## Linear Operators: Spectral operators |

### From inside the book

Results 1-3 of 56

Page 2044

On the real axis 0 = 0 and hence , since T ( t ) is

) T ( t ) l SK , Ost < 00 , a fact which also follows from ( 61 ) . The inequality ( 81 )

shows that T ( 5 ) is

On the real axis 0 = 0 and hence , since T ( t ) is

**strongly**continuous at t = 0 , ( 82) T ( t ) l SK , Ost < 00 , a fact which also follows from ( 61 ) . The inequality ( 81 )

shows that T ( 5 ) is

**strongly**continuous at $ = 0 provided remains in a sector - 4 ...Page 2194

attempt will be made to characterize the strong closure of a commutative algebra

of spectral operators . It has been observed ( cf . VI . 1 . 5 ) that a convex set in the

...

**Strongly**Closed Algebras and Complete Boolean Algebras In this section anattempt will be made to characterize the strong closure of a commutative algebra

of spectral operators . It has been observed ( cf . VI . 1 . 5 ) that a convex set in the

...

Page 2220

... Tư ) → R ( 1 ; T )

rational functions which are continuous on V . Then r ( Ta ) is a finite product of

terms of the form XI – Tą and R ( 2 ; T . ) . Thus , since the product of bounded ,

... Tư ) → R ( 1 ; T )

**strongly**. Now let r be an element of the class R ( V ) ofrational functions which are continuous on V . Then r ( Ta ) is a finite product of

terms of the form XI – Tą and R ( 2 ; T . ) . Thus , since the product of bounded ,

**strongly**...### What people are saying - Write a review

We haven't found any reviews in the usual places.

### Other editions - View all

### Common terms and phrases

algebra of projections analytic apply arbitrary assumed B-space Banach space Boolean algebra Borel sets boundary conditions bounded bounded Borel bounded operator Chapter clear closure commuting compact complete consider constant contained converges Corollary corresponding countably additive defined Definition denote dense determined differential operator domain elements equation equivalent established example exists extension fact finite follows formula given gives Hence Hilbert space hypothesis identity inequality integral invariant inverse Lemma limit linear linear operator manifold Math Moreover multiplicity norm normal positive preceding present problem projections PROOF properties prove range regular resolution resolvent respectively restriction Russian satisfies scalar type seen sequence shown shows spectral measure spectral operator spectrum strongly subset subspace sufficiently Suppose Theorem theory topology unbounded uniformly unique valued vector weakly zero