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ON SYMMETRIC BASIC SEQUENCES IN
LORENTZ SEQUENCE SPACES
BY

ZVI ALTSHULERt, P. G. CASAZZA AND BOR-LUH LIN

ABSTRACT

We examine the symmetric basic sequences in some classes of Banach spaces
with symmetric bases. We show that the Lorentz sequence space d(a,p) has a
unique symmetric basis and every infinite dimensional subspace of d(a,p) contains a subspace isomorphic to [P. The symmetric basic sequences in d(a,p) are
identified and a necessary and sufficient condition for a Lorents sequence space
with exactly two nonequivalent symmetric basic sequences in given.
conclude by exhibiting an example of a Lorentz sequence space having a subspace
with symmetric basis which is not isomorphic either to a Lorentz sequence
·
space or to an [P-space.

we

Introduction
A basis {xn} of a Banach space X is called symmetric if every permutation
{xa(n)} of {xn} is a basis of X, equivalent to the basis {xn}. In this paper we consider
the problem of constructing symmetric basic sequences in some Banach spaces
'

'

with symmetric bases.
Much of our work is done with the Lorentz sequence spaces d(a,p). Let

+ oo. For any a = (a 1 , a2 , • • ·) E c0 \1 1 , a 1 ~ a2 ~ • • • ~ 0, let d(a, p)
= {x = (a 1 ,a 2 ,···)ec 0 : SUPaen l:f- 1 } aa(i)IPan < + oo} where n is the set of all
permutations of the natural numbers. Then d( a, p) with the norm JJ x II
1~p<

= (supa en ~noo

1J

aa(n) JPan) 11 P for xEd (a, p) is a Banach space and the sequence

of unit vectors {xn} is a symmetric basis of d(a, p) [2,4]. For p = 1, these
spaces have been studied by W. L. C. Sargent [10], D. J. H. Garling [2], W.
Ruckle [9], and J. R. Calder and J. B. Hill [1]. For 1 < p <

+ oo,

Garling [4]

t This is part of the first author's Ph. D. thesis, prepared at the Hebrew University of
Jerusalem under the supervision of Dr. L. Tzafriri.
Received July 13, 1972 and in revised form January 5, 1973

140

141

SYMMETRIC BASIC SEQUENCES

Vol. 15, 1973 ·

showed that d( a, p) is a reflexive Banach space which, in general, is distinct from
the [P-spaces. See [1] for further references on other work on d(a, p).
Another class . of Banach spaces with symmetric basis is that of the Orlicz
sequence spaces. J. Lindenstrauss and · L. Tzafriri [ 6, 7] have shown that every
orlicz sequence space has a subspace isomorphic to some

[P.

They have also shown

that there are Orlicz sequence spaces which have at least two nonequivalent
symmetric bases. We show that d( a, p) has a unique symmetric basis for all a and p
and that every infinite.dimensional subspace X of d(a, p) has a subspace isomorphic
to [P which can be chosen to be complemented in X if X has a symmetric basis.
The Lorentz sequence spaces which have exactly t\vo nonequivalent symmetric
basic sequences are characterized. Finally, an example of a Lorentz sequence
space having a subspace with symmetric basis which is isomorphic neither to lP
nor to any Loret?-tz sequence space is given.
· .We introduce a new type of block basic sequence of a symmetric basis Vihich
has the property that it always has a symmetric subsequence. In the spaces d(a,p),
these are the only symmetric block basic sequences of the unit vector basis {xn}
of d(a,p) which are not equivalent to the unit vector basis of

[P.

The notations and terminology in this paper are essentially those of I. Singer

[12]. A sequence {xn} of a Banach space X is ca11ed a basis of X if every x EX
has a unique expansion of the form x

= Lf' 1 anxn- Let 1 ~ p < + oo; a basis

{xn} of X is called p-Hilbertian if 'L;' 1 tt11 Xn converges in X for every {an} E lP.
A basis {xn} is q- Besselian, 1 ~ q < + oo, if 'L;> 1 anxn converges in X implies
that {an} E lq.
If {xn} is a basis of a Banach space X, a sequence {y11 }in X is said to be a block
basic sequence of {x if there is an increasing sequence of natural numbers {Pn}
such that y n = 'Lfn ;~ + 1aixi for n = 1, 2, · · · . A block basic sequence {Yn} is said
to be bounded if 0 <; inf1 <n< +co II Yn ~ sup 1 <n< +co Yn j < + oo. We will denote
by [{Yn}J the closed linear span of the sequence {Yn}· If {x and {Yn} are bases
of X andY, respectively, we saythat {xn} ~ominates {Yn}, and write {xn} > {Y 11 }, if
11

}

11 }

:L:

1 tt11 Xn

00

converges in X implies that Ln

1

tt11 Yn converges in Y. The basis {X11 }

is equivalent to the bas'is {y11 }, and we write {xn} rv {Yn}, if {xn} > {Yn} and {Yn}
> {xn}. It is clear that a basis {xn} is equivalent to ·the unit vector basis . of [P. if
and only if {xn} is p- Hilbert ian and p- Bessel ian.
If {X 11 } and {Yn} are symmetric bases, it is easy to show that {xn}
.

