My Math Forum  

Go Back   My Math Forum > College Math Forum > Topology

Topology Topology Math Forum


Thanks Tree1Thanks
Reply
 
LinkBack Thread Tools Display Modes
May 30th, 2017, 05:39 AM   #1
Member
 
Joined: May 2017
From: France

Posts: 53
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.
Dattier is offline  
 
June 8th, 2017, 08:59 AM   #2
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

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"
zylo is offline  
June 9th, 2017, 04:04 AM   #3
Member
 
Joined: May 2017
From: France

Posts: 53
Thanks: 1

What about the sequence sin(nx) ?

Last edited by Dattier; June 9th, 2017 at 04:07 AM.
Dattier is offline  
June 9th, 2017, 04:29 AM   #4
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

sin(nx) is bounded for any n and x.
zylo is offline  
June 9th, 2017, 07:25 AM   #5
Member
 
Joined: May 2017
From: France

Posts: 53
Thanks: 1

In this case, can you give a sub-sequence which have a limits for any x ?
Dattier is offline  
June 9th, 2017, 07:54 AM   #6
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

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.
zylo is offline  
June 9th, 2017, 08:20 AM   #7
Member
 
Joined: May 2017
From: France

Posts: 53
Thanks: 1

Yes, a sub-sequence wich converges for any x.
Dattier is offline  
June 9th, 2017, 08:52 AM   #8
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

Quote:
Originally Posted by Dattier View Post
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?
zylo is offline  
June 9th, 2017, 09:13 AM   #9
Member
 
Joined: May 2017
From: France

Posts: 53
Thanks: 1

I'll let you look again
Dattier is offline  
June 19th, 2017, 06:22 AM   #10
Member
 
Joined: May 2017
From: France

Posts: 53
Thanks: 1

Hi,

The solution :

Let $\displaystyle a<x<y<b$, then

$\displaystyle \forall n\in \mathbb N, {-M_{a-1}-M_{a}}\leq f_n(a)-f_n(a-1)\leq \frac{f_n(x)-f_n(y)}{x-y} \leq f_n(b+1)-f_n(b)\leq M_{b+1}+M_b $

with the theorem of Ascoli we can conlued.

Cordially.
Dattier is offline  
Reply

  My Math Forum > College Math Forum > Topology

Tags
beautiful, convex, functions, result



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Beautiful music is supported by beautiful mathematics Eureka Art 9 December 12th, 2015 03:10 AM
Product of 2 convex functions Vasily Applied Math 1 June 23rd, 2012 01:11 PM
Convex functions Rak Real Analysis 1 December 1st, 2009 09:45 AM
intersection of 2 convex functions defined on a polytope vikram Applied Math 0 February 4th, 2009 03:27 PM





Copyright © 2017 My Math Forum. All rights reserved.