyperplan 01tH that is tangent to

the set ( , )B vσ in the points like 2

( , )B vσ does not have in common with

the set ( , )B vσ but the points like 2

( , )B vσ the set ( , )B vσ being located

completely on one side of this hyperplan 01tH that divides the Hilbert

space [ )2

1,L +∞ into two semispaces.

So, in any of the points 1

( , )B vσ an 2

( , )B vσ the set ( , )B vσ

admits different support hyperplans. So the situation in Figure 1 in which we have the hyperplan

0tH

tangent to the set ( , )B vσ in 1

( , )B vσ and 2

( , )B vσ is impossible. See

Figure 2. So the function g(z,v) cannot have other two zeroes

2 1 0z itσ= +

and 2 1 0

z itσ= + symmetrical to the critical axis 1/ 2σ = .

36

Therefore, according to Stage 1 the function ( )zζ cannot have other two zeroes

2 1 0 2 1 0 and z it z itσ σ= + = + symmetrical to the critical

axis 1/ 2σ = . Q.E.D.

Summary Stage 1 The putting of the problem. Stage 2 We define the Hilbert space complex [ )

21,L +∞

Stage 3 We consider two certain zeroes symmetrical 1

z and 2

z

symmetrical to the axis 12

σ = .

Thus 1 1 0z itσ= + and 2 2 0z itσ= + where 2 11σ σ= − . We define the vectors ( , )B vσ and A(t) and it is shown that

[ )2( ) 1,A t L∈ ∞ , t∀ ∈ and [ )2( , ) 1,B v Lσ ∈ ∞ ,

37

[ ]1 1( , ) , 0,102 2

t σ σσ ⎛ ⎞∀ ∈ ×⎜ ⎟⎝ ⎠

.

Stage 4 We show that the family of functions A(t), t ∈ is a continuous set in the space [ )2 1,L ∞ .

Stage 5 We show that the family of functions ( , )B vσ is a continuous set in the space [ )2 1,L ∞ depending on

[ ]1 1( , ) ,1 0,102 2

v σ σσ ⎛ ⎞∈ − ×⎜ ⎟⎝ ⎠

.

Stage 6 We show that ( ) 0 tA t ≠ ∀ ∈ . We show that when the vector A(t) passes continuously the set in the space [ )2 1,L ∞ then the hyperplan tH perpendicular on A(t) also varies continuously as geometric position.

Stage 7 The Putting of Riemann’s Problem in Terms of Functional Analysis.

Stage 8 The form of the set ( , ) ( , )B v B vσ σ= . Stage 9 Final considerations on the zeroes of ( )zζ .

Bibliography Dicţionar de Analiză matematică, The Scientific and Encyclopedic

Publishing House, Bucharest, 1989. Analiză complexă, aspecte clasice şi moderne, The Scientific and

Encyclopedic Publishing House, Bucharest, 1988.

38

Chapter 4

The problem of invariant Subspaces The problem of invariant Subspaces is a quite important problem.

In this chapter, the author tries to find subtle connections between the theory of fixed point, the spectral theory and the existence of invariant lines for continuous linear operators on Hilbert spaces.

October 2005 by Vasiliu Lucilius

In the book Theorems and problems of Mathematical Analysis of

brothers Marius Radulescu and Sorin Radulescu, we have the following very strong results at page 87:

3.63. Let it be an even number and nS the sphere by the radius 1 with the center in the origin of 1n+ . Let it show that for any continuous function : n nf S S→ there is 0 nx S∈ so that

{ }10 0 0 0( ) or ( ) 1n

nf x x f x x S x x+= = − = ∈ =

We demonstrate the following theorem:

Theorem 1 Let it be H a separable Hilbert space. We mark by { }1 .HS x H x= ∈ =

Let it be a function : H Hf S S→ . F is strictly continuous on HS and in the same time f is weakly continuous on HS . Let it show that there is 0 Hx S∈ so that 0 0 0 0( ) or ( )f x x f x x= = − .

