Topics on Mean Value Theorems
Gen-Bin Huang
January 19, 2001
Abstract
Let f be a continuous function on [a, x], then there exists a ξx in [a, x] such
that
Z
x
a
f (t)dt = f (ξx )(x − a) .
This is the mean value theorem for integrals. We call ξx an integral mean point
of f on [a, x]. Recently, there are a number of studies on the location of ξx .
In particular, Zhang [7] showed that if f is C r (r ∈ N ) near a point a, with
f 0 (a) = f 00 (a) = . . . = f (r−1) (a) = 0 but f (r) (a) 6= 0, then
lim
x→a
ξx − a
1
=
1 .
x−a
(r + 1) r
Note that there is also a differential mean point cx in the classical mean value theorem. Using a similar method, I give a detailed estimation of the location of cx .
Since the classical mean value theorem implies the mean value theorem for integrals, 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 Zhang’s result to consider the integral mean point of the generalized
mean value theorem, assuming f to behave like K + C1 (x − a)r near a (r > −1)
and g 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 if f 0 (c) is not a local extremum, then c
is a differential mean point. I obtain a parallel theorem for the converse of the
mean value theorem for integrals.
2
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
i
18
Chapter 1
Introduction
1.1
The mean value theorems
First, we introduce the mean value for integrals. Assume that f is continuous,
then
1 Zb
f (t)dt =
µ=
b−a a
Rb
a
f (t)dt
.
a dt
Rb
This µ is called the arithmetic average or the mean value of f in the interval
[a, b]. We state the mean value theorem for integrals:
Theorem 1.1 ([1, p.258] Mean value theorem for integrals).
If f is a continuous function on [a, x], then there exists a number ξx in [a, x] such
that
Z
x
a
f (t)dt = f (ξx )(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
1
one ξx in the interval for which f (ξx ) is equal to the average value of f but gives
no further information about the location ξx . In this thesis, we call ξx an integral
mean point of f on [a, x].
Instead of the simple arithmetic average we can form the weighted averages:
Rx
µ=
a
f (t)g(t)dt
a g(t)dt
Rx
g is the weight function and g(t) ≥ 0. We give the generalized mean value theorem
for integrals:
Theorem 1.2 ([1, p.257] Generalized mean value theorem for integrals).
If f is a continuous function on [a, x], g is integrable on [a, x] and g ≥ 0, then
there exists a number ξx in [a, x] such that
Z
x
a
f (t)g(t)dt = f (ξx )
Z
x
g(t)dt .
(1.2)
a
Obviously, Theorem 1.1 is the special case when g(t) = 1.
Theorem 1.3 ([1, p.197] Classical mean value theorem).
Suppose f is continuous on the closed interval [a, x] and differentiable on (a, x),
then for some cx between [a, x], we have
f 0 (cx ) =
f (x) − f (a)
.
x−a
(1.3)
We could estimate cx as ξx in the same way as in (1.2). And we call cx a
differential mean point of f on [a, x].
Theorem 1.4 ([1, p.198] Cauchy’s mean value theorem).
Suppose f , g are continuous on [a, x], differentiable on (a, x). If g 0 (t) 6= 0 for any
t ∈ (a, x), then there exists cx ∈ (a, x) such that
f (x) − f (a)
f 0 (cx )
= 0
.
g(x) − g(a)
g (cx )
2
(1.4)
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
point ξx , as x → a. B. Zhang [7], improving B. Jacobson’s result [4], showed that
if f is r (r ∈ N ) times differentiable at a, with f 0 (a) = f 00 (a) = . . . = f (r−1) (a) = 0
but f (r) (a) 6= 0, then
lim
x→a
1
ξx − a
=
1 .
x−a
(r + 1) r
Schwind-Ji-Koditschek [5] went on further. They allow f and g to have a ’singularity’ at a. Namely they showed that if
lim
t→a
f (t) − K
= C1
(t − a)r
and
g(t)
= C2 ,
t→a (t − a)s
lim
where r 6= 0,s > −1, r + s > −1, C1 , C2 6= 0, then
ξx − a
s+1 1
=(
)r .
x→a x − a
r+s+1
lim
On the other hand, Tong and Braza [6] studied the converse of the classical
mean value theorem. They showed that if f 0 (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 interrelated. 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 estimates of the differential mean point and the converse of the mean value theorem
for integrals.
