200
5.5
5.5.1
CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION
Deeper Properties of Continuous Functions
Intermediate Value Theorem and Consequences
When one studies a function, one is usually interested in the values the function
can have. In other words, one is interested in the range of the function. More
precisely, if f is a function with domain D, one tries to answer questions of the
type:
1. What kind of a set is the range R of f ?
2. Under which conditions is the range of f bounded?
3. Under which conditions is it closed?
4. Does f have a minimum? a maximum?
You should recognize some of these questions. You studied them in Calculus
I. We answer some of these questions in this section. We begin by stating and
proving a very important theorem: the intermediate value theorem. Intuitively,
this theorem says that the idea of a continuous function is that its graph can
be drawn without lifting the pencil. In other words, its graph has no holes or
breaks. Thus, it takes on all values between any two it achieves. More precisely:
Theorem 5.5.1 (Intermediate Value Theorem) Suppose that f is continuous on the closed interval I = [a; b], let A = f (a), and B = f (b) and suppose
that A 6= B. If C is a number between A and B, then there exists a number c
in (a; b) such that f (c) = C. If in addition f is strictly monotone on I then c
is unique.
Proof. The theorem has two parts: existence of c and uniqueness of c. We
prove each part separately.
Existence of c The reader should draw a picture corresponding to the situation
of the theorem and represent on the picture the various quantities involved
in the proof. Assume all the hypotheses of the theorem are satis…ed. If
A 6= B, then either A > B or A < B. We do the proof in the case A > B.
The case A < B is identical, and is left to the reader. We look at g (x) =
f (x) C. Since f is continuous on [a; b] and C is a constant, g is also
continuous on [a; b]. Furthermore, by our assumption, g (a) > 0. Since
g is continuous, by theorem 5.4.18, g will remain strictly positive to the
right of a. For the same reason, g is strictly negative at b, and therefore
to the left of b. Thus, the set S = fx 2 I such that f (x) C > 0g is not
empty. Furthermore, it is a subset of I and therefore is bounded. Hence S
has a supremum. Let c = sup S. The claim is that c is the number in the
theorem, in other words, f (c) = C. We prove this fact by contradiction.
Assume that f (c) 6= C. Then, either f (c) > C or f (c) < C.
5.5. DEEPER PROPERTIES OF CONTINUOUS FUNCTIONS
201
case 1: f (c) > C. By continuity of f , f (x) will remain strictly
greater than C for a while. In other words, there is a number x1 > c
such that f (x1 ) > C. Thus, x1 2 S and x1 > c. This contradicts the
fact that c is an upper bound for S.
case2: f (c) < C. Again, by continuity of f , f (x) < C to the left of
c. Thus, there exists x1 < c such that f (x1 ) < C. Thus, x1 < c is
an upper bound for S, which contradicts the fact that c = sup S.
Therefore, f (c) = C
Uniqueness of c We do a proof by contradiction. Assume c is not unique; that
is there exists at least two such numbers, call them c1 and c2 . Without loss
of generality, we may assume that c1 < c2 . f is either strictly increasing
or strictly decreasing.
case 1: f is strictly increasing. In this case, f (c1 ) < f (c2 ). Therefore,
we have:
C = f (c1 ) < f (c2 ) = C
which is a contradiction.
case 2: f is strictly decreasing. In this case, f (c1 ) > f (c2 ). Therefore,
we have:
C = f (c1 ) > f (c2 ) = C
which is also a contradiction.
This theorem is very important for both theoretical and practical reasons.
On the theoretical side, as we will see shortly, many important results depend
on this theorem. On the practical side, this theorem can be used to prove that
certain equations have solutions. It can also be used to …nd approximations of
these solutions. In fact, these approximations can be computed with as much
accuracy as one needs. We illustrate this by an example.
Example 5.5.2 Show that x4 + 10x3 207x2 70x + 1400 = 0 has a solution
between 2 and 3. Find an approximation of this solution correct to 2 decimal
places.
Let f (x) = x4 + 10x3 207x2 70x + 1400. Then, f is continuous for all real
numbers, so it is continuous on [2; 3]. In addition, f (2) = 528 and f (3) = 322.
So, 0 is between f (2) and f (3). By the intermediate value theorem, there exists
a number c between 2 and 3 such that f (c) = 0. To …nd an approximation,
we compute f (2:1) ; f (2:2) ; :::; f (2:9) until we …nd two consecutive values for
which f changes sign. In this case, f (2:6) = 40:138 and f (2:7) = 48:056. So,
repeating the same argument as above, we …nd that c is between 2:6 and 2:7.
