the diagonal conditions for laplace transformsopab.web.fc2.com/no.9.pdfpreface in this paper, i...

The diagonal conditions

for Laplace transforms

No.3

Takao Saito

Thank you for our world

The diagonal conditions

for Lapalace transforms

Preface

In this paper, I explain about the diagonal conditions for Laplace

transforms. Both Pascal’s triangle operator and the lattice forms are

generated from matrices conditions. On the contrary, Laplace transforms

are well-known as integral operations. In this time, if we operated on

the diagonal conditions of the matrices forms then we are able to obtain

“Now Laplace transforms” (T (0) operation).

Papers

About solvers of differential equations of relativety

for Laplace transforms

Operator algebras for Laplace transforms

Rings and ideal structures for Laplace transforms

Extension and contraction for transrated operators

Operator algebras for group conditions

Pascal’s triangle matrix for Laplace transforms

Lattice structures 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

(Sat) 5.May.2012

Takao Saito

Contents

Preface

Chapter 7

◦ H(a) operation · · · · · · 7

◦ Duality of H(a) operations · · · · · · 10

◦H(a) operation · · · · · · 14

◦ Duality of H(a) operations · · · · · · 19

◦ Some results · · · · · · 25

Chapter 8

◦ T (a) operation with diagonal conditions · · · · · · 26

◦ R(a) operation · · · · · · 29

◦ R(a) operation · · · · · · 32

◦ Some results · · · · · · 35

Chapter 9

◦ V(a) operation · · · · · · 36

◦ Duality of V(a) operation · · · · · · 42

◦ Duality of L(a) operation · · · · · · 43

◦ V (a) operation · · · · · · 47

◦ Some results · · · · · · 53

◦ Conclusions · · · · · · 54

◦ References · · · · · · 55

§ Chapter 7

In this chapter, I explain about the diagonal conditions for Pascal’s

matrix operation (F(a)). In fact, if a is zero then we have special diagonal

operation (H(a)).

©H(a) operation

The fundamental condition of T (a) operation is defined as following.

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt (a ≤ t ≤ ∞)

Especially, if f(t) is to power series then we have following conditions.

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

F(a)　　　　　　　　　　　

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

=

a0

a a0

a2 2a a0

a3 3a2 3a a0　　　　　　　　　

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0

T (0)1

T (0)t

T (0)t2

T (0)t3

...

T (0)tn

(see, p.21, Chapter 2, No.7, N o.2)

This F(a) operation is able to represented as above form. In this

time, F(a) operation is able to decomposed to the diagonal condition

(H(a) operation)and ideal structures (G(a) operation). Especially, the

diagonal condition has very important condition for Pascal’s matrix form

F(a). So we will have the property of Pascal’s matrix form by checking

the diagonal condition (H(a) operation).

7

This F(a) operation is able to separate following.

a0

a a0

.... . .

an · · · nCkan−k a0

=

00

00

00

00

+

a0 − 00

a a0 − 00

.... . . a0 − 00

an · · · nCkan−k a0 − 00

F(a) H(a) −G(a)

Now, H(a) operation is defined as above form. So projection operator

is defined as this condition. (see, p.53, Chapter 3, No.7, N o.2) This

definition of H(a) operation is treated on L2-spaces. So this condition

is little dificult conception. This condition is across conditions of ring

and ideal forms on a. As a whole, these ring (H(a)) and ideal (G(a))

operations have {1, 0} and {0, 1} conditions on a, resp. In fact, H(0)

operation generates the characteristics condition. So this 00 has four

kinds condition in L2-spaces. It’s {0, 0}, {0, 1}, {1, 0}, {1, 1}, resp. This

condition is included in projection operators.

Similarly, this F∗(a) operation is represented as following form.

a0 a · · · an

a0...

. . . nCkan−k

a0

=

00

00

00

00

+

a0 − 00 a · · · an

a0 − 00 . . ....

a0 − 00nCka

n−k

a0 − 00

F∗(a) H∗(a) −G∗(a)

In this case, 00 is represented as 00. And this projective condition is

represented as following

{0, 0}, {0, 1}, {1, 0} and {1, 1}, resp,

in L2-spaces.

8

Now, I’m considering this diagonal condition of F(a) operation. So

we are able to decompose as following conditions.

a0

a a0

......

. . .


=

a0

a0

a0

a0

+

a...

. . .

an · · · nCkan−k

(1)

F(a) H(a) −G(a)

In this case, the H(a) operation is defined on L1-spaces. This form

has pararel conditions of ring and ideal forms on a. If we consider the

infinite condition of a then we will have convergence of R(∞) on L1

and L2-spaces, resp. Consequntry, if we consider on L1-spaces then H(a)

operation obtains identity {1, 1} on multiplication. On the contrary, G(a)

is able to treat as {0, 0} conditions. If a is considered on L2-spaces then we

should also adopt {0, 1} and {1, 0} conditions. It’s projective condition

as P (a), too.

Similarly, this F∗(a) operation is represented as following form.

a0 a · · · an

a0...

. . . nCkan−k

a0

=

a0

a0

a0

a0

+

a · · · an

. . ....

nCkan−k

(2)

F∗(a) H∗(a) −G∗(a)

In this case, the H∗(a) operation is also defined on L1-spaces. So this

form has pararel conditions of ring and ideal forms on a. If we consider

the infinite condition of a then we will have convergence of R∗(∞) on L1

and L2-spaces, resp. Consequntry, if we consider on L1-spaces then H∗(a)

operation obtains identity∗ as {1, 1} on multiplication. On the contrary,

G∗(a) is able to treat as {0, 0} conditions. If a is considered on L2-spaces

9

then we should also adopt {0, 1} and {1, 0} conditions. It’s projective

condition as P ∗(a), too.

Now, the projection operator is wellknown as following formula.

P (0) = P ∗(0) = P 2(0)

N.B. Special case.

In this time, in general, the projection operator is able to extend to

following condition.

P (a)iso↔ P ∗(a)

iso↔ P 2(a)


©Duality of H(a) operations

Since previous forms, H(a) operations are able to define as following

condition.

H(a)def .=

a0

a0

a0

a0

= P (a) , H∗(a)def .=

a0

a0

a0

a0

= P ∗(a)

The property of these operations have following conditions.

If a is finite real number without zero then we have just 1. So it’s

P (a) = H(a) =

1

1

1

1

= H∗(a) = P ∗(a) (3)

So we have

P (a) = H(a) = H∗(a) = P ∗(a) (Hermitian)

10

iff a is finite real number without zero.

On the other hand, if a is infinite real number or zero then we should

have

P (0) = H(0) =

00

00

00

00

iso↔

00

00

00

00

= H∗(0) = P ∗(0)

So we have

P (0) = H(0)iso←→ H∗(0) = P ∗(0).

