# Uniform continuity on an open interval?

Suppose I want to check if $$f(x)$$ is uniform continuous on a bounded interval $$I$$ (for eg open interval $$(0,1)$$), given that it is continuous on $$I$$. How do I do that?

My approach: Take $$\bar{I}$$, then two case can happen:

Case I: If I can continuously extend the function, then $$f(x)$$ is uniformly continuous on $$I$$.

Case II: If I cannot extend the function continuously, then two sub cases are possible

Subcase II a: $$f(x)$$ is tends to an infinite limit i.e. it shoots up/down arbitrarily for eg functions like $$\frac{1}{x}$$. In which case I conclude that $$f$$ is not uniformly continuous on $$I$$.

Subcase II b: $$f(x)$$ doesn't have a limit i.e. function of the type sin$$\frac{1}{x}$$. In this case as well $$f(x)$$ is not uniform continuous on $$I$$.

So is my above classification of continuous function sufficient to determine which functions are uniform continuous and which are not? So far it had worked well for me.

• A function continuous function $f:(0,1)\rightarrow \Bbb R$ can be extended to a continuous function $\tilde f$ on $[0,1]$ if and only if $f$ is uniformly continuous on $(0,1)$. – Sumanta Das Dec 6 '18 at 12:14
• @UserS I was looking for a statement like that. Can you give me a specific source for that theorem? – henceproved Dec 6 '18 at 12:16

Yes, that is correct. In fact, assuming that the domain of $$f$$ is $$(a,b)$$:

1. If both limits $$\lim_{x\to a^+}f(x)$$ and $$\lim_{x\to b^-}f(x)$$ exist, then $$f$$ is uniformly continuous, because you can define$$\begin{array}{rccc}F\colon&[a,b]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}\lim_{x\to a^+}f(x)&\text{ if }x=a\\f(x)&\text{ if }x\in(a,b)\\\lim_{x\to b^-}f(x)&\text{ if }x=b.\end{cases}\end{array}$$Then $$F$$ is continuous and, since its domain is a closed and bounded interval, $$F$$ is unifomly continuous. In particular, $$f$$ is uniformly continuous.
2. If the limit $$\lim_{x\to a^+}f(x)$$ doesn't exist, then $$f$$ cannot be uniformly continuous because then either $$\lim_{x\to a^+}\bigl\lvert f(x)\bigr\rvert=+\infty$$ or there will two real numbers $$m$$ and $$M$$, with $$m, such that the inequalities $$f(x)>M$$ and $$f(x) will occur for values of $$x$$ arbitrarily close to $$a$$. It is easy to prove that each possibility is incompatible with the fact that $$f$$ is uniformly continuous.
3. The case in which the limit $$\lim_{x\to b^-}f(x)$$ doesn't exist is similar.

Let $$f:(0,1)\rightarrow \Bbb R$$ be uniformly continuous and consider a sequence $$\{x_n\}$$ converging to $$0$$ , so that $$\{x_n\}$$ is cauchy. Now a uniformly continuous function sends Cauchy sequence to Cauchy sequence i.e. $$\{f(x_n)\}$$ is also cauchy , hence $$\{f(x_n)\}$$ converges to some limit $$l\in \Bbb R$$. Again using uniform continuity you can show this limit is independent of choice of sequence converging to $$0$$ i.e. both $$x_n,y_n$$ converges to $$0$$ implies that $$|x_n-y_n|$$ can be made arbitrarily small for large $$n$$, so by uniform continuity $$|f(x_n)-f(y_n)|$$ can be made arbitrarily small for large $$n$$ and since both $$\{f(x_n)\},\{f(y_n)\}$$ are convergent ,the converges same limit. So that we can extend $$f$$ by defining $$f(0)=l$$. In a similar manner one can assign a value for $$f$$ on the point $$1$$. This gives a continuous extension of $$f$$ on $$[0,1]$$.

For the converse , let $$\tilde f$$ be a continuous extension of $$f$$ on $$[0,1$$, then $$\tilde f$$ is uniformly continuous as $$[0,1]$$ is compact , hence $$f$$ , being a restriction of $$\tilde f$$ is uniformly continuous on $$(0,1)$$.

More general statement--- Let $$D$$ be a dense subset of $$[0,1]$$ and $$f:D\rightarrow \Bbb R$$ is uniformly continuous then $$f$$ can be extended to whole $$[0,1]$$ continuously.

• I added these facts as the user henceproved asked for a source. – Sumanta Das Dec 6 '18 at 13:07