the negative condition for laplace transformsopab.web.fc2.com/no.2.pdfpreface this paper is builded...

The negative condition

for Laplace transforms

Takao Saito

Thank you for our world

The negative condition

for Lapalace transforms

Preface

This paper is builded by three parts. First is negative condition of op-

erator algebras in L1, L2-spaces. Second is operator algebras for Laplace

transforms. Finally, the property of semi-groups in 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

The projection operator for Laplace transforms

Now, let′s consider with me!

695-52 Chibadera-cho

Chuo-ku Chiba-shi

Postcode 260-0844 , Japan

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

(Mon)5.Oct.2009

Takao Saito

Contents

Preface

§ Chapter 4

◦AboutD(-a),O(-a)for matrix operator · · · · · ·8◦D(a) About (D(a),O(a))(D(−a),O(−a)) · · · · · ·9◦ Relation ofD(a) andO(a) · · · · · ·11

◦About < D(a),O(a) >< D(−a),O(−a) > · · · · · ·12

◦ About Y(−a),N (−a) for Laplace transforms · · · · · · · · · · · ·14

◦About (Y(a),N (a))(Y(−a),N (−a)) · · · · · · · · · · · ·14

◦About < Y(a),N (a) >< Y(−a),N (−a) > · · · · · ·16

◦ About F(−a),G(−a) for matrix operator · · · · · ·18

◦ About < F(a),G(a) >< F(−a),G(−a) > · · · · · ·19

T (a) (Laplace transforms)

◦ About T (−a),S(−a) for Laplace transforms · · · · · · · · · · · ·21

◦ Relation of T (a) and S(a) · · · · · · · · · · · ·22

◦ About < T (a),S(a) >< T (−a),S(−a) > · · · · · ·23

◦ Some results · · · · · ·25

§ Chapter 5

◦ Figure of basic operator algebras · · · · · ·28

◦ Figure of basic operator algebras (Duality) · · · · · · · · · · · ·29

◦ Basic operator algebras · · · · · ·30

◦ Basic operator algebras (Duality) · · · · · ·30

◦ Basic normed operator algebras · · · · · · 31

◦ Basic normed operator algebras (Duality) · · · · · · · · · · · ·31

◦ Figure of basic operator algebras (Duality) · · · · · · · · · · · · 45

◦ Some results · · · · · · · · · · · ·48

§ Chapter 6

◦ Associative and commutative of D(a) operations (L1-space)

· · · 49

◦ Associative and commutative of D(a) operations (L2-space)

· · · 50

◦ Semi-groups for D(a) operation · · · · · · · · · · · ·52

◦ Semi-group for O(a) operation · · · · · · · · ·52

◦ Associative and commutative of Y (a) operations (L1-space)

· · · 54

◦ Associative and commutative of Y (a) operations (L2-space)

· · · 55

◦ Semi-groups for Y (a) operation · · · · · · · · · · · ·56

◦ Semi-group for N (a) operation · · · · · · · · ·57

◦ About the property of eas · · · · · · · · · · · · 58

◦ Associative and commutative of Pascal’s matrix (L2-space)

· · · 58

◦ Semi-groups for F (a) operation · · · · · · · · · · · ·59

◦ Semi-group for G(a) operation · · · · · · · · ·59

◦ Associative and commutative of Laplace transforms (L2-space)

· · · 62

◦ Semigroups of translate operators and convolutions · · · 63

◦ Some results · · · · · · · · · · · · 66

◦ Conclusion · · · · · · · · · 67

◦ References · · · · · · · · · · · · 68

Chapter 4

In this chapter, we want to extend to negative conditions for ′′a′′ in

Laplace transforms T (a). The first step, I want to explain about D(a)

and Y(a) operations in L1, L2-spaces. The next step, I explain about

F(a) and T (a) operations in L2-spaces.

©About D(−a),O(−a) for matrix operator. (vector spaces)

D(a) operation is the simplest form for F (a) operations. F (a) op-

eration (matrix operator) is isomorphic with T (a) operations. (Laplace

transforms)

D(a) operations are defined as following.

D(a) = easD(a) = eas

, O(a) = easO(a) = 0 (annihilator) (1)

as a whole

D(a) + O(a) = H(a) = easH(a).

D(a) and D(a) operations have a property of ring conditions because

of it generats from matrix structures.

Especially, this operation has following important property.

(semi− group) D(−a) = D−1(a) (group or field)

Left side is generated as semi-groups. In this time, it’s able to extend

to group condition with inverse. So D(a) operation generats group struc-

tures. Furthermore, this group is able to extend to ring. In fact, D(a) is

On the other hand,

H(a) = D(a) +O(a) = D(−a) +O(−a) = H(−a) = Identity

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

In this time,

D(a) =

(−a)0

= D(−a). (2)

So we have following.

D(a) = D(−a) , O(a) = O(−a).

Therefore D(a) and O(a) operations has identical forms, respectively.

©About (D(a),O(a))(D(−a),O(−a))

If it’s defined on L1-space then I will define following.

(D(a),O(a))(D(−a),O(−a))def .= (D(a)D(−a),O(a)D(−a) +D(a)O(−a) +O(a)O(−a))

O(a) = D(0)−D(a) , O(−a) = D(0)−D(−a).

In this case,

O(a)D(−a) = (D(0)−D(a))D(−a) = D(−a)−D(0) = 0

D(a)O(−a) = D(a)(D(0)−D(−a)) = D(a)−D(0) = 0

O(a)O(−a) = (D(0)−D(a))(D(0)−D(−a)) = D(0)−D(a)−D(−a)+D(0)

= 2D(0)−D(a)−D(−a) = 0

Therefore

O(a)D(−a) +D(a)O(−a) +O(a)O(−a) = 0.

(D(a),O(a))(D(−a),O(−a)) = (D(0), 0) = (I, 0).

On the other hand,

D(a) = D(−a) = D−1(a).

D2(a) = I(a)

The involution of D(a) operation is given by

D∗(a) =

= D(a) = I(a) (3)

D(a) = D∗(a) = D2(a) = I(a)

Similarly

N (a) = N ∗(a) = N 2(a) = 0

In this time, the projection operator is able to represent following.

P = P ∗ = P 2 = {1}, {0} in L1 − spaces.

Moreover, since the property of group ring, we have following conditions

D(a) = D(na) = Dn(a) invariant− form.

I is identity as ring condition and 0 is satisfied ideal structures. So,

(I, 0)

↙ ↘ring ideal

(see, p.16, Chapter 1, No.1)

If this condition is presented on L2-spaces then the operator algebra

for D(a) is following.

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

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

Similarly

D(−a) +O(−a) = H(−a) = I(−a) (additive)

< D(−a),O(−a) >= 0 (Multiplicative)

The projection operator is able to represent as P = {1, 0} in L2-spaces.

Now, I want to extend finite conditions to infinite conditions. Fi-

nite conditions couud deal with field theory, however infinite space is two

kind of situations 1 and 0. First and second situations 1,0 are maked

O⊥(∞) and D⊥(∞) , respectively. If we deal with D⊥(∞) then we ob-

taine O⊥(∞) = T (0) = {1}. It commutes spectral dicomposition. Now,

D⊥(∞) = {0} on Hilbert spaces. Of caurse ‖T (0)‖ = 1. Since ideal

N (a) = {0} , then O(a) = {0} iff a is finite. If a is infinite then

‖O⊥(∞)‖ = ‖D(0)‖ = 1. These operations are bounded on Hilbert

spaces, respectively. In this time, the behaviour of projection operator

P (a), Q(a) are same with D(a),O(a) operations, respectively.

P (a) + Q(a) = {1} for all ′′a′′.

D(a) +O(a) = H(a) = D(0)

If a is infinite condition then we have following.

O⊥(−∞) = D(0) = O⊥(∞) = {1}.

D⊥(−∞) = O(0) = D⊥(∞) = {0}.

©About < D(a),O(a) >< D(−a),O(−a) > .

< D(a),O(a) >< D(−a),O(−a) >def .= < D(a)D(−a),O(a)D(−a)+D(a)O(−a)+O(a)O(−a) >

O(a) = D(0)−D(a) , O(−a) = D(0)−D(−a).