3
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
ξx . 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 theorems
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]).
Suppose f : [a, x] → R is continuous on [a, x]. Define F (t) =
Rt
a
f (s)ds. Then F
is continuous on [a, x] and F 0 = f on [a, x]. Hence by Theorem 1.3, there is some
cx ∈ (a, x) such that
F (x) − F (a) = F 0 (cx )(x − a) .
That means,
Z
x
a
f (t)dt = f (cx )(x − a) .
♦
Conversely, suppose F is C 1 and F 0 = f . So
F (x) − F (a) =
Z
x
a
4
f (t)dt .
By Theorem 1.1 and above formula, we conclude that
x
Z
F (x) − F (a) =
f (t)dt = f (ξx )(x − a) .
a
Since F is a primitive of f , we could replace f (ξx ) by F 0 (ξx ) and obtain
F (x) − F (a)
.
x−a
F 0 (ξx ) =
It is the same form as (1.4). Thus if F is C 1 , 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) =
Z
t
f (s)g(s)ds,
G(t) =
a
Z
t
g(s)ds .
a
F and G are continuous on [a, x] and differentiable on (a, x). Also
G(x) =
Z
x
g(t)dt > G(a) = 0 .
a
Hence we apply Theorem 1.4 to obtain some cx ∈ (a, x) such that
F (x) − F (a) =
F 0 (cx )
[G(x) − G(a)] .
G0 (cx )
Thus
Z
x
a
f (t)g(t)dt =
f (cx )g(cx ) Z x
g(t)dt
g(cx )
a
= f (cx )
5
Z
x
a
g(t)dt.
♦
Conversely, under additional assumption on f and g, then Theorem 1.2 implies
Theorem 1.4. If f and g are both C 1 , let φ =
f0
g0
continuous, θ = g 0 > 0 for
simplicity, then by Theorem 1.2, there exists cx ∈ (a, x) such that
x
Z
φ(t)θ(t)dt = φ(cx )
a
Z
x
θ(t)dt .
a
Therefore,
Z
x
a
f 0 (t)dt =
f 0 (cx ) Z x 0
g (t)dt .
g 0 (cx ) a
So
f (x) − f (a)
f 0 (cx )
= 0
.
g(x) − g(a)
g (cx )
♦
6
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 ξx .
Jacobson [4] described the following theorem may fall into the category of ”interesting facts we once knew, but have now forgotten”. The theorem says that as x
approaches a, the value of ξx approaches the midpoint between a and x.
Theorem 2.1 ([4]).
If f is continuous on [a, x], differentiable at a, f 0 (a) 6= 0 and ξx is taken as in
Theorem 1.1, then
lim
x→a
ξx − a
1
= .
x−a
2
7
Before proving Theorem 2.1, we give Lemma 2.2:
Lemma 2.2 If (t) → 0 as t → a and (t)(t − a)m is integrable, then
Rx
(a) limx→a
(b) limx→a
a
(t)(t−a)m dt
(x−a)m+1
(ξx )(ξx −a)m
(x−a)m
=0;
= 0, where m > 0 .
Proof.
(a) Given any δ > 0, there exist some x0 > 0 such that for any 0 < |t − a| < x0 ,
|(t)| < δ. Then take any x such that 0 < |x − a| < x0 . Without loss of
generality, let x > a. Then
Rx
|
a
(b) Since a < ξx < x, 0 <
(t)(t − a)m dt
δ ax (t − a)m dt
|
≤
(x − a)m+1
(x − a)m+1
δ
=
.
m+1
R
ξx −a
x−a
|
< 1. Therefore,
(ξx )(ξx − a)m
| ≤ |(ξx )|
(x − a)m
and so converges to 0 as x → a.
♦
Proof of Theorem 2.1.
By Taylor’s expansion,
f (t) = f (a) + f 0 (a)(t − a) + (t)(t − a) ,
(2.1)
where (t) → 0 as t → a. Integrate (2.1) from a to x, and obtain
Z
x
a
f (t)dt = f (a)(x − a) +
f 0 (a)(x − a)2 Z x
+
(t)(t − a)dt .
