Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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5.6 The Intermediate Value Theorem 133

K. We say a function f : [a, b] → R satisﬁes a Lipschitz condition of order α > 0 if

there is some positive constant M so that

|f(x

) − f(x

)| ≤ M|x

− x

Let Lipα denote the set of all functions satisfying a Lipschitz condition of order α.

(a) Prove that if f ∈ Lip α, then f is uniformly continuous.

(b) Prove that if f ∈ Lip α and α > 1, then f is constant.

belongs to Lipα.

5.6. The Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is another fundamental result often

used in the calculus without proof.

5.6.1. INTERMEDIATE VALUE THEOREM.

If f is a continuous real-valued function on [a, b] with f(a) < 0 < f(b), then there

exists a point c ∈ (a, b) such that f(c) = 0.

f(a)

f(b)

FIGURE 5.8. Applying IVT to a function f : R → R .

Figure 5.8 shows the meaning of this theorem graphically. Like the Extreme

Value Theorem, this result seems intuitively clear. However, it also depends on the

completeness of the real numbers. For example, if our functions were only deﬁned

on the set of rational numbers, then the function f(x) = x

− 2 for x ∈ Q would

satisfy f (1) = −1 < 0 and f(2) = 6 > 0, but f(x) would never take the value 0

for any rational value of x.

PROOF. Deﬁne a subset A of [a, b] by

A = {x ∈ [a, b] : f(x) < 0}.

Since a ∈ A, A is not empty. And since b is an upper bound for A, the Least Upper

Bound Principle allows us to deﬁne c = sup A; and it belongs to [a, b].

134 Functions

We claim that f(c) = 0. First, since c is the least upper bound for A, there is a

sequence (a

) in A such that c −

< a

≤ c. So c = lim

n→∞

. Therefore, since

f(a

) < 0 for all n ≥ 1 and f is continuous at c,

f(c) = lim

n→∞

f(a

) ≤ 0.

This means that c 6= b, and thus c < b. Choose any sequence c < b

≤ b such

that c = lim

n→∞

. Since c is the upper bound for A, it follows that b

6∈ A and so

f(b

) ≥ 0. Consequently,

f(c) = lim

n→∞

f(b

) ≥ 0.

Putting the two inequalities together yields f(c) = 0. ¥

5.6.2. COROLLARY. If f is a continuous real-valued function on [a, b], then

f([a, b]) is a closed interval.

PROOF. The Extreme Value Theorem (Theorem 5.4.4) shows that the range of f is

bounded, and the extrema are attained. Thus there are points c and d in [a, b] such

that

f(c) = m := inf

x∈[a,b]

f(x) and f(d) = M := sup

x∈[a,b]

f(x).

Suppose that c ≤ d (the case c > d may be handled similarly by considering the

function −f). Let y be any value in (m, M). The function g(x) = f(x) − y is

continuous on the interval [c, d]. Moreover,

g(c) = m − y < 0 < M − y = g(d).

Thus by the Intermediate Value Theorem, there is a point x in (c, d) such that

g(x) = 0, and so f(x) = y. This is true for every point y in (m, M), as well as the

two endpoints. Therefore, f ([a, b]) = [m, M]. ¥

A path in S ⊂ R

from a to b, both points in S, is the image of a continuous

function γ from [0, 1] into S such that γ(0) = a and γ(1) = b.

5.6.3. COROLLARY. Suppose that S ⊂ R

and f is a continuous real-valued

function on S. If there is a path from a to b in S and f(a) < 0 < f(b), then there

is a point c on the path so that f(c) = 0.

PROOF. Let γ : [0, 1] → S deﬁne the path from a to b. Consider the continuous

function g = f ◦ γ. Then g(0) < 0 < g(1). By the Intermediate Value Theorem,

there is a point x in (0, 1) so that g(x) = 0. Then c = γ(x) is the desired point. ¥

5.7 Monotone Functions 135

Exercises for Section 5.6

A. (a) Show that there is some x ∈ (0, π/2) so that cosx = x.

(b) Prove that this is the only real solution.