In this case,

O(a)D(−a) = (D(0)−D(a))D(−a) = D(−a)−D(0) = 0

D(a)O(−a) = D(a)(D(0)−D(−a)) = D(a)−D(0) = 0

O(a)O(−a) = (D(0)−D(a))(D(0)−D(−a)) = D(0)−D(a)−D(−a)+D(0)

= 2D(0)−D(a)−D(−a) = 0

Therefore

O(a)D(−a) +D(a)O(−a) +O(a)O(−a) = 0.

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

In this condition, we are better off considering on L1-spaces because

of O(a) is pure null operation for all a.

©About Y(−a),N (−a) for Laplace transforms.

This condition is basic integral operations for Laplace transforms. eas

term traet as norm for Y (a) operations. The matrix operator D(a) is iso-

morphic with integral operator Y (a). Similarly, there unitary operations

D(a) and Y(a) are isomorphic, respectivily.

Y (a) = easY(a) , N(a) = easN (a)

as a whole,

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

Y(a)f(t) =∫ ∞

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

0f(t)dt

Y(−a)f(t) =∫ ∞

−af(t)dt =

∫ ∞

0f(t)dt +

−af(t)dt

Y(−a)f(t) = W(−a)f(t)−∫ −a

0f(t)dt = W(−a)f(t)−N (−a)f(t)

Y(−a)f(t) +N (−a)f(t) = W(−a)f(t)

So Y(−a) +N (−a) = W(−a).

W(a) = Y(a) +N (a) = Y(−a) +N (−a) = W(−a) = Identity.

N.B. Y(0) is also identity.

Moreover, now Y(a) operation is defined on L1-spaces. So we have

following relations.

Y(a) = Y(−a) , N (a) = N (−a) , respectively.

In this time, we should be atention Y(0) operation. This operation is

isomorphic with D(0) operation. In fact, D(0) is generated as 00. This

condition is no define. Therefore Y(0) should be also defined on ”no

define”. Moreover, since Y(a) is isomorphic with D(a) then we have

following formura.

(semi− group) Y(−a) = Y−1(a) (group).

Similarly, we have following in integral operations.

Y(a) = Y(−a) = Y−1(a) = I(a)

Y2(a) = I(a).

Moreover, the involution for Y(a) operation has following.

Y∗(a)f(t) =∫ ∞

af(t)dt =

∫ ∞

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

when f(t) is real functional.

So we have

Y(a) = Y∗(a) = Y2(a) = I(a)

Similarly,

N (a) = N ∗(a) = N 2(a) = {0}On the other hand, the projection operator is represrnted following.

P = P ∗ = P 2 = {1}, {0} in L1 − spaces.

Moreover, since the property of group ring, we have following conditions

Y(a) = Y(na) = Yn(a) invariant− form.

©About (Y(a),N (a))(Y(−a),N (−a)).

Now, I want to define following

(Y(a),N (a))(Y(−a),N (−a))def .=

(Y(a)Y(−a),N (a)Y(−a) + Y(a)N (−a) +N (a)N (−a)).

N (a) = Y(0)− Y(a) , N (−a) = Y(0)− Y(−a)

N (a)Y(−a) = (Y(0)− Y(a))Y(−a) = Y(−a)− Y(0)

Y(a)N (−a) = Y(a)(Y(0)− Y(−a)) = Y(a)− Y(0)

N (a)N (−a) = (Y(0)−Y(a))(Y(0)−Y(−a)) = Y(0)−Y(a)−Y(−a)+Y(0)

= 2Y(0)− Y(a)− Y(−a).

Therefore

N (a)Y(−a) + Y(a)N (−a) +N (a)N (−a) = 0.

(Y(a),N (a))(Y(−a),N (−a)) = (Y(0), 0) = (1, 0) algebraic.

1 is identity as group condition and 0 is satisfied subgroup in this

group. Furthermore, this argument able to extend to ring condition from

group structures.

(1, 0)

↙ ↘ring ideal

Now, Y(a) operation define on L1-spaces. p.26. Since, the property of

O(a) operation in L1-spaces then N (a) is also better off considering as

pure null conditions. So I define N (a)f(t) as following.

N (a)f(t) =∫ a

0f(t)dt = 0 =

−af(t)dt = N (−a)f(t)

N.B f(t) = 0 for (−a ≤ t ≤ a).

This projection operator is able to represent following.

P = P ∗ = P 2 = {1, 0} in L2 − spaces.

If it’s defined on L2-spaces then the operator algebra for Y(a) operation

is presented following.

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

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

Similarly

Y(−a) +N (−a) = W(−a) = I(−a) (additive)

< Y(−a),N (−a) >= 0 (Multiplicative)

©About < Y(a),N (a) >< Y(−a),N (−a) > .

< Y(a),N (a) >< Y(−a),N (−a) >def .=

< Y(a)Y(−a),N (a)Y(−a) + Y(a)N (−a) +N (a)N (−a) > .

N (a) = Y(0)− Y(a) , N (−a) = Y(0)− Y(−a)

N (a)Y(−a) = (Y(0)− Y(a))Y(−a) = Y(−a)− Y(0)

Y(a)N (−a) = Y(a)(Y(0)− Y(−a)) = Y(a)− Y(0)

N (a)N (−a) = (Y(0)−Y(a))(Y(0)−Y(−a)) = Y(0)−Y(a)−Y(−a)+Y(0)

= 2Y(0)− Y(a)− Y(−a).

Therefore

N (a)Y(−a) + Y(a)N (−a) +N (a)N (−a) = 0.

< Y(a),N (a) >< Y(−a),N (−a) >=< Y(0), 0 >=< 1, 0 > algebraic.

Similarly

< Y (a), N(a) >< Y (−a), N(−a) >=< Y (a)Y (−a), Y (a)N(−a)+Y (−a)N(a)+N(a)N(−a) >

=< Y (0), 0 >=< 1, 0 >= 0 algebraic.

Reference

< D(a), O(a) >< D(−a), O(−a) >=< D(a)D(−a), D(a)O(−a)+O(a)D(−a)+O(a)O(−a) >

=< D(0), 0 >=< I(0), 0 >= 0 for matrix.

©About F(−a),G(−a) for matrix operator.

F (a) operation is builded as Pascal’s triangle. This form should con-

sider the ideal structures with D(a) operations. Furthermore these ma-

trices will be generated from power condition for Laplace transforms. In

this time, I want to explain the L2-space, only.

Let the following

F (a) = easF(a) = eas

a2 2a a0

.... . .

an · · · a0

G(a) = easG(a) = −eas

a2 2a 0...

an · · · 0

H(a) = easH(a) = eas

As a whole, we have

F (a) + G(a) = H(a) = D(a) + O(a).

F(a) + G(a) = H(a) = D(a) +O(a).

Similarly, this operation has following property.

F(−a) = F−1(a)

In this time, the F (a) operation is same with D(a) operation. Only,

F (a) operation has ideal structure.

On the other hand,

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

N.B. F(0)def .= Identity.

In this case, we have following conditions

F(a) 6= F(−a) , G(a) 6= G(−a) if a 6= 0.

Clearly,

L2 − spaces F(a)homo−→ D(a) L1 − spaces.

If F(a) operation is defined on Hilbert space then we have following

conditions.

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

Fundamental structure for projectin operator is following.

Q⊥(−∞) = P (0) = Q⊥(∞) = {1}

P⊥(−∞) = Q(0) = P⊥(∞) = {0}

The normed condition for F(a) and G(a) operations are following.

‖F(a)‖ = ‖F(−a)‖ = ‖F−1(a)‖ = ‖F∗(a)‖ = ‖F2(a)‖ = {1, 0}

Similarly

‖G(a)‖ = ‖G(−a)‖ = ‖G−1(a)‖ = ‖G∗(a)‖ = ‖G2(a)‖ = {0, 1}

On the other hand, the projection operator for normed condition is

following.

‖P (a)‖ = ‖P ∗(a)‖ = ‖P 2(a)‖ = {1, 0}

©About < F(a),G(a) >< F(−a),G(−a) > .