Demonstration We know that the ball unit is weakly compacted (Alaoglu’s

Theorem). It is considered an ε approximation of the sphere HS made up of

points that are taken on the sphere so as to make up spherical triangles with the side shorter than ε . This is possible because the sphere HS is a limited set.

39

The number of the points that make up for this ε approximation is finite. We shall mark by ix these points 1,i nε∈ . Let it be the points

( ) 1,i Hf x S i nε∈ ∈ . Considering the dimensional finite linear space generated by these

points and ( )i ix f x , and 1,i nε∈ and we should manage the conditions in such a way as to make this space to have an uneven dimension.

If it has an uneven dimension, we keep it like that. If not, let it be a point

oi Hx S∈ . We consider a spherical calotte

centered in 0i

x and with the diameter 10

d ε< . We show that there is a

point tx in this calotte that does not belong to the finite dimensional

linear space generated by the points ix and ( ), 1,if x i nε∈ . It would be impossible for all calotte points to belong to this space.

Why? Let us apply to the calotte the unit operators of the Hilbert space H. These operators have the property Ax x= . These operators applied to the spherical calotte from near to near built up the entire sphere HS . So the resulted infinite linear combinations generate the entire Hilbert space H. Therefore, a finite dimensional space would generate a infinite dimensional space. Which is absurd.

So, there is a point tx situated in a spherical calotte centered in 0i

x

and with the diameter 10

d ε< so that it does not belong to the linear

space generated by the points ix , ( )if x , 1,i nε∈ , dimensional finite space.

We mark this space by nHε

. The space nHε

is a finite dimensional

space of uneven dimension made up by points ix , ( ), 1,if x i nε∈ to

which was added eventually the point tx with 0 0, 1,

10t ix x i nεε

− < ∈ .

Now we make up a continuous spherical ε -approximation of a strongly continuous function : H Hf S S→ , f being in the same time a weakly continuous function from HS to HS . How shall we do it?

Let it be an ABCΔ with peaks on the sphere HS the points A, B, C

belonging to the ε -spherical approximation , 1,ix i nε∈ and the spherical sides being shorter than ε .

40

For the point 1 2 3A B Cα α α+ + with 2 2 3 1α α α+ + = and

1 2 3, , [0,1]α α α ∈ Considering 1 2 3( ) ( ) ( )f A f B f Cα α α+ + . Then we consider a center homotety HO that leads the points from ABCΔ on the spherical triangle ABC and the points of the flat triangle ( ) ( ) ( )f A f B f C on the spherical triangle ( ) ( ) ( )f A f B f C . In the case we also have the

point tx with 0 10i tx x ε

− < point that renders the uneven dimension of

the space nHε

we consider 0 0( ) ( ), 1,t if x f x i nε= ∈ .

We obtain in this way a continuous function :n nn H Hf S S

ε ε ε→ as

0ε → and nε → ∞ , we obtain a sequence of functions with the property

4( ) ( ) ( ), ( )n nf x f x x Hε ε

δ ε− ≤ ∈ ∗∗ nf εare strongly continuous

functions and they are also weakly continuous functions, because nHε

are finite dimensional spaces.

Let us work now on the estimation ( )∗∗ . We first notice that the norm on spaces nH

ε is the same as the norm on the Hilbert space H.

Let us be 1 2i i ix x x+ + a spherical triangle with the peaks 1 2, ,i i ix x x+ + situated on ε - spherical approximation.

Let it be x ∈ Δ spherical 1 2i i ix x x+ + the sides of the spherical triangle being shorter than ε .