‖ or ‖P (∞) = H(∞)

iso←→H∗(∞) = P ∗(∞)

N.B. In this time, H(∞) operations are represented as following con-

ditions.

H(∞)def .=

∞0

∞0

∞0

∞0

= P (∞) , H∗(∞)def .=

∞0

∞0

∞0

∞0

= P ∗(∞)

These H(0) operation and H(∞) operation have to obtain the same

properties, resp. Slightly different, H(0) does not have the ideal structure

and H(∞) has to consider with the ideal structure. So we have following

conditions.

H(0) = H(∞) , H∗(0) = H∗(∞), resp.

H(a) operation is just diagonal matrix. And this form has closely

relation with R(0) (Now Laplace transforms) operation. So the properties

of R(0) = R(a) operations just to check the properties of H(a) operation.

Similarly, the properties of H∗(a) operation have relations with R∗(0) =

R∗(a) operations.

11

Moreover, R(0) operation is also builded with isomorphic condition on

R∗(0) operation. And, in general,

R(∞) = R(0) 6= R∗(0) = R∗(∞).

iso l l iso

H(∞) = H(0) 6= H∗(0) = H∗(∞).

(see, p.11∼16, Chapter 1, No.7, N o.2)

N.B. In this time, R(0)=R(0).

Therefore we have following.

R(∞) = R(0)iso←→R∗(0) = R∗(∞)

iso l l iso

H(∞) = H(0)iso←→H∗(0) = H∗(∞)

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

In this view point, R(0) operation has special condition. This op-

eration has to consider on isomorphic condition. If we are able to treat

for all a from zero then R(a) operations are able to consider as identical

identity. On the contrary, T (0) operation needs to consider on set con-

dion for all projection operators. This phenomenon is reason that T (0)

operation (Now Laplace transforms) had better treat on L2-spaces from

first. So these operator algebras are based on Hilbert spaces.

(see, the paper dThe projection operator for Laplace transformsc)Site: http://opab.web.fc2.com/no.1.pdf

As a whole, we have following form.

Matrices operators

F(a)homo−→ H(a).

iso l l iso

T (a)homo−→ R(a) = T (0).

12

Integral operators

iso l l iso

Matrices operators∗

F∗(a)homo−→ H∗(a).

iso l l iso

T ∗(a)homo−→ R∗(a) = T ∗(0)

Integral operators∗

on L2-spaces.

In this time, T (a) and R(a) operations are defined as following.

T (a)f(t) =∫ ∞

af(t)e−stdt

R(a)f(t) =∫ ∞

0f(t)e−stdt, resp.

Especially, T (0) = T (0) = R(a).

Therefore, if H(a) operation is treated on real space then we have

H(a) = H∗(a) for all a.

On the contrary, if H(0) operation is treated on complex spaces then

we should have following form.

H(0)iso↔ H∗(0)

N.B. In general, 00 has been defined on complex space.

As a whole,

H(a)iso↔ H∗(a) for all a.

So

‖H(a)‖ = ‖H∗(a)‖

N.B. In this time, I have treated as a is real number.

13

The first step, H(a) operation has been represented on L1-space. If

it’s explained on L2-space then we have to treat as H(a) = 00 · I. In fact,

H(a) = a0 · I is easier than H(a) = 00 · I to understand this situation

because of a0 = 1 a.e. However if a = 0 then we will had better treat

on L2-space. Consequently, Now Laplace transforms R(0) = T (0) is

considering on this H(a) operation. And this R(0) operation is able to

treat on L1, L2 and Banach spaces. So we have that R(0) operation is

able to treat as identical identity.

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

©H(a) operation

The extended condition of T (a) operation is also defined as following.

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt (0 ≤ t ≤ ∞)

In this time, the norm of T (a) operation is defined as ‖eas‖.So ‖T (a)‖ = |eas|

Especially, if f(t) is to power series then we have following conditions.

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

F(a)

eas

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

= eas

a0

a a0

a2 2a a0

a3 3a2 3a a0　　　　　　　　　

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0

T (0)1

T (0)t

T (0)t2

T (0)t3

...

T (0)tn


14

In this time, we are able to rewrite as easT (a) −→ T (a).

So I defined the extended condition for F(a) operation.

F (a)def .= easF(a)

T (a)1

T (a)t

T (a)t2

T (a)t3

...

T (a)tn

= eas

a0

a a0

a2 2a a0

a3 3a2 3a a0　　　　　　　　　

......

.... . .

annC1a

n−1nC2a

n−2nCka

n−k a0

T (0)1

T (0)t

T (0)t2

T (0)t3

...

T (0)tn

This F (a) operation is able to represented as easF(a). In general,

F(a) operation is called Pascal’s triangle matrix. Now, F (a) operation is

defined as following.

F (a) = easF(a)

This operation is generated for comparing with T (a) (extended) oper-

ation.


In this time, F (a) operation is able to decomposed to the diagonal

condition (H(a) operation)and ideal structures (G(a) operation). Espe-

cially, the diagonal condition has very important condition for Pascal’s

matrix form F (a). In fact, F (a) operation has same property for H(a)

operation. In other words, we will have the property of Pascal’s matrix

form by checking the diagonal condition (H(a) operation).

This F (a) operation is able to have following conditions.

eas

a0

a a0

.... . .


= eas

00

00

00

00

15

F (a) H(a)

+eas

a0 − 00

a a0 − 00

.... . . a0 − 00

an · · · nCkan−k a0 − 00

−G(a)

in L2-spaces.

Now, H(a) operation is defined as above form. So projection oper-

ator is defined as this local condition. (see, p.53, Chapter 3, No.7, N o.2)

This definition of H(a) operation is treated on L2-spaces. So this con-

dition is little dificult conception. This condition is across conditions of

ring and ideal forms on a. As a whole, these ring H(a) and ideal G(a)

operations have the element of {1, 0} and {0, 1} conditions on a, resp. In

fact, H(0) operation generates the characteristics condition. Since a = 0

condition then we are able to use the following formula.

e0·s = 00

(see, p.8, Chapter 1, No.7, N o.2).

As a whole, this condition also has 00. So this 00 has four kinds con-

dition in L2-spaces. It’s {0, 0}, {0, 1}, {1, 0}, {1, 1}, resp. This condition

is included in projection operators.

So we have following condition.

H(0) = H(0) = P (0)

Therefore H(0) operation is able to treat samely with H(0) operation

iff a is zero.

16

Similarly, this F ∗(a) operation is represented as following form.

eas

a0 a · · · an

a0...

