Journal of Inequalities in Pure and
Applied Mathematics
ON SOME SPECTRAL RESULTS RELATING TO THE RELATIVE
VALUES OF MEANS
volume 4, issue 3, article 59,
2003.
C.E.M. PEARCE
School of Applied Mathematics,
Adelaide University,
Adelaide SA 5005, Australia.
EMail: cpearce@maths.adelaide.edu.au
Received 21 January, 2003;
accepted 22 July, 2003.
Communicated by: P. Cerone
Abstract
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2000
Victoria University
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Abstract
In the case of two positive numbers, the geometric mean is closer to the harmonic than to the arithmetic mean. We derive some spectral results relating to
corresponding properties with more than two positive numbers.
2000 Mathematics Subject Classification: 26E60, 26D15
Key words: AGH inequality, Means, Relative values, Spectrum.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
The Dichotomy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3
Comparison Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
4
Characterisation of εk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
References
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
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1.
Introduction
Let A, G, H denote respectively the arithmetic, geometric and harmonic means
of n positive real numbers x1 , . . . , xn , which are not all equal. It is well–known
that H < G < A. Scott [3] has shown in the case n = 2 that G is closer to H
than to A, so that
(1.1)
1
A−G
> .
A−H
2
He showed by a counterexample that this need not be the case when n > 2.
Subsequently Lord [1] and Pearce and Pečarič [2] addressed the question of
the behaviour of the quotient
fn (x1 , . . . , xn ) :=
A−G
A−H
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
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in the case of general n. Several generalisations and extensions of (1.1) were
obtained. The following are pertinent to the present article.
Since
fn (ax1 , . . . , axn ) = f (x1 , . . . , xn )
for a > 0, it suffices to consider the values taken by fn when x = (x1 , . . . , xn )
lies on the intersection
(
)
n
X
K := x ∈ Rn : xi ≥ 0 for 1 ≤ i ≤ n and
x2i = 1
i=1
of the nonnegative orthant and the surface of the unit hypersphere. The function
fn is clearly well–defined and continuous on the interior of K except at e :=
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√1 , . . . , √1
n
n
, where it is undefined since A, G and H all coincide. In fact
this singularity is removable. It is shown in [1] that defining fn (x) = 1 for
boundary points of K (where some but not all values xi vanish) and fn (e) = 12
makes fn continuous on the whole of K. Since K is compact, fn possesses and
realises an infimum αn . Further, the range of fn constitutes the interval [αn , 1],
1
the sequence (αn )∞
2 is strictly decreasing to limit zero and αn > n for n ≥ 3.
1
The seminal paper of Scott gives α2 = 2 .
In this article we continue the development of [1] and [2] and derive some
striking structural results, principally as follows. In Section 2, Theorem 2.1, we
show that if x is such that fn (x) = αn , then {x1 , . . . , xn } contains precisely two
distinct values. In Section 3, Theorem 3.3, we show that if fn (x) = αn , then the
smaller of the two distinct components of x must occur with multiplicity one.
We conclude in Section 4 by giving characterisations of αn and some related
infima arising naturally in our analysis.
We postpone consideration of asymptotics to a subsequent article.
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
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2.
The Dichotomy Theorem
Theorem 2.1. For n > 2, any set {x1 , . . . , xn } for which fn (x) = αn contains
precisely two distinct values.
P
Q
Proof. First suppose that S1 := ni=1 xi and Sn := ni=1 xi are fixed. P
Subject
to these constraints, the mimimum of fn correspond to an extremum of ni=1 x1i
and satisfies
∂L
= 0 for i = 1, . . . , n,
∂xi
where L denotes the Lagrangian
!
!
n
n
n
X
X
Y
1
L :=
−λ
x i − S1 − µ
x i − Sn .
x
i
i=1
i=1
i=1
Then
Y
1
∂L
=− 2 −λ−µ
xj = 0
∂xi
xi
j6=i
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
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(i = 1, . . . , n),
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that is,
1
+ λxi + µSn = 0 (i = 1, . . . , n).
xi
Hence each xi must be equal to one of the two solutions of the quadratic
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λx2 + µSn x + 1 = 0.
