# 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??

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.

• $$|f(x)| \leq |f(x)-f(x- \delta_0)|+|f(x-\delta_0)-f(x-2 \delta_0)|+...+|f(x-N \delta_0)-f(0)|+|f(0)|$$ So that I can use the uniform continuity, $$|x-(x- \delta_0)|=|(x-\delta_0)-(x-2 \delta_0)|=...=|\delta_0|<\delta$$ So do I have to take $\delta_0 <\delta$ or do I have to take a specific $\delta_0$? – Mary Star Feb 26 '14 at 12:41
• @MaryStar Well, there indeed is a small problem. I thought your definition of uniform continuity allowed for weak inequalities (i.e. $\leq$ rather than $<$). However it is easy to see that both definitions are equivalent, and there is no need to replace $\delta_0$ by a smaller $\delta$. Hope my comment is clear. – user1337 Feb 26 '14 at 12:50
• I am confused why we expressed $f(x)$ in this way...Since $x \geq 0$ we want write all the numbers that are between $x$ and $0$. Is this correct? Or is there an other reason? – Mary Star Feb 26 '14 at 13:13
• @MaryStar $f(x)=f(x)+0+0+0+ \dots + 0$. – user1337 Feb 26 '14 at 13:27
• @MaryStar Because we know something about the difference of the values of $f$ at sufficiently close points. – user1337 Feb 26 '14 at 13:54