. . . nCkan−k

a0

= eas

00

00

00

00

F ∗(a) H∗(a)

+eas

a0 − 00 a · · · an

a0 − 00 . . ....

a0 − 00nCka

n−k

a0 − 00

−G∗(a)

in L2-spaces.

In this case, if a = 0 condition then we are able to use the following

form.

e0·s = 00

So this projective condition is represented as following

{0, 0}, {0, 1}, {1, 0} and {1, 1}, resp,

in L2-spaces.

This condition is also included in projection operators.

In this case, we have following condition.

H∗(0) = H∗(0) = P ∗(0)

Therefore H∗(0) operation is able to treat samely withH∗(0) operation

iff a is zero.

N.B. In this case, the projection operator is defined on isomorphic

17

condition.

So

P (0)iso←→ P ∗(0).

Now, I’m able to consider the diagonal condition of F (a) operation.

So we are able to also decompose as following conditions.

eas

a0

a a0

......

. . .


= eas

a0

a0

a0

a0

+ eas

a...

. . .

an · · · nCkan−k

F (a) H(a) −G(a)

In this case, the H(a) operation is defined on L1-spaces. This form

has pararel conditions of ring and ideal forms on a. If we consider the

infinite condition of a then we will have convergence of T (∞) on L1 and

L2-spaces, resp. Consequntry, if we consider on L1-spaces then H(a) op-

eration obtains identity {1, 1} on multiplication. On the contrary, G(a) is

able to treat as {0, 0} conditions. If a is considered on L2-spaces then we

should also adopt {0, 1} and {1, 0} conditions. It’s projective condition

as easP (a), too.

Similarly, this F ∗(a) operation is represented as following form.

eas

a0 a · · · an

a0...

. . . nCkan−k

a0

= eas

a0

a0

a0

a0

+ eas

a · · · an

. . ....

nCkan−k

F ∗(a) H∗(a) −G∗(a)

In this case, the H∗(a) operation is also defined on L1-spaces. So this

form has pararel conditions of ring and ideal forms on a. If we consider

18

the infinite condition of a then we will have convergence of R∗(∞) on L1

and L2-spaces, resp. Consequntry, if we consider on L1-spaces then H∗(a)

operation includes identity∗ as {1, 1} on multiplication. On the contrary,

G∗(a) is able to treat as including {0, 0} conditions. If a is considered

on L2-spaces then we should also adopt {0, 1} and {1, 0} conditions. It’s

projective condition as easP ∗(a), too.

In this time, the projection operator should represent as following

conditions

e0·sP (0) = e0·sP ∗(0) = e0·sP 2(0)

N.B. In this time, e0·s = 00 = 00 = e0·s iff s is finite. (Special case)

In general, the projection operator is able to extend to following

condition.

ea·sP (a)iso↔ ea·sP ∗(a)

iso↔ e2a·sP 2(a)


©Duality of H(a) operations

Since previous forms, H(a) operations are able to define as following

condition.

H(a)def .= eas

a0

a0

a0

a0

= D(a) , H∗(a)def .= eas

a0

a0

a0

a0

= D∗(a)

The property of these operations have following conditions.

If a is finite real number without zero then we have just 1 in matrices.

So it’s

19

D(a) = H(a) = eas

1

1

1

1

6= eas

1

1

1

1

= H∗(a) = D∗(a)

So we have

D(a) = H(a) 6= H∗(a) = D∗(a).

In other words,

‖D(a)‖ = ‖H(a)‖ = ‖H∗(a)‖ = ‖D∗(a)‖ = |eas|.

On the other hand, if a is infinite real number or zero then we should

have

P (0) = H(0) =

00

00

00

00

iso↔

00

00

00

00

= H∗(0) = P ∗(0)

N.B. e0·s = 00

So we have

P ∗(0) = H(0)iso←→ H∗(0) = P ∗(0).

l or lH(∞)

iso←→ H∗(∞)

N.B. In this time, H(∞) operations are represented as following con-

ditions.

H(∞) = e∞·s

∞0

∞0

∞0

, H∗(∞) = e∞·s

∞0

∞0

∞0

20

These H(0) operation and H(∞) operation have to obtain the same

properties, resp. Slightly different, H(0) does not have the ideal structure

and H(∞) has to consider with the ideal structure. In general, we have

following conditions.

H(0) 6= H(∞) , H∗(0) 6= H∗(∞), resp.

H(a) operation is just diagonal matrix. And this form has closely

relation with R(0), T (0) (Now Laplace transforms) operations. So the

properties of R(0), T (0) operations just to check the properties of H(a)

operation. Similarly, the properties of H∗(a) operation has closed relation

with R∗(0), T ∗(0) operations.

Moreover, R(0) operation is also builded with isomorphic condition on

R∗(0) operation. And, in general,

T (0) = R(0) 6= R∗(0) = T ∗(0).

iso l l iso

F (0) = H(0) 6= H∗(0) = F ∗(0).

(see, p.11∼16, Chapter 1, No.7, N o.2)

Therefore we have following.

T (0) = R(0)iso←→ R∗(0) = T ∗(0)

iso l l iso

F (0) = H(0)iso←→ H∗(0) = F ∗(0)

In this view point, R(0) has special condition. This operation has

to consider on isomorphic condition for all a. If we are able to treat

on isomorphic condition then R(0) operations are able to consider as

identical identity. On the contrary, T (0) operation needs set condition.

This phenomenon is reason that T (0) operation (Now Laplace transforms)

had better treat on L2-spaces from first. So these operator algebras are

based on Hilbert spaces.

21

(see, the paper dThe projection operator for Laplace transformsc)Site: http://opab.web.fc2.com/no.1.pdf

As a whole, we have following form.

Matrices operators

F (a)homo−→ H(a).

iso l l iso

T (a)homo−→ R(a).

Integral operators

iso l l iso

Matrices operators∗

F ∗(a)homo−→ H∗(a).

iso l l iso

T ∗(a)homo−→ R∗(a)

Integral operators∗

on L2-spaces.

In this time, T (a) and R(a) operations are defined as following.

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt

R(a)f(t) =∫ ∞

0f(t)e−(t−a)sdt, resp.

Especially, T (0) = 00R(0).

Therefore, if H(a) operation is treated on real space then we have

H(a) 6= H∗(a) for all a.

On the contrary, if H(0) operation is treated on complex spaces then


H(0)iso↔ H∗(0)

22

N.B. In general, 00 has been defined on complex space.

As a whole,

H(a)iso↔ H∗(a) for all a.

So

‖H(a)‖ = ‖H∗(a)‖.

N.B. In this time, I have treated as a is real number.