For a minimum, these solutions must be distinct, since fn (e) =
for n ≥ 3.
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1
2
while αn <
1
2
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For j = 1, 2, . . . , n and fixed n > 2, define
Vj = {x : {x1 , . . . , xn } contains precisely j distinct values} ,
Vj∗ = {fn (x) : x ∈ Vj }
and
δj = inf Vj∗ .
An immediate implication of Theorem 2.1 is the following result.
On Some Spectral Results
Relating to the Relative Values
of Means
Corollary 2.2. We have
δ2 = δ3 = · · · = δn = αn .
For j > 1, the set Vj∗ contains its infimum only for j = 2.
Proof. If 1 ≤ j ≤ n − 1, any element of Vj can be approximated arbitrarily
closely by elements of Vj+1 , but not conversely. Since K is compact and fn
continuous, we must therefore have that δj+1 ≤ δj . Thus
1
δn ≤ δn−1 ≤ . . . ≤ δ2 ≤ δ1 = .
2
On the other hand, by Theorem 2.1
δ2 = αn = inf {fn (x)} = min{δ1 , δ2 , . . . , δn }.
The first part of the corollary follows.
The second part follows by invoking Theorem 2.1 again.
C.E.M. Pearce
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3.
Comparison Results
In the remaining sections of the paper we examine more closely the central case
when {x1 , . . . , xn } contains only two distinct values, that is x ∈ V2 . We may
assume without loss of generality an ordering
x1 ≤ x2 ≤ · · · ≤ xn .
We decompose
V2 =
n−1
[
On Some Spectral Results
Relating to the Relative Values
of Means
Uk ,
k=1
C.E.M. Pearce
where
Uk = {x : x1 = x2 = · · · = xk < xk+1 = · · · = xn }
(1 ≤ k < n).
For x ∈ Uk we have for the k equal points denoted by x and the rest by y
that
k
x + 1 − nk y − xk/n y 1−k/n
A−G
n
.
.
=
k
k
k
n−k
A−H
x+ 1− y−n
+
n
n
x
y
If we set β = k/n and u = x/y, this gives
fn (x) =
with β ∈
1
, 2 , . . . , n−1
n n
n
βu + 1 − β − uβ
βu + 1 − β − 1 βu + 1 − β
and 0 < u < 1.
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This may be rearranged as
(3.1)
fn (x) = 1 −
u
g(u, β),
(u − 1)2
where
(3.2)
g(u, β) =
uβ − 1 u−(1−β) − 1
+
.
β
1−β
We shall find it convenient to have alternative sets of variables and functions.
Set v = u1/n . Then for x ∈ Uk we put
k
.
hk (n, v) = fn (x) and φk (v) = g u,
n
Proposition 3.1. For fixed n ≥ 3 and v ∈ (0, 1), the sequence
strictly increasing.
n−1
(hk (n, v))k=1
is
Proof. By virtue of the representation (3.1), (3.2), it suffices to prove that the
sequence (φk (v))n−1
k=1 is strictly decreasing. To show that φk (v) > φk+1 (v), we
need to establish the inequality
v k − 1 v −(n−k) − 1
v k+1 − 1 v −(n−k−1) − 1
+
>
+
,
k
n−k
k+1
n−k−1
which on multiplication by v n−k becomes
Θ(v) < 0,
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
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where Θ is the polynomial
v n+1
vn
1
1
1
n−k 1
(3.3) Θ(v) =
−
+v
+
−
−
k+1
k
k n−k k+1 n−k−1
v
1
+
−
.
n−k−1 n−k
Since n + 1 > n > n − k > 1 > 0, (3.3) expresses Θ in descending powers
of v. The coefficients taken in sequence have exactly three changes in sign,
regardless of whether the expression in brackets is positive, negative or zero.
