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

Possible Duplicate:
Is a uniformly continuous function vanishing at $0$ bounded by $a|x|+c$?

On the wikipedia page for Modulus of Continuity, it states that for $f$ uniformly continuous on $\mathbb{R}$, $\exists a,b>0 \in \mathbb{R}$ such that $|f(x)| \leq a|x|+b$ for all $x \in \mathbb{R}$. Can someone please explain why this is?

Regards, MM.

share|cite|improve this question

marked as duplicate by Jonas Meyer, Davide Giraudo, David Mitra, Srivatsan, Asaf Karagila Jan 29 '12 at 21:36

This question was marked as an exact duplicate of an existing question.

@JonasMeyer: Yes, that proof can be adapted accordingly I think. Shall I remove this post? – Mathmo Jan 29 '12 at 21:29
up vote -1 down vote accepted

$f$ is uniformly continuous, thus given $\epsilon>0$ there exists $\delta>0$ such that $|x-y|\leq \delta$ yields $|f(x)-f(y)|\leq \epsilon$. By chosing any sequence $(x_k)_{k=0}^N$ such that $x_0=0$, $x_N=x$ and $|x_{k+1}-x_k|\leq \delta$, we have
$$ |f(x)-f(0)|\leq \sum_{k=0}^{N-1} |f(x_{k+1})-f(x_k)|\leq N\epsilon. $$ But since $|x|/\delta\geq N$, you indeed obtain

$$ |f(x)|\leq (\epsilon/\delta)|x|+|f(0)|.$$

share|cite|improve this answer
Oops, I've seen the Jonas Meyer's comment after posting. – Student Jan 29 '12 at 21:37
There is a mistake. It is not always possible to take $b=|f(0)|$. For example, consider $f(x)=\sqrt[3]{x}$, which is uniformly continuous, but for which $|f(x)|\leq a|x|$ is impossible. I suggest revisiting the assumption that $|x|/\delta\geq N$. – Jonas Meyer Jan 29 '12 at 22:12

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