< F(a),G(a) >< F(−a),G(−a) >def=< F(a)F(−a),G(a)F(−a) + F(a)G(−a) + G(a)G(−a) >

G(a) = F(0)−F(a) , G(−a) = F(0)−F(−a).

G(a)F(−a) = (F(0)−F(a))F(−a) = F(−a)−F(0)

F(a)G(−a) = F(a)(F(0)−F(−a)) = F(a)−F(0)

G(a)G(−a) = (F(0)−F(a))(F(0)−F(−a)) = F(0)−F(a)−F(−a)+F(0)

= 2F(0)−F(a)−F(−a)

Therefore

G(a)F(−a) + F(a)G(−a) + G(a)G(−a) = 0.

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

I is identity as ring condition and 0 is satisfied ideeal structures.

< I, 0 >

↙ ↘ring ideal

In this time, the operator algebra for F(a) operation is presented

following.

F(a) + G(a) = H(0) = I(a) (additive)

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

©About T (−a),S(−a) for Laplace transforms.

T (a) operation is my Laplace transforms. It’s defined as following

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt , S(a)f(t) =

0f(t)e−(t−a)sdt

R(a)f(t) =∫ ∞

0f(t)e−(t−a)sdt

As a whole,

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

In this time, the term of kernel do include with a term. It depends

with greatest lower bounded of integral operation. Similarly, I treated as

L2-spaces, only. This operation is isomorphic to F (a) operation. (matrix

operator)

Similarly

T (a) =∫ ∞

af(t)e−stdt , S(a) =

0f(t)e−stdt

So we have

T (−a)f(t) =∫ ∞

−af(t)e−stdt =

∫ ∞

0f(t)e−stdt +

−af(t)e−stdt

T (−a)f(t) = R(−a)f(t)−∫ −a

0f(t)dt = T (0)f(t)− S(−a)f(t)

T (−a)f(t) + S(−a)f(t) = R(−a)f(t)

So T (−a) + S(−a) = R(−a).

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

N.B. R(a) = T (0) = R(−a) = identity.

Now, I want to extend finite conditions to infinite conditions. Finite

conditions couud deal with field theory, however infinite space is two kind

of situations 1 and 0. First and second situations 1,0 are maked S(∞)

and T (∞) , respectively. If we deal with T (∞) then we obtaine S(∞) =

T (0) = {1}. It commutes spectral dicomposition. Now, T (∞) = {0}on Hilbert spaces. Of caurse ‖T (0)‖ = 1. Since ideal G(a) = {0} , then

S(a) = {0} iff a is finite. If a is infinite then ‖S(∞)‖ = ‖T (0)‖ = 1 This

operations are bounded on Hilbert spaces, respectively. In this time, the

behaviour of projection operator P (a), Q(a) are same with T (a),S(a)

operations, respectively.

P (a) + Q(a) = {1} for all ′′a′′.

T (a) + S(a) = R(a) = T (0)

The relations for T (a) and S(a) operations in Hilbert spaces are following.

S⊥(−∞) = T (0) = S⊥(∞) = {1}

T ⊥(−∞) = S(0) = T ⊥(∞) = {0}Fundamentally, the structure for projectin operator is following.

Q⊥(−∞) = P (0) = Q⊥(∞) = {1}

P⊥(−∞) = Q(0) = P⊥(∞) = {0}

The normed condition for T (a) and S(a) operations are following.

‖T (a)‖ = ‖T (−a)‖ = ‖T −1(a)‖ = ‖T ∗(a)‖ = ‖T 2(a)‖ = {1, 0}

Similarly

‖S(a)‖ = ‖S(−a)‖ = ‖S−1(a)‖ = ‖S∗(a)‖ = ‖S2(a)‖ = {0, 1}

On the other hand, the projection operator for normed condition is

following.

‖P (a)‖ = ‖P ∗(a)‖ = ‖P 2(a)‖ = {1, 0}

©About < T (a),S(a) >< T (−a),S(−a) > .

< T (a),S(a) >< T (−a),S(−a) >def .=

< T (a)T (−a),S(a)T (−a) + T (a)S(−a) + S(a)S(−a) > .

S(a) = T (0)− T (a) , S(−a) = T (0)− T (−a)

S(a)T (−a) = (T (0)− T (a))T (−a) = T (−a)− T (0)

T (a)S(−a) = T (a)(T (0)− T (−a)) = T (a)− T (0)

S(a)S(−a) = (T (0)−T (a))(T (0)−T (−a)) = T (0)−T (a)−T (−a)+T (0)

= 2T (0)− T (a)− T (−a).

Therefore

S(a)T (−a) + T (a)S(−a) + S(a)S(−a) = 0.

< T (a),S(a) >< T (−a),S(−a) >=< T (0), 0 >=< 1, 0 >= 0 algebraic.

1 is identity as ring condition and 0 is satisfied ideeal structures.

< 1, 0 >

↙ ↘ring ideal

See the paper “The projrction operator for Laplace transforms”

p.75,76 L2-spaces.

In this time, the operator algebra for T (a) operation is presented fol-

lowing.

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

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

Reference

< D(a), O(a) >< D(−a), O(−a) >=< D(a)D(−a), D(a)O(−a)+O(a)D(−a)+O(a)O(−a) >

=< D(0), 0 >=< I(0), 0 >= 0 for matrix.

< Y (a), N(a) >< Y (−a), N(−a) >=< Y (a)Y (−a), Y (a)N(−a)+N(a)Y (−a)+N(a)N(−a) >

=< I(0), 0 >= 0 for matrix.

< F (a), G(a) >< F (−a), G(−a) >=< F (a)F (−a), F (a)G(−a)+F (−a)G(a)+G(a)G(−a) >

=< I(0), 0 >= 0 for matrix.

Some results

◦ T (a) operation has following relation

T (−a) = T−1(a)

◦ T (a) operation has anti-symmetrical relation for axis of ordinate.

◦ The unitary operation for T (a) is following.

T (a) = T (−a) = T −1(a)

◦ Projection operator in L1-spaces are able to represent two forms. {1}, {0}

◦ Symmetrical concept for T (a) operation generats the elements that it

is separated by ring and ideal.

◦ T (a) and S(a) operations are acrossing on infinite conditions in L2-

spaces.

◦ The projection forms on L2-space is able to represent as {1, 0}.

◦ The relations for T (a) and S(a) operations in Hilbert spaces are follow-

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

The projection forms are following.

Q⊥(−∞) = P (0) = Q⊥(∞) = {1}

P⊥(−∞) = Q(0) = P⊥(∞) = {0}

◦ T (a) and F (a) operations will be necessary the idea of norm.

◦ The projection form for T (a) operation is following.

‖T (a)‖ = ‖T ∗(a)‖ = ‖T 2(a)‖ = {1, 0}

Chapter 5

In this chapter, I want to explain the negative conditions for Laplace

transforms. Since the property of projection operator, we have P =

P ∗ = P 2. So I add the *-algebras for Laplace transforms. Fundamental

concepts are L1 − space D(a)iso←→ Y (a)

homo←− F (a)iso←→ T (a) L2 − space

and it’s all satisfied the property of semi-groups.

D(a) and O(a) are defined following. It is the best simple operations.

O(a) = easO(a) = eas

1− a0

H(a) = D(a) + O(a) = easD(0) = eas

1. . .

Similarly

D∗(a) = easD∗(a) = eas

O∗(a) = easO∗(a) = eas

1− a0

H∗(a) = D∗(a) + O∗(a) = easD∗(0) = eas

1. . .

N.B. Since this D(a) operation is satisfied completed space then we

should define following.

lima→±∞ a0 def .

On the other hand, I define

00 def .= 1.

©Figure of basic operator algebras. (eigenvalues)

Moreover, I want to explain from eigenvalues.

D(a) + O(a) = eas · D(0) = H(a)

↖ extended ↗D(a) = easD(a)

D(0) is identity.

