Linear Operators: Spectral operators |
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Page 2300
... finite dimensional range for all p . Thus , by Lemma VII.6.7 , I — E , has finite dimensional range for all sufficiently large p . Since E is a count- ably additive spectral resolution , we have E ( μ ; T + P ) ( I − E „ ) = 0 if μ is ...
... finite dimensional range for all p . Thus , by Lemma VII.6.7 , I — E , has finite dimensional range for all sufficiently large p . Since E is a count- ably additive spectral resolution , we have E ( μ ; T + P ) ( I − E „ ) = 0 if μ is ...
Page 2333
... finite . that Since f is in L2 ( 0 , ∞ ) , it is sufficient by Schwarz's inequality to show 00 f ( s ) s + t ds | g ( t ) = √ ; = f ( ut ) du 0 1 + u belongs to L2 ( 0 , ∞ ) . Putting f ( x ) = f ( tx ) , we may use Theorem III.11.17 ...
... finite . that Since f is in L2 ( 0 , ∞ ) , it is sufficient by Schwarz's inequality to show 00 f ( s ) s + t ds | g ( t ) = √ ; = f ( ut ) du 0 1 + u belongs to L2 ( 0 , ∞ ) . Putting f ( x ) = f ( tx ) , we may use Theorem III.11.17 ...
Page 2441
... finite absolute constant c ' such that I ( α ) ≤ c ' ( 1+ | x | ) -n + 1 . Using this inequality and using ( 37 ) , we see that there exists a finite constant M " independent of ε such that av 4 ( 43 ) | V1 ( r , r ' ) | + ( r , r ...
... finite absolute constant c ' such that I ( α ) ≤ c ' ( 1+ | x | ) -n + 1 . Using this inequality and using ( 37 ) , we see that there exists a finite constant M " independent of ε such that av 4 ( 43 ) | V1 ( r , r ' ) | + ( r , r ...
Contents
SPECTRAL OPERATORS | 1924 |
Spectral Operators | 1925 |
Terminology and Preliminary Notions | 1928 |
Copyright | |
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A₁ algebra Amer analytic applications arbitrary B-space Banach Banach space Boolean algebra Borel sets boundary bounded Chapter clear closed commuting compact complex constant contains continuous converges Corollary corresponding defined Definition denote dense determined differential operator Doklady Akad elements equation equivalent established example exists extension finite follows formula function given gives H₁ Hence Hilbert space hypothesis identity integral invariant inverse Lemma limit linear operators Math multiplicity Nauk SSSR norm normal perturbation plane positive preceding present problem Proc projections PROOF properties prove range resolution resolvent restriction Russian satisfies scalar type sequence shown shows similar spectral measure spectral operator spectrum subset subspace sufficiently Suppose Theorem theory topology unbounded uniformly unique valued vector zero