C OMPOSITIO M ATHEMATICA
D. V AN D ULST
A certain subspace of characteristic zero of (l
1)
Compositio Mathematica, tome 28, n
o2 (1974), p. 195-201
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195
A CERTAIN SUBSPACE OF CHARACTERISTIC ZERO OF
(l1)*
D. van Dulst
Noordhoff International Publishing Printed in the Netherlands
We construct an
example
of asubspace ’
V of theconjugate
E* = l’of E =
11
with characteristicr(V)
= 0 andsatisfying
thefollowing
twoconditions:
(K1) if xn ~
xo for03C3(E, V),
thenlim ~xn~ ~ I xo I I (K2)
If xn ~ xo for03C3(E, V)
andIntroduction
Let E be a Banach space, E* its
conjugate
and Tl asubspace
of E*.The
unit ball ofE(E*,
Vrespectively)
we denoteby SE(SE*, SV
respec-tively).
Dixmier([2])
defined the characteristicr(V)
of Vas follows:r(V)
= sup{03B1 :
a~
0 andUSE-
CSV03C3(E*,E)}.
Clearly r( )
> 0implies
that Y is03C3(E*, E)-dense
inE*,
but the converseis not true
(see [2]
for anexample).
The
following
two results involve characteristics.PROPOSITION 1:
( [6, proposition
4.1]).
Let E be a Banach space and let Vbe a
separable subspace of
E*. Then(Kl)
isequivalent
tor(V)
= 1.PROPOSITION 2:
([3],
see also[9,
p.486])
Let E be aseparable
Banachspace and let V be a
subspace of
E* withr(V)
> 0. Then there exists anequivalent norm |||·|||
onE for
which(K1)
and(K2)
hold.Our
example
shows that inproposition
1 theseparability
of Y is essen-tial and also that in
proposition
2 the conditionr(V)
> 0 is not necessary.First we prove,
setting
E =11,
E* =l~,
that for each k E N there exists a(non-separable) subspace Vk
of E* such that(K1)
and(K2)
holdwhereas
1 Apparently the problem of the existence of such a subspace was raised by Kadec.
We thank Prof. Singer for communicating it to us and for some discussions resulting in the proof of proposition 1.
196
This
Yk
will be a suitablequasi-complement
of co inE*,
which we defineby modifying
a construction of Rosenthal([8]).
Thisleads, by
a proce- dure oftaking 11-sums,
to asubspace
V of E*satisfying
both(Kl )
and(K2)
and withr(V)
= 0.We
begin by sketching
aproof
ofproposition
1 which differs from theone
suggested by
Mil’man.PROOF OF PROPOSITION 1: We first observe that
(Kl )
isequivalent
to thesequential 03C3(E, V)-closedness
ofSE.
Since V isseparable,
thetopology 03C3(E, V)
is metrizable when restricted to bounded subsets of E. Hence thesequential Q(E, V)-closure
and the03C3(E, V)-closure
of,S’E
coincide.Thus
(K1)
means thatSE
is6(E, V)-closed
and this in turn isequivalent, by [2, Théorème 8],
tor(V)
= 1.Observe that
r(V)
= 1implies (K1)
also fornon-separable V, by [2, Théorème 8]. The separability
of V is neededonly
for theproof
of theconverse
implication.
One should also note that
(K1) implies
that Y isu(E*, E)-dense,
wheth-er Y is
separable
or not.Our
example
will be based on thefollowing
LEMMA: Let E =
ll,
E* = 100 and let V be a03C3(l~, ll)-dense quasi-com-
plement of
co in 100(We
assume co to be imbedded in l’ in the canonicalway).
Then we have:If xn ~ xo for a(ll, V)
and{xn}
isnorm-bounded, then ~xn - x0~ ~
0. Inparticular, (K1)
and(K2 )
aresatisfied.
Proof:
Let{xn’}
be anysubsequence
of{xn}.
Since11
is the dual of theseparable
space co,{xn’}
contains(see [1])
a03C3(l1, co )-convergent
sub-sequence
{xn"}.
Thus{xn"}
isu(ll, co)-Cauchy
as well as03C3(l1, V)-Cauchy
and therefore
a(Il,
co +V)-Cauchy.
Since co + V is norm-dense in100,
the boundedness of{xn"}
nowimplies
that{xn"}
is03C3(l1, l’)-Cauchy
andtherefore
norm-convergent (see [4,
p.281 ]),
say to x.V being 03C3(l~, l1)-
dense in
l~, u(ll, V)-limits
areunique.
Thisevidently implies
thatx = xo. We have now shown that any
subsequence
of{xn}
contains asubsequence converging
to xo in norm.Hence ~xn - x0~ ~
0.The statement
proved clearly implies (K2),
and also(Kl ),
since(K1)
isequivalent
to thesequential U(ll, V)-closedness
ofSl1.
In order to understand our
example
it is necessary to recallbriefly
Rosenthal’s construction of a
quasi-complement
of co in 100(cf. [8]).
This construction is based on the
following observations,
thecomplete proofs
of which can be found in[8].
(i)
Asubspace X
of a Banach space E isquasi-complemented
in E ifand
only
if there exists a6(E*, E)-closed subspace
Y of E* such that Yn Xl. = {0}
andYl
n X ={0}. Indeed,
if Y has theseproperties,
then
Y~
is aquasi-complement
of X in E.(ii)
If Y is a reflexivesubspace
ofE*,
then Y isQ(E*, E)-closed.
Thisfollows from the
Krein-Smulian
theorem.(iii)
If an infinitecompact topological
space S contains an infinite per- fectsubset,
thenC(S)*
contains asubspace isomorphic
to12.
