Question

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

score 216 · Answer 1 · 2026-04-09 15:20:29

216

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

Published on 09 Apr 2026 - 15:20

score 59 · Answer 2 · 2026-05-13 18:04:24

59

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Compact Operator Improves Regularity in Function Spaces

Published on 13 May 2026 - 18:04

score 139 · Answer 3 · 2026-03-27 03:41:03

139

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

Published on 27 Mar 2026 - 3:41

score 267 · Answer 4 · 2026-03-27 01:43:49

267

Views

Perturbing by a compact operator preserves closed range

Published on 27 Mar 2026 - 1:43

score 69 · Answer 5 · 2026-03-27 01:47:03

69

Views

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

Published on 27 Mar 2026 - 1:47

score 342 · Answer 6 · 2026-03-27 01:43:24

342

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Examples of compact operators, solution verification

Published on 27 Mar 2026 - 1:43

score 123 · Answer 7 · 2026-03-27 01:42:11

123

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

Published on 27 Mar 2026 - 1:42

score 368 · Answer 8 · 2026-03-27 01:42:27

368

Views

Non diagonalizable normal, linear and bounded operator

Published on 27 Mar 2026 - 1:42

score 166 · Answer 9 · 2026-03-27 01:50:45

166

Views

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

Published on 27 Mar 2026 - 1:50

score 75 · Answer 10 · 2026-03-27 01:48:10

75

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 27 Mar 2026 - 1:48

score 239 · Answer 11 · 2026-03-27 03:42:20

239

Views

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

Published on 27 Mar 2026 - 3:42

score 343 · Answer 12 · 2026-03-27 01:44:09

343

Views

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

Published on 27 Mar 2026 - 1:44

score 103 · Answer 13 · 2026-03-27 01:47:41

103

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 27 Mar 2026 - 1:47

score 280 · Answer 14 · 2026-03-27 01:42:34

280

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Hilbert-Schmidt operators

Published on 27 Mar 2026 - 1:42

score 327 · Answer 15 · 2026-03-27 03:41:08

327

Views

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

Published on 27 Mar 2026 - 3:41

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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