In the case of two positive numbers, the geometric mean is closer to the harmonic than to the arithmetic mean

http://jipam.vu.edu.au/

Volume 4, Issue 3, Article 58, 2003

ON SOME SPECTRAL RESULTS RELATING TO THE RELATIVE VALUES OF MEANS

C.E.M. PEARCE

SCHOOL OFAPPLIEDMATHEMATICS, ADELAIDEUNIVERSITY, ADELAIDESA 5005, AUSTRALIA. cpearce@maths.adelaide.edu.au

Received 21 January, 2003; accepted 22 July, 2003 Communicated by P. Cerone

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.

Key words and phrases: AGH inequality, Means, Relative values, Spectrum.

2000 Mathematics Subject Classification. 26E60, 26D15.

1. INTRODUCTION

LetA,G,Hdenote respectively the arithmetic, geometric and harmonic means ofnpositive real numbersx₁, . . . , x_n, which are not all equal. It is well–known thatH < G < A. Scott [3]

has shown in the casen = 2thatGis closer toHthan toA, so that

(1.1) A−G

A−H > 1 2.

He showed by a counterexample that this need not be the case whenn >2.

Subsequently Lord [1] and Pearce and Peˇcariˇc [2] addressed the question of the behaviour of the quotient

f_n(x₁, . . . , x_n) := A−G A−H

in the case of general n. Several generalisations and extensions of (1.1) were obtained. The following are pertinent to the present article.

Since

f_n(ax₁, . . . , ax_n) =f(x₁, . . . , x_n)

ISSN (electronic): 1443-5756

c 2003 Victoria University. All rights reserved.

008-03

for a > 0, it suffices to consider the values taken by f_n when x = (x₁, . . . , x_n) lies on the intersection

K:=

(

x∈Rⁿ: x_i ≥0for 1≤i≤n and

n

X

i=1

x²_i = 1 )

of the nonnegative orthant and the surface of the unit hypersphere. The function f_n is clearly well–defined and continuous on the interior of K except at e :=

√1

n, . . . ,^√1_n

, where it is undefined sinceA, GandH all coincide. In fact this singularity is removable. It is shown in [1] that definingfn(x) = 1for boundary points ofK(where some but not all valuesxi vanish) andfn(e) = ¹₂ makesfncontinuous on the whole ofK. SinceKis compact,fnpossesses and realises an infimumα_n. Further, the range of f_n constitutes the interval [α_n,1], the sequence (αn)^∞₂ is strictly decreasing to limit zero andαn > _n¹ forn ≥ 3. The seminal paper of Scott givesα₂ = ¹₂.

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 ifxis such thatf_n(x) = α_n, then {x₁, . . . , x_n} contains precisely two distinct values. In Section 3, Theorem 3.3, we show that iff_n(x) =α_n, then the smaller of the two distinct components ofxmust occur with multiplicity one. We conclude in Section 4 by giving characterisations ofα_nand some related infima arising naturally in our analysis.

We postpone consideration of asymptotics to a subsequent article.

2. THEDICHOTOMY THEOREM

Theorem 2.1. Forn > 2, any set{x1, . . . , xn}for which fn(x) = αn contains precisely two distinct values.

Proof. First suppose that S₁ := Pn

i=1x_i and S_n := Qn

i=1x_i are fixed. Subject to these con- straints, the mimimum off_ncorrespond to an extremum ofPn

i=1 1

xi and satisfies

∂L

∂x_i = 0 for i= 1, . . . , n, whereLdenotes the Lagrangian

L :=

n

X

i=1

1 xi

−λ

n

X

i=1

x_i−S₁

!

−µ

n

Y

i=1

x_i−S_n

! .

Then ∂L

∂x_i =−1

x²_i −λ−µY

j6=i

x_j = 0 (i= 1, . . . , n), that is,

1 xi

+λx_i+µS_n= 0 (i= 1, . . . , n).

Hence eachx_imust be equal to one of the two solutions of the quadratic λx²+µS_nx+ 1 = 0.

For a minimum, these solutions must be distinct, sincef_n(e) = ¹₂ whileα_n < ¹₂ forn ≥3.

Forj = 1,2, . . . , nand fixedn >2, define

V_j ={x: {x₁, . . . , x_n}contains preciselyj distinct values}, V_j^∗ ={f_n(x) : x∈ V_j}

and

δ_j = infV_j^∗.

An immediate implication of Theorem 2.1 is the following result.

Corollary 2.2. We have

δ₂ =δ₃ =· · ·=δ_n=α_n. Forj >1, the setV_j^∗ contains its infimum only forj = 2.

Proof. If1≤j ≤n−1, any element ofV_j can be approximated arbitrarily closely by elements ofV_j+1, but not conversely. SinceKis compact andf_ncontinuous, we must therefore have that δ_j+1 ≤δ_j. Thus

δ_n≤δ_n−1 ≤ . . . ≤δ₂ ≤δ₁ = 1 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.

