Approach from modulus for Laplace transformsopab.web.fc2.com/modulus.pdfApproach from modulus for Laplace transforms No.1 Takao Saito 1 Thank you for our world 2 Approach from modulus

Approach from modulus

for Laplace transforms

No.1

Takao Saito

1

Thank you for our world

2

Approach from modulus

for Lapalace transforms

3

Preface

This paper is builded by three parts. First is representations for

Laplace transforms by modulus. Second is the applications for opera-

tor algebras. Finally, it’s the σ-algebras for Laplace transforms.

Papers

About solvers of differential equations of relatively

for Laplace transforms

The projection operators for Laplace transforms

The negative condition for Laplace transforms

The ring conditions for Laplace transforms

The field conditions for Laplace transforms

The ring conditions for Now Laplace transforms

The field conditions for Now 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

(Wed)20.Nov.2013

Takao Saito

4

Contents

Preface

§ Chapter 1

◦ Representation of D(a) and O(a) operations by modulus · · · 7◦ Representation of Y (a) and N(a) operations by modulus · · ·11

◦ Representation of F (a) and −G(a) operations by modulus · · ·15

◦ Representation of T (a) and −S(a) operations by modulus · · ·19

◦ Some results · · · · · · · · · · · · 23

§ Chapter 2

◦ Application to D(a) and O(a) operations · · · · · ·24

◦ Application to Y (a) and N(a) operations · · · · · ·28

◦ Application to F (a) and G(a) operations · · · · · ·33

◦ Application to T (a) and S(a) operations · · · · · ·37

◦ Some results · · · · · · · · · · · · 42

§ Chapter 3

◦ Modulo in e0s or 00 for D(a) operations · · · · · ·43

◦ Modulo in e0s or 00 for Y (a) operations · · · · · ·44

◦ Modulo in e0s or 00 for F (a) operations · · · · · ·46

◦ Modulo in e0s or 00 for T (a) operations · · · · · ·48

◦ Some results · · · · · · · · · · · · 50

5

◦ Conclusion · · · · · · · · · · · · 51

◦ References · · · · · · · · · · · · 52

6

Chapter 1

In this chapter, I have been explained about representations of operator

algebras on modulus. We are able to obtain that these operator algebras

(D(a), Y (a), F (a) and T (a) operations) have equivalent conditions, resp.

Especially, T (a) operation is extended Laplace transforms.

© Representation of D(a) and O(a) operations by modulus.

Now, we have following form.

D(a) = eas ·D(0) + {−O(a)}

So

D(a) − { −O(a)} = eas ·D(0).

In this time, D(a), O(a) and D(0) operations are defined as following

form.

D(a) = easD(a) = eas

(a0

a0

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

(

0

)= −O(a) (1)

and

D(0) = e0s

(00

00

). (2)

Therefore, we are able to obtain two kinds representation by modulus.

D(a) ≡ O(a) (mod eas) or D(a) ≡ O(a) (mod D(0)).

N.B. Zero condition of matrix in O(a) operation is stronger than eas.

Moreover, since D(a) = easD(a) , O(a) = easO(a), resp, then we

have

D(a) ≡ O(a) (mod e0s = 00) or D(a) ≡ O(a) (mod D(0)).

7

So we have

D(a) = easD(0) + {−O(a)}.and

D(a) = D(0) + {−O(a)}Of course, D(a) operation has a property of ring condition. In this

time, Since −O(a) = O(a), we are able to say in other words that O(a)

operation has a property of ideal condition. So O(a) operation is subring

of D(a) operation. Both D(a) and O(a) operations have common term

on modulus. In this case, this common term is eas and it’s modulus.

In a words, eas is modulus. Therefore, D(a) and O(a) operations have

equivalence on mod eas.

D(a) ≡ O(a) (mod eas)

Similarly, D(a) and O(a) operations have equivalence on (mod e0s =

00).

D(a) ≡ O(a) (mod 1).

N.B. This mod 1 is able to extend to e0s = 00. In this case, this

modulus is no-defined.

Especially, since O(a) = 0, we have following conditions.

D(a) ≡ 0 (mod D(0))

In this time, we are able to be to H(a) operation from D(a) operation.

N.B. H(a) operation has been definded as following form.

H(a) = easD(0) = eas

(00

00

)(3)

So, H(a) operation has been preserved with obtaining the property of

D(0) operation. In a word, H(a) operation has a property of D(0) oper-

ation. This D(0) operation is able to relate with Now Laplace transforms

deeply and have characteristics condition. In this time, if we consider

σ-ring of D(0) operation then we will be able to understand, clearly. In

general, the relation of D(a) and H(a) operations are homomorphic. So

D(a) = easD(a)homo−→ easD(0) = H(a).

8

Similarly,

D(a)homo−→ D(0) = H(a).

Therefore, the previous special case has inclusion monomorphic be-

tween D(a) and H(a) operations. So

D(a)iso←→ H(a).

Similarly,

D(a)iso←→ H(a).

Clearly, now we have following conditions.

D(a)− {−O(a)} = H(a) = easD(0) or D(a) ≡ O(a) (mod D(0)).

Similarly

D(a) ≡ O(a) (mod D(0)).

This means that the property ofD(a) operation andO(a) operation are

same. In general, D(a) operation has identity. So we will have following

condition.

{1} ≡ {0} (mod D(0))

In this case, since O(a) = 0 then we have D(a) = H(a). This forms

are able to extend on Now Laplace transforms. Furthermore these op-

erations are able to also represent by characteristics condition. Because

D(a) (O(a) = 0) and H(a) operations are containing D(0) operation.

Especially, in this case, D(a) operation is able to consider on D(0) op-

eration. And D(0) operation is possible to be characteristics condition.

This condition has closed relation with Now Laplace transforms and σ-

algebras. And it will generate rings, fields and groups of D(a) and H(a)

operations. For example, the primitive ring (subring) of D(a) operation

is generated by D(0) operation, etc. This relation is able to represent by

projection operator. So we have following form.

PD(a) = D(0) or PH(a) = H(0) = D(0).

or

D(0) = H(0)P←− D(a)

P ↖ l iso

H(a)

9

N.B. It’s able to consider as isomorphic theorem because of O(a) = 0.

In general, D(a) operation is homomorphic for H(a) operation. In a

word,

D(a)homo−→ H(a).

So we have following formation.

D(0) = H(0)P←− D(a)

P ↖ ↓ homo

H(a)

Especially, if eas has irreducible conditions then we are able to only

consider both unitary matrix condition D(0) = H(0) and extended ma-

trices condition {D(a)} or {H(a)}. In fact,

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

and

D(0) = D(0)homo←− D(a).

