Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let the metric space $X=(0,\infty)$ and determine whether the following are uniformly continuous on $X$:
(1) $f(x)=\sqrt{x}$
(2) $f(x)=1/x$
(3) $f(x)=\ln(x)$
(4) $f(x)=x\ln(x)$

Since this isn't $\mathbb{R}$ I don't think I can use the usual method: showing that $|f'|$ is bounded.

Any tips on how to solve these problems?

share|cite|improve this question
Are these homework problems? – user38268 Feb 8 '12 at 9:24
Wikipedia says that "Every Lipschitz continuous map between two metric spaces is uniformly continuous. In particular, every function which is differentiable and has bounded derivative is uniformly continuous." – Rudy the Reindeer Feb 8 '12 at 9:29
@BenjaminLim Yes. Matt N. Unfortunately the lecturer specifically pointed out that we cannot use that property since we have not proven it. – Emir Feb 8 '12 at 9:33
In the first three cases the functions are monotone. Therefore (if you are trying to show uniform continuity) it suffices to explicitly compute $f(x+\delta) - f(x)$ and show that it is bounded by a function $g(\delta)$ independent of $x$ with $g(\delta)\to 0$ as $\delta \to 0$. – Willie Wong Feb 8 '12 at 9:47
OTOH, if you don't have bounded derivatives, do you know how to demonstrate a contradiction to uniform continuity? In particular, do you know how to explicitly construct a sequence $(x_n)$ such that the $\delta_n$ neighborhoods whose image are in $\epsilon$ neighborhoods necessarily have $\delta_n \to 0$ as $n\to\infty$? – Willie Wong Feb 8 '12 at 9:50
up vote 3 down vote accepted

You can prove it directly. For example, for (1) use that $|\sqrt{x} - \sqrt{y}| \leq |\sqrt{x} + \sqrt{y}|$:

Then you have $|\sqrt{x} - \sqrt{y}|^2 \leq |\sqrt{x} + \sqrt{y}||\sqrt{x} - \sqrt{y}| = |x - y|$

So for $\delta := \varepsilon^2$ you get $|\sqrt{x} - \sqrt{y}| < \varepsilon$.

share|cite|improve this answer

Your Answer


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.