2!
a
8
(2.2)
Equation (2.1) evaluated at ξx (a < ξx < x) yields
f (ξx ) = f (a) + f 0 (a)(ξx − a) + (ξx )(ξx − a),
where (ξx ) → 0 as ξx → a. Thus,
f (ξx )(x − a) = f (a)(x − a) + f 0 (a)(ξx − a)(x − a)
+(ξx )(ξx − a)(x − a) .
(2.3)
From (1.1), (2.2) and (2.3) we have
f 0 (a)(x − a)2 Z x
+
(t)(t − a)dt
2!
a
= f 0 (a)(ξx − a)(x − a) + (ξx )(ξx − a)(x − a) .
Hence,
ξx − a
f 0 (a)
=
+
f (a)
x−a
2
0
Rx
a
(t)(t − a)dt (ξx )(ξx − a)
−
.
(x − a)2
x−a
(2.4)
By Lemma 2.2,
Rx
lim
x→a
a
(t)(t − a)dt
(ξx )(ξx − a)
= lim
=0.
2
x→a
(x − a)
x−a
From (2.4) we obtain limx→a
ξx −a
x−a
= 12 .
♦
We consider the case f 0 (a) = 0 in Theorem 2.1 and obtain Theorem 2.3.
Theorem 2.3 ([7]).
If f is continuous on [a, x], and is twice differentiable at a, with f 0 (a) = 0,
f 00 (a) 6= 0. If ξx is taken as in Theorem 1.1, then
1
ξx − a
=√ .
x→a x − a
3
lim
9
In a similar way, we can establish the following general result.
Theorem 2.4 ([7]).
If f is continuous on [a, x], and is r times (r > 1, r ∈ N ) differentiable at a, with
f 0 (a) = f 00 (a) = . . . = f (r−1) (a) = 0 but f (r) (a) 6= 0. If ξx is taken as in Theorem
1.1, then
lim
x→a
ξx − a
1
=
1 .
x−a
(r + 1) r
If we consider a large function class, which need not assume f to be differentiable, to estimate ξx . And we take a transformation of f .
Theorem 2.5 ([5]).
If f is continuous on (a, x] and g is integrable on (a, b) with g(t) ≥ 0 for t ∈ (a, b).
If limt→a
f (t)−K
(t−a)r
= C1 , limt→a
g(t)
(t−a)s
= C2 for some K ∈ <, C1 , C2 6= 0, r 6= 0 and
s > −1 such that r + s > −1, then
(a) there exists ξx ∈ (a, x] such that
Z
x
a
f (t)g(t)dt = f (ξx )
Z
x
g(t)dt ,
(2.5)
a
(b) for any choice of ξx , we have
lim
x→a
ξx − a
s+1 1
=(
)r .
x−a
r+s+1
Proof.
(a) Since both
lim
t→a
f (t) − K
(t − a)r
and
lim
t→a
g(t)
(t − a)s
10
(2.6)
exist, then given the limits
f (t) = K + C1 (t − a)r + 1 (t)(t − a)r ,
(2.7)
g(t) = C2 (t − a)s + 2 (t)(t − a)s ,
(2.8)
and
where 1 (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 + [C1 C2 + C1 2 (t) + C2 1 (t) + 1 (t)2 (t)](t − a)r+s |
≤ |K(C2 + 2 (t))|(t − a)s + |C1 C2 + C1 2 (t) + C2 1 (t) + 1 (t)2 (t)|(t − a)r+s
≤ K1 (t − a)s + K2 (t − a)r+s
where K1 , K2 are some constant, then
Rx
a
f (t)g(t)dt exists. Now,we give the
proof of (a).
Case 1: 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, there
exists ξx ∈ (a, x], such that
Z
x
a
f (t)g(t)dt =
Z
x
a
F (t)g(t)dt = f (ξx )
Z
x
g(t)dt .
a
Case 2: r < 0.