B. How many solutions are there to tanx = x in [0, 11]?

C. Show that 2sinx + 3 cosx = x has three solutions.

D. Show that a polynomial of odd degree has at least one real root.

E. The temperature T (x) at each point x on the surface of Mars (a sphere) is a continuous

function. Show that there is a point x on the surface such that T (x) = T (−x).

HINT: Represent the surface of Mars as {x ∈ R

: kxk = 1}. Consider the function

f(x) = T (x) −T (−x), and compare f(x) and f (−x).

F. Let f be a continuous function from B = {x ∈ R

: kxk ≤ 1}, the closed ball in R

into R. Show that f cannot be one-to-one.

G. If f is a continuous real-valued function on (0, 1), what are the possibilities for the

range of f? Give examples for each possibility, and prove that your list is complete.

H. (a) Show that a continuous function on (−∞, +∞) cannot take everyreal value exactly

twice.

(b) Find a continuous function on (−∞, +∞) that takes every real value exactly three

times.

5.7. Monotone Functions

Since most functions we encounter are increasing or decreasing on intervals

contained in their domain, it makes sense to spend a short time tying down some

of the special properties of functions on such intervals. You have probably seen the

following deﬁnitions in calculus.

5.7.1. DEFINITION. A function f is called increasing on an interval (a, b) if

f(x) ≤ f(y) whenever a < x ≤ y < b. It is strictly increasing on (a, b) if

f(x) < f(y) whenever a < x < y < b. Similarly, we deﬁne decreasing and

strictly decreasing functions. All of these functions are called monotone.

Sometimes, monotone increasing and monotone decreasing are used as syn-

onyms for increasing and decreasing.

5.7.2. PROPOSITION. If f is an increasing function on the interval (a, b), then

the one-sided limits of f exist at each point c ∈ (a, b), and

lim

x→c

−

f(x) = L ≤ f(c) ≤ lim

x→c

f(x) = M.

For decreasing functions, the inequalities are reversed.

136 Functions

PROOF. Consider the case of f increasing. The set F = {f(x) : a < x < c} is a

nonempty set of real numbers bounded above by f(c). Therefore, L = sup

a<x<c

f(x)

is deﬁned by the Least Upper Bound Principle (2.5.3), and L ≤ f (c). We will show

that L is the left limit.

Indeed, let ε > 0. Since L − ε is not an upper bound for F , there is a point x

with a < x

< c such that f(x

) > L −ε. By deﬁnition of the supremum, we have

f(x) ≤ L for all x in (a, c). Hence for all x

< x < c, we have

L − ε < f(x

) ≤ f (x) ≤ L < L + ε.

From the deﬁnition of limit, it now follows that lim

x→c

−

f(x) = L.

The limit from the right and the decreasing case are handled similarly. ¥

Since monotone functions have limits from each side, this restricts the possible

discontinuities is several ways. The ﬁrst thing to observe is that if the function does

not jump at c, it is necessarily continuous there.

5.7.3. COROLLARY. The only type of discontinuity that a monotone function

on an interval can have is a jump discontinuity.

PROOF. Let lim

x→c

−

f(x) = L and lim

x→c

f(x) = M. Since L ≤ f(c) ≤ M, the

equality L = M implies that

lim

x→c

f(x) = L = M = f(c)

and thus f is continuous at c. ¥

5.7.4. COROLLARY. If f is a monotone function on [a, b] and the range of f

intersects every nonempty open interval in [f (a), f (b)], then f is continuous.

PROOF. Suppose that f is increasing and has a jump discontinuity at c. Then

(with notation as previously) the range of f is contained in (−∞, L] ∪ [M, ∞)

with the exception of at most one point, f (c), in between. Thus either (L, f (c)) or

(f(c), M) is a nonempty interval in [f (a), f (b)], which is disjoint from the range

of f. Consequently, if the range of f meets every open interval in [f (a), f(b)], then

f must be continuous. ¥

Here is a stronger conclusion. Recall that a set is countable if it is ﬁnite or can

be written as a list indexed by N (see Appendix 2.8).