.

rv

{Yn} if and

onl:y if for any sequence of scalars a 1 .~ a2 ~ • • • ~ 0, 'L~ 1 anxn converges in X if
and only · if · Ln~ 1 ttnYn converges in Y. We also note that if

142

Z. ALTSHULER ET AL.

Israel J. Math .,

P~'~+1

L

Yn =

CX·X·
z. z.

i = Pn+ 1

for n = 1, 2, ···,is a block basic sequence of a symmetric basis {xn}, and for each

n, an is a permutation of {Pn + 1,pn + 2, ···,Pn+ 1 }, then {Yn) ~ {zn} where
zn = l:f~;~+ 1 j cxa(i) jxi, n = 1, 2, · · ·. Therefore, when working with block basic
sequences {Yn} of a symmetric basis {xn} we will always assume that cxPn+t
~ex
+z ~
Pn
-

··· >
cx Pr. + 1
-

~

-

0 for n = 1' 2 ' · ··.

Let {xn} be a symmetric basis in a Banach space X. Define
n

Ill x Ill =

sup
aen

sup
1Pd:::1

L

f3 ifi(x )xa(i) ,

i= 1

XEX,

H

1<n<+oo

where {.fn} is the sequence of biorthogonal functionals of {xn} in X*. Then the
symmetric norm

Ill x Ill, x EX, is an equivalent norm on X. Throughout this paper,

we shall assume that every Banach space with symmetric basis is equipped with
the symmetric norm.

1. Preliminaries
In this section we state some simple and well-known facts on symmetric basic
sequences in Banach spaces.
PROPOSITION 1. Every sym1netric basic sequence in a Banach space is either

weakly convergent to zero or is equivalent to the unit vector basis of 11 •
It is known that in the lP spaces, 1 ~ p < oo, all symmetric bases are equivalent
[12, p. 573J.As a consequence of Proposition 1, we have
CoROLLARY 1. In the spaces X= c0 or lP, 1 ~ p <

+ oo,

all symmetric basic

sequences are equivalent.
PROPOSITION 2. Let X be a Banach space with a symmetric basis {xn}· If

every bounded block basic sequence of {xn} is symmetric, then {xn} is equivalent
to the natural basis of c0 or lP for some p, 1 ~ p < + oo.
PROOF. Let {y11 } be a bounded . block basic sequence of {X 11 }. Since {y11 } is
symmetric, {Yn} ~ {y 211 }. Choose a subsequence {xni} of {xn} such that

Yzi if n = 2i,
zn =

Xni

if n

= 2i + 1,

i

= 1 2 ...
' '

i = 1 2 ...

' '

'
'

is a bounded block basic sequence of {x11}. Then, since {zn} is symmetric,

SYMMETRIC BASIC SEQUENCES

Vol. 15, 1973

143

{xni} ~ {zn} ~ {Y2n} ~ {Yn}. Hence by a result of M. Zippin [13], {x11 } is
Q.E.D.
equivalent to the natural basis of c0 or lP, 1 ~ p < + oo.

{xn}

r>J

PROPOSITION

Yn :::::

3. Let {xn} be a symmetric basis of a Banach space X. If

Lf~ ~~ + 1 aixb

5up 1 :::n< + oo

PROOF.

n = 1,2, · · · is a bounded block basic sequence of {x,J and

(Pn+ 1 - Pn) <

+ oo,

then {Yn} is equivalent to {x11 }.

We may assume that

=

xn

yn

= 1 for n = 1' 2' · · · • Suppose

~:== 1 anxn converges in X. Since {xn} is symmetric and

< Yn jJ ~ 1,
~:== 1 janapn+ilxPn+iconvergesinXforeach i =1,2,···,MwhereM = sup 1 <n<+o:>
(Pn+ 1 - Pn). Since
M

L

n=l i=l

L:

Ianapn+i lxp __ +i

j aPn+d

'

anYn converges in X.
Conversely, if "L:_ 1 anYn converges in X, note that for each n = 1, 2, · · ·, there

the series

1

I I

exists kn such that Pn + 1 ~ kn ~ Pn + 1 and akn ~ 1 / M > 0. Hence,
converges in X and so "L:=l anxn converges in X.

L:_

1

anaknxkn
Q.E.D.

4. Let {xn} be a symmetric basis in a Banach space X. If

PROPOSITION

Yn = 'Lf";~ + 1 etixi, n = 1, 2, · · ·, is a bounded block basic sequence of {xn} such
that inf1 :;;;n<+oo supPn+l<i:::Pn+t jail> 0, then {Yn} dominates {xn}. However, in
general, {Yn} is not equivalent to {xn}.
PROOF.

