The existence in every separable Banach space of a

Séminaire d’analyse fonctionnelle
École Polytechnique
R. I. OVSEPIAN
A. P EŁCZY ŃSKI
The existence in every separable Banach space of a fundamental
total and bounded biorthogonal sequence and related constructions
of uniformly bounded orthonormal systems in L2
Séminaire d’analyse fonctionnelle (Polytechnique) (1973-1974), exp. no 20, p. 1-15
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THE EXISTENCE
iN EVERY SEPARABLE BANACH SPACE
OF
A FUNDAMENTAL TOTAL
AND BOUNDED BIORTHOGONAL SEQUENCE AND RELATED CONSTRUCTIONS
BOUNDED ORTHONORMAL SYSTEMS IN
OF UNIFORMLY
L2
by
R.I.
OVSEPIAN
(Erevan)
and
A.
PE0142CZY0144SKI (Warszawa)
*
Institute of Mathematics of the Armenian
Academy
of Sciences.
#0
Institute of Mathematics,
Polish Academy of Sciences.
Abstract.
1)
*
n
n
(x,)
Tn every
separable
Banach space X
biorthogonal sequence
a
is constructed such that linear combinations of the xl
""
dense in X,
for every
x
in X
if
x
(x)
s
are
n
4~
0 for all
n
then x = 0 and
n
CJ ,
n
11
11
2) Linear subspacps of
which admit
basis consisting of uniformly bounded functions
are
an
orthonormal
characteri-zed.
XX.1
The
present paper consists
In the first
and 4)
we
using
one
of three sections.
trick invented
a
Olevskii (~9~
by
Lemmas 3
prove
Theorem 1 :
In every
Banach space X there exists
separable
a
fundamental
*
and total
biorthogonal
(xn ,xn)
n
sequence
such that
*
Recall that
a
sequence
n
n
of
Banach space X and bounded linear functionals
a
of’
X -
the dual of X is said to be
X, i.e. elements
on
biorthogonal
if
x*(x )
n
~
n,m =
1,2,...
A
biorthogonal
combinations of the
v:’s
are
of elements of
pairs consisting
=
m
amn
for
sequence
(xn ,x n)
dense in
X, and is total if the condition
is fundamental
if linear
-.L
-M-
x*n(x) =
0 for
Theorem 1
1,2, ... implies that
n =
answers
question
a
weaker result has been
previously
([1J, p.238).
of Banach
obtained
by
A
slightly
Davis and Johnson
[4].
The main result of the second section is
Theorem 2 :
L2(03BC)
a
where 03BC
Let E be
is
separable
a
probability
e,
set S. Then E admits
an
bounded functions if and
Moreover if E n
orthonormal basis
can
linear
measure
subspace
on
a
orthonormal basis
a
si g me field
of subsets of
of
consisting
separable subspace
be constructed
space which is dense in the
Hilbert space
a
uniformly
if
only
is
of
norm
11 - 11
so
.0
then the
of
that
it spans
in E n
L-(I,) .
a
linear sub-
XX.2
As
corollary
we
obtain that every
subspace
codimension admits
an
orthonormal basis
consisting
a
infinitely
of H.
many times differentiable functions.
In the third section
an
we
isometric
that there exists
j(X)
ded
a
as
Using
a
question
a
answers
consider the class of such Banach spaces X
embedding,
scalar-valued continuous functions
ball of
uniformly bounded
of
This
of finite
[14J~
Shapiro
which admit
L ~ (0~1)
oi
a
is not
on
a
j,
into
a
space
a
L2(~),
space X has the above
(ii)
satisfies the condition
[13]
result of Rosenthal
i.e. j(X)
11
of all
S such
show that
we
regar-
of Theorem 2.
property if and only if it contains
to the space
of all
S such that the unit
on
probability measure V
totally bounded subset of
L2( )
C(S)
compact Hausdorff space
Borel
subspace of
recent profound
subspace isomorphic
say
a
Banach
closed linear
a
absolutely convergent series
of scalars.