3. COMPARISON RESULTS

In the remaining sections of the paper we examine more closely the central case when {x₁, . . . , x_n}contains only two distinct values, that isx∈ V₂. We may assume without loss of generality an ordering

x₁ ≤x₂ ≤ · · · ≤x_n. We decompose

V₂ =

n−1

[

k=1

U_k,

where

Uk={x:x1 =x2 =· · ·=xk < xk+1 =· · ·=xn} (1≤k < n).

Forx∈ U_kwe have for thekequal points denoted byxand the rest byythat A−G

A−H =

k

nx+ 1− ^k_n

y−x^k/ny^1−k/n

k

nx+ 1− ^k_n

y−n.

k

x +^n−k_y . If we setβ =k/nandu=x/y, this gives

f_n(x) = βu+ 1−β−u^β βu+ 1−β−1 _β

u + 1−β withβ ∈₁

n,²_n, . . . ,ⁿ⁻¹_n and0< u <1.

This may be rearranged as

(3.1) f_n(x) = 1− u

(u−1)²g(u, β), where

(3.2) g(u, β) = u^β −1

β + u^−(1−β)−1 1−β .

We shall find it convenient to have alternative sets of variables and functions. Setv = u^1/n. Then forx∈ U_kwe put

h_k(n, v) =f_n(x) and φ_k(v) = g

u,k n

.

Proposition 3.1. For fixedn≥3andv ∈(0,1), the sequence(hk(n, v))ⁿ⁻¹_k=1 is strictly increas- ing.

Proof. By virtue of the representation (3.1), (3.2), it suffices to prove that the sequence(φ_k(v))ⁿ⁻¹_k=1 is strictly decreasing. To show thatφ_k(v)> φ_k+1(v), we need to establish the inequality

v^k−1

k +v^−(n−k)−1

n−k > v^k+1−1

k+ 1 +v^{−(n−k−1)}−1 n−k−1 , which on multiplication byv^n−kbecomes

Θ(v)<0, whereΘis the polynomial

(3.3) Θ(v) = vⁿ⁺¹ k+ 1 − vⁿ

k +v^n−k 1

k + 1

n−k − 1

k+ 1 − 1

n−k−1

+ v

n−k−1− 1 n−k .

Since n+ 1 > n > n−k > 1 > 0, (3.3) expresses Θin 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

(3.4) Θ(w) = 0

has at most three positive solutions.

Now by elementary algebra we have that

Θ(1) = Θ⁰(1) = Θ⁰⁰(1) = 0,

so thatw = 1 is a triple zero ofΘ(w). HenceΘ(w)has no zeros on(0,1)and therefore must have constant sign on(0,1). BecauseΘ(0) <0, we thus haveΘ(w)<0throughout(0,1)and

we are done.

For1≤k < n, put

U_k^∗ ={f_n(x) : x∈ U_k} and

ε_k= inf U_k^∗. Lemma 3.2. For eachn ≥3we have

ε_k











< 1

2 for 1≤k < n 2

= 1

2 for n

2 ≤k ≤n−1.

Proof. Sincev = 1givesf_n = ¹₂ andv = 0givesf_n = 1, a necessary and sufficient condition thatε_k< ¹₂ is that there should existv ∈(0,1)for which

nvⁿ (1−vⁿ)²

v^k−1

k + v^−(n−k)−1 n−k

> 1 2 or

(3.5) Ω(v)<0,

where

Ω(v) = v²ⁿ−2n

k v^n+k+ 2vⁿ n

k + n

n−k −1

−v^k 2n n−k + 1

=v²ⁿ−2n

k v^n+k+ 2vⁿ n

k + k

n−k

−v^k 2n n−k + 1.

The polynomialΩhas four changes of sign in its coefficients, and so has at most four positive zeros. We may verify readily that

(3.6) Ω(1) = Ω⁰(1) = Ω⁰⁰(1) = 0,

while

(3.7) Ω⁰⁰⁰(1) = 2n²(n−2k).

If ⁿ₂ < k ≤ n−1, thenΩ(v)has a triple zero at v = 1and so can have at most one zero on (0,1). SinceΩ(0) > 0, condition (3.5) can thus be satisfied if and only if there is such a zero, in which caseΩ(1−∆) <0for all∆>0sufficiently small. But by Taylor’s theorem

Ω(1−∆) = Ω(1)−∆Ω⁰(1) + ∆²

2! Ω⁰⁰(1)−∆³

3! Ω⁰⁰⁰(1) + 0(∆⁴)

≈ −∆³

3 n²(n−2k), (3.8)

which is positive.

Hence we must haveε_k≥ ¹₂. But sinceecan be approximated arbitrarily closely by elements ofU_kby lettingv →1, we must haveε_k ≤f_n(e) = ¹₂. Thusε_k= ¹₂.

