C OMPOSITIO M ATHEMATICA
B. M AUREY
G. S CHECHTMAN
Some remarks on symmetric basic sequences in L1
Compositio Mathematica, tome 38, n
o1 (1979), p. 67-76
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67
SOME REMARKS ON SYMMETRIC BASIC
SEQUENCES
INL1
B.
Maurey
and G. Schechtman*COMPOSITIO MATHEMATICA, Vol. 38, Fasc. 1, 1979, pag. 67-76
Sijthoff & Noordhoff International Publishers-Alphen aan den Rijn
Printed in the Netherlands
Summary
Every subspace of LI
with an unconditional basis isisomorphic
to acomplemented subspace of
asubspace of Li
with asymmetric
basis.1. Introduction
The purpose of this paper is to show
that,
inspite
of thequite simple representation
ofsubspaces
ofLi
withsymmetric bases, given
in
[2],
such spaces can be verycomplicated
from thepoint
of view ofthe structure of their
complemented subspaces.
THE MAIN THEOREM: Let
(xi)i=l
be an unconditional basic sequence inLp,
1~p ~2.
There exists asymmetric
basic sequence(si)~i=1
1 inLp
such that
(xi)~i=1
1 isequivalent
to a block basis with constantcoefficients of (Si)~i=1.
Inparticular [xi]~i=1
1 isisomorphic
to a com-plemented subspace of [si]~i=1.
Theorems of such a nature were
previously
discoveredby
Linden-strauss
[8],
Szankowski[15]
and Davis[4].
Theproof
of the maintheorem, given
in§2,
uses thetechnique
of Davis’proof
as well assome results
concerning
the spaceXp
of Rosenthal[12].
Section 3 is devoted
mainly
to an alternativeproof
of thefollowing
reduction
(cf.
Dacunha-Castelle and Krivine[3]):
everysubspace
ofLi
contains some1,
if andonly
if everysubspace
ofLi
with asymmetric
basis contains one.Notions which are not
explained
here can be found in[9]
or[10].
* Supported in part by the U.S.-Israeli Binational Science Foundation.
2. The main result
We
begin
with two known lemmas:LEMMA 1
(Rosenthal [12]):
Let 1 p 2. There exists a constantKp
such thatif
m ~ 1 is a real number and(yi,m)~i=1
1 is a sequenceof independent, identically distributed, symmetric
random variables(on [0, 1])
eachtaking
three values in such a manner thatI/Yi,mllLp
=m-’
and
IIYi,m IIL2
= m, then :is a
projection
of normIIPII:5 Ap,
whereAp depends only
on p. I1follows that
(Yi,m)~ i=1
isequivalent,
with constantdepending
on ponly,
to sequence of the functionals
biorthogonal
to the unit vector basis inthe space
and the result follows..
Let
1~ p 2, m~ 1
and define a norm oné2 by
The unit vectors
(ei)i=l
ine2 clearly
form a1-symmetric
basis for69
In the
following
lemma we collect someproperties
of the03BBm(n)’s.
Fora
proof
see[4]
or[10,
lemma 3.b.3 and theproof
ofproposition 3.b.4].
and the maximum is attained at m =
m(n)
=n(1/p-l/2)(1/2)
(3)
For any sequenceof positive
numbers(~i)~i=1
1 there exists asubsequence (ni)i=l of
theintegers
suchthat, putting
mi=m(ni),
and
We are
ready
now for theproof
of the main theorem. Theproof
iswritten for the case p = 1
only,
the case 1 p 2requires only
minorchanges (and
the case p = 2 istrivial).
PROOF OF THE MAIN THEOREM: Let
(Xk)k=l
be a normalized un-conditional basic sequence in
Li
and let 1 p 2.(~i)~i=1
will denote asequence of
positive
numbers to bespecified
later and(ni)~i=1
1 willdenote
subsequence
of theintegers
with theproperties
stated inLemma 2.3.
Let
(Yi,k)~i=1,~k=1
be a double sequence ofindependent, symmetric,
three-valued random variables on
[0, 1]
such thatBy
Lemma 1:for all scalars
(ci)~i=1,
whereKp depends only
on p.Then:
and
si(r, t), i = 1, 2,... clearly
constitutes a1-symmetric
basicsequence in
L1(Lp).
Choose a
sequence(Tl)~l=1
ofdisjoint
subsets of theintegers
withTe
= ne(A
denotes thecardinality
of the setA),
and put(Wt)t=l
is a block basis with constant coefficients of(si)î=,;
we aregoing
to showthat,
if theEi’s
are chosenproperly, (Wl)~l=1
isequivalent
to(Xt)t=l.
Let
and note
that, by
thetriangle inequality, (1)
and Lemma 2:71
for all scalars
(at)t=t.
Andsimilarly:
Now
is a sequence of
symmetric independent
random variables and thusforms a 1-unconditional basic sequence in
Lp [cf. 12];
it follows that(VeYe=1
is a 1-unconditional basic sequence inLl(Lp),
thusHence,
if theEi’s satisfy
we get from
(2), (3)
and(4)
thatfor all choices of scalars
(al)~l=1
so it isenough
to prove that(Vl)~l=1
isequivalent
to(Xl)~l=1.