N.B. This homomorphism coincides with projection.

10

© Representation of Y (a) and N(a) operations by modulus.


Y (a) = eas · Y (0) + {−N(a)}

So

Y (a) − { −N(a)} = eas · Y (0).


Y (a) ≡ −N(a) (mod eas) or Y (a) ≡ −N(a) (mod Y (0)).

Moreover, since Y (a) = easY(a) , N(a) = easN (a),resp, then we have

Y(a) ≡ −N (a) (mod e0s = 00) or Y(a) ≡ −N (a) (mod Y(0)).

N.B. Y (a), N(a) and Y (0) operations are defined as following form.

Y (a)f(t) = easY(a)f(t) =∫ ∞

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

∫ a

00easdt = −N(a)f(t) = 0

and

Y (0)f(t) = e0sY(0)f(t) =∫ ∞

0f(t)e0sdt , resp.

In this time, especially, since f(t) = 0 on 0 ≤ t ≤ a then we have

following conditions.

Y (a) = easT (0) + {−N(a)} = easY(0) + N(a)

and

Y(a) = Y(0) + {−N (a)} = Y(0) +N (a)

So we have

Y (a)−N(a) = easY(0) and Y(a)−N (a) = Y(0).

Now, Y (a) operation has a property of ring condition. N(a) operation

has a property of ideal condition. So N(a) operation is subring of Y (a)

operation. Both Y (a) and N(a) operations have common term on mod-

ulus. In this case, this common term is eas and it’s modulus. In a words,

11

eas is modulus. Therefore, Y (a) and N(a) operations have equivalence

on mod eas. Therefore

Y (a) ≡ N(a) (mod eas).

Similarly, Y(a) andN (a) operations have equivalence on (mod e0s =

00).

Y(a) ≡ N (a) (mod 1).



Especially, Since N(a) = 0 then we have following conditions.

Y (a) ≡ 0 (mod Y (0))

In this time, we are able to be to W (a) operation from Y (a) operation.

N.B. W (a) operation has been definded as following form.

W (a)f(t) = easY (0)f(t) =∫ ∞

0f(t)easdt.

So, W (a) operation has been preserved with obtaining the property

of Y (0) operation. In a word, W (a) operation has a property of Y (0)

operation. This Y (0) operation is normal integral operation and it has

characteristics condition. In this time, if we consider σ-ring of Y (0) oper-

ation then we will be able to understand, clearly. In general, the relation

of Y (a) and W (a) operations are homomorphic. So

Y (a) = easY(a)homo−→ easY (0) = W (a).

Similarly,

Y(a)homo−→ Y (0) = W(a).


tween Y (a) and W (a) operations. So

Y (a)iso←→ W (a).

Similarly,

Y(a)iso←→W(a).

12


Y (a)−N(a) = W (a) = easY (0) or Y (a) ≡ N(a) (mod Y (0)).

In this time, since N(a) = 0 then we have Y (a) = W (a). This forms

are applied to extended Now Laplace transforms. Furthermore these op-


Y (a) (N(a) = 0) and W (a) operations are containing Y (0) operation. In

other words, normal integral operation is able to have the characteristics

condition. Especially, in this case, Y (a) operation is able to be Y (0)

operation. So, Y (a) operation has characteristics condition χ(a). This

condition has closed relation with σ-algebras. And it will generate rings,

fields and groups of Y (a) and W (a) operations. For example, the primi-

tive ring (subring) of Y (a) operation is generated by Y (0) operation, etc.

This relation is able to also represent by simple projection operator. So

we have following form.

PY (a) = Y (0) or PW (a) = W (0) = Y (0).

or

Y (0) = W (0)P←− Y (a)

P ↖ l iso

W (a)

N.B. It’s able to consider as isomorphic theorem by N(a) = 0.

In general, Y (a) operation is homomorphic for W (a) operation. In a

word,

Y (a)homo−→ W (a).


Y (0) = W (0)P←− Y (a)

P ↖ ↓ homo

W (a)

13


consider both unitary condition (Y (0) or Y(a)) for Now Laplace trans-

forms {T (0)} and {W (a)} or {Y (a)} operation, resp. In fact,

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

and

Y(0) = Y (0)homo←− Y(a).


14

© Representation of F (a) and −G(a) operations by modulus.


F (a) = eas · F (0) + {−G(a)}

So

F (a) − { −G(a)} = eas · F (0).


F (a) ≡ −G(a) (mod eas) or F (a) ≡ −G(a) (mod F (0)).

Moreover, since T (a) = easT (a) , S(a) = easS(a),resp, then we have

F(a) ≡ −G(a) (mod e0s = 00) or F(a) ≡ −G(a) (mod F(0)).

N.B. F (a), G(a) and F (0) operations are defined as following form.

F (a) = easF(a) = eas

(a0

a a0

), G(a) = −

(

a

)(4)

and

F (0) = F(0) =

(00

00

)= D(0) = D(0). (5)

→ refer No.7 paper.

So we have

F (a) = easF (0) + {−G(a)}.and

F(a) = F(0) + {−G(a)}Now, F (a) operation has a property of ring condition. −G(a) oper-

ation has a property of ideal condition. So −G(a) operation is subring

of F (a) operation. Both F (a) and −G(a) operations have common term

on modulus. In this case, this common term is eas and it’s modulus. In

a words, eas is modulus. Therefore, F (a) and −G(a) operations have


F (a) ≡ −G(a) (mod eas)

15

Similarly, F(a) and−G(a) operations have equivalence on (mod e0s =

00).

F(a) ≡ −G(a) (mod 1).



In general, since G(a) operation has been ideal then we have following

conditions.

F (a) ≡ 0 (mod F (0))

In this time, we are able to be to H(a) operation from F (a) operation.

N.B. H(a) operation has been definded as following form.

H(a) = easF (0) = eas

(00

00

)= easD(0) = H(a) (6)

So, H(a) operation has been preserved with obtaining the property

of F (0) operation. In a word, H(a) operation has a property of F (0) op-

eration. This F (0) operation is isomorphic with Now Laplace transforms

and it has characteristics condition. In this time, if we consider σ-ring of

F (0) operation then we will be able to understand, clearly. In general,

the relation of F (a) and H(a) operations are homomorphic. So

F (a) = easF(a)homo−→ easF (0) = H(a).

Similarly,

F(a)homo−→ F (0) = H(a).


tween F (a) and H(a) operations. So

F (a)iso←→ H(a) iff G(a) = {0}.

Similarly,

F(a)iso←→ H(a) iff G(a) = {0}.

