École Polytechnique
L. T ZAFRIRI
Symmetric structures in some Banach lattices
Séminaire d’analyse fonctionnelle (Polytechnique) (1977-1978), exp. n
o9, p. 1-14
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SEMINAIRE SUR LA GEOMETRIE
DES ESPACES DE BANACH
1977-1978
SYMMETRIC STRUCTURES
IN SOME
BANACH LATTICESL. TZAFRIRI
(Université
deJerusalem)
CENTRE DEMATHI;MATIQUES
PLATEAU DE PALAISEAU - 91128 PALAISEAU CEDEX
Téléphone : 941.82.00 - Poste No Télex : ECOLEX 691596 P
Expose NoIX
6 Janvier 1978The purpose of these notes is to
present
certain results which will appear in ajoint
paper with W.B.Johnson,
B.Maurey
andG. Schechtman
L4].
The
starting point
is a characterization ofsymmetric
basicsequences of finite
length
in a Banach lattice oftype
2. Insteadof
stating
this theorem in the framework of Banach lattices oftype
2 we
prefer
to use theequivalent
context of Banach lattices which,are 2-convex and q-convex for some
’
We recall that a Banach lattice Z is said to be r-convex
for if there exists a constant such
that,
for every choice of inZ,
1 1=
respectively
° The smallest
possible
value takenby
M is denotedby M (r) (Z)
and iscalled the
r-convexity
constant of Z. The dual notion ofr-concavity
is defined
by replacing
thesign £
with ~ andrequiring
that M > 0.Again,
~ the smallestpossible
value forM 1
is denotedby M (r) (Z)
andis called the
r-concavity
constant of Z.These notions were
essentially
introduced in[3]
but mostof their basic
properties
were described in[6] (see
also[8]).
Theequivalence
between the notion oftype
2 for alattice,
on onehand,
and that of
2-convexity together
with someq --concavity,
on theother,
follows
easily
from ageneralization,
due to B.Maurey [8],
of theclassical
inequality
of Khintchine. The actual value of q in thisequivalence
is of noimportance ;
whatreally
counts is to ensurethat the lattice does not contain
uniformly isomorphic copies of In
for all n.
We also recall that a sequence
[z.} 1
is calledK-symmetric
1 I:
provided that,
for everypermutation 71
of theintegers
and every choiceof ,
lm1
Theorem 1 : Let X be a Banach lattice which is 2-convex and 2m-concave for some
integer
m.Then,
for everyK~ 1,
there exists a constantD~ so
that,
for every normalizedK-symmetric
basic sequence in X and every choice of scalars 9 we have1 i=1 1 1=
where and E means summation over all n!
permutations
7’
T1: n
of the first n
integers.
The notationA ~
B is used instead ofwriting
The constant D
depends only
onK,
’ m, ’M(0) (X)
andM (2m) (X).
Proof : Let
111-111
be aK-equivalent
new norm on the linear span of suchthat,
endowed with the new norm,fx n
"
i i= " i i= i i=
is
1-symmetric
and still normalized.Clearly,
the2-convexity
constantM(2)(U)
of U endowed with111.111 depends only
on K and thetype
2-constantof X.
Moreover,
if E means summation over all thecyclic ’permutations
o
’
of
tl,2,...,nl then,
for every choice of(a.)..,
i 1=This of course proves
that,
for some constant and every choice ofThe above
inequality
is theonly part
in theproof
of Theorem 1which
actually requires
theassumption
that X is 2-convex. For the remain- der of theproof
it would suffice to assume that X is B-convex(i.e.
that it has some non-trivial
type).
We also
notice,
for everypermutation
n of the firstintegers,
j I
Hence, by averaging
in the senseof kni
2m over allpossible
n, it follows thatThis proves one side of the
inequality.
