The projection operator for Laplace transforms

The projection operator

for Laplace transforms

No.1

Takao Saito

1

Thank you for our world

2

The projection operator

for Lapalace transforms

3

Preface

This paper is builded by three parts. First is D(a) → T (a) operations

in Hilbert spaces. Second is irredusibility for Laplace transforms. Finally,

the property of projection for Laplace transforms.

Papers

About solvers of differential equations of relatively

for Laplace transforms

Operator algebras for Laplace transforms

Rings and ideal structures for Laplace transforms

Groups and matrix operator for Laplace transforms

Extension and contaction for transrated operators

Operator algebras for group conditions

Some matrices rings 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

(Tue)28.Jul.2009

Takao Saito

4

Contents

Preface

§ Chapter 1

D(a)

◦D(a) operations in Hilbert spaces · · · · · ·8◦D(a) operations in Hilbert spaces · · · · · ·10

◦ Representation of inner product in D(a) operation · · · · · ·12

◦ Example for L2-spaces · · · · · · · · · · · ·13


Y (a)

◦ Y(a) operations in Hilbert spaces · · · · · ·17

◦ Y (a) operations in Hilbert spaces · · · · · ·19

◦ Representation of inner product in Y (a) operation · · · · · ·21



F (a)

◦ F(a) operations in Hilbert spaces · · · · · ·27

◦ F (a) operations in Hilbert spaces · · · · · ·29

◦ Representation of inner product in F (a) operation · · · · · ·31


T (a) (Laplace transforms)

◦ T (a) operations in Hilbert spaces · · · · · ·35

◦ T (a) operations in Hilbert spaces · · · · · ·37

◦ Representation of inner product in T (a) operation · · · · · · 39

5


◦ Some results · · · · · · · · · · · ·43

§ Chapter 2

◦ Rotation operator A(a) based by Laplace transforms · · · 44

◦ Hyperbolic operator A1(a) based by Laplace transforms · · · 48

◦ About property of ebt · · · · · · · · · · · ·51

◦ Rotation operator A(a) based by Laplace transforms · · · · · ·51

◦ Hyperbolic operator A1(a) based by Laplace transforms · · ·54

◦ Some results · · · · · · · · ·59

§ Chapter 3

◦ Laplace transforms and Cauchy problem · · · · · · · · · · · ·60

◦ Projection of Laplace transforms PT (a) = T (0) · · · · · · · · ·62

◦ Ideal structure to ring conditions · · · · · ·64

◦ Homomorphic theorem and projection for equivalence · · · 65

◦ The property of projection and T (a),S(a) operations · · · 67

◦ Relation of T (a) and S(a) (unitary condition) · · · · · · 70

◦ Some results · · · · · · · · · · · · 71

◦ Conclusion · · · · · · · · · 72

◦ References · · · · · · · · · · · · 73

6

Chapter 1

In this chapter, let consider the relation of ring condition and ideal

structure. The ideal structures are generated as subring.

Now

ring1 = ring2 + ideal

p = q · r + s

At the first time, D(a), Y (a), F (a) and T (a) operations are included

in this ring1 condition. The next step, ring2 is generated by primitive

structure. What is called identity (r)? The coefficient of identity is

indicated as eas (q). (irreducible) D(0), Y (0), F (0) and T (0) operations

are included in this identical operations. Finally, the ideal structures are

represented by O(a), N(a), G(a) and S(a) operations, respectively (s). As

a whole, this ring2 and ideal structures are also ring conditions (ring1).

At the first step, I want to explain the simplest form D(a) operation.

D(a) → Y (a) → F (a) → T (a)

D(a), F (a) operations are matrices oprations. And Y (a), T (a) opera-

tions are integral operations.

Now, we have following condition ring1. (L2-spaces)

(L2 − spaces) D(a)iso←→ Y (a)

iso←→ F (a)iso←→ T (a) (L2 − spaces)

The ideal form is following relation ideal.

O(a)iso←→ N(a)

iso←→ G(a)iso←→ S(a)

As a whole, we have following condition rins2.

H(a)iso←→ W (a)

iso←→ H(a)iso←→ R(a)

7

If D(a) and Y (a) operations are defined on L1-spaces then we have

following.

(L1 − spaces) D(a)iso←→ Y (a)

homo←− F (a)iso←→ T (a) (L2 − spaces)

The ideal form is following relation ideal.

O(a)iso←→ N(a)

homo←− G(a)iso←→ S(a)

As a whole, we have following condition rins2.

D(a)iso←→ W (a)

iso←→ H(a)iso←→ R(a)

©D(a) operations in Hilebert spaces

At the first time, D(a) operation is defined following.

D(a) =

(a0

a0

)(1)

D(a) operation for inner product spaces in real condition is presented

following.

< D(a),D(a) >≥ 0

< D(a),D(a) >= 0 iff D(∞) = 0

λ < D(a),D(b) >=< λD(a),D(b) >

< D(a) +D(b), D(c) >=< D(a),D(c) > + < D(b),D(c) >

< D(a),D(b) >=< D(b),D(a) >

(2)

Proof. At the first time, since definition of inner product, then we

have √< D(a),D(a) > = ‖D(a)‖

So

< D(a),D(a) >= ‖D(a)‖2

In this time, the norm of D(a) operation is presented as the determim-

inant of matrix D(a). So we have, ‖D(a)‖ = 1 ≥ 0. N.B. ‖D(∞)‖ = 0.

Therefore

< D(a),D(a) >= ‖D(a)‖2 ≥ 0

8

< D(a),D(a) >≥ 0.

Second steps, If a →∞ then ‖D(a)‖ = 0.

Therefore

< D(a),D(a) >= 0 iff D(∞) = 0.

Third steps, clearly, λ < D(a),D(b) >= ‖λD(a)‖D(b)‖ cos θ =<

λD(a),D(b) >.

Fourth steps, now we have ‖D(a) + D(b)‖ ≤ ‖D(a)‖ + ‖D(b)‖. In

this time, D(a) and D(b) are positive operator then ‖D(a) + D(b)‖ =

‖D(a)‖+ ‖D(b)‖.

So

< D(a) +D(b),D(c) >= ‖D(a) +D(b)‖‖D(c)‖ cos θ

= ‖D(a)‖‖D(c)‖ cos θ + ‖D(b)‖‖D(c)‖ cos θ

=< D(a),D(c) > + < D(b),D(c) > .

Finally, if it’s real spaces then we have

< D(a),D(b) >= ‖D(a)‖‖D(b)‖ cos θ = ‖D(b)‖‖D(a)‖ cos θ =< D(b),D(a) > .

Therefore, the proof was completed.

This normed condition is following.

‖D(a)‖ ≥ 0

‖D(a)‖ = 0 iff D(∞) = 0

‖λD(a)‖ ≤ |λ|‖D(a)‖‖D(a) +D(b)‖ ≤ ‖D(a)‖+ ‖D(b)‖

(3)

Moreover it is able to complete as a →∞.

Therefore

D(a) operation is also able to define on Hilbert spaces.

9

©D(a) operations in Hilebert spaces

The next time, D(a) operation is defined following.

D(a) = eas

(a0

a0

)(4)

D(a) operation for inner product spaces in real condition is presented

following.

< D(a), D(a) >≥ 0

< D(a), D(a) >= 0 iff D(±∞) = 0

λ < D(a), D(b) >=< λD(a), D(b) >

< D(a) + D(b), D(c) >=< D(a), D(c) > + < D(b), D(c) >

< D(a), D(b) >=< D(b), D(a) >

(5)


have √< D(a), D(a) > = ‖D(a)‖

So

< D(a), D(a) >= ‖D(a)‖2

In this time, the norm of D(a) operation is presented as the deter-

miminant of matrix D(a). So we have, ‖D(a)‖ = ‖eas‖ ≥ 0.

Therefore

< D(a), D(a) >= ‖D(a)‖2 ≥ 0

< D(a), D(a) >≥ 0.

Second steps, If a → −∞ and s ≥ 0 then eas → 0.

Therefore

< D(a), D(a) >= 0 iff D(±∞) = 0.

Third steps, clearly, λ < D(a), D(b) >= ‖λD(a)‖D(b)‖ cos θ =<

λD(a), D(b) >.

Fourth steps, now we have ‖D(a) + D(b)‖ ≤ ‖D(a)‖ + ‖D(b)‖. In

10

this time, D(a) and D(b) are positive operator then ‖D(a) + D(b)‖ =

‖D(a)‖+ ‖D(b)‖.

So

< D(a) + D(b), D(c) >= ‖D(a) + D(b)‖‖D(c)‖ cos θ

= ‖D(a)‖‖D(c)‖ cos θ + ‖D(b)‖‖D(c)‖ cos θ

=< D(a), D(c) > + < D(b), D(c) > .


< D(a), D(b) >= ‖D(a)‖‖D(b)‖ cos θ = ‖D(b)‖‖D(a)‖ cos θ =< D(b), D(a) > .



‖D(a)‖ ≥ 0

‖D(a)‖ = 0 iff D(±∞) = 0

‖λD(a)‖ ≤ |λ|‖D(a)‖‖D(a) + D(b)‖ ≤ ‖D(a)‖+ ‖D(b)‖

(6)


Therefore

D(a) operation is also able to define on Hilbert spaces.

11

© Representation of inner product in D(a) operation

Let the following definition.

D(a) = easD(a) = eas

(a0

a0

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

(

0

)as “a′′ is finite. (7)

It’s the simplest form for considering about operation T (a) and S(a).

D(a) is identity. So D(a) = D∗(a). It’s called Hermitian. In this time,

D(∞) = 0 and O(∞) = I in L2-spaces.

D(a)v ≡ D(a)v + O(a)v = easD(a)v + easO(a)v

= eas[D(a)v +O(a)v] = easD(0)v

N.B. v is n-th vector.

Therefore

D(a) = D(a) + O(a) = eas[D(a) +O(a)] = easD(0)

⇓x = x1 + x2 = α < x′1 + x′2 >= αx0 N.B. x0 = e.

< α, x′1 > + < α, x′2 >=< α, x′1 + x′2 >=< α, x0 >

Hence

< eas,D(a) > + < eas,O(a) >=< eas, D(0) >

< D(a), 1 >=< D(a), 1 > + < O(a), 1 >=< easD(0), 1 >=< 1, D(a)∗ >

Similarly, a and b are generated by shift and norm ,respectivily.

