June 4th, 2017, 02:31 PM  #1 
Member Joined: May 2017 From: France Posts: 57 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. 
June 4th, 2017, 04:36 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,307 Thanks: 526 
I am not sure what you are asking, but the equation will hold for any c. Both integrals = .

June 4th, 2017, 05:06 PM  #3 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1 
Ok, for the first inclusion, but it would miss the other.

June 4th, 2017, 05:15 PM  #4 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1 
Another form : Let $\displaystyle n\in\mathbb N, n>1$ $\displaystyle \{f\in C^0([0,1],\mathbb R^*)\int_0^1f^n\text{ d}x=(\int_0^1 f\text{ d}x)^n\}=\{f\exists c\in \mathbb R, f=c\}$ ? 
June 5th, 2017, 06:01 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,307 Thanks: 526  
June 6th, 2017, 06:36 AM  #6 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1 
$\displaystyle \subset $ It's another form. 
June 6th, 2017, 05:32 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,307 Thanks: 526  
June 7th, 2017, 12:12 AM  #8 
Member Joined: May 2017 From: France Posts: 57 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 $\displaystyle \{f\in C^0([0.1],\mathbb R^*)  \int_0^1 f^{2n}\text{ d}x=(\int_0^1 f\text{ d}x)^{2n}\} \subset \{f\exists c\in \mathbb R^*, f=c\}$ 
June 7th, 2017, 05:30 PM  #9 
Global Moderator Joined: May 2007 Posts: 6,307 Thanks: 526 
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. 
June 9th, 2017, 04:02 AM  #10 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1 
No, $\displaystyle f\neq 0$, it can't change of sign. I think this equality is true. 

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