Continuously uniform function proof

aeronautical · Apr 16, 2009

1. The problem statement, all variables and given/known data

Let f : (0,1) → (0,1) be a strict growing continuous function. Does f have to be continuously uniform? Please note that its from (0,1) → (0,1) and NOT [0,1] → [0,1]. Please help me with the steps...I have no clue where to start...thanks...

dx · Apr 16, 2009

Consider the function [tex] f(x) = -1 / (x - 1) [/tex]. Prove that this is a counterexample.

aeronautical · Apr 16, 2009

Sorry but could you elaborate further? What theorem should I use?

dx · Apr 16, 2009

Sorry, the function in my previous post should have been [tex] f(x) = -1 / (x - 1) [/tex]

You don't need to use any special theorem. Just try drawing a graph of this function, and see if you can prove that it is not uniformly continuous on (0,1).

If you find that difficult, think about the following easier example. Prove that 1/x is not uniformly continuous on (0,1).

aeronautical · Apr 16, 2009

I find this rather difficult, so I started pursuing the second option you suggested, and found this link as well:
http://mathcs.org/analysis/reals/cont/answers/contuni1.html

I can't figure out how you linked my question to the function you were kind to write down for me.

dx · Apr 16, 2009

The function I wrote down is the same function except reflected in the y axis, so that it is strictly increasing (a condition in the question), and moved to the right so that the infinity is at 1 instead of at 0.

aeronautical · Apr 16, 2009

I drew the function. It goes to plus minus infinity at x=1. So what does this tell me? That it is not continuously uniform?

aeronautical · Apr 16, 2009

Here is my plot....

dx · Apr 16, 2009

aeronautical said: ↑

I drew the function. It goes to plus minus infinity at x=1. So what does this tell me? That it is not continuously uniform?

You will notice that it is the same as the function 1/x except moved and reflected. You posted a link to a website showing how to prove that 1/x in not uniformly continuous on (0,1), so you can use the same method with slight modifications to prove that this one is not uniformly continuous.

aeronautical · Apr 16, 2009

However, I do not understand why the specify that (0,1) → (0,1) is strict growing continuous function. So this proof is by contradiction?

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Continuously uniform function proof

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