Question

Integral inequality $\left| \int\limits_{-1}^{1} x f(x) dx \right| \le \left| \int\limits_{-1}^{1} f(x) dx \right|$

score 73 · Answer 1 · 2023-04-30 11:04:13

73

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Integral inequality $\left| \int\limits_{-1}^{1} x f(x) dx \right| \le \left| \int\limits_{-1}^{1} f(x) dx \right|$

Published on 30 Apr 2023 - 11:04

score 58 · Answer 2 · 2026-03-26 04:14:12

58

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Bellman-like integral inequality

Published on 26 Mar 2026 - 4:14

score 61 · Answer 3 · 2023-05-17 22:29:04

61

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Proving $\left\vert\int_n^{n+1}s\int_n^xt^{-s-1}dt\,dx\right\vert<\vert sn^{-s-1}\vert$, for $n \in \mathbb{N}$, $x \in [n;n+1]$, $s\in \mathbb{C}$

Published on 17 May 2023 - 22:29

score 107 · Answer 4 · 2023-05-28 18:31:07

107

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If $g:[a,b] \to \mathbb{R}^m$ is continuous and $x_0 \in [a,b]$, prove $|\int_a^b g(x)dx| \geq \int^a_b|g(x)|dx-2\int^a_b|g(x)-g(x_0)|dx$.

Published on 28 May 2023 - 18:31

score 45 · Answer 5 · 2023-05-29 09:35:18

45

Views

Proving lower bound for an integral

Published on 29 May 2023 - 9:35

score 79 · Answer 6 · 2023-06-03 15:40:02

79

Views

If $x:[0,1]\rightarrow \mathbb{R}$ verifies $0\leq x(t)\leq \int_0^te^sx(s)^2ds \ \forall t\in [0,1]$ then $x(t)=0 \ \forall t\in[0,1]$

Published on 03 Jun 2023 - 15:40

score 131 · Answer 7 · 2023-06-03 22:36:20

131

Views

Second order differential inequality

Published on 03 Jun 2023 - 22:36

score 353 · Answer 8 · 2026-03-26 04:14:08

353

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a functional inequality (positive functions on circle)

Published on 26 Mar 2026 - 4:14

score 695 · Answer 9 · 2023-06-26 11:42:15

695

Views

Upper bound on an integral

Published on 26 Jun 2023 - 11:42

score 175 · Answer 10 · 2026-02-23 17:25:31

175

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Understanding Proof of Talagrand's Inequality

Published on 23 Feb 2026 - 17:25

score 145 · Answer 11 · 2023-07-01 16:08:29

145

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Prove $\int_{0}^{\pi} \sqrt[n]{\prod_{k=1}^{n}\csc^2(x-\alpha_k)}dx\geq4π$ for reals $\alpha_1,\alpha_2,...\alpha_n$

Published on 01 Jul 2023 - 16:08

score 53 · Answer 12 · 2023-07-14 09:15:22

53

Views

Issue with a lemma used for proving Morrey's theorem.

Published on 14 Jul 2023 - 9:15

score 32 · Answer 13 · 2023-07-16 15:38:43

32

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On the validity of certain Gronwall-type inequality

Published on 16 Jul 2023 - 15:38

score 171 · Answer 14 · 2023-07-21 14:29:30

171

Views

On the proof of the Hölder inequality.

Published on 21 Jul 2023 - 14:29

score 580 · Answer 15 · 2023-08-01 13:07:18

580

Views

$\int_0^{\pi/2} f(t)\sin t dt =\int_{\pi/2}^\pi f(t)\sin t dt < f(\frac{\pi}{2}) \Rightarrow \int_0^\pi \sqrt{f(t)} dt < \pi \sqrt{f(\frac{\pi}{2})}$

Published on 01 Aug 2023 - 13:07

List Question

Integral inequality $\left| \int\limits_{-1}^{1} x f(x) dx \right| \le \left| \int\limits_{-1}^{1} f(x) dx \right|$

Bellman-like integral inequality

Proving $\left\vert\int_n^{n+1}s\int_n^xt^{-s-1}dt\,dx\right\vert<\vert sn^{-s-1}\vert$, for $n \in \mathbb{N}$, $x \in [n;n+1]$, $s\in \mathbb{C}$

If $g:[a,b] \to \mathbb{R}^m$ is continuous and $x_0 \in [a,b]$, prove $|\int_a^b g(x)dx| \geq \int^a_b|g(x)|dx-2\int^a_b|g(x)-g(x_0)|dx$.

Proving lower bound for an integral

If $x:[0,1]\rightarrow \mathbb{R}$ verifies $0\leq x(t)\leq \int_0^te^sx(s)^2ds \ \forall t\in [0,1]$ then $x(t)=0 \ \forall t\in[0,1]$

Second order differential inequality

a functional inequality (positive functions on circle)

Upper bound on an integral

Understanding Proof of Talagrand's Inequality

Prove $\int_{0}^{\pi} \sqrt[n]{\prod_{k=1}^{n}\csc^2(x-\alpha_k)}dx\geq4π$ for reals $\alpha_1,\alpha_2,...\alpha_n$

Issue with a lemma used for proving Morrey's theorem.

On the validity of certain Gronwall-type inequality

On the proof of the Hölder inequality.

$\int_0^{\pi/2} f(t)\sin t dt =\int_{\pi/2}^\pi f(t)\sin t dt < f(\frac{\pi}{2}) \Rightarrow \int_0^\pi \sqrt{f(t)} dt < \pi \sqrt{f(\frac{\pi}{2})}$

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