Prime conditions for Laplace transformsopab.web.fc2.com/no.25.pdfOperator algebras for Laplace transforms (No.1»4) Papers The projection operator for Laplace transforms Pascal’s

Prime conditions

for Laplace transforms

No.1

Takao Saito

Thank you for our world

Prime conditions

for Laplace transforms

No.1

Preface

Now, I explain about the prime conditions for the Laplace transforms.

In general, Now Laplace transform (T (0) operation) has been treated as

identities. So, we have been contained the property of prime number.

Textbook

Operator algebras for Laplace transforms (No.1∼4)

Papers

The projection operator for Laplace transforms

Pascal’s triangle matrix for Laplace transforms

The diagonal conditions for Laplace transforms

The ring conditions for Now Laplace transforms

The prime numbers for Laplace transforms

Now, let′s consider with me!

Address

695-52 Chibadera-cho

Chuo-ku Chiba-shi

Postcode 260-0844 Japan

URL: http://opab.web.fc2.com/index.html

(Mon) 5.Feb.2018

Takao Saito

Contents

Preface

§ Chapter 1

D(a) and Y (a) operation

◦ The relation of a0 and eas for D(a) and Y (a) operations · · · 7

F (a) and T (a) operation

◦ The relation of a0 and eas for F (a) and T (a) operations · · · 11

◦ Some results · · · · · · · · · 16

§ Chapter 2


◦ The relation of unitary and e0s for D(a) and Y (a) operations · · ·17


◦ The relation of unitary and e0s for F (a) and T (a) operations · · ·23

◦ Some results · · · · · · · · · 29

§ Chapter 3


◦ The relation of D(a) and D(0) operations · · ·30


◦ The relation of T (a) and T (0) operations · · ·33

◦ Some results · · · · · · · · · 37

◦ Conclusion · · · · · · · · · · · · 39

◦ References · · · · · · · · · · · · 41

§ Chapter 1

In this chapter, I explain about the relation of eas and a0 for Now

Laplace transform. In general, we have been obtained e0s = 00. So the

projective conditions of eas and a0 have this condition.

D(a) and Y (a) operations

Now, D(a) operations have defined as following forms.

D(a) = easD(a) = eas

(a0

a0

), O(a) = easO(a) = eas

(00 − a0

00 − a0

)

and

H(a) = D(a) + O(a) = eas

(00

00

)

Especially, D(0) operation has the following.

D(0) =

(00

00

)

On the other hand, Y (a) operations have defined as following forms.

Y (a)f(t) = easY(a)f(t) =∫ ∞

af(t)easdt , N(a)f(t) = easN (a)f(t) =

∫ a

0f(t)easdt

and

W (a)f(t) = Y (a)f(t) + N(a)f(t) =∫ ∞

0f(t)easdt

Especially, Y (0) operation has the following.

Y (0)f(t) =∫ ∞

0f(t)e0sdt

© The relation of a0 and eas of D(a) and Y (a) operations

7

Now, a0 has been existed as the element of the matrix of D(a) op-

eration. On the other hand, e0s has been existed as the kernel of Y(a)

operation. In this time, we have the following relation.

a0 = 00 = e0s

So, I am able to represent as follows.

D(a)iso←→ Y(a)

So I also have

D(a) = easD(a)iso←→ easY(a) = Y (a).

In a word,

D(a)iso←→ Y (a).

Especially, if a = 0 then we have

D(0)iso←→ Y (0).

In a word, this relation of

00 = e0s

has been lived in D(0) and Y (0) operations. And, we also have the

following by based the relation.

D(a)iso←→ Y (a)

In this time, this relation is able to extend to as follows.

a0 iso←→ eas

And, if we have eas = e0s then it also has

a0 = eas.

This relation has been applied to explain the jump up structures for

D(a) operations. In a word, a0 has been treated as D(0) operation. On

8

the contrary, eas has been treated as D(a) operation. Now, since a0 = eas

then we are able to have the following relation.

D(0) = D(a)

As a whole, we will have the following relations.

D(a)

P ↓ =

D(0) = Y (0)

= ↑ P

Y (a)

N.B. aP−→ 0.

P is projection operator.

In a word, D(0) operation is able to represent as Y (0) operation.

