Question

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

score 76 · Answer 1 · 2026-04-16 05:37:43

76

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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 16 Apr 2026 - 5:37

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 64 · Answer 3 · 2026-04-16 02:52:31

64

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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 16 Apr 2026 - 2:52

score 110 · Answer 4 · 2026-04-17 13:19:01

110

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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 17 Apr 2026 - 13:19

score 48 · Answer 5 · 2026-04-14 15:31:08

48

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Proving lower bound for an integral

Published on 14 Apr 2026 - 15:31

score 82 · Answer 6 · 2026-04-17 08:40:29

82

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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 17 Apr 2026 - 8:40

score 134 · Answer 7 · 2026-04-17 01:29:44

134

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Second order differential inequality

Published on 17 Apr 2026 - 1:29

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 697 · Answer 9 · 2026-04-08 05:21:58

697

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Upper bound on an integral

Published on 08 Apr 2026 - 5:21

score 178 · Answer 10 · 2026-04-11 17:17:39

178

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Understanding Proof of Talagrand's Inequality

Published on 11 Apr 2026 - 17:17

score 148 · Answer 11 · 2026-04-16 16:12:31

148

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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 16 Apr 2026 - 16:12

score 56 · Answer 12 · 2026-04-13 23:29:19

56

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

Published on 13 Apr 2026 - 23:29

score 35 · Answer 13 · 2026-04-16 08:16:43

35

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On the validity of certain Gronwall-type inequality

Published on 16 Apr 2026 - 8:16

score 174 · Answer 14 · 2026-04-15 10:56:57

174

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On the proof of the Hölder inequality.

Published on 15 Apr 2026 - 10:56

score 583 · Answer 15 · 2026-04-15 21:06:45

583

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

Published on 15 Apr 2026 - 21:06

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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