N.B. This {} means set condition. This set condition has related with

σ-algebras.

16


F (a)− {−G(a)} = H(a) = easF (0) or F (a) ≡ −G(a) (mod F (0)).

In this time, especially, since G(a) = {0} then we have F (a)homo→ H(a).

And F (a) operation is Pascal’s triangle matrix. Furthermore these op-


F (a) ({G(a)} = 0) and H(a) operations are containing F (0) operation.

Especially, in this case, F (a) operation is able to be F (0) operation. So,

this Pascal’s triangle matrix (F (a) operation) has characteristics condi-

tion χ(a). This condition has closed relation with σ-algebras. And it will

generate rings, fields and groups of F (a) and H(a) operations. For exam-

ple, the primitive ring (subring) of F (a) operation is generated by F (0)

operation, etc. This relation is able to represent by projection operator.

So we have following form.

PF (a) = F (0) or PH(a) = H(0) = F (0).

or

F (0) = H(0)P←− F (a)

P ↖ l iso

H(a)

N.B. It’s able to consider as isomorphic theorem by {G(a)} = 0.

In general, F (a) operation is homomorphic for H(a) operation. In a

word,

F (a)homo−→ H(a).


F (0) = H(0)P←− F (a)

P ↖ ↓ homo

H(a)

Especially, if eas has irreducible conditions then we are able to only con-

sider both unitary condition F (0) operation and Pascal’s triangle marix

{H(a)} or {F (a)}. In fact,

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

17

and

F(0) = F (0)homo←− F(a).


18

© Representation of T (a) and −S(a) operations by modulus.


T (a) = eas · T (0) + {−S(a)}

So

T (a) − { −S(a)} = eas · T (0).


T (a) ≡ −S(a) (mod eas) or T (a) ≡ −S(a) (mod T (0)).

Moreover, since T (a) = easT (a) , S(a) = easS(a), resp, then we have

T (a) ≡ −S(a) (mod e0s = 00) or T (a) ≡ −S(a) (mod T (0)).

N.B. T (a), S(a) and T (0) operations are defined as following form.

T (a)f(t) = easT (a)f(t) =∫ ∞

af(t)e−(t−a)sdt , S(a)f(t) = easS(a)f(t) =

∫ a

0f(t)e−(t−a)sdt

and

T (0)f(t) = e0sT (0)f(t) =∫ ∞

0f(t)e−(t−0)sdt , resp.

So we have

T (a) = easT (0) + {−S(a)}.and

T (a) = T (0) + {−S(a)}Now, T (a) operation has a property of ring condition. −S(a) oper-

ation has a property of ideal condition. So −S(a) operation is subring

of T (a) operation. Both T (a) and −S(a) operations have common term

on modulus. In this case, this common term is eas and it’s modulus.

In a words, eas is modulus. Therefore, T (a) and −S(a) operations have


T (a) ≡ −S(a) (mod eas)

19

Similarly, T (a) and−S(a) operations have equivalence on (mod e0s =

00).

T (a) ≡ −S(a) (mod 1).



Especially, if S(a) = 0 then we have following conditions.

T (a) ≡ 0 (mod T (0))

In this time, we are able to be to R(a) operation from T (a) operation.

N.B. R(a) operation has been definded as following form.

R(a)f(t) = easT (0)f(t) =∫ ∞

0f(t)e−(t−a)sdt.

So, R(a) operation has been preserved with obtaining the property

of T (0) operation. In a word, R(a) operation has a property of T (0)

operation. This T (0) operation is Now Laplace transforms and it has

characteristics condition. In this time, if we consider σ-ring of T (0) oper-

ation then we will be able to understand, clearly. In general, the relation

of T (a) and R(a) operations are homomorphic.

So

T (a) = easT (a)homo−→ easT (0) = R(a).

Similarly,

T (a)homo−→ T (0) = R(a).


tween T (a) and R(a) operations.

So

T (a)iso←→ R(a) iff S(a) = 0.

Similarly,

T (a)iso←→R(a) iff S(a) = 0.

20


T (a)− {−S(a)} = R(a) = easT (0) or T (a) ≡ −S(a) (mod T (0)).

In this time, especially, if S(a) = 0 then we have T (a) = R(a). This

forms are extended Now Laplace transforms. Furthermore these Laplace

transforms are able to also represent by characteristics condition. Because

T (a) (S(a) = 0) and R(a) operations are containing T (0) operation. Es-

pecially, in this case, T (a) operation is able to be T (0) operation. So, My

Laplace transforms (T (a) operation) has characteristics condition χ(a).

This condition has closed relation with σ-algebras. And it will generate

rings, fields and groups of T (a) and R(a) operations. For example, the

primitive ring (subring) of T (a) operation is generated by T (0) operation,

etc. This relation is able to represent by projection operator. So we have

following form.

PT (a) = T (0) or PR(a) = R(0) = T (0).

or

T (0) = R(0)P←− T (a)

P ↖ l iso

R(a)

N.B. It’s able to consider as isomorphic theorem iff S(a) = 0.

In general, T (a) operation is homomorphic for R(a) operation.

In a word,

T (a)homo−→ R(a).


T (0) = R(0)P←− T (a)

P ↖ ↓ homo

R(a)

21


consider both unitary condition (Now Laplace transforms {T (0)} and

extended Now Laplace transforms {R(a)} or {T (a)}.

In fact,

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

and

T (0) = T (0)homo←− T (a).


22

Some results

◦ Now, T (a) operations obtain two kinds representation by modulus.

T (a) ≡ −S(a) (mod eas) or T (a) ≡ −S(a) (mod T (0)).

N.B. T (0) operation is Now Laplace transforms. So ‖T (0)‖ = 1.

◦ Moreover, since T (a) = easT (a) , S(a) = easS(a), resp, then we have

T (a) ≡ −S(a) (mod e0s = 00) or T (a) ≡ −S(a) (mod T (0)).

◦ The properties of T (a) operation and S(a) operation have same condi-

tions.

◦ We are able to be T (a)iso←→ R(a) iff it’s inclusion monomorphic.

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

◦ The relation of between {1} and {0} is equivalence. Therefore

{1} ≡ {0} (mod T (0))

◦ (mod eas) will treat on C∗-algebras, etc.

23

Chapter 2

In this chapter, I have explained the application to these operator

algebras. In this time, these operator algebras have preserved on a =

0 → ∞ by using (mod e∞s) = (mod e0s). My Hilbert or Banach

spaces are also discussing on this area.

© Application to D(a) and O(a) operations

Now, we have following condition.

D(a) ≡ O(a) (mod e0s = 00)

If a = 0 then we have

D(0) ≡ O(0) (mod e0s = 00).

