## Linear Operators: General theory |

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Since the integral Ss / ( s ) u ( ds ) satisfies the inequality \\ $ t ( s ) ( ds ) < sup \ ( s ) [ 0 ( 4 , S ) , it is

Since the integral Ss / ( s ) u ( ds ) satisfies the inequality \\ $ t ( s ) ( ds ) < sup \ ( s ) [ 0 ( 4 , S ) , it is

**clear**that the integral is a continuous linear functional on C ( S ) . The following theorem is a converse to this ...Page 282

It is

It is

**clear**that T ( € ) CT ( 8 ) if ε < d and that -te T ( € ) whenever t e T ( E ) . The function f is said to be almost periodic if it is continuous and if for every e > 0 there is an L L ( E ) > 0 such that every interval in R of ...Page 292

It is

It is

**clear**that if F , and F , are elements of Eg , then F F2 € Eg . It is also**clear**that if Fi € Es , then S - F , € Eg , and that if F1 , F , E , with F , F , = $ , then F , UF , Eg . € Ez € It follows that Ez is a field .### What people are saying - Write a review

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### Contents

B Topological Preliminaries | 10 |

quences | 26 |

Algebraic Preliminaries | 34 |

Copyright | |

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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