We have:

1

2

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

n

n

n

i H

i H

i H

f x f x

f x f x

f x f x

ε

ε

ε

δ ε

δ ε

δ ε

+

+

− <

− <

− <

And ( ) 0 when 0δ ε ε→ → by the uniformly continuity of f on

nHSε. On the other hand 1 2 1 3 2( ) ( ) ( ) ( )n i i if x t f x t f x t f x

εα α α+ += + + and

1 2 3α α α= + +f( x ) f( x ) f( x ) f( x ) where 1 2 3 1α α α+ + = and

1 2 3, , [0,1]α α α ∈ , 1t > and 1t → when 0ε → , t being the homotety factor. It is obvious that from the continuity of f and from the fact that

1t → when 0ε → that

41

1 1 2

2 3

1 1( ) ( ) ( ), ( ) ( ) ( ) and

1( ) ( ) ( )

n n

n

i iH H

iH

f x f x f x f xt t

f x f xt

ε ε

ε

δ ε δ ε

δ ε

+

+

− < − <

− <

and 1 1 2 3( ) 0 ( ) and ( ) 0 when 0.δ ε δ ε δ ε ε→ → →

Therefore 4( ) ( ) ( ),n

n nHf x f x x H

ε εε

δ ε− < ∈ , where 4 ( ) 0δ ε →

when 0ε → . This is the estimation ( )∗∗ . This happens on each spherical triangle

with the side shorter than ε of ε -approximations. But all the functions ( ) :

n nn H Hf x S Sε ε ε

→ have the property that

0 nn Hx Sε ε

∃ so that 0 0 0 0( ) or ( )n n n n n nf x x f x xε ε ε ε ε ε

= = − .

Making 0 and n H Hx S Sε

∈ weakly compact, it results that we can extract from the sequence 0nx

ε a subsequence that converges weakly to a

point 0x α . But : H Hf S S→ is weakly continuous so passing to the limit on

this subsequence in the estimation ( )∗∗ we obtain that 0x Hα∃ ∈ so that

0 0 0 0( ) or ( )f x x f x xα α α α= = − . It is known that ( )nf x

ε is strongly continuous on the finite linear

space nHε

so it is also weakly continuous. Therefore we transported the result on Hilbert spaces. We now consider a continuous linear operator

:A H H→ . Being continuously linear, it transforms the lines that pass through the origin HO also in lines which pass by HO the transformation being continuous.

These lines intersect in the unit sphere { }1HS x H x= ∈ = .

Thus it is obtained a function :n H Hf S S→ , which is strongly continuous. But if A is linear strongly continuous it is also weakly continuous from the definition of weakly continuity. So the function

:A H Hf S S→ it is also a weakly continuous function. We are now able to demonstrate the following theorem:

42

Theorem 2 Let it be H a Hilbert separable space and :A H H→ a continuous

linear operator. Then this operator admits an unidimensional invariant subspace.

Demonstration Let it be the function :A H Hf S S→ , that is weakly continuous and

strongly continuous { }1HS x H x= ∈ = . We know according to the

Theorem 1 that 0 0 0 0 0 so that ( ) or ( )Hx S f x x f x xα α α α α∃ ∈ = = − . Therefore, there is a line 0 0Hx O xα α− that the operator A invaries.

We obtained therefore a line d that is invariant to the action of the continuous operator , ( )A A d d= . Q.E.D

Observation. By using estimation ( )∗∗ we can show that the sequence 0nx

ε is a Cauchy sequence in the norm of space H so passing to

the limit in the estimation ( )∗∗ we obtain that 0x Hα∃ ∈ , so that

0 0( )f x xα α= or 0 0( )f x xα α= − . In that manner we can renounce at the hypothesis of weakly continuity of the function f and we can renounce also at the using of the Alaoglu’s theorem.

43

Chapter 5

A study on Navier-Stokes equations. The Navier-Stokes equations represent a turning point in the field

of differential equations. The classical methods could not do much about it by now. Therefore it is necessary to find new efficient methods to reduce the difficulties of the problem. Such a method might be the using of the integral representation of self-adjoints operators’ formulae, which derives from a quite different domain from the one of differential equations of evolution, namely from the spectral theory of linear continuous operators and from the theory of measure. What this formula does? It manages to regularize the operator with continuous Lipschitz regularization, so that it shifts the problem in the frame of the classical theory which has been known until nowadays.

1st of September 2005, Vasiliu Lucilius

We shall try to demonstrate the following important theorem:

Theorem 1 Considering H a Hilbert space. Considering A an unlimited linear

self-adjoint operator. Then A can be approximated as good as possible with a series of Lipschitz continuous operators. It means that it exists a series of ,A Aλ λ continuous Lipschitz operators with the property that

when 0A Aλ λ→ → .