We repeat the same procedure with f (2:61) ; f (2:62) ; :::; f (2:69). We …nd that
f (2:64) = 5:0656 and f (2:65) = 3:7457. By the same argument as above, c
is between 2:64 and 2:65. So, we know that the 2:6 part is correct. We need
202
CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION
to repeat the procedure one more time since we want the solution correct to
2 decimal places. We compute f (2:641) ; f (2:642) ; :::; f (2:649). We …nd that
f (2:645) = :66205 and f (2:646) = :21917. Therefore, c is between 2:645 and
2:646. Thus, 2:64 is part of the exact answer.
Corollary 5.5.3 Let f be continuous on an interval I (any interval). Then,
the range of f is also an interval.
Proof. See homework.
This theorem tells us that if the domain of f , a continuous function, consists
of one piece, so will its range. Some of the problems below ask the reader to
sketch some graphs relating a function and its range. The reader should do them
before reading further, they will help in the understanding of these concepts.
Note in particular that the theorem does not say that the range and the domain
are the same type of intervals. One could be closed, the other could be open.
One could be bounded, the other not.
In the previous theorem, I was any interval. When we study f on a closed
interval, we can reach even more powerful conclusions, as the next theorems
show.
Corollary 5.5.4 Let f be continuous on a closed interval [a; b]. Then, the range
of f is bounded.
Proof. See homework.
Theorem 5.5.5 Let f be de…ned and continuous on [a; b]. Then, f attains the
supremum and the in…mum of its range.
Proof. We show that f attains the supremum of its range. Let R denote the
range of f . By the previous theorem, we know R is bounded, thus bounded above.
Since it is not empty (f (a) 2 R), it has a supremum. Let C = sup R. We need
to show there exists c 2 [a; b] such that f (c) = C. De…ne g (x) = C f (x) for
x 2 [a; b]. Then, either g (x) = 0 for some x in [a; b] or g (x) > 0 for all x in
[a; b].
case 1: g (x) = 0 for some x in [a; b]. Then, we are done.
case 2: g (x) > 0 for all x in [a; b]. We show this cannot happen. Suppose it
did. Then, g would be continuous and non zero on [a; b] (why?). Therefore,
1
would also be continuous on [a; b] (why?). Thus, by the
the function
g (x)
1
previous theorem,
would be bounded on [a; b]. Because C = sup R,
g
given > 0, one can …nd x 2 [a; b] such that 0 < g (x) < (why?), thus
1
1
1
> . It follows that cannot be bounded, which is a contradiction.
g (x)
g
Corollary 5.5.6 Let f be de…ned and continuous on [a; b], then the range of f
is a closed interval.
Proof. See exercises at the end of the section
5.5. DEEPER PROPERTIES OF CONTINUOUS FUNCTIONS
203
Corollary 5.5.7 Let f be de…ned and continuous on [a; b]. Then, there exists
x1 and x2 in [a; b] such that for all x 2 [a; b], we have:
f (x1 )
f (x)
f (x2 )
Proof. See exercises at the end of the section
We have seen some interesting properties of continuous functions. First, we
saw that if I is an interval and f is a continuous function, then f (I) is also
an interval. Thus, a continuous function preserves intervals. We also saw that
the image of a closed interval was a closed interval. More precisely, f ([a; b]) =
[m; M ] where m = inf (f ([a; b])) ; M = sup (f ([a; b])). We might ask ourselves
what other properties of sets are preserved by continuous functions. These are
important questions that have been asked by mathematicians. If you want to
know some of these answers, keep taking math courses!
5.5.2
Topological Interpretation of Continuity
If you have taken a topology class before real analysis, you studied a di¤erent
de…nition of continuity. In the theorem below, when we use the word continuous, we mean it according to the de…nition given a few sections above. The
second part of the statement in the theorem is what is used for the de…nition of
continuity in topology. The proof of the theorem shows the two de…nitions are
equivalent.
Theorem 5.5.8 Let E be a subset of R and f : E ! R a function. Then:
f continuous () f
1
(V ) open in E for every open subset V of R
Proof. We outline the proof and leave the details as homework.
Suppose f is continuous on E. Let V
R be open, show that f 1 (V )
1
is open. Pick a 2 f (V ), show that a is an interior point that is there
exists > 0 such that (a
;a + )
f 1 (V ). But f (a) 2 V which is
open, so there exists > 0 such that (f (a)
; f (a) + )
V . Since f
is continuous, for each , there exists a . See how the you get from
continuity can be used to show (a
; a + ) f 1 (V ).
