## Linear Operators, Part 2 |

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Page 861

Clearly if x - 1

Clearly if x - 1

**exists**then T : -17 . = T , T : -1 = 1 . If T ; '**exists**in B ( x ) , then Tc T_TX T { [ ( T = y ) z ] yz , ( T7'y ) z = T ?? ( yz ) , and if a Tile , then az = T : lz for every ze X. Also T : ' ( Tze ) = T ...Page 1057

By Lemma 2 , the integral ( tu )

By Lemma 2 , the integral ( tu )

**exists**if ( u )**exists**and t > 0 ; and the integral 0 ( Vu )**exists**and equals PS 2 ( x ) ei ( x , Vu ) dx S. 2 ( Vy ) \ y \ " e ( 9,4 ) du En lar " JE " if PSen 2 ( Vy ) \ y | - " pilv , u ) dy**exists**...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 T CT ,. I is called the closure of T. ( a ) There**exists**a densely defined ...### What people are saying - Write a review

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### Contents

BAlgebras | 859 |

Bounded Normal Operators in Hilbert Space | 887 |

Miscellaneous Applications | 937 |

Copyright | |

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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 complex Consequently constant contains continuous converges Corollary corresponding defined Definition denote dense 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 multiplicity Nauk neighborhood norm obtained partial positive preceding present problem projection proof properties prove range regular remark representation respectively restriction result Russian satisfies seen sequence singular solution spectral square-integrable statement subset subspace sufficiently Suppose symmetric Theorem theory topology transform unique vanishes vector zero