Without loss of generality, we take C1 and
Rx
a
g(t)dt > 0. From (2.7), then
limt→a f (t) = ∞, and there exists δ > 0 such that
Rx
f (t) >
a
f (t)g(t)dt
≡η
a g(t)dt
Rx
for t ∈ (a, a + η). Assume that there exists ξx satisfying (2.5), but ξx 6∈
(a, a + δ). Take a1 ∈ (a, a + δ), f is still continuous on [a1 , x], f (t) > η on
11
(a, a1 ), then
min f (t) = min f (t);
a1 ≤t≤x
a≤t<x
min f (t) ≤= min f (t) ≤ η < f (a1 ) ≤ max f (t) .
a1 ≤t<x
a1 ≤t<x
a≤t<x
Therefore, there exists ξx between a and x such that
Rx
f (ξx ) = η =
f (t)g(t)dt
.
a g(t)dt
a
Rx
(b) Since
Z
x
f (t)g(t)dt =
a
x
Z
K[C2 (t − a)s + 2 (t)(t − a)s ]dt + C1 C2
a
+C2
Z
+
Z
x
a
x
a
r+s
1 (t)(t − a)
dt + C1
Z
x
a
(x − a)r+s+1
r+s+1
2 (t)(t − a)r+s dt
1 (t)2 (t)(t − a)r+s dt
Also,
f (ξx )
Z
x
g(t)dt =
a
Z
x
a
K[C2 (t − a)s + 2 (t)(t − a)s ]dt + C1 C2
+C2
(ξx − a)r (x − a)s+1
s+1
Z x
1 (ξx )(ξx − a)r (x − a)s+1
+ C1 (ξx − a)r
2 (t)(t − a)s dt
s +Z 1
a
+1 (ξx )(ξx − a)r
x
a
2 (t)(t − a)s dt
Hence by (a), Lemma 2.2, we obtain the following result
C1 C2
(x − a)r+s+1
(ξx − a)r (x − a)s+1
(1 + o(1)) = C1 C2
(1 + o(1)) .
r+s+1
s+1
Then we have the result
s+1
ξx − a r
= lim (
) .
r + s + 1 x→a x − a
♦
Remark:
Obviously, Theorem 2.4 is a special case of Theorem 2.5.
12
2.2
Differential mean points
We have the same results as Theorem 2.4 of the classical mean value theorem.
Theorem 2.6 If f is continuous on [a, b] and differentiable on (a, b), f 0 is Rief 0 (t)−l
(t−a)r
mann integrable and limt→a
= C1 exists in < (r 6= 0, r > 1), then the
differential mean point cx satisfies
cx − a
1 1
=(
)r .
x→a x − a
1+r
lim
Proof.
Since limt→a
f 0 (t)−l
(t−a)r
= C1 ,
f 0 (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 cx
yields
f 0 (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 0 (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) =
Z
x
f 0 (t)dt
a
= l(x − a) + C1
(x − a)r+1 Z x
+
(t)(t − a)r dt .
r+1
a
(2.12)
Equating (2.11) and (2.12), we have
(x − a)r+1 Z x
+
(t)(t − a)r dt
C1 (cx − a) (x − a) + (cx )(cx − a) (x − a) = C1
r+1
a
r
r
13
Take x → a, then we get the following result by Lemma 2.2
cx − a
1 1
=(
)r .
x→a x − a
1+r
lim
♦
If we consider f is in C 2 [a, b], we can derive a more detailed asymptotic
expression for cx .
Theorem 2.7 If f is C 2 [a, b], and
f 0 (t) = l + C1 (t − a)r + C2 (t − a)r+1 + (t)(t − a)r+1 ,
(2.13)
where C1 6= 0, r > 0. Then l = f 0 (a), and
cx − a r
1
C2 r + 1
1 1
(
) =
1+
[
−(
) r ](x − a) + o(cx − a) .
x−a
r+1
C1 r + 2
r+1
Proof.
First by Taylor’s theorem, there is some η between a and t such that
f 0 (t) = f 0 (a) + f 0 (η)(t − a) .
Compare with (2.12) and take limit t → a, we obtain l = f 0 (a). Then multiply
(x − a) to (2.13) we get
f 0 (t)(x − a) = f 0 (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 cx
yields
f 0 (cx )(x − a) = f 0 (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)
14
(2.15)
where (cx ) → 0, as cx → a.
