## Linear Operators: Spectral theory |

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Page 1057

Thus , to show that the proper value integral ( 3 )

Thus , to show that the proper value integral ( 3 )

**exists**generally , it is sufficient for us to consider the cases u 0 and u ( 1 , 0 , ... , 0 ) .Page 1261

23 If an operator T has a closed linear extension there

23 If an operator T has a closed linear extension there

**exists**a unique closed linear extension T such that if T , is any closed linear extension of T then ...Page 1262

Then there

Then there

**exists**a Hilbert space H , 2 H , and an orthogonal projection Q in H , such that Ax PQx , X EH , P denoting the orthogonal projection of Hi on H.### What people are saying - Write a review

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### Contents

BAlgebras | 861 |

Commutative BAlgebras | 868 |

Commutative BAlgebras | 874 |

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additive Akad algebra Amer analytic assume Banach spaces basis belongs Borel boundary conditions boundary values bounded called clear closed closure coefficients compact complete Consequently constant contains continuous converges Corollary corresponding defined Definition denote dense derivatives determined domain eigenvalues element equal equation essential spectrum evident Exercise exists extension finite follows formal differential operator formula function given Hence Hilbert space identity independent indices inequality integral interval Lemma limit linear mapping Math matrix measure Nauk neighborhood norm obtained partial positive preceding present problem projection proof properties prove range regular remark representation respectively restriction result satisfies seen sequence singular solution spectral square-integrable statement subset subspace sufficiently Suppose symmetric Theorem theory topology transform unique vanishes vector zero