Similarly, if a = ∞ then we have

D(∞) ≡ O(∞) (mod e0s = 00).

So we are able to have following important condition.

{1} = D(0) ≡ O(0) = {0} = D(∞) ≡ O(∞) = {1} (mod e0s = 00)

N.B. see, No.1, No.11 papers.

Therefore, we obtain

{1} ≡ {0} (mod 00 = e0s).

Since this condition, I have been considered as set condition {1}, {0}.This form is special case for D(a) and O(a) operations. So

D(0) ≡ D(∞) (mod e0s = 00)

24

and

O(0) ≡ O(∞) (mod e0s = 00).

This means that D(0) operation has same property with D(∞). Sim-

ilarly, O(0) operation has same property with O(∞). In a word, D(a)

and O(a) operations are able to discuss on Hilbert spaces.

Similarly, we have following conditions.

D(a) ≡ O(a) (mod eas)

So

D(0) ≡ O(0) (mod e0s = 00)

and

D(∞) ≡ O(∞) (mod e∞s = ∞0)

Now, let’s e0s = e∞s. So we obtain

{1} = D(0) ≡ O(0) = {0} = D(∞) ≡ O(∞) = {1} (mod e0s = e∞s).

Therefore

D(0) ≡ D(∞) (mod e0s = e∞s)

and

O(0) ≡ O(∞) (mod e0s = e∞s).

This means that D(0) operation (a kind of Now Laplace transform)

is preserved by the property of D(∞) operation (a kind of Extended

Laplace transforms). In a words, D(0) operation is able to extend to

D(a) operation.

(see, p.10, No.11)

On the other hand, D(0) operation is able to represent as charactristic

condition χ(0). This characteristic condition has two forms as {1} and

{0}. However, now we have following condition.

{1} ≡ {0} (mod e0s = e∞s).

25

This means that characteristic conditions {1} and {0} are same prop-

erties each other. So we are able to treat two forms equally. For example,

projection operator P and Q have the properties, etc. In other words,

P ≡ Q (mod e0s = 00).

(see, Chapter 1, No.1)

N.B. Q ≡ −Q (mod e0s = 00). Because if P is {1} condition then we

have Q = {0} = −Q. So

P − {−Q} = P −Q = {1}.

Therefore we have previous condition.

Furthermore, we are able to have following conditions.

D(a)iso←→ P , O(a)

iso←→ Q , resp.

So we have,

D(a) ≡ O(a) (mod e0s)

and

D(a) ≡ O(a) (mod eas) or D(a) ≡ O(a) (mod e0s).

This means that D(a) operation has same property for O(a) operation

modulo e0s and D(0) operation.

N.B. H(a) = easD(0) , D(a) + O(a) = H(a) and

(mod eas) = (mod e0s).

Especially, if a is infinite conditions then we have following forms.

(mod e∞s) ≡ (mod e0s).

So we have

D(a) ≡ O(a) (mod e0s = e∞s) for all a.

26

and

D(a) ≡ O(a) (mod e0s = e∞s) for all a.

So we will have following forms.

D(a)iso←→ D(a) , O(a)

iso←→ O(a).

Now I have been considered to distinguished as D(a) is ring and O(a)

is ideal , resp. However these operations D(a and O(a) have a same

property modulo e0s. This thing means that modulo e0s has common

ring for D(a) and O(a) operations. This common ring is called primitive

ring. If the primitive ring is seen from ideal structure then the ideal

structure has been generated the maximal ideal or ring boundary. This

ideal structure (O(a) operation) and ring condition (D(a) operation)have

equivalent. This structure of D(a) operations are generated as following

forms.

D(a) =

H(a)︷︸︸︷eas · D(0) + {−O(a)}

p = q · r + s

↓ ↓ ↘ ↓ring coefficient primitive ring ideal


{1} = D(0) ≡ O(0) = {0} = D(∞) ≡ O(∞) = {1} (mod e0s = 00)

Therefore, these operations D(a) and O(a) will be treated on set con-

ditions. So, it’s algebras. Especially, we have following conditions.

D(0) ≡ O(∞) (mod e0s = 00).

Similarly,

{1} = D(0) ≡ O(0) = {0} = D(∞) ≡ O(∞) = {1} (mod e0s = e∞s).

So

D(0) ≡ O(∞) (mod e0s = e∞s).

27

N.B. D(0) = D(0).

On the contrary, we have following conditions.

D(∞) ≡ O(0) (mod e0s = 00).

and

D(∞) ≡ O(0) (mod e0s = e∞s).

N.B. D(∞) = D(∞).

This condition D(0) is applied to Now Laplace transforms. In general,

this operation is extended to ′′a′′ condition. So the operation D(a) is

extended for New Laplace transforms. Essentially, both D(0) and D(a)

operations have same properties. Now, the process of between D(0) and

D(a) operations have only explanation. It is that the operation of between

D(0) operation and D(a) operation has been null condition. In other

words, clearly, it means that O(a) operation has null condition. When it

has this condition, we are able to consider as irreducible condition. In this

case, it is dominated by how to treat the coefficient of D(0) operation.

So it’s that eas is prime number, etc. In this time, in general, we have

following condition.


Therefore the coefficient is characteristics. If e0s has closed condition

then we should that eas also consider on closed condition. On the con-

trary, if e0s has open condition then we should that eas also consider on

open condition. For example, if we consider D(0) and D(∞) operations

only then it will be irreducible condition. In this case, D(a) operations

are possible to generate irreducible conditions between D(0) and D(∞)

operations iff D(a) operation has prime condition.

© Application to Y (a) and N(a) operations


Y(a) ≡ N (a) (mod e0s)

28


Y(0) ≡ N (0) (mod e0s = 00).


Y(∞) ≡ N (∞) (mod e0s = 00).


{1} = Y(0) ≡ N (0) = {0} = Y(∞) ≡ N (∞) = {1}


{1} ≡ {0} (mod e0s = 00).

Since this condition, I have been considered as set condition {1}, {0}.This form is special case for Y(a) and N (a) operations. So

Y(0) ≡ Y(∞) (mod e0s = 00)

and

N (0) ≡ N (∞) (mod e0s = 00).

This means that Y(0) operation has same property with Y(∞). Sim-

ilarly, N (0) operation has same property with N (∞).


Y (a) ≡ N(a) (mod eas)

So

Y (0) ≡ N(0) (mod e0s = 00)

and

Y (∞) ≡ N(∞) (mod e∞s = ∞0)


{1} = Y (0) ≡ N(0) = {0} = Y (∞) ≡ N(∞) = {1} (mod e0s = e∞s).