By this condition, we are able to also have

D(0) = 00 = e0s = Y (0).

Therefore, D(0) operation has been obtained as no-definition. So,

D(0) operation is able to fix to D(a) operation. In this time, we are able

to also have inclusion monomorphism in this structure. In a word, we

also have the following relation.

D(0)homo−→ D(a)

and it’s possible.

Aa a whole, we are able to have the following relations.

D(0a)iso←→ D(a)

So we are able to also have the following.

D(0) = D(a)

9

In a word, the poperty of D(0) operation has been preserved to the

property of D(a) operation. Especially, if the norm of D(0) operation is

one then we have the unital conditions for D(a) operations. In general,

if we have the following condition

‖D(0)‖ = ‖D(a)‖then D(0) operation does not necessarily to have unital condition because

it has no-defined. However, this D(0) operation has 1 and 0, a.e. So, we

have to obtain as follows.

D(0) = 1, 0.

On the contrary, we are able to also have

D(0) = eas = D(a).

In a word, we do not have always the necessity as D(0) = 1, 0. But if

it’s able to represent as 1 and 0 for all numbers then we are able to treat

only 1 and 0 in D(0) operation.

Similarly, we also have the following relation by taking the inclusion

monomorphism.

Y (0) = Y (a)

And the poperty of Y (0) operation has been preserved to the property

of Y (a) operation. Especially, if the norm of Y (0) operation is one then

we have the unital conditions for Y (a) operations. In general, if we have

the following condition

‖Y (0)‖ = ‖Y (a)‖then Y (0) operation does not necessarily to have unital condition because

it also has no-definition by e0s. However, this Y (0) operation has 1 and

0, a.e. So, we have to obtain as follows.

Y (0) = 1, 0.


Y (0) = eas = Y (a).

In a word, we do not have always the necessity as Y (0) = 1, 0. But if


only 1 and 0 in Y (0) operation.

10

F (a) and T (a) operations

Now, F (a) operations have defined as following forms.

F (a) = eas

(a0

a a0

), G(a) = eas

(00 − a0

−a 00 − a0

)

and

H(a) = F (a) + G(a) = eas

(00

00

)

On the other hand, T (a) operations have defined as the following

forms.

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt , S(a)f(t) =

∫ a

0f(t)e−(t−a)sdt

and

R(a)f(t) = T (a)f(t) + S(a)f(t) =∫ ∞

0f(t)e−(t−a)sdt

© The relation of a0 and eas of F (a) and T (a) operations

Now, a0 has been existed as the element of the matrix of F(a) op-

eration. On the other hand, e0s has been existed as the kernel of T (a)

operation. In this time, we have the following relation.

a0 = 00 = e0s

So, I am able to represent as follows.

F(a)iso←→ T (a)

So I also have

F (a) = easF(a)iso←→ easT (a) = T (a).

In a word,

F (a)iso←→ T (a).

11

Especially, if a = 0 then we have

F (0)iso←→ T (0).

In a word, this relation of

00 = e0s

has been lived in F (0) and T (0) operations. And, we also have the

following by based the relation.

F (a)iso←→ T (a)

In this time, this relation is able to extend to as follows.

a0 iso←→ eas

And, if we have eas = e0s then it also has

a0 = eas.

This relation has been applied to explain the jump up structures for

the Laplace transforms. In a word, a0 has been treated as the kernel of

Now laplace transforms (T (0) operation). On the contrary, eas has been

treated as the kernel of Extended Laplace transforms (T (a) operation).

Now, since a0 = eas then we are able to have the following relation.

T (0) = T (a)

As a whole, we will have the following relations.

F(a)

P ↓ =

F (0) = T (0)

= ↑ P

T (a)

N.B. aP−→ 0.

P is projection operator.

12

In a word, T (0) operation (Now Laplace transform) is able to rep-

resent as F (0) operation. So we have

T (0) = 00 = e0s.

Therefore, T (0) operation (Now Laplace transform) has been obtained

as no-definition. So, T (0) operation (Now Laplace transform) is able to

fix to T (a) operation (Extended Laplace transform). In this time, we are

able to also have inclusion monomorphism in this structure. In a word,

we also have the following relation.

T (0)homo−→ T (a)


Aa a whole, we are able to have the following relations.

