Let $f: [0,\infty ) \mapsto \mathbb{R}\;$ be a uniformly continuous function.
Show there exist $a,b \; \in \mathbb{R}\;$ such that $\forall x \in dom(f)\; , \left | f(x) \right |\leq ax+b$ .
This seems rather intuitive for me, for example by looking at a linear function such as $x=y$.
However, I can't get the hang of how to prove it.
For all I know, by the raw defenition : $\forall \varepsilon >0, \exists \delta>0\;$ such that for each $x,y \in dom(f)\;$ such that $\left | x-y \right | < \delta \Rightarrow\; \left | f(x)-f(y) \right | < \varepsilon$
I would love to get a hint on how to approach this. thanks
By uniform continuity exists $\delta_0>0$ such that $|f(x)-f(y)|<1,\forall x,y \geq 0:|x-y|<\delta_0$
Choose $0<s<\delta_0$ and take an arbitrary $x \geq 0$ and let $n$ the unique non-negative integer such that $ns \leq x \leq(n+1)s$(this integer is $[\frac{x}{s}]$)
Then $$|f(x)| \leq |f(x)-f(0)|+|f(0)|$$ $$\leq |f(0)-f(s)|+|f(s)-f(2s)|+...+|f(ns)-f(x)|+|f(0)|$$ $$\leq n+1+|f(0)| \leq \frac{1}{s}x+|f(0)|+1$$