Demonstration In the book Initiations in the Theory of Linear Operators by

F.H.Vasilescu, we have the following theorem on page 109:

Theorem Considering ( )A H∈ an self-adjoint operator. It exists then a

spectral measure : ( ) ( )E Bor A H→ with the following properties:

2(1) ( ) : ( ) , :D A x H t d E t x x⎧ ⎫

= ∈ < ∞⎨ ⎬⎩ ⎭

∫

44

(2) ( ) ( ) , ( )A x tdE t x x D A⎛ ⎞

= ∈⎜ ⎟⎝ ⎠∫

The set ( )H is the set of closed operators. On page 103, we have the following Lemma: Lemma. Considering 1: ( )D H HΤ Τ ⊂ → a densely defined linear

operator. Then its ∗Τ adjunct is a closed linear operator. It results that every densely defined self-adjoint operator is closed.

Also, in the same book, on page 113, we have the formula ( ) ( ) ( ) ( ) ( ), where ( ) and ( )f A f t dE t L H A l H f B= ∈ ∗ ∈ ∈∫ .

We define the functions 1 2, :f f → :

1 2( ) sup( ,0) : ( ) inf( ,0)f t t f t t= = − . We mark with 1 2( ) : ( )A f A A f A+ −= = . We have 1 2( ) ( )f t f t t− = .

The operators A+ and A− are positive self-adjointss operators and A A A+ −= − . Because A A A+ −= − due to (∗ ) formula, we can approximate A and A+ − with the Yosida approximants. So we obtain a sequence of Aλ continuous Lipschitz linear operators so that

( ) ( )A x A xλ → when 0λ → . Q.E.D. In the book Functional Analysis by Romulus Cristescu , Didactical

and Pedagogical Publishing House, Bucharest, 1979, on page 170, we have the following

Lemma: Considering U an operator that applies in itself a complex Hilbert space. It is normal if and only if it can be represented as

1 2U V iV= + where 1 2 and V V are transferable self-adjoint operators. In

consideration of Theorem 1, we obtain

Theorem 2 Considering H a Hilbert space. Considering A an normal operator.

Then A can be approximated as good as possible with a sequence of continuous Lipschitz operators.

Demonstration The demonstration is relevant out of Theorem 1 and Lemma. We are trying to demonstrate Theorem 3.

45

Theorem 3 Considering H a Hilbert space. Considering A an unlimited linear

operator. Then A can be approximated as good as possible with a sequence of continuous Lipschitz linear operators.

Demonstration We are defining the operators

1 and A A A A∗ ∗Τ = − Τ = + we have 1 1A A A A∗ ∗ ∗∗ ∗Τ = + = + = Τ and A A A A∗ ∗ ∗∗ ∗Τ = − = − = −Τ .

So 1Τ is self-adjoints and Τ is normal. Applying theorems 1 and 2 it results that Τ and 1Τ each can be approximated with a sequence of continuous Lipschitz operators. So, also their sum can be approximated with a sequence of continuous Lipschitz operators. But Τ + 1Τ 2A= . So, A can be approximated with a sequence of continuous Lipschitz operators, meaning that it exists ( )A xλ continuous Lipschitz so that

( ) when 0A x Axλ λ→ → . We shall return now to the Navier-Stokes system.

(1)1

( , )

, 0

ni i

j i ij j i

n

u u pu u f x tt x x

x t

γ=

∂ ∂ ∂+ = Δ − +

∂ ∂ ∂

∈ ≥

∑

1

0 ( , 0) (2)n

n

i i

udivu x tx=

∂= = ∈ ≥

∂∑

with the initial conditions 0( ,0) ( ) ( ) (3)nu x u x x= ∈

Here, 0 ( )u x is a C∞ vector field of free divergence on n , ( , )if x t are the components of an applied given external force (for example,

gravitation, γ is a positive coefficient (viscosity) and 2

21

n

i ix=

∂Δ =

∂∑ is the

Laplacean. Euler’s equations are (1), (2), (3), with υ =0. For solutions which

are reasonable from a physical point of view, we have to be sure that ( , )u x t does not strongly increase when x → ∞ . This way we will focus

our attention to the f forces and to the initial 0u conditions.

