Question

If $m(X)<\infty$ and $f$ bounded and measurable, then $1\leq q<p<\infty$ implies $\|f\|_{p}^{p}\leq\|f\|_{q}^{q}\|f\|_{\infty}^{p-q}.$

score 37 · Answer 1 · 2019-06-12 07:15:49

37

Views

If $m(X)<\infty$ and $f$ bounded and measurable, then $1\leq q<p<\infty$ implies $\|f\|_{p}^{p}\leq\|f\|_{q}^{q}\|f\|_{\infty}^{p-q}.$

Published on 12 Jun 2019 - 7:15

score 111 · Answer 2 · 2019-06-12 18:20:37

111

Views

The Lebesgue measure on $\mathbb{T} = \Bbb{R}/2\pi\Bbb{Z}$

Published on 12 Jun 2019 - 18:20

score 183 · Answer 3 · 2019-06-13 17:39:45

183

Views

Show that $f: (0,1] \rightarrow \mathbb{R}$ given by $ f(x) = \sin{\frac{1}{x}} $ is Lebesgue-integrable and calculate its Lebesgue Integral

Published on 13 Jun 2019 - 17:39

score 98 · Answer 4 · 2019-06-16 16:11:47

98

Views

Prove that $ \int_{\mathbb R} g(x)d\mu(x)=\int_{\mathbb R} g(x)f(x)dx. $

Published on 16 Jun 2019 - 16:11

score 150 · Answer 5 · 2026-03-30 03:34:54

150

Views

Proof that function is not Lebesgue integrable.

Published on 30 Mar 2026 - 3:34

score 34 · Answer 6 · 2026-03-25 07:45:44

34

Views

$g: [0, 1] \rightarrow [0, \infty]$ such that $\int_{x}^{y} g \leq M(y-x)$ for all $x < y \in [0, 1]$ then $g \leq M$ a.e.

Published on 25 Mar 2026 - 7:45

score 130 · Answer 7 · 2019-06-17 06:38:48

130

Views

Is the Lebesgue integral simply an integral w.r.t dy instead of dx, with an appropriate measure-theoretic x?

Published on 17 Jun 2019 - 6:38

score 1.2k · Answer 8 · 2019-06-17 15:30:58

1.2k

Views

inverse triangle inequality in $L^p$

Published on 17 Jun 2019 - 15:30

score 153 · Answer 9 · 2019-06-17 15:39:48

153

Views

Riemann Lebesgues lemma in $L^p$

Published on 17 Jun 2019 - 15:39

score 158 · Answer 10 · 2019-06-17 21:55:24

158

Views

composite of measurable functions

Published on 17 Jun 2019 - 21:55

score 211 · Answer 11 · 2026-03-25 05:06:59

211

Views

Improper Lebesgue integral $\int_{\mathbb R}\frac{\sin(x)}x~\mathrm dx$

Published on 25 Mar 2026 - 5:06

score 64 · Answer 12 · 2019-06-18 16:26:11

64

Views

Very simple proof $f$ is integrable, small detail missing

Published on 18 Jun 2019 - 16:26

score 56 · Answer 13 · 2019-06-18 16:53:34

56

Views

Computing $\lim_{n\to\infty}\int_0^{\frac{n}{2}}\left(1-\frac{2x}{n}\right)^n d\lambda(x)$

Published on 18 Jun 2019 - 16:53

score 81 · Answer 14 · 2019-06-18 22:51:18

81

Views

$\Vert \cdot \Vert_2$ norm convergence of $f_n$ to $f$

Published on 18 Jun 2019 - 22:51

score 56 · Answer 15 · 2019-06-19 03:23:27

56

Views

Integral of $m(\{x\in[0,1]:|f(x)|>t\})$ is finite

Published on 19 Jun 2019 - 3:23

List Question

If $m(X)<\infty$ and $f$ bounded and measurable, then $1\leq q<p<\infty$ implies $\|f\|_{p}^{p}\leq\|f\|_{q}^{q}\|f\|_{\infty}^{p-q}.$

The Lebesgue measure on $\mathbb{T} = \Bbb{R}/2\pi\Bbb{Z}$

Show that $f: (0,1] \rightarrow \mathbb{R}$ given by $ f(x) = \sin{\frac{1}{x}} $ is Lebesgue-integrable and calculate its Lebesgue Integral

Prove that $ \int_{\mathbb R} g(x)d\mu(x)=\int_{\mathbb R} g(x)f(x)dx. $

Proof that function is not Lebesgue integrable.

$g: [0, 1] \rightarrow [0, \infty]$ such that $\int_{x}^{y} g \leq M(y-x)$ for all $x < y \in [0, 1]$ then $g \leq M$ a.e.

Is the Lebesgue integral simply an integral w.r.t dy instead of dx, with an appropriate measure-theoretic x?

inverse triangle inequality in $L^p$

Riemann Lebesgues lemma in $L^p$

composite of measurable functions

Improper Lebesgue integral $\int_{\mathbb R}\frac{\sin(x)}x~\mathrm dx$

Very simple proof $f$ is integrable, small detail missing

Computing $\lim_{n\to\infty}\int_0^{\frac{n}{2}}\left(1-\frac{2x}{n}\right)^n d\lambda(x)$

$\Vert \cdot \Vert_2$ norm convergence of $f_n$ to $f$

Integral of $m(\{x\in[0,1]:|f(x)|>t\})$ is finite

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