If a = 0 then diagonal is open conditions. It’s refered to irreducible

conditions, directry. Therefore it represents on a = 0 only. An this norm

is eas. Other condition of this matrix is satisfied pure null conditions.

Now, we obtain a01 = a0

2 = 1 iff a1 = a2 6= 0. So this matrix is generated

identity iff a 6= 0.

Reference

< eas,Y(a) > + < eas,N (a) >=< eas, Y (0) >

12

< W (a), 1 >=< Y (a), 1 > + < N(a), 1 >=< easY (0), 1 >=< 1,W ∗(a) >

< eas,F(a) > + < eas,G(a) >=< eas, F (0) >

< H(a), 1 >=< F (a), 1 > + < G(a), 1 >=< easF (0), 1 >=< 1, H∗(a) >

< eas, T (a) > + < eas,S(a) >=< eas, T (0) >

< R(a), 1 >=< T (a), 1 > + < S(a), 1 >=< easT (0), 1 >=< 1, R∗(a) >

On the other hand,

D(a) +O(a) = D(0) = I(a) additive

In this time, D(a) is identity as a is finite. Similarly, O(a) is an-

nihilator as a is finite. If a is infinite condition then we should define

as D(∞) = 0 and O(∞) = I. So this projection operator is treated as

P (a) = 1 and Q(a) = 0 as a is finite,respectively. If a is infinite then

P (∞) = 0 and Q(∞) = 1.

P (a) + Q(a) = {1}

Similarly, the representation for inner product is following.

< D(a), O(a) >= {0} (ideal) Multiplicative

= {0} (ideal)

As a whole, this form is generated ring condition.

Example 1. Now, we consider the relations of D(a) and O(a) opera-

tions in L2-spaces. This space is generated on orthogonal condition. So

we should attention to D(a) and O(a) operations are orthogonal, respec-

tively. By using the projection operator then we have following.

< D(0),O(a) >==< D(a), P ∗(a)O(a) >

=< D(a), P (a)O(a) >=< D(a),O(0) >= 0

So

< D(0),O(a) >=< D(a),O(0) >= 0.

13

Moreover

< D(a),O(0) >=< D(a), Q(a)O(a) >=< Q∗(a)D(a),O(a) >

=< Q(a)D(a),O(a) >=< 0 · D(0),O(a) >= 0

So

< D(0),O(a) >= 0.

On the contrary,

< D(a),O(∞) >=< D(a), Q(a)O(a) >=< Q∗(a)D(a),O(a) >

=< Q(a)D(a),O(a) >=< D(∞),O(a) >= 0

So

< D(a),O(∞) >=< D(∞),O(a) >= 0.

Therefore, we have following

< D(a),D⊥(0) >= 0.

N.B. D(a) and O(a) are orthogonal conditions, respectivily.

Moreover,

(P (a)D(a), P (a)O(a)) = (Q(a)O(a), Q(a)D(a))

(lima→0

D(a), lima→0

O(a)) = ( lima→∞O(a), lim

a→∞D(a)) = (I, 0)

(lima→0

P (a), lima→0

Q(a)) = ( lima→∞Q(a), lim

a→∞P (a)) = (I, 0)

lima→0

(P (a)D(a) + Q(a)O(a)) = lima→∞(Q(a)O(a) + P (a)D(a))

P (0)D(0) = Q(∞)O(∞) = I (ring)

Q(0)O(0) = P (∞)D(∞) = 0 (ideal)

14

If it’s inner product space then we have following.

=< D(a), P ∗(a)P (a)O(a) >

=< D(a), P (a)O(a) >=< D(a),O(0) >= 0

Similarly

< Q(a)O(a), Q(a)D(a) >=< O(a), Q∗(a)Q(a)D(a) >

=< O(a), Q(a)D(a) >=< O(a),D(∞) >= 0.

So we have

=< Q(a)O(a), Q(a)D(a) > .

< D(a), P (a)O(a) >=< O(a), Q(a)D(a) >= 0.

Example 2. If there operations are defined on L1-spaces then we

attention to that D(a) and O(a) operations are parallel conditions, re-

spectively. So we have following.

< D(0),O(a) >==< D(a), P ∗(a)O(a) >

=< D(a), P (a)O(a) >=< D(a),O(0) >= 0

So

< D(0),O(a) >=< D(a),O(0) >= 0

Moreover

< D(a),O(0) >=< D(a), Q(a)O(a) >=< Q∗(a)D(a),O(a) >

=< Q(a)D(a),O(a) >=< 0 · D(0),O(a) >= 0

So

< D(0),O(a) >= 0

On the contrary,

< D(a),O(∞) >=< D(a), Q(a)O(a) >=< Q∗(a)D(a),O(a) >

15

=< Q(a)D(a),O(a) >=< D(∞),O(a) >= 0

So

< D(a),O(∞) >=< D(∞),O(a) >= 0

Moreover,

(P (a)D(a), P (a)O(a)) = (Q(a)D(a), Q(a)O(a))

(lima→0

D(a), lima→0

O(a)) = ( lima→∞D(a), lim

a→∞O(a)) = (I, 0)

(lima→0

P (a), lima→0

Q(a)) = ( lima→∞P (a), lim

a→∞Q(a)) = (I, 0)

lima→0

(P (a)D(a) + Q(a)O(a)) = lima→∞(P (a)D(a) + Q(a)O(a))

P (0)D(0) = P (∞)D(∞) = I (ring)

Q(0)O(0) = Q(∞)O(∞) = 0 (ideal)


=< D(a), P ∗(a)P (a)O(a) >

=< D(a), P (a)O(a) >=< D(a),O(0) >= 0

Similarly

< Q(a)O(a), Q(a)D(a) >=< O(a), Q∗(a)Q(a)D(a) >

=< O(a), Q(a)D(a) >=< O(a),D(∞) >= 0.

So we have

=< Q(a)D(a), Q(a)O(a) > .

< D(a), P (a)O(a) >=< D(a), Q(a)O(a) >= 0.

16

©Y(a) operations in Hilebert spaces

At the first time, Y(a) operation is defined following.

Y(a)f(t) =∫ ∞

af(t)dt

Y(a) operation for inner product spaces in real condition is presented

following.

< Y(a),Y(a) >≥ 0

< Y(a),Y(a) >= 0 iff Y(∞) = 0

λ < Y(a),Y(b) >=< λY(a),Y(b) >

< Y(a) + Y(b),Y(c) >=< Y(a),Y(c) > + < Y(b),Y(c) >

< Y(a),Y(b) >=< Y(b),Y(a) >

(8)


have √< Y(a),Y(a) > = ‖Y(a)‖

So

< Y(a),Y(a) >= ‖Y(a)‖2

In this time, the norm of Y(a) operation is presented as the norm of

kernel. So we have, ‖Y(a)‖ = ‖ker‖ = 1 ≥ 0.

Therefore

< Y(a),Y(a) >= ‖Y(a)‖2 ≥ 0

< Y(a),Y(a) >≥ 0.

Second steps, Y(a) operation is able to rewrite following.

Y(a)f(t) =∫ ∞

af(t)dt.

If a →∞ then we have following form.∫ ∞

∞f(t)dt = 0.

17

So, Y(∞) is also converged to 0.

Therefore

< Y(a),Y(a) >= 0 iff Y(∞) = 0.

Third steps, clearly, λ < Y(a),Y(b) >= ‖λY(a)‖Y(b)‖ cos θ =<

λY(a),Y(b) >.

Fourth steps, now we have ‖Y(a) + Y(b)‖ ≤ ‖Y(a)‖ + ‖Y(b)‖. In

this time, Y(a) and Y(b) are positive operator then ‖Y(a) + Y(b)‖ =

‖Y(a)‖+ ‖Y(b)‖.

So

< Y(a) + Y(b),Y(c) >= ‖Y(a) + Y(b)‖‖Y(c)‖ cos θ

= ‖Y(a)‖‖Y(c)‖ cos θ + ‖Y(b)‖‖Y(c)‖ cos θ

=< Y(a),Y(c) > + < Y(b),Y(c) > .


< Y(a),Y(b) >= ‖Y(a)‖‖Y(b)‖ cos θ = ‖Y(b)‖‖Y(a)‖ cos θ =< Y(b),Y(a) > .



‖Y(a)‖ ≥ 0

‖Y(a)‖ = 0 iff Y(∞) = 0

‖λY(a)‖ ≤ |λ|‖Y(a)‖‖Y(a) + Y(b)‖ ≤ ‖Y(a)‖+ ‖Y(b)‖

(9)


Therefore

Y(a) operation is able to define on Hilbert spaces.

18

©Y (a) operations in Hilebert spaces

The next time, Y (a) operation is defined following.

Y (a)f(t) =∫ ∞

af(t)easdt

Y (a) operation for inner product spaces in real condition is presented

following.

< Y (a), Y (a) >≥ 0

< Y (a), Y (a) >= 0 iff Y (±∞) = 0

λ < Y (a), Y (b) >=< λY (a), Y (b) >

< Y (a) + Y (b), Y (c) >=< Y (a), Y (c) > + < Y (b), Y (c) >

< Y (a), Y (b) >=< Y (b), Y (a) >

(10)


have √< Y (a), Y (a) > = ‖Y (a)‖

So

< Y (a), Y (a) >= ‖Y (a)‖2

In this time, the norm of Y (a) operation is presented as the norm of

kernel. So we have, ‖Y (a)‖ = ‖k(s)‖ = ‖eas‖ ≥ 0.

Therefore

< Y (a), Y (a) >= ‖Y (a)‖2 ≥ 0

< Y (a), Y (a) >≥ 0.

Second steps, Y (a) operation is able to rewrite following.

Y (a)f(t) = eas∫ ∞

af(t)dt.

If a → −∞ then the integral condition is produced as unitary opera-

tion. So, as a whole, Y (−∞) is converged to 0. Simirarly, if a →∞ then

we have following form. ∫ ∞

∞f(t)dt = 0.

19

So, as a whole, Y (∞) is also converged to 0.

Therefore

< Y (a), Y (a) >= 0 iff Y (±∞) = 0.

Third steps, clearly, λ < Y (a), Y (b) >= ‖λY (a)‖Y (b)‖ cos θ =<

λY (a), Y (b) >.