The first step, H(a) operation has been represented on L1-space. If it’s

explained on L2-space then we have to treat as H(a) = ea·s00 · I. In fact,

H(a) = easa0 ·I is easier than H(a) = eas00 ·I to understand this situation

because of a0 = 1 a.e. However, in this case, H(a) 6= H∗(a). Moreover if

a = 0 then we had better treat on L2-space. Consequently, Now Laplace

transforms R(0) = T (0) is considering on this H(a) operation. And this

R(0) operation is able to treat on L1, L2 and Banach spaces. So we have

that R(0) operation is able to treat as identical identity.

Moreover, H(a) operation is isomorphic with R(a) operation. This

R(a) operation has following form.

R(a) = easT (0) = easT (0) = easR(a)

So if s is treated on real space then we have that eas is real constant.

Therefore R(a) is operation that it’s extended from unitary operation.

This operation is able to treat on C∗ or W ∗-algebras.

On the other hand, H(a) operation has diagonal condition. So the

property of diagonal conditions are related with F (a) operation.

Now, we have following conditions.

eas · PF (a) = easF (0) = H(a)

This form is projection from F (a) operation to H(a) operation. This is

things that the ring of F (a) is able to represent by using H(a) operation.

In general, F (a) operation is homomorphism to H(a) operation. And

the properties of F (a) operation is same with H(a) operation. So we

just check the property of H(a) operation then we are able to have the

property of F (a) operation. In this means, H(a) operation is important

23

operations for F (a) moreover T (a) operations.

In fact, R(a) operation is able to check the property of H(a) operarion.

Furthermore the property of T (a) operation just check the property of

H(a) operation.

24

Some results

◦ Pascal’s triangle matrix operation is able to generate the diagonal con-

dition (H(a) operation). ThisH(a) operation is defined as following form.

H(a) =

a0

a0

a0

in L1 − spaces

or

H(a) =

00

00

00

in L2 − spaces.

◦ This H(a) operation is isomorphic with R(a) operation. This R(a)

operation is Now Laplace transforms (T (0)). Therefore,

R(a)f(t) = e0·s∫ ∞

0f(t)e−stdt = T (0)f(t) N.B. e0·s = 00.

H(a)iso←→R(a).

◦ The properties of R(a) operation just check the properties of H(a)

operation. So T (0) operation also just check the property of H(a) oper-

ation. Therefore, Now Laplace transforms is able to represent as H(a)

operation.

◦ H(a) operation is defined as following conditions.

H(a) = easH(a) in L1, L2 − spaces.

◦ In general, in this case, we have to consider on following conditions.

H(a) 6= H∗(a). Therefore, H(a)iso←→ H∗(a).

◦ R(a) = easT (0) operations is also isomorphic with H(a) operation. So,

Now Laplace transforms are able to ride on the diagonal conditions.

25

§ Chapter 8

In this chapter, I want to explain about T (0) operation on diagonal

condition. Now Laplace transforms (T (0) operation) is a to zero condition

of T (a) operation. It has characteristics conditions.

©T (a) operation with diagonal conditions

In this time, T (a) operation is defined on following integral operations.

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt

R(a)f(t) = easR(a)f(t) = eas∫ ∞

0f(t)e−stdt = easT (0)f(t)

The space of H(a) operations are related with s-function of R(a)

operation. Therefore if s is traeated in real spaces then H(a) is also

treated in real spaces. On the other hand, if s is treated in complex

spaces then H(a) is also treated in complex spaces. This T (a) operation

is isomorphic with F (a) operation. In this case, F (a) operation has ideal

structure. So, T (a) operation also has ideal structure.

On the contrary, if a = 0 then we are able to have following conditions.

F(0) = D(0) = H(0)　　　　　　　　　　　

T (0)1

T (0)t

T (0)t2

...

T (0)tn

=

00

00

00

. . .

00

T (0)1

T (0)t

T (0)t2

...

T (0)tn

(4)

If the norm of ‖T (a)‖ operation has |eas| then we have

26

e0·s

T (0)1

T (0)t

T (0)t2

...

T (0)tn

= e0·s

00

00

00

. . .

00

T (0)1

T (0)t

T (0)t2

...

T (0)tn

Since e0·s = 00 then we also have first form.

Therefore

T (0) = 00T (0).

So “Now Laplace transforms T (0)” is able to represent as special

matrix conditions. This T (0) operation has relation with D(0) operations

(see, p.8, Chapter 1, No.1, No.1) via H(0) operation and this form has

following characterisic conditions.

T (0) = 00 (characteristics)

So if T (0) operaration is represented on real space then we have,

e0·s = 00 def .= 00 = e0·s.

as s is finite.

and

if T (0) operaration is represented on complex space then we have,

e0·s = 00 iso←→ 00 = e0·s.

as s is infinite.

Therefore

T (0) = T ∗(0) (Hermitian) (real spaces)

‖T (0)‖ = ‖T ∗(0)‖ (complex spaces).

Moreover, in this time,

‖T (0)‖ = ‖ker(s, t)‖ = ‖e−(t−0)s‖t=0 = ‖e0·s‖ = ‖00‖So we have

‖T (0)‖ = {1}, {0}

27

in L1, L2-spaces.

In general, T (0) operation is represented as following forms.

T (0) = {1}, {0}

in L1-spaces.

or

T (0) = {1, 1}, {1, 0}, {0, 1}, {0, 0}

in L2-spaces.

Both T (0) operations of L1 and L2-spaces are able to represent as

projection operator.

On the other hand, since the property of H(a) operations, then we

have

H(a) = H∗(a) (real spaces)

iso l l iso

R(a) = R∗(a) (real spaces)

and

‖H(a)‖ = ‖R(a)‖ = ‖R∗(a)‖ = ‖H∗(a)‖ (complex spaces).

In other words

H(a)iso↔ H∗(a)

iso l l iso

R(a)iso↔R∗(a)

in complex spaces.

(see, p.40,42, Chapter 8, No.3, N o.1)

So, if s is extended to infinite complex space then we have to treat on

normed conditions.

28

©R(a) operation

Now, R(a) operation is defined as following condition.

R(a)f(t) = e0·s∫ ∞

0f(t)e−stdt = 00 · T (0)f(t)

N.B. e0·s = 00.

Precisely speaking, R(a) operation is defined above condition with

e0·s. So R(a) operation also has characteristic condition because of e0·s =

00. And, in this time, R(a) operation is able to represent as {1, 1} or

{0, 0} conditions. In fact, these representations are given as identical

identity. Finally, the properties of R(a) operation is same as T (0) oper-

ation with characteristics.

Similarly, this -∗ algebras form is represented as

R∗(a)f(t) = e0·s∫ ∞

0f(t)e−stdt = 00T ∗(0)f(t).

for operations.

So, in this case, f(t) = f(t).

In this time, since a has been treated as real number, then we have

following conditions.

