# Uniform continuity on (0,1) implies boundedness

I need to prove that if $f: (0,1) \rightarrow \mathbb{R}$ is Uniformly continuous then it is bounded.

Thank you.

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What do you know about uniform continuous functions and continuous functions on compact sets? – Jonas Teuwen Feb 27 '11 at 19:45
I haven't studied yet about compact sets. – user6163 Feb 27 '11 at 19:46
What do you know about continuous functions on closed bounded intervals? Can you extend $f$ continuously to $[0,1]$? – Jonas Teuwen Feb 27 '11 at 19:49

Recall that if $f$ is uniformly continuous, then given $\epsilon>0$ we can find $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$.

To show it is bounded, it doesn't really matter what $\epsilon$ is, so let it be some fixed constant. Let $N=\lfloor\frac{1}{\delta}\rfloor$ and take $$x_1=\delta,\ x_2=2\delta,\dots,\ x_n=n\delta,\dots, x_N=N\delta.$$ Notice every $y\in (0,1)$ satisfies $|y-x_i|<\delta$ for some $i$. Then $|f|$ will be bounded by $$\text{max}_{1\leq i\leq N} \{|f(x_i)|+\epsilon \}$$

which follows by applying the definition of Uniformly Continuous.

Hope that helps,

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You're saying that you're walking on this range, in steps of 1/delta? can you clarify more this sentence Notice every y∈(0,1) satisfies |y−xi|<δ? Thank you so much. – user6163 Feb 27 '11 at 20:35
@Nir: Certainly. Lets find explicitly which $x_i$ to take depending on $y$ (this may be better than the intuitive explanation, I am not sure). Let $y\in(0,1)$. Set $r=\lfloor\frac{y}{\delta}\rfloor$, and then we can write $y=r\delta + e$ where $e<\delta$ (this follows from what the floor function is). Then as $x_r=\delta r$ we see $y-x_r=e$ and hence $|y-x_r|<\delta$ – Eric Naslund Feb 27 '11 at 20:43
Why is $y\in (0,1)$ ? – GinKin Jun 17 '14 at 18:41

Here's one approach. You can use uniform continuity (with $\varepsilon=1$, say) to show that $f$ is bounded on $(0,\delta)$ and $(1-\delta,1)$. You probably already know the theorem that implies that $f$ is bounded on $[\delta,1-\delta]$.

If you don't know about boundedness of continuous functions on $[a,b]$, then what you can do here is cover $(0,1)$ with a finite number of tiny intervals where you know (using uniform continuity) that $f$ can't vary by more than $1$.

Uniformly continuous functions can also be extended to the closure, so an approach that would actually do more would be to show that $\lim_{x\to1}f(x)$ and $\lim_{x\to 0}f(x)$ exist, so that you may consider $f$ to be a restriction of a continuous function on $[0,1]$.

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Hmm, that is what I said in the comments I guess. I did not really want to give the answer. – Jonas Teuwen Feb 27 '11 at 19:59
@Jonas T: You're right, the last paragraph in my answer has the same idea you gave in the comments. There was no intentional overlap, just a delay in finishing the writing. For just this problem, though, that approach would be slightly more work than the others I suggested. – Jonas Meyer Feb 27 '11 at 20:06
Yes, you're right, but the idea is more general I think. If you have the theorem that you have a continuous extension and that a continuous function is bounded on a compact set it is a one line proof. But probably he didn't have the first one. – Jonas Teuwen Feb 27 '11 at 20:18
@Jonas T: Yes, and that is actually the way I thought of it initially. To show that a uniformly continuous function $f$ on a totally bounded metric space is bounded, you can either cover the space with finitely many $\delta$ balls on which $f$ varies by no more than $1$, or you can argue that $f$ can be continuously extended to the completion, which is compact. The first is more direct, but the second is how I tend to think of it. – Jonas Meyer Feb 27 '11 at 20:23