Also,
f (x) − f (a) =
Z
x
f 0 (t)dt
a
= f 0 (a)(x − a) + C1
+
Z
x
(x − a)r+2
(x − a)r+1
+ C2
r+1
r+2
(t)(t − a)r+1 dt.
(2.16)
a
Equating (2.15) and (2.16), we have
(
cx − a r
C2
1
) [1 +
(cx − a) +
(cx )(cx − a)]
x−a
C1
C1
Z x
1
C2 x − a
1
=
+
(
)+
(t)(t − a)r+1 dt .
r + 1 C1 r + 2
C1 (x − a)r+1 a
By Theorem 2.6 and Taylor’s expansion, we have
(
cx − a r
C2
1
C2 x − a
) = [1 −
(cx − a) + o(cx − a)][
+
(
)
x−a
C1
r + 1 C1 r + 2
Z x
1
+
(t)(t − a)r+1 dt]
C1 (x − a)r+1 a
1
C2 x − a
C2 x − a
+
(
)
= [1 −
1 + o(cx − a)][
r + 1 C1 r + 2
C1 (1 + r) r
+o(x − a)
=
1
C2 r + 1
1 1
{1 +
[
−(
) r ](x − a)} + o(x − a) .
r+1
C1 r + 2
r+1
♦
Theorem 2.8 Suppose f , g are continuous on [a, b], and differentiable on (a, b),
such that f 0 , g 0 are integrable over [a, b] and g 0 (x) 6= 0 for all x ∈ (a, b). If
g 0 (t)
= C2 ;
t→a (t − a)s
lim
and
lim
t→a
f 0 (t)
g 0 (t)
−K
(t − a)r
15
= C1
where C1 , C2 6= 0, and r 6= 0, s > −1, r + s > −1. Then the Cauchy differential
mean point cx satisfies
lim
x→a
cx − a
s+1 1
=(
)r .
x−a
r+s+1
Proof.
Since
g 0 (t) = C2 (t − a)s + 2 (t)(t − a)s
and
f 0 (t)
= K + C1 (t − a)r + 1 (t)(t − a)r
g 0 (t)
(2.17)
where 1 (t), 2 (t) → 0 as t → a, we have
f 0 (t) = [K + C1 (t − a)r + 1 (t)(t − a)r ]g 0 (t)
= KC2 (t − a)s + C1 C2 (t − a)r+s + K2 (t)(t − a)s
+[C1 2 (t) + C2 1 (t) + 1 (t)2 (t)](t − a)r+s .
On the other hand, equation (2.17) evaluated at cx yields
f 0 (cx )
= K + C1 (cx − a)r + 1 (cx )(cx − a)r .
g 0 (cx )
Also,
f (x) − f (a) =
Z
x
f 0 (t)dt
a
Z x
KC2 (x − a)s+1 C1 C2 (x − a)r+s+1
+
+K
2 (t)(t − a)s dt
=
s+1
r+s+1
a
Z
x
+
a
[C1 2 (t) + C2 1 (t) + 1 (t)2 (t)](t − a)r+s dt
and
g(x) − g(a) =
Z
x
g 0 (t)dt
a
C2 (x − a)s+1 Z x
=
+
2 (t)(t − a)s dt .
s+1
a
16
Hence by (1.4) and Lemma 2.2, we obtain the following result
C1 C2 (x − a)r+s+1 Z x
+ [C1 2 (t) + C2 1 (t) + 1 (t)2 (t)](t − a)r+s dt
r+s+1
a
Z x
C1 C2 + C2 1 (cx )
r
s+1
r
=
(cx − a) (x − a) + [c1 + c (cx )](cx − a)
2 (t)(t − a)s dt .
s+1
a
Then we have
lim
x→a
cx − a
s+1 1
=(
)r .
x−a
r+s+1
♦
17
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 )
t2 −t1
=
t22 + t2 t1 + t21 > 0 for all t1 6= t2 , but f 0 (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 of f 0 and at certain accumulation points.
Definition 3.1 We say a function f is locally linear (horizontal) about a point
c in its domain I if there is some open interval (α, β) ⊂ I containing c such that
f (t) |(α,β) is a linear (constant) function.
18
Theorem 3.2 ([6]).