If a 6= 0 then

R(a) = R∗(a) (Hermitian).

iso l l iso

T (0) = T ∗(0) (Hermitian)

in real spaces.

If a = 0 then we should consider two forms.

One is

R(0) = R∗(0) (Hermitian).

29

in real spaces.

And other is

R(0)iso←→R∗(0)

in complex spaces.

In other words, if R(a) operation is treated on real space then we have

R(a) = R∗(a) for all a.

On the contrary, if R(0) operation is treated on complex spaces then


R(0)iso↔R∗(0)

As a whole,

00 · T (0) = R(a)iso↔R∗(a) = 00 · T ∗(0) for all a.

N.B. In this time, I have treated as that a is real number.

R(a) operation is independed with a. In other words, R(a) opera-

tion is “identical identity”.

This situation is represented on L1-space. If the first step is ex-

plained on L2-space then we have to treat as R(a) = e0·s = 00. In fact,

R(a) = e0·s = a0 is easier than R(a) = 00 to understand this situation

because of e0·s = a0 = 1 a.e. So R(a) operation is easy to be identical

identity {1, 1} or {0, 0}. However if a = 0 then we will had better treat

on L2-space. So “Now Laplace transforms (T (0) operation)” should con-

sider on L2-spaces. In this time, T (0) operation needs the conception

of projection operators with {1, 0} and {0, 1}. Therefore this operation

had better extend to Hilbert spaces from Riemann integration. Just this

Laplace transform (T (0)) must jump up the conception to L2-spaces, etc.

So it is easy seemingly but it needs wide conceptions. In this time, T (a)

operation is able to also use for this invitational conceptions.

On the other hand, since T (0) = R(a), T (0) (Now Laplace trans-

forms) also exists on H(a) operation.

30

Therefore, T (0) operation appears on diagonal condition of F(a) op-

eration.

N.B. T (a)iso↔ F(a)

As a whole, T (a) and F(a) operations have following properties.

F(a)homo−→ H(a).

iso l l iso

T (a)homo−→ R(a).

Similarly, the dual condition is represented as following forms.

F∗(a)homo−→ H∗(a).

iso l l iso

T ∗(a)homo−→ R∗(a).

Reference

In this time, My Laplace transform (T (a)f(t)) is represented as fol-

lowing.

T (a)f(t) = easT (a)f(t) =∫ ∞

af(t)e−(t−a)sdt

This homomorphic condition gives the projection form of T (a) oper-

ation.

Therefore

PT (a) = T (0) = T (0) = e0·sR(a).

The ring of T (a) operation has same property of T (0) “Now Laplace

transforms” operation. In fact, T (0) operation has characteristics condi-

tion as 00. However the ring condition of T (0) operation is same with

T (a) operation. Therefore if we understand this T (0) operation then

T (a) is able to also understand the structures. In this view point, T (a)

31

operation is easier than T (0) operation.

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

F(0) = H(0).

iso l l iso

T (0) = T (0) = R(0) = R(0).

Similarly,

F∗(0) = H∗(0).

iso l l iso

T ∗(0) = T ∗(0) = R∗(0) = R∗(0).

©R(a) operation

Now, R(a) operation is defined as following form.

R(a)f(t) =∫ ∞

0f(t)e−(t−a)sdt = eas

∫ ∞

0f(t)e−stdt = easT (0)f(t)

Since the definition of R(a) operation then we have following form.

R(a)f(t) = easT (0)f(t)

Especially, if a = 0 then we have “Now Laplace transforms” (T (0)

operation). In general, this R(a) operation exists on T (0) operation. So

R(a) operation is able to be treated on C∗-algebras.

If eas = eas then we have to consider on real space. And this R(a)

operation is able to be following form.

And if a 6= 0 then

R(a) = R∗(a) (Hermitian).

If a = 0 then we should consider two forms.

One is

R(0) = R∗(0) (Hermitian).

32

in real spaces.

And if eas iso←→ eas then we have

R(0)iso←→ R∗(0)

in complex spaces.

In other words, if R(a) operation is treated on real space then we

have

R(a) = R∗(a) for all a.

On the contrary, if R(0) operation is treated on complex spaces then


R(0)iso↔ R∗(0)

As a whole,

R(a)iso↔ R∗(a) for all a.

In general, we consider this form.

N.B. In this time, I have treated as that a is real number.

R(a) operation is not included ideal structure. So this operation

is treated as pure ring condition. In fact, T (0) operation “Now Laplace

transform” has ring condition without ideal structure on eigenspaces.

Therefore it’s interesting condition. In this time, R(a) operation has mul-

tiplicative condition for T (0) operation. So, essentially, R(a) operation

also has characteristics condition. R(a) operation is not T (a) operation,

however the properties of R(a) and T (a) operations need to be same con-

dition.

Consequently, this R(a) operation is able to appraise T (a) operation.

So we are able to obtain the property of T (a) operations by treating

T (0) operations (Now Laplace transforms). This operation is opposed to

projection operator and it’s important conditions for T (a) operation. In

this case, T (a) operation is isomorphic with T (0) operation. This situ-

ation has extended form from inclusion monomorphism. Therefore, we

will have

T (a)iso←→ T (0).

33

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

T (∞) = T (0) = {1}or

T (∞) = T (0) = {0}(5)

Similarly, we have

T ∗(∞) = T ∗(0) = {1}or

T ∗(∞) = T ∗(0) = {0}(6)

So this property of T (0) operation is used on identical element for ring

condition. R(a) operation is extended form from T (0) operation. The

property of T (0) operation is contained in T (a) operation. In this time,

this R(a) operation is perfect operation for T (a) operation. Therefore

T (a) operation is also perfect. T (a) operation has all properties of T (0)

operation (Now Laplace transforms). Now, T (0) operation is unitary

operator of conpactness. So T (a) operation (My Laplace transforms)

is also compact operator. Just moment, this T (a) operation is able to

extend to a →∞. Therefore this T (a) operation should converge to zero

on Banach spaces.

34

Some results

◦ T (0) operation “Now Laplace transforms” is able to represent as char-

acteristics conditions.

◦ T (0) operation is isomorphic with H(0) and R(a) operations. Precisely

spaking, we have following form.

T (0) = 00 · R(a)iso←→ H(0)

◦ R(a) and R∗(a) operations are able to represent as identical identity

{1, 1}, {0, 0} and {1, 1}, {0, 0}, resp. On the contrary, T (0) and T ∗(0)

operations also include {1, 0}, {0, 1} and {1, 0}, {0, 1}, resp. So it’s pro-

jective.

◦ R(a) operation is able to also ride on H(a) operation.

N.B. H(a) operation is diagonal matrix.

◦ In general, R(a) operation is isomorphic with R∗(a) operations. So

R(a)iso←→ R∗(a) for all a.

