# Show that there exist positive constants $a$, $b$ such that $|f(x)| \leq a |x| + b$ for every $x \in \mathbb{R}$.

Problem

For a uniformly continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$, show that there exist positive constants $a$, $b$ such that $|f(x)| \leq a |x| + b$ for every $x \in \mathbb{R}$.

Progress

This problem is confusing me greatly.

Suppose $f(0)=\alpha$. Then define $g(x)=f(x)-\alpha$ so we have $g(0)=0$.

$g$ is uniformly continuous on $\mathbb{R}$ also, and I feel it will be of more use to us, but I really have no idea where to go with this. Any help would be very appreciated.

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Do you know the definition of uniform continuity? – Eckhard Jan 22 '13 at 18:09
@Eckhard Yes, though I can't see how it would be applied in this context... – Mathmo Jan 22 '13 at 18:10

By uniform continuity there exists $\delta>0$ such that $$|x-y|\le\delta\implies|f(x)-f(y)|\le1.$$ If $x\in[0,\delta]$ then $$|f(x)-f(0)|\le1\implies|f(x)|\le1+|f(0)|.$$ If $x\in[\delta,2\,\delta]$ then $$|f(x)-f(\delta)|\le1\implies|f(x)|\le1+|f(\delta)|\le2+|f(0)|.$$ Use induction to prove that if $x\in[(n-1)\delta,n\,\delta]$ then $$|f(x)|\le n+|f(0)|\le \frac{x}{\delta}+1+|f(0)|.$$ A similar argument applies to the case $x<0$.