# Dense subset and extension of uniformly continuous function

1. Homework Statement .
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.

Office_Shredder
Staff Emeritus
Gold Member
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.

1 person

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}##.

1 person