A Rearrangement Inequality and the Permutahedron
Author(s): A. Vince
Reviewed work(s):
Source: The American Mathematical Monthly, Vol. 97, No. 4 (Apr., 1990), pp. 319-323
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2324517 .
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319
NOTES
1990]
Hence we have provedthe
THEOREM. If p is a prime greaterthan 3 thereexist nontrivialmorphisms
of
inequivalentextensionsof Zp byZp whichare notisomorphisms.
The authorwishesto thankW. Messingfordiscussionswhichled to thisnote.
REFERENCE
1. J. J. Rotman,The Theoryof groups,An Introduction,
3rd ed., Allynand Bacon.
andthePermutahedron
A Rearrangement
Inequality
A. VINCE
FL 32611
ofFlorida,Gainesville,
University
ofMathematics,
Department
and Polya
byHardy,Littlewood,
One chapterof theclassicbook"Inequalities"
The main
withterms
sequences
rearranged.
involving
[3] is dedicatedto inequalities
Let a1 < a2 < -*
examplein thatchapteris thefollowing.
an and b1 < b2 <
. . < bn be sequencesof real numbersand g any permutation
of the set
{1,2,..., n}. Then
n
n
n
Eaibn-i+l
i=l
<
(1
Eaibi
i=l
Eaib,.i
~i=l
alonga rodandthe
theai as fixeddistances
andPolyainterpret
Hardy,Littlewood,
with
moment
maximum
To
get
the
these
distances:
at
be
suspended
weights
to
as
bi
fromthatend;to get
furthest
weights
respectto an endoftherodhangtheheaviest
weights
closest.
hangtheheaviest
moment
theminimum
exist.Three
inequality
ofthisrearrangement
andgeneralizations
Manyvariations
appearbelow,moreat theendof thisnote,and Marshalland Olldn[5] containsa
in
In thisnoteall sequences{a i} and { bi} are increasing
recentsurvey.
relatively
the sense a1 < a2 < * < an and b1 < b2 < * * < bn,and all sumsandproducts
are from i = 1 to i = n unless otherwisestated. Also N = {1, 2, ..., n} and Sn
of N.
denotesthesetof all permutations
Example1. [7] For sequenceswithpositivetermsand forall 7T E Sn
Hl(ai + bn-i+l) > H(a1 +
b71)
> H(ai
+ bi).
Example2. [8] Forp > 1 and all 7TE Sn
Elai
- bilP < Elai
-
bjP
< Elan-i+1
-
bilP.
(1)
inequality
of theHardy-Littlewood-Polya
Example3. [4] One generalization
convex
is as follows.If thesequenceshavepositivetermsand f is an increasing
thenforall 7TE Sn
function
Ef(aibn-i+f
) < Ef(ay)we < Ef(aibi)y
r
Recall that f is increasing if f(x) >, f(y) whenever x > y, and f is convex if
320
[April
A. VINCE
f(ax + (1 - a)y) < af(x) + (1 - a)f(y) for all 0 < a < 1. In the differentiable
case, of course,convexis equivalentto f"(x) > 0.
The purpose of this note is to show how the permutahedron
leads to a very
simple generalizationof the Hardy-Littlewood-Polya
inequality,fromwhich the
inequalitiesabove and manyotherrearrangement
inequalitiesimmediatelyfollow.
In [5] rearrangement
inequalitiesare derivedusing majorization,
whichis a partial
orderon the set of vectorsin n-dimensionalEuclidean space. This elegantmethod
goes back at least to Schur [10] and is a unifyingprincipalfor many types of
inequalities.The intentionin thisnote is to use a muchsimplerpartialorderon Sn
to obtain the rearrangement
inequalities.
Let .f denote the identitypermutation_f(i) = i of N and .* the reverse
permutationf*(i) = n - i + 1, i = 1,..., n. An inversionof a permutation7 of N
is a pair (S7j, 7Tk) such that j < k and s7j > 7Tk. For example,(5, 3) is one of the
fourinversionsof g = 2 51 3 4. Now considerthe directedgraph Pn whose vertex
set is Sn,and thereis an edge (a, 7T)directedfroma to s7wheneververtexa is obtained fromvertexs7 by interchanging
the elementsof an inversionof the form
(n7j, 7T(j + 1)). Pn is sometimescalled a permutahedron[1] and an example is
shownin Figure1. The transitive
closureof Pn inducesa partialorderon theset Sn.
