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Topics on Mean Value Theorems

Gen-Bin Huang

January 19, 2001

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Abstract

Let fbe a continuous function on [a, x], then there exists a x in [a, x] such

that xa

f(t)dt= f(x)(x a) .

This is the mean value theorem for integrals. We call x an integral mean point

of f on [a, x]. Recently, there are a number of studies on the location of x.

In particular, Zhang [7] showed that if f is Cr (r N) near a point a, withf(a) =f(a) =. . .= f(r1)(a) = 0 but f(r)(a) = 0, then

limxa

x ax a =

1

(r+ 1)1r

.

Note that there is also a differential mean pointcxin the classical mean value the-

orem. Using a similar method, I give a detailed estimation of the location ofcx.

Since the classical mean value theorem implies the mean value theorem for inte-

grals, my result is more general. In fact the two theorems are roughly equivalent.

There is also some relationship between the Cauchy mean value theorem and the

generalized mean value theorem for integrals. In 1999, Schwind-Ji-Koditschek

generalized Zhangs result to consider the integral mean point of the generalized

mean value theorem, assuming fto behave like K+C1(x a)r near a (r > 1)andg to behave like C2(x a)s near a. I also proved an analogue for its relative,the Cauchy mean point.

On the other hand, Tong and Braza [6] studied the converse of the classical

mean value theorem. They showed that iff(c) is not a local extremum, then c

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is a differential mean point. I obtain a parallel theorem for the converse of the

mean value theorem for integrals.

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Contents

1 Introduction 1

1.1 The mean value theorems . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Relationships among the mean value theorems . . . . . . . . . . . 4

2 Estimation of integral mean points and differential mean points 7

2.1 Integral mean points . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Differential mean points . . . . . . . . . . . . . . . . . . . . . . . 13

3 Converses of mean value theorems 18

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Chapter 1

Introduction

1.1 The mean value theorems

First, we introduce the mean value for integrals. Assume thatfis continuous,

then

= 1b a

b

af(t)dt=

baf(t)dtbadt

.

Thisis called the arithmetic average or the mean value offin the interval

[a, b]. We state the mean value theorem for integrals:

Theorem 1.1 ([1, p.258] Mean value theorem for integrals).

Iffis a continuous function on[a, x], then there exists a numberx in[a, x]such

that xa

f(t)dt= f(x)(x a) . (1.1)

This is a simple but very important mean value theorem of integral calculus.

It states that the mean value theorem of a continuous function in an interval

belongs to the range of the function. And it asserts only the existence of at least

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one x in the interval for which f(x) is equal to the average value offbut gives

no further information about the location x. In this thesis, we callx anintegral

mean pointoff on [a, x].

Instead of the simple arithmetic average we can form the weighted averages:

=

xa f(t)g(t)dtx

a g(t)dt

gis the weight function and g(t) 0. We give the generalized mean value theoremfor integrals:

Theorem 1.2 ([1, p.257] Generalized mean value theorem for integrals).

Iff is a continuous function on [a, x], g is integrable on [a, x] andg 0, thenthere exists a numberx in [a, x] such that

xa

f(t)g(t)dt= f(x) xa

g(t)dt . (1.2)

Obviously, Theorem 1.1 is the special case when g(t) = 1.

Theorem 1.3 ([1, p.197] Classical mean value theorem).

Supposef is continuous on the closed interval [a, x] and differentiable on(a, x),

then for somecx between [a, x], we have

f(cx) =f(x) f(a)

x a . (1.3)

We could estimate cx as x in the same way as in (1.2). And we call cx a

differential mean pointoff on [a, x].

Theorem 1.4 ([1, p.198] Cauchys mean value theorem).

Supposef, g are continuous on[a, x], differentiable on(a, x). Ifg (t) = 0for anyt (a, x), then there existscx (a, x) such that

f(x) f(a)g(x) g(a) =

f(cx)

g(cx) . (1.4)

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Here we call cx a Cauchy differential mean point or simply Cauchy mean point.