Hence by Descartes’ rule of signs the polynomial equation
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
Θ(w) = 0
(3.4)
has at most three positive solutions.
Now by elementary algebra we have that
Θ(1) = Θ0 (1) = Θ00 (1) = 0,
so that w = 1 is a triple zero of Θ(w). Hence Θ(w) has no zeros on (0, 1) and
therefore must have constant sign on (0, 1). Because Θ(0) < 0, we thus have
Θ(w) < 0 throughout (0, 1) and we are done.
For 1 ≤ k < n, put
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Uk∗
= {fn (x) : x ∈ Uk }
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and
εk = inf Uk∗ .
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Lemma 3.2. For each n ≥ 3 we have

1
n


for 1 ≤ k <
 <
2
2
εk

1
n

 =
for
≤ k ≤ n − 1.
2
2
Proof. Since v = 1 gives fn = 12 and v = 0 gives fn = 1, a necessary and
sufficient condition that εk < 12 is that there should exist v ∈ (0, 1) for which
v k − 1 v −(n−k) − 1
nv n
1
+
>
n
2
(1 − v )
k
n−k
2
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
or
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(3.5)
Ω(v) < 0,
where
2n n+k
n
2n
2n
n n
Ω(v) = v − v
+ 2v
+
− 1 − vk
+1
k
k n−k
n−k
2n n+k
k
2n
2n
n n
=v − v
+ 2v
+
− vk
+ 1.
k
k n−k
n−k
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The polynomial Ω has four changes of sign in its coefficients, and so has at
most four positive zeros. We may verify readily that
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Ω(1) = Ω0 (1) = Ω00 (1) = 0,
J. Ineq. Pure and Appl. Math. 4(3) Art. 59, 2003
(3.6)
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while
Ω000 (1) = 2n2 (n − 2k).
(3.7)
If n2 < k ≤ n − 1, then Ω(v) has a triple zero at v = 1 and so can have at
most one zero on (0, 1). Since Ω(0) > 0, condition (3.5) can thus be satisfied
if and only if there is such a zero, in which case Ω(1 − ∆) < 0 for all ∆ > 0
sufficiently small. But by Taylor’s theorem
Ω(1 − ∆) = Ω(1) − ∆Ω0 (1) +
(3.8)
≈−
∆3 000
∆2 00
Ω (1) −
Ω (1) + 0(∆4 )
2!
3!
∆3 2
n (n − 2k),
3
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
which is positive.
Hence we must have εk ≥ 12 . But since e can be approximated arbitrarily
closely by elements of Uk by letting v → 1, we must have εk ≤ fn (e) = 12 .
Thus εk = 21 .
If k = n2 , then Ω(v) has exactly four positive zeros, all at v = 1, so Ω has
constant sign on (0, 1). Since Ω(0) > 0, we thus have Ω(v) > 0 on (0, 1).
Arguing as in the previous paragraph, we derive again that εk = 21 .
Finally, if k < n2 , we have by (3.8) that Ω(1 − ∆) < 0 for ∆ > 0 sufficiently
small, so that condition (3.5) is satisfied. This completes the proof.
Theorem 3.3. The sequence (εk )1≤k< n2 is strictly increasing.
Proof. The desired result is equivalent to (ξk )
being strictly decreasing,
where
u
ξk = sup
φ (u) = 1 − εk .
2 k
u∈(0,1) (1 − u)
1≤k< n
2
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By Proposition 3.1,
u
u
φk (u) >
φk+1 (u)
2
(1 − u)
(1 − u)2
for each u ∈ (0, 1), so that
ξk ≥ ξk+1
for
1 ≤ k ≤ n − 1.
Further, ξk is realised for some choice of u, for u = uk , say, and arguing as in
Lemma 3.2 we must have uk ∈ (0, 1) for 1 ≤ k < n2 .