1.
Proof of Theorem 1 .
O1 evski i’ s
tion of
Lemma 3 of
[A]
Banach space X, then
rated
by
Lemma 1 :
We
[9].
a
a
lemma which is
If A is
a
non-empty subset of
xo, x1,...,x
2n-l
2n-1
subspace
Banach space and let
of
n
be elements 0 f X and let
ill,
elements of X
exists
and
then
a
such that
unitary
real
a
subspace of X geneX generated by A.
be a
-
Let
a
denotes the closed linear
A and lin A - the linear
Let X be
modifica-
with
begin
X" 0
positive integer.
-lk
X*1
0 x1,
I
I
*
...,
...7
x 2n_1
x2n-1
bebe.
XX.3
Proof :
The conditions
2x2 -
matrix.
The
(3)
and
(4)
are
specific unitary
satisfied for every unitary
matrix for which
(1)
and
(2)
hold is defined to be the matrix which transform the unit vector basis
2 -dimensional
of the
12-n
Hilbert space
onto the Haar basis of this space.
We put
We have
Clearly (5) implies (1)
Proposition
---- -- -
-
..
---
--
-.
1 :
Let
-.
(x aAl
and
(2).
x 1* )
be
a
fundamental and total
n
I
sequence in
a
Banach
space X
such that there exists
-
nite sequences
(nk)
such that
an
biorthogonal
increasing inf
i-
1,
I
Then there exists
XX.4
a
fundamental and total biorthogonal sequence (e ,e )
n
n
in X such that
iix ))
Proof :
one may assume that
Without loss of
n
generality
f or all
n.
sequence
Next
Pick
(m r)
a
permutation p(.)
of the indices
so
of the indices and
that
an
=
1
increasing
if
put
we
m
akr.
k, j
verify
where
are
defined
as
in Lemma 1
that such defined sequence
for n = m r. Using
*
(e n ,e n)
Lemma 1
has the desired
we
easily
properties.
XX.5
We shall
Proof of Theorem 1 :
of X
implies
assume
that there exist sequences
3~
and
that dim X
F1 C F2 C .M,
of
subspaces
of X
E1 c E2
x =
0.
*M*
In view of
11
Proposition
1
it is
(x) =
enough
~.
C ..,
such that dim
is dense in X and if f
then
=
Then the
of
E~ _
separability
subspaces of X
dim
F~ ;
i for
0 for all
to construct
a
biortho-
XX.6
inductive
hypothesis implies that
and
to
we
a
inf er that
S 3. We define
Thus
linear functional on x.
Remark 1 :
Using
in the
C
and since dim ]
H
H ~F.
to be any
n
3° Day’s technique (cf. [3])
case
the Borsuk antipodal mapping theorem
one
can
choose
(both
g
extension of
prezerving
norm
which bases
in the
case
on
of
*
complex scalars)
real and of
for
s
Now the
1,2,...
=
and x3s
x-
3s
inspection
so
of the
that
proof
separable Banach space for every e&#x3E; 0
and bounded biorthogonal sequence (e
such
n
n
in every
en)
for all
ced
by
n.
1+
However,
c.
1 :
Corollary
norm
We do not know whether for every
III ~ 111
In every
was
separable
,e*)
with
n
by
n
and
sup
lle*ll
n
C.
there exists
yields
a
that lie n nil lien
(1+V2)2 +
n
.