29

Therefore

Y (0) ≡ Y (∞) (mod e0s = e∞s)

and

N(0) ≡ N(∞) (mod e0s = e∞s).

This means that Y (0) operation for Now Laplace transform is pre-

served by the property of Y (∞) operation for Extended Laplace trans-

forms. In a words, Y (0) operation (normal integral operation) is able to

extend to Y (a) operation.

(see, p.14, No.11)

On the other hand, Y (0) operation is able to represent as charactristic

condition χ(0). This characteristics condition is generated from e0s and it

has two forms as {1} and {0}. However, now we have following condition.

{1} ≡ {0} (mod e0s = e∞s).




P ≡ Q (mod e0s = 00).


N.B. Q ≡ −Q (mod e0s = 00). Because, now, we have N(a) = {0}.So we are able to have that Q puts as {0}. In a word,

P − {−Q} = P −Q = {1}.Therefore we have previous condition.

Now, we have following conditions.

Y(a)iso←→ P , N (a)

iso←→ Q , resp.

So we have,

Y(a) ≡ N (a) (mod e0s)

and

Y (a) ≡ N(a) (mod eas) or Y (a) ≡ N(a) (mod e0s).

30

N.B. In this time, N(a) = {0} = −N(a).

This means that Y (a) operation has same property for N(a) oper-

ation modulo e0s and Y (0) operation.

N.B. W (a) = easY (0) , Y (a) + N(a) = W (a) and




So we have

Y(a) ≡ N (a) (mod e0s = e∞s) for all a.

and

Y (a) ≡ N(a) (mod e0s = e∞s) for all a.


Y(a)iso←→ Y (a) , N (a)

iso←→ N(a).

Now I have been considered to distinguished as Y (a) is ring and N(a)

is ideal , resp. However these operations Y (a) and N(a) have a same


ring for Y (a) and N(a) operations. This common ring is called primitive


structure has been generated the maximal ideal or ring boundary.

Reference

Y (a) =

W (a)︷︸︸︷eas · Y (0) + {−N(a)}

p = q · r + s


31

In general, we are able to be

Y(0)f(t) = N (∞)f(t).

Because of

Y(0)f(t) =∫ ∞

0f(t)dt = N (∞)f(t).

If N (∞) to N(∞) operations then we must have following condition.

Y (0)f(t) = N(∞)f(t) iff e0s = e∞s.

Surely, we have following form.

Y (0)f(t) =∫ ∞

0f(t)e0sdt =

∫ ∞

0f(t)e∞sdt = N(∞)f(t).

Therefore, these operations Y (a) and N(a) will be treated on set

conditions. So, it’s algebras. Especially, in this case, we have following

conditions.

Y(0) = N (∞)

and

Y (0) = N(∞) iff e0s = e∞s.

N.B. Y(0) = Y (0).

This condition Y (0) is applied to Now Laplace transforms. In general,

this operation is extended to ′′a′′ condition. The operation Y (a) is applied

to New Laplace transforms. Essentially, both Y (0) and Y (a) operations

have same properties. Only difference is the process of between Y (0) and

Y (a) operations have two different explanations. One is the operation

of between Y (0) operation and Y (a) operation has been null condition.

In other words, N(a) operation has null condition. When it has this

condition, we are able to consider as irreducible condition. And other is

the operation of between Y (0) operation and Y (a) operation has been

dense condition. In other words, N(a) operation has dense condition.

When it has this condition, we are able to consider as reducible condition.

This classification is dominated by how to treat the coefficient of Y (0)

operation. So it’s eas. In this time, we have following condition.


32

Therefore the coefficient is characteristics. If it has closed condition

then we should consider on irreducible condition. On the contrary, if it

has open condition then we should consider on reducible condition. For

example, if we consider Y (0) and Y (∞) operations only then it will be

irreducible condition and if it also contains with Y (a) (0 < ‖a‖ < ∞)

operation then it will be redicible condition. Of course, Y (a) operation

is possible to also generate irreducible condition between Y (0) operation

iff Y (a) operation has prime conditions.

© Application to F (a) and G(a) operations


F(a) ≡ −G(a) (mod e0s)


F(0) ≡ −G(0) (mod e0s = 00).


F(∞) ≡ −G(∞) (mod e0s = 00).


{1} = F(0) ≡ −G(0) = {0} = F(∞) ≡ −G(∞) = {1}


{1} ≡ {0} (mod e0s = 00).

Since this condition, I have been considered as set condition {1}, {0}.This form is special case for F(a) and G(a) operations. So

F(0) ≡ F(∞) (mod e0s = 00)

and

−G(0) ≡ −G(∞) (mod e0s = 00).

This means that F(0) operation has same property with F(∞). Sim-

ilarly, −G(0) operation has same property with −G(∞).

33


F (a) ≡ −G(a) (mod eas)

So

F (0) ≡ −G(0) (mod e0s = 00)

and

F (∞) ≡ −G(∞) (mod e∞s = ∞0)


{1} = F (0) ≡ −G(0) = {0} = F (∞) ≡ −G(∞) = {1} (mod e0s = e∞s).

Therefore

F (0) ≡ F (∞) (mod e0s = e∞s)

and

−G(0) ≡ −G(∞) (mod e0s = e∞s).

This means that F (0) operation is preserved by the property of F (∞)

operation. In a words, unitary of Pascal’s triangle matrix is able to extend

to the F (a) operation.

(see, p.18, No.11)

On the other hand, F (0) operation is able to represent as charactristic



{1} ≡ {0} (mod e0s = e∞s).




P ≡ Q (mod e0s = 00).

34


N.B. Q ≡ −Q (mod e0s = 00). Because

Let Q =

(0

1

)(7)

So we have

Q− {−Q} =

(0

1

)−

(0

−1

)= 2

(0

1

)= 2

(00

00

). (8)

Therefore

Q ≡ −Q (mod e0s = 00).


F(a)homo−→ P , G(a)

homo−→ Q , resp.

Moreover, F(a) and G(a) operations have a property of inclusion

monomorphism for projection operators. So we have,

F(a) ≡ G(a) (mod e0s)

and

F (a) ≡ G(a) (mod eas) or F (a) ≡ G(a) (mod e0s).

This means that F (a) operation has same property for G(a) operation

modulo e0s and F (0) operation.

N.B. H(a) = easF (0) , F (a) + G(a) = H(a) and




So we have

F(a) ≡ G(a) (mod e0s = e∞s) for all a.

35

and

F (a) ≡ G(a) (mod e0s = e∞s) for all a.