Theproof
of the otherinequality
is stillquite elementary
buttechnically
more difficult. Let[a.} 1
i i=1 be a fixed sequence of scalars.Then,
for everypermutation n
of
11,2,...,nJ,
itfollows, by using
Khintchine’sinequality
and the2m-concavity
ofX,
that ,Hence, by expanding
the inner sum to the m-th power(here
is where weuse that m is an
integer)
andby averaging
in the sense 2m over allpossible
n, weget
thatwhere
and
The
expression S1
can be estimated as follows :We assume now
(without
loss ofgenerality)
that n satisfiesn/2 > m.
Inthis case n-m+1 >
n/2
and therefore weget
thatwhere A is the Khintchine coefficient in L
1 (0,1).
To evaluate the
expression S2
we first fix12m
andm1,...,m£
andput
,
Then, by applying
H81der’sinequality
in thelattice I ni (X) (see
e.g.-1 2m
and the vectors
we
get
thatBut )-
m and thus at least one of theintegers m . , j = 1,2,-..~,
sayM1’
isstrictly larger
than 1.Therefore, by estimating
the norm infollows that
Using
the estimate forS1
weget
thatHence,
if the maximum is not attained at the first termthen, by using
the estimate obtained for
S2,
it followsthat,
for some constantInB 2’
Consequently,
from
which, by applying
another variant of Htllder’sinequality (see
e.g.[6])
for the vectors ,i = 1,2,...,n
definedby
This
completes
theproof since,
as iseasily checked,
The
importance
of Theorem 1 can be illustratedby presenting
some of its
applications.
Forexample,
in theparticular
case whenx= L p (0,1)
forp > 2
Theorem 1 can be restated in thefollowing simpli-
fied form.
Theorem 2 : For every
p > 2
and K ~ 1 there exists a constantD= D(p,K)
such
that,
for every normalizedK-symmetric
basic sequence ini i I
L
p (0,1)
and every choice of scalarswhere
Proof : This is an immediate consequence of the
inequality
together
with the factthat,
forp > 2, L P (0,1)
is ofcotype
p whichimplies
thatTheorem 2
already
shows thatsymmetric
basicsequences
inL p (0,1), p>2
have a veryspecial
form. While in the infinite dimen- sional casethey
must beequivalent
to the unit vector basis of eithertp or t2 (cf. [5J)
it follows from Theorem 2that,
in the finite dimen- sional case,they generate what,
in theterminology
of H.P. Rosenthal[10J,
is called an Xp -space.
The results of H.P. Rosenthal from the above mentioned paper show e.g. that each such space isisomorphic
toa
complemented subspace
of LP (0,1) (with
theisomorphism
andprojection
constants
independent
of the dimension of thesubspace).
Theorem 2 can be also used to establish a series of results
on the
uniqueness
of somesymmetric
structures. One suchapplication
is related to the well-known fact that each of the
spaces 2 ,
P
has,up
toequivalence,
aunique symmetric
basis. D.R. Lewisasked,
inconnection with this
fact,
whether a similar result holds also for.the,p-spaces,
n=1,2,...
Togive
aprecise meaning
to the notion of uni- pqueness for
symmetric
bases in finite dimensional spaces we introduce thefollowing
definition.Definition 3 :
Let 3
be afamily
of finite-dimensional Banach spaces each of which has a normalized1-symmetric
basis. We say that each elementof llfl
has aunique symmetric
basis if there exists a functionD(K), K > 1,
so that whenever a spaceFE ð
has a normalizedK-symmetric
basis 1 1= then
ff n_ is D(K)-equivalent
to the1-symmetric
basis1 1=
of F.
In order to state the result on the
uniqueness
ofsymmetric
bases of finite
length
we also need thefollowing
verysimple
knownfact.