T (0a)iso←→ T (a)

So we are able to also have the following.

T (0) = T (a)

In general, since T (a)iso←→ F(a)

iso←→ F (a) then T (a) operation is able

to replace to as follows.

T (a) →(

1

a 1

)← F(a)

iso←→ F (a)

In a word, the poperty of F (0) operation has been preserved to the

property of F (a) operation. Especially, if the norm of F (0) operation is

one then we have the unital conditions for F (a) operations. In general,

if we have the following condition

‖F (0)‖ = ‖F (a)‖

then F (0) operation does not necessarily to have unital condition because

it has no-defined. However, this F (0) operation has 1 and 0, a.e. So, we

have to obtain as follows.

F (0) = 1, 0.

13


F (0) = eas = F (a).

In a word, we do not have always the necessity as F (0) = 1, 0. But if


only 1 and 0 in F (0) operation.

Similarly, we also have the following relation by taking the inclusion

monomorphism.

T (0) = T (a)

And the poperty of T (0) operation has been preserved to the property

of T (a) operation. Especially, if the norm of T (0) operation is one then

we have the unital conditions for Y (a) operations. In general, if we have

the following condition

‖T (0)‖ = ‖T (a)‖then T (0) operation does not necessarily to have unital condition because

it also has no-definition by e0s. However, this T (0) operation has 1 and

0, a.e. So, we have to obtain as follows.

T (0) = 1, 0.


T (0) = eas = T (a).

For example, we are able to obtain the following relation.

T (a)T (b) = easebs = e(a+b)s = T (a + b) (Semi− group)

In a word, we do not have always the necessity as T (0) = 1, 0. But if


only 1 and 0 in T (0) operation.

Since this flow of D(a) to T (a) operations then I have the following

relation,

T (a)homo−→ D(a).

14

In a word, the Extended Laplace transforms (T (a) operation) is able

to represent by D(a) operation. Therefore we are able to have the re-

lation of a0 and eas. Moreover, this D(a) operation is able to also be

D(0) operation. So we have the conception of T (a) operation has been

obtained the property of 00 = e0s. In other words, if we have T (a) opera-

tion (Extended Laplace transform) then we should have the property as

follows.

a0 = 00 = e0s = eas

In a word, these operators has been treated as D(a), D(0), T (0) and

T (a) operations, resp.

a0 ←→ D(a) , 00 ←→ D(0)

e0s ←→ T (0) , eas ←→ T (a)

In general, a0 has been obtained the isomorphic conditions with eas.

However, the norms of a0 and eas are not same. In a word, we should

have the following relations.

a0 6= 00a = e0as = eas as a 6= 0

N.B. ‖00a‖ = ‖e0as‖ = ‖eas‖

In this time, since a0 is able to be 1 then we are able to use to unitary

condition. And a0 and eas are able to treat as Now Laplace transform

(T (0) operation) and Extended Laplace transform (T (a) operation), resp.

On the contrary, if a = 0 of a0 then we should have the following relation.

T (0) = T (a)

15

Some results

◦ We have the following relation.

00 = e0s

◦ The 00 has been obtained as follows.

a0 = 00

◦ Since 00 has no-definition then we also have no-definition of e0s.

◦ So, we are able to be

eas = e0s = 00 = a0.

◦ In this time, we have the following condition.

T (a)iso←→ D(0)

◦ T (a) operation (Extended Laplace transform) is able to represent as

D(0) operation. In a word,

T (a)iso←→

(00

00

).

16

§ Chapter 2

In this chapter, I explain about the prime conditions for Now Laplace

transfors. The Now Laplace transforms are able to treat as unitary con-

ditions. So, we have one element. On the contrary, the one element does

not include in prime numbers.



D(a) = eas

(a0

a0

), O(a) = eas

(00 − a0

00 − a0

)

and

H(a) = D(a) + O(a) = eas

(00

00

)


D(0) =

(00

00

)


Y (a)f(t) =∫ ∞

af(t)easdt , N(a)f(t) =

∫ a

0f(t)easdt

and

W (a)f(t) = Y (a)f(t) + N(a)f(t) =∫ ∞

0f(t)easdt

© The relation of unitary and e0s for D(a) and Y (a) operations

The norm of umitary has 1. On the contrary, e0s is able to have any

numbers as e0s = 00. So, e0s also have 1. In a word, e0s also has unitary.