To show the inequalities are strict, suppose if possible that equality holds for
some value of k, so that
uk
uk+1
φk (uk ) =
φk+1 (uk+1 ).
2
(1 − uk )
(1 − uk+1 )2
(3.9)
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
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By Proposition 3.1,
uk+1
uk+1
φk (uk+1 ) >
φk+1 (uk+1 ),
2
(1 − uk+1 )
(1 − uk+1 )2
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so that by (3.9)
uk
uk+1
φk (uk+1 ) >
φk (uk ) = ξk ,
2
(1 − uk+1 )
(1 − uk )2
contradicting the definition of ξk .
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4.
Characterisation of εk
In the previous section we saw that for 1 ≤ k < n2 the supremum ξk is realised
for some u = uk ∈ (0, 1). We now consider the determination of uk . For
1/n
convenience we again employ vk = uk .
Theorem 4.1. (i) For 1 ≤ k < n2 , v = vk is the unique solution on (0, 1) of
the equation
(4.1)
Φk (v) = 0,
where
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
n
n
k
n
nn + k
n−k
Φk (v) = (v − 1) v
−v
+
+
k
k n−k
n−k
n
1
1
1
n v
n−k
− 2nv
−v
+
+
.
k
k n−k
n−k
(ii) If v ∈ (0, 1), then v < vk or v > vk according as Φk (v) < 0 or Φk (v) > 0.
Proof. Since fn achieves a minimum at v = vk ∈ (0, 1), we have that
k
d
nv n
v − 1 v −(n−k) − 1
·
+
=0
dv (v n − 1)2
k
n−k
for v = vk , this value of v corresponding to a local maximum of the differenti-
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ated expression. The left–hand side is the quotient of
n
n
k
n
2
n+k−1 n + k
n−1
k−1
n(v − 1) v
−v
+
+v
k
k n−k
n−k
n+k
1
1
vk
2 n
2 n−1 v
n
− 2n (v − 1) v
−v
+
+
k
k n−k
n−k
by (v n − 1)4 . Removing this denominator and the factor n(v n − 1)v k−1 from the
numerator gives that v = vk satisfies (4.1). Statement (i) will therefore follow
if it can be shown that (4.1) has a unique solution on (0, 1). Uniqueness gives
that the differentiated expression has positive gradient for v < vk and negative
gradient for v > vk . Statement (ii) will then follow, since the term cancelled is
negative.
It therefore remains only to show that Φk (v) has a unique zero on (0, 1). This
we do as follows. The polynomial Φk (v) may be written in descending powers
of v as
n
n
2n − k n + k
2n n − k
2n−k
n
−v
+v
+
−v
+
k
k n−k
n−k
k
n
k
n
n−k
+v
+
−
,
k n−k
n−k
the coefficients of which exhibit four changes of sign. Hence by Descartes’ rule
of signs, Φk (v) has at most four positive zeros.
By elementary algebra,
(4.2)
2
Φk (1) = Φ0k (1) = Φ00k (1) = 0, Φ000
k (1) = n (2k − n),
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
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so that Φk (v) has a triple zero at v = 1. Hence Φk (v) has at most one zero on
(0, 1).
Now Φk (v) < 0 and for ∆ > 0 small
Φk (1 − ∆) = −
∆3 000
Φ (1) + 0(∆4 ) > 0,
3! k
by Taylor’s theorem and (4.2). Hence Φk (v) has a zero on (0, 1) and this must
be unique.
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
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References
[1] N. LORD, More on the relative location of means II, Math. Gaz., 85 (2001),
114–116.
[2] C.E.M. PEARCE AND J. PEČARIĆ, More on the relative location of means
I, Math. Gaz., 85 (2001), 112–114.
[3] J.A. SCOTT, On the theorem of means: is G nearer A or H?, Math. Gaz.,
82 (1998), 104.
On Some Spectral Results
Relating to the Relative Values
of Means
C.E.M. Pearce
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