Linear Operators: Part III: Spectral Operators [by] Nelson Dunford and Jacob T. Schwartz, with the Assistance of William G. Bade and Robert G. Bartle, Volume 1 |
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Page 2078
25 If X is weakly complete , then A " SA 26 Prove that ASA and that A SA . 27 Prove that ASA + 1 . 28 Show that A SA . 29 Let f ( z ) = + anz " be analytic in a neighborhood of the unit circle .
25 If X is weakly complete , then A " SA 26 Prove that ASA and that A SA . 27 Prove that ASA + 1 . 28 Show that A SA . 29 Let f ( z ) = + anz " be analytic in a neighborhood of the unit circle .
Page 2190
... inequality are continuous functions off , it will suffice to prove it for every function f in a set dense in EB ( 1 , 2 ) . Thus , let f = { 1-1 Q1 Xor , where the sets 01 , ... , On are dis= = Qi joint sets in E whose union is 1.
... inequality are continuous functions off , it will suffice to prove it for every function f in a set dense in EB ( 1 , 2 ) . Thus , let f = { 1-1 Q1 Xor , where the sets 01 , ... , On are dis= = Qi joint sets in E whose union is 1.
Page 2459
If xn € Lac ( H ) and limno Xn = x , then , by what we have already proved , we may write x = y1 + y2 + ys , where yı € Lac ( H ) and Y2 , Yz are ... Using this last fact , it is easy to prove assertion ( c ) of the present lemma .
If xn € Lac ( H ) and limno Xn = x , then , by what we have already proved , we may write x = y1 + y2 + ys , where yı € Lac ( H ) and Y2 , Yz are ... Using this last fact , it is easy to prove assertion ( c ) of the present lemma .
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Contents
SPECTRAL OPERATORS | 1924 |
Introduction | 1927 |
Terminology and Preliminary Notions | 1929 |
Copyright | |
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adjoint operator Amer analytic apply arbitrary assumed B-space Banach space belongs Boolean algebra Borel set boundary conditions bounded bounded operator Chapter clear closed commuting compact complex constant contains continuous converges Corollary corresponding defined Definition denote dense determined differential operator domain elements equation equivalent established exists extension finite follows formal formula given gives Hence Hilbert space hypothesis identity inequality integral invariant inverse Lemma limit linear operator Math Moreover multiplicity norm perturbation plane positive preceding present problem projections PROOF properties prove range resolution resolvent restriction Russian satisfies scalar type seen sequence shown shows similar solution spectral measure spectral operator spectrum subset sufficiently Suppose Theorem theory topology unbounded uniformly unique valued vector zero