17

In other words, e0s has been extended from unitary. In general, we have

the following relation.

eas = e0s

In a word, D(0) operation is able to be to D(a) operation. In a word,

D(a) = D(0)

as a0 = 00.

And it’s possible.

So, in general, we don’t need to define as e0s = 1. In this time, eas

is able to take the prime number. So e0s dose not have prime number as

1. (N.B. Peas = e0s.) These things will be able to combine each other.

e0s also has prime number because of eas = e0s. However, if e0s = 1 then

it’s not prime number. On the contrary speaking, e0s has any numbers

without 1. In a word, prime numbers are not able to be unitary if it’s

now conditions. In this time, I recommend to decompose 1 as 1a. N.B.

eas is prime number. So, 0a conditions has been also existed each other.

In a word, we have the following relation.

1aiso←→ 0a

iso←→ a

Now, in general, e0s has been obtained as any numbers. So, e0s does

not need only one and zero. So e0s has been extended from unitary

conditions. If D(0) operation has defined on unitary conditions then

I do not need the relation of prime number. However, if it has been

defined on e0s then we should have the conception of prime numbers.

So, e0s also has “1” element. The D(0) operation of e0s has been also

called as “Now Laplace transform”. Similarly, e0s is able to also have

“0” element. And it’s possible. As a whole, e0s has been generated from

1 and 0 elements. In a word, these are identities and characteristics

condition. In this time, e0s has proper. Of course, e0s is able to also

have other numbers. In a word, e0s has been obtained representative

elements as 1 and 0. So, we should consider the relation of 1 of prime

number if 1 is able to be prime number and 1 of representative element

as multiplicative identity. Similarly, additive condition also has same. In

general, this 1 has a property of kernel. In a word, the prime number

18

has been generated itself without kernel. Similarly, if this number has 0

then we have annihilation. So, the prime number has been also generated

itself without this annihilation.

In this time, we have the following relation.

not prime number ≡ prime number

So, since 1 and 0 elements do not have prime number then 1 and 0 ele-

ments are able to also include in prime number. So, 1 and 0 elements are

able to be representative elements of prime numbers. In other words, the

all prime numbers are able to decompose at the representative elements.

In a word,

1aiso←→ 0a

iso←→ eas (A)

eas is prime number.

This relation are able to also explain with the following relation.

KerP = Im⊥P

Precisely speaking,

KerP = Im⊥P ∗

in complex spaces.

In this case, since P has prime numbers then we have P ∗ = P in

complex spaces. In a word, upper relation (A) has been treated in Banach

spaces.

Now D(a) operations have the following.

D(a) = easD(a) = eas

(a0

a0

)

So we have

D(a) =

(a0

a0

)

In this time, the D(a) operation also has

D(a) =

(1

1

)and

(0

0

)

19

as a0 = 1, 0.

In this time, it’s identities of D(0) operation.

In a word, the diagonal of D(a) operation has been extended from 1

with zero. In a word,

a0 = 1, 0 for all a.

In general, since 00 has been no defined then we are able to have all

numbers as a = 0,∞. In fact, we have 00 = ∞0 because it has the

following.

00 = 0−0 = {00}−1 =1

00=

10

00= {1

0}0 = ∞0

In a word, the previous form has special case. In general, we should

have as following relation.

a0 = 00


So I am able to represent as follows.

D(a) =

(00

00

)

In this time, the 00 are able to have independently. And this repre-

sentation has been extended from original form of D(a) operation. The

extension of this theory is the following.

a0 → 00

on D(a) operation.

Especially, if a = 0 then we should have the following form.

D(0) =

(00

00

)= F (0)

iso←→ T (0)

In a word, T (0) operation (Now Laplace teransform) is able to also

represent as D(0) = F (0) operation.

20

In this time, I also have the following forms.

(1

1

),

(1

0

),

(0

1

),

(0

0

)

These forms have been obtained the projective conditions. In a word,

D(0) operations are able to extend to the projective conditions.

N.B. In general, I’m able to have all numbers for 00.

On the other hand, Y(a) and Y(0) = Y (0) operations are able to

have the following relation.

