## Linear Operators: General theory |

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Page 61

From that theorem it is

From that theorem it is

**seen**that Tx exists for each x , and that T is continuous . Thus Lemma 4 shows that T is bounded . Now , when Tx exists , Tx = lim ...Page 254

Thus by forming the chain UB , UG , ... , Uz ' , Up . it is

Thus by forming the chain UB , UG , ... , Uz ' , Up . it is

**seen**that vg is equivalent to vg , and thus that up is in V. Since { vp } is a basis ...Page 449

Let B , be the dense subset of the boundary B of A at which there are non - zero functionals tangent to A. We have

Let B , be the dense subset of the boundary B of A at which there are non - zero functionals tangent to A. We have

**seen**that ( aA , + ( 1 - a ) An B = $ for ...### What people are saying - Write a review

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### Contents

Preliminary Concepts | 1 |

B Topological Preliminaries | 10 |

quences | 26 |

Copyright | |

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Akad algebra Amer analytic applied arbitrary assume B-space Banach Banach spaces bounded called clear closed compact complex Consequently contains converges convex Corollary countably additive defined DEFINITION denote dense determined differential disjoint domain 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 mean measure space metric space neighborhood norm o-field open set operator positive problem Proc PROOF properties proved range regular respect Russian satisfies scalar seen separable sequence set function Show shown sphere statement subset Suppose Theorem theory topological space topology transformations u-integrable u-measurable uniformly union unique unit valued vector weak zero