↙ decomposition ↘D(a) + O(a) = H(a) = D(0)

D(−a) + O(−a) = H(−a) = D(0)

↘ contracted ↙D(−a) = e−asD(−a)

D(−a) + O(−a) = e−as · D(0) = H(−a)

D(a)−1 + O(−a) = e−as · D(0) = H(a)−1

N.B. D(−∞) = 0 (annihilator).

Now,D(a) operation is the simplest form in these operator alge-

bras. D(a) is isomorphic to Y (a) operation. Since, Y(a) is Riemann or

Lebesgue integral opeerations.

D(a) is matrix operation that it’s generated only diagonal elements.

This condition is also extended by |eas|. O(a) is pure null operation. So

D(a) is influenced the right terms, directly. D(0) is identity. And this

D(0) is able to decompose D(a) and O(a). In fact, D(a) is able to define

as identity.

If a is negative condition then D(a) is able to be contraction opera-

tor. In this time, it’s represented by D(−a) (a is positive). The D(−a)

is same with D(a)−1. So the contraction operators are represenred two

forms. In this case, I could also obtain the following form.

D(−a) = D(a)−1.

Since O(a) is pure null operation then it’s satisfied the ideal condition.

©Figure of basic operator algebras. (Duality of eigenspaces)

D∗(a) + O∗(a) = eas · D∗(0) = H∗(a)

↖ extended ↗D∗(a) = easD∗(a)

D∗(0) is identity.

↙ decomposition ↘D∗(a) + O∗(a) = H∗(a) = D∗(0)

D∗(−a) + O∗(−a) = H∗(−a) = D∗(0)

↘ contracted ↙D∗(−a) = e−asD∗(−a)

D∗(−a) + O∗(−a) = e−as · D∗(0) = H∗(−a)

D∗(a)−1 + O∗(−a) = e−as · D∗(0) = H∗(a)−1

Essentially D∗(a) is same with D(a). D∗(a) is extended from unitary

operation by |eas|. D∗(0) is also identity. If D(a) is defined in real space

then we have D(a) = D∗(a).

D∗(0) is able to decomposeD∗(a) andO∗(a), respectively. The element

ofD(a) was difined as 00. So we haveD∗(a) = D(a) = identity. Therefore

D∗(a) has always Hermitian form.

D∗(a) = D(a) =

If a is negativ condition then D∗(a) is also contracted and have inverse

condition by |e−as| , simultaneously.

Precisely,

D∗(−a) = e−asD(−a) = e−as

((−a)0

(−a)0

= e−asI∗(−a) = e−asI∗(a)−1 = D∗(a)−1.

N.B. I∗(−a) = I∗(a)−1

Therefore

D∗(−a) = D∗(a)−1.

©Basic operator algebras. (eigenvalues)

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

D(a) = easD(a)

D(0) = I

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

D(−a) + O(−a) = e−asD(0) = H(−a)

D(a)−1 + O(−a) = e−asD(0) = H(a)−1

O(a) is also ideal. (simple)

©Basic operator algebras. (Dual eigenvalues)

D∗(a) + O∗(a) = easD∗(0) = H∗(a)

D∗(a) = easD∗(a)

D∗(0) = I

D∗(a) +O∗(a) = H∗(a) = I(a)

D∗(−a) + O∗(−a) = e−asD∗(0) = H∗(−a)

D∗(a)−1 + O∗(−a) = e−asD∗(0) = H∗(a)−1

O∗(a) is also ideal. (simple)

and normed condition is following.

©Basic normed operator algebras. (eigenspaces)

‖D(a) + O(a)‖ = ‖easD(0)‖ = ‖H(a)‖ = |eas|

‖D(a)‖ = |eas|‖D(a)‖

‖D(0)‖ = 1

‖D(a) +O(a)‖ = ‖D(0)‖ = 1

‖D(−a) + O(−a)‖ = ‖e−asD(0)‖ = ‖H(−a)‖ = |e−as|or

‖D(a)−1 + O(−a)‖ = ‖e−asD(0)‖ = ‖H(a)−1‖ = |e−as|

N.B. ‖T (a)‖ = ‖F (a)‖ = ‖D(a)‖ = |eas|

©Basic normed operator algebras. (Dual eigenvalues)

‖D∗(a) + O∗(a)‖ = ‖easD∗(0)‖ = ‖H∗(a)‖ = |eas|

‖D∗(a)‖ = |eas|‖D∗(a)‖

‖D∗(0)‖ = 1

‖D∗(a) +D∗(a)‖ = ‖D∗(0)‖ = 1

‖D∗(−a) + O∗(−a)‖ = ‖e−asD∗(0)‖ = ‖H∗(−a)‖ = |e−as|or

‖D∗(a)−1 + O∗(−a)‖ = ‖e−asD∗(0)‖ = ‖H∗(a)−1‖ = |e−as|

N.B.‖T ∗(a)‖ = ‖F ∗(a)‖ = ‖D∗(a)‖ = |eas|. Of course ‖D(a)‖ = ‖D∗(a)‖.

Similarly

T (a)f(t) = Y (a)f(t)e−st

Therefore

Y (a)f(t) =∫ ∞

af(t)easdt

N(a)f(t) =∫ a

0f(t)easdt

Y(a)f(t) =∫ ∞

af(t)dt

N (a)f(t) =∫ a

0f(t)dt

W (a)f(t) = Y (a)f(t) + N(a)f(t) = eas · Y (0)f(t)

∗-algebra is following.

Y ∗(a)f(t) =∫ ∞

af(t)easdt

N∗(a)f(t) =∫ a

0f(t)easdt

Y∗(a)f(t) =∫ ∞

af(t)dt

N ∗(a)f(t) =∫ a

0f(t)dt

W ∗(a)f(t) = Y ∗(a)f(t) + N∗(a)f(t) = eas · Y ∗(0)f(t)

N.B f(t) is real functional.

©Figure of basic operator algebras.

Y (a) + N(a) = eas · Y (0) = W (a)

↖ extended ↗Y (a) = easY(a)

Y (0) is identity.

↙ decomposition ↘Y(a) + N (a) = W(a) = Y (0)

Y(−a) + N (−a) = W(−a) = Y (0)

↘ contracted ↙Y (−a) = e−asY(−a)

Y (−a) + N(−a) = e−as · Y (0) = W (−a)

Y (a)−1 + N(−a) = e−as · Y (0) = W (a)−1

N.B. Y (−∞) = 0 (annihilator).

In this case, Y (a) is kind of the simplest form for integral operations.

‖Y (a)‖ = |eas| , ‖N(a)‖ = 0 and ‖W (a)‖ = |eas|. So there is extended

from unitary conditions. The unitary conditions are generated iff a = 0.

Therefore we could represent Y (0) = Y(0) and it’s unitary operators.

Moreover the unitary operation Y(0) is decomposed by Y(a) and N (a),

respectivrly. In this time, Y(a) is the best simple form for the integral

operation. And it’s following form.

Y(a)f(t) =∫ ∞

af(t)dt

iso↔ D(a) = I(a).

Similarly, N (a) is given as following.

N (a)f(t) =∫ a

0f(t)dt

iso↔ O(a) = 0(a).

Now, Y (−a) is contracted by |e−as|. So Y (−a) is contraction operator.

In this time, it’s generated Y (−a) = Y (a)−1. Therefore contraction form

has two types. However the ideal N(−a) have same conditon.

©Figure of basic operator algebras. (Duality)

Y ∗(a) + N∗(a) = eas · Y ∗(0) = W ∗(a)

↖ extended ↗Y ∗(a) = easY∗(a)

Y ∗(0) is identity.

↙ decomposition ↘Y∗(a) + N ∗(a) = W∗(a) = Y ∗(0)

Y∗(−a) + N ∗(−a) = W∗(−a) = Y ∗(0)

↘ contracted ↙Y ∗(−a) = e−asY∗(−a)

Y ∗(−a) + N∗(−a) = e−as · Y ∗(0) = W ∗(−a)

Y ∗(a)−1 + N∗(−a) = e−as · Y ∗(0) = W ∗(a)−1

Essentially, Y ∗(a) is samely with Y (a) operation. Y ∗(a) is conjugate

operator of Y (a) operation. So it’s satisfied the ring condition. If a = 0

then Y (0) = Y ∗(0). In this case, this form is satisfied Hermitian form.

