C OMPOSITIO M ATHEMATICA
Z. A LTSHULER
A Banach space with a symmetric basis which contains no `
por c
0, and all its symmetric basic sequences are equivalent
Compositio Mathematica, tome 35, n
o2 (1977), p. 189-195
<http://www.numdam.org/item?id=CM_1977__35_2_189_0>
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189
A BANACH SPACE WITH A SYMMETRIC BASIS WHICH CONTAINS
NO lp
OR c0, AND ALLITS SYMMETRIC BASIC
SEQUENCES
AREEQUIVALENT
Z. Altshuler*
Abstract
A Banach space
having
theproperties
described in the title of thispaper is constructed.
In this paper we
investigate
thesymmetric
basic sequences in aBanach space with a
symmetric
basis. It is well known that the unitvector basis in the spaces co and
tp(l ~
p~),
is asymmetric basis,
andevery
symmetric
basic sequence in each of these spaces isequivalent
toit. A natural
question
is whether there exists any other Banach spaceX,
with asymmetric
basis{en}:=,
which has the sameproperty.
Let usrecall that
by [1],
it turns out thatif,
in addition to theassumption
thatevery
symmetric
basic sequence in X isequivalent
to{en}:=,
we knowthat the same holds in
X*,
the dual ofX,
withrespect
to{fn}:=~’
thesequence of the
biorthogonal
functionals associated tolenl-=~,,
then{en}:=l
isequivalent
to the unit vector basis of co orep,
for some1 p oo.
We answer the
question
raised aboveaffimatively by proving
thefollowing
THEOREM: There exists a Banach space
X,
with asymmetric
basis{en}:=,
such that allsymmetric
basic sequences in X areequivalent
toeach
other,
and X is notisomorphic
to Co orép, for
any 1 ~ p ~.* This is part of the author’s Ph.D. thesis, prepared at the Hebrew University of Jerusalem under the supervision of Professor L. Tzafriri. The author wishes to thank Professor Tzafriri for his guidance.
Noordhoff International Publishing
Printed in the Netherlands
190
Clearly
a Banach spacehaving
theproperties
described in thetheorem above contains no
subspace isomorphic
to Co ortp (1 ~ p
oo).
A natural candidate for such anexample
is the space constructedby Figiel
and Johnson[2],
which has asymmetric
basis and nosubspace
of which isisomorphic
to co orep.
Ourexample
is obtainedby
a modification of their construction. Beforepassing
to theproof
ofthe theorem we need some definitions and notations.
DEFINITION: Let X be a Banach space with a
symmetric
basis{en}:=l’
LetNi
i =1, 2,
... be subsetsof
the setof
natural numbersN,
so that
Ni = N for
everyi,
N =UJ§=, Ni
andNi
flNj = 0 for
all i¥j.
For any 0 ¥ a =
li
aiei E X putui I Jb= ,
1 a.e.. whereNi
=f i, j)Jb= , 1 for
i = 1, 2, ....
The sequence{ua)}=l is
called a basic sequencegenerated by a.
Clearly
for any a E X{ua)}=l is
asymmetric
basic sequence in X.If
tUnln’=,
and{Vn}:=l
are basic sequences in Banach spacesX, respectively Y,
we say that{Un}:=l
andf vn}n=1
areequivalent
when aseries
liajui
converges if andonly
ifYiaivi
converges. We write in this caselunl - {Vn}.
We say that a basic sequence{Un}:=l
is bounded if there exists an M > 0 such thatM-’ inf n Il Un Il ~
SUPNIl Un
M.The first
example
of a Banach space which contains notp
or Co isdue to Tsirelson
[4]. Figiel
and Johnson[2],
described the dual of this space, which will be denotedby T,
and showed that T contains nosubsymmetric
basic sequence. We also recall that the unit vectorbasis
IXnln’=l
of T is an unconditional basis.We are
ready
now to construct ourexample.
