I am reviewing uniform continuity of functions to prepare for the upcoming GRE. I thought it would be useful to have a lot of examples/counterexamples with me. So basically I came up with the following list of true or false, but I am still missing some of the examples/counterexamples.

If $f$ is continuous and $\text{dom}(f)$ is compact, then $f$ is uniformly continuous.


If $f$ is continuous and $\text{dom}(f)$ is unbounded, then $f$ cannot be uniformly continuous.

FALSE. $f(x)=x^2$ on $\mathbb{R}$ is not uniformly continuous, but $f(x)=x$ is uniformly continuous.

If $f$ is continuous and $\text{dom}(f)$ is not compact but bounded, then $f$ cannot be uniformly continuous.

Not sure. $f(x)=\frac{1}{x}$ on $(0,1]$ is not uniformly continuous. I remember there's some theorem about "extending" a function but I forgot what it was about.

If $f$ is continuous and has unbounded derivative on $\text{dom}(f)$, then $f$ cannot be uniformly continuous.

I suspect it to be true, but not sure how to show it. First difficulty is that for any $c$, I need to write $|f'(c)|$ in terms of something like $|f(a)-f(b)|$. Mean Value Theorem looks promising, but the direction of implication seems wrong.

  • $\begingroup$ On your second answer, leave out $x^2.$ Your counterexample is simply $x.$ $\endgroup$ – zhw. Sep 5 '16 at 16:46

The last two are false:

For the second-to-last statement, observe that a constant function is uniformly continuous regardless of the domain.

For the last statement, $f(x)=\sqrt{x}$ is uniformly continuous on $(0,1)$ but has an unbounded derivative on this domain.

I believe the "extension" theorem you're thinking of is the fact that a continuous function on $(a,b)$ is uniformly continuous on this interval if and only if it can be extended to a continuous function on $[a,b]$.

  • $\begingroup$ Yeah, I think that's the extension theorem I learned. It's interesting though because I always thought that the reason why $x^2$ is not uniformly continuous is because its derivative grows too fast. But it seems like $\sqrt{x}$ also has unbounded derivative on $(0,1)$ but its fine? $\endgroup$ – 3x89g2 Sep 5 '16 at 3:54
  • $\begingroup$ Yes, it's fine because of the extension theorem. But in fact it even turns out to be uniformly continuous on $[0,\infty)$. $\endgroup$ – carmichael561 Sep 5 '16 at 3:55
  • $\begingroup$ I was thinking of the same counterexample for the last one, but is $\sqrt x$ really uniformly continuous on $(0,1)$? $\endgroup$ – Lentes Sep 5 '16 at 3:55
  • 1
    $\begingroup$ Alternately, note that $|\sqrt{x}-\sqrt{y}|^2\leq |\sqrt{x}-\sqrt{y}|(\sqrt{x}+\sqrt{y})=|x-y|$. $\endgroup$ – carmichael561 Sep 5 '16 at 3:56
  • $\begingroup$ This is a poor question to ask, but then what's the "fundamental difference" between $x^2$ and $\sqrt{x}$? They both are unbounded, they both increase like crazy, just at different points. $\endgroup$ – 3x89g2 Sep 5 '16 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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