0 this bound
Bessaga
we
be repla-
can
have
equivalent
biorthogonal
an
fundamental and total
1.
n
biorthogonal
sequence in X such
thatile n II
co ..
n
Remark 2 :
A similar
argument
Theorem 1 allows to prove the
that
fundamental
a
Banach space X there exists
=
n
is any fundamental and total
for all
observed
such that there exists in X
(e n
sequence
as
c &#x3E;
of Theorem 1
to that which is used in the
following
proof
of
=
1
c
XX.7
let T :
Theorem 11 : Let X and Y be Banach spaces and
one
bounded linear
operator.
compact, then
*
quences (x n’ ,x n) in X
is not
If X is
separable, T(X)
X -- - Ybe one-to-
is dense in Y and T
there exists fundamental and total
biorthogonal
se-
*
and
(y’ ny)
n
in Y such that
and
for all
2.
n.
Constructions of uniformly bounded
orthonormal sequences.
We
(=
a
employ the following notation. If ~
negative normalized measure) on a sigma
non
.
for any
p-absolutely
and
L2(J1)
is
probability
a
field of subsets of
~
_
as
usually
a
/~
square summable scalar valued functions
denote
measure
x
the Banach spaces of those
and y
on
S.
x
respectively.
The
proof
Instead of
([9J,
Lemma
of Theorem 2 is similar to the
Proposition 1,
a
set S.
product
apply
the
following
of Theorem 1.
result due to
Olevskil
4).
Proposition 2 :
of
we
proof
Let
. , .
Let p be aprobability measure on a sigma field of subsets
be an infinite orthonormal (with respect to the inner
(xn)
n
&#x3E;) sequence of functions in
Then there exists
an
orthonormal sequence
Loo(03BC)
(e n)
such that 1 im inf
such that
n
oo.
XX.8
and
Proposition 2 can be obtained by a non essential modification of the proofs of Lemma 1 and Proposition 1. Actually 01evskil stated Proposition 2 f or the Lebesgue measure on [0,I].
The
of
proof
To prove Theorem 2 it is convenient to
Let
Lemma 2 :
gn
L4(p))
converges to
dimensional
no &#x3E; k
(g n)
and
a
subspace
be
a
use
normalized sequence in
the
following simple
L 2(p)
which
fact.
weaklyY (in
..
zero
Then for every finite
and let
of
say
function h in the
F, and for k&#x3E;0
orthogonal complement
there exist
index
an
of F such that
and
Let
Proof :
20132013201320132013
p = dim F.
Let
l
2
p
any orthonormal basis for F.
’
Pick e&#x3E;0so that
Since
(gI
) conver g es
Clearly
h
belongs
weakl y
to 0 in
L 2(g),
there exists
I
to the
orthogonal complement
of
an
index
n°
and
&#x3E; k
XX.9
We have
and
(i)
It follows from
Proof of Theorem 2 :
that there exists in E
creasing sequence of finite dimensional subspaces
00
U
dim F = p and
~
is dense in E.
F
p
00
subset of
Loo(p)
Loo(03BC)
is dense in E n
there exists in E
In view of
can
one
sequence
a
Proposition 2,
in
(h )
mal sequence
(Fp)
choose the sequence
in the
L°(p)
it
L’(p)
(gn)
is
P E
norm.
is
so
that the
p
a
in-
such that
separable
union U
the
to define
that for
s=
(ii) yields that
assumption of Lemma 2.
inductively
an
orthonor-
1,2, ..,
where I
We define
h1
as
F1
with
the functions
some
the conditions
I,et
any element of
us
consider
(7)
and
separately
~
Suppose
that for
have been defined to
(8) and
two
so
that
cases.
F
The condition
satisfying
enough
so
if
Clearly
F112
C:F2 c ..,.
an
satisfy
XX.10
complete the induction and the proof of the sufficiency
conditions (i) and (ii). The necessity is trivial.
This
A similar
Remark 1 :
Banach space X into
a
and T is not
sup
n
jjx j)
and
Remark 2 :
and
E
by
x2m(t)
F.G.
by
=
0
Arutunian
T(x),x n
denotes the inner
&#x3E;H
It is
2m-1
If E
in X such that
(x )
sequence
n
is
a
an
orthonormal
m=2
enough
where
and
nor
E
H, then
(xn)
biortho g onal
decomposition
E1
for x E X and for
H
admit
of
L 2 0, 1]
(x )
x2
can
sequen-
onto
uniformly bounded
and
to define
n
x.