Recently, there are a number of studies on the location of the integral mean

pointx, asx a. B. Zhang [7], improving B. Jacobsons result [4], showed thatiffisr(r N) times differentiable ata, withf(a) =f(a) =. . .= f(r1)(a) = 0but f(r)(a) = 0, then

limxa

x ax a =

1

(r+ 1)1r

.

Schwind-Ji-Koditschek [5] went on further. They allowfand g to have a singu-

larity at a. Namely they showed that if

limta

f(t) K(t a)r =C1

and

limta

g(t)

(t a)s =C2 ,

where r = 0,s > 1, r+s > 1,C1, C2= 0, then

limxa

x ax a = (

s+ 1

r+s+ 1)1r .

On the other hand, Tong and Braza [6] studied the converse of the classical

mean value theorem. They showed that iff(c) is not a local extremum, that c

is a differential mean point.

It seems to be well-known that the four mean value theorems above are in-

terrelated. In fact, Theorem 1.3 implies Theorem 1.1. Under some additional

assumptions, Theorems 1.1 also implies Theorem 1.3. Similarly, Theorem 1.2

and Theorem 1.4 are roughly equivalent. Thus it is natural to study the esti-

mates of the differential mean point and the converse of the mean value theorem

for integrals.

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In the next section, we shall discuss the relationships among the mean value

theorems. In Chapter 2, we shall first discuss the works of B. Jacobson, B. Zhang

and Schwind-Ji-Koditschek on the limiting position of the integral mean point

x. Then we shall prove corresponding theorems for the differential mean point

cx of the classical mean value theorem and Cauchy mean value theorem.

In Chapter 3, we shall discuss the converse of the classical mean value theorem

studied by Tong and Braza. Then, we prove a parallel theorem for the mean value

theorem for integrals.

1.2 Relationships among the mean value theo-

rems

It is known that Theorem 1.3 implies Theorem 1.1. There is a proof in the

Calculus book written by Campbell and Dierker [3].Proof of Theorem 1.1 ([3, p.209]).

Supposef : [a, x] R is continuous on [a, x]. Define F(t) =taf(s)ds. Then Fis continuous on [a, x] andF =f on [a, x]. Hence by Theorem 1.3, there is some

cx (a, x) such thatF(x) F(a) =F(cx)(x a) .

That means, xa

f(t)dt= f(cx)(x a) .

Conversely, suppose F isC1 and F =f. So

F(x) F(a) = xa

f(t)dt .

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By Theorem 1.1 and above formula, we conclude that

F(x) F(a) = xa

f(t)dt= f(x)(x a) .

Since F is a primitive off, we could replace f(x) byF(x) and obtain

F(x) =F(x) F(a)

x a .

It is the same form as (1.4). Thus ifF isC1, then Theorem 1.1 implies Theorem

1.3.

Assume g is continuous and g > 0, then Theorem 1.4 implies Theorem 1.2.

The proof is as follows:

Define

F(t) = t

af(s)g(s)ds, G(t) =

t

ag(s)ds .

F and G are continuous on [a, x] and differentiable on (a, x). Also

G(x) = xa

g(t)dt > G(a) = 0 .

Hence we apply Theorem 1.4 to obtain some cx (a, x) such that

F(x) F(a) =F(cx)

G(cx)[G(x) G(a)] .Thus

xa

f(t)g(t)dt = f(cx)g(cx)

g(cx)

xa

g(t)dt

= f(cx) xa

g(t)dt.

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Conversely, under additional assumption onfandg, then Theorem 1.2 implies

Theorem 1.4. If f and g are both C1, let = f

g continuous, = g > 0 for

simplicity, then by Theorem 1.2, there exists cx (a, x) such that xa

(t)(t)dt= (cx) xa

(t)dt .

Therefore, x

af(t)dt= f

(cx)g(cx)

x

ag(t)dt .

So

f(x) f(a)g(x) g(a) =

f(cx)

g(cx) .

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Chapter 2

Estimation of integral mean

points and differential mean

points

2.1 Integral mean points

In this section, we study the limiting position of an integral mean point x.