Ifk = ⁿ₂, thenΩ(v)has exactly four positive zeros, all atv = 1, soΩhas constant sign on (0,1). SinceΩ(0)>0, we thus haveΩ(v)>0on(0,1). Arguing as in the previous paragraph, we derive again thatε_k= ¹₂.

Finally, ifk < ⁿ₂, we have by (3.8) thatΩ(1−∆) <0for∆> 0sufficiently small, so that

condition (3.5) is satisfied. This completes the proof.

Theorem 3.3. The sequence(ε_k)1≤k<ⁿ

2 is strictly increasing.

Proof. The desired result is equivalent to(ξ_k)1≤k<ⁿ₂ being strictly decreasing, where ξ_k = sup

u∈(0,1)

u

(1−u)²φ_k(u) = 1−ε_k. By Proposition 3.1,

u

(1−u)²φ_k(u)> u

(1−u)²φ_k+1(u) for eachu∈(0,1), so that

ξ_k ≥ξ_k+1 for 1≤k ≤n−1.

Further,ξk is realised for some choice ofu, foru = uk, say, and arguing as in Lemma 3.2 we must haveuk ∈(0,1)for1≤k < ⁿ₂.

To show the inequalities are strict, suppose if possible that equality holds for some value of k, so that

(3.9) u_k

(1−uk)²φ_k(u_k) = u_k+1

(1−uk+1)²φ_k+1(u_k+1).

By Proposition 3.1,

u_k+1

(1−u_k+1)²φ_k(u_k+1)> u_k+1

(1−u_k+1)²φ_k+1(u_k+1), so that by (3.9)

u_k+1

(1−u_k+1)²φ_k(u_k+1)> u_k

(1−u_k)²φ_k(u_k) =ξ_k,

contradicting the definition ofξ_k.

4. CHARACTERISATION OFε_k

In the previous section we saw that for 1 ≤ k < ⁿ₂ the supremum ξ_k is realised for some u =u_k ∈ (0,1). We now consider the determination ofu_k. For convenience we again employ v_k =u^1/n_k .

Theorem 4.1. (i) For1≤k < ⁿ₂,v =v_kis the unique solution on(0,1)of the equation

(4.1) Φ_k(v) = 0,

where

Φ_k(v) = (vⁿ−1)

vⁿn+k

k −v^n−k n

k + n

n−k

+ k

n−k

−2nvⁿ vⁿ

k −v^n−k 1

k + 1

n−k

+ 1

n−k

.

(ii) Ifv ∈(0,1), thenv < v_k orv > v_kaccording asΦ_k(v)<0orΦ_k(v)>0.

Proof. Sincef_nachieves a minimum atv =v_k∈(0,1), we have that d

dv

nvⁿ (vⁿ−1)² ·

v^k−1

k +v^−(n−k)−1 n−k

= 0

forv =v_k, this value ofv corresponding to a local maximum of the differentiated expression.

The left–hand side is the quotient of n(vⁿ−1)²

v^n+k−1n+k

k −vⁿ⁻¹ n

k + n

n−k

+v^k−1 k n−k

−2n²(vⁿ−1)²vⁿ⁻¹ v^n+k

k −vⁿ 1

k + 1

n−k

+ v^k n−k

by (vⁿ−1)⁴. Removing this denominator and the factor n(vⁿ −1)v^k−1 from the numerator gives thatv =v_k 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 forv < v_kand negative gradient forv > v_k. Statement (ii) will then follow, since the term cancelled is negative.

It therefore remains only to show that Φ_k(v) has a unique zero on (0,1). This we do as follows. The polynomialΦ_k(v)may be written in descending powers ofv as

−v²ⁿn−k

k +v^2n−k n

k + n

n−k

−vⁿ

2n−k

n−k +n+k k

+v^n−k n

k + n

n−k

− k n−k , the coefficients of which exhibit four changes of sign. Hence by Descartes’ rule of signs,Φ_k(v) has at most four positive zeros.

By elementary algebra,

(4.2) Φ_k(1) = Φ⁰_k(1) = Φ⁰⁰_k(1) = 0, Φ⁰⁰⁰_k(1) =n²(2k−n),

so thatΦ_k(v)has a triple zero atv = 1. HenceΦ_k(v)has at most one zero on(0,1).

NowΦ_k(v)<0and for∆>0small

Φ_k(1−∆) =−∆³

3! Φ⁰⁰⁰_k(1) + 0(∆⁴)>0,

by Taylor’s theorem and (4.2). HenceΦ_k(v)has a zero on(0,1)and this must be unique.

REFERENCES

[1] N. LORD, More on the relative location of means II, Math. Gaz., 85 (2001), 114–116.

[2] C.E.M. PEARCEANDJ. PE ˇCARI ´C, More on the relative location of means I, Math. Gaz., 85 (2001), 112–114.

[3] J.A. SCOTT, On the theorem of means: isGnearerAorH?, Math. Gaz., 82 (1998), 104.