Recall that the transitiveclosure is the "smallest" directedgraph with the same
vertexset as Pn and withthe propertythatif (7T, T) and (T, a) are directededges
then so is (7T,a); the partialorderon Sn is definedby a > s7 if thereis an edge
directedfroma to S in the transitiveclosure of Pn. A real valued functiong:
if g(a) > g( g) whenevera > g. This is all thatis
Sn-* R is called orderpreserving
needed to provethefollowing
theorem.
1234
^_2134
-
<
~~~~~324
4321
FIG. 1. Permutahedron
P4.
1990]
321
NOTES
THEOREM. Let gl,...,
gnbe real valuedfunctions
definedon an intervalI. Then
'< Egi(b,,Tj)
Egi(bn-i+J)
for all sequencesb, < b2 < *
< bnin I and all
<
Egi(bi)
'iT
E
on I,
gi is increasing
gi+-
(2)
Snifand onlyif
1 < i < n.
(3)
thenit is clearthat
Remark.If thefunctions
gi aredifferentiable,
g1(x) < g2(x) <
gn(x)
forall x
E- I
(3')
is equivalentto condition
(3).
< bnand let g: Sn-- R be definedby
Proof. ('-) Fix a sequenceb1 < b2 < *
g( g) = Yg,(bb). To show that g is order preservingit sufficesto show that
g(a) > g('r) whenever(a, 7T)is an edge of thepermutahedron:
g(a)
-
g('g) = [gj(bqj)
= (gj+l
-
-
+ gj+?(ba(j+?))]
gj)(b,,j)
-
(gj+l
-
+
gj+?(bT(j+
[gj(brj)
gJ)(br(j+1))
)
> 0.
The last inequalityfollowsfrombj < b,,(j+1)and theassumptionthat gj+1 - gj is
increasing.
Since.4 < ST<. f foranypermutation
gi, also g(.f*) < g(r) < g(,),
whichis precisely
inequality
(2).
(=:) By way of contradiction
assume that gm+1- g,,,is not increasingforsome
m. Then there exists x > y such that (g,,,+1- gm)(x) < (gm+1 - gm)(y). Now
choose any sequence b, < b2 < *- - < bnin I withbm= x and bmi+= y. Let 'T be
the transposition(m m + 1). Then Egi(b,i) =
E
>
gi(bi) + g.(x)
L
i *in,m+?
+ g.+,(y)
gi(bi) + gm(y) + g+l(x)
=
Egi(bi)
contradicting
inequality
(2). a
The inequalities
ofExamples1,2, and 3, as wellas thosebelow,resultbysimply
For example,choosinggi(x) = aix
theappropriate
substituting
gi in thetheorem.
yields the classic Hardy-Littlewood-Polya
inequality(1). Choosing gi(x) =
-log (ai + x) yieldsExample1; and choosinggi(x) = f(aix) givesExample3. In
condition
each case it is an exerciseto showthatthegi satisfy
(3) or (3') in the
For
For theHardy-Littlewood-Polya
condition
theorem.
inequality
(3) is immediate.
butstillelementary.
ofcondition
Example3 theverification
(3) is a littletricky,
Example 4. Let gi(x) = f(ai
Ef (ai-
-
x). If f is convexthenforall
bi) < Ef(ai
-bi)
< Ef(an-i+1
XTE
Sn
- bi).
Withf(x) = IxIP thisis Example2.
Example 5. [2] Take gi(x) = g(ai, x) whereg is a real valued functionof two
variablesdefinedon a domain D = [a, b] X [c, d]. If
g(x1, Y)
-
g(x1, Y2)
-
g(x2, YI) + g(x2, Y2) > 0
(4)
322
[April
A. VINCE
forall xl > x2, y1 > y2 in thedomainthen
bn-i+,) < Eg(aj,
Eg(aj,
bvi
g(i
i
(5)
for all 7TE S,7.Note thatfora functionwithcontinuoussecond derivativescondition (4) can be replacedby
d2g(x,y)
dxdy
>
forall(x, y) E D.
(4')
Example 6. Many inequalitiescan be generalizedto more than two sequences.
1 < i < n, be not necessarilyincreasingreal sequences
Let {a'}, {a2}, ...,{am},
order.
and let a (1), a (2),..*, a(n) denotethesequencea1, a2, ..., an in increasing
Suppose g(x1, . ., xm)satisfiescondition(4) or (4') foreverypair of variables.Then
ai2.
Eg(a,
aim)< J:g(a1j),a(2)..
a(m)).
