Schinazi R.B. From Calculus to Analysis

170 5 Continuity, Limits, and Differentiation

We construct a subsequence of b

as follows. Take N = 1 above. There is n

≥ 1

such that |f

−1

) −f

−1

(b)|≥.Next,weuseN =n

+1 to get n

such

that |f

−1

) −f

−1

(b)|≥, and so on. We get a subsequence b

such that



−1

) −f

−1

(b)



≥ for all k ≥1.

That is, for every natural k,wehave

−1

) −f

−1

(b) ≥ or f

−1

) −f

−1

(b) ≤−.

Note that either there are inﬁnitely many k for which the ﬁrst inequality holds or

there are inﬁnitely many k for which the second inequality holds. Assume that the

ﬁrst inequality holds for inﬁnitely many k



(the other possibility is treated in a simi-

lar way). Let b



a subsequence of b

(and a subsubsequence of b

)forwhichwe

have

−1



) −f

−1

(b) ≥ for all k ≥1.

Hence, for all k ≥1,

−1



) ≥f

−1

(b) +>f

−1

(b).

Since f

−1



) and f

−1

(b) are in I (which is the range of f

−1

) and since I is an

interval, f

−1

(b) + is also in I .Iff is strictly increasing, we have



−1



)



≥f



−1

(b) +





−1

(b)



Hence,



≥f



−1

(b) +



>b.

Since b



is a subsequence of b

, it must converge to b. By letting k go to inﬁnity

we have

b ≥f



−1

(b) +



>b.

That is, b>b, a contradiction. If we assume that f is strictly decreasing, we also get

a contradiction. Therefore, if b

converges to b, then f

−1

) converges to f

−1

(b).

That is, f

−1

is continuous at b.

Example 5.30 In this example we show that a discontinuous function may have a

continuous inverse. Deﬁne the function f on I =[0, 3] by

f(x)=x for 0 ≤x ≤2 and f(x)=x +1for2<x≤3.

We start by showing that f is not continuous at 2. Let a

= 2 − 1/n. Then a

converges to 2. For every n ≥1, a

is in I , and a

< 2. Hence, f(a

) = a

, which

converges to 2. Deﬁne now b

=2 +1/n; it also converges to 2 and is in I . Since

> 2, we have f(b

) = b

+1, which converges to 3. Since a

and b

converge

to2butf(a

) and f(b

) converge to different limits, f is not continuous at 2.

It is clear that f is strictly increasing on I . Hence, it is one-to-one and has an

inverse function f

−1

. It is also easy to see that the range of f is J =[0, 2]∪(3, 4]

(note that J is not an interval) and that

−1

(x) =x for 0 ≤x ≤2 and f

−1

(x) =x −1for3<x≤4.

5.4 Intervals, Continuity, and Inverse Functions 171

We now show that f

−1

is continuous on J . We start with x = 2. Assume that a

in J and converges to 2. Take  =1/2 > 0. There exists a natural N such that

−2|< 1/2 for all n ≥N.

Hence, a

< 5/2 for every n ≥ N . Since a

is also in J =[0, 2]∪(3, 4],wemust

have a

≤ 2 for every n ≥ N . Therefore, f

−1

) = a

for every n ≥ N, and so

−1

) converges to 2 =f

−1

(2). Hence, f

−1

is continuous at 2.

Consider now a point b in J different from 2. Either b is in [0, 2) or in (3, 4].

Assume that it is in [0, 2) (the other case is similar). Let b

be a sequence in J that

converges to 2. Take  =

2−b

> 0. There exists N in N such that if n ≥N, then

−b|<=

2 −b

Therefore,

<b+ =

2 +b

< 2.

Hence, f

−1

) = b

for n ≥ N . This sequence converges to b, which is also

−1

(b). Thus, f

−1

is continuous at b. Therefore, even though f is not continu-

ous on I , its inverse function f

−1

is continuous on J .

As pointed out in the preceding section, given α =0, the function f deﬁned by

f(x)=x

=exp(α ln x)

is continuous and differentiable on (0, ∞). However, these results rely on properties

of the exp function that will only be proved in Chap. 7. In the application below we

show that if α is rational, then we can prove that this function is continuous. We will

also prove below that it is differentiable.

Application 5.2 Let r>0 be a rational. The function f deﬁned on I =[0, ∞) by

f(x)=x

is continuous on I .

Let r = p/q for p and q naturals. We know that the function g : x → x

continuous on I for every natural q. We also know that it is a strictly increasing

function and that its inverse function g

−1

:x →x

1/q

is deﬁned on J = I =[0, ∞)

(see Sect. 1.4). Since g is strictly monotone on the interval I , we have that g

−1

continuous on J .Leth :x →x

; this is also a continuous function on I .Wehave

f =h ◦g

−1

Hence, as a composition of continuous functions, f is continuous at a.