If f is continuous on [a, b] and differentiable on (a, b), and c is given in (a, b).
Then
(a) Weak form: If f 0 (c) 6= sup{f 0 (t)|t ∈ (a, b)} and f 0 (c) 6= inf{f 0 (t)|t ∈ (a, b)},
then there is some interval (a1 , b1 ) ⊂ (a, b) such that f 0 (c) = (f (b1 ) −
f (a1 ))/(b1 − a1 ).
(b) Strong form: If f 0 (c) is not a local extremum value of f 0 on (a, b), and
c is not an accumulation point of Ac = {t ∈ (a, b)|f 0 (t) = f 0 (c)}, then
there is a sub-interval (a1 , b1 ) ⊂ (a, b) such that c ∈ (a, b) and f 0 (c) =
(f (b1 ) − f (a1 ))/(b1 − a1 ).
(c) If f 0 (c) is a local extremum value of f 0 on (a, b) (i.e, f 0 (c) is a total extremum
value of f 0 on some sub-interval (a∗ , b∗ ) ⊂ (a, b) containing c), then f is
either locally linear about c or f 0 (c) 6= (f (β) − f (α))/(β − α) wherever
α < c < β and (α, β) ⊂ (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 an
integral mean point on (a1 , b1 ), that is,
1 Z b1
f (c) =
f (t)dt ?
b 1 − a1 a 1
(3.1)
We give a result similar to Theorem 3.2.
Theorem 3.3 If f is continuous on [a, b], and c is given in (a, b). Then
(a) Weak form: If f (c) 6= sup{t ∈ (a, b)|f (t)} and f (c) 6= inf{t ∈ (a, b)|f (t)},
then there is some interval (a1 , b1 ) ⊂ (a, b) such that (3.1) hold.
19
(b) Strong form: If f (c) is not a local extremum value of f on (a, b), and c is
not an accumulation point of Ac = {t ∈ (a, b)|f (t) = f (c)}, then there is a
sub-interval (a1 , b1 ) ⊂ (a, b) such that c ∈ (a, b) and (3.1) hold.
(c) If f (c) is a local extremum value of f 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 horizontal about c or f (c) 6=
Rβ
α
f (t)dt/(β − α) wherever α <
c < β and (α, β) ⊂ (a∗ , b∗ ).
Proof.
(a) If f (c) is not a total extremum value on (a, b), then there are c1 , c2 ∈ (a, b)
(assume c1 < c2 ) such that f (c1 ) < f (c) < f (c2 ). Since
1 Zy
f (t)dt
f (ci ) = lim
x,y→ci y − x x
for i = 1, 2, there are sub-intervals (x1 , y1 ), (x2 , y2 ) ⊂ (a, b) such that
Z y1
Z y2
1
1
f (t)dt < f (c) <
f (t)dt .
y 1 − x 1 x1
y 2 − x 2 x2
Without loss of generality, suppose x1 < x2 . Consider the value K =
1
y2 −x1
R y2
x1
f (t)dt. If K = f (c), we are done. We need consider only the cases
K > f (c) and K < f (c).
(1) K > f (c). Since g(y) =
1
y−x1
Ry
x1
f (t)dt is continuous on (x1 , b) and
g(y1 ) < f (c) < g(y2 ), there is a point ȳ between y1 and y2 such that
g(ȳ) = f (c).
(2) K < f (c). Since h(x) =
1 R y2
y2 −x x
f (t)dt is continuous on (a, y2 ) and
h(x1 ) < f (c) < h(x2 ), there is a point x̄ between x1 and x2 such that
h(x̄) = f (c).