5.7.5. THEOREM. A monotone function on [a, b] has at most countably many

discontinuities.

PROOF. Assumethat f is increasing, and let A = f(b)−f (a). If f is discontinuous

at c, deﬁne the jump at c to be J(c) = f(c

) −f(c

−

). This is the length of the gap

(f(c

−

), f(c

)) in the range of f . Since these intervals are disjoint, the sum of all

the jumps is at most A.

5.7 Monotone Functions 137

Let’s estimate how many discontinuities with jump at least 2

−k

are possible. If

there are N

of them, we see that 2

−k

≤ A or N

≤ 2

A. In particular, N

is ﬁnite. Therefore the points with J(c) ≥ 1 can be listed as c

, . . . , c

, where

≤ A. Then the points with 2

−1

≤ J(c) < 1 may be listed as c

, . . . , c

; and

≤ 2A. Continuing in this manner, the discontinuities with J(c) ∈ [2

−k

, 2

1−k

)

may be listed as c

n−1

, . . . , c

with k

≤ 2

A. In this way, we construct a

sequence containing all of the discontinuities of f. Thus the set of discontinuities

is countable. ¥

The relation to inverse functions is important. Recall, from Corollary 1.2.4,

that if f maps a set S one-to-one and onto a set T , then there is a unique function,

denoted f

−1

, mapping T back to S such that f

−1

(f(s)) = s for all s ∈ S and

f(f

−1

(t)) = t for all t ∈ T .

5.7.6. THEOREM. Let f be a continuous strictly increasing function on [a, b].

Then f maps [a, b] one-to-one and onto [f(a), f (b)]. Moreover the inverse function

−1

is also continuous and strictly increasing.

PROOF. Since f is strictly increasing, it is clearly one-to-one. By the Intermediate

Value Theorem (Theorem 5.6.1), the range of f contains [f(a), f(b)]. Again, since

f is monotone, there are no other values in the range. So f maps [a, b] one-to-one

and onto [f(a), f(b)].

Let g be the inverse function. Suppose that f (a) ≤ s < t ≤ f(b), and let

x = g(s) and y = g(t). Then we must have x < y, for if x ≥ y then f (x) ≥ f(y),

contrary to fact. Hence g is strictly increasing. The range of g is [a, b]. Thus by

Corollary 5.7.4, g is continuous. ¥

5.7.7. EXAMPLE. Consider the function tanx. This function has period π and

so clearly is not monotone. However it is monotone on subintervals of its range. A

natural choice is the interval (−

). Note that tanx is strictly increasing on this

interval and that lim

x→−π/2

tanx = −∞ and lim

x→π/2

−

tanx = +∞. So tanx maps

(−

) one-to-one and onto R. The inverse function tan

−1

is the uniquely deﬁned

function that assigns to each real y the value x ∈ (−

), satisfying tanx = y.

Hence lim

y→+∞

tan

−1

(y) =

and lim

y→−∞

tan

−1

(y) = −

. Figures 5.9 and 5.10 give

the graphs of these two functions.

5.7.8. EXAMPLE. We construct a function associated to the Cantor set knownas

the Cantor function. Recall from Example 4.4.8 that the Cantor set is constructed

by successively removing the middle thirds from the unit interval and obtaining the

138 Functions

–10

–4 –2 2 4

FIGURE 5.9. The graph of tan(x).

–1

–4 –2 2 4

FIGURE 5.10. The graph of tan

−1

(x).

set as the intersection of the sets S

obtained from this procedure:

= [0, 1/3] ∪ [2/3, 1],

= [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1],

= [0, 1/27] ∪ [2/27, 1/9] ∪ [2/9, 7/27] ∪ [8/27, 1/3]

∪ [2/3, 19/27] ∪ [20/27, 7/9] ∪ [8/9, 25/27] ∪ [26/27, 1].

