Question

Proving an operator operator is compact on $L^p$, using Kolmogorov's theorem

score 131 · Answer 1 · 2019-12-28 11:22:52

131

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Proving an operator operator is compact on $L^p$, using Kolmogorov's theorem

Published on 28 Dec 2019 - 11:22

score 136 · Answer 2 · 2019-12-29 16:24:12

136

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How to show That T is not compact

Published on 29 Dec 2019 - 16:24

score 1.2k · Answer 3 · 2019-12-30 03:59:38

1.2k

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A compact operator that is the limit of finite rank operators

Published on 30 Dec 2019 - 3:59

score 59 · Answer 4 · 2020-01-02 02:04:10

59

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Compact operators on $L^2([a,b])$: Are the dimensions of eigenspaces the same on dense subspaces?

Published on 02 Jan 2020 - 2:04

score 601 · Answer 5 · 2020-01-12 17:39:57

601

Views

How to show that finite rank operator on Banach space is compact

Published on 12 Jan 2020 - 17:39

score 448 · Answer 6 · 2020-01-15 04:45:07

448

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Finite rank operator are dense in trace class with trace class norm .

Published on 15 Jan 2020 - 4:45

score 891 · Answer 7 · 2020-01-20 08:02:53

891

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spectrum of an operator and its point and continuous spectrum

Published on 20 Jan 2020 - 8:02

score 101 · Answer 8 · 2020-01-27 10:40:02

101

Views

If $T$ is compact there is some $n$ such that $\operatorname{range}(T-\alpha I)^n=\operatorname{range}(T-\alpha I)^{n+1}$

Published on 27 Jan 2020 - 10:40

score 717 · Answer 9 · 2020-02-03 17:43:58

717

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Operators that improve strong convergence to norm convergence?

Published on 03 Feb 2020 - 17:43

score 69 · Answer 10 · 2026-03-25 22:26:39

69

Views

Continuous group of automorphisms of $K(H)$

Published on 25 Mar 2026 - 22:26

score 83 · Answer 11 · 2020-02-08 17:53:49

83

Views

Show that $T$ is compact

Published on 08 Feb 2020 - 17:53

score 564 · Answer 12 · 2026-03-25 19:01:38

564

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Equivalence between closed range + finite dimensional kernel, and a statement about sequences.

Published on 25 Mar 2026 - 19:01

score 48 · Answer 13 · 2020-02-12 21:56:38

48

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Prove the compactness of an operator in a Hilbert space

Published on 12 Feb 2020 - 21:56

score 427 · Answer 14 · 2020-02-13 20:14:44

427

Views

Is the operator $T:C([0,1])\to C([0,1])$ defined by $(Tf)(x)=\int_0^x h(y)f(y) dy$ where $h\in L^2([0,1])$ compact?

Published on 13 Feb 2020 - 20:14

score 162 · Answer 15 · 2020-02-13 21:14:59

162

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$Tx=\sum_{k=0}^{\infty}\lambda_k(x,e_k)e_k$ bounded and compact iff $\lambda_k\to 0$

Published on 13 Feb 2020 - 21:14

List Question

Proving an operator operator is compact on $L^p$, using Kolmogorov's theorem

How to show That T is not compact

A compact operator that is the limit of finite rank operators

Compact operators on $L^2([a,b])$: Are the dimensions of eigenspaces the same on dense subspaces?

How to show that finite rank operator on Banach space is compact

Finite rank operator are dense in trace class with trace class norm .

spectrum of an operator and its point and continuous spectrum

If $T$ is compact there is some $n$ such that $\operatorname{range}(T-\alpha I)^n=\operatorname{range}(T-\alpha I)^{n+1}$

Operators that improve strong convergence to norm convergence?

Continuous group of automorphisms of $K(H)$

Show that $T$ is compact

Equivalence between closed range + finite dimensional kernel, and a statement about sequences.

Prove the compactness of an operator in a Hilbert space

Is the operator $T:C([0,1])\to C([0,1])$ defined by $(Tf)(x)=\int_0^x h(y)f(y) dy$ where $h\in L^2([0,1])$ compact?

$Tx=\sum_{k=0}^{\infty}\lambda_k(x,e_k)e_k$ bounded and compact iff $\lambda_k\to 0$

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