# If $f$ is continuous on $[0,\infty)$ and differentiable on $(0,\infty)$ and if $lim_{x\to\infty}f'(x)=0$ Then $f$ uniformly continuous on $[0,\infty)$

I got this problem:

Let $f$ be a continuous function on $[0,\infty)$ and differentiable function on $(0,\infty)$ such that $\lim_{x\to\infty}f'(x)=0$.

(1) Prove that for each $0<\epsilon$ there exist $0<M$ such that if $x$ and $y$ are numbers that satisfy the inequality $M<y<x<y+1$, Then $|f(x)-f(y)|<\epsilon$.

(2) Prove that $f$ is uniformly continuous on $[0,\infty)$ using (1).

I managed to prove (1) (by using the mean value theorem for derivatives), But when I tried to prove (2) I got stuck.

Any help on how to prove (2) by using (1) will be appreciated.

## 4 Answers

As for (2), you are almost done. Split $[0,\infty) = [0,M] \cup (M,\infty)$ and recall that $f$ is UC on $[0,M]$ since this is a compact set. So, pick $\epsilon>0$ and its companion $\delta>0$ provided by the UC of $f$ on $[0,M]$. Then, using (1), define $\tilde{\delta}=\min\{\delta,1\}$ and deduce that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\tilde{\delta}$.

Indeed, if $x$, $y \in (M,\infty)$ and $|x-y|<1$, then (1) yields $|f(x)-f(y)|<\epsilon$.

• Bah, you beat be by 11 seconds. – Trevor J Richards Sep 15 '14 at 14:25

I'm not an expert but if M can be any number, and y and x are proven to exist for any number M as y,x > M, then can't you say that given infinitesimally small numbers of M, (and differences between x and y) that all values for f in the first quadrant exist?

I haven't done this type of problem before I think but maybe that will push you in the right direction.

Sketch of a Proof: Fix $\epsilon>0$. Let $M$ be the value guaranteed to exist by part $(1)$. On $[0,M]$, define $\delta_\min(x)$ to be the greatest $\delta>0$ such that if $y\in[0,M]$ and $|y-x|<\delta$, then $|f(y)-f(x)|<\epsilon$. Show that $\delta_\min$ exists on $[0,M]$, and is continuous and non-zero on $[0,M]$. Since $[0,M]$ is compact, this implies that $\Delta:=\displaystyle\min_{x\in[0,M]}(\delta_\min(x))$ exists and is non-zero. This $\Delta$ is the value needed for uniform continuity. That should get you started, but leave plenty of work for the homework!

My approach is similar to that from Simiore or Trevor with one difference: given $\epsilon>0$, bound $|x-y|$ by $1$ for $x,y>M$ and bound $|x-y|$ by $\delta'$ for $x,y\leq M\color{blue}{+2}$. The reason for the $+2$ is it facilitates the second part of the proof after you set $\delta=\min\{1,\delta'\}$. Indeed, suppose that $|x-y|<\delta$, then $$x,y\in [0,M+2]\bigcup(M,\infty)$$ so that either $x$ and $y$ both belong to one of the sets on the RHS above, in which case you are done, or $x$ belongs to one set and $y$ belongs to the other. But this latter scenario is impossible because $|x-y|<1$.