Exercises on Uniform Continuity

page 216

1. Is it true that if is an unbounded set of real numbers and for every number then the function fails to be uniformly continuous?
The assertion given here is false. Every function defined on the set of integers must be uniformly continuous.

2. Given that prove that is continuous but not uniformly continuous on the set .
Since the number does not belong to the domain of , the function is constant in a neighborhood of every number in its domain. Therefore is continuous on the set . To see why fails to be uniformly continuous we observe that given any positive number we can find a number and a number such that , and for any such choice of numbers and we must have

3. Given that for all real numbers prove that is not uniformly continuous on the set .

We defineandfor every positive integer . Since and for every we know that does not approach as . Nowas and so it follows from the relationship between limits of sequences and uniform continuity that the function fails to be uniformly continuous.

4. Ask Scientific Notebook to make some 2D plots of the function defined by the equation for . Plot the function on each of the intervals , , and . Revise your plot and increase its sample size if it appears to contain errors. Why do these graphs suggest that fails to be unformly continuous on the interval ? Prove that this function does, indeed, fail to be uniformly continuous.

Solution: To prove that fails to be uniformly continuous we shall show that for every number there exist two positive numbers and such that and . We begin by choosing a number such that whenever we haveNow choose a positive integer such thatSince for sufficiently large , we can use the Bolzano intermediate value theorem to choose a number such that . Now since we can use the Bolzano intermediate value theorem again to choose a number such thatWe now observe thatand so the proof is complete.

1. A function is said to be Lipschitzian on a set if there exists a number such that the inequality holds for all numbers and in . Prove that every Lipschitzian function is uniformly continuous.
Suppose that is a function defined on a set , that is a positive number and that the inequalityholds for all numbers and in the set . Suppose that . We define and observe that, whenever and belong to and we have

2. Given that for all prove that is uniformly continuous but not lipschitzian on .

Solution: The fact that is uniformly continuous on the closed bounded set follows at once from the fact that is continuous there. Now, to prove that fails to be Lipschitzian, suppose that is any positive number. Given we see thatand this exceeds whenever .

1. Suppose that is uniformly continuous on a set that is a sequence in the set and that has a partial limit Prove that it is impossible to have as .
Using the fact that is uniformly continuous on we choose a number such that the inequality holds whenever and belong to and . Since is a partial limit of the sequence we know that the conditionholds for infinitely many positive integers . Choose a positive integer such thatFor every one of the infinitely many positive integers for which the conditionholds, since we haveTherefore the sequence of numbers is frequently in the interval and so it cannot approach .
Of course, this sequence cannot approach either.

2. Did you assume that in Part a? If you did, go back and do the exercise again. You have no information that . If you didn't assume you can sit this question out.

3. Suppose that is uniformly continuous on a bounded set and that is a sequence in Prove that it is impossible to have as .
The assertion follows at once from Part a and the fact that every bounded sequence of numbers has a partial limit in .

4. Prove that if is uniformly continuous on a bounded set then the function is bounded.

Solution: To obtain a contradiction, assume that is unbounded above. Choose a sequence in such that for each . Now use part c of this exercise.
Well, of course, such a choice of would make which we know now to be impossible.

1. Given that is a set of real numbers, that and that for all , prove that is continuous on but not uniformly continuous.
Hint: Use the preceding exercise. Choose a sequence in that converges to . Note that as .

2. Given that is a set of real numbers and that fails to be closed, prove that there exists a continuous function on that fails to be uniformly continuous on .
Choose a number and use Part a.

3. Is it true that if is an unbounded set of real numbers then there exists a continuous function on that fails to be uniformly continuous on ?
No, as we remarked after Exercise 1, every function on the set of integers must be uniformly continuous.

5. Is it true that the composition of a uniformly continuous function with a uniformly continuous function is uniformly continuous?
Yes, this assertion is true. Suppose that is a uniformly continuous function on a set and that is a uniformly continuous function on a set that includes the range of . To show that the function is uniformly continuous on , suppose that .
Using the uniform continuity of on the set we choose a number such that the inequality holds whenever and belong to and . Now, using the uniform continuity of on the set we choose a positive number such that the inequality will hold whenever and belong to and . Then whenever and belong to and we have .

1. Suppose that is uniformly continuous on a set and that is a convergent sequence in . Prove that the sequence cannot have more than one partial limit.
We know from the result proved in Exercise 6 d that the sequence is bounded. We choose a partial limit of and we want to prove that is the only partial limit of . Suppose that is any number unequal to . We define Choose a number such that the inequalityholds whenever and belong to and .
We write the limit of as and choose an integer such that the inequality holds whenever . Using the fact that is a partial limit of the sequence we choose an integer such that . Now given any integer , sincewe must haveand consequently

Thus if , the number cannot lie in the neighborhood of and we conclude that fails to be a partial limit of the sequence .

2. In Part a, did you assume that the limit of the sequence belongs to ? If so, go back and do the problem again.

3. Prove that if is uniformly continuous on a set and is a convergent sequence in then the sequence is also convergent. Do not assume that the limit of belongs to .
In view of Part a, the present result follows at once from an earlier theorem on limits of sequences.

4. Suppose that is uniformly continuous on a set that is a real number and that and are sequences in that converge to the number Prove that The existence of these limits was guaranteed in Part c. Now since as we deduce from the relationship between uniform continuity and limits of sequences that as .

5. Suppose that is uniformly continuous on a set and that . Explain how we can use Part d to extend the definition of the function to the number in such a way that is continuous on the set .
We know that there exists a sequence in that converges to and we know that there is a single limit for all of the sequences that can be made in this way. We define to be this common limit value. This extension of the function to the set is uniformly continuous. The proof will be given in the more extended case that we consider below in Part f.

6. Prove that if is uniformly continuous on a set then it is possible to extend to the closure of in such a way that is uniformly continuous on .
For every number we define by the method described in Part e. To show that the extension of to is uniformly continuous, suppose that . Using the uniform continuity of on we choose such that the inequality holds whenever and belong to and . We shall now observe that whenever and belong to and we must have . To make this observation, suppose that and belong to and that .
choose a sequence in that converges to and a sequence in that converges to . Sincewe know that the inequality must hold for all sufficiently large and therefore, sinceand sincefor all sufficiently large we must have

6. Suppose that is a continuous function on a bounded set . rove that the following two conditions are equivalent:

1. The function is uniformly continuous on .

2. It is possible to extend to a continuous function on the set .

The fact that condition a implies condition b follows from Exercise 9. On the other hand, if has a continuous extension to the set then, this extension, being a continuous function on a closed bounded set, must be uniformly continuous; and so must be uniformly continuous on .

7. Given that is a function defined on a set of real numbers, prove that the following conditions are equivalent:

1. The function fails to be uniformly continuous on the set .

2. There exists a number and there exist two sequences and in such that as andfor every .

At the suggestion of my good friend Sean Ellermeyer this exercise was upgraded to a theorem. I have left in the exercise. Sometimes I find it interesting to see which of my students recognize that an item is the same as one they have already seen.