◦ R(a) operation is special form for T (a) operation. And it’s able to rep-

resent as diagonal conditions for Pascal’s triangle operations. In general,

R(a) operation is following form.

R(a) = eas · T (0)

Therefore R(a) operation is able to extend to C∗-algebras, but we

have to attention to treat it because of T (0) operation is characteristics

condition.

35

§ Chapter 9

In this chapter, I want to explain about diagonal conditions of lattice

structures for Laplace transforms. This condition is generalized to certain

matrix conditions. And these matrices forms will be isomorphic with T (a)

operations.

©V(a) operation

T (a) operation is defined as following condition.

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt

Now, this T (a) operations is able to move to following polynomial

form.

T (a)f(t) =∞∑

n=0

n∑

k=0

f (n)(0)

(n− k)!

an−k

sk+1

as ‖T (a)‖ = 1.

In general, the lattice form is represented following

L(a)f(t) =

f(0)s

a0

f ′(0)s

a f ′(0)s2 a0

f ′′(0)2!

1sa2 f ′′(0)

s2 a f ′′(0)s3 a0

f (3)(0)3!

1sa3 f (3)(0)

2!1s2 a

2 f (3)(0)s3 a f (3)(0)

s4 a0

......

. . .

(7)

In this case, ‖L(a)‖ = 1.

Especially, I define V(a) operation as following.

36

V(a)f(t) =

f(0)s

a0

f ′(0)s2 a0

f ′′(0)s3 a0

f (3)(0)s4 a0

. . .

(8)

=

f(0)0!

1sa0

f ′(0)1!

1!s2 a

0

f ′′(0)2!

2!s3 a

0

f (3)(0)3!

3!s4 a

0

. . .

(9)

In this time, if f(t) = tn, then we have f (n)(0) = n!. Therefore,

V(a)

1

t

t2

t3

...

=

a0

a0

a0

a0

. . .

T (0)

1

t

t2

t3

...

= H(a)T (0)

1

t

t2

t3

...

(10)


V(a)tn = H(a)T (0)tn.

In general, it’s able to represent as

V(a) = H(a)T (0) = R(0) = R(a).

So we have

V(0) = H(a)T (0).

This operator algebra means the space of s-function. In other words,

V(a) operation is operator algebras of s-function. In this view point, V(a)

operation differes from H(a) operation. So Pascal’s matrix operation is

operated as special form for this lattice structures.

37

N.B. In general, V(a) operation is isomorphic with R(a) operation.

As a whole,

Matrices operators

F(a)iso←→ L(a)

homo−→ V(a)iso←→ H(a)

iso l l iso

T (a)homo−→ R(a)

Integral operators

N.B. Now, R(a) operation defines following.

R(a)f(t) =∫ ∞

0f(t)e−stdt = T (0)f(t).

So R(a) operation has been existed as diagonal condition on V(a)

operation.

Therefore, T (0) operation (Now Laplace transforms) has been also

existed on diagonal condition because of R(a) = T (0). So T (a) operation

needs to distinguish from T (0) = T (0) operation. This fundamental

concept is generated from following operator algebra.

T (a) = {−S1(a)} + T (0)

Moreover, by using the projection operator, we have following condi-

tions.

PL(a) = L(0) = V(0) and PT (a) = T (0) = R(0), resp.

Therefore

L(a)P−→ L(0) = V(0)

homo ↘ ↑ P

V(a)

38

Similarly,

T (a)P−→ T (0) = R(0)

homo ↘ ↑ P

R(a)

N.B. In this case, ‖T (a)‖ = ‖L(a)‖ = 1.

On the other hand, if f(t) = tn then we are able to represent as follow-

ing.

V(a)

1

t

t2

t3

...

= D(a)T (0)

1

t

t2

t3

...

(11)

Therefore

V(a) = D(a)T (0) if f(t) = (1 t t2 · · ·)T .

So

D(a)iso←→ V(a)

homo−→ V(0)iso←→ T (0).

For example, since

V(a)tn−1 =f (n−1)(0)

sna0 and V(a)tn =

f (n)(0)

sn+1a0.

In this time, let V(a)tn−1 = V(a)tn. In other words,

f (n−1)(0)

sna0 =

f (n)(0)

sn+1a0.

In general, it’s represented as

sf (n−1)(t) = f (n)(t).

39

This form be to Cauchy problems.


So we have,

f(t) = est.

So

f (n)(0) = sn

In this conditions, V(a) operations is able to represent as following.

V(a)est =

e0·ss

a0

e0·ss

a0

e0·ss

a0

e0·ss

a0

. . .

=

a0

sa0

sa0

sa0

s. . .

(12)

N.B. In this case, e0·s def .= a0.

(see. p.8, Chapter 1, No.7, N o.2)

So we have,

V(a)est =

a0

a0

a0

. . .

T (0)1 (13)

= D(a)T (0)1

Therefore

V(a) = D(a)T (0).

In general, V(a) operation is able to represent as following form.

L(a)f(t) = {−Z1(a)f(t)} + V(a)f(t)

40

N.B. In this time,

−Z1(a)f(t) =

f ′(0)s

af ′′(0)

2!1sa2 f ′′(0)

s2 af (3)(0)

3!1sa3 f (3)(0)

2!1s2 a

2 f (3)(0)s3 a

......

. . .

(14)

It has an ideal structure.

So we have following operator algebras.

N.B. V(a) = V(0) = L(0)

L(a) = {−Z1(a)} + L(0)

↓ ↓ ↓

Lower triangle matrix Ideal Diagonal condition

l l l

My Laplace transforms Ideal Now Laplace transtorms

↑ ↑ ↑

T (a) = {−S1(a)} + T (0)

N.B. In this case, ‖T (a)‖ = ‖L(a)‖ = 1.

These operator algebras are created to compare with following condi-

tion.

T (a)f(t) =∞∑

n=1

n−1∑

k=0

f (n)(0)

(n− k)!

an−k

sk+1

︸︷︷︸−S1(a)f(t)

+T (0)f(t)


41

This T (0) operation is “Now Laplace transforms” and it’s appeared

on diagonal condition on the lattice form V(a) operation. Especially,

since this “Now Laplace transforms” is unitary conditions, then the lattice

form is represented as V(a) operation. In this time, since V(a) operation is

same with V(0) operation, then we are able to have L(0) = V(0) operation

instead of T (0) operation. The term of∑ ∑

is able to represent as ideal

structures as −Z1(a) operation. As a whole, T (a) operation is isomorphic

with L(a) operation.

In this time, L(a) operation is able to preserve with V(a) operation.

This V(a) operation is able to extend as following conditions.

