Uniform continuous

Let $f:[0, +\infty) \rightarrow \mathbb{R}$ be a continuous function such that for each $h>0$ it holds

$\lim_{x \rightarrow +\infty} \left( f(x+h) - f(x) \right)=0$

Prove that $f$ is uniformly continuous.

Solution

Firstly, let us define the following set:

$S_{\varepsilon,X} = \{t \in \mathbb R_{\geq 0} : \text{if }x\geq X\text{ then }|f(x+t)-f(x)| \leq \varepsilon\}$

By hypothesis, for any fixed $\varepsilon$ , we have

$\bigcup_{X\in\mathbb R_{\geq 0}}S_{\varepsilon,X} = \mathbb R_{\geq 0}$

Note that each set $S_{\varepsilon,X}$ is closed because for a fixed $x$ , the set of values of $t$ such that $|f(x+t)-f(x)|\leq \varepsilon$ is closed and $S_{\varepsilon,X}$ is an intersection of these closed sets over all $x\geq X$ .

Note that we could also say that

$\bigcup_{X\in\mathbb Z_{\geq 0}}S_{\varepsilon,X} = \mathbb R_{\geq 0}$

since the sets $S_{\varepsilon,X}$ increase with $X$ – giving a countable union of closed sets whose union is the whole space.

We can then apply the Baire Category Theorem to say that since a countable union of closed sets has non-empty interior, some element of the union must have an interior! In particular, for any $\varepsilon>0$ , there must be some $X$ such that some interval $(a,b)$ is a subset of $S_{\varepsilon,X}$ . However, then if we have that $x,y \geq X + a$ and $|x-y| < |b-a|$ we could choose some pair $a',b'\in (a,b)$ with $x-a' = y-b'$ and then observe that

$f(x)=f(x-a' + a')$

$f(y)=f(y-b' + b')$

Then the distance from $f(x)$ to $f(x-a')$ is at most $\varepsilon$ as is the distance from $f(y)$ to $f(y-b')$ since $a',b'\in \subseteq S_{\varepsilon,X}$ . Thus we find that if $x,y \geq X+a$ and $|x-y| < |b-a|$ we have $|f(x)-f(y)| \leq 2\varepsilon$ – and this works out for any choice of $\varepsilon>0$ . This fact suffices to establish that $f$ is uniformly continuous with a small bit of further work.