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 May 30th, 2017, 06:39 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 sub-sequence simply converging on $\displaystyle \mathbb{R}$. Cordially.
 June 8th, 2017, 09:59 AM #2 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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 10:35 AM. Reason: add: "in R," "for each x"
 June 9th, 2017, 05:04 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 05:07 AM.
 June 9th, 2017, 05:29 AM #4 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 sin(nx) is bounded for any n and x.
 June 9th, 2017, 08:25 AM #5 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 In this case, can you give a sub-sequence which have a limits for any x ?
 June 9th, 2017, 08:54 AM #6 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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, 09:20 AM #7 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 Yes, a sub-sequence wich converges for any x.
June 9th, 2017, 09:52 AM   #8
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Quote:
 Originally Posted by Dattier 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 sub-sequence simply converging on $\displaystyle \mathbb{R}$. Cordially.
So what is the proof?
Example?

 June 9th, 2017, 10:13 AM #9 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 I'll let you look again
 June 19th, 2017, 07:22 AM #10 Member   Joined: May 2017 From: France Posts: 57 Thanks: 1 Hi, The solution : Let \$\displaystyle a

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