So I understand the definition of uniform continuity, but wanted some suggestions to prove that a function is or isn't uniformly continuous.

I have looked ahead and have seen that if a function is continuous on a compact domain, then the function is also uniformly continuous. But for the purposes of the examples below, I cannot use compactness, or the fact that a function is uniformly continuous if it is continuous and has a bounded derivative.

Here's one example: prove $f(x)= \sqrt{x}$ is uniformly continuous on $(0, \infty)$. I have seen this question asked before on this forum, and the respondents used the traditional definition. Ie. they found a delta such that if $|x-y|<\delta$, then $|\sqrt{x}-\sqrt{y}|< \epsilon$.

I was wondering if I could try this in a slightly different way. Suppose I prove that $f$ is continuous at $x_0$, and the choice of $\delta$ doesn't depend on $x_0$.

After some scratch work, I proved that $f(x)$ is continuous at $x_0$ provided that $|x-x_0|< \min\{1,(\sqrt{x_0+1} + \sqrt{x_0})*\epsilon\}$. Since $x_0$ is always greater than $0$, we can say that $\delta= \min\{1, \epsilon\}$. Because I proved that our choice of $\delta$ doesn't depend on $x_0$, is this a valid proof that $f(x)$ is uniformly continuous?

Here's another example: I want to prove that $f(x)=x^2$ is not uniformly continuous on $(0, \infty)$. After some work, I proved that $f(x)$ is continuous at $x_0$ (in dom $f$), provided that $\delta< min \{1, \frac {\epsilon}{ (2x_0+1)}\}$. Here, since our choice of delta does depend on $x_0$, we can say the function is not uniformly continuous.

So am I approaching these problems correctly? I realize that I have not been using the definition of uniformly continuity. (that if $f$ is uniformly continuous, then if $|x-y|<\delta$, then$|f(x)-f(y)|< \epsilon$.

Also, sorry about the formatting. I realize I should learn proper formatting soon!. Also, I have read the following post regarding this topic: $\sqrt x$ is uniformly continuous

  • 1
    $\begingroup$ Regarding formatting, there's nothing to learn: you literally just need to enclose your math in dollar signs and add backslashes before character sequences like epsilon, delta, and sqrt. $\endgroup$ – Kevin Arlin Oct 19 '14 at 0:36
  • $\begingroup$ sometimes to show that $f$ is uniformly continuous is easier to show that for any $x_n-y_n\to 0\Rightarrow f(x_n)-f(y_n)\to 0$ $\endgroup$ – Jose Antonio Oct 19 '14 at 0:39
  • $\begingroup$ @JoseAntonio how does your statement imply uniform continuity, as opposed to the function $ g(x)= x_n-y_n$ being continuous at x=0 $\endgroup$ – Zslice Oct 19 '14 at 2:46
  • $\begingroup$ It is equivalence to uniformly continuous @Zslice $\endgroup$ – Jose Antonio Oct 19 '14 at 2:50

First of all, you make a mistake many beginners make: when talking about uniform continuity, it is essential to tell on what set you are studying it. For instance, $x^2$ is uniformly continuous on $[0,r]$ for every $r>0$, but is not so on $[0,\infty)$. Similarly, $\sin \frac 1 x$ is uniformly continuous on $[r,\infty)$ for every $r>0$, but is not so on $(0,\infty)$. The domain of definition is crucial.

Second, there are two main alternative techniques to plainly using the definition:

  • the first one is quick: if $f$ is defined on a compact set and is continuous, then it is uniformly continuous

  • the second one is subtler: if $f$ is derivable with $f'$ continuous and you manage to find a $M>0$ such that $\sup |f'(x)| \le M$ on your domain of definition, then $f$ will be Lipschitz and, in particular, uniformly continuous.


This would be a perfectly fine argument, if your $\delta$ were correct. You really are using the definition of uniform continuity, which is precisely that one can pick for any $\epsilon$ a $\delta$ independent of $x_0$. Unfortunately, your $\delta$ is wrong for $\sqrt x$. If $x=10^{-2},y=10^{-4},$ then certainly $|x-y|<10^{-2}$ and $|x-y|<1$, but $|\sqrt(x)-\sqrt(y)|>10^{-2}$. Your numbers are fine on $[1,\infty)$, but that's by far the easier part!

As for $x^2$, all you've done is show that there is some choice of $\delta$ which depends on $x_0$. That's a far cry from showing that every choice of $\delta$ depends on $x_0$, so you have not shown that $x^2$ fails to be uniformly continuous.

  • $\begingroup$ To show $f(x)=x^2$ is not uniformly continuous, I need to fix $/epsilon$ $>0$ and show that for all $/delta$ $>0$, there is a counter example. How do I do that? $\endgroup$ – Zslice Oct 19 '14 at 2:59
  • $\begingroup$ The first step in such a question is to search this site! See the link. Also, you need to use the other kind of slash for your deltas and epsilons. math.stackexchange.com/questions/503093/… $\endgroup$ – Kevin Arlin Oct 19 '14 at 21:06

As you have an answer for the first case, I only give you a sketch for $x^2$.

I leave you to check: $f$ is uniformly continuous iff for all $x_n-y_n\to 0\implies f(x_n)-f(y_n)\to 0$. For the case $x^2$ is as simple as pick two sequences whose difference converges to $0$ but the difference of its images does not.

Consider $x_n=n+1/n$ and $y_n=n$. So $x_n-y_n\to 0$ but $x_n^2-y_n^2=2+1/n^2\not\to0$. Thus $x^2$ is not uniformly continuous.


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