Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm starting out university math and I'm struggling with understanding how to prove uniform continuity. I think I understand the concept of finding a $|x-x_0|<\delta$ for $|f(x)-f(x_0)|<\epsilon$ but all the examples I have found so far have been very vague in explaining how they relate to each other.

I have figured out that the smallest $\delta$ has something to do with the steepest part of the $f()$ function in such way that if $\delta$ satisfies $\epsilon$ in the steepest climb or descent, it will satisfy it everywhere else too.

But the problem I'm facing is that I don't always understand how I'm supposed to figure out the relation between these two variables.

I am able to solve an example like $f(x) = 5x+8$ like so: $x \geq 0, x=x_0+\delta, |f(x)-f(x_0)| = |5(x_0+\delta)+8-5x_0-8| = 5|\delta|$ and thus $5|\delta|<\epsilon$ so the solution is $\delta < \frac{\epsilon}{5}$. This seems easy and reasonable.

Here is an example that I can't crack: $f:[0,\infty[\rightarrow\mathbb{R}, f(x)=x^2$

So what I did first was define $x_0 \geq 0, x=x_0+\delta$

Then I wrote $|f(x)-f(x_0)|<\epsilon$ where $|f(x)-f(x_0)|$ is $|(x_0+\delta)^2 - x_0^2| = |x_0^2 + 2x_0\delta + \delta^2 - x_0^2| = |2x_0\delta + \delta^2|$

At this point many of the examples on the net are saying that I can break this in two parts, $|2x_0\delta| < \frac{\epsilon}{2}$ and $|\delta^2| < \frac{\epsilon}{2}$ and solve them separately. So I get $\delta < \sqrt{\frac{\epsilon}{2}}$ and $\delta < \frac{\epsilon}{4x_0}$

So what am I supposed to do with these two deltas I got? And why am I supposed to break it in parts? Shouldn't I get a single value for the $\delta$?

I understand that because $x^2$ grows at an increasing speed, no $\delta$ can satisfy all $\epsilon$ (and thus it's not uniformly continuous). But I don't know how I'm supposed to get there.

Also, if i confine the $f(x)=x^2$ to $f:[0,5]\rightarrow\mathbb{R}$, how can I then show that it's uniformly continuous?

Many of the documents i've found by googling "uniform continuity" seem to take shortcuts and I get lost.

If someone can explain this in a "layman way" clearly I would be very grateful!

share|improve this question
3  
+1 for explaining all you have done. –  user17762 Jun 8 '11 at 22:15
1  
I'll try to write something up tonight, if there are no further answers than ncmathsadist's. –  Michael Chen Jun 9 '11 at 1:08
    
Do you understand how to find an expression for delta, given epsilon and $x$, in non-uniform continuity? –  Michael Chen Jun 9 '11 at 11:54

3 Answers 3

Here is another big cache of special cases. Suppose a function $f$ is differentiable an interval and that its derivative is bounded. If $$M = \sup |f'|,$$ then for any $x, y$ in the interval we have $$|f(x) - f(y)| \le M|x - y|.$$
So if you specify an $\epsilon > 0$, $\delta = \epsilon / M$ works for the definition of continuity.

More generally, if $f$ is continuous on a closed bounded interval, it is uniformly continuous there. This is a consequence of the Heine-Borel theorem.

share|improve this answer
1  
So this sounds like the thing I wrote in my post: the $\delta$ that applies to all $\epsilon$ is related to the supremum of $f$'s derivative. Unfortunately this definition does not explain how I can find the $\delta$ for some equation (like in my post). :( –  supset Jun 8 '11 at 21:54
2  
Observe that what you're actually proving is that a differentiable function with bounded derivative is Lipschitz-continuous (and that Lipschitz-continuity implies uniform continuity -- see en.wikipedia.org/wiki/Lipschitz_continuity ) –  lentic catachresis Jun 8 '11 at 22:09

Let's start with showing continuity for $f(x)=x^2$, and then trying to show uniform continuity.

With thanks to Theo Buehler, you want to find an expression for $\delta$ such that for all $h$ where $|h|<\delta$, $|f(x_0) - f(x_0+h)| < \epsilon$. This way $x_0+h$ stands for all values that are within $\delta$ of $x_0$.

What you are doing for $f(x) = x^2$ is a valid way of doing things: with $x_0 \geq 0$, you find that $|f(x_0) - f(x_0 + h)| = |2x_0h + h^2|$. The goal is to make this value less than epsilon.

To do this, we can use the triangle inequality: in general, $|a| + |b| \geq |a+b|$. Set $a := 2x_0h$ and $b := h^2$, and you'll have $|2x_0h + h^2| \leq |2x_0h| + |h^2|$.

If $|2x_0h| + |h^2| < \epsilon$, then $|2x_0h + h^2|$ is less than epsilon. A quick way to make this true is to set $|2x_0h| < \epsilon/2$ and $|h^2| = \delta^2 < \epsilon/2$. If both of parts of are less than $\epsilon/2$, then their sum will be less than $\epsilon$, and you can trace the argument back to say that $|f(x_0 + h) - f(x_0)|<\epsilon$.

