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How to prove: if $f: (a,b) \rightarrow R$ is uniformly continuous, then $f$ can be extended to a continuous function $F: [a,b] \rightarrow R$.

It's suffice to show that $f$ can be extended to a continuous function $G: [a,b) \rightarrow R$.

I saw similar theorem (Tietze extension theorem) in topology books, but I'm not very familiar with lots of topology concepts. Also I've read one proof for this, it uses Cauchy sequence and Cauchy Completeness Theorem, which I'm also not that comfortable with. Can you think of any other proof for this using methods from uniform continuity, limits, etc. in real analysis? Thanks! :)

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Draw a picture. Can you imagine what can go wrong when you extend $f$ to $a$ if $f$ is merely continuous? Can you then think about a limit condition that would make the extension possible? Can you then use uniform continuity to help you verify the limit condition? – Soarer Mar 12 '11 at 0:42
AFAIK, there isn't any way around using completeness (which necessarily involves Cauchy sequences) in this proof. If you replace $\mathbb{R}$ by a non-complete space this fact can cease to be true. But if you have questions about completeness you could post them on this site. – Nate Eldredge Mar 12 '11 at 1:00
Also, the Tietze extension theorem is a red herring here. It's another theorem that gives a condition for a function to have a continuous extension, but it's not a condition that will be useful here. – Nate Eldredge Mar 12 '11 at 1:01
The question had been removed for some reason, so I restored it to an earlier version. – Jonas Meyer Mar 12 '11 at 3:42
Dear Lindsay, please do not keep removing the question. I've rolled back to an earlier version (with the question in it). – Akhil Mathew Mar 15 '11 at 22:17
up vote 5 down vote accepted

There are two parts to this statement. The first half does not require completeness.

Claim If $f:(a,b)\to\mathbb{R}$ is uniformly continuous, $\limsup_{x\to a^+} f \leq \liminf_{x\to a^+} f < \infty$.

Sketch of proof First we show that $\limsup f$ must be bounded. Suppose it were not. Let $\delta$ be some small number such that $\delta < b-a$. Consider $x_0 = a + \delta$, and $y_0 = f(x_0)$. Construct a sequence $(x_i,y_i)$ in the following manner. Pick $x_i < a + \frac12 (x_{i-1} - a)$ such that $y_i = f(x_i) > 2 y_{i-1}$. (Notice that if this sequence terminates after finitely many terms and cannot be continued, you have will have that $f|_{(a,\frac12(x_N + a))}$ is bounded.) However, we have that $|x_{i+1} - x_{i}| < \frac{\delta}{2^i}$ while $|y_{i+1} - y_{i}| > 2^i y_0$. So it is easy to see that by taking $i\nearrow \infty$ you get a contradiction to uniform continuity.

Similarly $f$ must be bounded below. Now suppose the $\limsup f > \liminf f$ are not equal. Then by a similar argument, you can construct an alternating sequence $x_1, x'_1, x_2, x'_2$ such that $2 |x'_i - a| < |x_i - a| < \frac12 |x'_{i-1} - a|$ and $f(x_i) \to \limsup f$ while $f(x'_i) \to\liminf f$. Then you have that for $0 < \epsilon < \limsup f -\liminf f$, for any $\delta$, there exists some $x_i$ such with a point $x'_i$ within $\delta$ of it taking value more than $\epsilon$ away under $f$, contradicting uniform continuity. Q.E.D.

Like Nate said in his comments, now you are required to use the completeness of $\mathbb{R}$. By using the completeness you can say that since $\limsup f \leq \liminf f$, the limit $\lim_{x\to a^+}f$ exists, so setting $f(a) = \lim_{x\to a^+} f$ you get a continuous function.

For illustration, imagine instead of you have the function $f:(0,1)\to \mathbb{R}^2\setminus \{ 0\}$ given by

$$ f(s) = (s \cos s, s\sin s) $$

This function is uniformly continuous on $(0,1)$, but does not have continuous extension to $[0,1)$ with codomain $\mathbb{R}^2\setminus \{ 0\}$.

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Note: without the completeness of $\mathbb{R}$, the notions of $\limsup$ and $\liminf$ does not make rigorous sense. In which case you need to change the statement in the claim to something like, instead of $\limsup \leq \liminf$: for any two decreasing sequences $x_i\to a$ and $x'_i\to a$ with the property that $f(x_i)$ is increasing and $f(x'_i)$ decreasing, one must have $f(x_i)\leq f(x'_j)$ for all $i,j$. – Willie Wong Mar 12 '11 at 2:02
Is there a reason why you write $\limsup \le \liminf$ and not $\limsup = \liminf$? Isn't $\limsup$ always $\ge\liminf$? – joriki Mar 12 '11 at 7:26
@joriki: (1) because in sketching the proof I didn't actually write out that step (and that's with an eye toward the case where $\limsup$ and $\liminf$ don't necessarily exist, in which case it is easier to prove that a maximizing sequence will eventually be smaller than or equal to a minimizing sequence) (2) yes. – Willie Wong Mar 12 '11 at 22:09

If you have Cauchy's criterion for functions at your disposal there is a simpler proof: Given an $\epsilon>0$, by uniform continuity there is a $\delta>0$ with $|f(x)-f(x')|<\epsilon$ for all $x, x'\in\ ]a,a+\delta[\>$. Since $\epsilon$ was arbitrary it follows by Cauchy's criterion that $\lim_{x\to a+} f(x)=:f(a)$ exists, and similarly at $b$.

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could you please explain why $\lim{f(x)}=f(a)$ when $x \rightarrow a+$ without using Cauchy's criterion? I'm not familiar with that. thanks! – Lindsay Duran Mar 17 '11 at 15:51

$\Rightarrow$ Suppose that $f$ is uniformly continuous on $I=(a,b)$. Let $(x_n)$ be a sequence in $(a,b)$ that converges to $a$. Then since $(x_n)$ is a convergent sequence, it must also be a Cauchy sequence. which implies that $(f(x_n))$ is also a Cauchy sequence, and so $(f(x_n))$ must converge in $\mathbb{R}$, that is $\lim _{n\to \infty} f(x_n)=L$ for some $L\in \mathbb{R}$.

Let $y(n)$ be any other sequence converging to $a$. then $\lim _ {n\to \infty} (x(n)-y(n))=0$ So by using uniform continuity of $f$ we have $\lim _{n\to \infty} f(y_n)=\lim [f(y_n)-f(x_n)+f(x_n)]=\lim [f(y_n)-f(x_n)]+\lim f(x_n)=0+L=L$.

So for every sequence $(y_n)$ in $(a,b)$ that converges to $a$, we have that $(f(y_n))$ converges to $L$. Therefore by the Sequential Criterion for Limits, we have that $f$ has the limit L at the point $a$. Therefore, define $f(a)=L$ and so $f$ is continuous at $a$. Using same argument for $b$ we can prove the claim of theorem.

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Welcome to Math.SE! In posting a new answer to such an old Question, one with previously Accepted and upvoted answers, it might be helpful to Readers to highlight any new information being added by your post. – hardmath Jan 9 at 7:36
thank you hardmath – Prajakta Jan 9 at 7:44

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