# Questions tagged [integral-inequality]

For questions inequalities which involves integrals, like Cauchy-Bunyakovsky-Schwarz or Hölder's inequality. To be used with (inequality) tag.

76 questions
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### Prove $\int _0^\infty f^2 dx \leq \cdots$ for $f$ convex

Prove $$\int _0^\infty f^2(x) dx \leq \frac{2}{3}\cdot \max_{x \in \mathbb R^+} f(x) \cdot \int _0^\infty f(x) dx$$ for $f(x) \geq 0$ and convex. I know via Holder's we can get without the ...
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### Hölder inequality from Jensen inequality

I'm taking a course in Analysis in which the following exercise was given. Exercise Let $(\Omega, \mathcal{F}, \mu)$ be a probability space. Let $f\ge 0$ be a measurable function. Using Jensen's ...
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### Is the Riemann integral of a strictly smaller function strictly smaller?

We all know that if $f\leq{}g$ in $[a,b]$ then $$\int_a^bf\,dx\leq\int_a^bg\,dx$$ now, imagine that we have $f<g$, is it true that $$\int_a^bf\,dx<\int_a^bg\,dx$$
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### Prove an integral inequality: $\left(\int|f|^2dx\right)^2\le 4\left(\int|xf(x)|^2dx\right)\left(\int|f'|^2dx\right)$

If $f$ is real-valued and continuously differentiable on $\mathbb{R}$, prove that $$\left(\int|f|^2dx\right)^2\le 4\left(\int|xf(x)|^2dx\right)\left(\int|f'|^2dx\right)$$ Attempt: I tried the ...
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### Is $L^p \cap L^q$ dense in $L^r$?

It is known that $L^p \cap L^q \subset L^r$, where $1 \le p \le r \le q \le \infty$. Are all of these inclusions dense? I.e., do we have \begin{equation*} \overline{L^p \cap L^q} = L^r \end{equation*}...
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### Asymptotic behaviour of a multiple integral on the unit hypercube

A few days ago I found an interesting limit on the "problems blackboard" of my University: $$\lim_{n\to +\infty}\int_{(0,1)^n}\frac{\sum_{j=1}^n x_j^2}{\sum_{j=1}^n x_j}d\mu = 1.$$ The correct claim, ...
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### Inequality of numerical integration $\int _0^\infty x^{-x}\,dx$.

There was a friend asking me how to prove $$\int_0^\infty x^{-x}\,dx<2$$ Mathematica showed that its approximate value is 1.99546, so I think it isn't easy to solve it, can you provide me some ...
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### Prove Friedrichs' inequality

I'm trying to show that the theorem (Friedrichs' inequality) in my book: Assume that $\Omega$ be a bounded domain of Euclidean space $\Bbb R^n$. Suppose that $u: \Omega \to \Bbb R$ lies in the ...
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### How to prove $\int_{-\pi}^\pi (f(x))^2dx\le \int_{-\pi}^\pi (f'(x))^2dx$ [duplicate]

Let $f$ be $C^1$ in $[-\pi, \pi]$ and satisfies $\int_{-\pi}^\pi f(x)dx=0$, periodic boundary condition. Then, prove that $\int_{-\pi}^\pi (f(x))^2dx\le \int_{-\pi}^\pi (f'(x))^2dx$. I try to prove ...
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### $f$ convex, $g$ concave and increasing, $\int_0^1 f = \int_0^1 g$, then $\int_0^1(f)^2 \geq \int_0^1(g)^2$

Let $f,g:[0,1] \to [0, \infty)$ be two continuous functions such that $$f(0) = g(0) = 0,$$ $f$ is convex, $g$ is concave and increasing and $$\displaystyle \int_0^1f(x)dx = \int_0^1g(x)dx.$$ Prove ...
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I got the following question by mail from someone I don't know from Adam. (Quoted in part.) if $f(t)$ continuously diff. on $[0,1]$ and a) $\int_0^1f(t)\ dt=0$ b) $m\le f\,'\le M$ on $... 2answers 139 views ### About the (non-trivial, this time) zeroes of an almost-periodic function This is a follow-up on my previous question that turned out to be almost-trivial. Let$\varphi(t)=\sin(t)+\sin(t\sqrt{2})+\sin(t\sqrt{3})$. Such function is not periodic, but it is bounded, Lipshitz-... 8answers 387 views ### How can we prove that$\pi > 3$using this definition I've been trying to prove that$\pi > 3$by using the following definition: $$\pi = 2\int_{-1}^1{ {\sqrt{1-t^2}}}\, dt$$ Which comes from finding what the area of the unit circle is. (This path ... 2answers 1k views ### Inequality for Expected Value of Product Let$(\Omega, \mathbb{P}, \mathcal{F})$be a probability space, and let$\mathbb{E}$denote the expected value operator. Consider the random variables$f: \Omega \rightarrow \{0,1,2\}$and$g: \Omega ...
Given an arbitrary measure space (of possibly infinite measure), if $f \in L^1 \cap L^\infty$, then by Hölder's inequality, $f^2 \in L^1$, so $f \in L^2$. Intuition suggests that $f \in L^p$ even for ...