Question

$K(X,Y)$ a closed subset of $B(X,Y)$ for normed spaces $X,Y$

score 205 · Answer 1 · 2013-01-10 17:54:55

205

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$K(X,Y)$ a closed subset of $B(X,Y)$ for normed spaces $X,Y$

Published on 10 Jan 2013 - 17:54

score 289 · Answer 2 · 2013-01-16 14:34:44

289

Views

A Linear map $‎u : X ‎\longrightarrow ‎Y‎‎$ ‎ ‎is ‎not ‎bounded ‎below ‎‎iff ‎there ‎is ...

Published on 16 Jan 2013 - 14:34

score 1.2k · Answer 3 · 2013-01-17 10:36:06

1.2k

Views

Eigenvalues integral operator - general case

Published on 17 Jan 2013 - 10:36

score 562 · Answer 4 · 2013-01-18 12:24:15

562

Views

What is the spectral radius of the operator $T_k:C[0,1]\to C[0,1]$ defined as $T_k x (t)= \int_0^1 k(t,s) x(s) ds$?

Published on 18 Jan 2013 - 12:24

score 141 · Answer 5 · 2013-01-23 15:34:39

141

Views

If limit of $f(n)$ is zero then the operator is compact

Published on 23 Jan 2013 - 15:34

score 95 · Answer 6 · 2013-01-28 11:49:00

95

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Given $F:X \to X$ on $X$:Hilbert space satisfying some properties, prove that $F$ is surjective.

Published on 28 Jan 2013 - 11:49

score 2.9k · Answer 7 · 2013-01-28 15:22:52

2.9k

Views

There are compact operators that are not norm-limits of finite-rank operators

Published on 28 Jan 2013 - 15:22

score 208 · Answer 8 · 2013-02-02 14:10:58

208

Views

Show compactness of an operator and calculate its SVD

Published on 02 Feb 2013 - 14:10

score 1.3k · Answer 9 · 2013-02-02 19:35:02

1.3k

Views

Fredholm and Compact Operators

Published on 02 Feb 2013 - 19:35

score 761 · Answer 10 · 2013-02-03 20:05:20

761

Views

Compactness of multiplication operator on $C[0,1]$

Published on 03 Feb 2013 - 20:05

score 357 · Answer 11 · 2013-02-04 03:30:46

357

Views

Proof of compactness of bounded linear operator

Published on 04 Feb 2013 - 3:30

score 91 · Answer 12 · 2013-02-05 18:35:38

‎‎$‎T:\ell^{‎2‎} \rightarrow ‎\ell^{‎2‎}‎$ ‎‎defined ‎by ‎‎$‎T(\{x_{n}\})=\{2^{-n}x_{n}\}$ ‎is ‎compact

score 153 · Answer 13 · 2013-02-06 16:58:10

153

Views

Continuous, selfadjoint and compact?

Published on 06 Feb 2013 - 16:58

score 133 · Answer 14 · 2013-02-16 10:07:42

133

Views

1.4.5 ‎Theorem ‎of‎ ‎Murphy's ‎book

Published on 16 Feb 2013 - 10:07

score 2k · Answer 15 · 2013-02-22 16:41:55

2k

Views

compact and self adjoint square root of an operator

Published on 22 Feb 2013 - 16:41

List Question

$K(X,Y)$ a closed subset of $B(X,Y)$ for normed spaces $X,Y$

A Linear map $‎u : X ‎\longrightarrow ‎Y‎‎$ ‎ ‎is ‎not ‎bounded ‎below ‎‎iff ‎there ‎is ...

Eigenvalues integral operator - general case

What is the spectral radius of the operator $T_k:C[0,1]\to C[0,1]$ defined as $T_k x (t)= \int_0^1 k(t,s) x(s) ds$?

If limit of $f(n)$ is zero then the operator is compact

Given $F:X \to X$ on $X$:Hilbert space satisfying some properties, prove that $F$ is surjective.

There are compact operators that are not norm-limits of finite-rank operators

Show compactness of an operator and calculate its SVD

Fredholm and Compact Operators

Compactness of multiplication operator on $C[0,1]$

Proof of compactness of bounded linear operator

‎‎$‎T:\ell^{‎2‎} \rightarrow ‎\ell^{‎2‎}‎$ ‎‎defined ‎by ‎‎$‎T(\{x_{n}\})=\{2^{-n}x_{n}\}$ ‎is ‎compact

Continuous, selfadjoint and compact?

1.4.5 ‎Theorem ‎of‎ ‎Murphy's ‎book

compact and self adjoint square root of an operator

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