The sequence
is,
as wasalready mentioned,
a bounded away from zero 1-un- conditionalorthogonal
sequenceand, for l = 1, 2, ...
by
Lemma 2.It follows
easily
thatis
equivalent,
inLp,
to the unit vector basis oft2
and thus there existsa constant
Kp
such thatFinally, by
theunconditionality
of(xl)~l=1
andby
Khinchine’sinequality,
theexpression
CONCLUSION: Let 1 s p s 2 there exists a
1-symmetric
sequencesi)’
1 inLp
such that every unconditional basic sequence(xi)’
1 inLp
is
equivalent,
with constantdepending only
on theunconditionality
constant
of (xi)~i=1,
to a block basis with constantcoefficients of (Si)~i=1.
PROOF:
By [14]
thereexists,
inLp,
a 1-unconditional basic sequence such that any other unconditional basic sequence inLp
isequivalent,
with constant
depending
on theunconditionality
constantonly,
to asubsequence
of it.Apply
the main theorem to this universal basicsequence..
3. Dacunha-Castelle and Krivine’s reduction
This section is devoted to the
proof
of thefollowing proposition.
PROPOSITION 3:
If
everysubspace of Li
with asymmetric
basiscontains an
isomorphic
copyof tp for
some1 ~p ~ 2
then everysubspace of Li
contains such asubspace.
We
begin by recalling
two definitions.DEFINITION 1
([6], [7]) :
An unconditional basis(Xi)~i=1 1 is
said to beq-concave
(1 ~
q(0)
if there exists a constant c > 0 such that for73
every N and every N elements
we have:
DEFINITION 2
([5]):
Given 1~p ~2,
an 1-unconditional normalized be the space of all sequences of scalars a et2
such(|.|)mn
is defined in thepreceding section). (ei)~i=1
1 will denote the unit vectors in this space: note thatthey
form a1-symmetric
basis for Y.PROPOSITION 4: Let q 2 and let
(xi)~i=1
1 be a q-concave 1-un- conditional basic sequence inL 1,
let q p2,
and let(mn)~n=l
and(si)~i=1 be
as in theproof of
the main theorem. Then(si)~1=1 is equivalent
to
(ei)~i=1.
PROOF: We use the same notation as in the
proof
of the maintheorem.
By
Khinchine’sinequality
and thetriangle inequality
int2lp
we
get
that:AP
denotes here the Khinchine’s constant ofLp :
note that we haveused the fact that for a
fixed t, (~~i=1 aiYi,k(r)xk(t))~k=1
constitutes a 1-unconditional basic sequence inLp.
For the other side
inequality
we need the fact[cf. 1]
that there exists anisometry (into)
T :Lp - Li
and a constant C such thatfor all
f E Lp. Using
this fact and theq-concavity
of(xi)~i=1
l(with
constant c,
say)
weget that,
We shall need also the
following
standard.LEMMA 5 : Let U be an
infinite
dimensionalsubspace of
Y =Y(l2, tp, (xi)~i=1, (Mn)~n=1).
Then U contains asubspace
which embedsisomorphically in [xi]~i=1.
PROOF: Using
standardarguments (compare
forexample
theproof
of
proposition
3.b.4 in[10])
theproof
reduces toshowing
that theidentity
map from Y toe2
isstrictly singular.
Now any infinite dimensional
subspace
ofe2
contains a norm onevector with
arbitrarily
small l~ norm. Fix n and let x be a norm onevector in
e2
such that~x~l~~
m;2p/(2-p).
Let y and z be elements oft2
such that
and
so that
75
Hence every infinite dimensional
subspace
of Y contains vectors withnorm one in
e2
and witharbitrarily big
norm in Y. ·PROOF OF PROPOSITION 3: Assume that every
subspace
ofLi
with asymmetric
basis contains somee,.
Notice first that if(xi)~i=1
1 is a 1-unconditional basic sequence inLi
which is q-concave for someq 2
then, by
the maintheorem, Proposition 4,
Lemma 5 and the fact[cf. 11]
that every infinite dimensionalsubspace
ofe,
contains anisomorph
oftp, [xi]~i=1
1 containse, isomorphically
for some p(neces- sarily
1 «5 p ~q ).
Now let X be a
subspace
ofLi.
If X does not containé,
thenby
Rosenthal’s theorem
[13]
X isisomorphic
to asubspace
ofLr
forsome 1 r - 2 and since
Lr
has an unconditional basis theimage
of Xin
Lr
contains an unconditional basic sequence(yi)~i=1.
It isclearly enough
to show that[yi]~i=1 contains
somet,.
Let
(gi)~i=1
be a sequence inLi isometrically equivalent
to the unitvector basis of
t2l, (for
instance a sequence ofL1-norm
one,independent 2I r
stable randomvariables),
anddefine,
for i =1, 2,
...and 0 ~S, t ~ 1
Then
(xi)~i=1
1 is 1-unconditional inL1«0,1)2)
andby
thetriangle
in-equality
inL2/r,
i.e.
(xi)~i=1
1 is2/r
concave.Thus,
it follows from the firstpart
of theproof
that[xi]~i=1
1 contains asubspace isomorphic
tot,
for some1 ~ p ~
2/r. Passing
to a smallersubspace
we may assume that there existdisjoint
blocksof
(xi)~i=1
which areequivalent
to the usual basis oftp.
It iseasily
checked that the blocks
1 are
equivalent
to the usual basis ofREFERENCES
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(Oblatum 1-VI-1977) B. Maurey
École
Polytechnique Paris, France andThe Institute for Advanced Studies The Hebrew University
Jerusalem Israel
G. Schechtman
The Institute for Advanced Studies The Hebrew University
Jerusalem Israel