First we define asequence of norms,
1.ln,
on co,by
fâilî=,
1 is therearrangement
innon-increasing
order offlaillî=l,
W -
i-1 and s
=E=t
(oi. Notice that sincewe have that for all n =
1, 2,..., l’In
isequivalent
to the sup norm on co. We put now Y =f y
E Co;Illn 1YInXnIIT ce)
where{Xn}:=l
is the unit vector basis of T. The space Y is asubspace (called
thediagonal)
ofthe direct sum
Z = (En=, (D (c,, 1-1n»T-
Since for any unitvector
j
=1, 2, ...
weget 1 ejln= (2n + 2-n)-1,
we deduce that the sequence ofunit vectors
{ej}j=l
1belong
toY,
andthey clearly
form asymmetric basis,
withsymmetric
constant 1. We also remark that we may assume, without loss ofgenerality,
that everysymmetric
basicsequence in Y is
equivalent
to asymmetric
block basic sequence of{en}:=l’
So in order to prove the theorem it suffices to check the block bases of{en}:=l’
LEMMA 1: Let ym =
le,-+’ +,
aiei be a normalized block basis in Y.If limi--
ai = 0 then there exists asubsequence {Ymj}j=l of {Ym}:=l
whichis
equivalent
to a block basisof (x;)J§=i,
the unit vector basisof
T.PROOF: For fixed m and N we have
by (2)
thatTherefore we can construct
inductively
twoincreasing
sequences ofintegers lmjlî=,,
andINjlî=l
such that111,’ =Ij IY,jl,,X,,IIT 2-j-’
for allj
> 1 and[[I)t-,i’ 1 Yj I.X. Il T 2-j-’
forall j
> 1. The block basisf y,jl’
can be identified with the basic sequence
(§)Jt=,
in ZNi where+gmi =
(y Y.j, - - ., Ymjl * , »)
eZ j
=1 2,.... Put v
=(0, 0,..., 0, > N;-’ y m y N;_’+’ m .
Yym. , 0, 0, ...) E Z j = 1, 2,...
and notice that foreach j,’
Hence the basic sequence
{Ymj}7=1 in
Y iséquivalent
to{Vj}7=1 which,
in turn, is
equivalent
to the block basis Zj_ n e N;_, IYmjlnxn j
=1, 2, ...
of{jcj:=,.
We can
already
state some consequences of Lemma 1.PROPOSITION 1: Let Y and
(e;)J’J=,
I be as above. Thenthe following
assertions are true :
(i)
There is nosymmetric
block basis ym =INT/i+,
aieim = 1, 2, ...
of (e;)J’J=,
1 such that thecoefficients (a;)J’J=,
1 tend to zero.(ii)
y contains nosubspace isomorphic to
Co ortp for
any 1 p00
PROOF: The first assertion follows from Lemma 1 and the fact that T contains no
subsymmetric
basic sequence. To prove the second assertion we assume first that there is a block basis{Uj}7=1
1 of(e;)J’J=i
1which is
equivalent
to the unit vector basis oftp,
for some p ? 1.Since
IIj=l ujlly 00
as n -OJ it is easy to construct a block basis{Vm}:=l
of{Uj}7=1
1 withcoefficients,
in theexpansion
with respect to192
teili=l, tenulng lU GCrU. The pi oui ui this case can uc then completed
by using (i).
Suppose
now that there is a block basisJujlî=,
1 ofleilî=l
1 which isequivalent
to the unit vector basis of co. If the coefficients of theyj’s
form a sequence
tending
to zero, then wecomplete
theproof
of(ii) by (i). Otherwise,
it followseasily
that(e;)J’J=i
1 itself isequivalent
to theunit vector basis of co, hence for all
k = 1, 2, ... Illk i=, eill,:5 M,
forsome M > 0. On the other hand for
any k -
4 wepick
aninteger
n =
n(k)
such thatsk/2 2’n -- 2sk.