Here is another connection between strict monotonicity, continuity, and intervals.

172 5 Continuity, Limits, and Differentiation

Continuity, strict monotonicity, and intervals

(a) Let f be continuous on an interval I . Then the range of f

J =



f(x):x ∈I



is an interval.

(b) Let f be strictly monotone on an interval I . If the range of f is an inter-

val, then f is continuous on I .

We prove (a) ﬁrst. Assume that the function f is continuous on I .Letx<y

in J . Assume that z is strictly between x and y. We want to show that z is in J

as well. By the deﬁnition of J , there are x

and y

in I such that f(x

) = x and

f(y

) = y. In particular, f(x

)<z<f(y

). Since f is continuous on I , it is con-

tinuous on [x

]. By the intermediate value theorem, there is z

in [x

] such

that f(z

) =z. That is, z is also in J . This proves that J is an interval.

We now turn to the proof of (b). Since f is strictly monotone on I , f has an

inverse function f

−1

deﬁned on the range of f , J . We now show that the function

−1

is strictly monotone. Assume that x<yare in J . There are x

and y

in I such

that x =f(x

) and y = f(y

). Assume that f is strictly increasing (the decreasing

case is similar). By contradiction, assume that

≥y

Since f is increasing, we get

f(x

) ≥f(y

That is, x ≥ y, a contradiction. We must have x

. Since x

= f

−1

(x) and

= f

−1

(y), we get that f

−1

is strictly increasing on J , which is an interval by

assumption. Hence, since the inverse function of a function deﬁned on an interval

is continuous, we have that (f

−1

)

−1

is continuous on I .But(f

−1

)

−1

=f . Hence,

f is continuous. This concludes the proof of (b).

Application 5.3 Let f be continuous and strictly monotone on the open interval I .

Then, the range of f , J , is an open interval as well.

Since f is continuous, we know that J is an interval. We need to prove that J is

open. In order to prove that J is open on the right, it is enough to show that either

J is not bounded above, or it is bounded above, but its least upper bound does not

belong to J . Assume that J is bounded above by some M. Since J is not empty

(intervals are not empty according to our deﬁnition), by the fundamental property

of the reals, J has a least upper bound b. By contradiction, assume that b belongs

to J . Then b is in the range of f , and there is a in I such that f(a)=b. Assume

that f is strictly increasing (the decreasing case is similar). Since I is open, there is

c>ain I , and therefore f(c)>f(a)=b. Hence, f(c)is in J and is strictly larger

than the upper bound b of J . We have a contradiction. The l.u.b. b does not belong

5.4 Intervals, Continuity, and Inverse Functions 173

to J , and J is open on the right. Similarly, one can show that it is open on the left

as well.

Inverse function and differentiability

Let f be continuous and one-to-one on the open interval I . Assume that f

is differentiable at a in I and that f



(a) = 0. Let b = f(a). Then f

−1

differentiable at b =f(a), and



−1





(b) =



(a)



−1

(b))

We know that J , the range of f , is an interval. In fact, J is an open interval (see

Application 5.3). This is important because it tells us that b is not a boundary point

of J , and therefore there is an interval centered at b where f

−1

is deﬁned. Let k

be a nonzero sequence converging to 0 such that b +k

is in J for all n ≥1. Let the

sequence h

be deﬁned by

−1

(b +k

) −a.

Note that if h

= 0, then f

−1

(b + k

) = a = f

−1

(b) and k

= 0, which is not

possible. Hence, h

is also a nonzero sequence. Since f

−1

is continuous (why?),

−1

(b +k

) converges to f

−1

(b) =a. Thus, h

converges to 0. Since f is differ-

entiable at a,

f(a+h

) −f(a)

converges to f



(a). Note that

f(a+h

) −f(a)=f



−1

(b +k

)



−f



−1

(b)



=b +k

−b =k

Hence,

f(a+h

) −f(a)

−1

(b +k

) −f

−1

(b)

which converges to f



(a). Since f



(a) is not 0, we have that

−1

(b +k

) −f

−1

(b)

converges to 1/f



(a). This shows that f

−1

is differentiable at b and the stated for-

mula.

