Question

Weak $L^{p}$ spaces are quasi-normed?

score 1.9k · Answer 1 · 2013-05-21 23:06:54

1.9k

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Weak $L^{p}$ spaces are quasi-normed?

Published on 21 May 2013 - 23:06

score 246 · Answer 2 · 2013-10-08 10:21:03

246

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Upper Bound for Operator Norm in Marcinkiewicz Interpolation Theorem

Published on 08 Oct 2013 - 10:21

score 86 · Answer 3 · 2017-05-02 15:14:30

86

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Do we have $\lVert f\rVert_{L^p(X)}=\int_0^\infty p\alpha^{p-1}d_f(\alpha)\,\mathrm{d}\alpha$ for non $\sigma$-finite $X$?

Published on 02 May 2017 - 15:14

score 528 · Answer 4 · 2026-03-22 20:17:19

528

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Intuition regrading the weak $L^p$ functions

Published on 22 Mar 2026 - 20:17

score 523 · Answer 5 · 2017-11-24 15:30:06

523

Views

Intuition of the Lorentz norm in an Wikipedia article

Published on 24 Nov 2017 - 15:30

score 1.5k · Answer 6 · 2017-11-27 19:46:55

1.5k

Views

Normability of weak $L^p$-spaces

Published on 27 Nov 2017 - 19:46

score 111 · Answer 7 · 2017-12-02 06:27:13

111

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the additivity for space of weak $L^p$

Published on 02 Dec 2017 - 6:27

score 111 · Answer 8 · 2017-12-03 10:38:23

111

Views

some inequality for weal-$L^p$

Published on 03 Dec 2017 - 10:38

score 118 · Answer 9 · 2017-12-04 12:42:30

118

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Show that$\,\,\int_{|f|\leq s}|f\,|^{q}\,d\mu\leq\frac{q}{q-p}s^{q-p}\lVert f\,\rVert_{L^{p,\infty}}^p\,\,,$ where $0<p<q<\infty$

Published on 04 Dec 2017 - 12:42

score 225 · Answer 10 · 2017-12-05 06:36:59

225

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Show that $f$ is belongs to weak $L^{1}(E)$ and $f$ is bounded on $E$. Then $f$ is $L^{p}(E)$ for each $1<p<\infty$

Published on 05 Dec 2017 - 6:36

score 72 · Answer 11 · 2017-12-06 14:54:30

For $0<p<\infty.$Show that $\left\lVert\max_{1\leq j\leq m}|f_j|\right\rVert_{L^{p,\infty}}^{p}\leq\sum_{j=1}^{m}\lVert f_j\rVert_{L^{p,\infty}}^p$

score 133 · Answer 12 · 2017-12-08 12:22:09

133

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$\lVert f_{1}+\cdots+f_{m}\rVert_{L^{p,\infty}}^{p}\le \frac{2-p}{1-p}\sum_{j=1}^{m}\lVert f_{j}\rVert_{L^{p,\infty}}^{p}$

Published on 08 Dec 2017 - 12:22

score 163 · Answer 13 · 2020-09-23 21:19:25

163

Views

Convolution and weak $L^p$ spaces

Published on 23 Sep 2020 - 21:19

score 108 · Answer 14 · 2020-11-20 22:02:09

108

Views

Bounded gradient in weak $L^1$ implies compactness in $L^1$ (?) Relation to fractional Sobolev compactness.

Published on 20 Nov 2020 - 22:02

score 66 · Answer 15 · 2020-12-11 19:04:25

66

Views

Weak convergence in $L^{p,\infty}$

Published on 11 Dec 2020 - 19:04

List Question

Weak $L^{p}$ spaces are quasi-normed?

Upper Bound for Operator Norm in Marcinkiewicz Interpolation Theorem

Do we have $\lVert f\rVert_{L^p(X)}=\int_0^\infty p\alpha^{p-1}d_f(\alpha)\,\mathrm{d}\alpha$ for non $\sigma$-finite $X$?

Intuition regrading the weak $L^p$ functions

Intuition of the Lorentz norm in an Wikipedia article

Normability of weak $L^p$-spaces

the additivity for space of weak $L^p$

some inequality for weal-$L^p$

Show that$\,\,\int_{|f|\leq s}|f\,|^{q}\,d\mu\leq\frac{q}{q-p}s^{q-p}\lVert f\,\rVert_{L^{p,\infty}}^p\,\,,$ where $0<p<q<\infty$

Show that $f$ is belongs to weak $L^{1}(E)$ and $f$ is bounded on $E$. Then $f$ is $L^{p}(E)$ for each $1<p<\infty$

For $0<p<\infty.$Show that $\left\lVert\max_{1\leq j\leq m}|f_j|\right\rVert_{L^{p,\infty}}^{p}\leq\sum_{j=1}^{m}\lVert f_j\rVert_{L^{p,\infty}}^p$

$\lVert f_{1}+\cdots+f_{m}\rVert_{L^{p,\infty}}^{p}\le \frac{2-p}{1-p}\sum_{j=1}^{m}\lVert f_{j}\rVert_{L^{p,\infty}}^{p}$

Convolution and weak $L^p$ spaces

Bounded gradient in weak $L^1$ implies compactness in $L^1$ (?) Relation to fractional Sobolev compactness.

Weak convergence in $L^{p,\infty}$

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