# Homework Help: Dense subset and extension of uniformly continuous function

1. Sep 30, 2013

### mahler1

1. The problem statement, all variables and given/known data.
Let $(X,d)$ be a metric space, $D \subset X$ a dense subset, and $f: D→ℝ$ a uniformly continuous function. Prove that f has a unique extension to all $X$.

3. The attempt at a solution.

I have some ideas but not the complete proof. If $x \in D$, then I define $f'(x)=f(x)$, so let $x \in X \setminus D$. $D$ is a dense subset, so there exists $(x_n)_{n≥1}$ $\subset D$ such that $x_n→x$. I define $f'(x)=\lim_{n\rightarrow +\infty} {x_n}$. Observe that this limit makes sense since $(f(x_n)_n≥1$ is a Cauchy sequence (to prove this one has to use the fact that f is uniformly continuous) in $ℝ$. $ℝ$ is complete, so the sequence is convergent. Then I should check that $f'$ is well defined, in other words, if there are two different sequences $(x_n)_{n≥1}$ and $(y_n)_{n≥1}$in $D$ converging to a point $x$, then it must be $\lim_{n\rightarrow +\infty} {x_n}$=$\lim_{n\rightarrow +\infty} {y_n}$ (this is easy to prove). Now I want to prove that f' is uniformly continuous, this means that for every $ε>0$ $\exists δ_ε>0$:
$d(x,y)<δ_ε$ implies $|f'(x)-f'(y)|<ε$. Here I got totally stuck. This is an exercise belonging to the topic of compact metric spaces so I suppose I have to use something about compact spaces, the problem is that $ℝ$ is not compact, so I don't know where should I use some information about compactness.

2. Sep 30, 2013

### Office_Shredder

Staff Emeritus
You passed over the key point of existence, which is that for f' to be well-defined, you need
$$\lim_{n\to \infty} f(x_n) = \lim_{n\to \infty} f(y_n)$$.

And I don't understand what this has to do with compact metric spaces, the uniform continuity of f' will come from f regardless of what kind of space X is.

3. Oct 1, 2013

### brmath

A function can be uniformly continuous on all of R -- consider for example f(x) = x. So you don't need to worry about compactness. But you do need to focus on what uniformly continuous means.

With ordinary continuity people say : for every $\epsilon$ you can find a $\delta$ blah, blah. A lot of fuss is made about the fact that the $\delta$ depends on $\epsilon$. What no one EVER says, is that $\delta$ also depends on x. So if you look at a function like f(x) = 1/x on (0,1), it is continuous there, but $\delta$ gets smaller and smaller as you approach zero -- that is, it depends on x.

Uniform continuity means that the $\delta$ does NOT depend on x. That's it. 1/x will never qualify on (0,1).

Now take your extension and compare it to what you know about the original function.

By the way, I wish you had called the extension something other than $f' \text{ like } f^+\text{ or} f_{ext}$.