## Linear operators: Spectral operators |

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Results 1-3 of 83

Page 2159

It

subset of p(T) is in £f[T) and hence in a set measurable T. Thus, since a(T) £ ro,

every Borel set in the plane is measurable T. Q.E.D. In view of

behooves ...

It

**follows**that every Borel subset of ro is measurable T. By**Lemma**3.9 everysubset of p(T) is in £f[T) and hence in a set measurable T. Thus, since a(T) £ ro,

every Borel set in the plane is measurable T. Q.E.D. In view of

**Lemma**10, itbehooves ...

Page 2308

Since T^t) is closed by Theorem XIII.2.10 and by

Definition XIII. 2. 17 and the remark preceding Definition XIII. 2. 29, that S is

closed. From this, it

Since T^t) is closed by Theorem XIII.2.10 and by

**Lemma**XII. 1.6(a), it**follows**fromDefinition XIII. 2. 17 and the remark preceding Definition XIII. 2. 29, that S is

closed. From this, it

**follows**immediately that S — XI is closed, and hence from ...Page 2364

exists and is given by the formula K(r]) = R(X0;T)(I—nPR(X0;T))-K Since R(X0 ; T)

is compact by Lemma 2.2, it

that K(rj) is a compact operator which depends analytically on -q for \rj\ < 1 + Sj. It

...

exists and is given by the formula K(r]) = R(X0;T)(I—nPR(X0;T))-K Since R(X0 ; T)

is compact by Lemma 2.2, it

**follows from Lemma**VII. 6. 6 and Theorem VI. 5.4that K(rj) is a compact operator which depends analytically on -q for \rj\ < 1 + Sj. It

...

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### Contents

SPECTRAL OPERATORS | 1924 |

Spectral Operators | 1925 |

Terminology and Preliminary Notions | 1928 |

Copyright | |

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