Jacobson [4] described the following theorem may fall into the category of inter-

esting facts we once knew, but have now forgotten. The theorem says that as x

approaches a, the value ofx approaches the midpoint between a and x.

Theorem 2.1 ([4]).

Iff is continuous on [a, x], differentiable ata, f(a)= 0 and x is taken as inTheorem1.1, then

limxa

x ax a =

1

2 .

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Before proving Theorem 2.1, we give Lemma 2.2:

Lemma 2.2 If(t) 0 ast a and(t)(t a)m is integrable, then

(a) limxa

x

a (t)(ta)mdt

(xa)m+1 = 0 ;

(b) limxa(x)(xa)m

(xa)m = 0, wherem >0 .

Proof.

(a) Given any >0, there exist some x0 > 0 such that for any 0< |t a| < x0,|(t)| < . Then take anyx such that 0 a. Then

|xa (t)(t a)mdt

(x a)m+1 | xa(t a)mdt

(x a)m+1

=

m+ 1 .

(b) Since a < x < x, 01, r N) differentiable ata, withf(a) =f(a) =. . .= f(r1)(a) = 0 butf(r)(a) = 0. Ifx is taken as in Theorem1.1, then

limxa

x ax a =

1

(r+ 1)1r

.

If we consider a large function class, which need not assume fto be differen-

tiable, to estimate x. And we take a transformation off.

Theorem 2.5 ([5]).

Iffis continuous on(a, x]andg is integrable on(a, b)withg(t) 0fort (a, b).If limta

f(t)K(ta)r

=C1, limtag(t)

(ta)s =C2 for someK , C1, C2= 0, r = 0 and

s > 1 such thatr+s > 1, then

(a) there existsx (a, x] such that xa

f(t)g(t)dt= f(x) xa

g(t)dt , (2.5)

(b) for any choice ofx, we have

limxa

x ax a = (

s+ 1

r+s+ 1)1r . (2.6)

Proof.

(a) Since both

limta

f(t) K(t a)r

and

limta

g(t)

(t a)s

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exist, then given the limits

f(t) =K+C1(t a)r +1(t)(t a)r , (2.7)

and

g(t) =C2(t a)s +2(t)(t a)s , (2.8)

where1(t) 0, 2(t) 0.First, from equations (2.7) and (2.8), we have

|f(t)g(t)

| =

|[K(C2+2(t))](t

a)s + [C1C2+C12(t) +C21(t) +1(t)2(t)](t

a)r+s

| |K(C2+2(t))|(t a)s + |C1C2+C12(t) +C21(t) +1(t)2(t)|(t a)r+s

K1(t a)s +K2(t a)r+s

whereK1, K2 are some constant, thenxa f(t)g(t)dtexists. Now,we give the

proof of (a).

Case1: r >0.

Assume F(a) =K, and F(t) =f(t) for t (a, b], then F is continuous on[a, b]. Since g is integrable and g(t) 0, by applying Theorem 1.2, thereexists x (a, x], such that

xa

f(t)g(t)dt= xa

F(t)g(t)dt= f(x) xa

g(t)dt .

Case2: r 0. From (2.7), then

limtaf(t) = , and there exists >0 such that

f(t)>

xa f(t)g(t)dtx

a g(t)dt

for t (a, a+ ). Assume that there exists x satisfying (2.5), but x(a, a+). Takea1 (a, a+), f is still continuous on [a1, x], f(t)> on

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(a, a1), then

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2.2 Differential mean points

We have the same results as Theorem 2.4 of the classical mean value theorem.

Theorem 2.6 Iff is continuous on [a, b] and differentiable on(a, b), f is Rie-

mann integrable and limtaf(t)l(ta)r

= C1 exists in (r= 0, r > 1), then thedifferential mean pointcx satisfies

limxa

cx a

x a = (

1

1 +r

)1r .

Proof.

