# Unexpectedly uniformly continuous functions

The other day in a exam, I was given the following exercise:

Given $f : [0,1] \to \mathbb{R}$ continuous and such that $f(0) = 0, f(1) = 1$, let $g : \mathbb{R} \to \mathbb{R}$ be $g(x) = [x] + f(x - [x])$. Prove that $g$ is uniformly continuous.

I'm looking for more examples of this kind of exercise to practice with (i.e. functions with uniform continuity that are not as straightforward to prove that they are).

## 1 Answer

Any continuous function $f$ from a closed interval $[a, b]$ to $[a, b]$ is uniformly continuous. See https://en.wikipedia.org/wiki/Heine–Cantor_theorem.

If $f(a) = a$ and $f(b) = b$ then extending $f$ to a function $\Bbb{R} \to \Bbb{R}$ by taking $f(x) = x$ for $x \not\in [a, b]$ still gives a uniformly continuous function.

If you want practice on proving uniform continuity from first principles, you can create your own examples from this.

• This is a well known result. I'm looking for practical examples. Notice that in the example, the theorem can not be applied. Also, properties about sum and composition of uniformly continuous functions do not apply either. – Misguided Dec 2 '15 at 0:45
• The theorem can be applied: you prove $g$ is continuous on $[0, 1]$ and conclude that it is uniformly continuous on $[0, 1]$. – Rob Arthan Dec 2 '15 at 1:06
• You are supposed to prove that it is uniformly continuous on $\mathbb{R}$, which is not compact. – Misguided Dec 2 '15 at 1:10
• @Julián: I have revised my answer to address your comments. – Rob Arthan Dec 2 '15 at 1:19