This topic concerns a very classical question: *extend of a function between two metric spaces to obtain a new function enjoying certain properties*. I am interested in the following three properties:

- Continuity,
- Uniformly continuity,
- Pointwise equi-continuity, and
- Uniformly equi-continuity.

Throughout this topic, by and we mean metric spaces with metrics and respectively.

**CONTINUITY IS NOT ENOUGH**. Let us consider the first situation where the given function is *only* assumed to be continuous. In this scenario, there is no hope that we can extend such a continuous function to obtain a new continuous function . The following counter-example demonstrates this:

Let and let be any continuous function on such that there is a positive gap between and . For example, we can choose

Since is monotone increasing, we clearly have

Hence any extension of cannot be continuous because will be discontinuous at . Thus, we have just shown that continuity is not enough. For this reason, *we require to be uniformly continuous*.

**SIMPLE OBSERVATIONS**. We start with the following basic results.

Claim 1. Let be uniformly continuous. Then if is a Cauchy sequence in , then is a Cauchy sequence in .

This is a very classical and simple result. The proof simply makes use of the definition of uniform continuity and Cauchy’s sequence. It is interesting to note that if is closed, then we can drop “uniformly”. This is because continuous function on a bounded set is always uniformly continuous.

In reverse, the statement is also true under certain conditions, for example, any real-valued function from a bounded set is uniformly continuous if it sends any Cauchy sequence to a Cauchy sequence. It is worth noting that the boundedness is essential. We do not consider the situation of functions between metric spaces in this post.

Claim 2. Assume that the hypotheses of Claim 1 hold. Then if both and converge to , then

This is also a classical and simple result. The proof goes as follows: we mix the two sequences: which also converges to . From this the two limits and must be equal.

**UNIFORMLY CONTINUOUS EXTENSION**. Since continuity of is not enough, in general, to extend, we are forced to assume a uniform continuity. We hope that this is enough to prove that the extended function is continuous. However, we can do more. The following claim confirms that the extended function, in fact, is also uniformly continuous.

Claim 3. Let be complete and is uniformly continuous. Then there exists a unique continuous extension . Moreover, is also uniformly continuous.

Proof. The proof of Claim 3 goes as follows. Let be arbitrary. Then there exists a sequence converging to , namely, as . Clearly, is a Cauchy sequence in , which by Claim 1, implies that is Cauchy in . Therefore, we can defineThe preceding limit exists because is complete. By Claim 2, the value is independent of the choice of .

Continuity of . We now claim that is continuous. Indeed, fix a point and assume that converges to . It suffices to show that

To see this, given , there is such that

For each with , let be a sequence converging to . One one hand, by the definition of , there holds

On the other hand, given , by Claim 1, there is some such that

which, by triangle inequality, gives

Hence

which implies that

This tells us that

However, from the choice of , we can also estimate

Hence,

Thus we have shown that

The conclusion follows.

Uniqueness of . This is obvious because is dense in .

Uniform continuity of . Let be given and let be such that

whenever . Let be arbitrary such that

By definition, there are two points such that

Notice that

Hence

proving the uniform continuity of .

**UNIFORMLY EQUI-CONTINUOUS EXTENSION**. We now turn out attention to a family of uniformly continuous functions.

Claim 4. Let be complete and is a family of uniformly continuous functions . If the family is uniformly equi-continuous on , then the uniquely extended family of uniformly continuous functions on obtained from Claim 3 is also uniformly equi-continuous.

Proof. Clearly, the unique family consists of uniformly continuous functions on . We need to verify the uniformly equip-continuity of the family. By the definition, it suffices to show that given any , there is some depending only on such thatfor any and for any with

Keep in mind that the family is already uniformly equi-continuous on . Using this fact, let be such that

whenever

Now we let be arbitrary such that

As before, for an arbitrary but fixed function , there are two points such that

We use the sub index to denote the dependence on the function being consider. Notice that

which then implies that

Putting all estimates above together we deduce that

for any , whenever

proving the uniform equi-continuity of .

**AN APPLICATION**. To conclude the note, we consider the well-known Arzelà-Ascoli theorem which states that any pointwise bounded sequence of functions in , where is compact, which is also uniformly equi-continous, is precompact in . There are counter-examples in the literature saying that the compactness of is essential. For example, any translation of a smoothly compactly supported function on would work. We shall, however, demonstrate that we cannot construct any similar counter-example on bounded sets which are not closed.

Since the Arzelà-Ascoli theorem concerns the pointwise boundedness, in the following claim, we address this property through the extending process as before.

Claim 5. Let be complete and is a family of uniformly equi-continuous functions . If the family is pointwise bounded on , then the uniquely extended family obtained from Claim 3 is also pointwise bounded on .

Proof. Let be arbitrary. By definition, we need to prove thatFix any and by the uniform equi-continuity of , we can let be such that

whenever with

By way of contradiction, there is a sequence of such that

Now we choose some in such a way that and fix it. Clearly,

Since the right hand side is bounded, we obtain a contradiction. Hence the extended family is also pointwise bounded.

Combining Claims 4 and 5 we deduce that on any pre-compact subset , a family of functions in which is also pointwise bounded and uniformly equiv-continuous is pre-compact in . Therefore, we are in position to conclude that one cannot construct a counter-example of the well-known Arzelà-Ascoli theorem against the compactness of , however, only on any bounded subset . An unbounded set is the only possibility that we can construct.

I thank Nguyễn Đức Bảo for useful discussion during the preparation of this note.

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