F(a)iso←→ F (a) , G(a)

iso←→ G(a).

Now I have been considered to distinguished as F (a) is ring and G(a)

is ideal , resp. However these operations F (a) and G(a) have a same


ring for F (a) and G(a) operations. This common ring is called primitive


structure has been generated the maximal ideal or ring boundary.

Reference

F (a) =

H(a)︷︸︸︷eas · F (0) + {−G(a)}

p = q · r + s


In this time, we are able to be following condition.

F(0) = G(∞)

and

F (0) = G(∞) iff e0s = e∞s.

N.B. T (0) = T (0).

For example,

‖F(0)‖ = ‖(

00

00

)‖ = ‖00‖ = 1 = ‖ −

(

∞

)‖ = ‖G(∞)‖. (9)

So we have

F(0) = G(∞).

36

N.B. F(0) = F (0) = D(0) = D(0).

This condition F (0) is treated on diagonal conditions for Pascal’s

triangle matrix. In general, this operation is extended to ′′a′′ condition.

The operation F (a) is Pascal’s triangle matrix. Essentially, both F (0) and

F (a) operations have same properties. Only difference is the process of

between F (0) and F (a) operations have two different explanations. One

is the operation of between F (0) operation and F (a) operation has been

null condition. In other words, G(a) operation has null condition (G(a) =

{0}). When it has this condition, we are able to consider as irreducible

condition. And other is the operation of between F (0) operation and F (a)

operation has been dense condition. In other words, G(a) operation has

dense condition. When it has this condition, we are able to consider as

reducible condition. This classification is dominated by how to treat the

coefficient of F (0) operation. So it’s eas. In this time, we have following

condition.





example, if we consider F (0) and F (∞) operations only then it will be

irreducible condition and if it also contains with F (a) (0 < ‖a‖ < ∞)

operation then it will be redicible condition. Of course, F (a) operation

is possible to also generate irreducible condition between F (0) operation

iff F (a) operation has prime conditions.

© Application to T (a) and S(a) operations


T (a) ≡ −S(a) (mod e0s)


T (0) ≡ −S(0) (mod e0s = 00).


T (∞) ≡ −S(∞) (mod e0s = 00).

37


{1} = T (0) ≡ −S(0) = {0} = T (∞) ≡ −S(∞) = {1} (mod e0s = 00)


{1} ≡ {0} (mod e0s = 00).

Since this condition, I have been considered as set condition {1}, {0}.This form is special case for T (a) and S(a) operations. So

T (0) ≡ T (∞) (mod e0s = 00)

and

−S(0) ≡ −S(∞) (mod e0s = 00).

This means that T (0) operation has same property with T (∞). Sim-

ilarly, −S(0) operation has same property with −S(∞). In a word, T (a)

and S(a) operations are able to discuss on Hilbert spaces.


T (a) ≡ −S(a) (mod eas)

So

T (0) ≡ −S(0) (mod e0s = 00)

and

T (∞) ≡ −S(∞) (mod e∞s = ∞0)


{1} = T (0) ≡ −S(0) = {0} = T (∞) ≡ −S(∞) = {1} (mod e0s = e∞s).

Therefore

T (0) ≡ T (∞) (mod e0s = e∞s)

and

−S(0) ≡ −S(∞) (mod e0s = e∞s).

This means that T (0) operation (Now Laplace transform) is preserved

by the property of T (∞) operation (Extended Laplace transforms). In a

words, Now Laplace transform is able to extend to T (a) operation.

38

(see, p.22, No.11)

On the other hand, T (0) operation is able to represent as charactristic



{1} ≡ {0} (mod e0s = e∞s).




P ≡ Q (mod e0s = 00).


N.B. Q ≡ −Q (mod e0s = 00). Because

Let Q =

(0

1

)(10)

So we have

Q− {−Q} =

(0

1

)−

(0

−1

)= 2

(0

1

)= 2

(00

00

). (11)

Therefore

Q ≡ −Q (mod e0s = 00).


T (a)homo−→ P , S(a)

homo−→ Q , resp.

Moreover, T (a) and S(a) operations have a property of inclusion

monomorphism for projection operators. So we have,

T (a) ≡ S(a) (mod e0s)

and

T (a) ≡ S(a) (mod eas) or T (a) ≡ S(a) (mod e0s).

39

This means that T (a) operation has same property for S(a) operation

modulo e0s and T (0) operation.

N.B. R(a) = easT (0) , T (a) + S(a) = R(a) and




So we have

T (a) ≡ S(a) (mod e0s = e∞s) for all a.

and

T (a) ≡ S(a) (mod e0s = e∞s) for all a.


T (a)iso←→ T (a) , S(a)

iso←→ S(a).

Now I have been considered to distinguished as T (a) is ring and S(a) is

ideal , resp. However these operations T (a and S(a) have a same property

modulo e0s. This thing means that modulo e0s has common ring for T (a)

and S(a) operations. This common ring is called primitive ring. If the

primitive ring is seen from ideal structure then the ideal structure has

been generated the maximal ideal or ring boundary.

(see, P.55, No.10)

Clearly,

T (0)f(t) = S(∞)f(t).

Because of

T (0)f(t) =∫ ∞

0f(t)e−stdt = S(∞)f(t).

If S(∞) to S(∞) operations then we must have following condition.

T (0)f(t) = S(∞)f(t) iff e0s = e∞s.

40

Surely, we have following form.

T (0)f(t) =∫ ∞

0f(t)e−(t−0)sdt =

∫ ∞

0f(t)e−(t−∞)sdt = S(∞)f(t).

Therefore, these operations T (a) and S(a) will be treated on set con-

ditions. So, it’s algebras. Especially, in this case, we have following

conditions.

T (0) = S(∞)

and

T (0) = S(∞) iff e0s = e∞s.

N.B. T (0) = T (0).

This condition T (0) is treated on Now Laplace transforms. In gen-

eral, this operation is extended to ′′a′′ condition. The operation T (a) is

extended to New Laplace transforms. Essentially, both T (0) and T (a) op-

erations have same properties. Only difference is the process of between

T (0) and T (a) operations have two different explanations. One is the

operation of between T (0) operation and T (a) operation has been null

condition. In other words, S(a) operation has null condition. When it

has this condition, we are able to consider as irreducible condition. And

other is the operation of between T (0) operation and T (a) operation has

been dense condition. In other words, S(a) operation has dense condi-

tion. When it has this condition, we are able to consider as reducible

condition. This classification is dominated by how to treat the coefficient

of T (0) operation. So it’s eas. In this time, we have following condition.





example, if we consider T (0) and T (∞) operations only then it will be

irreducible condition and if it also contains with T (a) (0 < ‖a‖ < ∞)

operation then it will be redicible condition. Of course, T (a) operation

is possible to also generate irreducible condition between T (0) operation

iff T (a) operation has prime conditions.

