Is it True that if f is Uniformly Continuous and Unbounded Review my work please

[mmmboh]

[PLAIN]http://img820.imageshack.us/img820/7729/3iiin.jpg

So I gave it a go, and I just want to make sure my argument is convincing:

If f is uniformly continuous and unbounded on [0,∞), then for some c in [0,∞], lim _x->cf(x) = ±∞(notice the closed brackets, I wanted to leave the option that c can be infinity, is this how I should write it? if not, how?), because if this wasn't true, then lim _x->cf(x)=A (A is finite) for all c in [0,∞], but then f would be bounded (is this convincing?). Now suppose c is in [0,∞) (open bracket, so c isn't infinity), then lim _x->cf(x) = ±∞, but since c is finite and f is continuous, then by the definition of continuity f is defined at c, but then f(c)= ±∞, but ±∞ isn't even in the reals so then f wouldn't be undefined at c, so it wouldn't be continuous. So c=∞, and lim _x->∞f(x) = ±∞.

Thoughts?

 

[micromass]

because if this wasn't true, then lim _x->cf(x)=A (A is finite) for all c in [0,∞], but then f would be bounded (is this convincing?)

I disagree with this. If the limit is not infinity, then the possibility also exists that the limit does not exist. For example f(x)=x\sin(x) (note that this function is not uniform continuous, so it doesn't contradict our statement.

 

[mmmboh]

Ok well, the limit may be not exist at infinity, but it will still be bounded, no? (if it is uniformly continuous). How do you think I should approach this?

Note: If f is uniformly continuous and unbounded on [0,∞), then for some c in [0,∞], lim _x->cf(x) = ±∞. This still remains true right? I mean this is the definition of unbounded, so maybe I can just erase the part you disagreed with?

Basically if I wrote this: If f is uniformly continuous and unbounded on [0,∞), then for some c in [0,∞], lim _x->cf(x) = ±∞. Now suppose c is in [0,∞), then lim _x->cf(x) = ±∞, but since c is finite and f is continuous, then by the definition of continuity f is defined at c, but then f(c)= ±∞, but ±∞ isn't even in the reals so then f wouldn't be undefined at c, so it wouldn't be continuous. So c=∞, and lim _x->∞f(x) = ±∞.

Edit: Hm come to think of it, what I wrote might imply that lim _x->∞xsinx=∞ (ignoring the uniformly continuous part), so I need to justify it for uniform continuity somehow, assuming it is true.

 

[micromass]

, lim _x->cf(x) = ±∞. This still remains true right?

Hmm, I think that you should prove this. I don't think it's evident (I don't even know if it's true)...

 

[mmmboh]

Well that's almost what the question is asking me to prove. Do you have an idea on how to do it for me? I know that lim _x->cf(x) = ±∞ could not happen anywhere but at infinity, but I'm not sure how to show it must happen at infinity in the case of uniform continuity (assuming it is true, I believe it is).

I guess I can show that if f is unbounded and continuous on [0,∞) and lim _x->∞f(x)≠±∞, then it is not uniformly continuous, but still, any ideas?

 

[micromass]

If you don't find anything fast, then maybe you can start searching after counterexamples...
A counterexample will look like f(x)=xsin(x), but this function goes to fast up and down. Maybe you can find a function which will go up and down a bit slower...

 

[ystael]

You can be much cruder; try a piecewise-linear function of bounded slope.

 

[micromass]

Yes, that's a very good counterexample!

 

[mmmboh]

Thanks. You think it's ok to just leave it defined as I did? because I can't think of a nicer way to do it, you know without having to write ...

 

[micromass]

Well, I think the idea of the counterexample is clear. You could probably make the definition more formal, but this would be rather ugly and imo doesn't help in grasping the concept. So I'd say to leave it like it is...