For these values of k and n =n(k)
we have
We consider now block bases
generated by
one vector in Y.LEMMA 2:
Every
block basis{ua)}=l of {ej}7=1 generated by
a vectora E
Y, is equivalent to (e;)J’J=,.
PROOF: Let
{ua)}=l
be a block basisgenerated by
a vector 0 # a =Li aiei E Y. Then,
for any{3 = Li {3iei E Y,
we haveIILi {3iua)1I (SUPj lajl)lli {3ieill,
so in order to prove that{ua)} --- {e}
we have toshow that
Li (3iua)
converges, for anyQ
E Y. We first observe that it isenough
to prove this for(3
= a i.e. to show thatX, aiUa)
is a con-vergent series for any 0 ¥ a =
Li
aiei E Y.Indeed,
if this is true forany a = Li aiei E
Y then for anyQ = Li {3iei E
Y we wouldget
thatLi (ai + (3;)ua+),
and therefore also thatI;Q;u)",
is aconvergent
séries. Fix a =(a;)J’J= , E co
with 1 > a , & a2 >_ > ai * ...> 0,
and notice that in order to check whetherE, aiUa)
converges in Y we haveto compute the
l’In-norms
of the double sequence{aiaj}j=’
which isthe
expansion
ofI; a;u)"
with respect to{ei}=l’
Leta(t)
be a non-increasing
function on[l,oo)
such thata(i) == ai
for all i.If,
for someinteger m, 1 . j
= m then at least one of theintegers i or j
is greater than orequal
tom 1/2,
and thereforeaiaj:5 a(m 1/2).
It follows that thenon-increasing
rearrangement of{aiai}j=l (as
a one indexedsequence)
is
majorated by
the sequenceQ = (Q;)J’J=,
whoseexplicit
form iswhere
T(m )
is the number of distinct divisors of m.Thus,
for every n, we haveIli aiua)ln
~1{3ln.
For each
integer
m, letcp(m)
be the firstplace
wherea (m )
appears in the sequence(3. Then,
forp (m ) k ç(m
+1)
we haveSince Slp(m) ~
log cp(m)
we get thatTo estimate further the norm of
/3
we use thequite
known fact(see
e.g.
[3,
p.118]) that I§=i T(1)
= klog k
+(2 y -1 )k
+0(kl/2)
where y = 0.57721 ... is the Euler constant. Notice thatç(1)
=1,
andfor j
> 1we have
cp(j) = {-/ r(i) = (j2 - 1) log (j2 - 1) + (21’ - 1)(j2 - 1) + O(j),
consequently
(4) ç(j)
~ 1 +cj2 . log j
forall j ~ 1,
and some constant ci > 0.We also have
Using
the fact that sm behavesasymptotically
aslog m
and sub-stituting (4)
and(5)
in(3)
we deduce that’
for all n and some c3 > 0.
Since a E Y
implies
thata E co
we get that[[Q[[y
~c31lally,
i.e.IIi aiua)lly ~ c311ally
for all a E Y.We are
ready
togive
theproof
of the theorem. Let ym =le-+,
aiesbe a
symmetric
normalized block basic sequence in Y. We mayassume without loss of
generality
thatapm+1 ~ apm+2 ~ ... apm+l ~ 0,
forall
m = 1, 2,....
If sup.(Pm+1 - Pm)
+ 00 thenclearly {Ym} ’"’-’ leml,
194
hence we may assume also that supm
(Pm+1 - pm) =
+00.Suppose
now that for any e > 0 there exists aninteger
N =N(e)
such that
[]INI/i +ç a;e;]] e
for all m withPm+1 - Pm N.