20
(b) If f (c) is not a local extremum of f on (a, b), then there is a sub-interval
(a∗0 , b∗0 ) ⊂ (a, b) such that c ∈ (a∗0 , b∗0 ) and f (c) is not a total extremum value
of f on (a∗0 , b∗0 ). Let
a∗i+1 = (a∗i + c)/2, b∗i+1 = (b∗i + c)/2
for i = 0, 1, 2, 3... Then
a∗0 < a∗1 < ... < a∗i < ... < c < ... < b∗i < ... < b∗1 < b∗0
and limi→∞ a∗i = limi→∞ b∗i . By part (a), for each sub-interval (a∗i , b∗i ) ⊂
(a∗0 , b∗0 ), (i > 0), there are ai , bi such that (ai , bi ) ⊂ (a∗i , b∗i ) and f (c) =
R bi
ai
f (t)dt/(bi − ai ). If c ∈ (ai , bi ) for some i, the theorem is proved. Hence
we suppose c ∈
/ (ai , bi ) for all i ∈ N . By Theorem 1.1 for f on [ai , bi ], there
is some ci ∈ (ai , bi ) such that f (ci ) =
1
bi −ai
R bi
ai
f (t)dt. Hence f (c) = f (ci ).
Notice that since ci ∈ (ai , bi ) ⊂ (a∗i , b∗i ), limi→∞ (b∗i − a∗i ) = 0, and ci 6= 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) Let f (c) = 0 be a local maximum of f since we can replace f by f (t) − f (c).
There is a sub-interval (α, β) ⊂ (a∗ , b∗ ) such that α < c < β and
1 Zβ
f (c) =
f (t)dt .
α−β α
So
Rβ
α
f (t)dt = 0 and f (t) ≤ 0 for all t ∈ (α, β). As f is continuous, f (t) ≡ 0
on (α, β). Thus f is locally horizontal about c.
♦
21
Alternative proof.
Rx
Define F (x) =
a
f (t)dt. Then F is continuous on [a, b] and F 0 (x) = f (x) for any
x ∈ (a, b). Then
f (c) 6= sup{t ∈ (a, b)|f (t)} (6= inf{t ∈ (a, b)|f (t)})
if and only if
F 0 (c) 6= sup{F 0 (t)|t ∈ (a, b)} (6= inf{F 0 (t)|t ∈ (a, b)}) .
Thus, by Theorem 3.2 (a), there exists some interval (a1 , b1 ) ⊂ (a, b) such that
F 0 (c) =
F (b1 ) − F (a1 )
,
b 1 − a1
that is
1 Z b1
f (c) =
f (t)dt .
b 1 − a1 a 1
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 of f 0 (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.
22
Example:
Define




x3 ,








2(1/4)3 + (x − 1/2)3 ,







2(1/4)3 + 2(1/8)3 + (x − 3/4)3 ,




x ∈ [0, 1/4]
x ∈ [1/4, 5/8]
x ∈ [5/8, 13/16]
f(x) =  ...




1


)3 ] + (x −
2[(1/4)3 + (1/8)3 + ... + ( 2k+1








...





P∞

1 3
3

2
k=2 ( 2k )
+ (x − 1) .
2k −1 3
)
2k
,
k+2
k+1
x ∈ [ 2 2k+1−3 , 2 2k+2−3 ]
x ∈ [1, 2]
The function f is continuous on [0, 2] and differentiable on (0, 2). It is monotonk
0
ically increasing, with f 0 ( 2 2−1
k ) = 0 for k = 0, 1, 2, · · · and f (1) = 0, but
f (x) − f (y)
>0
x−y
for x, y ∈ [0, 2], with x 6= y. Thus f has 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.
23
Bibliography
[1] R.G. Bartle and D.R. Sherbert, An Introduction to Real Analysis, second
edition, Wiley, New York (1992).
[2] J.M. Borwein and Xianfu Wang, The converse of the mean value theorem
may fail generically, Amer. Math. Monthly 105 (1998), 847-848.
[3] H.E. Campbell and P.F. Dierker, Calculus, second edition, Prindle, Weber
Schmidt (1978).
[4] B. Jacobson, On the mean value theorem for integrals, Amer. Math. Monthly
89 (1982), 300-301.
[5] W.J. Schwind, J. Ji, and D.E. Koditschek, A physically motivated further
note on the mean value theorem for integrals, Amer. Math. Monthly 106
(1999), 559-564.
[6] J. Tong and P.A. Braza, A converse of the mean value theorem, Amer. Math.
Monthly 104 (1997), 939-942.
[7] B. Zhang, A note on the mean value theorem for integrals, Amer. Math.
Monthly 104 (1997), 561-562.
24