Converge of supremum of a uniformly continuous function.

Let $f:[0, \infty) \to \mathbb{R}$ be a uniformly continuous function. Prove that $$\limsup _{x \to \infty} \frac {f(x)}{x}<\infty$$

Since $f$ is uniformly continuous, by definition, for every $\epsilon >0$, there exists $\delta >0$ such that if $|p-q|<\delta$, then $|f(p)-f(q)|<\epsilon$, where $p,q \in [0, \infty)$. This is an exam review problem so a full solution would be appreciated.

1 Answer

Let $\delta$ be such that for any $x, y\in[0,\infty)$ such that $|x-y|\leqslant \delta$, $\left|f(x)-f(y)\right|\leqslant 1$. For $t\in [0,\infty)$, let us denote $[t]$ its floor function. Then for any $x\gt 0$, \begin{align} \left|f\left(x\right)\right|&\leqslant \left|f\left(x\right)-f\left([x /\delta]\delta\right)\right| + \left|f\left([x /\delta]\delta\right)\right|\\ &\leqslant 1+\sum_{i=1}^{[x\delta]}\left| f(i\delta)- f\left((i-1)\delta\right)\right| +\left|f(0)\right|\\ &\leqslant 1 + [x\delta]+\left|f(0)\right| \end{align}
and the result follows.