Proposition
4 : Let be aK-symmetric
normalized basis of a , i-iProof : The
right
handside inequality
iseasily proved by using
theformal
identity
map from Yonto , n
2(use
the2-convexity
of(y In
i i=11as in the
proof
of Theorem 1 and the fact that Y is oftype
2 withconstant
equal
to the Khintchine constant B). Conversely,
if T is an. p
1 n
invertible
operator
from Y intoL (0,1)’so
2 thatJITIJ.IIT-
2then, by transforming
the vectors into aK-equivalent
sequencei i=1
of
symmetrically exchangeable
random variables in a sequence of n random variables whosejoint
distribution inRn
is in- variant underpermutations
orchanges
ofsigns),
we conclude that, We leave the details to the reader. 0
Theorem 5 :
i )
Each member of thefami ly q
p
of all
subspaces
ofL
p
(0,I), p > 2,
which have a normalizedI-symmetric . basis,
has aunique
symmetric
basis.ii)
Each member of thefamily 1 S; P s: 00
hasa
unique symmetric
basis.Proof : Part
(i)
is an immediate consequence of Theorem 2. If{x.}?
1- i i=l
is an
arbitrary
normalizedK-symmetric
basis of anX E Q , p>2 then,
.
- p
a
priori;
theexpression
,appearing
when weapply
Theorem 2 for
(x 3n
1 1= ,depends
on theparticular K-symmetric
basisused.
However,
in view ofProposition 4, wn
isessentially equal
toi.e. we
get
the same value for w(up
to a constant which2 n
depends only
on K andp)
no matter which normalizedK-symmetric
basisof X has been used.
Part
(ii)
for is an obviouscorollary
of(i)
while for p = 1 and this fact has beenproved
in[7].
oRemark :
By using
a morecomplicated argument
it ispossible
to showthat,
infact,
even each member of thefamily U 3
p
has a
unique
sym- pmetric basis i.e. that the constants
appearing
in theproof
of Part(ii)
of Theorem 5 can be chosen as to be
independent
of p. Whether a similar assertion is true also for UQ
p
is not known.
Actually,
it iseasily
seen that a
positive
answer to thisquestion
would beequivalent
tothe fact that each member of the
family
of 11 finite dimensionalspaces with a
1-symmetric
basis has aunique symmetric
basis.Although,
as mentioned
above,
thisproblem
is still open, some results of aposi-
tive nature were
proved
in[4].
It was shown therethat,
forgiven q 2
and each member of the class G of all finite dimensional Banach spaces with a normalized1-symmetric
basis which hasq-concavity
constant
M,
has aunique symmetric
basis.Theorem 2 has additional
applications
to somequestions
con-cerning rearrangement
invariant(r.i.)
function spaces. If we restrict ourselves tocountably generated
measure spaces without atoms thenwe have to consider
only
the cases of r.i. function spaces on[0,1J
or on An r.i. function space X on an interval
I,
where I iseither
[0,1]
or will be a Banach lattice of measurable functionson I
(endowed
with thepointwise order) which,
forsimplicity,
issupposed
tosatisfy
thefollowing
two conditions.(i)
Theintegrable simple
functions on Ibelong
to X and forma dense set there.
(ii)
The functionX 0 1
has norm one.Besides the class of L
p
(0,I)
and the bestknown class of r.i. function spaces is that of Orlicz function spaces.
For
instance,
it is known(cf. ~1~)
that any Lp (0,1), 1s;p2
containsa
large
class of Orlicz function spaces as well as other r.i.function
spaces on
~0,1J
or on Forp > 2
the situation is howevercomple- tely different,
as is shownby
thefollowing
result.Theorem 6 : Assume that an r.i. function space X on an interval
I,
where I is
[0,1J
or isisomorphic
to asubspace
of LP
(0,1), p > 2. Then,
up to anequivalent norm, X
isequal
toL p (I), L2 (1)
orto
L P (I)(1L2(I),
the lastpossibility being
distinctonly
in the casewhen
Proof : Consider first the case
1=~0,1]
and let T be anisomorphism
from X into L
p (0,1, n p > 2.
