# Uniform Continuity Proof

• Apr 18th 2012, 08:31 AM
MarkJacob
Uniform Continuity Proof
Hey guys,

I'm having a little trouble trying to prove a question on uniform continuity that is in one of my textbooks (which doesn't have answers, unfortunately).

"Is the function f(x) = x3+x uniformly continuous on R? Prove your answer"

I am almost certain that the answer is no for the simple reason that as x gets increasingly large the gaps get bigger so the same epsilon value won't work, but I'm not sure as to how to go about writing this in mathematical terms.

I know that for a function to be uniformly continuous |f(x) - f(x')|<epsilon whenever |x - x'|<delta for ALL x, x' that are elements of the subset, S, which in this case is R.

I would appreciate any help although preferably a proof that I can follow so that I can make the link to what I know to how to write it, if that makes sense.

Mark
• Apr 18th 2012, 09:12 AM
FernandoRevilla
Re: Uniform Continuity Proof
• Apr 18th 2012, 06:35 PM
MarkJacob
Re: Uniform Continuity Proof
Thanks for this, it did help me with understanding it.
Although I am now wondering how the '+ x' at the end would adjust this proof.

I'm guessing that I would write "y3 + y - x3 - x",

Which would then simplify to (y-x)(1 + (y2 + yx + x2)). Correct?

But then when you pick values for x and y you cannot do so without there being a delta left over in your answer (ie., you can show the contradiction as there is a delta left over)

Does this mean that it is, indeed, uniformly continuous?

Or do I need to choose x and y so that y = x + d/(d+3) (d = delta)
This would then cancel the deltas out and give 1 (as it is the multiplicative inverse), which contradicts the statement that |y2 + yx + x2| < 1.
Although my problem with this is that I'm not sure if this fits with the previous restrictions that were made on x and y.

Any help here would be great!

Thanks,

Mark