Y(a) ≈ Y(0)

Because of

Y(a) =∫ ∞

ae0s1dt ≈

∫ ∞

0e0s2dt = Y(0).

Precisely speaking, it’s able to be

Y(a) = Y(0).

In a word, we are able to have the following.

I(a) = Y(a) =∫ ∞

ae0s1dt =

∫ ∞

0e0s2dt = Y(0) = I(0)

N.B. I(a), I(b) are identities.


If we consider on Y(a) operations then we should have the following

relation, at least.

Y(a)iso←→ Y(0a)

In a word, a element has been existed in zero. So we are able to also

have

Y(a) = Y(0a)


21

In this time, if we consider the D(a) operations then we have

D(a) = D(0)

The background of this relation has the following form.

Y(a)iso←→ D(a)

Moreover, we have been obtained

Y (a)iso←→ D(a).

22



F (a) = eas

(a0

a a0

), G(a) = eas

(00 − a0

−a 00 − a0

)

and

H(a) = F (a) + G(a) = eas

(00

00

)


forms.

T (a)f(t) =∫ ∞


∫ a

0f(t)e−(t−a)sdt

and

R(a)f(t) = T (a)f(t) + S(a)f(t) =∫ ∞

0f(t)e−(t−a)sdt

© The relation of unitary and e0s for F (a) and T (a) operations

The norm of umitary has 1. On the contrary, e0s is able to have any

numbers as e0s = 00. So, e0s also have 1. In a word, e0s also has unitary.

In other words, e0s has been extended from unitary. In general, we have


eas = e0s

In a word, T (0) operation (Now Laplace transforms) is able to be to

T (a) operation (Extended Laplace transforms). In a word,

T (a) = T (0)


So, in general, we don’t need to define as e0s = 1. In this time, eas

is able to take the prime number. So e0s dose not have prime number as

23

1. (N.B. Peas = e0s.) These things will be able to combine each other.

e0s also has prime number because of eas = e0s. However, if e0s = 1 then

it’s not prime number. On the contrary speaking, e0s has any numbers

without 1. In a word, prime numbers are not able to be unitary if it’s

now conditions. In this time, I recommend to decompose 1 as 1a. N.B.

eas is prime number. So, 0a conditions has been also existed each other.


1aiso←→ 0a

iso←→ a

Now, in general, e0s has been obtained as any numbers. So, e0s does

not need only one and zero. So e0s has been extended from unitary

conditions. If the Lapalce transform has defined on unitary conditions

then I do not need the relation of prime number. However, if it has been

defined on e0s then we should have the conception of prime numbers.

So, e0s also has “1” element. The Laplace transform of e0s has been

also called as “Now Laplace transform”. Similarly, e0s is able to also

have “0” element. And it’s possible. As a whole, e0s has been generated

from 1 and 0 elements. In a word, these are identities and characteristics

condition. In this time, e0s has proper. Of course, e0s is able to also

have other numbers. In a word, e0s has been obtained representative

elements as 1 and 0. So, we should consider the relation of 1 of prime

number if 1 is able to be prime number and 1 of representative element

as multiplicative identity. Similarly, additive condition also has same. In

general, this 1 has a property of kernel. In a word, the prime number

has been generated itself without kernel. Similarly, if this number has 0

then we have annihilation. So, the prime number has been also generated

itself without this annihilation.

In this time, we have the following relation.

not prime number ≡ prime number

So, since 1 and 0 elements do not have prime number then 1 and 0 ele-

ments are able to also include in prime number. So, 1 and 0 elements are

able to be representative elements of prime numbers. In other words, the

all prime numbers are able to decompose at the representative elements.

24

In a word,

1aiso←→ 0a

iso←→ eas (A)

eas is prime number.

This relation are able to also explain with the following relation.

KerP = Im⊥P

Precisely speaking,

KerP = Im⊥P ∗

in complex spaces.

In this case, since P has prime numbers then we have P ∗ = P in

complex spaces. In a word, upper relation (A) has been treated in Banach

spaces.

Now F (a) operations have the following.

F (a) = easF(a) = eas

(a0

a a0

)

So we have

F(a) =

(a0

a a0

)

In this time, the original form of F(a) operation has

F(a) =

(1

a 1

)

In a word, the diagonal of F(a) operation has been extended from 1

to a0 with zero. In a word,

a0 = 1, 0 for all a.