For everyinteger
n the sequence*
rt
is
K-symmetric
inL P (0,1)
withTherefore,
once this sequence is normalized we are in aposition
toapply
Theorem 2. It follows that there exists a constant inde-pendent
of n, sothat,
for everystep
function T of the formwhere For a fixed T this formula is
clearly .
valid even when n is
replaced by
any otherinteger
m>n. Hence, we mayas well
put
a = lim infan
instead ofan (observe that, by setting
’If-= 1n
in the above
formula,
weget
aC
for alln)
andconclude
thatfor any
step
function IF as above. Since the set of thesestep
functionsis dense in X it follows that
for all f E X.
Consequently,
if a = 0 thenX=L 2(0,1)
while for a. > 0 wehave that
X= L P (0,1),
up to anequivalent
norm.The case I=
[0,(0)
can be deduced from that of I =[0, 1] .
Indeed,
observe firstthat,
for everyinteger
n, the restriction of X to the interval[0,n]
can be contracted to an r.i. function space on[0,1].
It follows that thereexist (3
andy n
suchthat,
for everyfE X which is
supported by
the interval[0,n].
and, again,
for a fixedf,
thenumbers p
andy n
can bereplaced by
respectively
Hence, for any I’EX,
The
and y
= 0 lead to X =respectively
X= L while when
both and y are strictly positive
weget that,
up to an
equivalent renorming,
ThatL p (0,oo)QL ~ (0,~)
is
actually isomorphic
to asubspace
of Lp (0,1)
for everyp > 2
canbe seen
by considering
thesubspace
of L(which
isclearly isomorphic
to LP (0,1)) consisting
of thosepairs (f,g)
forwhich
f = g.
Corollary
7 :(i)
The space LP (0,1), lpoo has
aunique
represen-tation as an r.i. function space on
~0,,1~.
(ii)
The space L has two distinctrepresentations
as an r.i. function space on~0,~), namely
Lp (0,00)
Assertion
(i)
is a direct consequence of Theorem 6 for Forp = 1
and theuniqueness
wasessentially proved
in
[9] ;
theproof being
based on[7].
The sameargument
also shows that and have each aunique representation
as an r.i.function space on
The space
appearing
in the statement(ii),
is the dual of L where
I/p+ I/q= I.
Theproof
of(ii )
follows also from Theorem 6 and the fact
that,
e.g. whenp> 2,
is
isomorphic
to L This latter fact can beproved by using
thedecomposition
methodprovided
we know that L P(0,00)
contains a
complemented
copy of To check this factone can use either Poisson processes as in
[4]
or thefact,
due toH.P. Rosenthal
[10],
that Xp -spaces
areisomorphic
tocomplemented subspace
of Lp
(O,co) ,
for allp > 2.
This suffices sinceis
clearly
acomplemented subspace
of anyultrapower
of thesubspaces generated by Ix n2n
(i-1)2 _ n,i2 _ n) n= 1,2,...
and each suchsubspace
is an
X p -space (and
theirultrapowers
arecomplemented
in some LAssertion
(i)
ofCorollary 7
isactually
valid for alarger
class of r.i. function spaces on
[0,I].
In[4]
Section 5 it isshown,
for
example,
that anyq 2
concave r.i. function space on[0,I]
hasa
unique representation
as an r.i. space on[0,I].
Without the assump- tion ofq 2 concavity
this result is not true ingeneral.
A 2-concave r.i. function space on[0,1]
withuncountably
manymutually non-equi-
valent
representations
as an r.i. function space on[0,1]
was constructedin
[4]
Section 10. Infact,
such anexample
can be built for every1 p 2
as tohave,
inaddition,
theproperty that,
on onehand,
itembeds
isomorphically
inL P (0,1) and,
on theother,
it contains acomplemented
copy ofL P (0,1).
We conclude with two more
applications
of Theorem 1.Theorem 8 :
Suppose
that a Banach lattice X isisomorphic
to a sub-space of a Banach lattice Y which is of
type
2and,
inaddition,
satis-fies an upper r>2 estimate
(i.e.
for someconstant and for every sequence of
pairwise disjoint
vectors1 in
Y).