Fourth steps, now we have ‖Y (a) + Y (b)‖ ≤ ‖Y (a)‖ + ‖Y (b)‖. In

this time, Y (a) and Y (b) are positive operator then ‖Y (a) + Y (b)‖ =

‖Y (a)‖+ ‖Y (b)‖.

So

< Y (a) + Y (b), Y (c) >= ‖Y (a) + Y (b)‖‖Y (c)‖ cos θ

= ‖Y (a)‖‖Y (c)‖ cos θ + ‖Y (b)‖‖Y (c)‖ cos θ

=< Y (a), Y (c) > + < Y (b), Y (c) > .


< Y (a), Y (b) >= ‖Y (a)‖‖Y (b)‖ cos θ = ‖Y (b)‖‖Y (a)‖ cos θ =< Y (b), Y (a) > .



‖Y (a)‖ ≥ 0

‖Y (a)‖ = 0 iff Y (±∞) = 0

‖λY (a)‖ ≤ |λ|‖Y (a)‖‖Y (a) + Y (b)‖ ≤ ‖Y (a)‖+ ‖Y (b)‖

(11)


Therefore

Y (a) operation is able to define on Hilbert spaces.

20

© Representation of inner product of Y(a) for Laplace transforms.

Let the following

Y (a)f(t) =∫ ∞

af(t)easdt = eas

∫ ∞

af(t)dt = easY(a)f(t)

N(a)f(t) =∫ a

0f(t)easdt = eas

∫ a

0f(t)dt = easN (a)f(t).

W (a)f(t) ≡ Y (a)f(t) + N(a)f(t) = easY(a)f(t) + easN (a)f(t)

= eas[Y(a)f(t) +N (a)f(t)] = easY (0)f(t)

So

W (a) = Y (a) + N(a) = eas[Y(a) +N (a)] = easY (0)

⇓x = x1 + x2 = α < x′1 + x′2 >= αx0 N.B. x0 = e.

< α, x′1 > + < α, x′2 >=< α, x′1 + x′2 >=< α, x0 >

Hence

< eas,Y(a) > + < eas,N (a) >=< eas, Y (0) >

< W (a), 1 >=< Y (a), 1 > + < N(a), 1 >=< easY (0), 1 >=< 1,W ∗(a) >

Reference

< eas,D(a) > + < eas,O(a) >=< eas, D(0) >

< D(a), 1 >=< D(a), 1 > + < O(a), 1 >=< easD(0), 1 >=< 1, D∗(a) >

< eas,F(a) > + < eas,G(a) >=< eas, F (0) >

< H(a), 1 >=< F (a), 1 > + < G(a), 1 >=< easF (0), 1 >=< 1, H∗(a) >

< eas, T (a) > + < eas,S(a) >=< eas, T (0) >

< R(a), 1 >=< T (a), 1 > + < S(a), 1 >=< easT (0), 1 >=< 1, R∗(a) >

21

On the other hand,we have following.

Y(a) +N (a) = W(a)

In this case, the projection operator have two kinds form ,respectively.

Since, Y(0) is identity and Y(∞) is kind of annihilator in L2 spaces

then the projection operator have P (0) = 1 and P (∞) = 0 ,respectively.

Similarly, N (0) is annihilator and N (∞) is kind of identity in L2 spaces

then we have Q(0) = 0 and Q(∞) = 1 (maximal ideal). As a whole,

P (a) + Q(a) = {1}


< Y(a),N (a) >= Y ′(a) or N ′(a) (ideal)

= P ′(a) or Q′(a) (ideal)

or

< Y(a),N (a) >= {0} (ideal)

= {0} (ideal)

Example 1. Now, we consider the relations of Y(a) and N (a) operations

in L2-spaces. Similarly, Y(a) and N (a) operations are orthogonal condi-

tions, respectively. And by using the projection operator then we have

following.

< Y(0),N (a) >==< Y(a), P ∗(a)N (a) >

=< Y(a), P (a)N (a) >=< Y(a),N (0) >= 0

So

< Y(0),N (a) >=< Y(a),N (0) >= 0.

Moreover

< Y(a),N (0) >=< Y(a), Q(a)N (a) >=< Q∗(a)Y(a),N (a) >

=< Q(a)Y(a),N (a) >=< 0 · Y(0),N (a) >= 0

22

So

< Y(0),N (a) >= 0.

On the contrary,

< Y(a),N (∞) >=< Y(a), Q(a)N (a) >=< Q∗(a)Y(a),N (a) >

=< Q(a)Y(a),N (a) >=< Y(∞),N (a) >= 0

So

< Y(a),N (∞) >=< Y(∞),N (a) >= 0.


< Y(a),Y⊥(0) >= 0.

N.B. Y(a) and N (a) are orthogonal conditions, respectivily.

Moreover,

(P (a)Y(a), P (a)N (a)) = (Q(a)N (a), Q(a)Y(a))

(lima→0

Y(a), lima→0

N (a)) = ( lima→∞N (a), lim

a→∞Y(a)) = (1, 0)

(lima→0

P (a), lima→0


a→∞P (a)) = (1, 0)

lima→0

(P (a)Y(a) + Q(a)N (a)) = lima→∞(Q(a)N (a) + P (a)Y(a))

P (0)Y(0) = Q(∞)N (∞) = {1} (ring)

Q(0)N (0) = P (∞)Y(∞) = {0} (ideal)


=< Y(a), P ∗(a)P (a)N (a) >

=< Y(a), P (a)N (a) >=< Y(a),N (0) >= 0

23

Similarly

< Q(a)Y(a), Q(a)N (a) >=< Y(a), Q∗(a)Q(a)N (a) >

=< Y(a), Q(a)N (a) >=< Y(a),N (∞) >= 0.

So we have

=< Q(a)N (a), Q(a)Y(a) > .

< Y(a), P (a)N (a) >=< N (a), Q(a)Y(a) >


Y(a) +N (a) = W(a)


Since, Y(0) is identity and Y(∞) is also identity in L1 spaces. The

projection operator have P (0) = 1 and P (∞) = 0 . Similarly, N (0)

is annihilator and N (∞) is also annihilator in L1 spaces. The projection

operator Q(0) = 0 and Q(∞) = 1 (maximal ideal). As a whole,

P (a) + Q(a) = {1}


< Y(a),N (a) >= Y ′(a) or N ′(a) (ideal)


or

< Y(a),N (a) >= {0}= {0}

Example 2. If this relations are defineded on L1-spaces then we have

to attention to that Y(a) and N (a) operations are parallel conditions,

respectively. So we have following.

< Y(0),N (a) >==< Y(a), P ∗(a)N (a) >

24

=< Y(a), P (a)N (a) >=< Y(a),N (0) >= 0

So

< Y(0),N (a) >=< Y(a),N (0) >= 0

Moreover

< Y(a),N (0) >=< Y(a), Q(a)N (a) >=< Q∗(a)Y(a),N (a) >

=< Q(a)Y(a),N (a) >=< 0 · Y(0),N (a) >= 0

So

< Y(0),N (a) >= 0

On the contrary,

< Y(a),N (∞) >=< Y(a), Q(a)N (a) >=< Q∗(a)Y(a),N (a) >

=< Q(a)Y(a),N (a) >=< Y(∞),N (a) >= 0

So

< Y(a),N (∞) >=< Y(∞),N (a) >= 0

Moreover,

(P (a)Y(a), P (a)N (a)) = (Q(a)Y(a), Q(a)N (a))

(lima→0

Y(a), lima→0

N (a)) = ( lima→∞Y(a), lim

a→∞N (a)) = (1, 0)

(lima→0

P (a), lima→0

Q(a)) = ( lima→∞P (a), lim

a→∞Q(a)) = (1, 0)

lima→0

(P (a)Y(a) + Q(a)N (a)) = lima→∞(P (a)Y(a) + Q(a)N (a))

P (0)Y(0) = P (∞)Y(∞) = {1} (ring)

Q(0)N (0) = Q(∞)N (∞) = {0} (ideal)


=< Y(a), P ∗(a)P (a)N (a) >

25

=< Y(a), P (a)N (a) >=< Y(a),N (0) >= 0

Similarly

< Q(a)Y(a), Q(a)N (a) >=< Y(a), Q∗(a)Q(a)N (a) >

=< Y(a), Q(a)N (a) >=< Y(a),N (∞) >= 0.

So we have

=< Q(a)Y(a), Q(a)N (a) > .

< Y(a), P (a)N (a) >=< Y(a), Q(a)N (a) >

26

©F(a) operations in Hilebert spaces

In this time, F(a) operation is based on Pascal’s triangle. So I have

F(a) operation is defined following.

F(a) =

(a0

a a0

)(12)

F(a) operation for inner product spaces in real condition is presented

following.

< F(a),F(a) >≥ 0

< F(a),F(a) >= 0 iff F(∞) = 0

λ < F(a),F(b) >=< λF(a),F(b) >

< F(a) + F(b), F (c) >=< F(a),F(c) > + < F(b),F(c) >

< F(a),F(b) >=< F(b),F(a) >

(13)


have √< F(a),F(a) > = ‖F(a)‖

So

< F(a),F(a) >= ‖F(a)‖2

In this time, the norm of F(a) operation is presented as the determim-

inant of matrix F(a). So we have, ‖F(a)‖ ≥ 0 in L2-spaces.

Therefore

< F(a),F(a) >= ‖F(a)‖2 ≥ 0

< F(a),F(a) >≥ 0.

Second steps, if a → ∞ then the matrix condition is converged to 0.

So, as a whole, F(∞) is converged to 0.

Therefore

< F(a),F(a) >= 0 iff F(∞) = 0.

27

Third steps, clearly, λ < F(a),F(b) >= ‖λF(a)‖‖F(b)‖ cos θ =<

λF(a),F(b) >.

Fourth steps, now we have ‖F(a) + F(b)‖ ≤ ‖F(a)‖ + ‖F(b)‖. In

this time, F(a) and F(b) are positive operator then ‖F(a) + F(b)‖ =

‖F(a)‖+ ‖F(b)‖.

So

< F(a) + F(b),F(c) >= ‖F(a) + F(b)‖‖F(c)‖ cos θ

= ‖F(a)‖‖F(c)‖ cos θ + ‖F(b)‖‖F(c)‖ cos θ

=< F(a),F(c) > + < F(b),F(c) > .