&#x3E;
of
product
fundamental and total
such that neither
such that the functions
and
a
n n *
orthonormal bases.
2
T(X).
operator from
is separable
#
is defined
There exists
E1
bounded linear
orthonormal basis f or E. Moreover if X is
n
.,
E2 =
an
*
E X
n=1,2..., where . , .
be chosen so that (x
subspaces
is
one
Let E =
Hilbert space H.
a
(T(x ) )
*
and
to
one
a
then there exists
compact
oo
separable
argument gives
Let T : X -- -~H be
Theorem 2’ :
of the
is any orthonormal basis for
are
unbounded,
(m= 19 2y
(unpublished) we have
However
x 2m-1 0
as
was
for
0t2
observed earlier
XX.11
Corollary 2 : If E i s a linear subspace of a separable
where p is a non purely atomic probability measure and
complement
of E is finite
orthonormal baxis.
can
a
Moreover if E --,
be chosen from elements of E Q
Proof :
It
is
to show that
enough
space
the
L-(J.1)
orthogonal
uniformly bounded
in E then the basis
is
L~(~.) .
rE]
(ii) of Theorem 2 To check (i) first observe that the density
r
regarded
integer
as
L2(~)
of
subspace
a
p and for every
in
L2( )
implies
linearly independent
"
00
there exist
(i)
satisfies the conditions
of
that for every
f1,f9,...,f
2
1
.1
p+1
and
L Ð (~)
positive
in
such that the matrix
is invertible.
Let
for i =
1, 2, ... , p+1.
thogonal complement
existence of
to any basis of the or-
The above observation
applied
of E and any
element f of
y in
an
be the inverse matrix and let
i,k
L 00 (J.1)
non
such that
zero
y~f&#x3E; ~
1 and
[E] yields
the
~r, g &#x3E; ~ 0 for all g
orthogonal complement of E. The last condition means that y E
Hence there is no f/0 in [E] which is orthogonal to all y E
Lco(g)
is dense in [EJ- Hence (E] satisfies (i).
equivalently
in the
The"moreover" part of the Corollary follows from the observation
that if
[E]
satisfies
(ii)
than E also satisfies
An immediate consequence of
Corollary 3 :
(ii).
Corollary 2 is
Let f be an y unbounded function in
Then the ortho-
uniformly bounded orthonormal basis consisting of trigonometrical polynomials. This basis has no extension to any
uniformly bounded orthonormal basis for
gonal complement
Corollary
of f admits a
3 answers
a
question of Shapiro
[l4]
XX.12
Fat
3.
Definition :
space S.
to
a
A closed linear
a
spaces.
subspace
Loop) - L2(~)
I
fat with
:
or
Z of
regarded
as
measure
C(S)
a
Hausdorff
compact
a
on
is said to be fat with
respect
L2( )
subset of the Hilbert space
denote the natural
respect to p iff
operator
Borel
probability
bounded set.
totally
Let
rem
be
Let 03BC
C(S)
if the unit ball of Z
is not
Z is
of
subspaces
equivalently
injection.
A
It is clear that
the rest.riction of I
if E T I
to Z is not
p
satisfies the condition (ii)
(Z)
compact
a
of Theo-
2n
Our next result characterizes Banach spaces which admit fat isometric
embeddings
in terms of
C(S) spaces. Some of the equivalent conditions are stated
2-absolutely summing operators, i.e. such bounded linear ope-
into
rators which admit
C12’
measure (cf.
factorization
a
and
through
a
equivalent :
(a) there exists
such that
injection
follwing
conditions
uniformly bounded sequence (x n) of elements
subsequence of (x n) is a weak Cauchy sequence,
no
there exists
a
2-absolutely summing operator from
(d)
there exists
a
2-absolutely summing
y
non
X onto
1~
y
compact operator from
X
12,
y
for every for
there exists
is fat with
Proof :
to
of X
11,
(c)
subspace isomorphic
a
some
are
a
X contains
(e)
for
p
e
(b)
into
I
[8]).
