Question

Functional calculus for compact operators using singular value decomposition and regularisation filters

score 215 · Answer 1 · 2026-02-23 08:30:32

215

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Functional calculus for compact operators using singular value decomposition and regularisation filters

Published on 23 Feb 2026 - 8:30

score 57 · Answer 2 · 2026-03-25 09:51:18

57

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Compact Operator Improves Regularity in Function Spaces

Published on 25 Mar 2026 - 9:51

score 138 · Answer 3 · 2020-03-17 01:29:06

138

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Multiplier algebra of the group $C^*$-algebra for a discrete group

Published on 17 Mar 2020 - 1:29

score 266 · Answer 4 · 2020-03-22 23:56:47

266

Views

Perturbing by a compact operator preserves closed range

Published on 22 Mar 2020 - 23:56

score 68 · Answer 5 · 2020-03-24 20:15:28

68

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if $K$ is a compact operator then there is $K_\epsilon$ with finite dimensional range within the Ball radius $\epsilon$.

Published on 24 Mar 2020 - 20:15

score 341 · Answer 6 · 2020-03-25 22:19:22

341

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Examples of compact operators, solution verification

Published on 25 Mar 2020 - 22:19

score 122 · Answer 7 · 2020-03-26 04:58:36

122

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if $\Lambda = \Lambda ^2$ be a compact operator on a banach space , then $Range(\Lambda)$ is finite dimensional

Published on 26 Mar 2020 - 4:58

score 367 · Answer 8 · 2020-03-26 17:31:43

367

Views

Non diagonalizable normal, linear and bounded operator

Published on 26 Mar 2020 - 17:31

score 165 · Answer 9 · 2020-03-29 13:51:09

165

Views

Integral kernel of Hilbert-Schmidt operator on $L^2(\mathbb{R})$

Published on 29 Mar 2020 - 13:51

score 74 · Answer 10 · 2020-04-03 16:18:16

74

Views

Proving $R_xy=(x_1y_1,x_2y_2,\dots)$ with $R_x$ over $\ell^{2}$, is compact if, and only if $\lim_{x \rightarrow \infty} x_n = 0.$

Published on 03 Apr 2020 - 16:18

score 238 · Answer 11 · 2020-04-03 16:41:16

238

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A finite-dimensional invariant subspace for the left regular representation implies that the group is finite

Published on 03 Apr 2020 - 16:41

score 342 · Answer 12 · 2020-04-06 18:25:33

342

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Fourier-like operator on $L^2$ is compact

Published on 06 Apr 2020 - 18:25

score 102 · Answer 13 · 2020-04-07 03:23:25

102

Views

examples of a compact operator $K: l^2 \to l^2$ with point spectrum $\sigma_p(K)=\{0,1,\frac{1}{2},\frac{1}{3},...\}$, with finite and infinite kernel

Published on 07 Apr 2020 - 3:23

score 279 · Answer 14 · 2020-04-07 08:38:38

279

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Hilbert-Schmidt operators

Published on 07 Apr 2020 - 8:38

score 326 · Answer 15 · 2020-04-07 22:37:18

326

Views

isolated point of spectrum of compact self-adjoint linear operator on infinite-dimensional separable Hilbert space.

Published on 07 Apr 2020 - 22:37

List Question

Functional calculus for compact operators using singular value decomposition and regularisation filters

Compact Operator Improves Regularity in Function Spaces

Multiplier algebra of the group $C^*$-algebra for a discrete group

Perturbing by a compact operator preserves closed range

if $K$ is a compact operator then there is $K_\epsilon$ with finite dimensional range within the Ball radius $\epsilon$.

Examples of compact operators, solution verification

if $\Lambda = \Lambda ^2$ be a compact operator on a banach space , then $Range(\Lambda)$ is finite dimensional

Non diagonalizable normal, linear and bounded operator

Integral kernel of Hilbert-Schmidt operator on $L^2(\mathbb{R})$

Proving $R_xy=(x_1y_1,x_2y_2,\dots)$ with $R_x$ over $\ell^{2}$, is compact if, and only if $\lim_{x \rightarrow \infty} x_n = 0.$

A finite-dimensional invariant subspace for the left regular representation implies that the group is finite

Fourier-like operator on $L^2$ is compact

examples of a compact operator $K: l^2 \to l^2$ with point spectrum $\sigma_p(K)=\{0,1,\frac{1}{2},\frac{1}{3},...\}$, with finite and infinite kernel

Hilbert-Schmidt operators

isolated point of spectrum of compact self-adjoint linear operator on infinite-dimensional separable Hilbert space.

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