We deﬁne a function on [0, 1] as follows. Set f(0) = 0 and f(1) = 1; next set

f(x) = 1/2 on [1/3, 2/3]; then f(x) = 1/4 on [1/9, 2/9] and 3/4 on [7/9, 8/9];

and so on. The road to understanding this function is the ternary (base 3) expansion.

We may write a number in [0, 1] in base 3 as

x = (0.x

. . . )

base 3

k≥1

−k

5.7 Monotone Functions 139

where each x

belongs to {0, 1, 2}. The ﬁrst time a 1 appears as some x

, the point

belongs to one of the 2

− 1 closed intervals on which f is assigned a value of

the form m/2

More precisely, for x = (0.x

. . . x

k−1

k+1

. . . )

base 3

, where x

∈ {0, 2}

for 1 ≤ i < k, we set f(x) = (0.

. . .

k−1

1000. . . )

base 2

. For the remaining

points in C that have a ternary expansion using only 0s and 2s, we have f(x) =

(0.

. . .

. . . )

base 2

. The restriction of f to C was considered in Example 4.4.8 in

order to show that C was very large. Some numbers have two ternary expansions.

Rather than verify that both expansions lead to the same function value (which they

do, as the interested reader can verify), we merely choose the expansion which

contains a 1. This leads to a well-deﬁned function.

FIGURE 5.11. An approximation to the Cantor function.

It is evident from our construction that f is increasing. Moreover, it is con-

stant on each of the intervals removed from C. So the function has a rather ﬂat

appearance. The interesting thing is that f is continuous. To see this, it sufﬁces

by Corollary 5.7.4 to show that the range of f has no gaps coming from a jump

discontinuity. This follows since every number of the form m/2

for 0 ≤ m ≤ 2

belongs to the range of f. Any interval contains many of these points. In particular,

there is no gap in the range of f, and thus f is continuous and maps [0, 1] onto

itself.

By construction, f is constant on each interval component of [0, 1] \ S

. Thus

f is differentiable with derivative 0 on each of these intervals. In Example 4.4.8,

we showed that the lengths of the intervals removed from [0, 1] to form C summed

to 1. So f is continuous on [0, 1] and has derivative 0 at “almost all” points. That

is, f has derivative 0 except on a set of measure zero. Nevertheless, it cannot be

differentiable everywhere because the Mean Value Theorem would then show it

to be constant, which it clearly is not. In fact, it can be shown that f fails to be

differentiable at every point of C.

140 Functions

For each component interval of [0, 1]\C, the two endpoints lie in C and f takes

the same value at both. Thus it follows that f(C) = [0, 1]. Recall the discussion

of cardinality in Appendix 2.8 and at the end of Example 4.4.8. This function f

provides another way to show that the cardinality of C is greater than (and thus the

same as) the real line.

Exercises for Section 5.7

A. Let f and g be decreasing functions deﬁned on R.

(a) Is the composition f(g(x)) monotone?

(b) Is the sum f(x) + g(x) monotone?

B. Show that if f is continuous on [0, 1] and one-to-one, then it is monotone.

C. What is the inverse function of f(x) = x

on (−∞, 0]?

D. (a) Show that the restriction f of cos x to [0, π] has an inverse function, and graph them

both.

(b) Why do we choose the interval [0, π]?

between f

−1

and g

−1

E. Show that the cubic f (x) = ax

+ bx

+ cx + d, with a 6= 0, is one-to-one and thus

has an inverse function if and only if 3ac ≥ b

HINT: Compute the derivative and its discriminant.

F. Verify that the formula for the Cantor function in terms of the ternary expansion yields

the same answer for both expansions of a point x when two expansions exist.

