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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.

TRUE

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.

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

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  • $\begingroup$ 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
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    $\begingroup$ 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

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