## Linear Operators: General theory |

### From inside the book

Results 1-3 of 91

Page 182

Thus da / du is

) m ( ds ) , Ee2 . ... additive set function My on E , the equation 41 ( 8 - 1 ( E ) ) = uz

( E )

Thus da / du is

**defined**u - almost everywhere by the formula ( d2 1 ( E ) = S ( ( 19) m ( ds ) , Ee2 . ... additive set function My on E , the equation 41 ( 8 - 1 ( E ) ) = uz

( E )

**defines**an additive set function με on Σ . Moreover ( a ) if My is countably ...Page 240

It is evident that if we

and uld ) = 0 ... The space B ( S ) is

all bounded scalar functions on S . The norm is given by W = sup \ | ( s ) . SES

SES ...

It is evident that if we

**define**the set function u on & by placing u ( E ) = 0 if E + $and uld ) = 0 ... The space B ( S ) is

**defined**for an arbitrary set S and consists ofall bounded scalar functions on S . The norm is given by W = sup \ | ( s ) . SES

SES ...

Page 516

38 ( Markov ) Let S be a non - void set and a function on S to S . A function y

- 1E ) , ECS , where 6 - 1E = [ sløs e E ] . Show that there is a non - negative

bounded ...

38 ( Markov ) Let S be a non - void set and a function on S to S . A function y

**defined**on the family of subsets of S is said to be - invariant in case u ( E ) = u ( 0- 1E ) , ECS , where 6 - 1E = [ sløs e E ] . Show that there is a non - negative

bounded ...

### What people are saying - Write a review

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

### Contents

Preliminary Concepts | 1 |

B Topological Preliminaries | 10 |

Algebraic Preliminaries | 34 |

Copyright | |

31 other sections not shown

### Other editions - View all

### Common terms and phrases

algebra Amer analytic applied arbitrary assumed B-space Banach spaces bounded called clear closed compact operator complex condition Consequently constant contains continuous functions converges convex convex set Corollary countably additive defined DEFINITION denote dense determined differential dimensional disjoint domain element equation equivalent everywhere Exercise exists extension field finite follows formula function defined function f given Hence Hilbert space identity implies inequality integral interval Lebesgue Lemma limit linear functional linear operator linear space Math neighborhood norm operator operator topology problem projection PROOF properties proved range reflexive representation respect satisfies scalar seen semi-group separable sequence set function Show shown statement subset subspace sufficient Suppose Theorem theory topology u-measurable uniform uniformly unique unit sphere valued vector weak weakly compact zero