## Linear Operators, Part 1 |

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Results 1-3 of 82

Page 61

Nelson Dunford, Jacob T. Schwartz. ( i ) and ( ii ) are satisfied , Lemma 3 shows

that the hypotheses of Theorem 1 . 18 are satisfied . From that theorem it is

that Tx exists for each X , and that T is continuous . Thus Lemma 4 shows that T is

...

Nelson Dunford, Jacob T. Schwartz. ( i ) and ( ii ) are satisfied , Lemma 3 shows

that the hypotheses of Theorem 1 . 18 are satisfied . From that theorem it is

**seen**that Tx exists for each X , and that T is continuous . Thus Lemma 4 shows that T is

...

Page 254

Now since Uge U there is a finite chain of the form given in [ * ] above in which

successive vectors have non - zero scalar products . Thus by forming the chain V

2 , Uq , . . . , Uq ' , Vg , it is

V ...

Now since Uge U there is a finite chain of the form given in [ * ] above in which

successive vectors have non - zero scalar products . Thus by forming the chain V

2 , Uq , . . . , Uq ' , Vg , it is

**seen**that ug is equivalent to v g , and thus that vp is inV ...

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

. Since aA , + ( 1 - a ) A is open , ( aA , + ( 1 - a ) A ) n B = $ for 0 < a < 1 . Let pe A

...

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 ) A ) B = $ for 0 < a < 1. Since aA , + ( 1 - a ) A is open , ( aA , + ( 1 - a ) A ) n B = $ for 0 < a < 1 . Let pe A

...

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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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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