Since limtaf(t)l(ta)r

=C1 ,

f(t) =l+C1(t a)r +(t)(t a)r (2.9)

where (t) 0 as t a. On the other hand, equation (2.9) evaluated at cxyields

f(cx) =l+C1(cx a)r +(cx)(cx a)r (2.10)

where (cx) 0 as cx a. Multiply (x a) to (2.10) we get

f(cx)(x a) =l(x a) +C1(cx a)r(x a) +(cx)(cx a)r(x a) . (2.11)

Also,

f(x) f(a) = x

a f

(t)dt

= l(x a) +C1 (x a)r+1

r+ 1 +

xa

(t)(t a)rdt . (2.12)

Equating (2.11) and (2.12), we have

C1(cx a)r(x a) +(cx)(cx a)r(x a) =C1 (x a)r+1

r+ 1 +

xa

(t)(t a)rdt

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Takex

a, then we get the following result by Lemma 2.2

limxa

cx ax a = (

1

1 +r)1r .

If we consider f is in C2[a, b], we can derive a more detailed asymptotic

expression forcx.

Theorem 2.7 Iff isC2[a, b], and

f(t) =l+C1(t a)r +C2(t a)r+1 +(t)(t a)r+1, (2.13)

whereC1= 0, r >0. Thenl= f(a), and

(cx ax a)

r = 1

r+ 1

1 +

C2C1

[r+ 1

r+ 2 ( 1

r+ 1)1r ](x a)

+o(cx a) .

Proof.

First by Taylors theorem, there is some between a and t such that

f(t) =f(a) +f()(t a) .

Compare with (2.12) and take limit t a, we obtain l = f(a). Then multiply(x a) to (2.13) we get

f(t)(x a) = f(a)(x a) +C1(t a)r(x a) +C2(t a)r+1(x a)

+(t)(t

a)r+1(x

a) (2.14)

where (t) 0, as t a. On the other hand, equation (2.14) evaluated at cxyields

f(cx)(x a) = f(a)(x a) +C1(cx a)r(x a) +(cx)(cx a)r(x a)

+C2(cx a)r+1(x a) +(cx)(cx a)r+1(x a) (2.15)

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where (cx)

0, as cx

a.

Also,

f(x) f(a) = xa

f(t)dt

= f(a)(x a) +C1 (x a)r+1

r+ 1 +C2

(x a)r+2r+ 2

+ xa

(t)(t a)r+1dt. (2.16)

Equating (2.15) and (2.16), we have

(cx ax a)

r[1 +C2C1

(cx a) + 1C1

(cx)(cx a)]

= 1

r+ 1+

C2C1

(x ar+ 2

) + 1

C1(x a)r+1 xa

(t)(t a)r+1dt .

By Theorem 2.6 and Taylors expansion, we have

(cx ax a)

r = [1 C2C1

(cx a) +o(cx a)][ 1r+ 1

+C2C1

(x ar+ 2

)

+ 1

C1(x a)r+1 x

a

(t)(t

a)r+1dt]

= [1 C2C1

x a(1 +r)

1r

+o(cx a)][ 1r+ 1

+C2C1

(x ar+ 2

)

+o(x a)

= 1

r+ 1{1 + C2

C1[r+ 1

r+ 2 ( 1

r+ 1)1r ](x a)} +o(x a) .

Theorem 2.8 Supposef, g are continuous on[a, b], and differentiable on(a, b),

such thatf, g are integrable over [a, b] andg(x) = 0 for allx (a, b). If

limta

g(t)

(t a)s =C2 ;

and

limta

f(t)g(t)

K(t a)r =C1

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whereC1, C2

= 0, andr

= 0, s >

1, r+s >

1. Then the Cauchy differential

mean pointcx satisfies

limxa

cx ax a = (

s+ 1

r+s+ 1)1

r .

Proof.

Since

g(t) =C2(t a)s +2(t)(t a)s

and f(t)

g(t) =K+C1(t a)r +1(t)(t a)r (2.17)

where 1(t), 2(t) 0 as t a, we have

f(t) = [K+ C1(t a)r +1(t)(t a)r]g(t)

= KC2(t a)s +C1C2(t a)r+s +K2(t)(t a)s

+[C12(t) +C21(t) +1(t)2(t)](t a)r+s .