In general, since 00 has been no defined then we are able to have all

numbers as a = 0,∞. In fact, we have 00 = ∞0 because it has the

following.

00 = 0−0 = {00}−1 =1

00=

10

00= {1

0}0 = ∞0

25

In a word, the previous form has special case. In general, we should

have as following relation.

a0 = 00


So I am able to represent as follows.

F(a) =

(00

a 00

)

In this time, the 00 are able to have independently. And this repre-

sentation has been extended from original form of F(a) operation. The

extension of this theory is the following.

1 → a0 → 00

on the diagonal of F(a) operation.

Especially, if a = 0 then we should have the following form.

F (0) =

(00

00

)= D(0)

iso←→ T (0)

In a word, T (0) operation (Now Laplace teransform) is able to also

represent as D(0) = F (0) operation.

In this time, I also have the following forms.

(1

1

),

(1

0

),

(0

1

),

(0

0

)

These forms have been obtained the projective conditions. In a word,

Now Laplace transforms (T (0) operations) are able to extend to the pro-

jective conditions.

If 0 → a then I present as follows.

(1

a 1

),

(1

a 0

),

(0

−a 1

),

(0

−a 0

)

26

These forms have the projective conditions for T (a) operations (Ex-

tended Laplace transforms). And most left form has been obtained Pas-

cal’s triangle matrix. For example, if it’s 3 orderd matrix condition then

we have

F(a) =

1

a 1

a2 2a 1

.

In general, if I cosider the n-th orderd Pascal’s matrix conditions then

we have been obtained as follows.

F(a)

F(a) =

1

a 1

a2 2a 1

a3 3a2 3a 1...

......

.... . .

an−1n−1C1a

n−2n−1C3a

n−3n−1Cka

n−1−k 1

In general, T (a) and T (0) = T (0) operations are able to have the

following relation.

T (a) ≈ T (0)

Because of

T (a) =∫ ∞

ae−(t−0)sdt ≈

∫ ∞

0e−(t−0)sdt = T (0).

Precisely speaking, it’s able to be

T (a) = T (0).

In a word, we are able to have the following.

I(a) = T (a) =∫ ∞

ae−(t−0)sdt =

∫ ∞

0e−(t−0)sdt = T (0) = I(0)

N.B. I(a), I(b) are identities.

27

If we consider on F(a) operations then we should have the following

relation, at least.

F(a)iso←→ F(0a)

In a word,

F(a) =

(1

a 1

)iso←→

(1

0a 1

)= F(0a)

In a word, a element has been existed in zero. So we are able to also

have

F(a) = F(0a)


Ref. Since we have this relation then we are able to operate the rank

form. In a word, this content does not change in spite of the form has

been changed.

Especially, if this has been been treated on eigenspaces then we are

able to have equivalence. In this time, this relation has been represented

as follows.

F(a) = F(0)

The background of this relation has the following form.

T (a)iso←→ F(a)

Moreover, we have been obtained

T (a)iso←→ F (a).

28

Some results

◦ Now we have

eas = e0s.

◦ T (a) operation has the following relation.

T (a) = T (0)

◦ Not prime number ≡ Prime number

◦ In general,

aiso←→ 0a

iso←→ 1a

◦ Especially, if eas is prime number then we have

eas iso←→ 0aiso←→ 1a

◦ T (a) operation also has the following relation.

T (a) = T (0)

29

§ Chapter 3

In this chapter, I explain about the relation of T (a) and T (0) opera-

tions. In this time, we have been obtained as D(a) = D(0). So we will

have T (a) = T (0) and it’s possible.



D(a) = eas

(a0

a0

), O(a) = eas

(00 − a0

00 − a0

)

and

H(a) = D(a) + O(a) = eas

(00

00

)


D(0) =

(00

00

)


Y (a)f(t) =∫ ∞

af(t)easdt , N(a)f(t) =

∫ a

0f(t)easdt

and

W (a)f(t) = Y (a)f(t) + N(a)f(t) =∫ ∞

0f(t)easdt

© The relation of D(a) and D(0) operations

Now, D(a) operation has been obtained as follows.

D(a) =

(a0

a0

)

30

In this time, if a < 0 and a > 0 in finite numbers then we have as

follows.

a0 = 1.

