How to prove this function is uniformly continuous?

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  • #1
quasar987
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1. Prove that [itex] f(x) = \frac{x^2}{1+(x- 1)^2}[/itex] is uniformly continuous on [itex]\mathbb{R}[/itex]


Homework Equations

: [itex] x^2 - y^2 = (x+y)(x-y)[/itex] [/B]


3. After using the above identity, I am left with

[tex]|f(x)-f(y)| = |x-y| \left| \frac{x+y-xy}{(1+(x- 1)^2)(1+(y- 1)^2)}\right| [/tex]

and I do not know how to make progress.

edit: oops, looks like I accidentally hit the "post" button while I was trying to fix my last equation. Why the heck is it not showing right?!
 

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  • #2
RUber
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If your right hand side (less the |x-y| term) has a maximum over the reals, I'll refer to it as it max(r(x)) , then the function is uniformly continuous.
##\forall x, y \in \mathbb{R}, \text{ for any } \epsilon>0, \exists \delta>0 s.t. |x-y|< \delta \implies |f(x) - f(y) | < \epsilon ##
With ##\delta = \frac{\epsilon}{\max(r(x) ) }.##
 
  • #3
pasmith
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1. Prove that [itex] f(x) = \frac{x^2}{1+(x- 1)^2}[/itex] is uniformly continuous on [itex]\mathbb{R}[/itex]


Homework Equations

: [itex] x^2 - y^2 = (x+y)(x-y)[/itex] [/B]


3. After using the above identity, I am left with

[tex]|f(x)-f(y)| = |x-y| \left| \frac{x+y-xy}{(1+(x- 1)^2)(1+(y- 1)^2)}\right| [/tex]

and I do not know how to make progress.

edit: oops, looks like I accidentally hit the "post" button while I was trying to fix my last equation. Why the heck is it not showing right?!

Clearly [itex]f[/itex] is continuous, and its limits as [itex]x \to \pm \infty[/itex] both exist and are finite. This is sufficient for [itex]f[/itex] to be uniformly continuous on the reals.

Proving this statement for arbitrary [itex]f : \mathbb{R} \to \mathbb{R}[/itex] is probably a better use of your time than struggling with the specific function in your question.
 
  • #4
quasar987
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If your right hand side (less the |x-y| term) has a maximum over the reals, I'll refer to it as it max(r(x)) , then the function is uniformly continuous.
##\forall x, y \in \mathbb{R}, \text{ for any } \epsilon>0, \exists \delta>0 s.t. |x-y|< \delta \implies |f(x) - f(y) | < \epsilon ##
With ##\delta = \frac{\epsilon}{\max(r(x) ) }.##
Of course. It goes without saying that it is in bounding the RHS (divided by |x-y|) that I am having trouble.
 
  • #5
quasar987
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Clearly [itex]f[/itex] is continuous, and its limits as [itex]x \to \pm \infty[/itex] both exist and are finite. This is sufficient for [itex]f[/itex] to be uniformly continuous on the reals.

Proving this statement for arbitrary [itex]f : \mathbb{R} \to \mathbb{R}[/itex] is probably a better use of your time than struggling with the specific function in your question.

The reason I ask this question is because it is from a final exam from a first year analysis class. The goal of these exams is to check that the students understand and are able to solve problems (aka answer basic questions) using the definitions and theorems seen in class. The solution to these exam questions is never to prove a Lemma such as the one you mention. (Maybe at Princeton but not where this exam is from :P)

Therefor, the question remains open: does anyone sees how to bound the RHS in my original post? Thanks!
 
  • #6
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Check for errors but I get

##|f(x)-f(y)| = 2 |x-y| \frac{|1-(x-1)(y-1)|}{(1+(x-1)^2)(1+(y-1)^2)} \le 2 |x-y| (\frac{1}{(1+(x-1)^2)(1+(y-1)^2)}+\frac{|(x-1)(y-1)|}{(1+(x-1)^2)(1+(y-1)^2)}) \le 2|x-y| (1 +\frac{1}{2}(\frac{(x-1)^2+(y-1)^2}{(1+(x-1)^2)(1+(y-1)^2)}) \le 4|x-y| ##

So if one takes ##\delta = \frac{\varepsilon}{4} ## you get uniform continuity
 
  • #7
quasar987
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Good job geoffrey, you are a master!
 
  • #8
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come on you make me blush
 

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