## Linear Operators: Spectral theory |

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Page 1190

Of course if T is a bounded

DT and T* = T are equivalent and thus a bounded operator is symmetric if and

only if it is self adjoint. If T is an

DT ...

Of course if T is a bounded

**everywhere**defined operator then the statements ToDT and T* = T are equivalent and thus a bounded operator is symmetric if and

only if it is self adjoint. If T is an

**everywhere**defined symmetric operator then ToDT ...

Page 1212

F)() (d.) = s.s.)(F(T))(), (d) = s.so)(A.P)() (d.) =s. (B.f)(A)P3)4(a) Thus, (Barif)(A) = (B

,f)(A) u-almost

, ), Aer. This fact, together with the uniqueness of the kernel W., shows that W,(s, ...

F)() (d.) = s.s.)(F(T))(), (d) = s.so)(A.P)() (d.) =s. (B.f)(A)P3)4(a) Thus, (Barif)(A) = (B

,f)(A) u-almost

**everywhere**on ea. Consequently (B.f)(a) = s.s.)"...(s.3) (ds), feliss.,, ), Aer. This fact, together with the uniqueness of the kernel W., shows that W,(s, ...

Page 1402

(t) is void, and it follows from Theorem 6.13 that W,(', 'A)e L2(a,b) u,-almost

modifications to show that if B(f) = 0 is a boundary condition satisfied by all fe? (T)

, we have ...

(t) is void, and it follows from Theorem 6.13 that W,(', 'A)e L2(a,b) u,-almost

**everywhere**in A. The proof of Theorem 5.4 will then apply with evident slightmodifications to show that if B(f) = 0 is a boundary condition satisfied by all fe? (T)

, we have ...

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### Contents

SPECTRAL THEORY | 858 |

Bounded Normal Operators in Hilbert Space | 887 |

Miscellaneous Applications | 937 |

Copyright | |

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