41

Some results

◦ Now we have following condition.

{1} = T (0) ≡ −S(0) = {0} = T (∞) ≡ −S(∞) = {1} (mod e0s = e∞s).

So

{1} ≡ {0} (mod e0s = e∞s).

◦ Projection operators have following property.

P ≡ Q (mod e0s = 00).

◦ Laplace transforms have modulo with the ideal structures. So

T (a) ≡ S(a) (mod eas).

◦ Especially, we have following conditions, now.

(mod eas) = (mod e0s = 00).

◦ Clearly, T (0) operation is equal with S(∞). (L2-spaces)

T (0) ≡ S(∞) (mod e0s = 00).

and

S(0) ≡ T (∞) (mod e0s = 00).

◦ Similarly,

T (0) ≡ S(∞) (mod e0s = e∞s).

and

S(0) ≡ T (∞) (mod e0s = e∞s).

42

Chapter 3

In this chapter, I want to explain about the modulo in e0s = 00. This

form is able to be characteristics conditions. This condition is related with

σ-algebras, deeply close. These operator algebras was extended from this

condition.

© Modulo in e0s or 00 for D(a) operations

In general, (mod e0s) or (mod 00) includes projective condition.

Since this thing, now, let we choose that e0s = 00 is unit element. So

we are able to have proper sub-algebras in projective condition. These

proper sub-algebras are identities of D(0) operation. In this case, we

have {0} and {1}. N.B. {0} is unit element of additive condition on

D(0) operations. Similarly {1} is unit element of multiplicative condition

on D(0) operations. In a word, {0} and {1} are sub-algebras for D(0)

operation (projective conditions). So e0s in modulo is defined as only unit

condition in sub-algebras. This sub-algebras are most primitive form in

sub-algebras. And D(a) operations are dominated on this sub-algebras.

In other words, projection operator is also dominated on identities. So

now projection operator includes (closed condition) or contains (open

condition) identities. For example,

(1

0

),

(0

1

),

(1

1

),

(0

0

). (12)

It’s all projection operator.

In this time, this projection operator is able to define as followoing.

P,Q =

(e0s1

e0s2

)=

(00

00

)= D(0) (13)

Reference, No.1 and No.7 papers.

43

For example, F(a) operation has defined as following form.

F(a) =

(a0

a a0

)(14)

If a → 0 then we have projective condition. This thing means that

F(0) = D(0) generates as σ-algebras. In a word, D(0) is a kind of σ-

algebras. N.B. F(0) = D(0). This D(0) operation has deep relartion

with Now Laplace transforms. In a word, D(0) has been generated σ-

algebras and especially, Now Laplace transform has been generated iden-

tities in this σ-algebras. These identities are called primitive or proper

sub-algebras. The σ-algebras are dominated on these identities. For ex-

ample,

{0} =

(0

0

),

(1

0

),

(0

1

),

(1

1

)= {1} (15)

N.B. It’s all σ-algebra. Especially, {0}, {1} are identities.


D(0) = {1} = O(∞) and O(0) = {0} = D(∞)

I was treated these conditions for convenience until now. If this iden-

tities treat on set condition of {0} and {1} then we have following con-

ditions, easier.

D(0) ≡ D(∞) (mod e0s)

and

O(∞) ≡ O(0) (mod e0s).

By {0} ≡ {1} (mod e0s).

These things are treated on L2 and L1-spaces, etc.

Refer No.1 and No.11 papers.

© Modulo in e0s or 00 for Y (a) operations


44

Since this thing, now, let we choose that e0s = 00 is unit element. So

we are able to have proper sub-algebras in projective condition. These

proper sub-algebras are normal integral operator Y (0) for Now Laplace

transforms. In this case, we have {0} and {1}. N.B. {0} is unit element

of additive condition on Y (0) operation. Similarly {1} is unit element of

multiplicative condition on Y (0) operation. In a word, {0} and {1} are

sub-algebras for Y (0) operation (projective conditions). So e0s in modulo

is defined as only unit condition in sub-algebras. This sub-algebras are

most primitive form in sub-algebras. And Y (a) operation is dominated on

this sub-algebras. In other words, projection operator is also dominated

on identities. So now projection operator includes (closed condition) or

contains (open condition) identities. For example,

(1

0

),

(0

1

),

(1

1

),

(0

0

). (16)



P, Q =

(e0s1

e0s2

)=

(00

00

)iso↔ Y (0) (17)


For example, D(a) operation has defined as following form.

D(a) =

(a0

a0

)(18)


D(0) generates as σ-ring. In a word, Y (0) is a kind of σ-ring. N.B.

D(0)iso↔ Y (0). Hahn’s decomposition is able to apply to these Y (0) or

Y (a) operations. On condition that these forms generate the characteris-

tics condition on Y (0) or Y (a) operations. N.B. Y (0) or Y (a) operations

are defined as following forms, resp.

Y (0)f(t) =∫ ∞

0f(t)e0sdt or Y (a)f(t) =

∫ ∞

af(t)easdt.

45

If t = 0 then the kernel of Y (0) operation generates the characteristics

condition. In general, if t = a condition then we have N(a) operation on

0 ≤ t ≤ a and Y (a) operation on a ≤ t ≤ ∞. Now, since N(a) ≡ Y (a)

(mod e0s) then N(a) and Y (a) operations at t = a also have important

conditions (characteristics). These things means that Y (0) and Y (a) op-

erations have a same properties on 0 and a, resp. So Y (a) operation is

same condition with Y (0) operation (normal integral operators). There-

fore it has same situations even if extended conditions Y (a). In other

words, all operations that it’s generated on a = 0 are able to treat on ′′a′′

condition, samely.


Y(0) = {1} = N (∞) and N (0) = {0} = Y(∞)

I was treated these conditions for convenience until now. If this identities

treat on set condition of {0} and {1} then we have following conditions,

easier.

Y(0) ≡ Y(∞) (mod e0s)

and

N (∞) ≡ N (0) (mod e0s).

By {0} ≡ {1} (mod e0s).



© Modulo in e0s or 00 for F (a) operations


Since this thing, now, let we choose that e0s = 00 is unit element. So we

are able to have proper sub-algebras in projective condition. These proper

sub-algebras are identities for Pascal’s triangle matrix (F (a) operation).