V (a) = easV(a)

Therefore L(a) operation is able to have various condition.

This condition is applied to T (a) operation with using Hahn-Banach

theory.


©Duality of V(a) operation

Now, V(a) operation is generated by s-function. So, in this time, I

have been considered as real condition of s-function.

V(a) operation defines as following.

V(a)f(t) =

f(0)s

a0

f ′(0)s2 a0

f ′′(0)s3 a0

f (3)(0)s4 a0

. . .

(15)

=

f(0)0!

1sa0

f ′(0)1!

1!s2 a

0

f ′′(0)2!

2!s3 a

0

f (3)(0)3!

3!s4 a

0

. . .

(16)

42

In this case, since s-function is generated as real space, then we have

following condition.

V(a)f(t) = V∗(a)f(t)

Especially, if f(t) is same with f(t) then we have

V(a) = V∗(a).

Therefore V(a) operation has Hermitian form if it’s real spaces. In

this time, V(a) operation has following form.

V(0) = V(a)iso←→R(a) = T (0)

So we have

T (0) = T ∗(0).

So “Now Laplace transforms” is generated by Hemitian form, too iff

it’s real space.

©Duality of L(a) operation

Now, L(a) operation is able to represent as following.

L(a)f(t) =

f(0)s

a0

f ′(0)s

a f ′(0)s2 a0

f ′′(0)2!

1sa2 f ′′(0)

s2 a f ′′(0)s3 a0

f (3)(0)3!

1sa3 f (3)(0)

2!1s2 a

2 f (3)(0)s3 a f (3)(0)

s4 a0

......

. . .

(17)

In this case, ‖L(a)‖ = 1.

If s-function is generated on real condition then we have

L∗(a)f(t) =

f(0)s

a0 f ′(0)s

a f ′′(0)2!

1sa2 f (3)(0)

3!1sa3 · · ·

f ′(0)s2 a0 f ′′(0)

s2 a f (3)(0)2!

1s2 a

2 · · ·f ′′(0)

s3 a0 f (3)(0)s3 a

f (3)(0)s4 a0

. . .

(18)

43

In this case, ‖L∗(a)‖ = 1.

In this time, the projection operator for L(a) operations are following

condition.

PL(a) = L(0) = V(a) = V∗(a) = L∗(0) = P ∗L∗(a).

The ideal∗ structure of L∗(a) operation is represented as following.

Z∗1(a)f(t) =

f ′(0)s

a f ′′(0)2!

1sa2 f (3)(0)

3!1sa3 · · ·

f ′′(0)s2 a f (3)(0)

2!1s2 a

2 · · ·f (3)(0)

s3 a. . .

(19)

As a whole, the −∗ algebras for L(a) operations are following.

N.B. V∗(a) = V∗(0) = L∗(0)

L∗(a) = {−Z∗1(a)} + L∗(0)

↓ ↓ ↓

Upper triangle matrix Ideal∗ Diagonal condition

l l l

My Laplace transforms Ideal∗ Now Laplace transtorms

↑ ↑ ↑

T ∗(a) = {−S∗1(a)} + T ∗(0)

N.B. In this case, ‖T ∗(a)‖ = ‖L∗(a)‖ = 1.

44

In this time, T ∗(a) operation is represented as following condition.

T ∗(a)f(t) =∞∑

k=0

k∑

n=0

f (k)(0)

(k − n)!

ak−n

sn+1

−S∗1(a)f(t) =∞∑

k=1

k−1∑

n=0

f (k)(0)

(k − n)!

ak−n

sn+1

and

T ∗(0)f(t) =∞∑

k=0

f (k)(0)

sk+1a0 =

∞∑

n=0

f (n)(0)

sn+1a0 = T (0)f(t)

as s-function is real spaces.

N.B. In this case, f(t) = f(t).

This T ∗(0) operation is also appeared on diagonal condition of V∗(a)

operation. In general, if s-function is complex number then it is not able

to extend to complex structures on Hermitian forms. So we have

T ∗(0) 6= T (0) (complex spaces).

So, I have been considered on real space, only iff it has Hermitian

form.

On the other hand, in general, we have

T ∗(0) = αT−1(0).

So we have

T (0) = αT−1(0).

iff it’s Hemitian form (real space).

Especially, when α = 1, we are able to have

T (0) = T−1(0).

This formula becomes to generate the following condition.

T−1(a) = T (−a)

45


If T (0) (Now Laplace transform) is defined on complex space then

I have following condition.

T−1(a)iso←→ T (−a)

because of T−1(0)iso←→ T (0).

Since this conditions, we have following relation.

‖T ∗(a)‖ = ‖T (a)‖

This condition is able to treat on complex spaces in spite of Hermite

form. This -∗ algebra becomes to treate on C∗- or W ∗-algebras.

So,

Bare ring −→ on real spaces.

Normed ring −→ on complex spaces, resp.

Since this conditions, we had better to define the complex form of 00.

Precisely speaking, we should redefine as the following form.

e0·s = 00 iso←→ 00 = e0·s

in complex spaces.

So , in general, the dual spaces should be separeted condition in com-

plex spaces.

On the contrary, if we are considering on real spaces then we are able

to define as following condition.

e0·s = 00 def= 00 = e0·s

In this time, we are able to treat on real spaces for all operations.

46

©V (a) operation

Now, T (a) operations for polynomial form is following.

T (a)f(t) =∞∑

n=0

n∑

k=0

f (n)(0)

(n− k)!

an−k

sk+1

as ‖T (a)‖ = |eas|.

In general, the lattice form is represented following

L(a)f(t) =

f(0)s

a0

f ′(0)s

a f ′(0)s2 a0

f ′′(0)2!

1sa2 f ′′(0)

s2 a f ′′(0)s3 a0

f (3)(0)3!

1sa3 f (3)(0)

2!1s2 a

2 f (3)(0)s3 a f (3)(0)

s4 a0

......

. . .

(20)

In this case, ‖L(a)‖ = |eas|.

Especially, I define V(a) operation as following.

V(a)f(t) =

f(0)s

a0

f ′(0)s2 a0

f ′′(0)s3 a0

f (3)(0)s4 a0

. . .

(21)

=

f(0)0!

1sa0

f ′(0)1!

1!s2 a

0

f ′′(0)2!

2!s3 a

0

f (3)(0)3!

3!s4 a

0

. . .

(22)

In this time, if f(t) = eastn, then we have f (n)(0) = easn!. Therefore,

47

V (a)

1

t

t2

t3

...

= eas

a0

a0

a0

a0

. . .

T (0)

1

t

t2

t3

...

(23)

N.B. V (a) = easV(a).



V (a)tn = easa0T (0)tn.

In general, it’s able to represent as

V (a) = a0R(a).

