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. $\endgroup$ Oct 19, 2014 at 0:36
  • $\begingroup$ sometimes to show that $f$ is uniformly continuous is easier to show that for any $x_n-y_n\to 0\Rightarrow f(x_n)-f(y_n)\to 0$ $\endgroup$ Oct 19, 2014 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, 2014 at 2:46
  • $\begingroup$ It is equivalence to uniformly continuous @Zslice $\endgroup$ Oct 19, 2014 at 2:50

3 Answers 3


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 several 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 differentiable with $f'$ continuous, and if $\sup |f'(x)| < \infty$ on the domain of definition, then $f$ will be Lipschitz and, in particular, uniformly continuous;

  • if $I$ and $J$ are intervals such that $I \cup J$ is also an interval, and if the restrictions $f|_I : I \to \mathbb C$ and $f|_J : J \to \mathbb C$ are uniformly continuous, then $f : I \cup J \to \mathbb C$ is also uniformly continuous;

  • if $I \subseteq J$ are intervals, and if $f : J \to \mathbb C$ is uniformly continuous, then $f|_I : I \to \mathbb C$ is also uniformly continuous.

Let us see all these four statements in action on your example $\sqrt {(\cdot)} : (0, \infty) \to \mathbb R$:

  • using the first statement, $\sqrt {(\cdot)} : [0,1] \to \mathbb R$ is continuous on a compact interval, therefore uniformly continuous;
  • using the second statement, the derivative of $\sqrt {(\cdot)} : (1, \infty) \to \mathbb R$ is $\frac 1 {2 \sqrt {(\cdot)}}$, which on $(1, \infty)$ is bounded by $\frac 1 2$, therefore $\sqrt {(\cdot)} : (1, \infty) \to \mathbb R$ is also uniformly continuous;
  • using the third statement, it follows that $\sqrt {(\cdot)} : [0, \infty) \to \mathbb R$ is uniformly continuous, because $[0,1] \cup (1, \infty) = [0, \infty)$, which is an interval;
  • using the fourth statement, it follows that $\sqrt {(\cdot)} : (0, \infty) \to \mathbb R$ is uniformly continuous, too.

As you can see, the trick was to split the problem into two subproblems: one for which the derivative is unbounded, but then the underlying domain of definition is compact, and one for which the underlying domain of definition is not compact, but then the derivative is bounded, and then combine the two.


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, 2014 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/… $\endgroup$ Oct 19, 2014 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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