Y ∗(0) is also existed on ring boundary. Of caurse, it’s unitary operator.

In general, if Y (a) is existed in real space then Y (a) is satisfied Her-

mitian form. Therefore Y (a) = Y ∗(a) and Y(a) = Y∗(a) in real spaces

(see p.118,119, Chapter 5, reprinted, paper-1).

In this case, the contraction operator Y ∗(−a) is same as Y ∗(a)−1.

Therefore inverse for Y ∗(a) is represented as negative condition for Y ∗(a).

If it’s unitery condition for Y ∗(a) (Y∗(a)) then it’s same with translated

operator.

Finally, N∗(a) is kind of ideal. This ideal structure is isomorphic with

kernel for semi-group. Therefore, in general, the ideal N∗(a) is not able

to be semi-groups. So I obtain

N∗(a)−1 6= N∗(−a).

Therefore the ideal structure is not able to change to N∗(a)−1 in this

inverse formation.

©Basic operator algebras.

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

Y (a) = easY(a)

Y (0) is unitary

Y(a) +N (a) = W(a) = Y (0)

Y (−a) + N(−a) = e−asY (0) = W (−a)

Y (a)−1 + N(−a) = e−asY (0) = W (a)−1

©Basic operator algebras. (Duality)

Y ∗(a) + N∗(a) = easY ∗(0) = W ∗(a)

Y ∗(a) = easY∗(a)

Y ∗(0) is unitary.

Y∗(a) +N ∗(a) = W∗(a) = Y ∗(0)

Y ∗(−a) + N∗(−a) = e−asY ∗(0) = W ∗(−a)

Y ∗(a)−1 + N∗(−a) = e−asY ∗(0) = W ∗(a)−1

Since F (0) = I and F (a)iso←→ Y (a) then I have Y (0) = I. It is clear

that Y (0) is able to represent “1”.

and normed condition is following

©Basic normed operator algebras.

‖Y (a) + N(a)‖ = |eas|‖Y (0)‖ = ‖W (a)‖ = |eas|

‖Y (a)‖ = |eas|‖Y(a)‖ = ‖eas‖

‖Y (0)‖ = 1

‖Y(a) +N (a)‖ = ‖Y (0)‖ = 1

‖Y (−a) + N(−a)‖ = |e−as|‖Y (0)‖ = ‖W (−a)‖ = |e−as|or

‖Y (a)−1 + N(−a)‖ = |e−as|‖Y (0)‖ = ‖W (a)−1‖ = |e−as|

N.B. ‖Y(a)‖ = 1 for all “a”.

‖Y ∗(a) + N∗(a)‖ = |eas|‖Y ∗(0)‖ = ‖W ∗(a)‖ = |eas|

‖Y ∗(a)‖ = |eas|‖Y∗(a)‖ = |eas|

‖Y ∗(0)‖ = 1

‖Y∗(a) +N ∗(a)‖ = ‖Y ∗(0)‖ = 1

‖Y ∗(−a) + N∗(−a)‖ = |e−as|‖Y ∗(0)‖ = ‖W ∗(−a)‖ = |e−as|or

‖Y ∗(a)−1 + N∗(−a)‖ = |e−as|‖Y ∗(0)‖ = ‖W ∗(a)−1‖ = |e−as|

N.B. ‖Y∗(a)‖ = 1 for all “a”.

In this case, |eas| = |eas|. Therefore, ‖Y (a)‖ = ‖Y ∗(a)‖.

©Definition of F (a) operations

a2 2a a0

......

annC1a

n−1nCka

n−k a0

G(a) = easG(a) = −eas

a2 2a 0

......

annC1a

n−1nCka

n−k 0

F ∗(a) = easF∗(a) = eas

a0 a a2 · · · an

a0 2a · · · ...

a0 · · · nCkan−k

. . ....

G∗(a) = easG∗(a) = −eas

0 a a2 · · · an

0 2a · · · ...

0 · · · nCkan−k

. . ....

©Figure of basic operator algebras. (Pascal′s triangle)

F (a) + G(a) = eas · F (0) = H(a)

↖ extended ↗F (a) = easF(a)

F (0) is identity.

↙ decomposition ↘F(a) + G(a) = H(a) = F (0)

F(−a) + G(−a) = H(−a) = F (0)

↘ contracted ↙F (−a) = e−asF(−a)

F (−a) + G(−a) = e−as · F (0) = H(−a)

F (a)−1 + G(−a) = e−as · F (0) = H(a)−1

N.B. F (−∞) = 0 (annihilator).

Now, F (a) and H(a) are able to extend from unitary operation. And

F (a) operator is isomorphic with T (a) operation. G(a) is kind of ideal.

F (0) is identity. Therefore it’s satisfied unitary operation. Precisely, the

diagonal on there operations F (0) and H(0) are defined as 00 for the

elements. Since, 00 = e0s, there operations F (0) and H(0) are difined as

F (0) =

)= H(0). (32)

Moreover this F (0) is decomposed F(a) and G(a) operations. F(a) is

satisfied the ring condition and G(a) is kind of ideal.

If a is negative condition then the operarions are generated the con-

traction operators by e−as. So, in this time, F (−a) and H(−a) are rep-

resented two forms by F (−a) = F (a)−1, H(−a) = H(a)−1 ,respectively.

Since, in general, G(−a) 6= G(a)−1 then the ideal G(−a) is common

for two forms. In this case, I could obtain

H(−a) = H(a)−1.

©Figure of basic operator algebras. (Duality of Pascal′s triangle)

F ∗(a) + G∗(a) = eas · F ∗(0) = H∗(a)

↖ extended ↗F ∗(a) = easF∗(a)

F ∗(0) is identity.

↙ decomposition ↘F∗(a) + G∗(a) = H∗(a) = F ∗(0)

F∗(−a) + G∗(−a) = H∗(−a) = F ∗(0)

↘ contracted ↙F ∗(−a) = e−asF∗(−a)

F ∗(−a) + G∗(−a) = e−as · F ∗(0) = H∗(−a)

F ∗(a)−1 + G∗(−a) = e−as · F ∗(0) = H∗(a)−1

Essentially, the structure F ∗(a) is same with F (a) operation. Now,

F ∗(a) and H∗(a) are able to extend from unitary operation. So, in this

case, we have Hermitian form as F (0) = F ∗(0). Now, F ∗(a) operator is

isomorphic with T ∗(a) operation. G∗(a) is kind of ideal. F ∗(0) is also

identity. Precisely, the diagonal on there operations F ∗(0) and H∗(0) are

defined as 00 for the elements. Therefore 00 = e0s. So there operations

F ∗(0) and H∗(0) are difined as

F ∗(0) = F (0) =

)= H(0) = H∗(0) (34)

Moreover this F ∗(0) is decomposed F∗(a) and G∗(a) operations. F∗(a)

is satisfied the ring condition and G∗(a) is kind of ideal.

If a is negative condition then the operarions are generated the con-

traction operators by e−as. So, in this time, F ∗(−a) and H∗(−a) are

represented two forms by F ∗(−a) = F ∗(a)−1, H∗(−a) = H∗(a)−1 ,respec-

tively. Since, in general, G∗(−a) 6= G∗(a)−1 then the ideal G∗(−a) is

common for two forms. In this case, I could obtain

H∗(−a) = H∗(a)−1.

Now,since Pascal’s matrix, I could have following.

©Basic operator algebras. (Pascal′s triangle)

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

F (a) = easF(a)

F (0) = I

F(a) + G(a) = F (0) = I

F (−a) + G(−a) = e−asF (0) = H(−a)

F (a)−1 + G(−a) = e−asF (0) = H(a)−1

S(a) and G(a) are ideals.

©Basic operator algebras. (Dual Pascal′s triangle)

F ∗(a) + G∗(a) = easF ∗(0) = H∗(a)

F ∗(a) = easF∗(a)

F ∗(0) = I

F∗(a) + G∗(a) = F ∗(0) = I

F ∗(−a) + G∗(−a) = e−asF ∗(0) = H∗(−a)

F ∗(a)−1 + G∗(−a) = e−asF ∗(0) = H∗(a)−1

S∗(a) and G∗(a) are ideals.

and normed condition is following.

