## Linear Operators: Spectral operators |

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Page 2017

and the fraction in the

- $ ) * + 16s1 s } 1 / 2 which is bounded on all of Rn . Thus As has a resolution of

the identity . Furthermore , the set Sy is the null set { s | 81 = 82 = 0 } , so that Ag ...

and the fraction in the

**inequality**( i ) of Theorem 6 is 1si - s ? l2 + 2 | 818212 { ( si- $ ) * + 16s1 s } 1 / 2 which is bounded on all of Rn . Thus As has a resolution of

the identity . Furthermore , the set Sy is the null set { s | 81 = 82 = 0 } , so that Ag ...

Page 2190

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

the present theorem it will suffice to prove that fe = S ( F ) . Since both terms in this

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

**inequality**ofthe present theorem it will suffice to prove that fe = S ( F ) . Since both terms in this

**inequality**are continuous functions of f , it will suffice to prove it for every ...Page 2221

Since \ f ( TQ ) | S 4M sup 15 ( 1 ) | SL , LEV the

Ta ) gn ( Twx | SL \ gn ( T ) x — 9n ( Ta ) x | shows that lim f ( T ) gn ( Ta XVII . 4 . 3

2221 STRONG LIMITS OF SPECTRAL OPERATORS Strong Limits of Spectral ...

Since \ f ( TQ ) | S 4M sup 15 ( 1 ) | SL , LEV the

**inequality**\ f ( Ta ) gn ( T ) x – f (Ta ) gn ( Twx | SL \ gn ( T ) x — 9n ( Ta ) x | shows that lim f ( T ) gn ( Ta XVII . 4 . 3

2221 STRONG LIMITS OF SPECTRAL OPERATORS Strong Limits of Spectral ...

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