![]() 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. Putting all estimates above together we deduce thatĪN APPLICATION. We use the sub index to denote the dependence on the function being consider. Using this fact, let be such thatĪs before, for an arbitrary but fixed function, there are two points such that Keep in mind that the family is already uniformly equi-continuous on. By the definition, it suffices to show that given any, there is some depending only on such that We need to verify the uniformly equip-continuity of the family. Clearly, the unique family consists of uniformly continuous functions on. 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. ![]() Let be complete and is a family of uniformly continuous functions. We now turn out attention to a family of uniformly continuous functions.Ĭlaim 4. Let be given and let be such thatīy definition, there are two points such that However, from the choice of, we can also estimate On the other hand, given, by Claim 1, there is some such that One one hand, by the definition of, there holds It suffices to show thatįor each with, let be a sequence converging to. Indeed, fix a point and assume that converges to. By Claim 2, the value is independent of the choice of. ![]() The preceding limit exists because is complete. Clearly, is a Cauchy sequence in, which by Claim 1, implies that is Cauchy in. Then there exists a sequence converging to, namely, as. Then there exists a unique continuous extension. Let be complete and is uniformly continuous. The following claim confirms that the extended function, in fact, is also uniformly continuous.Ĭlaim 3. ![]() We hope that this is enough to prove that the extended function is continuous. Since continuity of is not enough, in general, to extend, we are forced to assume a uniform continuity. From this the two limits and must be equal. The proof goes as follows: we mix the two sequences: which also converges to. This is also a classical and simple result. Assume that the hypotheses of Claim 1 hold. We do not consider the situation of functions between metric spaces in this post.Ĭlaim 2. It is worth noting that the boundedness is essential. 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. This is because continuous function on a bounded set is always uniformly continuous. It is interesting to note that if is closed, then we can drop “uniformly”. The proof simply makes use of the definition of uniform continuity and Cauchy’s sequence. This is a very classical and simple result. Then if is a Cauchy sequence in, then is a Cauchy sequence in. We start with the following basic results.Ĭlaim 1. For this reason, we require to be uniformly continuous. Thus, we have just shown that continuity is not enough. Hence any extension of cannot be continuous because will be discontinuous at. Since is monotone increasing, we clearly have Let and let be any continuous function on such that there is a positive gap between and. The following counter-example demonstrates this: In this scenario, there is no hope that we can extend such a continuous function to obtain a new continuous function. Let us consider the first situation where the given function is only assumed to be continuous. Throughout this topic, by and we mean metric spaces with metrics and respectively.ĬONTINUITY IS NOT ENOUGH. I am interested in the following three properties: This topic concerns a very classical question: extend of a function between two metric spaces to obtain a new function enjoying certain properties.
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