In this case{Ym}:=l
iséquivalent
to a block basisgenerated by
one vector andthus, by
Lemma2, {Ym}""" {em}. Indeed,
for any e> 0,
let um =i fmjiPm ai+Pmei
i andu ’ m = f:(f) ai+Pmei m
=1 , 2 , .. .. Using
the fact thatfor all m =
1, 2,... u m
have at most the firstN(E)
coordinates distinct from zero, andI/um - u1I E,
we deduce that{Um}:=l
is arelatively
compact
set in Y. Hence there exists asubsequence {U::J:=l
of{um} :=1
such thatlimm u m = Q = ~i {3iei
E Y.Clearly,
asubsequence
{Umj}j=l
1 of{u::J
can be chosen such thatIlumj -1311 2-j,
and sinceIIYmlly = IIumlly
=1,
for all m, we have[[Q[[y
= 1. Notice that{Umj}j=l is
a"translation" of
{Ymj}j=l’
Hence{Ymj}j=l is equivalent
to a block basisgenerated by Q.
We treat now the case when such an
N(e)
does not exist for alle > 0. In this case there exists an e > 0 and an
increasing
sequence ofintegers {Pm.}j=l such
thatpm+1- Pm; > j
andlli?21’+ +, a;e;lly =
e for allj.
Put Vj -~’= 1 ’+1 m;
aiei and Uj -~’= 1 +1 1 P ml
aiei’Notice
thatap +j>
c forsome constant c and
every j imply
1 ~[[uj[[y
~c[[I)=i e;[[ j = 1, 2, ...
i.e. c = 0.
Thus, lim j-+oo
ap+j=
0 which means that(vj)Jll=i
is a boundedblock basis of
(e;)J’J-, with
coefficientstending
to zero.By
Lemma 1(and passing
to asubsequence
ifnecessary)
we can assume that{Vj}j=l
1is
équivalent
to a block basis{Zj}j=l
of{xn}:=,
the unit vector basis ofT. The definition of the norm in T
implies
the existence of a constantA, > 0
suchthat,
forevery k,
we have[[I$i+, z;[]v ± A,k.
It followsthat for every
integer k
and some constantA2 > 0 [[1§b, y;[[y ± lli§£i vll y = Alll§£i zjllT = AiAk.
Since{ym; };
1 is asymmetric
basicsequence we
get
that{Ymj}j=,
and therefore{Ym}:=’
isequivalent
tothe unit vector basis of
fI, contrary
toProposition
1. Thiscompletes
the
proof
of the theorem.REMARK: One can check that the unit balls determined
by
thenorms
1.ln
n =1, 2, ...,
are the sets2-"Bo+ 2nBd,
whereBo
andBd
arethe unit balls of Co,
respectively
of the Lorentz spaced(i-1, 1). (Recall
that
d(i-t, 1)
is the space of all sequences(a;)J’J=, OE co
such thatIIalid
=1£ , âii-1
00, where{âi}=l
is thenon-increasing
rearrangement of{laïl}=l)’ Similarly,
it can be shown that the sequence of normsIl.IIn
n = 1, 2, ...
definedby Figiel
and Johnson in[2]
can begiven explictly
by
the formulasIn this case it is not true any more that for any a =
{ai}=l E
co we havellfail,=111.
~clllaicïj}î,i=lll,,
for all n and some c > 0.Indeed,
for thesequence ai =
i-1/2
we can find constantsA,
B > 0 such thatIlli-1121 î=lln :5
A but111(ij)-"21î,j=llln - Rn 1/2
for all n =1, 2,....
REFERENCES
[1] Z. ALTSHULER: Characterization of c0 and lp among Banach spaces with
symmetric basis. Israel J. of Math. 24(1) (1976) 39-44.
[2] T. FIGIEL and W. B. JOHNSON: A uniformly convex Banach space which contains
no lp. Compositio Math. 29(2) (1974) 179-190.
[3] W. J. LEVEQUE: Topics in number theory I. Addison-Wesley Publishing Company.
[4] B. S. TSIRELSON: Not every Banach space contains an imbedding of c0 or lp.
Functional analysis and its application 8 (1974) 138-141.
(Oblatum 6-VI-1976) Institute of Mathematics
Hebrew University
Jerusalem