On the other hand, equation (2.17) evaluated at cx yields

f(cx)

g(cx) =K+C1(cx a)r +1(cx)(cx a)r .

Also,

f(x) f(a) = xa

f(t)dt

= KC2(x a)s+1

s+ 1 +

C1C2(x a)r+s+1r+s+ 1

+K xa

2(t)(t a)sdt

+ x

a[C12(t) +C21(t) +1(t)2(t)](t a)r+sdt

and

g(x) g(a) = xa

g(t)dt

= C2(x a)s+1

s+ 1 +

xa

2(t)(t a)sdt .

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Hence by (1.4) and Lemma 2.2, we obtain the following result

C1C2(x a)r+s+1r+s+ 1

+ xa

[C12(t) +C21(t) +1(t)2(t)](t a)r+sdt

=C1C2+C21(cx)

s+ 1 (cx a)r(x a)s+1 + [c1+c(cx)](cx a)r

xa

2(t)(t a)sdt .

Then we have

limxa

cx ax a = (

s+ 1

r+s+ 1)1r .

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Chapter 3

Converses of mean value

theorems

It would be interesting to study the converse of the classical mean value

theorem. That is, if c is a point in [a, b], can c be a differential mean point of

f over some sub-interval of [a, b] ? The following example shows the converse

problem does not always hold.

Example: If f(t) = t3 on [1, 1] and [t1, t2] (1, 1), then f(t2)f(t1)t2t1 =t22+t2t1+t

21>0 for all t1=t2, but f(0) = 0. Therefore (1.4) does not hold.

Tong and Braza [6] give the following theorem to show the converse problem

may fail at extremum values off and at certain accumulation points.

Definition 3.1 We say a functionf is locally linear (horizontal) about a point

c in its domainIif there is some open interval(, ) I containingc such thatf(t) |(,) is a linear (constant) function.

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Theorem 3.2 ([6]).

Iff is continuous on [a, b] and differentiable on (a, b), and c is given in (a, b).

Then

(a) Weak form: Iff(c) = sup{f(t)|t (a, b)} andf(c) = inf{f(t)|t (a, b)},then there is some interval (a1, b1) (a, b) such that f(c) = (f(b1)f(a1))/(b1 a1).

(b) Strong form: If f

(c) is not a local extremum value of f

on (a, b), andc is not an accumulation point of Ac ={t (a, b)|f(t) = f(c)}, thenthere is a sub-interval (a1, b1) (a, b) such that c (a, b) and f(c) =(f(b1) f(a1))/(b1 a1).

(c) Iff(c)is a local extremum value off on(a, b)(i.e,f(c)is a total extremum

value of f on some sub-interval (a, b) (a, b) containing c), then f is

either locally linear about c or f

(c)= (f()f())/( ) wherever < c < and(, ) (a, b).

We consider the converse of mean value theorem for integrals. If we give a

number c is in (a, b), could we find an interval (a1, b1)(a, b) such that c is anintegral mean point on (a1, b1), that is,

f(c) = 1

b1

a1

b1a1

f(t)dt ? (3.1)

We give a result similar to Theorem 3.2.

Theorem 3.3 Iff is continuous on [a, b], andc is given in(a, b). Then

(a) Weak form: Iff(c)= sup{t (a, b)|f(t)} andf(c)= inf{t (a, b)|f(t)},then there is some interval(a1, b1) (a, b) such that (3.1) hold.

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(b) Strong form: Iff(c) is not a local extremum value off on(a, b), andc is

not an accumulation point ofAc = {t (a, b)|f(t) =f(c)}, then there is asub-interval(a1, b1) (a, b) such thatc (a, b) and (3.1) hold.

(c) Iff(c)is a local extremum value offon(a, b)(i.e. f(c)is a total extremum

value of f on some sub-interval (a, b) (a, b) containing c), then f iseither locally horizontal aboutc orf(c)= f(t)dt/( ) wherever f(c) and K < f(c).

(1) K > f(c). Since g(y) = 1yx1

yx1

f(t)dt is continuous on (x1, b) and

g(y1) < f(c) < g(y2), there is a point y between y1 and y2 such that

g(y) =f(c).

