Uniformly continuous functions on $\mathbb{R}$. 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 $\mathbb{R}$, $\exists a,b&gt;0 \in \mathbb{R}$ such that $|f(x)| \leq a|x|+b$ for all $x \in \mathbb{R}$. Can someone please explain why this is?

Regards, MM.
$f$ is uniformly continuous, thus given $\epsilon&gt;0$ there exists $\delta&gt;0$ 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

$$
|f(x)|\leq (\epsilon/\delta)|x|+|f(0)|.$$
$f$ is uniformly continuous, thus given $\epsilon&gt;0$ there exists $\delta&gt;0$ 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

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