Definitions 10/24/17

During the lecture on Tuesday we continued our discussion of uniform continuity by discussing Theorem 4.4.7. This theorem seems as if it will be very useful in the future and thus I would like to dig deeper into it.

1. Preliminaries

First let us recall definitions and theorems that will be mentioned.

Compact Set: a set K \subseteq \mathbb{R} is compact if every sequence x_n \subseteq K has a subsequence in K that converges to a limit l \in K.


Uniform Continuity: A function f:A \rightarrow R is uniformly continuous on A if for every \epsilon > 0 there exists a \delta >0 such that for all x,y \in A, \mid x-y \mid < \delta implies \mid f(x)-f(y) \mid < \epsilon.

Theorem 4.4.5 : A function f:A \rightarrow R is NOT uniformly continuous on A if there exists a \epsilon_0 > 0 and two sequences x_n and y_n in A satisfying \mid x_n-y_n \mid \rightarrow 0 but \mid f(x_n)-f(y_n) \mid \geq \epsilon_0.

2. Uniform Continuity on Compact Sets

Theorem 4.4.7 builds on the topic of compact sets by stating an interesting observation about continuous functions on compact sets.

Theorem 4.4.7: A function that is continuous on a compact set K is uniformly continuous on K.

3. Proof in Steps

While the specifics of this proof were discussed in lecture, it would be valuable to recall the basic steps of the proof.

Step 1: Negation of the Theorem. By negating the entire theorem, we are able to set up a proof by contradiction. The negation is as follows

By assuming these three statement, one can begin a proof BWOC.

Step 2: Implement Theorem 4.4.5. Using this theorem, one can say that for some \epsilon_0 > 0, there exists two sequences x_n and y_n in K satisfying \mid x_n-y_n \mid \rightarrow 0 but \mid f(x_n)-f(y_n) \mid \geq \epsilon_0.

Step 3: Implement the definition of Compact. Since x_n is in set K, then there exists a convergent subsequence x_nk such that x_nk \rightarrow x and x \in K.

Step 4: Implement Algebraic Limit Theorem. By once again using the definition of Compact, we can see that the subsequence y_nk is a convergent subsequence and its limit is an element of set K. However, by implementing the Algebraic Limit Theorem along with the assumption that\mid x_n-y_n \mid \rightarrow 0, one can see that y_nk \rightarrow x.

Step 5: Recall the continuity of f. Since x_nk and y_nk both converge to x and we assumed that f is continuous at x, it must also be the case that lim(f(x_nk)-f(y_nk)) = 0. However, this presents a problem as we assumed that \mid f(x_n)-f(y_n) \mid \geq \epsilon_0 for some \epsilon_0 > 0. With that, there exists a contradiction. Therefore, function that is continuous on a compact set must be uniformly continuous on that set.

4. Conclusion

With Theorem 4.4.7 and its proof presented, one could ask about the importance of this theorem. I would argue that one good example of this was presented in the case of h(x) = \sqrt x being uniformly continuous on the set [0, \infty ) . By splitting the domain into sets [0, 1] and [1, \infty ) , we were able to use Theorem 4.4.7 to quickly show that since h(x) is continuous on the compact set [0, 1], h(x) must also be uniformly continuous on the set. With this proven, we were then left with proving that h(x) is uniformly continuous [1, \infty ) and then going through the different cases regarding what set x and y are in. Remembering this example may be a good idea as implementing this technique of splitting the domain may prove to be useful in future proofs.

Posted on by Jonathan Rodriguez
Categories: Daily Blogs, Definitions

One Response

  1. Jeremy LeCrone says:
    October 26, 2017 at 12:53 pm

    This is a great exploration of this very important result, Jonathan.

    In Step 4, you state that “By once again using” compactness… however, one needs to be slightly more careful than this. Notice that the subsequence you get for (y_nk) is NOT NECESSARILY the same sequence of indices n_1, n_2, n_3, … as those returned for (x_nk)… If these indices do not align, we cannot conclude that lim_{k \to \infty}(x_{nk} – y_{nk}) = 0. So, we actually avoid a second application of compactness, and simply use algebraic limit theorem to conclude (y_{nk}) also converges AND it happens to also converge to x.

    I like the observations you make about the proof structure: Breaking apart domains to build up a proof is indeed a very useful tool to keep in mind; I would further observe that this “Constructive” nature of argument is ubiquitous in the field of analysis. So, very good observation!

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