(2) K < f(c). Since h(x) = 1y2x

y2x f(t)dt is continuous on (a, y2) and

h(x1)< f(c)< h(x2), there is a point xbetween x1 and x2 such that

h(x) =f(c).

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(b) Iff(c) is not a local extremum off on (a, b), then there is a sub-interval

(a0, b0) (a, b) such thatc (a0, b0) andf(c) is not a total extremum value

off on (a0, b0). Let

ai+1= (ai +c)/2, b

i+1= (b

i +c)/2

for i= 0, 1, 2, 3...Then

a0< a1 < ... < a

i < ... < c < ... < b

i < ... < b

1 < b

0

and limi ai = limi b

i . By part (a), for each sub-interval (a

i , b

i )

(a0, b0), (i > 0), there are ai, bi such that (ai, bi) (ai , bi ) and f(c) =bi

aif(t)dt/(bi ai). Ifc (ai, bi) for some i, the theorem is proved. Hence

we suppose c / (ai, bi) for alli N. By Theorem 1.1 for f on [ai, bi], thereis some ci (ai, bi) such that f(ci) = 1biai

biai

f(t)dt. Hencef(c) = f(ci).

Notice that since ci (ai, bi) (ai , bi ), limi(bi ai ) = 0, and ci= c,these ci cannot coincide infinitely often. This implies there is an infinite

discrete sequence cik such that limk cik = c. This is contradiction since

it implies that c is an accumulation point of the set Ac= {t (a, b)|f(t) =f(c)}.

(c) Letf(c) = 0 be a local maximum offsince we can replacefbyf(t)f(c).There is a sub-interval (, ) (a, b) such that < c < and

f(c) = 1

f(t)dt .

So f(t)dt= 0 andf(t) 0 for allt (, ). Asfis continuous,f(t) 0

on (, ). Thus f is locally horizontal about c.

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Alternative proof.

DefineF(x) =xa f(t)dt. ThenFis continuous on [a, b] andF

(x) =f(x) for any

x (a, b). Then

f(c) = sup{t (a, b)|f(t)} (= inf{t (a, b)|f(t)})

if and only if

F(c)

= sup

{F(t)

|t

(a, b)

} (

= inf

{F(t)

|t

(a, b)

}) .

Thus, by Theorem 3.2 (a), there exists some interval (a1, b1) (a, b) such that

F(c) =F(b1) F(a1)

b1 a1 ,

that is

f(c) = 1

b1 a1 b1a1

f(t)dt .

Part (a) is proved. Similarly, part (b) and part (c) also follow from Theorem 3.2

(b) and (c) respectively.

Theorem 3.2 shows that the converse to the classical mean value theorem may

fail at extremum values off(x) and at certain accumulation points. In [6], Tong

and Braza also studied the number of bad points (i.e. points which are neither

strong nor weak differential mean points). They gave two examples. The below

is one of them.

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Example: Define

f(x) =

x3 , x [0, 1/4]2(1/4)3 + (x 1/2)3 , x [1/4, 5/8]2(1/4)3 + 2(1/8)3 + (x 3/4)3 , x [5/8, 13/16]...

2[(1/4)3 + (1/8)3 +...+ ( 12k+1

)3] + (x 2k12k

)3 , x [2k+132k+1

, 2k+232k+2

]

...

2

k=2( 12k

)3 + (x 1)3 . x [1, 2]

The function fis continuous on [0, 2] and differentiable on (0, 2). It is monoton-

ically increasing, with f(2k12k

) = 0 for k= 0, 1, 2, and f(1) = 0, but

f(x) f(y)x y >0

for x, y [0, 2], with x =y. Thusfhas a countable number of bad points.Could these bad points be dense? Is it possible to have an uncountable number

of such points? In [2], Borwein and Wang answered these questions affirmatively.

They gave an example of a function having an uncountable number of bad points.

These points are dense in the nondegenerate interval [a, b]. However the Lebesque

measure is still zero.

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