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I'm trying to understand this whole concept of uniform continuity. I've read the wikipedia article, I've gone through the section about it in my textbook, I've copied down the definition a few times, and i've gone through this pdf: http://www.math.wisc.edu/~robbin/521dir/cont.pdf

Technically speaking, I think I know what to do:

When asked to prove that:

$f(x) = x^2$ is uniformly continuous on $[0,3]$

I went through the steps of the definition and chose my $\delta$ to be $\epsilon/6$ (which is positive) and said:

Ok, assuming I choose an $x$ and $x_0$ from $[0,3]$ so that $|x_0 - x|<\delta$ then I must show that $|f(x_0) - f(x)|<\epsilon$ for every $\epsilon>0$.

$|x_0^2 - x^2| = (x_0 + x)|x_0-x| < \delta* (x_0+x) = \epsilon /6 *(x_0+x)< \epsilon /6 * (3 + 3) = \epsilon $

this part I don't really understand...am I just choosing the largest possible values for $x$ and $x_0$ on the interval? I'm also not sure what's the point of this whole thing, and it took me a bit of playing around with different values for $\delta$ until I got one that works and then realized there's a systematic way to calculate $\delta$ and then I wondered then what's the point of this whole proof delta/epsilon structure?

I've also read some of the previous questions and answers on uniform continuity and some of the explanations went over my head. Is there a simpler way to understand this aside from just memorizing what steps to go through when I see a problem like this?

I understand that uniform continuity is a "stricter" form of continuity than the regular one, since it describes the attribute of continuity for an entire interval and not merely one point.

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Well posed question. Keep it up! +1 –  user21436 Feb 6 '12 at 16:15
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1 Answer

up vote 2 down vote accepted

I'm not sure this addresses your question directly; but, for what it's worth:

You need to show that given $\epsilon>0$, there is a corresponding $\delta>0$ such that $$\tag{1}|f(x)-f(y)|<\epsilon,\text{ whenever } |x-y|<\delta.$$
The point of the whole thing is to show that the above definition is satisfied: given $\epsilon>0$, you have to demonstrate that there is a $\delta$ such that $(1)$ holds. (There aren't any golden rules that always allow you to find $\delta$. Each problem generally requires different analysis.)

Note, please, in the above, you have $\epsilon$ fixed at the outset. Then you need to find and specify $\delta$, and then show that $(1)$ holds.

Also, keep in mind that this $\delta$ does not depend on the $x$ and $y$ values. This is the distinction between uniform and "plain ol'" continuity. For continuity of $f$ at $x_0$ you need to show that given $\epsilon>0$, there is a corresponding $\delta_{x_0}>0$ such that $$\tag{1}|f(x)-f(x_0)|<\epsilon,\text{ whenever } |x-y|<\delta_{x_0}.$$
Note that the $\delta_{x_0}$ here could change when $x_0$ changes.


For your problem, yes, you're essentially choosing the largest $x$ and $x_0$; it turns out you can do this and it "works". If $x$ and $x_0$ are in $[0,3]$ then $$|x_0^2-x^2|=|x+x_0||x-x_0|\le3\cdot|x-x_0|.$$

So, looking at the above, you think "aha, I can make $|x^2-x_0^2|$ as small as I like, independent of what $x_0$ is, as long as $x$ is sufficiently close to $x_0$. So, $f(x)=x^2$ is uniformly continuous on $[0,3]$.

Formally: Suppose you're given $\epsilon>0$. Set $\delta={\epsilon/6}$. Then if $|x-x_0|<\delta$, you have $$|x^2-x_0^2|\le3\cdot\delta < \epsilon/2.$$ This shows that, by definition, $f(x)=x^2$ is uniformly continuous over $[0,3]$.

It might be illustrative to prove that $f(x)=x^2$ is uniformly continuous over $[0,a]$. You would argue as above, but you'd take $\delta$ to be $\epsilon/2a$. Note that, the larger $a$ is, the smaller your $\delta$ will have to be.

Here's a question: is $f(x)=x^2$ uniformly continuous on $[0,\infty)$?

The answer is no. Let $\epsilon=1$. Draw the graph of $y=x^2$. Looking at the graph, you should be able to see that no matter how small $\delta$ is, you can find $x$ and $y$ within $\delta$ of each other (by taking $x$ big), with $|x^2-y^2|>1$.

$f$ is continuous over $[0,\infty)$ but not uniformly continuous over $[0,\infty)$.


An interactive version of the diagram below can be found here (click on the green point and drag).

$\delta$ is fixed below. Note that as $x$ gets larger, $\Delta f$ grows larger. enter image description here

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Nice answer. I would add that in order to gain a better understanding, one should try to apply the same reasoning to a function like $g(x)=\sqrt x$ which is not uniformly continuous on any interval $[0,a]$ with $a>0$, and see where it fails. –  Julián Aguirre Feb 6 '12 at 16:34
    
@DavidMitra, thanks a lot for your response. At the end though, when you mentioned how $f(x) = x^2$ is not uniformly continuous on $[0,\infty]$ I drew the graph like you said, and didn't succeed in making that closing point of understanding. What does it mean taking x big? Are both x and y "x-values" ? What is the significance of $\epsilon=1$ or is that arbitrary? –  nofe Feb 6 '12 at 17:31
    
@nofe $\epsilon=1$ is fixed, but was arbitrarily chosen. In the diagram above, note that, "going off to the right", you can find points $x$ and $x+\delta$ with $\Delta f$ big. And, you can do this no matter how small the fixed $\delta$ is. –  David Mitra Feb 6 '12 at 18:21
    
@nofe The idea of uniform continuity in there is that an arbitrary $\delta$ won't satisfy the realtion everywhere. (however for a closed finite interval you will find such $\delta$). I reccommend Spivak's chapter on UC, from his Calculus. He explains this topic really well. –  Pedro Tamaroff Feb 6 '12 at 19:36
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