## Linear Operators: General theory |

### From inside the book

Results 1-3 of 85

Page 424

Since each projection is a continuous map , each o the sets A ( x , y ) and B ( 0 , x ) is

Since each projection is a continuous map , each o the sets A ( x , y ) and B ( 0 , x ) is

**closed**. Hence tK = n , vex A ( x , y ) n naeg , xe x B ( a ...Page 488

If the adjoint of an operator U in B ( X , Y ) is one - to - one and has a

If the adjoint of an operator U in B ( X , Y ) is one - to - one and has a

**closed**range , then UX = Y. Proof . Let 0 #ye Y and define I ' = { y * y * € Y * ...Page 489

since the range of U * is

since the range of U * is

**closed**, a * U * y * for some y * € Y * . If z * is the restriction of y * to Z , then x * U * z * .### What people are saying - Write a review

User Review - Flag as inappropriate

i want to read

### Contents

B Topological Preliminaries | 10 |

quences | 26 |

Algebraic Preliminaries | 34 |

Copyright | |

80 other sections not shown

### Other editions - View all

### Common terms and phrases

Acad algebra Amer analytic applied arbitrary assumed B-space Banach Banach spaces bounded called clear closed compact complex condition contains continuous functions converges convex Corollary countably additive defined DEFINITION denote dense determined differential disjoint Doklady Akad element equation equivalent everywhere Exercise exists extension field finite follows function defined function f given Hence Hilbert space implies inequality integral interval isomorphism Lebesgue Lemma limit linear functional linear operator linear space mapping Math meaning measure space metric neighborhood norm operator positive measure problem Proc PROOF properties proved respect Russian satisfies scalar seen semi-group separable sequence set function Show shown sphere statement subset sufficient Suppose Theorem theory topological space topology transformations u-measurable uniform uniformly unique unit valued vector weak weakly compact zero