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May 30th, 2017, 05:39 AM  #1 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1  A new beautiful result on convex functions
Hi, Let $\displaystyle (f_n)_n$ a sequence of real convex functions on $\displaystyle \mathbb{R}$, simply bounded : $\displaystyle \forall x \in \mathbb{R}, \exists M>0, \forall n \in\mathbb{N}, f_n(x)\leq M$. So we can extract a subsequence simply converging on $\displaystyle \mathbb{R}$. Cordially. 
June 8th, 2017, 08:59 AM  #2 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
Any bounded infinite sequence u$\displaystyle _{n}$(x) in R has at least one limit point for each x.
Last edited by zylo; June 8th, 2017 at 09:35 AM. Reason: add: "in R," "for each x" 
June 9th, 2017, 04:04 AM  #3 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1 
What about the sequence sin(nx) ?
Last edited by Dattier; June 9th, 2017 at 04:07 AM. 
June 9th, 2017, 04:29 AM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
sin(nx) is bounded for any n and x.

June 9th, 2017, 07:25 AM  #5 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1 
In this case, can you give a subsequence which have a limits for any x ?

June 9th, 2017, 07:54 AM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
Any infinite subsequence has a limit point for any x. Do you mean a unique limit point for any x, ie, converges for any x? I don't have the slightest idea. 
June 9th, 2017, 08:20 AM  #7 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1 
Yes, a subsequence wich converges for any x.

June 9th, 2017, 08:52 AM  #8  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
Example?  
June 9th, 2017, 09:13 AM  #9 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1 
I'll let you look again

June 19th, 2017, 06:22 AM  #10 
Member Joined: May 2017 From: France Posts: 57 Thanks: 1 
Hi, The solution : Let $\displaystyle a<x<y<b$, then $\displaystyle \forall n\in \mathbb N, {M_{a1}M_{a}}\leq f_n(a)f_n(a1)\leq \frac{f_n(x)f_n(y)}{xy} \leq f_n(b+1)f_n(b)\leq M_{b+1}+M_b $ with the theorem of Ascoli we can conlued. Cordially. 

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