So we have

a0 = 1.

On the contrary, if a = 0 and a →∞ then we have

a0 = 00 = ∞0

and it’s no-defined. This form has been obtained as follows.

D(0) =

(00

00

)

In a word, D(0) operation has a part of D(a) operation. So we have


D(0) ⊆ D(a)

On the other hand, since 00 has no-definition then 00 is able to adopt

all numbers. So we also have

D(0) ⊇ D(a).


D(0) = D(a)

In a word,

D(a) = D(0).

In this time, the norm of D(a) operations are able to have one. In a

word,

‖D(a)‖ = ‖D(0)‖ = 1.

Moreover, we also have

a0 = 00 (a 6= 0).

And it has the following relation.

D(a) = easD(a) = easD(0) = H(a)

So we have

D(a) = H(a)

31

as a0 = 00.

Moreover, if we have eas = e0s then we also have

D(a) = D(0).

So, the core form for D(a) operations has the following conditions.

D(a) = D(0)

Similarly, since Y(a)iso←→ D(a) then we should have the following

relation.

Y(a) = Y(0)

So we have

Y(a) =∫ ∞

ae0sdt =

∫ ∞

0e0sdt = Y(0).

In a word,

I(a) =∫ ∞

adt =

∫ ∞

0dt = I(0)

Therefore,

I(a) = I(0).

N.B. ‖I(a)‖ = ‖I(0)‖ = 1.

I(a) and I(0) are identities.

In a word, if we consider Y(a) operation then we should have the

following condiion.

I(a) = I(0)

Moreover, this relation has been applied to as follows.

Y (a) = easY(a) = easY(0) = W (a)

So we have

Y (a) = W (a)

as I(a) = I(0).


Y (a) = Y (0).

So, the core form for Y (a) operations has the following conditions.

Y(a) = Y(0)

32



F (a) = eas

(a0

a a0

), G(a) = eas

(00 − a0

−a 00 − a0

)

and

H(a) = F (a) + G(a) = eas

(00

00

)


forms.

T (a)f(t) =∫ ∞


∫ a

0f(t)e−(t−a)sdt

and

R(a)f(t) = T (a)f(t) + S(a)f(t) =∫ ∞

0f(t)e−(t−a)sdt

© The relation of T (a) and T (0) operations

Now, F(a) operation has been obtained as follows.

F(a) =

(a0

a a0

)

In this time, if a < 0 and a > 0 in finite numbers then we have as

follows.

a0 = 1.

So we have

a0 = 1.

On the contrary, if a = 0 and a →∞ then we have

a0 = 00 = ∞0

33

and it’s no-defined. This form has been obtained as follows.

F(0) =

(00

00

)

In a word, F(0) operation has a part of F(a) operation. So we have


F(0) ⊆ F(a)

On the other hand, since 00 has no-definition then 00 is able to adopt

all numbers. So we also have

F(0) ⊇ F(a).

N.B. F (a) operations have been dominated by the diagonal.


F(0) = F(a)

In a word,

F(a) = F(0).

In this time, the norm of F(a) operations are able to have one. In a

word,

‖F(a)‖ = ‖F(0)‖ = 1.

Moreover, we also have

a0 = 00 (a 6= 0).

And it has the following relation.

F (a) = easF(a) = easF(0) = H(a)

So we have

F (a) = H(a)

as a0 = 00.


F (a) = F (0)

34

N.B. a0 = 00.

So, the core form for F (a) operations has the following conditions.

F(a) = F(0)

This F (a) operation has been originated from Pascal’s triangle matrix.

Similarly, since T (a)iso←→ F(a) then we should have the following

relation.

T (a) = T (0)

In fact, we have as follows.

T (a) =∫ ∞

ae−(t−0)sdt = e−as = e0s =

∫ ∞

0e−(t−0)sdt = T (0)

N.B. eas = e0s.

So we have

T (a) = T (0).

In a word,

I(a) =∫ ∞

adt =

∫ ∞

0dt = I(0)

Therefore,

I(a) = I(0).

N.B. ‖I(a)‖ = ‖I(0)‖ = 1.

I(a) and I(0) are identities.

In a word, if we consider T (a) operation then we should have the

following condiion.

