Schinazi R.B. From Calculus to Analysis

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160 5 Continuity, Limits, and Differentiation

To prove the chain rule, we approach f and g linearly near a. Deﬁne the function

v for h =0by

v(h) =

f(a+h) −f(a)

−f



(a)

and v(0) =0. Recall that since f is deﬁned on an open interval, v is deﬁned on an

interval (−δ,δ) for some δ>0 (see the observation following the deﬁnition of the

derivative in the preceding section). Since f is differentiable at a, the limit of v at 0

is 0 =v(0). Hence, v is continuous at 0. Moreover,

f(a+h) =f(a)+hf



(a) +hv(h).

Similarly, we deﬁne the function w as

w(h) =

g(f (a +h)) −g(f (a))

−g





f(a)



for h =0

and w(0) =0. Again, w is well deﬁned near 0. Since g is differentiable at f(a),the

limit of w at 0 is 0. Thus, w is continuous at 0. We have



f(a+h)





f(a)



+hg





f(a)



+hw(h).

Let h

be a sequence converging to 0 and never equal to 0. We have



f(a+h

)





f(a)+h



(a) +h

v(h

)



Let



(a) +h

v(h

Hence,



f(a+h

)





f(a)+k





f(a)







f(a)



w(k

Thus,

g(f (a +h

)) −g(f (a))





f(a)



w(k

Note that



(a) +v(h

Since h

converges to 0, v(h

) converges to 0, and k

converges to 0 as well. Thus,

converges to f



(a). Since w is continuous at 0 and k

converges to 0, w(k

)

converges to w(0) =0, and

w(k

) converges to 0. Hence,

lim

n→∞

g(f (a +h

)) −g(f (a))



(a)g





f(a)



This proves that g ◦f is differentiable at a and the chain rule.

Example 5.23 Given a real α = 0, consider the function f deﬁned on D = (0, ∞)

by f(x)=x

. Show that f is differentiable everywhere on D and that



(b) =αb

α−1

for all b>0.

5.3 Algebra of Derivatives and Mean Value Theorems 161

In Sect. 3.4 we deﬁned

=exp(α ln x).

We also stated that exp is differentiable on R and its derivative is itself. In Chap. 7

we will show that power series are differentiable on their open interval of conver-

gence and that they can be differentiated term by term. Given this result, it is not

difﬁcult to show that the inverse function of exp, ln, is differentiable on (0, ∞) and

its derivative is 1/x. This will be proved in the next section. Now we use these facts.

First observe that if g is deﬁned by g(x) =α lnx, then

f =exp ◦g.

Since exp is differentiable everywhere and g is differentiable on D, the chain rule

maybeappliedatanyb>0 to show that f is differentiable on D and that



(b) =g



(b) exp



g(b)



exp(α ln b) =α

=αb

α−1

Example 5.24 Given a real α>0, consider the function f deﬁned on R by f(x)=

. Show that f is differentiable everywhere and that



(b) =α

ln α for all b.

We use the same formula as in the preceding example:

=exp(x lnα).

Let g be deﬁned by g(x) =x ln α. Since exp and g are differentiable everywhere

and g



(x) =lnα, by the chain rule we have



(b) =ln α exp



g(b)



=lnα exp(b lnα) =α

lnα.

We now turn to important applications of the notion of derivative.

Local extrema

Let f be deﬁned on an open interval I . The function f is said to have a local

minimum at a in I if there is a δ>0 such that f(a)≤ f(x) for every x in

(a −δ, a +δ). Likewise, f is said to have a local maximum at a if there is a

δ>0 such that f(a)≥f(x)for every x in (a −δ,a +δ).

The following is a useful result.

Local extrema and differentiability

Let f be deﬁned on an open interval I . Assume that it has a local extremum

at a and that f is differentiable at a. Then f



(a) =0.

162 5 Continuity, Limits, and Differentiation

The proof is simple. Since I is open, there is a δ>0 such that (a −δ,a +δ) ⊂I .

Let h

. Then a + h

belongs to I for every n ≥ 1. We assume that f has a

local maximum at a (the local minimum case is similar). There is a δ

> 0 such that

if x belongs to (a −δ

,a+δ

), then f(x)≥ f(a). There is N such that if n ≥N,

then h

<δ

(why?). Hence,

f(a+h

) ≤f(a)

for n ≥N . This yields

f(a+h

) −f(a)

≤0.