For every Banach space X the
Proposition 3 :
natural
some
a
probability
Let T ba
embedding j
of X into
Borel measure 03BC
on
a
C(S)
space
j(X)
S such that
to p.
respect
(a) ~ (b). This is
(b) ~ (c).
isometric
profound
a
a
recent result of Rosenthal
bounded linear
operator from
11
[13].
onto
12
XX.13
(cf. ~2~
for the existence of such
~7] (cf.
Grothendieck
T admits
extension to
an
a
T is
Then
by
a
result of
2-absolutely summing. Hence, by
2-absolutely summing operator from
I)
(c) = (d).
Obvious.
(d) ~ (e).
Let T :
operator
and let S be
[11],
and Pietsch
exists
a
Borel
image
j(X)
(e) =~ (a).
(x n)
A :
L 2(p) __ .12.
of j(X)
is
under
(x n) does
Cauchy
(e)
with
n
uniformly
a
1 for
Cauchy sequences
strong Cauchy sequences.
3
was
subset
bounded
n/m (n,m=1, 2,...)
not contain weak
Proposition
the
respect to .
2
m
some
because
recently independently
by Weis [16J.
discovered
Our last result is related to
result of Sidon
Corollary 4 :
tends
a
that there exists
sequences into
our
is not
Ip
A l 03BC j for
is not compact,
totally bounded
Since T
subspace of C(S)
fat
a
A similar result to
Let
result of Persson
a
S such that T=
on
2
in X such that
takes weak
compact 2-absolutely summing
non
2
It follows from
Thus the sequence
of S.
a
for every isometric embedding j : X -- -~ C( S) there
operator
Thus
sequence
be
compact Housdorff space. By
a
of the unit ball
of
-- ~12
X
probability measure p
bounded linear
-vjea.’,,-.1y
to
infinite
of
a
[15].
I,et p
( gn)
Gaposhkin’s [6] generalization
be
a
zero
be
a
probability
uniformly
in
L2(p)
measure
on
a
sigme
bounded sequence in
Loo(J.1)
and lim sup
0.
field of subsets
such that
subsequence
and c &#x3E; 0 such that
nk
for every finite sequence of scalars
(gn)
Then there exists
n
an
X
12.
onto
I f.1
F8])
also
operator).
an
c ,c ,...,c (p~l,2,...).
XX.14
Without loss of
Proof :
Then
(g)
weakly
does not have
conver g es in
converges to
zero.
generality
Cauchy (in
L2(p)
Thus
g~ ~
regarded
L ( ~) ---L 2(p)
inj ect ion
L 2(p))
but
zero
as
that
a
because
subsequences
subse q uence
no
of
(g )
(g )
strongly
L’(p)
sequence of elements of
sequences because the natural
takes weak
in
stron g Cauchy sequences
assume
may
Lm(p))
does not contain weak
into
to
we
L 2(p).
Cauchy
sequences in
Since su p
complete
n
the 1 -,,,roof
it is
for the real
enough
case
to
and Dor
apply Rosenthal’s
[5]
criterion
(cf. Rosenthal
[13]
complex case).
for the
***
*
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[2]
S.
[3]
M.M.
[4]
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BANACH und S. MAZUR :
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DOR :
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GAPOSHKIN :
spanning
On the existence of fundamental
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complex
l1
J.
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LINDENSTRAUSS and A.
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Usp.
Mat.
des produits tenSoc. Matem. Sao Paulo
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Absolutely summing operators
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M.
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[10]
A.
PE0142cZYNSKI :
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