List questions from normed-spaces

Published on 09 Apr 2026 - 20:50

82

Views

Operator norm inequality $\|XY\|\geq\frac{\|X\|}{\|Y^{-1}\|}$

Published on 25 May 2020 - 20:12

58

Views

If closure of $C$ is a subvectorspace then so is $C$

Published on 26 May 2020 - 0:29

37

Views

Justification for squared L2 data and smoothness term as an error bound

Published on 09 Apr 2026 - 7:29

370

Views

Strange SVD Bound with Frobenius Norm

Published on 27 Mar 2026 - 1:44

47

Views

Prove an equivalent norm

Published on 31 Mar 2026 - 23:47

73

Views

Prove a bounded operator and compute norm

Published on 11 Apr 2026 - 17:36

120

Views

Show an operator is bounded and calculate the norm

Published on 11 Apr 2026 - 15:35

261

Views

Prove an inner product space

Published on 07 Apr 2026 - 8:43

18

Views

vector norm based on a semi-definite matrix

Published on 27 May 2020 - 9:20

155

Views

Tringular inequality proof for norm

Published on 10 Apr 2026 - 2:17

132

Views

The norm of a multilinear transformation between finite-dimensional vector spaces is always finite

Published on 28 May 2020 - 21:07

42

Views

Arbitrarily close approximation for complex square matrix

Published on 12 Apr 2026 - 6:54

208

Views

Why isn't it obvious, from the definition of the essential supremum, that $\lvert f \rvert \leq \|f\|_{\infty}, \ \mu-$a.e?

Published on 29 May 2020 - 10:11

96

Views

$f$ admits a continuous extension to $X$?

Published on 03 Apr 2026 - 16:54

List Question

Frobenius norm and tensor product

Operator norm inequality $\|XY\|\geq\frac{\|X\|}{\|Y^{-1}\|}$

If closure of $C$ is a subvectorspace then so is $C$

Justification for squared L2 data and smoothness term as an error bound

Strange SVD Bound with Frobenius Norm

Prove an equivalent norm

Prove a bounded operator and compute norm

Show an operator is bounded and calculate the norm

Prove an inner product space

vector norm based on a semi-definite matrix

Tringular inequality proof for norm

The norm of a multilinear transformation between finite-dimensional vector spaces is always finite

Arbitrarily close approximation for complex square matrix

Why isn't it obvious, from the definition of the essential supremum, that $\lvert f \rvert \leq \|f\|_{\infty}, \ \mu-$a.e?

$f$ admits a continuous extension to $X$?

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