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