I ran into the following exercise: Define $f:(0,c)\to\mathbb{R}$, for some $c\in\mathbb{R}$ with $00$. Moreover, let $\epsilon=2c\delta$. Then, because $0

Question

I ran into the following exercise:

Define $f:(0,c)\to\mathbb{R}$, for some $c\in\mathbb{R}$ with $0<c<\infty$, by$$f(x)=x^2.$$Then $f$ is uniformly continuous on $(0,c)$.

This is what I have done:

Pick $x,x_0\in(0,\infty)$, and assume that $|x-x_0|<\delta$ for some $\delta>0$. Moreover, let $\epsilon=2c\delta$. Then, because $0<x<c$ and $0<x_0<c$, it follows that $0<x+x_0<2c$. Hence, we have that $|x^2-x_0^2|=(x+x_0)|x-x_0|<2c|x-x_0|<2c\delta=\epsilon$, and $f$ is uniformly continuous on $(0,c)$, because it is independent of $x$.

Does this seem ok?

anonymous · Answer

Does the fact that we are reducing the domain makes the function uniformly continuous? I don't think that f(x) = x^2 is UC in general. But I don't remember. Anyway, I can't see an error on your proof. Maybe someone with a sharper eye can see it.

anonymous · Answer

You are correct, $f$ is not continuous over an unbounded domain. That's exactly the motivation behind this question.

anonymous · Answer

Seems right to me. Considering the example 6 in this pdf http://www.google.co.in/url?sa=t&rct=j&q=prove%20that%20f(x)%3Dx%5E2%20%20is%20uniformly%20continuous%20on%20(0%2Cc)&source=web&cd=1&ved=0CCEQFjAA&url=http%3A%2F%2Fwww.math.wisc.edu%2F~robbin%2F521dir%2Fcont.pdf&ei=Wp-hT4XMGrCyiQeU07T8CA&usg=AFQjCNE5tQ8AgRdWmtqbYDOamRUpHockCQ&cad=rja

anonymous · Answer

I forgot to mention the page no.  It is page 2 ,example 6

anonymous · Answer

lol, sorry for my lack of knowledge but how do you read the definition of UC function?

anonymous · Answer

Oh, nice. I didn't know this was a canonical exercise. With that clarified, could I assert that $f$ is uniformly continuous as $c\to\infty$? This would yield a contradiction, however.

anonymous · Answer

See eg 10 for your answer in the same pdf  :D

anonymous · Answer

Gee, I need to learn how to read! Thanks, shivam. :)

anonymous · Answer

@pre-algebra , LOL, trust me, I have never encountered this stuff ever before.  But still, I can understand something of what they wrote :D

anonymous · Answer

Suppose it is. Let $\epsilon = 1$ Pick $\delta > 0 $(by uniform continuity) such that $ |x - y| < \delta$ implies $ |x^2 - y^2 | < 1 $ But $ |x^2 - y^2| = |x - y| * |x + y| $ and if we let x = n, y = n + $ \delta /2 $, for some $n \in \mathbb{N}$, that implies that |x - y | = $ \delta/2 < \delta$ but $ |x^2 - y^2 | = d/2 * (2n + d/2)$ which is > 1 for a big enough n. So f cannot be uniformly continuous. I think this is a sufficient proof for $ f$ On a note, see that d is fixed, so the latter tends to infinity as n gets bigger.

anonymous · Answer

Typo, those d/2 should've been $\delta/2 * (2n + \delta/2) $

anonymous · Answer

And latter here refers to |x^2 - y^2|. Damn, my English is so ambiguous