< F(a),F(b) >= ‖F(a)‖‖F(b)‖ cos θ = ‖F(b)‖‖F(a)‖ cos θ =< F(b),F(a) > .



‖F(a)‖ ≥ 0

‖F(a)‖ = 0 iff F(∞) = 0

‖λF(a)‖ ≤ |λ|‖F(a)‖‖F(a) + F(b)‖ ≤ ‖F(a)‖+ ‖F(b)‖

(14)


Therefore

F(a) operation is also able to define on Hilbert spaces.

28

©F (a) operations in Hilebert spaces

Similarly, F (a) operation is defined following.

F (a) = eas

(a0

a a0

)(15)

F (a) operation for inner product spaces in real condition is presented

following.

< F (a), F (a) >≥ 0

< F (a), F (a) >= 0 iff F (±∞) = 0

λ < F (a), F (b) >=< λF (a), F (b) >

< F (a) + F (b), F (c) >=< F (a), F (c) > + < F (b), F (c) >

< F (a), F (b) >=< F (b), F (a) >

(16)


have √< F (a), F (a) > = ‖F (a)‖

So

< F (a), F (a) >= ‖F (a)‖2

In this time, the norm of F (a) operation is presented as the determim-

inant of matrix F (a). So we have, ‖F (a)‖ = ‖eas‖ ≥ 0.

Therefore

< F (a), F (a) >= ‖F (a)‖2 ≥ 0

< F (a), F (a) >≥ 0.

Second steps, If a → −∞ then eas → 0. Simirarly, if a →∞ then the

matrix condition is converged to 0. So, as a whole, F (∞) is converged to

0.

Therefore

< F (a), F (a) >= 0 iff F (±∞) = 0.

29

Third steps, clearly, λ < F (a), F (b) >= ‖λF (a)‖F (b)‖ cos θ =< λF (a), F (b) >.

Fourth steps, now we have ‖F (a) + F (b)‖ ≤ ‖F (a)‖ + ‖F (b)‖. In

this time, F (a) and F (b) are positive operator then ‖F (a) + F (b)‖ =

‖F (a)‖+ ‖F (b)‖.

So

< F (a) + F (b), F (c) >= ‖F (a) + F (b)‖‖F (c)‖ cos θ

= ‖F (a)‖‖F (c)‖ cos θ + ‖F (b)‖‖F (c)‖ cos θ

=< F (a), F (c) > + < F (b), F (c) > .


< F (a), F (b) >= ‖F (a)‖‖F (b)‖ cos θ = ‖F (b)‖‖F (a)‖ cos θ =< F (b), F (a) > .



‖F (a)‖ ≥ 0

‖F (a)‖ = 0 iff F (±∞) = 0

‖λF (a)‖ ≤ |λ|‖F (a)‖‖F (a) + F (b)‖ ≤ ‖F (a)‖+ ‖F (b)‖

(17)


Therefore

F (a) operation is also able to define on Hilbert spaces.

30

© Representation of inner product in F(a) operation.

Let the following

F (a) = eas

a0

a a0

a2 2a a0

.... . .

an · · · a0

, G(a) = −eas

a

a2 2a...

. . .

an · · · na

(18)

, respectively.

It’s isomorphic for operation T (a) and S(a), respectively. Now,G(a)

is satisfied ideal structure. So I will be preserved operation by G(a).

H(a)v ≡ F (a)v + G(a)v = easF(a)v + easG(a)v

= eas[F(a) + G(a)]v = easF (0)v

N.B. v is n+1-th vector.

Therefore

H(a) = F (a) + G(a) = eas[F(a) + G(a)] = easH(0)

⇓x = x1 + x2 = α < x′1 + x′2 >= αx0 N.B. x0 = e.

< α, x′1 > + < α, x′2 >=< α, x′1 + x′2 >=< α, x0 >

Hence

< eas,F(a) > + < eas,G(a) >=< eas, F (0) >

< H(a), 1 >=< F (a), 1 > + < G(a), 1 >=< easF (0), 1 >=< 1, H∗(a) >

Reference

< D(a), eas > + < O(a), eas >=< D(0), eas >

< D(a), 1 >=< D(a), 1 > + < O(a), 1 >=< easD(0), 1 >=< 1, D∗(a) >

31

< Y(a), eas > + < N (a), eas >=< Y (0), eas >

< W (a), 1 >=< Y (a), 1 > + < N(a), 1 >=< Y (0)eas, 1 >=< 1,W ∗(a) >

< T (a), eas > +S(a), eas >=< T (0), eas >

< R(a), 1 >=< T (a), 1 > + < S(a), 1 >=< easT (0), 1 >=< 1, R∗(a) >

On the other hand, we have following.

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


Since, F(0) is identity and F(∞) is kind of annihilator then the projection

operator have P (0) = 1 and P (∞) = 0 ,respectively. Similarly, G(0) is

annihilator and G(∞) is kind of identity then we have Q(0) = 0 and

Q(∞) = 1 (maximal ideal). As a whole,

P (a) + Q(a) = {1}


< F(a),G(a) >= F ′(a) or G ′(a) (ideal)


or

< F(a),G(a) >= {0} (ideal)

= {0} (ideal)

Example. F(a), F (a) operations are same behaviour with D(a), D(a)

operations on L1 and L2-spaces, respectivily. In this time, I adopt the

behaviour on L2-spaces. So we have following.

< F(0),G(a) >==< F(a), P ∗(a)G(a) >

=< F(a), P (a)G(a) >=< F(a),G(0) >= 0

So

< F(0),G(a) >=< F(a),G(0) >= 0.

32

Moreover

< F(a),G(0) >=< F(a), Q(a)G(a) >=< Q∗(a)F(a),G(a) >

=< Q(a)F(a),G(a) >=< 0 · F(0),G(a) >= 0

So

< F(0),G(a) >= 0.

On the contrary,

< F(a),G(∞) >=< F(a), Q(a)G(a) >=< Q∗(a)F(a),G(a) >

=< Q(a)F(a),G(a) >=< F(∞),G(a) >= 0

So

< F(a),G(∞) >=< F(∞),G(a) >= 0.


< F(a),F⊥(0) >= 0.

N.B. F(a) and G(a) are orthogonal conditions, respectivily.

Moreover,

(P (a)F(a), P (a)G(a)) = (Q(a)G(a), Q(a)F(a))

(lima→0

F(a), lima→0

G(a)) = ( lima→∞G(a), lim

a→∞F(a)) = (I, 0)

(lima→0

P (a), lima→0


a→∞P (a)) = (I, 0)

lima→0

(P (a)F(a) + Q(a)G(a)) = lima→∞(Q(a)G(a) + P (a)F(a))

P (0)F(0) = Q(∞)G(∞) = I (ring)

Q(0)G(0) = P (∞)F(∞) = 0 (ideal)

33

=< F(a), P ∗(a)P (a)G(a) >

=< F(a), P (a)G(a) >=< F(a),G(0) >= 0

Similarly

< Q(a)G(a), Q(a)F(a) >=< G(a), Q∗(a)Q(a)F(a) >

=< G(a), Q(a)F(a) >=< G(a),F(∞) >= 0.

So we have

=< Q(a)G(a), Q(a)F(a) > .

< F(a), P (a)G(a) >=< G(a), Q(a)F(a) >= 0

34

©T (a) operations in Hilebert spaces

At the first time, T (a) operation is defined following.

T (a)f(t) =∫ ∞

af(t)e−stdt

T (a) operation for inner product spaces in real condition is presented

following.

< T (a), T (a) >≥ 0

< T (a), T (a) >= 0 iff T (∞) = 0

λ < T (a), T (b) >=< λT (a), T (b) >

< T (a) + T (b), T (c) >=< T (a), T (c) > + < T (b), T (c) >

< T (a), T (b) >=< T (b), T (a) >

(19)


have √< T (a), T (a) > = ‖T (a)‖

So

< T (a), T (a) >= ‖T (a)‖2

In this time, the norm of T (a) operation is presented as the norm of

kernel. So we have, ‖T (a)‖ = ‖e−st‖ ≥ 0.

Therefore

< T (a), T (a) >= ‖T (a)‖2 ≥ 0

< T (a), T (a) >≥ 0.

Second steps, since the definition of T (a) operation, If a →∞ then

the integral condition is following∫ ∞

∞f(t)e−stdt = 0.

So T (∞) is converged to 0.

35

Therefore

< T (a), T (a) >= 0 iff T (∞) = 0.

Third steps, clearly, λ < T (a), T (b) >= ‖λT (a)‖T (b)‖ cos θ =<

λT (a), T (b) >.

Fourth steps, now we have ‖T (a) + T (b)‖ ≤ ‖T (a)‖ + ‖T (b)‖. In

this time, T (a) and T (b) are positive operator then ‖T (a) + T (b)‖ =

‖T (a)‖+ ‖T (b)‖.

So

< T (a) + T (b), T (c) >= ‖T (a) + T (b)‖‖T (c)‖ cos θ

= ‖T (a)‖‖T (c)‖ cos θ + ‖T (b)‖‖T (c)‖ cos θ

=< T (a), T (c) > + < T (b), T (c) > .


< T (a), T (b) >= ‖T (a)‖‖T (b)‖ cos θ = ‖T (b)‖‖T (a)‖ cos θ =< T (b), T (a) > .



‖T (a)‖ ≥ 0

‖T (a)‖ = 0 iff T (∞) = 0

‖λT (a)‖ ≤ |λ|‖T (a)‖‖T (a) + T (b)‖ ≤ ‖T (a)‖+ ‖T (b)‖

(20)


Therefore

T (a) operation is able to define on Hilbert spaces.

36

©T (a) operations in Hilebert spaces

The next time, T (a) operation is defined following.

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt

T (a) operation for inner product spaces in real condition is presented

following.

< T (a), T (a) >≥ 0

< T (a), T (a) >= 0 iff T (±∞) = 0

λ < T (a), T (b) >=< λT (a), T (b) >

< T (a) + T (b), T (c) >=< T (a), T (c) > + < T (b), T (c) >

< T (a), T (b) >=< T (b), T (a) >

(21)


have √< T (a), T (a) > = ‖T (a)‖

So

< T (a), T (a) >= ‖T (a)‖2

In this time, the norm of T (a) operation is presented as the norm of

kernel. So we have, ‖T (a)‖ = ‖e−(t−a)s‖ ≥ 0.

