operator theory

OPERATOR THEORY LECTURE NOTES UNIT IV FOR M.SC., MATHEMATICS INVITATION TO LINEAR OPERATORS FROM MATRICES TO BOUNDED LINEAR OPERATORS ON A HILBERT SAPCE BY TAKAYUKI FURUTA Dr. D. SENTHILKUMAR DEPARTMENT OF MATHEMATICS GOVT. ARTS COLLEGE

RELATION AMONG SEVERAL CLASSES OF NON-NORMAL OPERATORS:

2.6.1 PARANORMAL OPERATORS:  Definition:

An operator T on a Hilbert space H is said to be a paranormal operator if for any unit vector x H.  Definition: An operator T on a Hilbert space H is said to be a subnormal operator if T has a normal extension N, that is, there exists a normal operator N on a larger Hilbert space K H such that Tx = Nx for all x H.

 Therom 1:

If T is a paranormal operator, then the following inequalities hold for any x H. +1 x

4x

3x

2x

‖ x‖ -----------------(1) ‖ x‖ ‖ 3x ‖ ‖ 2x‖ ‖ x‖ ‖x ‖ If T is an invertible paranormal operator then the following inequalities hold for any vector x H.

…….

…….

-----------------(2) ‖x ‖ ‖ −1 x ‖

−1 x

‖ −2 x ‖

−2 x

‖ −3 x ‖

……

− +1 x

‖ − x‖

……..

1 ‖ −1 ‖

2.6.2 IMPLICATION RELATION AMONG SEVERAL CLASSES OF NON NORMAL OPERATORS:

To prove: Quasi normal Subnormal. For this first we show that Quasi normal Hyponormal. Let T be quasi normal and T = U ITI be the polar decomposition of T. We know that, “ T is quasi normal  UI T I = I T IU”. Let T*T - TT* = (UI T I)* (UI T I) - (UI T I) (UI T I)* = I T IU* UI T I - UI T I I T IU* = - UI T I I T IU* = I T I (I - UU*) I T I T*T - TT* 0.

Since UI T I = I T IU holds. Hence T is hyponormal. Hence Quasi normal Hyponormal. Next we show that Quasi normal Subnormal. Define [T] = T*T – TT*. Then, [T] 0. Since T is quasi normal, [T] T = T*TT – TT*T = T*TT – T*TT [T]T = 0. So that, T*[T] = 0 by taking adjoint [T]T = 0. T= T* = 0 for all natural number n. Since the square root ‘S’ of [T] is approximated uniformly by polynomials of [T] without constant terms, so that ST = T*S.

[since T is quasi normal and [T] = T*T – TT*, ST = T*S]. Therefore T is subnormal. Hence Quasi normal Subnormal Next we prove that Subnormal Hyponormal. Let T be subnormal then there is a normal extension N.

2.7. CHARACTERIZATION OF CONVEXOID OPERATORS AND RELATED EXAMPLES 2.7.1.CHARCTERIZATION OF CONVEXOID OPERATORS

Definition:

An operator T is said to be a convexoid operator if (T) = co (T) where (T) means the convex hull of the spectrum (T) of T. Definition: An operator T is said to be a condition ‖

‖ =

the spectrum (T) of T.

for all



operator

(T) where (T) means

. Definition: An operator T is said to be a transoloid operator if T - is normaloid for any c. Lemma 1: If x is any bounded closed set in the complex plane, then (i)

co x = {the intersection of all circles which is contain the set X} =



.



(ii) co x = {the intersection of all closed half planes which contain the set X} =





.

. Theorem 1: An operator T is convexoid if and only if T all complex number .

is spectraloid for

Proof: Taking X = (T) and since X = (T) is convex. Since,

, then the following (1) and (2) hold

(T) =





--------(1)

=





--------(2)

co From (1) and (2),

The proof follows from the relation co

-

= co

- ).

Theorem 2: −1

An operator t is convexoid iff ‖ all

‖

1 ( ,

(T))

for

 co

We prove this theorem, we have to prove the following (5) lemma. Proof: The resolvent of any operator T satisfies a first order growth with respect to (T) by ( iii) of theorem For any operator T, the following (i),(ii) and (iii) hold (i)

co (iii)

(iv) ‖

, (ii)

1 ( ,

(T)) −1

‖

w(T) −1

‖ 1 ( ,

( ))

‖

for all

‖ T‖ 1 ( ,

( ))



(T).

We have only to show if part since the opposite implication follows easily from this fact. The condition (*) is equivalent to the following. ‖

x‖

 co

and ‖x‖ = 1.

So that, 2

– Re(Tx, x)

+

2

2

2

Taking = | | to ∞, we obtain

−( + )

Re(Tx, x)

and dividing by |

Re s

| and transferring |

for ‖x‖ = 1.

|

This implies (T)  co inclusion always holds.

by (ii) of lemma 1 and the opposite

So we obtain (T) = co

.

