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Prove inequality

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Source: SEEMOUS

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seby97

#1 h Feb 14, 2017, 1:00 PM • 2 Y

Y by Adventure10, Mango247

Let $f:[0,1]\rightarrow R$ be a continuous function and n be an integer number,n>0.Prove that $\int_0^1f(x)dx \le (n+1)*\int_0^1 x^n*f(x)dx$

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Kent Merryfield

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Kent Merryfield

#2 h Feb 14, 2017, 2:56 PM • 3 Y

Y by adityaguharoy, AHKR, Adventure10

As stated, this is false.

Let $f(x)=1-x.$ Then the alleged inequality would read $\frac12\le\frac1{n+2}.$

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#3 h Feb 14, 2017, 5:23 PM • 1 Y

Y by Adventure10

seby97 wrote:

Let $f:[0,1]\rightarrow R$ be a continuous function and n be an integer number,n>0.Prove that $\int_0^1f(x)dx \le (n+1)*\int_0^1 x^n*f(x)dx$

Will it be :
$f: [0,1] \to (0,1)$

This post has been edited 1 time. Last edited by adityaguharoy, Feb 14, 2017, 5:24 PM

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jmerry

#4 h Feb 14, 2017, 6:27 PM • 2 Y

Y by Adventure10, Mango247

No, by essentially the same example.

What added condition would do it? Make $f$ ~~nonnegative and~~ increasing on $[0,1]$ . Then this is a case of the rearrangement inequality.

[Edit: OK, don't need it to be positive.]

This post has been edited 1 time. Last edited by jmerry, Feb 15, 2017, 5:50 PM

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#5 h Feb 15, 2017, 5:12 PM • 1 Y

Y by Adventure10

seby97 wrote:

Let $f:[0,1]\rightarrow R$ be a continuous function and n be an integer number,n>0.Prove that $\int_0^1f(x)dx \le (n+1)*\int_0^1 x^n*f(x)dx$

SEEMOUS 2011:

For a given integer $n\ge 1$ , let $f:[0,1]\to\mathbb{R}$ be a non-decreasing function. Prove that
$\int_0^1f(x)dx\le (n+1)\int_0^1x^nf(x)dx.$ Find all non-decreasing continuous functions for which equality holds.

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#6 h Feb 15, 2017, 7:43 PM • 1 Y

Y by Adventure10

Note that $\int_0^1f(x)dx\le (n+1)\int_0^1x^nf(x)dx$ is equivalent to
$\int_0^{1}\int_0^1f(x)y^n dx dy\le \int_{0}^{1}\int_0^1x^nf(x) dx$ or $0\le \frac{1}{2}\int_{0}^{1}\int_0^1x^nf(x)+y^nf(y)-x^nf(y)-y^nf(x) dx dy.$ Factoring this is $\frac{1}{2}\int_{0}^{1}\int_0^1(x^n-y^n)(f(x)-f(y)) dx dy\ge 0$ which is obvious as $f(x)$ is nondecreasing.

This post has been edited 1 time. Last edited by mssmath, Feb 15, 2017, 7:43 PM

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wer

#7 h Feb 15, 2017, 7:52 PM • 2 Y

Y by adityaguharoy, Adventure10

OR CEBASHEV INEQUALITY: $\frac{1}{b-a}\int_{a}^{b}f(x)g(x)dx\ge\int_{a}^{b}f(x)dx\int_{a}^{b}g(x)dx$ , white $f,g$ the same monotony

This post has been edited 1 time. Last edited by wer, Feb 16, 2017, 11:16 AM
Reason: ......

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jmerry

#8 h Feb 15, 2017, 8:14 PM • 2 Y

Y by Adventure10, teomihai

I know the original name is Russian, and there are several sensible ways to transliterate it into a Latin alphabet - but that's not one I've seen before.

Oh, and I prefer calling this inequality by a more descriptive name - the rearrangement inequality, as I posted earlier in this thread.

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lambda

#9 h Feb 16, 2017, 2:09 AM • 2 Y

Y by Adventure10, Mango247

An analytic method is to check the indicator functions $\chi_{[a,1]}$ . It holds for increasing simple functions which is the sum of such indicator functions with positive coefficients (plus a constant). Thus it is true for increasing measerable functions.

This post has been edited 1 time. Last edited by lambda, Feb 16, 2017, 2:22 AM

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#10 h Feb 16, 2017, 3:06 AM • 2 Y

Y by Adventure10, Mango247

wer wrote:

CEBASEV

jmerry wrote:

I know the original name is Russian, and there are several sensible ways to transliterate it into a Latin alphabet - but that's not one I've seen before.