Example 5.31 Let n be a natural number. Then the function g deﬁned by

g(x) =x

1/n

is differentiable on (0, ∞). Moreover, g



(x) =

1/n−1

174 5 Continuity, Limits, and Differentiation

Let f be deﬁned by

f(x)=x

The function f is differentiable on the open interval I =(0, ∞). Moreover, it is one-

to-one, and its inverse function is g (see Sect. 1.4). Since a>0, we have f



(a) =

n−1

> 0. Hence, we may apply the formula above to get



(b) =



(g(b))

n(b

1/n

)

n−1

1/n−1

The derivative of x

Let r be a rational. Then the function f deﬁned by f(x)=x

is differentiable

at any a in (0, ∞). Moreover, f



(a) =ra

r−1

We will prove the case r>0 and leave r ≤0 to the exercises. There are natural

numbers p and q such that r =p/q.Letg and h be deﬁned on (0, ∞) by

g(x) =x

1/q

and h(x) =x

Note that f(x)=x

=h(g(x)).Leta>0. Then g(a) > 0. Hence, g is differen-

tiable at a, and h at g(a). Moreover, by the chain rule and Example 5.30 we have



(a) =h





g(a)





(a) =pg(a)

p−1

1/q−1



1/q



p−1

1/q−1

=ra

r−1

Example 5.32 The function ln is differentiable on (0, ∞).

We use the following facts about exp. It is differentiable everywhere, its deriva-

tive is itself, and its range is J = (0, ∞). Since the derivative of exp is strictly

positive on the interval R = (−∞, ∞), we know that it is strictly increasing and

therefore one-to-one. Its inverse function ln is deﬁned on the range of exp. For any

b in J , there is a unique a in R such that b = exp(a). Since exp is never 0, ln is

differentiable at b, and

(ln)



(b) =

exp



(lnb)

Hence, ln is differentiable on J .

Exercises

1. Assume that f is continuous and one-to-one on the interval I . Suppose that

x<y<z, f(x)>f(y), and f(y)<f(z)for x,y,z in I . Find a contradiction.

2. Give an example of a function which is one-to-one on D ={−1, 0, 1} but not

monotone.

3. Consider the function sin on I = (−π/2,π/2). Use the following facts about

sin on I . It is continuous, differentiable, and its derivative cos is strictly positive

on I .

(a) Show that sin has an inverse function denoted by arcsin.

5.4 Intervals, Continuity, and Inverse Functions 175

(b) Find the interval J where arcsin is deﬁned.

4. Consider the function f deﬁned on D =(−∞, 0) ∪(0, ∞) by f(x)=1/x.

(a) Show that f is one-to-one on D.

(b) Show that f is not monotone on D.

an interval is one-to-one if and only if it is strictly monotone?

5. Consider the function f deﬁned on I =[0, 3] by

f(x)=x for 0 ≤x ≤2 and f(x)=x +1for2<x≤3.

(a) Show that f is strictly increasing on I .

(b) Show that f is one-to-one.

(d) Show that f

−1

is deﬁned on J by

−1

(x) =x for 0 ≤x ≤2 and f

−1

(x) =x −1for3<x≤4.

(e) Let 3 <b≤4. Prove that f

−1

is continuous at b.

6. Draw the graph of a function which is deﬁned on an interval I , whose range is

an interval J , but which is not continuous.

7. In this exercise we show that if a function f is continuous on a closed and

bounded interval [a,b], then its range J is also a closed and bounded interval

[c, d].

(a) Show that J is an interval.

(b) Explain why J has a minimum and a maximum.

8. In Application 5.2 we proved that for every rational r>0, the function x →x

is continuous on [0, +∞). Show that if r<0, the function x → x

is continu-

ous on (0, ∞).

9. Assume that I =∅ is an interval.

(a) If I is not bounded below nor above, show that I is all of R.

(b) If I is bounded below but not above, show that there is a real a such that

I =[a,∞) or I =(a, ∞).

10. Let r<0 be a rational. Show that f deﬁned by f(x)=x

is differentiable on

(0, ∞) and that f



(a) =ra

r−1

for a>0.

11. Assume that the function f is differentiable on R, its range is (0, ∞), and its

derivative is itself.

(a) Show that f is one-to-one.

(b) Show that f

−1

is differentiable on (0, ∞) and ﬁnd its derivative.

12. Let f be one-to-one and differentiable on the open interval I . Assume that



(a) =0forsomea in I . Prove that f

−1

is not differentiable at f(a).(Doa

proof by contradiction: assume that f

−1

is differentiable at f(a)and show that

this implies that (f

−1

)



(f (a)) ×f



(a) =1.)

13. Consider the function f deﬁned by f(x)=

1+x

for all x in R.

(a) Show that the range of f is not open.

176 5 Continuity, Limits, and Differentiation

(b) We know that the range of a continuous strictly monotone function on an

open interval is open (see Application 5.3). Which hypothesis does not hold

here?

