page 216
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.
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
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.
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.
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
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 .
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.
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.
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
.
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.
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
.
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.
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.
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
.
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
.
In Part a, did you assume that the limit of the sequence belongs to ? If so, go back and do the problem again.
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.
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 .
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.
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
Suppose that is a continuous function on a bounded set . rove that the following two conditions are equivalent:
The function is uniformly continuous on .
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 .
Given that is a function defined on a set of real numbers, prove that the following conditions are equivalent:
The function fails to be uniformly continuous on the set .
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.