Therefore

< T (a), T (a) >= ‖T (a)‖2 ≥ 0

< T (a), T (a) >≥ 0.

Second steps, T (a) operation is able to rewrite following.

T (a)f(t) = eas∫ ∞

af(t)e−stdt.

If a → −∞ then the integral condition is produced as unitary opera-

tion. So, as a whole, T (−∞) is converged to 0. Simirarly, if a →∞ then

we have following form.∫ ∞

∞f(t)e−stdt = 0.

37

So, as a whole, T (∞) is also converged to 0.

Therefore

< T (a), T (a) >= 0 iff T (±∞) = 0.

Third steps, clearly, λ < T (a), T (b) >= ‖λT (a)‖T (b)‖ cos θ =< λT (a), T (b) >.

Fourth steps, now we have ‖T (a) + T (b)‖ ≤ ‖T (a)‖ + ‖T (b)‖. In

this time, T (a) and T (b) are positive operator then ‖T (a) + T (b)‖ =

‖T (a)‖+ ‖T (b)‖.

So

< T (a) + T (b), T (c) >= ‖T (a) + T (b)‖‖T (c)‖ cos θ

= ‖T (a)‖‖T (c)‖ cos θ + ‖T (b)‖‖T (c)‖ cos θ

=< T (a), T (c) > + < T (b), T (c) > .


< T (a), T (b) >= ‖T (a)‖‖T (b)‖ cos θ = ‖T (b)‖‖T (a)‖ cos θ =< T (b), T (a) > .



‖T (a)‖ ≥ 0

‖T (a)‖ = 0 iff T (±∞) = 0

‖λT (a)‖ ≤ |λ|‖T (a)‖‖T (a) + T (b)‖ ≤ ‖T (a)‖+ ‖T (b)‖

(22)


Therefore

T (a) operation is able to define on Hilbert spaces.

38

© Representation of inner product in Laplace transforms.

Now, a has shifted element and it has norm condition ,coincidentally.

I am able to represent the graph condition or inner product on Hilbert

space. Now, a has a property of norm condition then we could represent

for adjoint operations. Therefore

< T (a)f, g >=< f, T ∗(a)g >

Especially

< T (a)f, g >= α < f, T (−a)g >

In this case, ”a” is shifted term for operation T (a). So this norm is

preserved. This situation is generated unitary operators. Therefore T (a)

is represented only a for norm condition.

Let the following

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt = eas

∫ ∞

af(t)e−stdt = easT (a)f(t)

S(a)f(t) =∫ a

0f(t)e−(t−a)sdt = eas

∫ a

0f(t)e−stdt = easS(a)f(t).

R(a)f(t) ≡ T (a)f(t) + S(a)f(t) = easT (a)f(t) + easS(a)f(t)

= eas[T (a)f(t) + S(a)f(t)] = easT (0)f(t)

So

R(a) = T (a) + S(a) = eas[T (a) + S(a)] = easT (0)

⇓x = x1 + x2 = α < x′1 + x′2 >= αx0 N.B. x0 = e.

< α, x′1 > + < α, x′2 >=< α, x′1 + x′2 >=< α, x0 >

Hence

< eas, T (a) > + < eas,S(a) >=< eas, T (0) >

39

< R(a), 1 >=< T (a), 1 > + < S(a), 1 >=< easT (0), 1 >=< 1, R∗(a) >

Reference

< eas,D(a) > + < eas,O(a) >=< eas, D(0) >

< D(a), 1 >=< D(a), 1 > + < O(a), 1 >=< easD(0), 1 >=< 1, D∗(a) >

< eas,Y(a) > + < eas,N (a) >=< eas, Y (0) >

< W (a), 1 >=< Y (a), 1 > + < N(a), 1 >=< easY (0), 1 >=< 1,W ∗(a) >

< eas,F(a) > + < eas,G(a) >=< eas, F (0) >

< H(a), 1 >=< F (a), 1 > + < G(a), 1 >=< easF (0), 1 >=< 1, H∗(a) >


T (a) + S(a) = R(a)


Since, T (0) is identity and T (∞) is kind of annihilator then the projection

operator have P (0) = 1 and P (∞) = 0 ,respectively. Similarly, S(0) is

annihilator and S(∞) is kind of identity then we have Q(0) = 0 and

Q(∞) = 1 (maximal ideal). As a whole,

{P (a) + Q(a) = {1}P (a)Q(a) = {0} (23)


< T (a),S(a) >= T ′(a) or S ′(a) (ideal)


or

< T (a),S(a) >= {0} (ideal)

= {0} (ideal)

Example. T (a), T (a) operations are also same behaviour with Y(a), Y (a)

40

operations on L1 and L2-spaces, respectively. In this time, I adopt the

behaviour on L2-spaces.

< T (0),S(a) >==< T (a), P ∗(a)S(a) >

=< T (a), P (a)S(a) >=< T (a),S(0) >= 0

So

< T (0),S(a) >=< T (a),S(0) >= 0.

Moreover

< T (a),S(0) >=< T (a), Q(a)S(a) >=< Q∗(a)T (a),S(a) >

=< Q(a)T (a),S(a) >=< 0 · T (0),S(a) >= 0

So

< T (0),S(a) >= 0.

On the contrary,

< T (a),S(∞) >=< T (a), Q(a)S(a) >=< Q∗(a)T (a),S(a) >

=< Q(a)T (a),S(a) >=< T (∞),S(a) >= 0

So

< T (a),S(∞) >=< T (∞),S(a) >= 0.


< T (a), T ⊥(0) >= 0.

N.B. T (a) and S(a) are orthogonal conditions, respectivily.

Moreover,

(P (a)T (a), P (a)S(a)) = (Q(a)S(a), Q(a)T (a))

(lima→0

T (a), lima→0

S(a)) = ( lima→∞S(a), lim

a→∞ T (a)) = (1, 0)

(lima→0

P (a), lima→0


a→∞P (a)) = (1, 0)

41

lima→0

(P (a)T (a) + Q(a)S(a)) = lima→∞(Q(a)S(a) + P (a)T (a))

P (0)T (0) = Q(∞)S(∞) = {1} (ring)

Q(0)S(0) = P (∞)T (∞) = {0} (ideal)


=< T (a), P ∗(a)P (a)S(a) >

=< T (a), P (a)S(a) >=< T (a),S(0) >= 0

Similarly

< Q(a)S(a), Q(a)T (a) >=< S(a), Q∗(a)Q(a)T (a) >

=< S(a), Q(a)T (a) >=< S(a), T (∞) >= 0.

So we have

=< Q(a)S(a), Q(a)T (a) > .

< T (a), P (a)S(a) >=< S(a), Q(a)T (a) >= 0

42

Some results

◦ D(a), Y (a), F (a) and T (a) operations could define on Hilbert space.

◦ D(a),Y(a) are most basic form as unitary operators.

◦ This unitary condition is separated to L1 and L2-spaces.

L1 − space

D(∞) =

( ∞0

∞0

)= I(∞) =

∫ ∞

∞dt = Y(∞). (24)

L2 − space

D(∞) =

( ∞0

∞0

)= {0} =

∫ ∞

∞dt = Y(∞). (25)

◦ The projection operator P (a) and Q(a) is able to define following.

lima→0

P (a) = {1} = lima→∞Q(a) ring

lima→0

Q(a) = {0} = lima→∞P (a) ideal.

◦ In this case, F(a) or T (a) is same form on L2-space.

◦ D(a), Y (a), F (a) and T (a) have a property of semi-groups.

Example

F (a)F (b) = F (a + b)

F (0) = I(0)

s− lima→0 F (a)v = v

(26)

43

Chapter 2

In this chapter, I want to explain the eigenspaces for Laplace trans-

forms. “Now Laplace transforms” is defined on eigenspaces. If the “Now

Laplace transforms” is to “My Laplace transforms”?

©Rotation operators A(a) based by Laplace transforms.

( L(a) cos ωt

L(a) sin ωt

)=

(cos aω − sin aω

sin aω cos aω

) ( L cos ωt

L sin ωt

)(27)

⇓(

T (a) cos ωt

T (a) sin ωt

)=


sin aω cos aω

) (T (0) cos ωt

T (0) sin ωt

)(28)

T (a)

(cos ωt

sin ωt

)=


sin aω cos aω

)T (0)

(cos ωt

sin ωt

)(29)

Let T (a) = A(a) · T (0),

by A(a) ≡(

cos aω − sin aω

sin aω cos aω

)(30)

A(a) +A(a)−1 =


sin aω cos aω

)+

(cos aω sin aω

− sin aω cos aω

)= 2

(cos aω 0

0 cos aω

)(31)

A(a) A(a)−1 =

(1 0

0 1

)(32)

On the next time,

44

Let A(a) · T (0) = λT (0). A(a)−1 = λ−1

{λ + λ−1 = 2 cos aω (1)

λ · λ−1 = 1 (2)(33)

Since (1) λ−1 = 2 cos aω − λ

λ(2 cos aω − λ) = 1 by(2)

λ2 − 2 cos aω · λ + 1 = 0

So we obtain λ = cos aω ± i sin aω = e±iaω.

Hence

{T (a) = eiaω · T (0)

T (a)−1 = e−iaω · T (0)−1 iff cos ωt, sin ωt.(34)

Moreover T (a)−1 = T (−a) iff T (0) = T (0)−1.

If we are able to consider T (0) operation under group rings then we

have

Xn(g) = X(gn) = X(ng) = X−n(g)

In this case,

T−1(0) = T (0 · (−1)) = T (0) = T1(0)

So I am able to have following

T−1(0) = T (0) = T1(0)

Therefore

T−1(a) = T (−a) = T1(a).

In general, T (a) has following relation with F (a).

T (a)iso←→ F (a)

as

F (a) = easF(a)def .= eas

(1

a 1

). (strong) (35)

45

So we have F (a)−1 = F (−a). F (0) = F (0)−1.

Therefore

T (a)−1 = T (−a) for all a.

Of caurse, T (0) = T (0)−1.

Especially, in this case, T (a)iso←→ D(a) as

D(a) = easD(a)def .= eas

(1

1

). (strong) (36)

So we have D(a)−1 = D(−a). Of caurse, D(0) = D(0)−1.

