List questions from compact-operators

Published on 26 Mar 2026 - 1:02

65

Views

Given $T \in B(X)$, find a compact operator $K$ with rank $1$ such that $\operatorname{rank}(TK - KT) \leq 2$

Published on 05 Jun 2020 - 7:38

104

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Can any Hilbert space be expressed as countable union of unit balls?

Published on 10 Jun 2020 - 4:34

78

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Continuation of compact operator to completion of the space

Published on 13 Jun 2020 - 16:58

142

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Condition for an operator to be compact

Published on 14 Jun 2020 - 10:43

557

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Compact integral operator on $L^\infty(\mathbb{R})$

Published on 18 Jun 2020 - 14:15

231

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Schauder Basis and Approximation property

Published on 19 Jun 2020 - 7:25

72

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On ranges of nuclear operators

Published on 22 Jun 2020 - 14:22

295

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How to find a finite-dimensional invariant subspace for a self-adjoint operator?

Published on 23 Jun 2020 - 14:33

59

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about properties of the operator $T_f(g) := f\cdot g$.

Published on 27 Jun 2020 - 4:06

221

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Inclusion map $i : C^{0,\beta}[0, 1] \rightarrow C^{0,\alpha}[0, 1] $ is linear, and therefore compact

Published on 25 Mar 2026 - 12:26

229

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Is a function of a compact operator a compact operator?

Published on 23 Feb 2026 - 12:09

70

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Define the map $T(f)=a(x)f(x)$ where $a(x) \in C[-\pi,\pi]$ and $f \in L^2$, show that $T$ is not compact unless $a=0$

Published on 26 Mar 2026 - 1:12

150

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Compactness of the trace operator in dimension 1

Published on 25 Mar 2026 - 6:19

50

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Show that $X= \ker((u-\lambda)^p) \oplus (u-\lambda)^p(X)$ if $u$ is compact.

Published on 11 Jul 2020 - 20:16

List Question

Is a nonlinear compact operator still compact if multiplied by a scalar function

Given $T \in B(X)$, find a compact operator $K$ with rank $1$ such that $\operatorname{rank}(TK - KT) \leq 2$

Can any Hilbert space be expressed as countable union of unit balls?

Continuation of compact operator to completion of the space

Condition for an operator to be compact

Compact integral operator on $L^\infty(\mathbb{R})$

Schauder Basis and Approximation property

On ranges of nuclear operators

How to find a finite-dimensional invariant subspace for a self-adjoint operator?

about properties of the operator $T_f(g) := f\cdot g$.

Inclusion map $i : C^{0,\beta}[0, 1] \rightarrow C^{0,\alpha}[0, 1] $ is linear, and therefore compact

Is a function of a compact operator a compact operator?

Define the map $T(f)=a(x)f(x)$ where $a(x) \in C[-\pi,\pi]$ and $f \in L^2$, show that $T$ is not compact unless $a=0$

Compactness of the trace operator in dimension 1

Show that $X= \ker((u-\lambda)^p) \oplus (u-\lambda)^p(X)$ if $u$ is compact.

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