## Linear Operators: Spectral Theory : Self Adjoint Operators in Hilbert Space, Volume 2 |

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Page 2017

and the fraction in the

* + 1681 sj } 1 / 2 which is bounded on all of RM . Thus As has a resolution of the

identity . Furthermore , the set S , is the null set { 8181 = 82 = 0 } , so that As is a ...

and the fraction in the

**inequality**( i ) of Theorem 6 is si - 2 + 2 878212 { ( sỉ — sî )* + 1681 sj } 1 / 2 which is bounded on all of RM . Thus As has a resolution of the

identity . Furthermore , the set S , is the null set { 8181 = 82 = 0 } , so that As is a ...

Page 2190

Thus for some constant K we have | S ( f ) | SK \ f | and to prove the final

of the present theorem it will suffice to prove that Ifle SIS ( f ) . Since both terms in

this

Thus for some constant K we have | S ( f ) | SK \ f | and to prove the final

**inequality**of the present theorem it will suffice to prove that Ifle SIS ( f ) . Since both terms in

this

**inequality**are continuous functions of f , it will suffice to prove it for every ...Page 2403

First we shall prove an

is elementary in the sense that it relates only to the norms of the integral kernels

involved . We then use this

First we shall prove an

**inequality**for integral operators ( Lemma 5 below ) whichis elementary in the sense that it relates only to the norms of the integral kernels

involved . We then use this

**inequality**to apply Theorem 1 in an illustrative but ...### What people are saying - Write a review

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adjoint operator analytic applications arbitrary assumed B-space Banach space belongs Boolean algebra Borel sets boundary bounded bounded operator Chapter clear closed commuting compact complex condition consider constant contained continuous converges Corollary corresponding countably additive defined Definition denote dense determined differential operator domain elements equation equivalent established example exists extension fact finite follows formula given gives Hence Hilbert space hypothesis identity inequality integral invariant inverse Lemma limit linear linear operator manifold Math Moreover multiplicity norm normal perturbation plane positive preceding present problem projections PROOF properties proved range regular resolution resolvent respectively restriction Russian satisfies scalar type seen sequence shown shows similar spectral measure spectral operator spectrum subset subspace sufficiently Suppose Theorem theory topology unbounded uniformly unique valued vector weakly zero