Since upper condition, we have

T (a)

(cos ωt

sin ωt

)= eiaωT (0)

(cos ωt

sin ωt

). (37)

T (a)

(cos ωt

sin ωt

)=

(T (a) cos ωt

T (a) sin ωt

)(38)

=

s

s2 + ω2(cos aω − ω

ssin aω)

ω

s2 + ω2(cos aω +

s

ωsin aω)

. (3) (39)

On the other hand

eiaωT (0)

(cos ωt

sin ωt

)= eiaω

(T (0) cos ωt

T (0) sin ωt

)(40)

= (cos aω + i sin aω)

s

s2 + ω2

ω

s2 + ω2

(41)

46

=

s

s2 + ω2(cos aω + i sin aω)

ω


. (4) (42)

Since (3) = (4), we have

s

s2 + ω2(cos aω − ω

ssin aω) =

s


− ω

s2 + ω2sin aω =

is

s2 + ω2sin aω

−ω = is. So s = iω.

Similarly

ω

s2 + ω2(cos aω +

s

ωsin aω) =

ω


s

s2 + ω2sin aω =

iω

s2 + ω2sin aω.

So s = iω.

Hence

T (a)

(cos ωt

sin ωt

)= eiaωT (0)

(cos ωt

sin ωt

)and s = iω. (43)

T (a) = easT (0)

In this time,

(cos ωt

sin ωt

)is eigenvector. (44)

Moreover

T (a)−1 = T (−a)

47

The matrix condition is represented following.

F (a)−1 = F (−a) and D(a)−1 = D(−a)

©Hyperbolic operator A1(a) based by Laplace transforms

( L(a) cosh ωt

L(a) sinh ωt

)=

(cosh aω sinh aω

sinh aω cosh aω

) ( L cosh ωt

L sinh ωt

). (45)

⇓(

T (a) cosh ωt

T (a) sinh ωt

)=

(cosh aω sinh aω

sinh aω cosh aω

) (T (0) cosh ωt

T (0) sinh ωt

)(46)

T (a)

(cosh ωt

sinh ωt

)=

(cosh aω sinh aω

sinh aω cosh aω

)T (0)

(cosh ωt

sinh ωt

)(47)

Let T (a) = A1(a) · T (0),

by A1(a) ≡(

cosh aω sinh aω

sinh aω cosh aω

)(48)

A1(a) +A1(a)−1 =

(cosh aω sinh aω

sinh aω cosh aω

)+

(cosh aω − sinh aω

− sinh aω cosh aω

)= 2

(cosh aω 0

0 cosh aω

)(49)

A1(a) A1(a)−1 =

(1 0

0 1

)(50)

Let A1(a) · T (0) = λT (0). A1(a)−1 = λ−1

{λ + λ−1 = 2 cosh aω (1′)

λ · λ−1 = 1 (2′)(51)

Since (1′), λ−1 = 2 cosh aω − λ

48

λ(2 cosh aω − λ) = 1 by (2′)

λ2 − 2 cosh aω · λ + 1 = 0

So we obtain λ = cosh aω ± sinh aω

Hence

{T (a) = (cosh aω + sinh aω)T (0)

T (a)−1 = (cosh aω − sinh aω)T (0)−1iff cosh ωt, sinh ωt.(52)


So we have

T (a)

(cosh ωt

sinh ωt

)= (cosh aω + sinh aω)T (0)

(cosh ωt

sinh ωt

). (53)

T (a)

(cosh ωt

sinh ωt

)=

(T (a) cosh ωt

T (a) sinh ωt

)(54)

=

s

s2 − ω2(cosh aω +

ω

ssinh aω)

ω


s

ωsinh aω)

(3′) (55)

On the other hand

(cosh aω + sinh aω)T (0)

(cosh ωt

sinh ωt

)(56)

= (cosh aω + sinh aω)

(T (0) cosh ωt

T (0) sinh ωt

)(57)

= (cosh aω + sinh aω)

s

s2 − ω2

ω

s2 − ω2

(58)

49

=

s

s2 − ω2(cosh aω + sinh aω)

ω


. (4′) (59)

Since (3′) = (4′), we have

s


ω

ssinh aω) =

s


s

s2 − ω2

ω

ssinh aω =

s

s2 − ω2sinh aω

ω

s= 1. So s = ω.

Similarly

ω


s

ωsinh aω) =

ω


s

ω= 1. So s = ω.

Hence

T (a)

(cosh ωt

sinh ωt

)= (cosh aω + sinh aω)T (0)

(cosh ωt

sinh ωt

)and s = ω.(60)

T (a) = easT (0)

In this time,

(cosh ωt

sinh ωt


Moreover

T (a)−1 = T (−a)


F (a)−1 = F (−a) and D(a)−1 = D(−a)

50

©About property of ebt.

L(a)ebt =eab

s− b= eab 1

s− b= eabLebt

L(a)ebt = eabLebt

⇓T (a)ebt = eabT (0)ebt

So T (a) = eabT (0) iff ebt.

© Rotation operator A(a) based by Laplace transforms (eas)

Now we have following

( L(a)ebt cos ωt

L(a)ebt sin ωt

)= eab


sin aω cos aω

) ( Lebt cos ωt

Lebt sin ωt

)(62)

⇓(

T (a)ebt cos ωt

T (a)ebt sin ωt

)= eab


sin aω cos aω

) (T (0)ebt cos ωt

T (0)ebt sin ωt

)(63)

T (a)

(ebt cos ωt

ebt sin ωt

)= eab


sin aω cos aω

)T (0)

(ebt cos ωt

ebt sin ωt

)(64)

Let T (a) = eab · A(a) · T (0).

e−abT (a) = A(a) · T (0)

by A(a) ≡(

cos aω − sin aω

sin aω cos aω

)(65)

A(a) +A(a)−1 =


sin aω cos aω

)+

(cos aω sin aω

− sin aω cos aω

)= 2

(cos aω 0

0 cos aω

)(66)

51

A(a) A(a)−1 =

(1 0

0 1

)(67)

Let A(a) · T (0) = λT (0). A(a)−1 = λ−1

{λ + λ−1 = 2 cos aω (1)

λ · λ−1 = 1 (2)(68)

Since (1), λ−1 = 2 cos aω − λ.

λ(2 cos aω − λ) = 1 by (2)

λ2 − 2 cos aω · λ + 1 = 0

So we obtain λ = cos aω ± i sin aω = e±iaω.

e−abT (a) = e±iaωT (0)

Hence

{T (a) = eabeiaω · T (0)

T (a)−1 = e−abe−iaω · T (0)−1 iff ebt cos ωt, ebt sin ωt.(69)


So we have

T (a)

(ebt cos ωt

ebt sin ωt

)= eabeiaωT (0)

(ebt cos ωt

ebt sin ωt

). (70)

T (a)

(ebt cos ωt

ebt sin ωt

)=

eab(s− b)

(s− b)2 + ω2(cos aω − ω

s− bsin aω)

eabω

(s− b)2 + ω2(cos aω +

s− b

ωsin aω)

(3)(71)

On the other hand

eabeiaωT (0)

(ebt cos ωt

ebt sin ωt

)= eab

eiaω(s− b)

(s− b)2 + ω2

eiaωω

(s− b)2 + ω2

(72)

52

=

eab(s− b)

(s− b)2 + ω2(cos aω + i sin aω)

eabω


. (4) (73)

Since (3) = (4), we have

eab(s− b)

(s− b)2 + ω2(cos aω − ω

s− bsin aω) =

eab(s− b)


cos aω − ω

s− bsin aω = cos aω + i sin aω

−ω = i(s− b). So iω = s− b.

Similarly

eabω

(s− b)2 + ω2(cos aω +

s− b

ωsin aω) =

eabω


cos aω +s− b

ωsin aω = cos aω + i sin aω

s− b = iω. So iω = s− b.

Hence T (a)

(ebt cos ωt

ebt sin ωt

)= eabeiaωT (0)

(ebt cos ωt

ebt sin ωt

)and iω = s− b.(74)

T (a) = easT (0)

In this time,

(ebt cos ωt

ebt sin ωt


Moreover

T (a)−1 = T (−a)

53


F (a)−1 = F (−a) and D(a)−1 = D(−a)

© Hyperbolic operator A1(a) based by Laplace transforms (eas)

( L(a)ebt cosh ωt

L(a)ebt sinh ωt

)= eab

(cosh aω sinh aω

sinh aω cosh aω

) ( Lebt cosh ωt

Lebt sinh ωt

). (76)

⇓(

T (a)ebt cosh ωt

T (a)ebt sinh ωt

)= eab

(cosh aω sinh aω

sinh aω cosh aω

) (T (0)ebt cosh ωt

T (0)ebt sinh ωt

)(77)

T (a)

(ebt cosh ωt

ebt sinh ωt

)= eab

(cosh aω sinh aω

sinh aω cosh aω

)T (0)

(ebt cosh ωt

ebt sinh ωt

)(78)

Let T (a) = eab · A1(a) · T (0).

e−abT (a) = A1(a) · T (0)

by A1(a) ≡(

cosh aω sinh aω

sinh aω cosh aω

)(79)

A1(a) +A1(a)−1 =

(cosh aω sinh aω

sinh aω cosh aω

)+

(cosh aω − sinh aω

− sinh aω cosh aω

)= 2

(cosh aω 0

0 cosh aω

)(80)

A1(a) A1(a)−1 =

(1 0

0 1

)(81)

Let A1(a) · T (0) = λT (0). A1(a)−1 = λ−1

{λ + λ−1 = 2 cosh aω (1′)

λ · λ−1 = 1 (2′)(82)

Since (1′), λ−1 = 2 cosh aω − λ

54

λ(2 cosh aω − λ) = 1 by (2′)

λ2 − 2 cosh aω · λ + 1 = 0

So we obtain λ = cosh aω ± i sinh aω.

