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Some further results on majorization in integers

Martn Egozcue

Department of Economics, University of Montevideo, and FCS Universidad de la Repub-

lica del Uruguay, Montevideo, Uruguay

Luis Fuentes Garca

Departamento de Mtodos Matemticos e de Representacin, Escola Tcnica Superior de

Enxeeiros de Camios, Canais e Portos, Universidade da Corua, 15001 A Corua,

Spain.

Abstract.

Keywords and phrases:

1 Introduction

The theory of majorization has conceited considerable attention in many applied areas of

Mathematics, see Marshall and Olkin (1979) and Goswami (2010) and references therein.

In particular, one research of interest has focused on the majorization in integers. That

is, this partial order is apply only to compare vectors with integer-valued components.In this paper we derive new results on this directionh. Our paper continues as follows.

In the next section we present some denitions and previous results. Our main theorems

are developed in section 3. We provide some examples in the last section.

2 Denitions and previous results

In order to complement the general overview of the relevant literature, and to make explicit

the expected contribution of this paper, I will present a more specic review of literature on

majorization theory in this section. First, we dene the concept of majorization between

two vectors, see Marshall and Olkin (1979).

Denition 2.1 Letx = (x1; x2;:::;xn) 2 Rn andy = (y1; y2;:::;yn) 2 R

n, withx1 x2

::: xn andy1 y2 ::: yn thenx majorizesy , and is denoted byx M y, if

i=kX

i=1

xi

i=kX

i=1

yi for allk= 1; 2;:::;n 1 andi=nX

i=1

xi =i=nX

i=1

yi

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Fulkerson (1969) dene what is called a transfer.

Denition 2.2 Letx1 x2 ::: xn be integers andxi > xj ,then the transformation

x0i

= xi 1

x0j = xj+ 1

x0k = xk fork 6=i;j;

is called a transfer fromi to j.

Muirhead (1903) prove that a majorize vector can be derived from a majorized vector

with a nite number of transfers. We can state this result as follows.

Theorem 2.1 Letx= (x1; x2;:::;xn

) andy = (y1; y2;:::;yn

) be vectors of integers, such

thatx1 x2 ::: xn andy1 y2 ::: yn. Ifx M y theny can be derived fromx

by succesive applications of a nite number of transfers.

Fulkerson y Ryser (1962) characterize the conditions under which majorization be-

tween two vectors is preserved if a unit is substracted from a component of each vector.

Theorem 2.2 Letx= (x1; x2;:::;xn) andy = (y1; y2;:::;yn) be vectors of integers, such

thatx1 x2 ::: xn andy1 y2 ::: yn.Ifx M y andi j then

x ej M y ei

whereek 2 Rn fork = i; j is a vector where thekth component is one and all the other

components are zeros.

3 Main results

In many applications in Economics, see Egozcue et.al. (2012), it is of interest to study themajorization of two vectors X andY , where Yis obtained by adding two components

ofX:

Theorem 3.1 Let x = (x1; x2;:::;xn) be a vector of integers, such that x1 x2 :::

xn 0. Let 1 i < j n. Let y = (y1; y2;:::;yn) with y1 y2 ::: yn where

its elements are the rearranged elements, in decreasing order of the following vector:

(x1; x2;:::;xi+xj ;:::;xn; 0) then

y M x

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Proof. Let i0 such that xk xi+ xj fork < i0 andxk < xi+ xj fork i0. We can see

thati0 i. We have

1. If1 < k i0 then yk = xk

2. Ifk= i0 thenyk =xi+xj xi0 =xk

3. Ifi0 < k ithenyk = xk1 xk

4. Ifi < k < j then yk =xk xk

5. Ifj k < nthen yk =xk+1 xk

6. Ifk= n then yn= 0 xn

Ifk < j from points 1; 2; 3 and 4 we have xi yi for all i k:

Ifk j we see that

kX

i=1

xi= SnX

i=k+1

xi;

kX

i=1

yi= SnX

i=k+1

yi

Then we need to show thatnX

i=k+1

xi

nX

i=k+1

yi

This last inequality holds, as can be deduced by points 5; 6:This nishes the entire proof

of this Theorem.

To establish some further results, we need the following technical Lemma.

Lemma 3.1 Let x = (x1; x2;:::;xn) and y = (y1; y2;:::;yn) be vectors of integers, such

thatx1 x2 ::: xn andy1 y2 ::: yn andPk

i=1 xi =Pi=n

i=1yi. If there is j >1

such that : xi yi for all1 i < j andxi yi for allj i n then

x M y

Proof. By hypothesisi=kX

i=1

xi

i=kX

i=1

yi for1 k j

Now we need to prove that

i=kX

i=1

xi

i=kX

i=1

yi for all n 1 k > j

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As:

S=i=nX

i=1

xi=i=kX

i=1

xi+i=nX

i=k

xi=i=nX

i=1

yi=i=kX

i=1

yi+i=nX

i=k

yi

we can write:i=kX

i=1

xi = Si=nX

i=k

xi

andi=kX

i=1

yi = Si=nX

i=k

yi

But as k > j theni=nX

i=k

yi

i=nX

i=k

xi

from this i=kX

i=1

xi = Si=nX

i=k

xi

i=kX

i=1

yi= Si=nX

i=k

yi for all k > j

This nishes the proof of Lemma

In the following theorem we show that majorization holds if the largest component is

sum with the smallest negative componente, provided that the sum of its components is

positive.

Theorem 3.2 Let x = (x1; x2;:::;xn) be a vector of integers, such that x1 x2 ::: 0 ::: xn with

Pi=ni=1xi > 0. Let y = (y1; y2;:::;yn) with y1 y2 ::: yn where

its elements are the rearranged elements, in decreasing order of vector(x1+ xn; x2; :::; 0).

Then

x M y

Proof. We are going to prove that the assumptions of the above Lemma hold. Let

yh= 0

and

yg =x1+xn

where1 h nand 1 g n.

Ifx1+ xn 0 therefore xi yi for all i such that 1 i < g andxi yi for all i

such that g i nthen

x M y

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If x1+xn 0 therefore xi yi for all isuch that 1 i < h and xi yi for all i

such that h i nthen

x M y

Theorem 3.3 Letx= (x1; x2;:::;xn) be a vector of integers wherex1 x2 ::: xn

0. Lety = (y1; y2;:::;yn) with y1 y2 ::: yn; where its elements are the rearranged

elements, in decreasing order of vector(x1; x2;:::;xi+xj;:::; 0) with1 i < j n. Then

y M x

Proof. Is obvious thati=n

Xi=1

xi=

i=n

Xi=1

yi

Assume that

yh= xi+xj

where1 h < n:Hencei=kX

i=1

xi=i=kX

i=1

yi for all k < h

We also havei=h

Xi=1

xi

i=h

Xi=1

yi =

i=h

Xi=1

xi+xj

And for any k; nsuch that n k > hwe will have

i=kX

i=1

xi

i=kX

i=1

yi =i=kX

i=1

xi+xj

Both inequalities hold as xj >0:

4 Examples

Acknowledgements

References

Goswami and Aouf. 2010 Majorization properties for certain classes of analytic functions

using the Salagean operator. Applied Mathematics Letters 23 (2010) 1351 1354

Marshall and Olkin (1979) Inequalities theory