14. Let g be a continuous strictly decreasing function on [0, 1). Show that the range

of g is an interval (a, b] for some reals a and b or an interval (−∞,b].

Chapter 6

Riemann Integration

6.1 Construction of the Integral

We start with some notation. Consider a bounded function f on a closed and

bounded interval [a,b].ThesetP is said to be a partition of [a,b] if there is a

natural n such that

P ={x

,...,x

}

where x

=a<x

< ···<x

=b. In words, P is a ﬁnite collection of ordered reals

in [a,b] that contains a and b.

The function f is said to be bounded on [a,b] if the set



f(x);x ∈[a,b]



has a lower bound and upper bound. Since this set is not empty, we may apply

the fundamental property of the reals to get the existence of a least upper bound

M(f,[a,b]) and a greatest lower bound m(f, [a,b]). In particular,



f,[a,b]



≤f(x)≤M



f, [a,b]



for all x ∈[a,b].

Throughout this chapter we will assume that f is bounded. For a ﬁxed partition

P ={x

,...,x

} and for all i =1,...,n,theset



f(x);x ∈[x

i−1

]



has a greatest lower bound m(f, [x

i−1

]) and a least upper bound M(f,[x

i−1

])

(why?). In particular,



f, [x

i−1

]



≤f(x)≤M



f,[x

i−1

]



for all x ∈[x

i−1

Intuitively the integral of f between a and b is the area between the graph of f and

the x axis. See Fig. 6.1. However, the construction of the integral is involved and

will require several steps.

We are now ready for our ﬁrst deﬁnition.

R.B. Schinazi, From Calculus to Analysis,

177

178 6 Riemann Integration

Fig. 6.1 The shaded area represents the integral of the function f between a and b

Lower and Upper Darboux sums

Let f be deﬁned and bounded on the closed and bounded interval [a,b].Let

P ={x

,...,x

} be a partition of [a,b]. The upper Darboux sum corre-

sponding to P is deﬁned by

U(f,P)=



i=1



f, [x

i−1

]



−x

i−1

The lower Darboux sum corresponding to P is deﬁned by

L(f, P ) =



i=1



f, [x

i−1

]



−x

i−1

For every partition P ,wehave

L(f, P ) ≤U(f,P).

The inequality L(f, P ) ≤U(f,P) is a direct consequence of



f, [x

i−1

]



≤M



f, [x

i−1

]



for i =1,...,n.

The Riemann integral of a function f on [a,b] is the shaded area between the

graph of f and the x axis in the graph below. Intuitively, the integral may be com-

puted as a limit of upper or lower Darboux (do not pronounce the ‘x’ in this name)

sums as the number of points in the partition increases to inﬁnity. These limits exist

and are equal, provided that the function f is not too irregular. In fact, it can be

shown that this is the case if and only if the set of x where f is not continuous is

negligible (in a sense to be made precise). See, for instance, the Riemann–Lebesgue

theorem in ‘An introduction to analysis’ by J.R. Kirkwood, second edition, PWS

Publishing Company.

6.1 Construction of the Integral 179

Example 6.1 Consider the function f deﬁned on [0, 1] by f(x)=x

.LetP be the

partition {0, 1/2, 1}. This is an increasing function, and we can compute easily



f, [0, 1/2]



=0,m



f, [1/2, 1]



=1/4,



f, [0, 1/2]



=1/4,M



f, [1/2, 1]



=1.

This yields

L(f, P ) =0 ×1/2 +1/4 ×1/2 =1/8

and

U(f,P)=1/4 ×1/

2 +1 ×1/2 =5/8.

With

more points in our partition, we would get, of course, a better approximation

of the area between the x axis and the graph of f . See the exercises.

We are now ready to deﬁne Riemann integrability.

Riemann integrability

Let f be deﬁned and bounded on the closed and bounded interval [a,b].The

function f is Riemann integrable if for every >0, it is possible to ﬁnd a

partition P of [a,b] such that

0 ≤U(f,P)−L(f, P ) < .

Note that this deﬁnes only the notion of integrability, not the integral (which

will be deﬁned later on). The notion of integrability relates to the regularity of the

function, it is similar and related to continuity and differentiability.

In the examples below we will use several times the following lemma on tele-

scoping sums.

Lemma 6.1 Let a

be a sequence of reals. For every n ≥1, we have



k=1

−a

k−1

) =a

−a

We do an induction on n.Wehave



k=1

−a

k−1

) =a

−a

and the formula holds for n = 1. Assume that it holds for n. Splitting the following

sum into two, we get

n+1



k=1

−a

k−1

) =



k=1

−a

k−1

) +(a

n+1

−a

Schinazi R.B. From Calculus to Analysis

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