I'm having trouble with a real analysis exercise which asks me to prove that a uniformly continuous function on $[0, \infty)$ is somewhat sublinear in the computer science way. At a first glance I thought it was simple, but I can't seem to get it done. Any help would be appreciated. The exact statement is the following:
Suppose $f: [0, \infty] \to \mathbb R$ is uniformly continuous. Show there are constants $c, k \in \mathbb R$ such that $|f(x)| \leq cx + k, ~~\forall x.$
I think I can prove such a statement on the interval $\left[1,\infty\right)$ (or any other interval bounded away from $0$) but it seems unclear to me why such a statement should be true near $0$.
The proof for $\left[1,\infty\right)$:
For some $\epsilon$ there is a $\delta$ such that $\mid x-y\mid\le \delta$ implies $\mid f(x)-f(y)\mid\le \epsilon$.
Now subdivide $\left[0,\infty\right)$ into intervals of length $\delta$.
Then $x\le\delta$ implies $\mid f(x)-f(0)\mid\le\epsilon$, thus $f(x)\le \epsilon +f(0)$. By induction on $k$ you get for $x\in\left[(k-1)\delta,k\delta\right]$ that $f(x)\le k\epsilon +f(0)$.
So for $x\ge \delta$ (i.e., $k\ge 2$) you can use $k\epsilon\le \frac{k}{k-1}\frac{\epsilon}{\delta}x\le 2\frac{\delta}{\epsilon}x$ to conclude $$f(x)\le 2\frac{\epsilon}{\delta}x+f(0).$$
We may have assumed $\delta<1$, so this shows the wanted inequality on the interval $\left[1,\infty\right)\subset \left[\delta,\infty\right)$.