## Linear Operators: General theory |

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Page 22

This contradiction proves that A is

that A is

. Let p0 be an arbitrary point of A, and let d0 = sup g(p0, p). The number d0 is

finite ...

This contradiction proves that A is

**sequentially compact**. Conversely, supposethat A is

**sequentially compact**and closed. It will first be shown that A is separable. Let p0 be an arbitrary point of A, and let d0 = sup g(p0, p). The number d0 is

finite ...

Page 314

By part (b), v(U(fa), E) = \V{M\ = |^| = vfa, Q.E.D. 12 Theorem. .4 subset K of ba(S,

Z) is weakly

(S, Z) such that lim X(E) = 0 uniformly for XeK. Proof. Let K Q ba(S, Z) be weakly ...

By part (b), v(U(fa), E) = \V{M\ = |^| = vfa, Q.E.D. 12 Theorem. .4 subset K of ba(S,

Z) is weakly

**sequentially compact**if and only if there exists a non-negative u in ba(S, Z) such that lim X(E) = 0 uniformly for XeK. Proof. Let K Q ba(S, Z) be weakly ...

Page 511

Then if A is compact in the weak operator topology, so is the weak operator

closure of co(A). If A is compact in the strong operator topology, so is the strong

operator closure of co(^). 4 If a set A C B(3t, St) ) is

weak ...

Then if A is compact in the weak operator topology, so is the weak operator

closure of co(A). If A is compact in the strong operator topology, so is the strong

operator closure of co(^). 4 If a set A C B(3t, St) ) is

**sequentially compact**in theweak ...

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### Contents

Preliminary Concepts | 1 |

B Topological Preliminaries | 10 |

Algebraic Preliminaries | 34 |

Copyright | |

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