# another uniform continuity problem (is this a “so simple” so that it seems complex scenario?)

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:

$f(x) = x^2$ 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

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.

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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