Is Splitting the Interval a Valid Approach to Prove Uniform Continuity?

[mmmboh]

[PLAIN]http://img258.imageshack.us/img258/78/52649134.jpg

So I've thought of a few ideas on how to prove this, but only one so far that I've sort of figured out what to do. What I want to do is split the interval up in two, so from [0,b] and from (b, ∞), for some b in the reals. Now since f is continuous on a close bounded interval [0,b], it is uniformly continuous in that interval, so now I have to show it is uniformly continuous in (b, ∞). What I thought I could do was show that this interval is Lipschitz continuous since lim_x->∞f(x)=A...right now I just intuitively believe this to be true, I am trying to prove it. I said suppose f wasn't lipschitz continuous in (b, ∞) for any b, then there doesn't exist a k>0 such that |f(x)-f(u)|<k|x-u| for all x,u in (b, ∞). Then by the archimedian property, there exists x,u in (b, ∞) such that |f(x)-f(u)|> M for all M in the naturals. Now what I want to show is that this implies that there is an c in (b, ∞) such that lim_x->cf(x)= ∞, which is impossible since f is continuous in this interval...but I am having trouble doing this.

Can anyone help? And does anyone know for sure if what I am doing is right? if my hypothesis that there is an interval (b, ∞) for some b such that the interval is Lipschitz continuous is wrong, I would like to know!

By the way, for the converse, I said that f(x)=x is uniformly continuous on [0, ∞), but lim_x->∞f(x)=∞, so the converse isn't true.

Thanks!

 

[micromass]

So you want to prove that there exists a c such that \lim_{x\rightarrow c}{f(x)}=\infty. I don't really see why you want this, but what you are trying to do isn't always true, sadly...

Take for example the function f(x)=\sin(x^2)... This isn't uniformly continuous and not Lipschitz, but there also does not exist a c such that \lim_{x\rightarrow c}{f(x)}=\infty.
But note that this f does not contradict our theorem, since the limit of x does not exist...

 

[snipez90]

Hmm, I'm not sure of your approach, but I think there is a much simpler route, namely make direct estimates, that doesn't involve showing Lipschitz continuity, which is a stronger condition.

Take e > 0. The basic idea is that the limit at infinity imposes a growth condition, so that for all x > M, some M > 0, |f(x) - A| < e/3. Now you already have the correct idea for how to deal with [0, M], namely that compactness gives us uniform continuity on this interval. It remains to show that the delta you obtain from the uniform continuity condition over [0,M] works on [0, infinity), and this should just be casework (consider possibilities for |x-y| when x, y are chosen to be nonnegative).

 

[mmmboh]

No sorry (to micromass), I think you misunderstood. What I am trying to prove is that if |f(x)-f(u)|> M for all M in the naturals, then there is a c in (b, ∞) such that lim_x->cf(x)= ∞...in your example you couldn't find an an x,u such that |f(x)-f(u)|> M for all M in the naturals. I believe if I can prove this then I can prove my problem.

Snipez90:Ok thanks! I'll try that.

 

[micromass]

Yes, you are correct. But in your problem, you can find for every M an x,u such that |f(x)-f(u)|>M. But I don't think these x and u are necessairily the same for every M...

 

[micromass]

Yes, that seems mighty fine

 

[mmmboh]

And did I understand what I'm suppose to do for the converse statement right? as in f(x)=x disproves it?

 

[micromass]

Yes, f(x)=sin(x) disproves the converse statement.

 

[mmmboh]

I said f(x)=x actually, but I take it it's right. Thanks!

 

[micromass]

Oh sorry, I meant f(x)=x. Sometimes my hands write crazy things