Hence

{T (a) = eab(cosh aω + sinh aω)T (0)

T (a)−1 = e−ab(cosh aω − sinh aω)T (0)−1 iff ebt cosh ωt, ebt sinh ωt.(83)


So we have

T (a)

(ebt cosh ωt

ebt sinh ωt

)= eab(cosh aω + sinh aω)T (0)

(ebt cosh ωt

ebt sinh ωt

). (84)

T (a)

(ebt cosh ωt

ebt sinh ωt

)=

eab(s− b)

(s− b)2 + ω2(cosh aω +

ω

s− bsinh aω)

eabω

(s− b)2 + ω2(cosh aω +

s− b

ωsinh aω)

(3′)(85)

On the other hand

eab(cosh aω + sinh aω)T (0)

(ebt cosh ωt

ebt sinh ωt

)(86)

= eab(cosh aω + sinh aω)

s− b

(s− b)2 − ω2

ω

(s− b)2 − ω2

(87)

=

eab(s− b)

(s− b)2 + ω2(cosh aω + sinh aω)

eabω

(s− b)2 + ω2(cosh aω + sinh aω)

. (4′) (88)

55

Since (3′) = (4′), we haveω

s− b= 1. So ω = s− b.

Ors− b

ω= 1. So ω = s− b.

Hence T (a)

(ebt cosh ωt

ebt sinh ωt

)= eab(cosh aω + sinh aω)T (0)

(ebt cosh ωt

ebt sinh ωt

)(89)

and ω = s− b.

T (a) = easT (0)

In this time,

(ebt cosh ωt

ebt sinh ωt


Moreover

T (a)−1 = T (−a)


F (a)−1 = F (−a) and D(a)−1 = D(−a)

Thereby we have following.

Eigenvalue of T (a) is eas.

So this form generates following.

(eas)−1 = e(−a)s.

In this case

< R(a), 1 >=< T (a), 1 >=< easT (0), 0 > iff its eigenspace.

N.B. < R(a), 1 >=< T (a), 1 > + < S(a), 1 >

56

N.B. A(a) is depended on a. So A = A(a). A(a) have property

of transrated operator. Therefore A(a)−1 = A(−a). Similarly,A1(a) is

depended on a. So A1 = A1(a). A1(a) have property of transrated

operator. Therefore A1(a)−1 = A1(−a).

In this case, all matrix operator have following property.

T (a) = easT (0)


F (a) = easF (0) = easD(0)

It’s important property for irreducibility.

In general,

T (a) = easT (a).

It’s the special case for T (a). T (a) → T (0) is homomorphic. And the

projection form is following.

PT (a) = T (0)

The T (a) and T (0) have same properties. We are not able to under-

stand the a = 0 conditions. However it’s important things for continuous

structures. In fact, element T (a) disperses at T (0) however two elements

have same properties. So “Now Laplace transforms” is miracle transform

in eigenspaces.

In this case, the ideal structure S(a) has pure null condition. So

S(a) = 0, since

S(a) = easS(0) = eas · 0 = 0.

As a whole, in this case, this T (a) operation is R(a) operation.

Therefore

T (a) = R(a)iso←→ D(a).

So T (a) operation has various forms.

57

Now, let extend to the two-sided form for ′′a′′. Since this conditions

we have following. N.B. (0 ≤ t ≤ ∞). In this time, ‖T (a)‖ is ex-

tended to |eas|. This condition is same with Hahn-Banach theorem. Since

P (a)iso↔ T (a)

iso↔ T (a), then I could extend to Banach spaces.

Now, I want to generate that T (∞) = 0 in Banach spaces. In this

space, S(∞) is infinite, however, as a whole, it’s satisfied bounded con-

ditions. If s is pure complex then we have Fourier transforms and it’s

converged to 1. On the contrary, S(−∞) is converged to zero and T (a)

operation is generated as contraction operator iff a ≤ 0. In this case, we

have (eas)−1 = e(−a)s. Therefore

T (a)−1 = T (−a) if a ≤ 0.

If a ≥ 0 then we should understand for positive operations. Therefore

it’s generated normed conditions larger than unitary condition. So we

could consider the C∗-algebras forms.

If a = 0 then T (0) is “Now Laplace transforms”. It’s generated with

identity. Let this identity condition is I(0). Left condition in eigenspace

is

e−asT (a) = e−as · easT (0) = T (0) = I(0)

Inverse condition is

T−1(a)T (a) = T (−a)T (a) = T (0) = I(0)

and projection operator is represented following

PT (a) = T (0) = T (0) = I(0).

This condition is all “Now Laplace transforms” I(0). I(0) operation

is represented following.

I(0)f(t) = T (0)f(t) =∫ ∞

0f(t)e−stdt.

The matrix condition is following.

F(0) =

(00

00

)= I(0) (91)

58

Some results

◦ T (a) operation could represent by using T (0) operation. In this case, I

need the rotation operator in these representations.

◦ We could have A(a) + A−1(a) = 2C(a) and A(a)A−1(a) = I(a). This is

generated as ring condition.

◦ In this case, this eigenvalue for T (a) operation is represented as eas.

◦ The properties of T (a) operation is applicable to F (a), Y (a) and D(a)

operations.

◦ If translated function is generated as cosh ωt, sinh ωt then we have to

base the following operation.

(cosh aω sinh aω

sinh aω cosh aω

)(92)

◦ This operation is able to have ring condition.

◦ As a result, the eigenvalue of T (a) also generates eas.

◦ T (0) operation is treated as greatest invariant.

T n(0) = T (0 · n) = T (0)iso←→ P n(0) = P (0).

◦ We could represent for all T (a) operations.

T (a) = easT (0)

◦ I could have following formula for all operations

Example.

T (a)−1 = T (−a).

59

Chapter 3

In this chapter, I want to explain the projection operator for Laplace

transforms. This concept of projection operator is generated from “My

Laplace transforms”. The Laplace transforms is defined as following.

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt

This factor of ′′a′′ generated the projection operator as P (a), Q(a).

© Laplace transform and Cauchy problem.

Now, we consider the following equations.

{u′(t) = Au(t) (a ≤ t)

u(a) = x(93)

A is closed linear operator on Banach space.

And x is element of Banach space.

T (a)u(t) =∫ ∞

ae−(t−a)λu(t)dt ‖T (a)‖ ≤ +∞

=

[−e−(t−a)λ

λu(t)

]∞

a

+∫ ∞

a

e−(t−a)λ

λu′(t)dt

=1

λu(a) +

1

λ

∫ ∞

ae−(t−a)λAu(t)dt

λT (a)u(t) = u(a) + A∫ ∞

ae−(t−a)λu(t)dt

λT (a)u(t) = u(a) + AT (a)u(t)

(λ− A)T (a)u(t) = u(a) = x

Therefore T (a)u(t) = (λ− A)−1x

λ is resolvent set of A.

60

T (a)u(t) is Laplace transforms of ′′a′′ for u(t).

Similarly, we have

T (a)u(t) = e−as(λ− A)−1x

λ is resolvent set of A.

T (a)u(t) is Laplace transforms of ′′a′′ for u(t).

On the other hand, since the projection form , we have following.

P =1

2πi

∫

X(λ− A)−1dλ

P is kind of projection and A is bouded linear operator. Now, we

are able to represent that λ is resolvent set of A. This form (λ−A)−1 is

same with it for T (a) operation.

If a →∞ then λ = A. This matter is extended to ring conditions and

maximal ideal structures on infinite spaces.Of caurse, it’s Banach spaces.

In this time, the operation A generate the resolvent. If the resolvent λ

equals the operation A then we are able to consider the irreducibility.

Perhaps, I will be able to define λ0 in all eigenvalues λ.

Let A operation is T̂ . So T̂ has many elements in this set condition.

In this time, this T (a) correspond to λa. So I am able to chose the one

representative element in this set. The element is R(a) operation and it’s

irreducible.

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

N.B. T (a) = easT (a)

Especially, if we consider the unitary space of T (a) then it’s existed

T (a) for corresponding to λa. Therefore I could chose the one element in

this set T̂ . This element is T (0) or λ0. Of caurse, it’s primitve ring. (sub-

ring). This theory will be treated in spectral decompositions. Therefore it

has represented spector in all spectral decompositions. This represented

spector has same property of T (0) operation. R(a) is R(a, 0, s) to be

precise. This zero condition is correspond to zero of T (0) operation.

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© Projection of Laplace transforms. PT (a) = T (0)

In this section, I want to explain the representation of projection for

T (a) operation.

The basic formura is following.

L(a)f(t) = eas[Lf(t)−∫ a

0f(t)e−stdt]

⇓T (a)f(t) = eas[T (0)f(t)−

∫ a

0f(t)e−stdt] (1)

Now, we should attend how to treat the functional f(t) as (0 ≤ t ≤ a).

If f(t) = 0 for 0 ≤ t ≤ a

We refer with ideal structures and want to clear operations S(a), T (a)

and functional f(t). Which is generated zero condition operators or func-

tional?

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

Let

T (0)f(t)−∫ a

0f(t)e−stdt = U(a)f(t)

Since T (0) is unitary, U(a) is also unitary.

So T (a)f(t) = easU(a)f(t) by (1).

N.B. f(t) = 0 for 0 ≤ t ≤ a.

So e−as could use the idea of projection.

Confirmation

Now, let e−as homo−→ P then

e−asT (a)homo−→ PT (a) = T (0)

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e−asT (a)homo−→ T (0)

Especially, if it’s irreducible then we have

T (a) = easT (0). (see“Reference′′)

On the other hand, the matrix condition for T (a) operation is repre-

sented following.

F (a) = easF (0)

Therefore irreducibility has relation with projection.

In general

Since(1), T (a)f(t) = easT (0)f(t)−∫ a

0f(t)e−(t−a)sdt

easT (0)f(t) = T (a)f(t) +∫ a

0f(t)e−(t−a)sdt

Hence

T (a)f(t) = easT (0)f(t)− S(a)f(t) for all f(t).

N.B. S(a)f(t) =∫ a0 f(t)e−(t−a)sdt

This condition is embedded in Hahn-Banach theorem by generated the

functional f(t) on interval 0 ≤ t ≤ a.

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

F (a) = easF (0)−G(a)

Reference

This concept is very important for eigenspaces and irreducibility.

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

eas ↑ ↑ eas

T (a)homo−→ T (0)

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Now, we consider the relations R(a) operation and T (a) operation,

respectively. T (a) is able to generate the set condition. I define that set

condition is T̂ . Now, we obtain that T (a) is homomorphic to T (0). So this

form is represented from set condition to only one element condition. This

concept is similar as kernel conditions. This T (0) is multipled by ‖eas‖.Therefore the generated R(a) is neccesarry the spectral decomposition.

