Bounding uniformly continuous functions

Question

Show that if $f \colon \mathbb{R} \to \mathbb{R}$ is uniformly continuous, then there is an $L > 0$ such that $$|f(x)| \leq L \cdot (1 + |x|).$$

I'm quite stuck here. Any hints would be much appreciated.

Answer 1

Suppose $f$ is uniformly continuous, and take $\varepsilon=1$. There exists some $\delta>0$ such that, for all $x,y\in\mathbb{R}$, $$|f(x)-f(y)| \leq 1$$ whenever $|x-y|\leq \delta$.

Fix any $x\geq0$ (the negative case is identical), and let $n = \left\lfloor\frac{x}{\delta}\right\rfloor$, so that $|x-n\delta| \leq \delta$. Then $$\begin{align*} |f(x)|&\leq |f(0)|+|f(x)-f(0)| \leq |f(0)|+|f(x)-f(n\delta)|+\sum_{k=0}^{n-1}|f((k+1)\delta)-f(k\delta)|\\ &\leq|f(0)| + 1+n \end{align*}$$ But we know that $n\leq \frac{x}{\delta}$, so we get $$\begin{align*} |f(x)|&\leq|f(0)| + 1+ \frac{x}{\delta} \leq \max(1/\delta, |f(0)| + 1)\cdot (1+x) \end{align*}$$ and we can just take $L=\max(1/\delta, |f(0)| + 1)$ to conclude.