By letting n go to inﬁnity and using that f is differentiable at a, we get



(a) ≤0.

We are now going to show that f



(a) ≥ 0, and this will conclude the proof. Let

=−h

. We have that |k

|=h

<δ

for n ≥N . Hence,

f(a+k

) ≤f(a)

for n ≥N . Using that k

< 0, we get

f(a+k

) −f(a)

≥0.

Letting n go to inﬁnity, we get f



(a) ≥0. This concludes the proof.

We use the result above to prove the following generalized mean value theorem.

Cauchy Mean Value Theorem

Let f and g be continuous on the closed interval [a,b] and differentiable on

the open interval (a, b). Then there is a real c in (a, b) such that



f(b)−f(a)







g(b) −g(a)





(c).

We deﬁne h on [a,b] by

h(x) =



f(b)−f(a)



g(x) −



g(b) −g(a)



f(x).

Note that h is continuous on [a,b] and differentiable on (a, b) (why?). We have



(x) =



f(b)−f(a)





(x) −



g(b) −g(a)





(x).

To prove the theorem, we only need to ﬁnd c in (a, b) such that h



There are two cases. If h is constant on [a,b], then we know that h



=0on(a, b).

Hence, c maybeanyrealin(a, b).

We now turn to the case where h is not constant. Note that

h(a) =h(b) =f(b)g(a)−f(a)g(b).

If h is not constant on [a,b], there is at least one x

in (a, b) such that h(x

) =h(a).

Assume that h(x

)<h(a)(the other case is similar). Since h is continuous on [a,b],

5.3 Algebra of Derivatives and Mean Value Theorems 163

which is closed and bounded, the extreme value theorem applies, and there is c in

[a,b] such that for all x in [a,b],wehave

h(x) ≥h(c).

Hence, h(c) ≤h(x

)<h(a)=h(b). Therefore, c cannot be a or b, and it must be

in the open interval (a, b). Since a minimum occurs at c for h and h is differentiable

on the open interval (a, b),wemusthaveh



Cauchy mean value theorem.

An easy consequence of the preceding theorem is the following:

Lagrange Mean Value Theorem

Let f be continuous on the closed interval [a,b] and differentiable on the

open interval (a, b). Then there is a real c in (a, b) such that

f(b)−f(a)=(b −a)f



(c).

We apply the Cauchy mean value theorem with g(x) = x, which is continuous

on [a,b] and differentiable on (a, b). Hence, there is c in (a, b) such that



f(b)−f(a)







g(b) −g(a)





(c).

Since g(x) =x and g



(x) =1, we get

f(b)−f(a)=(b −a)f



(c).

This completes the proof of the Lagrange mean value theorem (LMVT).

Geometrically, the LMVT has a nice interpretation: there is at least one c in (a, b)

where the line tangent to the graph of f (with slope f



(c)) has the same slope as the

line passing through (a, f (a)) and (b, f (b)). Draw a picture!

We will apply the LMVT to obtain a number of important results from Calculus.

Rolle’s Theorem

Let f be continuous on the closed interval [a,b] and differentiable on the

open interval (a, b). Assume also that f(a)= f(b). Then there is a real c in

(a, b) such that



This is an immediate application of the LMVT. There is c such that

f(b)−f(a)=(b −a)f



(c).

Since f(a)=f(b),wehavef



The following is a typical application of Rolle’s theorem.

164 5 Continuity, Limits, and Differentiation

Example 5.25 Let f be a function twice differentiable everywhere and such that f



is never 0. Then the equation f(x)=0 has at most two solutions.

By contradiction, assume that there are three solutions a<b<c. Since f(a)=

f(b), by Rolle’s theorem there is a c

in (a, b) such that f



) =0. There is also c

in (b, c) such that f



) =0. Note that c

. Since f



is differentiable, we may

apply Rolle’s theorem to f



. Using that f



) =f



) =0, there is c in (c

)

such that f



is never 0. We have a contradiction. The equation has

at most two solutions.

Null derivative

Let f be differentiable on the open interval I . Then the function f is a con-

stant on I if and only if f



(x) =0 for all x in I .