Therefore it’s generated the relations between T (a) and R(a) operations.

If T̂ is generated as one element then R(a) is also one element for ′′a′′.It is represented as T (a) operation. So this T (a) operation includes R(a)

condition. This operator algebra is following.

T (a) = R(a) + {−S(a)}

N.B. S(a) is ideal.

©Ideal structure to ring conditions.

Now I am considering ideal and ring conditions. Some of ring condi-

tions is represented following.

Ring =

eas3

a eas3

0 a eas3

(94)

The power conditions of this ring is represented following.

Power =

eas3

a eas3

0 a eas3

eas3

a eas3

0 a eas3

(95)

=

(eas3 )2

2aeas3 (e

as3 )2

a2 2aeas3 (e

as3 )2

(96)

So ideal structure to ring conditions are continuous by eas3 . If we have

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the power of this rings then it has continuous conditions. Now, we have

Hermitian form iff s is pure complex. This condition is represented that

T (a) ←→ T ∗(a) is continuous through diagonal T (0) or R(0). The lattice

conditions of T (0) is represented following.

T (0)f(t) =∞∑

n=0

f(0)(n)00

sn+1

I assert that T (a) goes to T ∗(a) through T (0) condition. Now, 00 is

no-define. T (0) has a miracle condition as Hermitian form. However,

now we have the knowledge of Hermitian form for T (a) operations. (see.

[6]) So we will should define 00. Furthermore, the 00 is existed on ring

boundary.

©Homomorphic theorem and projection for equivalence

Now, homomorphic theorem is following.

Ghomo−→ G′ (factor group)

homo ↘ l iso

G/N

( see “the first isomorphism theorem” )

m

So if G is projection then we have following

P n = Phomo−→ P 2 = P (factor group)

homo ↘ l iso

P 2 = P

On this time, I want to refer my equivalence.

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(power) (semi− groups)

T n(a) = T (na) ←→ T (a)T (b) = T (a + b)

homo ↘ ↙ homo

T (0)T (0) = T (0)

(projection)

Hence

P n(a) = P (na) ←→ P (a)P (b) = P (a + b)

homo ↘ ↙ homo

P (0)2 = P (0)

Piso←→ T̂

There power condition is able to be represented by invariant integral

forms. In general,

T n(a) = T (an) = T (na) = T−n(a)

Similarly, the projection operator will be represented following.

P n(a) = P (an) = P (na) = P−n(a)

Furthermore the normed condition is following.

‖T (a)‖ = ‖T (a)∗‖ = ‖T (a)‖2

Of caurse, the normed projection operator is satisfied following property.

‖P (a)‖ = ‖P (a)∗‖ = ‖P (a)2‖.

So T (a) have also a property of projection. And my operarator is

‖T (a)‖ = ‖T (a)∗‖ = |eas|

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In this time, it will be necesarry to define the “normed projection”

denoted by “Pnorm”. If “a′′ is finite number then I have ‖T (a)‖ = 1.

However if “a′′ is infinite number then I must have ‖T (∞)‖ = 0. (see

spector decomposition or projection operator)

Reference

Identity = ‖F(a)‖ = ‖F(a)∗‖ = ‖F(a)‖2

‖F (a)‖ = ‖F (a)∗‖ = |eas|Similarly ,in this case, If “a′′ is finite number then I have ‖F(a)‖ = 1.

However if “a′′ is infinite number then I must have ‖F(∞)‖ = 0. (see

spector decomposition or projection operator)

Further

Identity = ‖D(a)‖ = ‖D(a)∗‖ = ‖D(a)‖2

‖D(a)‖ = ‖D(a)∗‖ = |eas|

In this case, ideal structure is null condition. So ,in spite of “a′′ of

“finite” or “infinite” , we could have identity. (primitive ring)

In this section, I want to explain the unitary conditions.

© The property of projection and T (a),S(a) operations

Now, I want to consider the projection operator with spector. So

P (∞) = 0 and P (0) = 1 or Q(∞) = 1 and Q(0) = 0, respectively.

P(a),Q(a) have same properties on projection.

On this time, let

P (a) ←→ T (a) and Q(a) ←→ S(a).

So we can find T (∞) = 0. Therefore equivalence for T (a) will be

following.

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Hence


T (a)n = T (na)homo−→ T (0)2 = T (0) (factor group)

homo ↘ l iso

T (0)2 = T (0)

(projection)

Reference


F(a)n = F(na)homo−→ F(0)2 = F(0) (factor group)

homo ↘ l iso

F(0)2 = F(0)

(projection)

So we have

T (0) ←→ T (∞) S(0) ←→ S(∞)

l l l lP (0) = {1} ←→ P (∞) = {0} = Q(0) = {0} ←→ Q(∞) = {1}

l l l lF(0) ←→ F(∞) G(0) ←→ G(∞)

Now, left side is dominant for ideal structures and righi side is dom-

inant for ring conditions,respectively.It is not symmetric ideal and ring

for “a”. T (a) ↔ P (a) and S(a) ↔ Q(a).Therefore P (a) and Q(a) can be

represented ring conditions and ideal structures, respectively.

[ T (a) + S(a) = T (0) = {1}T (a) · S(a) = S ′(a) = {0} (97)

68

m[

P (a) + Q(a) = {1}P (a) ·Q(a) = {0} (98)

m[ F(a) + G(a) = F(0) = {1}F(a) · G(a) = G ′(a) = {0} (99)

N.B. P(a),Q(a) are orthogonal projections.

In this time, the cones are represented as P, Q, respectively. P, Q are

prjection. So

PT (a) = T (0) , QS(a) = S(0), respectively.

Now, if this graph condition is represented for vector space then we

have

eas(T (a),S(a)) = (easT (a), easS(a)) = (T (a), S(a)) = 0

So, in general, PT (a) is able to be represented by using projection.

Since this situation, we should extend to normed algebras (C∗-algebras).

Extended normed algebra will explain later.

Similarly, the cones are represented as P,Q, respectively. P, Q are

prjection. So

PF(a) = F(0) , QG(a) = G(0), respectively.

Now, if this graph condition is represented for vector space then we

have

eas(F(a),G(a)) = (easF(a), easG(a)) = (F (a), G(a)) = 0

So, in general, PF (a) is able to be represented by using projection.

Since this situation, we should extend to normed algebras (C∗-algebras).

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Of caurse, F(a) operation is able to extend to normed algebras too.

©Relation of T (a) and S(a) (unitary condition)

T (a) and S(∞) are satisfied ring conditions and S(a) and T (∞) are

satisfied ideal conditions. One is changed like threw the

T (a) −→ S(∞) = {1}Similarly, the other is changed like threw the

S(a) −→ T (∞) = {0}Therefore ring and ideal conditions are necessary the two forms T (a)

and S(a), respectively. However if it’s clear finite or infinite space then

we are able to treat as T (a) or S(a). This condition is satisfied the

behavior of projection operators iff it’s unitary operations. In general, if

this condition is extended to T (a), S(a) operations then we should add

to the general normed conditions.

In general, T (a) of eigenspase is able to extend to generalized T (a)

operation. So we have

e−asT (a) = e−as · easT (a) = T (a) = I(a)

Inverse condition is

T−1(a)T (a) = T (−a)T (a) = T (0) = I(0)

and projection operator is represented following

PT (a) = T (0) = T (0) = I(0).

This condition is representing extendness for I(a) operation in the

center of “Now Laplace transforms” I(0). In fact, this ′′a′′ condition is

operating as ideal. I(a) operation is represented following.

I(a)f(t) = T (a)f(t) =∫ ∞

af(t)e−stdt.

The matrix condition is following.

F(a) =

(a0

a a0

)= I(a) (100)

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Some results

◦ T (a) operation is able to extend to eas condition by Hahn-Banach the-

orem.

◦ We could regard that T (a),F(a) and S(a),G(a) are kind of projection

oprator.

◦ If we could extend to T (a), F (a) operations then we will consider the

negative condition for a.

◦ eas is able to become kind of generator for T (a), F (a) operations.

◦ Especially I could have following relation

T (a) = easT (0) (irreducible)

The matrix condition is

F (a) = easF (0) (irreducible)

◦ In this case, the norm of T (a), F (a) are obtained as |eas| and equal to

extended form by Hahn-Banach theorem.

◦ Projection operator is isomorphic with inverse of T (a) and F (a) oper-

ations. (field conditions)

Piso←→ P−1 iso←→ T (a)

iso←→ F (a).

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Conclusion

In this time,D(a), Y (a), F (a) and T (a) operations are defined on Hir-

bert spaces. L1-space does not have orthogonal condition. On the con-

trary, since this result, we have following condition. For example,

< T (a),S(∞) >=< T (∞),S(a) >= 0 in L1 − spaces.

However, if this space is builded on L2-spaces then we should have

< T (a),S(∞) >=< T (a), T ⊥(0) >= 0

So L2-spaces should become orthogonal spaces. This condition fits to

condition of Hilbert spaces. In this time, maximal ideal and ring condi-

tions are orthogonal, respectivily when a is infinite condition.

If T (a) operation is represented on eigenspaces then we have two im-

portant forms.

{T (a) = easT (0)

T (a)−1 = T (−a)(101)

In this case, T (a) operation has irreducible condition for T (0) oper-

ation. The norm of T (a) is given as ‖eas‖. T (0) is unitary operator.

Moreover, since the property of group-rings, I could have the condition

that a is negative. This condition is continue to the theory of semi-groups.

D(a), Y (a) and F (a) operations are samely.

Finally, T (a) operation becomes extended form by Hahn-Banach the-

ory. P (a) and T (a) have same proprties, respectively. Similalry, Q(a)

and S(a) have same it. Since, the Chapter 1, we have

T (0) = S⊥(∞) = {1} , T (∞) = S⊥(0) = {0}

P (0) = Q⊥(∞) = {1} , P (∞) = Q⊥(0) = {0}

in Hilbert spaces.

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References

[1] Bryan P.Rynne and Martin A.Youngson, Linear functional anal-

ysis, 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

Approximation, Springer, 1967

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

1957.

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

ory of Operator Algebras, AMS

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

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

Nostrand Company,Inc,1960.

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

Nostrand Company,Inc, 1961.

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The projection operator for Laplace transforms

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