©Basic normed operator algebras. (Pascal′s triangle)

‖F (a) + G(a)‖ = ‖easF (0)‖ = ‖H(a)‖ = |eas|

‖F (a)‖ = |eas|‖F(a)‖

‖F (0)‖ = 1

‖F(a) + G(a)‖ = ‖F (0)‖ = 1

‖F (−a) + G(−a)‖ = ‖e−asF (0)‖ = ‖H(−a)‖ = |e−as|or

‖F (a)−1 + G(−a)‖ = ‖e−asF (0)‖ = ‖H(a)−1‖ = |e−as|

N.B. ‖T (a)‖ = ‖F (a)‖ = |eas|

©Basic normed operator algebras. (Dual Pascal′s triangle)

‖F ∗(a) + G∗(a)‖ = ‖easF ∗(0)‖ = ‖H∗(a)‖ = |eas|

‖F ∗(a)‖ = |easF∗(a)‖

‖F ∗(0)‖ = 1

‖F∗(a) + G∗(a)‖ = ‖F ∗(0)‖ = 1

‖F ∗(−a) + G∗(−a)‖ = ‖e−asF ∗(0)‖ = ‖H∗(−a)‖ = |e−as|or

‖F ∗(a)−1 + G∗(−a)‖ = ‖e−asF ∗(0)‖ = ‖H∗(a)−1‖ = |e−as|

N.B.‖T ∗(a)‖ = ‖F ∗(a)‖ = |eas|. Of course ‖F (a)‖ = ‖F ∗(a)‖.

In this place , I reserch the operator T (a) from the structures of F (a).

Background is following.

Definitions

T (a)f(t) =∫ ∞

af(t)e−(t−a)sdt

S(a)f(t) =∫ a

0f(t)e−(t−a)sdt

T (a)f(t) =∫ ∞

af(t)e−stdt

S(a)f(t) =∫ a

0f(t)e−stdt

R(a)f(t) = T (a)f(t) + S(a)f(t) = eas · T (0)f(t)

∗-algebra is following.

T ∗(a)f(t) =∫ ∞

af(t)e−(t−a)sdt

S∗(a)f(t) =∫ a

0f(t)e−(t−a)sdt

T ∗(a)f(t) =∫ ∞

af(t)e−stdt

S∗(a)f(t) =∫ a

0f(t)e−stdt

R∗(a)f(t) = T ∗(a)f(t) + S∗(a)f(t) = eas · T ∗(0)f(t)

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

we have following. N.B. (−∞ ≤ t ≤ ∞). In this time, ‖T (a)‖ is extended

to |eas| by Hahn-Banach theorem. Since P (a)iso↔ T (a)

iso↔ T (a), then I

could extend to Banach spaces. I want to generate that T (∞) = 0 in

Banach spaces. In this space, S(∞) is infinite, however, as a whole, it’s

satisfied bounded conditions. 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.

©Figure of basic operator algebras.

T (a) + S(a) = eas · T (0) = R(a)

↖ extended ↗T (a) = easT (a)

T (0) is identity.

↙ decomposition ↘T (a) + S(a) = R(a) = T (0)

T (−a) + S(−a) = R(−a) = T (0)

↘ contracted ↙T (−a) = e−asT (−a)

T (−a) + S(−a) = e−as · T (0) = R(−a)

T (a)−1 + S(−a) = e−as · T (0) = R(a)−1

N.B. T (−∞) = 0 (annihilator).

This structure is represented as T (a) of before figure. R(a) is not

always unitary. So T (a), R(a) is able to extend from unitary operations.

Since ‖T (a)‖ = |eas| then we have ‖T (0)‖ = 1. Therefore T (0) is able

to be unitary operator. In this case, if a = 0 then we have 00 = e0s.

In general, 00 is not defined. So I am not clear the structure for T (0)

operation in this view point.

In this case, T (0) is able to decompose T (a) and S(a) conditions.

T (a) is generated as ring and S(a) is situated as ideal structures. N.B.

If f(t) = 0 (0 ≤ t ≤ a) then we have S(a)f(t) = 0, clearly.

This T (a) is able to contract by the factor of e−as. So R(−a) and

R(a)−1 are generated as contraction operators. Therefore we can obtain

T (−a) = T (a)−1.

Since this things, There contraction operators T (a), R(a) are repre-

sented two forms.

Now, S(−a) has also ideal structure. In general, S(−a) 6= S(a)−1. So

ideal structure is not able to be semi-groups.

©Figure of basic operator algebras. (Duality)

T ∗(a) + S∗(a) = eas · T ∗(0) = R∗(a)

↖ extended ↗T ∗(a) = easT ∗(a)

T ∗(0) is identity.

↙ decomposition ↘T ∗(a) + S∗(a) = R∗(a) = T ∗(0)

T ∗(−a) + S∗(−a) = R∗(−a) = T ∗(0)

↘ contracted ↙T ∗(−a) = e−asT ∗(−a)

T ∗(−a) + S∗(−a) = e−as · T ∗(0) = R∗(−a)

T ∗(a)−1 + S∗(−a) = e−as · T ∗(0) = R∗(a)−1

Essentially, T ∗(a) is samely with T (a) operation. T ∗(a) is conjugate

operator of T (a) operation. So it’s satisfied the ring condition. If a = 0

then T (0) = T ∗(0). In this case, this form is satisfied Hermitian form.

T ∗(0) is also existed on ring boundary. Of caurse, it’s unitary operator.

In general, if T (a) is existed in real space then T (a) is satisfied Her-

mitian form. Therefore T (a) = T ∗(a) and T (a) = T ∗(a) in real spaces

(see p.88,89, Chapter 5, reprinted, paper-1).

In this case, the contraction operator T ∗(−a) is same as T ∗(a)−1.

Therefore inverse for T ∗(a) is represented as negative condition for T ∗(a).

If it’s unitery condition for T ∗(a) (T ∗(a)) then it’s same with translated

operator.

Finally, S∗(a) is kind of ideal. This ideal structure is isomorphic with

kernel for semi-group. Therefore, in general, the ideal S∗(a) is not able

to be semi-groups. So I obtain

S∗(a)−1 6= S∗(−a).

©Basic operator algebras.

Now, “Basic oprator algebra” for Laplace transforms are following.

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

T (a) = easT (a)

T (0) = 1

T (a) + S(a) = T (0) = 1

T (−a) + S(−a) = e−asT (0) = R(−a)

T (a)−1 + S(−a) = e−asT (0) = R(a)−1

©Basic operator algebras. (Duality)

T ∗(a) + S∗(a) = easT ∗(0 = R∗(a)

T ∗(a) = easT ∗(a)

T ∗(0) = 1

T ∗(a) + S∗(a) = T ∗(0) = 1

T ∗(−a) + S∗(−a) = e−asT ∗(0) = R∗(−a)

T ∗(a)−1 + S∗(−a) = e−asT ∗(0) = R∗(a)−1

Since F (0) = I and F (a)iso←→ T (a) then I have T (0) = 1. However it

is not clear that T (0) is able to represent as “1”.

and normed condition is following

‖T (a) + S(a)‖ = |eas|‖T (0)‖ = ‖R(a)‖ = |eas|

‖T (a)‖ = |eas|‖T (a)‖ = ‖eas‖

‖T (0)‖ = 1

‖T (a) + S(a)‖ = ‖T (0)‖ = 1

‖T (−a) + S(−a)‖ = |e−as|‖T (0)‖ = ‖R(−a)‖ = |e−as|or

‖T (a)−1 + S(−a)‖ = |e−as|‖T (0)‖ = ‖R(a)−1‖ = |e−as|

N.B. ‖T (a)‖ = 1 for all “a”.