Suppose f 1 (V ) open in E for every open subset V of R, show f is continuous on E. We need to show f is continuous at every point of E. The
steps are similar to those of the other direction.
5.5.3
Exercises
x4 + 2x5 + 5 x6 + 2x4 + 6
+
= 0 has a solution between 1 and
x 1
x 7
7. Approximate the solution correct to 2 decimal places.
1. Prove that
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CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION
2. Same question as the previous problem for g (x) =
between 1 and 3.
5
x
1
+
7
x
2
+
16
x
3
3. Prove corollary 5.5.3.
4. Prove corollary 5.5.4.
5. In the proof of theorem 5.5.5 there are three statements you were asked to
justify. They are the statements with (why?). Justify these statements.
6. Prove corollary 5.5.6.
7. Prove corollary 5.5.7.
8. Sketch the graph of a continuous function on an open interval whose range
is also an open interval.
9. Sketch the graph of a continuous function on an open interval whose range
is a closed interval.
10. Sketch the graph of a function on an interval of the form [a; 1) whose
range is a closed interval.
11. Illustrate with an example the fact that if f is not continuous at a point,
then its range may not be an interval, even if its domain is a closed interval.
12. Prove that if f : [a; b] ! [a; b] is continuous then there exists x 2 [a; b]
such that f (x) = x.
13. Finish proving theorem 5.5.8.
5.5. DEEPER PROPERTIES OF CONTINUOUS FUNCTIONS
5.5.4
205
Hints for the Exercises
x4 + 2x5 + 5 x6 + 2x4 + 6
+
= 0 has a solution between 1 and
x 1
x 7
7. Find an approximation of this solution correct to 2 decimal places.
Hint: As in the example in the notes, you will have to use the intermediate
value theorem. However, you will have to be a little more creative as the
expression on the left of the equal sign is not de…ned at 1 and 7.
Answer: The solution correct to 2 decimal places is: 3:24
1. Prove that
2. Same question as the previous problem for g (x) =
between 1 and 3.
Hint: Same as previous problem.
5
x
1
+
7
x
2
+
16
x
3
3. Prove corollary 5.5.3.
Hint: Use the de…nition of an interval and the intermediate value theorem.
4. Prove corollary 5.5.4.
Hint: We do the proof in several steps, outlined here.
(a) Justify why f is bounded on some neighborhood of a.
(b) De…ne S = fc 2 [a; b] such that f is bounded on [a; c)g. Prove that
sup S exists using the completeness axiom.
(c) Let d = sup S. Prove by contradiction that d = b.
(d) Explain why part c proves the result.
5. In the proof of theorem 5.5.5 there are three statements you were asked to
justify. They are the statements with (why?). Justify these statements.
Hint: none
6. Prove corollary 5.5.6.
Hint: Use the theorems of this section.
7. Prove corollary 5.5.7.
Hint: Use the theorems of this section.
8. Sketch the graph of a continuous function on an open interval whose range
is also an open interval.
Hint: none.
9. Sketch the graph of a continuous function on an open interval whose range
is a closed interval.
Hint: none.
10. Sketch the graph of a function on an interval of the form [a; 1) whose
range is a closed interval.
Hint: none.
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CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION
11. Illustrate with an example the fact that if f is not continuous at a point,
then its range may not be an interval, even if its domain is a closed interval.
Hint: none.
12. Prove that if f : [a; b] ! [a; b] is continuous then there exists x 2 [a; b]
such that f (x) = x.
Hint: De…ne g (x) = f (x) x and apply the intermediate value theorem
to g.
13. Finish proving theorem 5.5.8.
Hint: See the hints in the notes.
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Bartle, G. Robert, The elements of Real Analysis, Second Edition, John
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[C2] Courant, R., Di¤ erential and Integral Calculus, Second Edition, Volume 2,
Wiley, 1937.
[DS] Dangello, Frank & Seyfried Michael, Introductory Real Analysis, Houghton
Mi- in, 2000.
[F]
Fulks, Watson, Advanced Calculus, an Introduction to Analysis, Third Edition, John Wiley & Sons, 1978.
[GN] Gaskill, F. Herbert & Narayanaswami P. P., Elements of Real Analysis,
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[LL] Lewin, J. & Lewin, M., An Introduction to Mathematical Analysis, Second
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459