We already know that if f is a constant on I , then f



= 0onI . We prove the

converse. Assume that f



=0onI .Leta<bin I . Then f is continuous on [a,b]

(why?) and differentiable on (a, b). The LMVT applies, and we get

f(b)−f(a)=f



(c)(b −a)

for some c in (a, b). Since f



any a and b, f is constant on I . We now turn to an important deﬁnition.

Increasing functions

Let f be deﬁned on D. The function f is said to be increasing on D if for all

x<yin D,wehave

f(x)≤f(y).

It is said to be strictly increasing on D if for all x<yin D,wehave

f (x) < f (y).

We have a symmetric deﬁnition for decreasing functions.

Decreasing functions

The function f is said to be decreasing on D if for all x<yin D,wehave

f(x)≥f(y).

It is said to be strictly decreasing on D if for all x<yin D,wehave

f (x) > f (y).

A function which is increasing or decreasing on D is said to be monotone on D.

5.3 Algebra of Derivatives and Mean Value Theorems 165

Monotone differentiable functions

Let f be differentiable on the open interval I .

(a) If f



> 0onI , then f is strictly increasing on I .

(b) If f



< 0onI , then f is strictly decreasing on I .

We prove (a), and we leave (b) to the reader. Take x<yin I . The LMVT may

be applied to f on [x,y] (why?). Hence, there is c in (x, y) such that

f(x)−f(y)=(x −y)f



(c).

Since f



> 0, we see that (x −y)f



true for all x<yin I , the proof of (a) is complete.

Remark If f



≥0 on an open interval, the function f may be increasing or strictly

increasing. See the exercises.

Example 5.26 The function g(x) =1/x is differentiable on D =(−∞, 0) ∪(0, ∞).

The derivative of g is g



(x) =−1/x

< 0 for all x in D. However, g is not decreas-

ing on D. For instance, g(−1)<g(1). This does not contradict the preceding result:

D is not an interval. The preceding result does show that g is strictly decreasing on

(−∞, 0) and on (0, ∞).

Example 5.27 Let f be differentiable on the open interval I .Leta be in I , and

δ>0 be such that (a −δ, a +δ) ⊂I . Assume that if a<x<a+δ, then f



(x) < 0

and if a −δ<x<a, then f



(x) > 0. Show that f has a local maximum at a.

The proof is a simple application of the LMVT. Take x in (a −δ,a). Then there

is c in (x, a) such that

f(a)−f(x)=f



(c)(a −x) > 0.

Hence, f(a)>f(x) for x in (a −δ, a).Takey in (a, a +δ). There is d in (a, y)

such that

f(a)−f(y)=f



(d)(a −y) > 0.

Hence, f(a)>f(x)for all x =a in (a −δ,a +δ). A local maximum occurs at a.

Exercises

1. Let n be a natural. Let f

,...,f

be deﬁned on an open interval I and

differentiable at a in I. Show that g =f

+···+f

is differentiable at a.

2. Assume that f is deﬁned on an open interval I , is differentiable at a in I , and

f(a)= 0. Show that 1/f is differentiable at a and ﬁnd a formula for (1/f )



(a).

3. Assume that f and f



are differentiable on R and that for every x in R, f(x)+



(x) =0. Show that g =f

+(f



)

is a constant.

166 5 Continuity, Limits, and Differentiation

4. Assume that f and g are differentiable on an open interval I. Assume also that



on I . Show that there is a constant c such that for all x in I ,

f(x)=g(x) +c.

5. Let f be differentiable on an open interval I . Show that if f



≥0onI , then f

is increasing on I .

6. Give an example of a differentiable function which is strictly increasing on R

but whose derivative is not strictly positive on R.

7. (a) Assume that f



(a) =0. Does this necessarily imply that a local extremum

occurs at a?

(b) Draw the graph of a continuous function that has a local maximum at 1 but

is not differentiable at 1.

8. Show that the equation x

+5x +3 =0 has exactly one real solution.

9. (a) Let M>0. Let f be differentiable on an open interval I and such that



(x)|≤M for each x in I . Show that for all x and y in I ,wehave



f(x)−f(y)



≤M|x −y|.