I(a) = I(0)

Moreover, this relation has been applied to as follows.

T (a) = easT (a) = easT (0) = R(a)

So we have

T (a) = R(a)

as I(a) = I(0).

35


T (a) = T (0).

In a word, we are able to treat as follows.

Extended Laplace transform︸︷︷︸T (a) operation

= Now Laplace transform︸︷︷︸T (0) operation

as I(a) = I(0).

And, the core form for T (a) operations has the following conditions.

T (a) = T (0)

36

Some results

◦ The fundamental conception of T (a) operations has the following rela-

tion.

T (a) = T (0)

N.B. It is not necessary to be always a = 0.

◦ In this time, we need the followig relation.

I(a) =∫ ∞

adt =

∫ ∞

0dt = I(0)

N.B. I(a) and I(0) are identities.

◦ This T (a) operation has prime conditions for Extended Laplace trans-

forms (T (a) operation)

◦ This T (0) operation has prime conditions for Now Laplace transforms

(T (0) operation)

◦ T (a) operation (Extended Laplace transform) has the following relation.

T (a) = R(a)

N.B. R(a) = easT (0).

◦ Moreover, if we have eas = e0s then we also have

T (a) = T (0).


In a word, we are able to treat as follows.



as I(a) = I(0).

37

　

38

Conclusion

Now, we have the following relation.

eas = e0as

In this time, eas is part of the kernel for Extended Laplace transform

(T (a) operation) and e0as is part of the kernel for Now Laplace transform

(T (0a) operation). In a word, we have the following condition.

T (a)iso←→ T (0a)

The next step, we have the jump up condition from T (0) operation

(Now Laplace transform) to T (a) operations (Extended Laplace trans-

form). In this time, we have the following relation.

T (a) = easT (0)

In this time, since eas has been defined as prime number then we were

able to have the following relation.

eas iso←→ 1aiso←→ 0a

By this means, we should consider that the prime numbers include the

1 and 0 elements.

Finally, we were able to have the following relation.

T (a) = T (0)

Since this fundamental form then we have

T (a) = easT (a) = easT (0) = e0sT (0) = T (0)

N.B. eas = e0s.

So, we are able to treat as follows.



as I(a) = I(0).

(Sat) 17.Feb.2018

Now let′s go to the next papers with me!

39

　

40

References

[1] Bryan P.Rynne and Martin A.Youngson, Springer, SUMS, Linear

functional analysis, Springer, SUMS, 2001.

[2] Micheal O Searcoid, Elements of Abstract Analysis, Springer, SUMS,

2002.

[3] Israel Gohberg and Seymour Goldberg, Basic operator theory,

Birkhauser, 1980.

[4] Prime Numbers and the Riemann Hypothesis, Barry Mazur, William

Stein, Cambridge University Press, 2016.

[5] P.M.Cohn, Springer, SUMS, An Introduction to Ring Theory.

[6] M.A.Naimark, Normed Rings, P.Noordhoff,Ltd, 1959.

[7] Gareth A.Jones and J.Mary Jones, Elementary Number THeory,

Springer, SUMS, 2006.

Furthermore

[8] Kosaku Yoshida, Functional Analysis, Springer, 1980.

[9] Irina V.Melnikova, Abstract Cauchy problems, Chapman, 2001.

[10] Paul L.Butzer Hubert Berens, Semi-groups of Operators and Ap-

proximation, Springer, 1967

[11] Hille and Philips, Functional analysis and semi-Groups, AMS, 1957.

[12] Richard V. Kadison John R.Ringrose Fundamentals of the theory of

Operator Algebras, AMS

[13] Dunford & Schwartz Linear operators I,II,III, Wiley.

[14] Charles E.Rickart, General theory of Banach algebras, D.Van Nos-

trand Company,Inc,1960.

[15] J.L.Kelley · Isaac Namioka Linear topological spaces, D.Van Nos-

trand Company,Inc, 1961.

[16] Gareth A.Jones and J.Mary Jones, Elementary number theory,

Springer, 2006.

[17] Paul R.Halmos, Measure Theory, D.Van Nostrand Company,Inc,

1950.

[18] R.B.J.T. Allenby, Rings,Fields and Groups, Edward Arnold, 1989.

41

Documents

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