46

30

3

( ) (1 ) on (4)

for any and k and

( , ) (1 ) on [0, ], for any ,m,k (5)

kx k

m kx t mk

u x C x

f x t C x t

α

α

αα

α

α

−

−

∂ ≤ +

∂ ∂ ≤ + + × ∞

We accept a solution of (1), (2), (3) as physically reasonable as long as it satisfies

, ( [0, )) (6)np u C∞∈ × ∞ and

n

2u(x,t) , for all t 0 (limited energy) (7)dx C< ≥∫

We have the problem (A) taken from The Clay Institute (5). Existence and smoothness of Navier-Stokes equation on 3 . Considering 0γ > and 3n = . Considering 0 ( )u x smooth, a vector

field of zero divergence satisfying (4). Considering ( , )f x t identical null. Then it exists the smooth function [ )3( , ), ( , ) on 0,ip x t u x t × ∞ , satisfying (1), (2), (3), (6), (7).

We define the Hilbert space in which we are operating. Consi-dering K a compact, nK ⊂ . Let be H²(K) and ( ) 32

1H H K⎡ ⎤= ⎣ ⎦ to which is added the divu=0 condition.

So

{ }22 1 2 3( , , ) , ( ), 1,3, 0t

iH u u u u u H K i divu= = ∈ ∈ = .

Now we consider p a given vector. We work on system (1).

We consider each operator type i

j

ux

∂∂

.

Taking into consideration Theorem 3 it exists series of continuous Lipschitz operators that approximates these operators for , 1,3i j ∈ . But,

by meanings of regularization, the ij

j

uux

∂∂

quantities become differen-

tiable on the space 2H . Applying the first theorem of finite increasing for these operators on Hilbert space 2H , we can deduce that these regulator operators are Lipschitz continuous on a set

{ }2 2( , ) . ( , ) ,B O R H B O R u H u⊂ = ∈ ≤ . So, equation (1) becomes a equation type

47

0

' ( )

( ,0) ( )λ= − ∇

=

u A u p

u x u x

where 1 2 3

, , , [0, ]t

p p pp tx x x

⎛ ⎞∂ ∂ ∂∇ = ∈ Τ⎜ ⎟∂ ∂ ∂⎝ ⎠

, p a given vector and A uλ

is a continuous Lipschitz regularization of operator

1

, 1,3n

ii j

j j

uu u i jx

γ=

∂Δ − ∈

∂∑ on set B(0,R)

2 , (0, )A u A v L u v u v B R Hλ λ− ≤ − ∀ ∈ ⊂ But these equations have an unique solution (0, )u B Rλ ∈ , which

can be obtained as a fixed-point of an contracting application defined on a metrical space that was conveniently chosen. This is a well-known classical theory. (4) page 352

But , 0A uλ λ → regulator operators form a Cauchy sequence.

(0, ) when - 0A u A u u B Rλ μ ε λ μ− < ∀ ∈ → . Out of Banach’s Theorem it results that even the fixed-points of

A uλ operators form a Cauchy sequence in ( , )B o R . So, when 0.u uλ λ∗→ → (3) page 29. So, Navier-Stokes system admits an

unique solution on [ ] { }20, (0, ) where (0, ) ,B R B R u H uΤ × = ∈ ≤ .

This unique solution extends on [ ) 20, H+∞ × . So, the system (1), (2), (3)

admits an unique solution on compact K taken arbitrary ⊂ nK . So, extending on compacts, it results that if we are to consider the p∇ vector given, the system (1), (2), (3) admits an unique solution u that depends on p .