‖T ∗(a) + S∗(a)‖ = |eas|‖T ∗(0)‖ = ‖R∗(a)‖ = |eas|

‖T ∗(a)‖ = |eas|‖T ∗(a)‖ = |eas|

‖T ∗(0)‖ = 1

‖T ∗(a) + S∗(a)‖ = ‖T ∗(0)‖ = 1

‖T ∗(−a) + S∗(−a)‖ = |e−as|‖T ∗(0)‖ = ‖R∗(−a)‖ = |e−as|or

‖T ∗(a)−1 + S∗(−a)‖ = |e−as|‖T ∗(0)‖ = ‖R∗(a)−1‖ = |e−as|

N.B. ‖T ∗(a)‖ = 1 for all “a”.

In this case, |eas| = |eas|. Therefore, ‖T (a)‖ = ‖T ∗(a)‖.

Some results

◦ If a is positive then T (a) is extended from unitary conditions. On the

contrary, if a is negative then T (a) is contracted from it. There conditions

will be treated in C∗-algebras.

◦ The contraction forms from unitary operator is able to represent fol-

lowing.

T (−a) + S(−a) = e−as · T (0) = R(−a)

T (a)−1 + S(−a) = e−as · T (0) = R(a)−1

◦ In general, T (a) and R(a) operations are satisfied ring conditions and

S(a) is ideal.

◦ In the ring conditions, we have following relations.

T (−a) = T (a)−1 , R(−a) = R(a)−1.

◦ T (0) operation is able to decompose T (a) and S(a) operations, respec-

tively.

◦ Essentially, T (0) operation and T (a) operation have same property.

◦ ∗-algebra for Lapace transforms is same with T (a) operation. Especially,

T (0)∗ = T (0) (Hemitian form).

Chapter 6

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

forms. As this application, I assert that it is able to treat the group-rings

for semi-groups.

(L1-spaces)

It’s the simplest operation for Pascal’s matrix.

D(a)v is defined following

If it’s treated on L1-spaces then we have following.

Since definition of D(a) operations, we have following.

[easD(a)]v = eas[D(a)v] = D(a)[easv]

Similarly

[easO(a)]v = eas[O(a)v] = O(a)[easv]

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

[easD(a)]v = eas[D(a))v] = D(a)[easv] (1)

For example, let following.

eas D(a) v

↓ ↓ ↓α x y

Since (1), then we have

(αx)y = α(xy) = x(αy). (semi− groups)

Therefore, D(a) operation has a property of seni-groups.

In general, the operation D(a) is able to extend to negative condition.

[e−asD(−a)]v = e−as[D(−a)v] = D(−a)[e−asv]

⇓(αx)y = α(xy) = x(αy). (semi− groups)

Similarly

It has a property of all semi− groups for all a.

(αx)y = α(xy) = x(αy).

If it’s treated on L2-space then we have following.

(L2-spaces)

Similarly

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

[easH(a)]v = eas[H(a)v] = H(a)[easv] (i)

eas H(a) v

↓ ↓ ↓α x y

Since (i), then we have

In general, the operation H(a) is able to extend to following condition.

[e−asH(−a)]v = e−as[H(−a)v] = H(−a)[e−asv]

Similarly

[easH(a)]v = eas[H(a)v] = H(a)[easv]

D(a)v is defined following

D(a) = eas

So we have

D(a)D(b) = D(a + b)

D(0) = I

s− lima→0 D(a)v = v

Therefore

D(a) is generated with semi− groups.

Essentially, O(a) operation is able to be ideal. So this semi-group

is treated iff it’s maximal ideal. If I defined on L2-space then O(∞) is

following.

O(∞) = −

a0 − 1

a0 − 1. . .

a0 − 1

def .= Identity as a → ±∞.(49)

So we have

O(±∞)O(±∞) = O(±∞)

O(±∞) = I

s− lima→±∞O(a)v = v

Therefore

O(a) is generated with semi− groups iff a is ±∞.

Moreover we will have following equivalence.

D(a)D(b) = D(a + b) −→ D(a)n = D(na)

↘ ↙

D(0)2 = D(0)

This condition is also semi-groups. Lower form is able to consider as

projection operator. And right is able to generate invariant form.

This condition is able to represent only one by using the group-ring.

D(a)nD(b)m = D(an + bm) = D−n(a)D−m(b)

If D(a) is norm condition , then we have following.

‖D(a)‖‖D(b)‖ = ‖D(a + b)‖ −→ ‖D(a)‖n = ‖D(na)‖

↘ ↙

‖D(0)‖2 = ‖D(0)‖ = 1

(L1-spaces)

Y (a) oparations are defined following.

Y (a)f(t) = easY(a)f(t) =∫ ∞

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

0f(t)easdt

W (a)f(t) = Y (a)f(t) + N(a)f(t) = easW(a)f(t) =∫ ∞

0f(t)easdt

So we have following conditions.

[easY(a)]f(t) = eas[Y(a)f(t)] = Y(a)[easf(t)]

Similarly

[easN (a)]f(t) = eas[N (a)f(t)] = N (a)[easf(t)]

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

[easY (0)]f(t) = eas[Y (0)f(t)] = Y (0)[easf(t)] (1).

Now, let following.

eas Y (0) f(t)

↓ ↓ ↓α x y

Similarly

[easY (0)]f(t) = eas[Y (0)f(t)] = Y (0)[easf(t)]

It has property of semi− groups.

If we consider on L2-space then we have following.

(L2-space)

Similarly

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

[easW(a)]f(t) = eas[W(a)f(t)] = W(a)[easf(t)] (i)

Now, let following.

eas W(a) f(t)

↓ ↓ ↓

α x y

Since (i), then we have

Similarly

[easW(a)]f(t) = eas[W(a)f(t)] = W(a)[easf(t)]

It has property of semi− group.

N.B. Essentially, we have Y (0) = W(a).

T (a)f(t) is defined following

Y (a)f(t) =∫ ∞

af(t)easdt

So we have

Y (a)Y (b) = Y (a + b)

Y (0) = 1

s− lima→0 Y (a)f = f

Therefore

Y (a) is generated with semi− groups.

Essentially, N (a) operation is able to be ideal. So this semi-group

is treated iff it’s maximal ideal. If I defined on L2-space then N (∞) is

following.

N (∞)f(t) = Y(0)f(t) =∫ ∞

0f(t)dt = Identity.

So we have

N (±∞)N (±∞) = N (±∞)

N (±∞) = I

s− lima→±∞N (a)f = f

Therefore

N (a) is generated with semi− groups iff a is ±∞.

Y (a)Y (b) = Y (a + b) −→ Y (a)n = Y (na)

↘ ↙Y (0)2 = Y (0)

Y (a)nY (b)m = Y (an + bm) = Y−n(a)Y−m(b)

If Y (a) is norm condition , then we have following.

‖Y (a)‖‖Y (b)‖ = ‖Y (a + b)‖ −→ ‖Y (a)‖n = ‖Y (na)‖↘ ↙‖Y (0)‖2 = ‖Y (0)‖ = 1

Let E(a) = eas separated three types E(a) , E−1(a) and E∗(a),respectively.Now,

I want to explain the property of E(a).

E(a) have a following properties.

E(a)E(b) = E(a + b)

E(0) = 1 (identity)

s− lima→0 E(a)T (0) = T (0)

Therefore

E(a) have properties of semi− groups.

Furthermore we will find that E(a) have ideal structures.

(L2-space)

In this time, I want to explain L2-space for F (a) operations.

F (a) is defined following

.... . .

an · · · a0

G(a) = −easG(a) = −eas

a0 − 1

a a0 − 1...

an · · · a0 − 1

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

Since this definitions then we have following.

[easF(a)]v = eas[F(a)v] = F(a)[easv]

Similarly

[easG(a)]v = eas[G(a)v] = G(a)[easv]

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

[easH(a)]v = eas[H(a))v] = H(a)[easv] (1)

eas H(a) v

↓ ↓ ↓α x y

Therefore, Pascal’s matrix H(a) has a property of seni-groups. In this

case, ‖H(a)‖ = |eas| and it’s irreducible form.

In general, the operation F(a) is able to extend to following condition.

[e−asF(−a)]v = e−as[F(−a)v] = F(−a)[e−asv]

[easF(a)]v = eas[F(a)v] = F(a)[easv]

Similarly

[easG(a)]v = eas[G(a)v] = G(a)[easv]

[easH(a)]v = eas[H(a)v] = H(a)[easv]

Since definition of Pascal’s matrix, we have following.