G. Deﬁne f on S = [0, 1] ∪ (2, 3] by f(x) =

(

x for 0 ≤ x ≤ 1

x − 1 for 2 < x ≤ 3.

(a) Show that f is continuous and strictly increasing on S.

(b) Show that f maps S one-to-one and onto [0, 2].

−1

is not continuous.

(d) Why is this not a contradiction to Theorem 5.7.6?

H. For x ∈ [0, 1], express it as a decimal x = x

. . . . Use a ﬁnite decimal

expansion without repeating 9s when there is a choice. Then deﬁne a function f by

f(x) = x

.0x

. . . .

(a) Show that f is strictly increasing.

(b) Compute lim

x→1

−

f(x).

x→a

f(x) = f(a) for 0 ≤ a < 1.

(d) Find all discontinuities of f.

CHAPTER 6

Differentiation and Integration

In this chapter, we examine the mathematical foundations of differentiation and

integration. The theorems of this chapter are useful not only to make calculus work

but also to study functions in many other contexts. We do not spend any time on the

important applications that typically appear in courses devoted to calculus, such as

optimization problems. Rather we will highlight those aspects that either depend

on or apply to results in real analysis.

After developing the basic properties of differentiation, we focus on the most

important approximation theorem of differential calculus, the Mean Value Theo-

rem. Integration, although based on the natural idea of area, requires some care to

deﬁne precisely. We concentrate on the proof of existence of the Riemann integral

and on the appropriately named Fundamental Theorem of Calculus. This theo-

rem, which ties differential and integral calculus together, makes the calculation of

integrals an algorithmic (if sometimes tricky) process.

As we assume that the reader has already seen calculus, we dive right in with

key deﬁnitions, assuming that the motivating examples are familiar.

6.1. Differentiable Functions

6.1.1. DEFINITION. A real-valued function f : (a, b) → R is differentiable at

a point x

∈ (a, b) if

lim

h→0

f(x

+ h) − f(x

)

= lim

→x

f(x) − f(x

)

x − x

exists. In this case, we write f

) for this limit.

If a function is deﬁned on a closed interval [a, b], then we say it is differentiable

at a or b if the appropriate one-sided limit exists. The function f is differentiable

on an interval [a, b] if it is differentiable at each point x

in the interval.

When f is differentiable at x

, we deﬁne the tangent line to f at x

to be the

linear function T (x) = f (x

) + f

)(x −x

). Figure 6.1 shows a typical tangent

line.

141

142 Differentiation and Integration

The phrase linear function has two different meanings. In linear algebra, a

linear function is one satisfying f(αx+βy) = αf(x)+βf(y); in calculus, a linear

function is one whose graph is a line [i.e., a linear function (in the linear algebra

sense) plus a constant]. To keep the language unambiguous, sometimes this second

family of functions are called afﬁne functions, and we do this in studying convexity

(Chapter 16). In this chapter, we use linear function in its usual calculus sense.

y = f(x)

y = T (x)

FIGURE 6.1. The tangent line T to f at a.

An immediate consequence of differentiability is continuity.

6.1.2. PROPOSITION. If f is differentiable at x

, then it is continuous at x

So every differentiable function is continuous.

PROOF. We compute

lim

x→x

f(x) = lim

x→x

f(x

) + (x − x

)

f(x) − f(x

)

x − x

= f (x

) + 0f

) = f (x

Notice that when f is differentiable at x

, the tangent line T passes through

the point (x

, f(x

)) with slope f

). The point of the derivative is that, locally,

the line T has the same slope as the function f . Thus T (x) is the best linear

approximation to f (x) for x very close to x

. Before making this precise, notice

that T is the best approximation to f near x using a ﬁrst order polynomial. In

Chapter 10, we will look at approximations using higher-order polynomials.

6.1.3. LEMMA. Let f be a function on [a, b] that is differentiable at x

. Let T (x)

be the tangent line to f at x

. Then T is the unique linear function with property

that

lim

x→x

f(x) − T (x)

x − x

= 0.