(b) Give an example of a function f deﬁned on the reals with the following

properties. There is M such that for all x and y,



f(x)−f(y)



≤M|x −y|,

but f is not differentiable everywhere.

10. Let f and g be differentiable on R. Assume that for all x in I , f



(x) ≤g



(x).

(a) Show that if x ≥0, we have

f(x)−f(0) ≤g(x) −g(0).

(b) Show that if x ≤0, we have

f(x)−f(0) ≥g(x) −g(0).

11. Let r>1 be a rational, and 0 ≤x<y. Show that

−x

<ry

r−1

(y −x).

12. Use the LMVT for f(x)=

√

1 +x to show that if x>0, then

√

1 +x<1 +x/2.

13. Use the LMVT to show that for all x in [0, 1),wehave

ln(1 −x) ≤−x.

14. Use the LMVT to show that for all x in [0, 1),wehave

ln(1 −x) ≥

−x

1 −x

15. Let f be differentiable on an open interval I .Leta be in I , and δ>0 such

that (a − δ,a +δ) ⊂ I . Assume that if a<x<a+δ, then f



(x) > 0 and if

a −δ<x<a, then f



(x) < 0. Show that f has a local minimum at a.

5.4 Intervals, Continuity, and Inverse Functions 167

16. In this problem we prove the second derivative test. Let f and f



be differen-

tiable on an open interval I . Assume that a is in I , f



(a) =0, and f



(a) > 0.

Since I is open, there is δ>0 such that (a −δ,a +δ) ⊂I . Deﬁne the function

φ on (−δ,δ) by

φ(h) =



(a +h) −f



(a)

for h =0

and φ(0) =f



(a).

(a) Show that φ is continuous on (−δ,δ).

(b) Show that there is δ

> 0 such that φ>0on(−δ

,δ

). (See Application 5.1

in Sect. 5.1.)

,wehavef



(a +h) > 0, and for −δ

<h<0, we

have f



(a +h) < 0.

(d) Conclude that f has a local minimum at a. (Use Exercise 15.)

17. Assume that f is differentiable everywhere except possibly at 0. Assume also

that f is continuous everywhere.

(a) If f



has a limit at 0, show that f is differentiable at 0.

(b) Use the function g(x) = x

sin(1/x) for x =0 and g(0) =0 to show that

the converse of (a) is not true.

18. Give an example of a function deﬁned on [0, 1], differentiable on (0, 1), and

such that the LMVT does not hold.

5.4 Intervals, Continuity, and Inverse Functions

An interval is a subset of R that has no holes. More formally:

Intervals

A subset I =∅ of the reals is said to be an interval if for all x<yin I ,anyz

such that x ≤z ≤y is also in I .

A typical example of interval is a subset of the reals such as

{x ∈R :−2 <x≤1}

that is denoted by (−2, 1].

Example 5.28 Let I = ∅ be an interval. Assume that I is bounded above but not

below. Show that there is a in R such that either I =(−∞,a) or I =(−∞,a].

Since I is not empty and is bounded above, it has a least upper bound (by the

fundamental property of the reals) that we denote by a. There are two possibilities.

The material in this section may be difﬁcult for a beginner. It is not essential for the sequel and

may be omitted.

168 5 Continuity, Limits, and Differentiation

Either a belongs to I , or it does not. Assume ﬁrst that a belongs to I . Since a is an

upper bound of I if x belongs to I , we have that x ≤a. Hence, I ⊂ (−∞,a].We

want to show that in fact I = (−∞,a].Lety<a. Since I is not bounded below,

there must be a z in I such that z<y. Hence, z<y<awhere z and a are in I .

Since I is an interval, y is in I. This proves that (−∞,a]⊂I , and so (−∞,a]=I .

The second possibility is that a, the least upper bound of I , is not in I .Wehave

that I ⊂ (−∞,a).Lety<a. Since a is the least upper bound of I , y cannot be

an upper bound of I . Thus, there is z

in I such that y<z

<a. On the other

hand, since I is not bounded below, there is a z

in I such that z

<y. Hence,

<y<z

with z

and z

in I . Using the deﬁnition of an interval, we get that y is

in I . Therefore, (−∞,a)⊂I and I =(−∞,a).