Then either X itself satisfies an upper r-estimate or1 i=1
2
isdisjointly finitely representable
in X.In the case when
Y= Lr(0,1),
r > 2 the theorem wasoriginally proved
in[2].
The statement isquite simple
in thispart,icular
case.Corollary
9 : :Suppose
that a Banach lattice X islinearly isomorphic
toa
subspace
ofLr(0,1),
r>2. Then either X islinearly isomorphic
andorder
equivalent
to anLr(v)-space
for a suitable vor ~2
isdisjointly
finitely representable
in X.A Banach lattice X as above
clearly
satisfies a lower r-estimate fordisjoint
elements(since Lr(0,1)
is ofcotype r).
On the otherhand,
i f IZ2is
notdisjointly finitely representable
inX,
thenby
Theorem8,
X satisfies also an upper r-estimate.
Thus, by [ 9],
it islinearly isomorphic
and orderequivalent
to anLr(v)-space.
The other
application
of Theorem 1 is a variant of Theorem 8 which is valid in thespecial
case when X is an r.i. function space onTheorem 10 :
Suppose
that an r.i. function space X isisomorphic
toa
subspace
of a Banach lattice Y which is oftype
2 and r > 2 convex.Then either X itself is r-convex or
X= L2(0,1),
up to anequivalent
renorming.
Since the
proof
of the Theorems 8 and 10 use the same basicideas we shall
present
hereonly
that of Theorem 10. Bothproofs rely
on the
following
observation.Lemma 11 : Let 1 1= be an
arbitrary
sequence of vectors in ann r-convex Banach lattice Y and let q r. Let
111.111
be a norm onR
defined
by
Then, Rn
endowed with the normIII. III
forms an r-convex Banach lattice whoser-convexity
constant is £M"(Y).
Proof : Since q > r, it
clearly
suffices to show thatRn
endowed with the normis r-convex with
r-convexity
constantk=
1,2,...,m
is a sequence of vectors inRn
thenProof of Theorem 10 : As
already
observedabove,
one can find aninteger
m so that Y is 2m-concave. Let T be an
isomorphism
from X into Y and fix n.By applying
Theorem 1 to theIITII . liT-in -symmetric
sequencein Y we conclude the existence of a constant
independent
of n sothat,
for everystep
funct i on f E X of the formIf,
for everyinteger
n,where M is a constant
exceeding
both ther-convexity
and2m-concavity
constants of Y
then,
for everystep
function f as above(with
narbitrary),
we have
On the other
hand, (+)
shows that alsoThis proves
that,
in this case, X isequal
toL2(o,1),
up to anequi-
valent
renorming.
Suppose
now that there exists aninteger
k so thatThen,
aseasily verified,
when we evaluate the norm of- I
, n ~ k
by using
the formula(+)
then the maximum is attained
by
the first term i.e.By
Lemma 11 theexpression
defines a lattice norm on
2
whoser-convexity
constant does not exceed M. Hencefor every which means that if
an
arbitrary step
functionentirely supported by
the interval- 1-
then,
for 1L - 7 - , -
functions
This means that if we evaluate the norm in X
of entirely supported by
the interval
[O,2-k) by using
the formula(+)
then the maximum isalways
attained in the first term.Thus, by
Lemma11,
the restriction of X to~0,2 k)
is r-convex and so is all of X. 0REFERENCES
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J.Bretagnolle
et D.Dacunha-Castelle,
Ann. E. N.S.,
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437-480.
[2]
L.E. Dor and T.Starbird, Projections
of Lp
onto
subspaces spanned by independent
random variables.[3]
E.Dubinski,
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and H.P.Rosenthal,
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W.B.Johnson,
B.Maurey,
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M.I. Kadec and A.Pe0142czynski ,
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