He meant Chebyshev inequality.

And oh yes it is essentially the Rearrangement inequality just a little refined ly written in case you figure out a difference.

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#11 h Jun 18, 2021, 5:27 AM

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For a given integer $n\ge 1$ , let $f:[0,1]\to\mathbb{R}$ be a non-decreasing function. Prove that
$\int_0^1f(x)dx\le(n+1)\int_0^1x^nf(x)dx.$ Find all non-decreasing continuous functions for which equality holds.

This post has been edited 2 times. Last edited by jasperE3, Jun 18, 2021, 5:27 AM
Reason: for the collection, see #5

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#12 h Oct 8, 2022, 12:42 PM

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wer wrote:

OR CEBASHEV INEQUALITY: $\frac{1}{b-a}\int_{a}^{b}f(x)g(x)dx\ge\int_{a}^{b}f(x)dx\int_{a}^{b}g(x)dx$ , white $f,g$ the same monotony

Take $g(x) = x^n,$ to get the solution.

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#14 h Mar 2, 2023, 3:57 PM

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Prove that $(n+1)\int_0^1x^nf(x)dx$ is in the range of $f$ .

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#15 h May 5, 2023, 6:24 PM

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peace09 wrote:

Prove that $(n+1)\int_0^1x^nf(x)dx$ is in the range of $f$ .

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Write $f(m)\le f(x)\le f(M)\implies x^nf(m)\le x^nf(x)\le x^nf(M)$ , whence the result follows upon integration. $\blacksquare$

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#16 h Jan 24, 2025, 7:47 PM

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jasperE3 wrote:

For a given integer $n\ge 1$ , let $f:[0,1]\to\mathbb{R}$ be a non-decreasing function. Prove that
$\int_0^1f(x)dx\le(n+1)\int_0^1x^nf(x)dx.$ Find all non-decreasing continuous functions for which equality holds.

Proof: (Trivial by Chebyshev's inequality, but we will keep it simpler)

We know ∀ $x \in [0, 1]$ , the inequality
$x \geq x^{n+1}$ holds.

claim: $x - x^{n+1}$ is increasing on $[0, 1)$ .

Define $h(x) = x - x^{n+1}$ .

derivative of $h(x)$ :
$h'(x) = 1 - (n+1)x^n \quad \text{for} \quad x \in [0, 1).$
We need to show that $h'(x) \geq 0$ for all $x \in [0, 1)$ . equality holds when $x = \frac{1}{(n+1)^{1/n}}$ , and for $x \in [0, 1)$ , we have
$1 - (n+1)x^n \geq 0.$ Thus,
$1 \geq (n+1)x^n.$
This implies
$f(x) \geq (n+1) f(x) x^n$ and integrating both sides from 0 to 1:
$\int_0^1 f(x) \, dx \geq (n+1) \int_0^1 f(x) x^n \, dx.$
The inequality returns equality when $f(x) = c$ , where $c \in \mathbb{R}$ is any constant function.

$\boxed{\text{Q.E.D.}}$

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chev

#17 h Mar 6, 2025, 1:00 AM

Y by

Levieee wrote:

jasperE3 wrote:

For a given integer $n\ge 1$ , let $f:[0,1]\to\mathbb{R}$ be a non-decreasing function. Prove that
$\int_0^1f(x)dx\le(n+1)\int_0^1x^nf(x)dx.$ Find all non-decreasing continuous functions for which equality holds.

Proof: (Trivial by Chebyshev's inequality, but we will keep it simpler)

We know ∀ $x \in [0, 1]$ , the inequality
$x \geq x^{n+1}$ holds.

claim: $x - x^{n+1}$ is increasing on $[0, 1)$ .

Define $h(x) = x - x^{n+1}$ .

derivative of $h(x)$ :
$h'(x) = 1 - (n+1)x^n \quad \text{for} \quad x \in [0, 1).$
We need to show that $h'(x) \geq 0$ for all $x \in [0, 1)$ . equality holds when $x = \frac{1}{(n+1)^{1/n}}$ , and for $x \in [0, 1)$ , we have
$1 - (n+1)x^n \geq 0.$ Thus,
$1 \geq (n+1)x^n.$
This implies
$f(x) \geq (n+1) f(x) x^n$ and integrating both sides from 0 to 1:
$\int_0^1 f(x) \, dx \geq (n+1) \int_0^1 f(x) x^n \, dx.$
The inequality returns equality when $f(x) = c$ , where $c \in \mathbb{R}$ is any constant function.

$\boxed{\text{Q.E.D.}}$

I believe
$1 \geq (n+1)x^n$ happens almost everywhere except a local at 1. Could you address it further?

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