If $f(x)$ is uniformly continuous for $x \ge 0$, then it is bounded above by a straight line.

Question

Explicitly: If $f(x)$ is uniformly continuous in $[0,+\infty)$, then there exists positive numbers $a, b$ such that $f(x) \lt ax+b$ for any $x \ge 0$.

Answer 1

We write the definition of the uniform continuity:

$$\text{for}\; \epsilon=1\quad \exists\alpha>0\colon |x-y|\le\alpha \Rightarrow|f(x)-f(y)|<1$$

Now if $x\in[0,\alpha]$ we have $f(x)<1+f(0)$ and if $x\in [\alpha,2\alpha]$ we have $f(x)\le f(\alpha)+1<f(0)+2$ and so on by induction we find: if $x\in [n\alpha,(n+1)\alpha]$ we have $f(x)<f(0)+n+1$ but $x\ge n\alpha$ so $\frac 1 \alpha x\ge n$ and then $$f(x)<n+f(0)+1\le \frac 1 \alpha x +(f(0)+1)$$ so take $a=\frac 1 \alpha$ and $b=f(0)+1$ and you have the desired result.