(6)
This inequalityfollowsdirectlyby inductionusingtherightinequalityof (5) as the
first step. Choosing g(x1, . . ., Xm) = x1 x2 *.. xm and g(x1, * *, xm)=
- log (x1 + *** + xm),respectively,
in (6) results,forsequencesof positiveterms,in
inequalitiesanalogous to (1) and Example 2 [9]:
'
12
M
E ala 2 . . . a m <
V12
E
a (i) a (2i) * * * aaM
and
H(al
+ a
2
+am)
+
>H H(a1
+ a(2
+
+am
Example 7. Choose g(xl,..., xm) = f[min(x1,..., xm)]in (6). If f is an increasing functionthen
Ef
Taking f(x)
=
x and f(x)
=
mnai
< Ef( minai
leads to inequalitiesof Minc [6]:
log(x), respectively,
E mina
< E minai
and
Sfrt
Hminai
maxfc< tHmna-).
min.
holdforthemaxfunction.
Similarinequalities
REFERENCES
AcademicPress,New York and London, 1971.
1. C. Berge,Principlesof Combinatorics,
inequalities,Canad. J. Math,24 (1972) 930-943.
2. P. W. Day, Rearrangement
3. G. H. Hardy, J. E. Littlewood,and G. P6lya, Inequalities,CambridgeUniversityPress,London,
1952.
inequalitiesinvolvingconvex functions,PacificJ. Math., 34 (1970)
4. D. London, Rearrangement
749-753.
5. A. W. Marshall and I. Olkin,Inequalities:Theoryof Majorizationand Its Applications,Academic
Press,New York, 1979.
Trans. Amer. Math. Soc., 159 (1971) 497-504.
6. H. Minc, Rearrangements,
323
NOTES
1990]
7. A. Oppenheim,InequalitiesconnectedwithdefiniteHermitianformsII, Amer. Math. Monthly.61
(1954) 263-266.
inequalitieswith some applications to
8. Y. Rinott, Multivariatemajorizationand rearrangement
probabilityand statistics,IsraelJ. Math.,15 (1973) 60-77.
59 (1952) 29-32.
9. H. D. Ruderman,Two new inequalities,Amer. Math. Monthly,
mit Anwendungendie Determinanten,Theorie
10. I. Schur, Uber eine Klasse von Mittelbildungen
22 (1923) 9-20.
Sitzungsber.,Berlin. Math. Gesellschaft,
Musingson thePrimeDivisorsofArithmetic
Sequences
PATRICK MORTON
Department
ofMathematics,
Wellesley
College,Wellesley,
MA 02181
Of the early proofs one usually sees in a number theorycourse, the most
beautifulis the proof,due to Euclid, that thereare infinitely
many primes.This
theoremmay be formulated
as follows.
If { a,n}?=1 is any sequenceof integers,and p is a primeforwhichpI an forsome
n, p is called a primedivisorof the sequence {an } n=,. (See [2].) Euclid's theorem
says that the sequence {n)}?. 1 has an infiniteset of primedivisors.What other
sequences have thisproperty?
For example,considerthe sequenceswhose termsare definedby the following
formulas:
1) an = f(n), wheref E Z[x] is a nonconstantpolynomial;
2) bn= [gn2], wherebracketsdenotethegreatestinteger;
3) Cn = [Tn2]2 - [ n21[V + [j2;
4) dn = 211+ 1.
Which of thesesequenceshas an infinitenumberof primedivisors?
The answeris, of course,thattheyall do. The fact that {an)n'= does was first
provedin an elementary
way by Schur[8], and is usuallystatedas follows.(See [2]
and [3] forotherproofsand moreon theprimedivisorsof polynomials.)
THEOREM 1. Let f(x) e Z[x] be nonconstant.
Thenthecongruence
f (x)
O(modp)
has a solutionx E Z for infinitely
manyprimesp. In otherwords,infinitely
many
primesdividethetermsof thesequence { f(n)})n=l.
The purpose of this note is to give a surprisingproof of this resultusing a
well-choseninfiniteseries,and thento see wherethe proofleads. It will turnout
that the proofcan be generalizedto show thatthe sequences {bn)n1 and {cn)}' =
both have infinitesetsof primedivisors,but thatthesame proofcannotdecide this
questionforthe simplersequence { dn}n= 1! The proofwill also be a surprisein that
it hides an algebraicstructure.
1. A proofof Schur'stheorem.Assumetheorem1 is falseforsome non-constant
f(x) E Z[x], and let m = degf. Then forn E Z, f(n) = 0 or (by the fundamental
theoremof arithmetic)
f(n)
=
?palpa2
...
pak