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

F (0) = I

s− lima→0 F (a)v = v

Therefore

F (a) is generated with semi− groups.

Essentially, G(a) operation is able to be ideal. So this semi-group

is treated iff it’s maximal ideal. If I defined on L2-space then G(∞) is

following.

G(∞) = −

a0 − 1

a a0 − 1...

an · · · a0 − 1

def .= Identity as a → ±∞.(57)

So we have

G(±∞)G(±∞) = G(±∞)

G(±∞) = I

s− lima→±∞ G(a)v = v

Therefore

G(a) is generated with semi− groups iff a is ±∞.

F (a)F (b) = F (a + b) −→ F (a)n = F (na)

↘ ↙F (0)2 = F (0)

F (a)nF (b)m = F (an + bm) = F−n(a)F−m(b)

If F (a) is norm condition , then we have following.

‖F (a)‖‖F (b)‖ = ‖F (a + b)‖ −→ ‖F (a)‖n = ‖F (na)‖↘ ↙‖F (0)‖2 = ‖F (0)‖ = 1

(L2-space)

T (a) operations are defined following

T (a)f(t) = easT (a)f(t) =∫ ∞

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

0f(t)e−(t−a)sdt

R(a)f(t) = T (a)f(t) + S(a)f(t) = easR(a)f(t) =∫ ∞

0f(t)e−(t−a)sdt

So we have following conditions

[easT (a)]f(t) = eas[T (a)f(t)] = T (a)[easf(t)]

Similarly

[easS(a)]f(t) = eas[S(a)f(t)] = S(a)[easf(t)]

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

[easR(a)]f(t) = eas[R(a)f(t)] = R(a)[easf(t)] (1)

Now, let following.

eas R(a) f(t)

↓ ↓ ↓α x y

[easT (a)]f(t) = eas[T (a)f(t)] = T (a)[easf(t)]

Similarly

[easS(a)]f(t) = eas[S(a)f(t)] = S(a)[easf(t)]

[easR(a)]f(t) = eas[R(a)f(t)] = R(a)[easf(t)]

It has property of all semi− groups for all a.

Since definition of Laplace transforms, we have following.

T (a + b)∫ t−b

af(τ)g(t− τ)dτ = T (a)

af(τ)g(b + t− τ)dτ =

= T (a)T (0)∫ t

af(τ)g(b + t− τ)dτ = T (a)

af(τ)T (0)g(b + t− τ)dτ =

= T (a)∫ t

af(τ)T (b)g(t− τ)dτ = T (a)T (b)

∫ t−b

af(τ)g(t− τ)dτ =

= T (a)f(t)T (b)g(t) =

= T (a)T (b)∫ t−a

bf(t− τ)g(τ)dτ = T (b)

bT (a)f(t− τ)g(τ)dτ =

= T (b)∫ t

bT (0)f(a + t− τ)g(τ)dτ = T (b)T (0)

bf(a + t− τ)g(τ)dτ =

= T (b)∫ t

bf(a + t− τ)g(τ)dτ = T (a + b)

∫ t−a

bf(t− τ)g(τ)dτ

T (a+b)∫ t−b

af(τ)g(t−τ)dτ = T (a)f(t)T (b)g(t) = T (a+b)

∫ t−a

bf(t−τ)g(τ)dτ

T (a+b){f(t)∗g(t)}(a,b) = {T (a)f(t)}{T (b)g(t)} = T (a+b){g(t)∗f(t)}(b,a)

Moreover

T (a + b)∫ t−b

af(τ)g(t− τ)dτ = T (a)

af(τ)g(b + t− τ)dτ ≡ T (a)h(t)

N.B. h(t) =∫ ta f(τ)g(b + t− τ)dτ

T (a)h(t) = T (0)h(t + a)

T (a)T (0)h(t) = T (0)T (0)h(t+a) = T (0)h(t+a) iff T (0)2 = T (0)

T (a)T (b)h(t− b) = T (a)h(t)

T (a)T (b)∫ t−b

af(τ)g(b+t−b−τ)dτ = T (a)

af(τ)g(b+t−τ)dτ = T (a+b)

∫ t−b

af(τ)g(t−τ)dτ

T (a)T (b)∫ t−b

af(τ)g(t− τ)dτ = T (a + b)

∫ t−b

af(τ)g(t− τ)dτ

Hence T (a)T (b) = T (a+b) iff T (0)2 = T (0)

Now, T (0) is able to regard as kind of projection operator.

So, since P 2 = P , then we have

T (a)T (b) = T (a + b).

T (a) operation is satisfied following conditions.

T (a)T (b) = T (a + b)

T (0) = I

s− lima→0 T (a)f = f

Therefore T (a) has a property of semi-groups.

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

↘ ↙

T (0)2 = T (0)

See p.76 appendix for ′′Rings and ideal structures for Laplace transforms′′

(reprint).

T (a)nT (b)m = T (an + bm) = T−n(a)T−m(b)

If T (a) is norm condition , then we have following.

‖T (a)‖‖T (b)‖ = ‖T (a + b)‖ −→ ‖T (a)‖n = ‖T (na)‖

↘ ↙

‖T (0)‖2 = ‖T (0)‖ = 1

Finally

In this time, all operations that T (a), Y (a), F (a) and D(a) are all

semi-groups. Ideal operations that S(a),N (a),G(a) and O(a) are also all

semi-groups iff a → ∞. If eas is kind of ideal then it’s satisfied dense

condition. Therefore S(a), N(a), G(a) and O(a) will have semi-groups.

So ideal structure is able to have a property of semi-groups. Especially,

the semi-groups (ring operations) have following properties.

T (a)−1 = T (−a) , F (a)−1 = F (−a) , etc

Some results

◦ The property of semi-group for T (a) operation is able to extend to the

L2-space.

◦ Since the convolution theorem, we could have following form.

T (a)T (b) = T (a + b).

◦ L1-space to L2-space condition for T (a) operation is continue as follow-

ing form.

(L1 − space) T (0) = R(a) (L2 − space)

◦ In this case, this operator algebras for T (a) operation is following rela-

tions.

Integral operator

T (a)homo−→ Y (a)

iso l l iso

F (a)homo−→ D(a)

Matrix operator

◦ Ideal operation is able to be semi-group iff a → ±∞ in Hilbert space.

◦ All operations are able to represent on semi-groups.

◦ This semi-groups are able to apply to group-rings.

Conclusion

In previous paper, we explained about the projection for Laplace trans-

forms in positive condition. In this time, I extend for “a′′ of T (a) to

negative condition. Essential conception does not change with positive

condition. However, in general, T (a) operation is not unitary operation.

So it’s generated anti-symmetrical form.

‖T (−a)‖ = ‖T (a)−1‖ = |e−as|.The projection operator in L2-space should consider the set condition

{1, 0}. If this projection operator in L1-space then we should consider

the separated form {1} and {0}, respectively.

The operator algebras for negative condition is realized in contracted

form. Rings are able to change to negative conditions. On the other

hand, this ideal structure is not able to change to negative except for

a → ±∞ in L2-spaces.

As a whole, There operator algebras for Laplace transforms have fol-

lowing condition.

Ring + Ideal = Ring.

Similarly the ∗−algebras have identical conceptions.

Finally, in this case, all ring operations have following relations.

T (a)T (b) = T (a + b).

So we have the property of semi-groups. Therefore the fundamental

condition for Laplace transforms may consider it. In this time, I extend

to group-ring for this application. So we have following.

T n(a)Tm(b) = T (na + mb) = T−n(a)T−m(b).

Similarly, the projection form is following.

P n(a)Pm(b) = P (na + mb) = P−n(a)P−m(b).

(Tue)6.Oct.2009

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

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

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

Approximation, Springer, 1967

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

[10] Takao Saito, Operator algebras for Laplace transforms (reprint),

(Mon)8.Oct.2007.

Furthermore

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

[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.

the negative condition for laplace transformsopab.web.fc2.com/no.2.pdfpreface this paper is builded...

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