My Math Forum  

Go Back   My Math Forum > College Math Forum > Real Analysis

Real Analysis Real Analysis Math Forum


Reply
 
LinkBack Thread Tools Display Modes
June 4th, 2017, 02:31 PM   #1
Member
 
Joined: May 2017
From: France

Posts: 50
Thanks: 1

A fun result on integral

Hi,


$\displaystyle n \in \mathbb N, n>1, \{f\in C^0([0.1],\mathbb R) | \int_0^1 f^{2n}\text{ d}x=(\int_0^1 f\text{ d}x)^{2n}\}=\{f|\exists c\in \mathbb R, f=c\}$ ?

Cordially.

Last edited by Dattier; June 4th, 2017 at 02:46 PM.
Dattier is offline  
 
June 4th, 2017, 04:36 PM   #2
Global Moderator
 
Joined: May 2007

Posts: 6,255
Thanks: 507

I am not sure what you are asking, but the equation will hold for any c. Both integrals = .
mathman is offline  
June 4th, 2017, 05:06 PM   #3
Member
 
Joined: May 2017
From: France

Posts: 50
Thanks: 1

Ok, for the first inclusion, but it would miss the other.
Dattier is offline  
June 4th, 2017, 05:15 PM   #4
Member
 
Joined: May 2017
From: France

Posts: 50
Thanks: 1

Another form :

Let $\displaystyle n\in\mathbb N, n>1$

$\displaystyle \{f\in C^0([0,1],\mathbb R^*)|\int_0^1f^n\text{ d}x=(\int_0^1 f\text{ d}x)^n\}=\{f|\exists c\in \mathbb R, f=c\}$ ?
Dattier is offline  
June 5th, 2017, 06:01 PM   #5
Global Moderator
 
Joined: May 2007

Posts: 6,255
Thanks: 507

Quote:
Originally Posted by Dattier View Post
Ok, for the first inclusion, but it would miss the other.
What is first inclusion? What is other?

Your last statement still looks like equality ().
mathman is offline  
June 6th, 2017, 06:36 AM   #6
Member
 
Joined: May 2017
From: France

Posts: 50
Thanks: 1

$\displaystyle \subset $

It's another form.
Dattier is offline  
June 6th, 2017, 05:32 PM   #7
Global Moderator
 
Joined: May 2007

Posts: 6,255
Thanks: 507

Quote:
Originally Posted by Dattier View Post
$\displaystyle \subset $

It's another form.
Please try to be less cryptic. I don't understand your point.
mathman is offline  
June 7th, 2017, 12:12 AM   #8
Member
 
Joined: May 2017
From: France

Posts: 50
Thanks: 1

$\displaystyle \{f|\exists c\in \mathbb R^*, f=c\}\subset \{f\in C^0([0.1],\mathbb R^*) | \int_0^1 f^{2n}\text{ d}x=(\int_0^1 f\text{ d}x)^{2n}\}$

and

$\displaystyle \{f\in C^0([0.1],\mathbb R^*) | \int_0^1 f^{2n}\text{ d}x=(\int_0^1 f\text{ d}x)^{2n}\} \subset \{f|\exists c\in \mathbb R^*, f=c\}$
Dattier is offline  
June 7th, 2017, 05:30 PM   #9
Global Moderator
 
Joined: May 2007

Posts: 6,255
Thanks: 507

I finally get your point. The first statement is true. The second is not (for 2n exponent) since only |f|=c is required. f may be a step function oscillating between 2 values of opposite sign, but same magnitude.

For odd n, the first is true. The second could have a function (like $\displaystyle sin2\pi x$) where negative part is mirror image of positive part.

Last edited by mathman; June 8th, 2017 at 12:48 PM.
mathman is offline  
June 9th, 2017, 04:02 AM   #10
Member
 
Joined: May 2017
From: France

Posts: 50
Thanks: 1

No, $\displaystyle f\neq 0$, it can't change of sign.

I think this equality is true.
Dattier is offline  
Reply

  My Math Forum > College Math Forum > Real Analysis

Tags
fun, integral, result



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
get the result as 100 jzh Number Theory 4 June 15th, 2012 06:09 AM
Best possible result Solarmew Applied Math 7 May 30th, 2012 07:58 PM
integral result formula ZardoZ Real Analysis 7 June 2nd, 2011 05:39 AM
looking for help to prove a result bigsnail Advanced Statistics 0 January 6th, 2010 08:04 AM
please explain the result bgbgbg Calculus 4 October 28th, 2009 04:07 PM





Copyright © 2017 My Math Forum. All rights reserved.