By using the method of Example 5.28 it is possible to show (see the exercises)

that all intervals are of the type above. More precisely, if I is an interval, then either

there is a real a such that I =(−∞,a)or (−∞,a],or(a, ∞) or [a,∞), or there are

reals a and b such that I =(a, b), [a,b), (a, b],or[a,b]. Finally, R =(−∞, +∞)

is also an interval.

Intervals such as (−∞,a), (a, ∞), and (a, b) are said to be open. Intervals such

as (−∞,a], [a,∞), and [a,b]

are said to be closed. Intervals such as (a,

b] are

neither open nor closed.

Recall that a function f is said to be one-to-one on D if for all x and y in D,

f(x)=f(y)implies x =y.

The next result makes a link between several notions.

One-to-one and strictly monotone functions

Let f be continuous on an interval I . The function f is one-to-one on I if

and only if it is strictly monotone.

One direction is immediate. Assume that f is strictly monotone. Then if f(x)=

f(y),wemusthavex = y. For if there is a strict inequality between x and y, there

must be a strict inequality between f(x) and f(y) (since f is strictly monotone).

Hence, f is one-to-one.

Assume now that f is one-to-one and continuous on the interval I . We want to

show that f is strictly monotone. We do a proof by contradiction. If f is not strictly

monotone, there must be three reals x,y,z in I with x<y<zsuch that either

(f (x) < f (y) and f (y) > f (z)) or (f (x) > f (y) and f (y) < f (z)). Both cases

will lead to a contradiction. We treat the ﬁrst one. The second one will be done in

the exercises.

Assume that x<y<z, f(x)<f(y), and f(y)>f(z). There are two subcases.

Either f(z)<f(x) or f(z)> f(x). Consider ﬁrst f(z)< f(x).Letc = f(x),

then f(z)<c<f(y). Since f is continuous on [y,z], by the Intermediate Value

Theorem, there is x

in (y, z) such that f(x

) = c.Butf(x)=c as well, and f is

one-to-one. Hence, x =x

. Since x<yand x

>y, we get a contradiction. We can-

not have f(z)<f(x). Since f(z)= f(x)(why not?), we must have f(z)>f(x).

5.4 Intervals, Continuity, and Inverse Functions 169

Let d =f(z).Wehavef(x)<d<f(y). By the IVT there is z

in (x, y) such that

d =f(z

). Since d =f(z)and f is one-to-one, we have z =z

. Since z

<y<z,

we have a contradiction. Hence, we cannot have f(x)<f(z)either. Therefore, as-

suming that x<y<z, f(x)<f(y), and f(y)>f(z) has led to a contradiction.

Similarly, if we assume that x<y<z, f(x)>f(y), and f(y)<f(z), we also get

a contradiction (see the exercises). Thus, f must be strictly monotone.

The preceding theorem depends crucially on the continuity and interval hypothe-

ses. We illustrate the importance that the function be continuous on an interval in

the next example.

Example 5.29 Consider the function g deﬁned on D =(−∞, 1) ∪(1, ∞) by

g(x) =

1 −x

Note that g is continuous on D (why?). Assume that g(x) =g(y). Then

1 −x

1 −y

Hence, x − xy = y − xy and x = y. That is, g is one-to-one. Note now that

g(0) = 0 >g(2) =−2, but g(2) =−2 <g(3) =−3/2. Hence, g is not monotone

on D. However, this does not contradict the fact that a one-to-one function which is

continuous on an interval is monotone. The set D is not an interval!

We now turn to the continuity of inverse functions.

Inverse functions and continuity

Let f be strictly monotone on an interval I . The function f has an inverse

−1

deﬁned on the range of f which is denoted by J . Moreover, f

−1

continuous on J .

What is remarkable here is that the inverse function of a function not necessarily

continuous is continuous! We will give such an example after the proof.

As argued before, it is clear that a strictly monotone function f is one-to-one.

Hence, it has an inverse function f

−1

deﬁned on the range of f ,

J =



f(x):x ∈I



We now show that f

−1

is continuous on J . By contradiction, assume that f

−1

is not

continuous at some b in J . Hence, there must be a sequence b

in J that converges

to b but such that f

−1

) does not converge to f

−1

(b). By taking the negation of

convergence we get the following. There is >0 such that for all N in N, there is

n ≥N such that



−1

) −f

−1

(b)



≥.