Linear Operators: General theory |
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Page 244
... Hence the sequence { y } converges to the vector [ ~ 1 , . in En . = xn ] 1 LEMMA . A finite dimensional normed linear space is complete and hence it is a B - space . PROOF . Let { b1 , · .... bn } be a Hamel basis for the finite dimen ...
... Hence the sequence { y } converges to the vector [ ~ 1 , . in En . = xn ] 1 LEMMA . A finite dimensional normed linear space is complete and hence it is a B - space . PROOF . Let { b1 , · .... bn } be a Hamel basis for the finite dimen ...
Page 423
... Hence y ( Tx ) < ε , so that Ta e N ( 0 ; y * , . . . , y * , ε ) . Therefore , Tis weakly continuous at the origin , and hence at every point . Conversely , suppose that T is weakly continuous , and y * € Y * . Then y * T is a linear ...
... Hence y ( Tx ) < ε , so that Ta e N ( 0 ; y * , . . . , y * , ε ) . Therefore , Tis weakly continuous at the origin , and hence at every point . Conversely , suppose that T is weakly continuous , and y * € Y * . Then y * T is a linear ...
Page 441
... hence a compact , subset of co ( Q ) . Hence i co ( Q ) = co ( K1 U. UK „ ) = co ( K1U ...... .UKn ) , by an easy induction on Lemma 2.5 . It follows readily that p has the form p Σak , a ≥0 , Σa , = 1 , k , e K ; and , since p is ex ...
... hence a compact , subset of co ( Q ) . Hence i co ( Q ) = co ( K1 U. UK „ ) = co ( K1U ...... .UKn ) , by an easy induction on Lemma 2.5 . It follows readily that p has the form p Σak , a ≥0 , Σa , = 1 , k , e K ; and , since p is ex ...
Contents
A Settheoretic Preliminaries | 1 |
B Topological Preliminaries | 10 |
Algebraic Preliminaries | 34 |
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A₁ Acad additive set function algebra Amer analytic arbitrary B-space B₁ ba(S Banach spaces Borel sets Cauchy sequence compact operator complex numbers contains continuous functions continuous linear converges convex set Corollary countably additive DEFINITION denote dense differential equations disjoint Doklady Akad domain E₁ element exists f₁ finite dimensional finite number function defined function f Hausdorff space Hence Hilbert space homeomorphism inequality integral L₁ L₁(S Lebesgue Lemma Let f linear functional linear map linear operator linear topological space measurable function measure space metric space Nauk SSSR N. S. neighborhood non-negative o-field open set operator topology positive measure space Proc PROOF proved real numbers Riesz Russian S₁ scalar semi-group sequentially compact Show spectral strong operator topology subset subspace Suppose T₁ theory topological space u-integrable u-measurable uniformly unit sphere valued function weakly compact zero ΕΕΣ