## Linear Operators, Part 1 |

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Page 269

Since the integral Sst ( s ) u ( ds ) satisfies the inequality Sst ( s ) u ( ds ) < sup | t (

s ) [ 0 ( 4 , S ) , it is

) . The following theorem is a converse to this statement . 2 THEOREM .

Since the integral Sst ( s ) u ( ds ) satisfies the inequality Sst ( s ) u ( ds ) < sup | t (

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 statement . 2 THEOREM .

Page 282

It is

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 length L contains at least one ...

It is

**clear**that T ( E ) CT ( 8 ) if € < 8 and that - te T ( E ) whenever te T ( € ) . Thefunction 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 length L contains at least one ...

Page 292

It is

if Fi e Ez , then S - F , € £3 , and that if F1 , F , € Ez with F , F , = $ , then Fi UF , €

Eg . It follows that Ez is a field . If { Fx } is a sequence of disjoint elements of Ez ...

It is

**clear**that if F , and F are elements of Ez , then F _ F , € £g . It is also**clear**thatif Fi e Ez , then S - F , € £3 , and that if F1 , F , € Ez with F , F , = $ , then Fi UF , €

Eg . It follows that Ez is a field . If { Fx } is a sequence of disjoint elements of Ez ...

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### Contents

Preliminary Concepts | 1 |

B Topological Preliminaries | 10 |

Algebraic Preliminaries | 34 |

Copyright | |

25 other sections not shown

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### 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 obtained 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