Bibliography Initiation into the theory of linear operators, by Florian Horia

Vasilescu, Technical Publishing House. Functional Analysis by Romulus Criestescu, The Didactical and

Pedagogical Publishing House, Bucharest, 1979. Principle and Applications of Fixed Point Theory by Ioan A. Rus,

Dacia Publishing House, 1979. Partial Differential Equations by Marin Marin, Tehnical

Publishing House, 1998. The web address of Clay Institute www.claymath.org

48

Chapter 6

A STUDY ON FERMAT’S GRAND THEOREM

We consider the space 4 . Let be the hyperplan 8 4 2 0+ + − =x y z t 4in . We show that in this hyperplan there can’t exist points such as ( ), , ,p p p pA B C D where 5, , , ∗≥ ∈p A B C and

{ }0∗∈ ∪D . We assume, by reduction to the absurd, that the hyperplan

contains points of the form ( ), , , , , , ∗∈p p p pA B C D A B C and

{ }0 , 5∗∈ ∪ ≥D p , prime number.

We have 8 4 2 0p p p pA B C D+ + − = . Hence results that 12D D= . We have 18 4 2 2 0+ + − ⋅ =p p p p pA B C D or 1

14 2 2 0−+ + − ⋅ =p p p p pA B C D . Results that 12C C= . Hence 1 2

1 12 2 2 0p p p p p pA B C D− −+ + − = . Hence 12B B= . Results that

1 2 31 1 12 2 2 0.− − −+ + − ⋅ =p p p p p p pA B C D Hence 12=A A .

So 1 2 31 1 1 12 2 2 2 0p p p p p p p pA B C D− − −⋅ + + − = . By dividing with 32 p− ,

we obtain: 1 1 1 18 4 2 0, 5p p p pA B C D p+ + − = ≥ . Analogously we show that 1A is divided by 2, 1B is divided by 2,

1C is divided by 2, 1D is divided by 2. Recursively, we find that A is divided by 2 , k B is divided by 2k , C

is divided by 2k , D is divided by 2k , , hence 0k A B C D∀ ∈ = = = = . If D=0, then following the same reasoning, we show that

8 4 2 0p p pA B C+ + = implies 0A B C= = = , where we have considered 5 and , ,p A B C ∗≥ ∈ . Hence the hyperplan 48 4 2 0 in x y z t+ + − = cannot contain

points like ( ) { } , , , where 5, , , and p p p pA B C D p A B C D o∗ ∗≥ ∈ ∈ ∪ . In absolute analogy we show that the hyperplan

4 8 5 2 0x y z t− − − ⋅ − = does not contain points such as

( ), , ,p p p pA B C D , where 5p ≥ , prime number , , ∗∈A B C and

{ }0∗∈ ∪D .

49

But the intersect of these two hyperplan is a set ( ), , ,0x y z with the

property that ( )12 0x y z+ + = . But the intersect does not contain points

such as ( ), , ,p p p pA B C D , where 5p ≥ , { }, , and 0A B C D∗ ∗∈ ∈ ∪ .

Hence the plane ( )12 0x y z+ + = in 3 4⊂ does not contain

points such as ( ), ,p p pA B C 5p ≥ , ,A B C ∗∈ . As a result, the plan 0x y z+ + = can not contain points like

( ), ,p p pA B C , , ,A B C ∗∈ . Hence the equation p p pA B C+ = can not

have solutions in for 5p∗ ≥ . In an analogue manner, using the same type of reasoning, we can

study very many equations of the type 1

0,=

=∑s

pk k

ka x where ∗∈kx ,

α ∈k and 1,∈k s . It can be seen that this method, although it is elementary, surpasses

in mathematical power and deepens the extremely complicated algebraic methods, based on the group theory.

50

Chapter 7

WE WILL DEMONSTRATE THAT EULER-MASCHERONI’S CONSTANT IS AN IRRATIONAL NUMBER

We have the sequence 1 1 11 ... ln 2 3n n n

nγ ∗= + + + + − ∈ .

We know that 0,57nγ γ→ = … We will note:

[ ] { }1 1 11 ...2 3

+ + + + = = +n n ns s sn

and [ ] { }ln ln ln= +n n n .