In this case, we have {0} and {1}. N.B. {0} is unit element of additive

condition on Pascal’s triangle matrix. Similarly {1} is unit element of

multiplicative condition on Pascal’s triangle matrix. In a word, {0} and

46

{1} are proper sub-algebras for it if a = 0 (projective conditions). So

minimum e0s in modulo is defined as only unit condition in sub-algebras.

This sub-algebras are most primitive form in sub-algebras. And Pascal’s

triangle matrix is dominated on this sub-algebras. In other words, projec-

tion operator is also dominated on identities. So now projection operator

includes (closed condition) or contains (open condition) identities. For

example,

(1

0

),

(0

1

),

(1

1

),

(0

0

). (19)



P, Q =

(e0s1

e0s2

)=

(00

00

)= F (0). (20)


In this time, F(a) operation has defined as following form.

F(a) =

(a0

a a0

)(21)

If a → 0 then we have projective condition. This thing means

that F(0) generates as σ-ring. Furthermore, T (0) is a kind of σ-ring.

N.B. F(0)iso↔ T (0). Hahn’s decomposition is able to apply to these

F (0) or F (a) operations. On condition that these forms generate the

characteristics condition on F (0) or F (a) operations. N.B. F (0) or F (a)

operations are defined as following forms, resp.

F (0) =

(00

00

), F (a) = eas

(a0

a a0

)(22)

Now, we have

F (0) ≡ G(∞) (mod e0s = e∞s)

So the property of F (0) and G(∞) is same. In this case, the charac-

teristics condition exists on 0 ≤ a ≤ ∞.

47

Now, we are able to obtain following conditions.

F(0) = {1} = G(∞) and G(0) = {0} = F(∞)



ditions, easier.

F(0) ≡ F(∞) (mod e0s)

and

G(∞) ≡ G(0) (mod e0s).

By {0} ≡ {1} (mod e0s).



© Modulo in e0s or 00 for T (a) operations


Since this thing, now, let we choose that e0s = 00 is unit element. So we

are able to have proper sub-algebras in projective condition. These proper

sub-algebras are Now Laplace transforms. In this case, we have {0} and

{1}. N.B. {0} is unit element of additive condition on Now Laplace

transforms. Similarly {1} is unit element of multiplicative condition on

Now Laplace transforms. In a word, {0} and {1} are sub-algebras for

extended Laplace transforms (projective conditions). So e0s in modulo

is defined as only unit condition in sub-algebras. This sub-algebras are

most primitive form in sub-algebras. And extended Laplace transforms

are dominated on this sub-algebras. In other words, projection operator is

also dominated on identities. So now projection operator includes (closed

condition) or contains (open condition) identities. For example,

(1

0

),

(0

1

),

(1

1

),

(0

0

). (23)


48


P,Q =

(e0s1

e0s2

)=

(00

00

)(24)


Moreover, F(a) operation has defined as following form.

F(a) =

(a0

a a0

)(25)


F(0) generates as σ-ring. In a word, T (0) is a kind of σ-ring. N.B.

F(0)iso↔ T (0). So F(0) operation has a characteristics condition. Hahn’s

decomposition is able to apply to these T (0) or T (a) operations on condi-

tion that these forms also generate the characteristics condition at t = 0

or t = a, resp. Because of the kernel of T (0) or T (a) operations generates

e0s = 00. N.B. T (0) or T (a) operations are defined as following forms.

T (0)f(t) =∫ ∞

0f(t)e−(t−0)sdt or T (a)f(t) =

∫ ∞

af(t)e−(t−a)sdt.

And if t = a condition then we have S(a) operation on 0 ≤ t ≤ a and T (a)

operation on a ≤ t ≤ ∞. Now, since S(a) ≡ T (a) (mod e0s) then S(a)

and T (a) operations at t = a have important conditions (characteristics).


T (0) = {1} = S(∞) and S(0) = {0} = T (∞)



ditions, easier.

T (0) ≡ T (∞) (mod e0s)

and

S(∞) ≡ S(0) (mod e0s).

By {0} ≡ {1} (mod e0s).



49

Some results

◦ Now Laplace transforms have been generated from identities {1}, {0}.

◦ Now Laplace transform is the smallest algebras in operator algebras.

◦ e0s and 00 have characteristics conditions, resp. The situations generate

Now Laplace transforms.

◦ σ-algebra is able to represent the projection operator.

◦ Extended Laplace transforms are isomorphic with the projection oper-

ator.

◦ Now Laplace transforms and Extended Laplace transforms have same

properties. So we have following conditions.

T (0) ≡ T (∞) (mod e0s)

by {1} ≡ {0} (mod e0s).

50

Conclusion

Now, T (a) operations are able to represent as algebraic forms. So

we are able to have certain forms by modulo conditions. This modulo

conditions have two kinds forms. One is eas and other is T (0). eas of

modulo conditions has been used on the assertion of extended conditions

for operator algebras. On the other hand, T (0) of modulo conditions has

been used on assertion of Now Laplace transforms. The characteristics

of T (0) operation are represent as e0s. So, both eas and T (0) have same

modulus by


In this time, we have following conditions.

T (a) ≡ S(a) (mod eas) or T (a) ≡ S(a) (mod T (0)).

This means that extended Laplace transforms (T (a) operation) and

the ideal structures (S(a) operation) have same properties by modulo eas

or T (0) operation. In a word, T (0) operation becomes sub-algebras on

T (a) and S(a) operations. In fact, T (0) operation is the smallest algebras

in T (a) and S(a) operations. So T (0) operation (Now Laplace transform)

is identity. But, now, we should attend to T (0) operation. In fact, if we

extend our consideration then T (0) operation has chatacteristics condi-

tion. So, we are able to extend to σ-algebras. In this time, T (a) operation

and T (0) operation are same conditions. Precisely speaking,

PT (a) = T (0).

The next time, we have

{1} = T (0) ≡ −S(0) = {0} = T (∞) ≡ −S(∞) = {1} (mod e0s = e∞s).

So we have,

{1} ≡ {0} (mod e0s = e∞s).

In this time, if we consider as T (a) and S(a) operations are set con-

ditions then we are able to obtain the algebraic forms. These forms are

able to be ring, field, and group on Laplace transforms.

(Wed)20.Nov.2013

Now, let′s go to next papers with me!

51

References

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

tional 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] Harry Hochtadt, Integral Equations, John & Sons,Inc, 1973.

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

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

[7] Emil G.Milewski, Rea’s Problem Solvers Topology, 1998.

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.

52

Documents

Approach from modulus for Laplace transformsopab.web.fc2.com/modulus.pdfApproach from modulus for Laplace transforms No.1 Takao Saito 1 Thank you for our world 2 Approach from modulus