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This is how far I know:

$f$ has a continuous extension to $[0,1]$ if and only if $f$ is uniformly continuous on $(0,1)$.

If $f$ has a continuous extension $G:[0,1]\to[0,1]$, then by compactness of $[0,1]$ we have the uniform continuity of $G$. Since every restriction of a uniformly continuous function is uniformly continuous (If $|x−y|<\delta$ implies $|G(x)−G(y)|<\varepsilon$, then the same implication holds for the restrictions of $F$), this implies the uniform continuity of $f$. And if $f$ is uniformly continuous, then the limits of $f(x)$ at $0$ and at $1$ exist, and

\begin{equation*} G(x)= \begin{cases} \lim_{y\to0} f(y) & \text{if } x=0, \\ f(x) & \text{if } 0 < x < 1, \\ \lim_{y\to1} f(y) & \text{if } x=1. \end{cases} \end{equation*}

is the unique (why?) continuous extension of $f$ to $[0,1]$. Once the existence of the limits is established (how?), the continuity of $G$ at $0$ and at $1$ follows directly from the definitions, and the continuity of $G$ at all points of $(0,1)$ is just the assumed continuity of $f$.

So my problem is that it is already established in the question that $f$ is uniformly continuous on $(0,1)$, which means I can find some sort of an extension to $[0,1]$. But what about a uniformly continuous extension to all of $\Bbb R$?

I am looking for a complete proof of the question.

Sangchul Lee
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  • Good question, though it could benefit from some MathJax formatting :) – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jul 05 '19 at 08:21
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    The reason why it can be extended to $[0,1]$ is fully explained here: https://math.stackexchange.com/questions/245237/extension-of-a-uniformly-continuous-function-between-metric-spaces This extension is unique. As for why it can be extended to whole $\mathbb{R}$: simply define $f(x)=f(0)$ for $x<0$ and $f(x)=f(1)$ for $x>1$. This extension is not unique, that claim is false. For example consider $\sin(x)$ and the "constant outside of $[0,1]$" variant of $\sin(x)$ I've just proposed. – freakish Jul 05 '19 at 08:23
  • There is a unique uniformly continuous extension to $[0,1]$ (basically for the reasons you note although it needs more rigor) whereas there is definitely not a unique uniformly continuous extension to all of $\mathbb R$. The simplest way to obtain such an extension is by making the function take the constant value $f(0)$ on $(-\infty,0)$ and $f(1)$ on $(1,\infty)$. – pre-kidney Jul 05 '19 at 08:23

1 Answers1

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Uniqueness of extension is not true. For example, if $f(x)=x$ for all $x \in (0,1)$ the we can define two extensions $g$ and $h$ to $\mathbb R$ as follows: $g(x)=x$ for all $x$, $h(x)=x$ for $0\leq x \leq 1$, $h(x) =0$ for $x <0$ and $h(x) =1$ for $x >1$. Both of these are uniformly continuous extensions to $\mathbb R$.

Existence: once you extend $f$ to $[0,1]$ you can simply define $f(x)=f(0)$ for $x <0$ and $f(x)=f(1)$ for $x>1$.

Hints for extending to $[0,1]$: let $x_n \to 0$ Then $f(x_n)$ is Cauchy so it has a limit $l$. If $y_n$ is another sequence converging to $0$ then $(x_1,y_1,x_2,y_2,...)$ tends to $0$ so $(f(x_1),f(y_1),f(x_2),f(y_2),...)$ is convegent. This implies that the odd and even terms of the sequence converge to the same limit. Hence $l$ is independent of $(x_n)$. We can define $f(0)$ as $l$. I leave the rest to you.

Kavi Rama Murthy
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