Rosenthal’s construction
([8])
of aquasi-complement
of Co now pro- ceeds as follows. We mayidentify
100 withC(03B2N),
where03B2N
denotes theStone-Cech
compactifation
of N. Thenci
can be identified withC(PNIN)*.
Since03B2NBN
is an infiniteperfect compact
Hausdorff space,(iii) implies
thatcô
contains11 isomorphically.
Let H cci
beisomorphic
to
l’
and let{03BC1, ···, 03BCn, ···}
be a basis of Hequivalent
to the ortho-normal basis of
1’.
We assumethat ~03BCn~
= 1(n
=1, 2, ... ).
For eachn E N let
bn
be the Dirac measure on N concentrated at n.Then
the closed linear span of{03B4n :
n~ N}
in(l~)*
can be identified withl’, by
the ca-nonical map. Now let G be the closed linear span of
It is
easily
verified that G isisomorphic
to H and therefore03C3((l~)*, l~)- closed, by (ii). Finally,
G nci = Gl
n co ={0},
so Tl =Gl
is aquasi- complement
of coby (i).
Since,
in thisconstruction, Vol
n11
= G n11
={0},
V is03C3(l~, l1)-
dense in
l~,
so the lemmaapplies.
EXAMPLE: We now show that
by
aslight
modification of the construc- tion described above we can obtain for each k ~ N a03C3(l~, 11 )-dense quasi-complement Vk
of co withr(Vk) ~ 1/k.
Let k E N be
arbitrary
and letGk
be the closed linear span ofk03B41 + 111
1and
Clearly Gk
isisomorphic
to H and thereforeu«1’)*, l’)-closed, by (ii).
Again,
as before it iseasily
verified thatTherefore Vk
=(Gk)~
is aquasi-complement
of co inl~, by (i).
Alsoso that
Vk
is03C3(l~, l1)-dense
in l~.Next we show that
198
By [2, Théorème 9]
it suffices to prove that(Here (X, Y),
forarbitrary subspaces X
and Y of a Banach spaceE,
de-notes the inclination of X to
Y,
i.e. the distance of the unitsphere
of Xto Y
(cf. [9]). Clearly, since 03B41
ESI,
andwe have
which proves our
claim,
sinceG,
=V/.
Now,
foreach k E N,
letEk
=11, E*k =
100 and letVk
be the03C3(E*k, Ek)-
dense
quasi-complement
of Co inE*k
withthat was constructed above.
Then, putting
we have
We will show that
satisfies
(Ki )
and(K2)
whereasr(V) =
0.To prove
(K1),
it suffices to show thatSE
issequentially a(E, V)-closed.
Let
{x(n)}~n= 1, with x(n) = (x(n)1, x(n)2···)
EE(n
EN),
be a sequence inSE
which converges for
u(E, V)
tox(0) = (x(0)1, x(0)2, ···)
E E. We mustshow
that ~x(0)~ ~
1. For this it isenough
to prove that for an arbi-trary k
e Nwhere 1tk is the natural
projection
of E onto(El
E9...E9 Ek E9 {0}
~ ···)l1,
which weidentify
with(El
E9E2
E9 ... E9Ek)l1. Clearly
the sequence
converges to
7rk(x(o»
for03C3(03C0k(E), n:(V» = U((E1
~ ... ~Ek)l1,
(V1
~ ··· OVk)l~).
Since~03C0k(x(n))~ ~
1 for all n ~N, (El ~ ···
OEk)l1 (which
is isometric tol1)
is isometric to the dual of theseparable
space
and
(V1
~ ... OVk)l~
is aquasi-complement
of(co
O ’ ’ ’ Oc0)l~
inthe
Lemma applies
hereand yields that ~03C0k(x(0))~ ~ 1. Hence ~x(0)~ ~ 1,
since k e N wasarbitrary.
To
show that(K2) holds,
let us assume thatx (n)
~x(0)
foru(E, V)
andthat
~x(n)~ ~ ~x(0)~.
We may also assumethat ~x(0)~
= 1. Let e > 0 bearbitrary
and let k E N be such thatAs in the
proof
of(Kl )
it follows from the Lemma that(n - oo ).
Hence there exists anno E N
such thatand
therefore, by (1),
We may also assume that
Thus
It follows now from
(1), (2), (3), (4)
and(5)
thatThis proves
(K2).
200
Finally,
let us show thatr(V)
= 0. We haveand it is
easily
seen thatBy
the definition ofr(Vk)
It follows that
Thus
r(V)
= 0. Thiscompletes
theexample.
We conclude with a
general
result onquasi-complements
of co in 100.All such
quasi-complements
obtainedby
Rosenthal’s construction are03C3(l~, 11 )-dense
in 100. This may not be the case ingeneral. However,
all
quasi-complements of co
are ’almost’03C3(l~, l1)-dense
inl~,
as we showin the
following
PROPOSITION 3: Let V be a
quasi-complement of
co in l~. Then the03C3(l~, l1)-
closure V’of
V in 100 hasfinite
codimension in 100.PROOF:
Suppose
that diml~/V’
= oo. Then wehave,
sinceYl - V’~,
that
dim V~ =
ooand,
of course, diml1/V~ =
oo.By [7,
Lemma2]V,-
contains a
subspace
L with dim L = oo which iscomplemented
in1’.
Let M be a
complement
of L in11.
Then 100 =L~
QMl. By [5] both L~
andM~
areisomorphic
to 100. Inparticular M~
isnon-separable.
SinceL c
V~
we have Y ~L~. Furthermore, l~/V
isseparable, by
the defi-nition
of V,
whereasl~/L~ ~ Ml
is not. This is acontradiction,
sincethe canonical map
l~/V ~ l~/L~
is a continuoussurjection.
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[4] G. KÖTHE: Topological vector spaces I. Springer Verlag (1969).
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