Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Given a uniformly continuous function $f(x)$ on the real numbers $\Bbb R$, then by the definition of uniform continuity this means: for any $\epsilon>0$ there exists $\delta >0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$. My question is: Given $\epsilon_{n} =\frac{1}{3^{n}}$ (for example), and $y_{n}=x_{n}+w_{n}$ so that $|x_{n} -y_{n}|=|w_{n}|$, then we know that there is $\delta_{n}>0$, how can we find the corresponding $\delta_{n}$? I don't know if my question makes sense.

share|cite|improve this question
In your first sentence you have written the definition of uniform continuity, not of continuity. – Mariano Suárez-Alvarez Feb 28 '11 at 3:38
OK, I fixed it. – user7586 Feb 28 '11 at 3:41
Existence does not imply constructibility. Uniform continuity guarantees that $\delta_n$ exists, it does not guarantee that there is an effective way of finding it. – Arturo Magidin Feb 28 '11 at 3:44
What Arturo said is correct, but many functions that are uniformly continuous provide a way to find the corresponding $\delta_n$. Intuitively, there is a maximum first derivative, $L$. Then $\delta_n$ would be $\epsilon_n/L$ but you need to worry about the higher derivatives. Often cutting down $\epsilon$ a little works, as the effect of the higher derivatives falls off faster than the first as $\delta$ goes to 0. If they don't exist, or blow up too badly, you need to work harder. – Ross Millikan Feb 28 '11 at 4:00
@Monica: First: you presented a very general question, with an arbitrary $f$. My comment is simply that there is no way to give an algorithm in such generality (specific functions may have algorithms that tell you how to find $\delta_n$). Second: $\delta_n$ is completely independent of $x_n$, $y_n$, and $w_n$; in the uniformly continuous case, $\delta$ depends only on $\epsilon$, so the points are utterly irrelevant. In fact, there is absolutely no point in mentioning them at all, except to clutter up the question. – Arturo Magidin Feb 28 '11 at 4:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.