## Extending Uniformly Continuous Functions

**Point of post: **In this post I will prove a theorem which, informally, says that every uniformly continuous function where are metric spaces, complete, may be extended to a uniformly continuous function . I do this not because this theorem in particularly useful (although, it has its uses) or difficult but because I have not been thrilled with any of the proofs I have seen so far, as far as the explanation of the motivation and technique go. I will forgo the usual structures for proofs for a more didactic approach.

**Theorem: **Let be a subspace of some metric space and a complete metric space. Then, if is a uniformly continuous function there exists a unique extension which is uniformly continuous.

**Motivation**

So what’s the idea? What kind of function can we, in general, construct from this which will be uniformly continuous on . Well, believe it or not the idea is simple enough. Look at the points we want to define the function on, namely the points in . Each of these points, say, has a sequence which converges to it. But, note that in particular each such sequence is Cauchy. But, by the uniform continuity of the sequences is Cauchy. But! Since is complete we know that this sequence converges to some point . So, what if we said that the extension of at is this limit ? This is the general approach. We will define this idea rigorously, show that it’s a well-defined function, prove it’s uniformly continuous, and show that in essence this is the *only *way to do it.

So, you might ask “Why does have to complete? Why can’t just be continuous?” These are legitimate questions. While I can’t claim that there aren’t weaker conditions on and for which this is true (although, the following is related, though more parallel than a generalization viz. Tietze’s Extension Theorem) I can show that we can’t forgo either completely. In particular (consider all domain spaces as subspaces of with the usual metric/topology) we can’t get rid of uniform continuity since the function

is continuous, and has a complete codomain space yet can’t be continuous extended to since we’d have to “define” to be such that

We also can’t forget the completeness of the codomain space. For this consider the map

this is evidently uniformly continuous. But, we can’t extend to a uniformly continuous mapping from otherwise we’d have that the codomain is the continuous image of a compact set and thus compact, but this is impossible. Thus, neither of the two conditions may be entirely removed.

**Proof: **

** ****Defining the function**

*We take the intuitive idea we discussed above and put this into rigorous form exactly what this function should be.*

* * We may clearly assume that is not closed otherwise we’re done. So, let . Then, there exists a sequence in such that . So, clearly since is convergent in it must be Cauchy. Thus, since is continuous we know that is Cauchy. So, since is complete we know that for some . So, let

*Showing that the function is well-defined*

*As I state in the first sentence of the following paragraph it’s not at all clear (initially at least) that is well-defined. In other words, what’s to say we couldn’t take take sequences in converging to some point and get two different values with our “formula”? Saying this one more way, what’s to say that the value at of is independent of the choice of sequence as our formulation of it suggests? This is precisely what we will show below. Namely that even if one starts with two different sequences converging to the “formula” will give the same value for .*

** **It’s not at all apparent that is independent of the choice of sequence . So, suppose that are two sequences in with . Then, using the process described above we see that that each sequence admits limits respectively. We aim to show that . To do this let be arbitrary. Since there exists some such that

.

Similarly, since there exists some such that

Lastly, we note that since is uniformly continuous there exists some such that

and since there exists some such that

So, let then we see that for

and since was arbitrary it follows that and so as required. Thus, the function is well-defined.

*The function is actually uniformly convergent*

*Ok, so we have a mapping . So, now what? We need to show that this mapping is actually uniformly continuous as we promised. Clearly this is non-obviously (initially) either. Every extension need not be continuous. What if we just said that for all and for some fixed . This would admit an extension of , but one which is almost undoubtedly discontinuous.*

We now claim that is uniformly continuous. To see this let be arbitrary. Then, there exists some such that

for all . We claim that is sufficient. To prove this we break it into three cases

*Case 1: *

If this this is the case we’re clearly done by construction.

*Case 2: *. Since we may choose some sequence such that . So, choose such that and (this is possible since we may choose such that since and we may choose such that since . So, we may take ). Then,

and so since . Thus,

*Case 3: *Suppose that . Then, there exists sequences in with . So, note that we may choose such that and . Next note that by construction

and thus

Thus,

Thus, since all the cases have been covered we may conclude definitively that is uniformly continuous.

