Show that there are $ a,b \geq 0 $ so that $ |f(x)| \leq ax+b, \forall x \geq 0.$

I have the following exercise: $$f:[0, +\infty) \rightarrow \mathbb{R} \text{ uniformly continuous } .$$ $$\text{Show that there are } a,b \geq 0 \text{ so that } |f(x)| \leq ax+b, \forall x \geq 0.$$ $$$$ $f:[0, +\infty) \rightarrow \mathbb{R} \text{ uniformly continuous } :$ $\forall \epsilon >0 \exists \delta >0 \text{ such that } \forall x,y \in [0, + \infty) \text{ with } |x-y|< \delta \Rightarrow |f(x)-f(y)|<\epsilon$ $$$$ How can I continue??

Solution 1:

Fix some $\epsilon_0>0$, and consider the corresponding $\delta_0$.

Now, for any $x>0$ express $f(x)$ as $$f(x)-f(x-\delta_0)+f(x-\delta_0)-f(x-2\delta_0)+f(x-2 \delta_0)-+ \dots -f(x-N \delta_0)+f(x-N \delta_0)-f(0)+f(0) $$

Where $N$ is the largest integer for which $x-N \delta_0$ remains positive. Next take the absolute value of the above, and using the triangle inequality break it down to pairs. To finish use the uniform continuity for each pair.

In more detail: It is easy to see that $$N=\lfloor x/\delta_0 \rfloor \leq x/\delta_0.$$ Thus $$\vert f(x) \rvert= \left| \left(\sum_{n=1}^{N} f(x-(n-1) \delta_0)-f(x-n \delta_0) \right)+f(x-N \delta_0)-f(0)+f(0) \right| \leq \\ \sum_{n=1}^N \left|f(x-(n-1) \delta_0)-f(x-n \delta_0) \right| +|f(x-N \delta_0)-f(0)|+|f(0)| \leq \\ N \epsilon_0+\epsilon_0+|f(0)| \leq \frac{\epsilon_0}{\delta_0} x+(\epsilon_0+|f(0)|). $$ We find that $a=\epsilon_0/\delta_0$ and $b=\epsilon_0+|f(0)|$ get the job done.