It cab be easily observed that [ ] [ ] { }ln 0,1ns n− ∈ . So, in order to demonstrate the irrationality of γ we have to demonstrate first that the sequence { } { }lnn nu n s= − converges to an irrational number. We know

that ln n is an irrational number, so { }ln n is also an irrational number. It is known the following famous theorem that we will note it as

Th [1]. Th[1] Any rational number from the [0,1] interval can be written as

1 where 0, 1

!

ki

ii

a a ii

α=

= ∈ −∑

So any irrational number out of the [0,1] interval can be written as:

11 !

∞

=

= ∑α i

i

ai

where 0, 1ia i∈ − , and the sum is infinite.

So { } { } i1

ln where a 1!i

n ni

au n s ii

∞

=

= − = < −∑

Because [ ] [ ] { }ln 0,1ns n− ∈ , the nu sequence will also have two limits, taking into consideration the convergence of nγ .

We will demonstrate that these tow limits are irrational. We will

generally demonstrate than any 1

with 1!i

ii

a a ii

∞

=

< −∑ series (which

obviously is convergent) it converges to an irrational number. Let there be

51

1

1!

ni

n ii

ax a ii=

= < −∑

1 1

!

n mi

n m ii

ax a ii

+

+=

= < −∑

( ) ( ) ( ) ( ) ( ) ( )1 221 ... ...

1 ! 2 ! ! 1 ! 2 ! !n n n mn n m

n m n

a a aa aax xn n n m n n n m

+ + ++ +++ − = + + + ≤ + + + <

+ + + + + +

( ) ( )1 1 1 1 1. . ... .! 1 1 ! 2 1 !

+ + −≤ + + + ≤

+ + + + + +n n n m

n n n n n m n m

( ) ( )21 1 1 1 11 ... ...! 1 1 1

⎛ ⎞+ −≤ + + + + + =⎜ ⎟⎜ ⎟+ + + +⎝ ⎠

mn m

n n m n n n

1 1 1 1 1 1 1. . .1! ! !11

n m n m nn n m n n m n n

n

⎛ ⎞⎜ ⎟+ − + − +

= = <⎜ ⎟+ +⎜ ⎟−+⎝ ⎠

For the fixed n and m → +∞ we obtain:

110!nl x

n< − < , so, noting 10 11

!

nn

l x

n

θ −< = < , we obtain that:

1 21

1... .1! 2! ! !

nn

aa aln n

θ= + + + +

Where ( )n0 1 and 0,1i ia i aθ≤ < − ∈ ∈

If we assume that 1l ∈ , so 1 , ,ml m nn

∗ ∗= ∈ ∈ , then

1 2 1... .1! 2! ! !

nn

aa amn n n

θ= + + + + , where ( )0,1 , 1, 1n ia i nθ ∈ ∈ ∈ − .

By multiplying this equality with n! we obtain:

( ) 1 21 ! ! ...1! 2! !

⎛ ⎞− − + + + =⎜ ⎟⎝ ⎠

nn

a a am n nn

θ .

But this final equality is impossible because ( )0,1nθ ∈ and the left member represents an integer. So, the supposition that 1l ∈ has lead to an absurdity. So 1 \l ∈ . The same we proceed when it comes to show that 2 \l ∈ .

So, \γ ∈ .

52

TABLE OF CONTENTS PREFACE ...................................................................................... 5 Chapter 1. Fractional powers with rational exponent of the

monotonous operators of class C¹ on . .......................................... 7 Chapter 2. Using the symmetry group in demonstrating the

radiality of several semilinear biharmonic equations’ solutions. ... 13 Chapter 3. A study on zeroes of the function ( )zζ of

Riemann................................................................................................. 18 Chapter 4. The problem of invariant Subspaces ................... 38 Chapter 5. A study on Navier-Stokes equations.................... 43 Chapter 6. A study on Fermat’s grand theorem..................... 48 Chapter 7. We will demonstrate that Euler-Mascheroni’s

constant is an irrational number......................................................... 50

53

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