*Uniqueness*

*So we have proved that admits a uniformly continuous extension which is uniformly continuous. But, we claimed that such an extension is unique. As always, there is no (immediately apparent) reason why we couldn’t have extended this function in two ways, say by a constant difference, and arrive at two distinct uniformly continuous extensions. In fact, we may extend to by declaring that for we have that for some (assuming that , but this is a pretty safe assumption) distinct . Clearly then are two distinct extensions of . But, there is no guarantee (and as we shall show no possible way) that both of these (let alone either of them) are uniformly continuous.*

We shall now prove that such an extension is unique. In fact, we will show that if are as above and is *continuous *there exists a*t most one continuous *extension . To prove this let, be two continuous extensions of . To prove that it suffices to show that for a fixed but arbitrary . Since is dense in there exists a sequence in with . Thus, by continuity.

from where it follows that

*Remark: *This follows the more general principle that there can exist at most one extension from to if is an arbitrary topological space and is Hausdorff. This follows since the ‘agreement’ set is dense (since it contains ) and closed (since where and is the diagonal.

*Conclusion*

We have thus proved that admits a well-defined extension as described in the motivation section, that this extension is uniformly continuous, and that such an extension is unique. Thus, the theorem follows.

**References:**

For a proof see page 78 of the first reference. For a general discussion of the subject matter above as a whole see either reference.

1. Simmons, George Finlay. *Introduction to Topology and Modern Analysis*. Malabar, FL: Krieger Pub., 2003. Print.

2. Rudin, Walter. *Principles of Mathematical Analysis*. New York: McGraw-Hill, 1976. Print.

in the first step “Defining the function”, when you say

“Thus, since f is continuous we know that {f(x_n)}_n is Cauchy”

aren’t you actually using the _uniform_ continuity of f..??

thank you

tom

Comment by july | September 15, 2011 |

Hello Tom! No, that actually isn’t a case. You are probably thinking of the fact that if is uniformly continuous then carries Cauchy sequences to Cauchy sequences, or with some lingo, that every uniformly continuous function is ‘Cauchy continuous’. That said, Cauchy continuity is not EQUIVALENT to uniform continuity. In particular, try proving that if $f:M\to N$ is a continuous map between metric spaces and $M$ is complete try to prove that $f$ is Cauchy complete. Does that help?

Best,

Alex

Comment by Alex Youcis | September 15, 2011 |

Thank you for the kind reply… but I still do not understand…

I agree when you say that “if f is uniformly continuous, then f carries Cauchy sequences to Cauchy sequences” and I still think that this is what you are using to show that {f(x_n)}_n is Cauchy…

uniformly continuous -> cauchy continuous and the converse is false…

but now the question is: does continuity imply cauchy continuity… I think the answer is no…

I do not understand “… to prove that $f$ is Cauchy complete “? do you mean Cauchy continuous?

I am sorry if I took so long… 😦

ttys

Comment by July | November 14, 2011

Tom, you are one-hundred percent correct that and the converse is not true. But, what I am saying is to prove the following theorem:

If this is still not clear, try looking here http://en.wikipedia.org/wiki/Cauchy-continuous_function#Properties

Let me know how it goes!

Best,

Alex

Comment by Alex Youcis | November 15, 2011 |

ok… the thm is trivial, isn’t it? x_n cauchy –> x_n converges by completeness –> f(x_n) converges by continuity –> f(x_n) Cauchy…

but this does not explain why you don’t need _uniform_ continuity when you say “Thus, since f is continuous we know that \{f(x_n)\}_{n\in\mathbb{N}} is Cauchy”… in any case this theorem relies heavily on the completeness of the domain, whereas we have completeness of the codomain in the extension thm….

I still think you must use the full strength of your hypothesis there…. tell why I am wrong…

best

tom

Comment by July | November 18, 2011 |

Oh Tom, I am so sorry. It seems all this time I was misinterpreting your question after all. You are one-hundred percent correct. We need the uniform continuity of on to guarantee that a Cauchy sequence in gets mapped to a Cauchy sequence in .

I must apologize for all of this confusion. I will change the wording of my post right now!

Thanks for catching that! I hope everything is ok now.

Best,

Alex

Comment by Alex Youcis | November 19, 2011 |

ahahah.. now I am a happier person…. :))

Congratulations for the blog, anyways… It’s very interesting… I will be around from time to time 😉

best,

tom

Comment by July | November 20, 2011 |

Haha! Glad to hear it my friend, and thanks!

Best,

Alex

Comment by Alex Youcis | November 20, 2011 |