So we have

V (a) = R(a).

This operator algebra means the space of s-function. In other words,

V (a) operation is operator algebras of s-function.

N.B. In general, V (a) operation is isomorphic with R(a) operation.

As a whole,

V (a)iso←→ R(a)

(Matrices operators) homo ↑ ↑ homo (Integral operators)

L(a)iso←→ T (a)

N.B. Now, R(a) operation defines following.

R(a)f(t) = easR(a)f(t) =∫ ∞

0f(t)e−(t−a)sdt.

So R(a) operation has been existed as diagonal condition on V (a)

operation.

48

Therefore, T (0) operation (Now Laplace transforms) has been also ex-

isted on diagonal condition because of R(a) = easT (0). So T (a) operation

needs to distinguish from T (0) operation. This fundamental concept is

generated from following operator algebra.

T (a) = {−S(a)} +

R(a)︷︸︸︷eas · T (0)

and

L(a) = {−Z(a)} +

V (a)︷︸︸︷eas · L(0)

Moreover, by using the projection operator, we have following condi-

tions.

PL(a) = L(0) = V (0) and PT (a) = T (0) = R(0), resp.

Therefore

L(a)P−→ L(0) = V (0)

homo ↘ ↑ P

V (a)

Similarly,

T (a)P−→ T (0) = R(0)

homo ↘ ↑ P

R(a)

For example, since

V (a)tn−1 =f (n−1)(0)

sna0 and V (a)tn =

f (n)(0)

sn+1a0.

In this time, let V (a)tn−1 = V (a)tn. In other words,

f (n−1)(0)

sna0 =

f (n)(0)

sn+1a0.

49

In general, it’s represented as

sf (n−1)(t) = f (n)(t).

This form be to Cauchy problems.


In this case, we are able to have following condition.

f(t) = easest.

So

f (n)(0) = eassn

In this conditions, V(a) operations is able to represent as following.

V(a)f(t) = eas

e0·ss

a0

e0·ss

a0

e0·ss

a0

e0·ss

a0

. . .

= eas

a0

sa0

sa0

sa0

s. . .

(24)

N.B. In this case, e0·s def .= a0.

(see. p.8, Chapter 1, No.7, N o.2)

So we have,

V (a)est = eas

a0

a0

a0

. . .

T (0)1 (25)

= easD(a)T (0)1

Therefore

V (a) = D(a)T (0).

50

In general, V (a) operation is able to represent as following form.

L(a)f(t) = {−Z(a)f(t)} + V (a)f(t)

N.B. In this time,

−Z(a)f(t) = eas

f ′(0)s

af ′′(0)

2!1sa2 f ′′(0)

s2 af (3)(0)

3!1sa3 f (3)(0)

2!1s2 a

2 f (3)(0)s3 a

......

. . .

(26)

It has an ideal structure.

So we have following operator algebra.

N.B. V (0) = L(0)

L(a) = {−Z(a)} +

V (a)︷︸︸︷eas · L(0)

↓ ↓ ↓

Lower triangle matrix Ideal Diagonal condition

l l l

My Laplace transforms Ideal Now Laplace transtorms

↑ ↑ ↑

T (a) = {−S(a)} +

R(a)︷︸︸︷eas · T (0)

This T (0) operation is “Now Laplace transforms” and it’s appeared on

diagonal condition on the lattice form V (a) operation. Especially, since

51

this “Now Laplace transforms” is unitary conditions, then the lattice

form is represented as V(a) operation. In this time, since V(a) operation

is same with V(0) operation, then we are able to have L(0) = V (0) =

V(0) operation instead of T (0) operation. The term of −Z(a) is able to

represent as ideal structures. As a whole, T (a) operation is isomorphic

with L(a) operation as extended conditions.

N.B. V (a) = easV(a).


‖V (a)‖ = ‖R(a)‖ = |eas|.

In this time, L(a) operation is able to preserve with V (a) operation.

This condition is applied to T (a) operation with using Hahn-Banach the-

ory.


Reference

T (a)f(t) =∞∑

n=1

n−1∑

k=0

f (n)(0)

(n− k)!

an−k

sk+1

︸︷︷︸−S1(a)f(t)

+T (0)f(t)

N.B. In this time, ‖T (a)‖ = 1.


52

Some results

◦ V(a) operation is diagonal conditions for L(a) operation. So the prop-

erty of V(a) operation is same with L(a) operation.

N.B. L(a) operation is matrix condition for polynomial form of T (a)

operation.

◦ V(a) operation is isomorphic with R(a) operations. So

V(a)iso←→R(a) = 00 · T (0)

Therefore V(a) operation is able to also consider as the characteristics

conditions.

◦ If s is real condition then we have Hermitian form. So

V(a) = V∗(a).

On the contrary, if s is complex condition then we should consider on

following form.

‖V(a)‖ = ‖V∗(a)‖ = {1} or {0}.

◦ V (a) operation is isomorphic with R(a) operation. So we have

V (a)iso←→ R(a) = easT (0).

◦ V (a) operation is able treat on C∗-algebras and it also has characteris-

tics form.

◦ V (a) operation is special matrix of the diagonal conditions for T (a)

operation.

53

Conclusion

We have researched about the diagonal conditions for T (a) opera-

tions. This condition is special conditions for T (a) operations. However,

it’s Now Laplace transforms (T (0) operation). In fact, T (0) operation

has many properties for T (a) operation.

In this case, we researched from two kinds of matrices forms. One

is Pascal’s triangle matrix form and other is generalised form as lattice

conditions.

Both matrices conditions are able to generate the T (a) operation. Es-

pecially, if a is zero conditon then we have diagonal conditions of these

matrices. One is H(a) operations in Pascal’s matrix form and other is

V(a) operations in lattice condition of T (a) operation.

These two operations have closed relations with Now Laplace trans-

forms (T (0) operation). Now, if we check the property of H(a) or V(a)

operation then we also obtain the property of F(a) or L(a) operation,

resp. So, the property of T (0) operation exists onto T (a) operation.

Moreover, we have following conditions.

F(a)iso←→ T (a)

iso←→ L(a)

↓ ↓ ↓H(0)

iso←→ T (0)iso←→ V(0)

In this time, if a is to zero condition then we have 00 form (characteris-

tics condition) in these diagonal. This diagonalization is T (0) operation.

Therefore, now laplace transform will be represent as following.

T (0)f(t) =∫ ∞

0f(t)e−(t−0)sdt

N.B. e0·s = 00.

Therefore, this structure is not easy but it has wide considerations.

(Sun) 6.May.2012

Now, let′s go to the next papers with me.

54

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56

the diagonal conditions for laplace transformsopab.web.fc2.com/no.9.pdfpreface in this paper, i...

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