Take the statements and rearrange them to get $|h| < \frac{\epsilon}{4|x_0|}$ and $|h| < \sqrt{\epsilon/2}$. To make both of these things true, let $|h|$ be less than the smaller of the two bounds, $|h| < \min\left(\frac{\epsilon}{4|x_0|},\sqrt{\epsilon/2}\right)$. Set the right hand side of the inequality to be your $\delta$. Then for $|h| < \delta$, $|f(x_0+h) - f(x_0)|<\epsilon$, which is the definition of (general, not uniform) continuity.

(If $x_0 = 0$, then $|2x_0h|=0$, so you only need to make $|h| < \sqrt{\epsilon/2}$.

To understand why this implies continuity but not uniform continuity, review your statement that $|f(x_0) - f(x_0 + h)| = |2x_0h + h^2|$. You want to find a delta such that the expression $|f(x_0) - f(x_0 + h)|$ is bounded by epsilon for all $x_0$ and $|h| < \delta$. But what happens as you make $x_0$ bigger? No matter what $\delta > 0$ you try, I can take $h = \delta/2$, and $x_0=2\frac{\epsilon}{\delta}-\frac{\delta}{4}$ to make $|f(x_0) - f(x_0 + \delta)| = |2x_0h + h^2| = 2\epsilon$, which exceeds the epsilon bound. In other words, the delta that was found in the above paragraph is dependent on $x_0$. If you make $x_0$ larger, then $|f(x_0) - f(x_0 + h)|$ will also tend to infinity.

In general, if you are proving (general or uniform) continuity from the definition, you are trying to manipulate inequalities to find $\delta$ in terms of $\epsilon$ and $x_0$. It can seem a bit counter-intuitive, but it gets easier with practice.

For your instance of restricting $f(x)=x^2$ to the set $[0,5]$, you can appeal to the Heine-Cantor theorem, which states that in metric spaces, continuous functions whose domain is a compact set are uniformly continuous.

You can also use the machinery that you created above. Given an $x_0 \neq 0$, $\delta = \min\left(\frac{\epsilon}{4|x_0|},\sqrt{\frac{\epsilon}{2}}\right)$, and for $x_0 = 0$, $\delta = \sqrt{\frac{\epsilon}{2}}$. On $[0,5]$, you can minimize $\frac{\epsilon}{4|x_0|}$ by maximizing $x_0$, so let $x_0 = 5$. Therefore you can find a delta that works for all $x_0 \in [0,5]$ by setting $\delta := \min \left(\frac{\epsilon}{20},\sqrt{\frac{\epsilon}{2}}\right)$.

share|improve this answer
    
Here's one thing that's bothering me a bit in your answer and the question. The reasoning is easy to make correct but it could be made much cleaner. In general, it doesn't suffice to just estimate $|f(x_0 + \delta) - f(x_0)|$ but rather you need to consider $|f(x_0 + h) - f(x_0)|$ for all $h$ with $|h| \lt \delta$ (and all $x_0$ in the case of uniform continuity). I'm sure this is clear to you, but I'm not sure that this really transpires from your answer. –  t.b. Jun 9 '11 at 11:58
    
@Theo Buehler: Right, sorry. I have been accustomed to take that shortcut, but your way is better. I've reworked my answer to make that true. –  Michael Chen Jun 9 '11 at 14:01
    
That's much nicer, thank you. This should be really helpful. –  t.b. Jun 9 '11 at 14:09

Firstly, you must understand that uniform continuity, unlike continuity, is a global condition on the function on its domain. That is, given an $\epsilon>0$ , there exists a $\delta>0$ which only depends on that $\epsilon$ so for any pair of points $x$ and $y$ in the domain, if $|x-y|<\delta$ then $|f(x)-f(y)|<\varepsilon$. An equivalent way to look at this -- that brings this global-ness more to the front -- is to see that, if two sequences (need not be bounded or anything) in the domain: $\langle x_n \rangle$ and $\langle y_n \rangle$ are equivalent (that is, the limit of their difference tends to $0$), then so is $\langle f(x_n) \rangle$ and $\langle f(y_n) \rangle$. An advantage of invoking sequential characterizations of certain properties of functions, is that they are easier to prove than resorting to epsilonics.

Now consider $f(x)=x^2$ on $\mathbb R$, it's easy to see that $\langle n \rangle$ and $\langle n+\frac{1}{n} \rangle$ are quivalent. But, when we apply $f$, we get the sequences $\langle n^2 \rangle$ and $\langle n^2+\frac{1}{n^2}+2 \rangle$ and their difference tends to $2$, that is, they are not equivalent.

Also, remember it helps to understand certain facts such as:

  1. A continuous function on a compact interval is uniformly continuous.
  2. A continuous function thats lipschitz is uniformly continuous.
  3. If a function is uniformly continuous, it'll map a cauchy sequence to a cauchy sequence.
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.