Theorem 3: Let T be a hyponormal operator,then the following properties hold. (i) (ii) (iii) (iv)

is also hyponormal for any c. is a transaloid operator. is also a hyponormal operator if exists. is a condition operator.

Proof: (i) To prove:

is also hyponormal for any

Since T is hyponormal

T*T

c.

TT*

Consider, ∗

(

= (T*T -

)-( T - T* +

= T*T - TT*

∗

) 2

) – (TT* -

T - T* +

0.

Therefore, is also a hyponormal for any

c.

2

)

(ii)To prove:

is a transaloid operator.

is hyponormal for any c, so that is normaloid c. Since a hyponormal operator is normaloid.

for any

Hence

is a transaloid operator.

(iii)To prove: Since T*T

−1

is also a hyponormal operator if −1

TT*, we have

*

−1 −1

This is equivalent to, I

*

−1

−1∗ −1

Hence

−1

−1∗ −1

−1∗

is also a hyponormal operator if

−1

exists.

exists. I.

(iii)To prove: is a condition

operator.

We know that, II

x‖

=r

‖ ‖ =r holds by (i) and (iii) and by the result that a hyponormal operators is normaloid. Therefore, ‖

‖ =

Hence is a condition Hence proved.

operator.

Theorem 4: The following properties hold. (i) (ii)

If T is a transaloid operator,then T is a convexoid operator. If T is a condition 1 operator, then T is a convexoid operator.

Proof: (i)To prove: If T is a transaloid operator,then T is a convexoid operator. Let T be a transaloid operator. (i.e)

is normaloid for any

c.

Then, is spectraloid for any

c.

Since a normaloid operator is spectraloid by theorem 3(2.6.1)

So that, T is convexoid by theorem 2 Hence If T is a transaloid operator,then T is a convexoid operator. (ii)To prove: If T is a condition operator. Let T be a condition

1

1

operator, then T is a convexoid

operator.

Then, we have ‖

−1

‖ =

1 ( ,

1 (T ))

( ,

(T))

for any

Because the inequality follows by the definition of condition operator. Hence T is a convexoid operator. Hence proved.

 co

1

Corollary 5: (i) (ii)

Every hyponormal operator is convexoid. Every normal operator is convexoid.

Proof: (i)To prove: Every hyponormal operator is convexoid. If T is hyponormal, then the transloid by (ii) of theorem 3. So that T is convexoid by theorem 4. Hence every hyponormal operator is convexoid. (ii)To prove: Every normal operator is convexoid. We know that, Every normal operator is a hyponormal operator and also every hyponormal operator is convexoid. Therefore, Every normal operator is convexoid. Hence proved.

Theorem 6: An operator T is convexoid iff for any 0 2 where (s) denotes co

Re ( .

)=

(Re(

))

Proof: We know that, Every self adjoint operator is convexoid, so that the condition ( - ) means Re (

)=

(Re(

=

(Re(

= Re Re ( Therefore, Then,

( )=

))

(

) = Re (

)) ) ( )) for any

with 0

( )

The reverse is obvious.

Hence proved.

2 .

Theorem

′

: ′

Any operator T is convexoid iff co (Re( )) for any with 0

co Re (

)=

2 .

Proof: We know that, co ( ) = co (Re(

)) =

( ) (Re(

))

= Re ( (

))

= Re ( co (Re(

)) = co(Re

( )) (

)) for any 0

2 .

Theorem 7: If an operator T is convexoid such that both ( ) and (Re( )) are connected, then Re ( ) = (Re( )) hold. Proof: We know that, “T is convexoid then ′ [by theorem 6]. (i.e) is Therefore,

′

′

at

condition is true”. = 0 is true.

Re ( ) and (Re( )) are both closed convex sets.

Since connected sets coinsides with a convex set in the real line, ′ then the eliminate co in . Re (

)=

Hence Re ( ) =

(Re( (Re( )).

Hence proved.

))

Corollary 8: Let T be any convexoid operator. If [ 0 , interval containing Re ( ),then 0 , 0 (Re( )).

0]

is the smallest

Proof: Suppose T is convexoid operator and [ 0 , 0 ] is the smallest interval containing Re ( ( )). By theorem 6, we have Re ( ( )) = Therefore,

0,

(Re( )). 0

.

Hence proved. Corollary 9: Suppose T is a convexoid operator and ( ) is connected. If [ , ] is the smallest interval containing (Re( )), then (Re( ))  [ , ]  Re ( ( )).

Proof: Suppose T is a convexoid operator and ( ) is connected. If the smallest interval [ , ]



(Re( )) then Re ( ( ))



[ , ]

[by theorem 6’] By using corollary 8, we have

(Re( ))



[ , ]



Re ( ( ))

Hence proved. 2.7.2 Examples: Example 1: An example of non convexoid, non-pranormal and normaloid operator.

Let T =

where M =

Solution: 2

Here

2

‖

2

= T.T =

2

x‖

2

2

does not hold.

Thus T is non paranormal operator. Hence proved.