Since {xn} is symmetric, we may assume that there exist e > 0 and

0 ~ kn ~ Pn+ 1 - Pn such that aPn+kn ~ e for n = 1, 2, · · ·. Suppose

L:':: 1 anYn

converges in X. Then
n

n

L aixpi+ki

L aiapi+kixPi+ki

i==l

n

i=1

Thus L:':: 1 anxpn+kn converges in X, so that L:_ 1 a11 X 11 converges in X.
Now, let {x11 } be any nonshrinking symmetric basis which is not equivalent to
the unit vector basis {en} of 11 (e.g., the unit vector basis of the spaced [12, p. 361 ]).
Since {xn} is nonshrinking, there is a bounded block basic sequence

= I:f"~~+l aixi for n =

Z 11

1,2, ···,which is of type l+ [12, p. 369]. Hence {z11 } ~{en}

and is a symmetric basic sequence. Let Yn = XP 2 n + z 211 for n = 1, 2, ···. Then {Yn}
is a bounded block basic sequence of {xn} and it is clear that {Yn} satisfies the
hypothesis of Proposition 4. However, {yn} "' {z 2u} "' {en}, so {yu} is not equivalent
Q.E.D
to {X 11 }.

Israeli. ·Math-.,

Z. ALTSHULER ET AL.

144

2. The Lorentz sequence spaces d(a,p)
Let 1 ~ p <

+ oo. For any sequence a= (a 1 , a2 , ···) E c0 \1 , a 1 ~ a 2 ~ ••• ~ 0
in the Lorentz sequence space d(a,p), the unit vector basis {x,J is symmetric [2,4]

For any x = (cx 1 ,cx2 ,···)Ed(a,p), let x = (& 1 ,& 2 ,···) where {&n} is an enumeration
1

)

I I I

of the nonzero elements of {cxn} such that &1 ~ &2 j ~ · · ·. Then it can be proved
11

= (L:= 1 j &n jPan) P. ·In the rest of the paper, we shall assume that
a= (a 1 ,a 2 ,···)Ec 0 \1 1 , 1 ~ a 1 ~ a2 ~ ••• ~ 0 and 1 :S p < + oo. It is clear that

that

j

x

the norm in d(a, p) is a symmetric norm.
PROPOSITION

5. If {xn} is the unit vector basis of d(a,p) then all bounded

block basic sequences of {xn} are p-Hilbertian. In particular, all synzmetric
basic sequences in d( a, p) are p-H ilbertian.
PROOF.

Let Yn

{xn} such that

=

'Lf~;~+ 1

cxixi, n = 1,2, ···, be a block basic sequence of
Yn = 1, n = 1,2,···. For any nonnegative scalars b 1 ,b 2 ,···,bn
_,

-

n

n

Pi+ 1

Z: biYi =

L

L:

.

bfj cxi jPai,j

1/p -

i= 1

where {ai,j}i =Pi+ 1

Pi+ 1 ,i = 1 , 2 ... ,n is an enumeration of {a 1 , a2, · · ·, ak} for some k.
For each i = 1,2, ···, n, L)i+;n+tl cxi Pai,i ~ Yi j P= 1. Hence I 'L?= 1 biYi I
~ (2:~= 1 bf) 11 P and {Yn} is p-Hilbertian..
Q.E.D.
...

1. Let {xn} be the unit vector basis ofd(a,p). lfyn= l:fn;!+ 1 aixi,
n = 1, 2, · · ·, is a bounded block basic sequence of {xn} such that limn-+--co cxn = 0,
then there exists a subsequence of {Yn} which is equivalent to the unit vector
basis of [P.
LEMMA

PROOF.

Since {xn} is a symmetric basis, and limn-+-oocxn = 0, by switching to a

subsequence if necessary, we may assume that cxPz + 1 ~ ap 1 + 2 ~ • • • ~ cxn ~ · · · ~ 0,
Pn+ 2 - Pn+l ~ Pn+ 1 - Pn _and . Yn = 1 for n = 1,2, ···. We shall construct a
block basic sequence zn = L[~~~ + 1 f3ixi for n = 1, 2, · · · of {xn} with the following
two properties:

I

-

(1)

\1

zn II = 1 and 'Lf:_+q~+ 1 f3f ai ~ 1 for n = 1, 2, · · ·;

(2) {z 11 } is equivalent ~o a subsequence of {y11 }.
We may assume that p 1

I

y 1 ., Then z 1 satisfies (1). Assume now

f3ixi with the required properties. Since
there exists ·_a positive _integer k such _that "Lf +kq "ai < 1 /2~.

we have constructed

a = {an} E c0 ,

= 1 and let z 1 ,=

,,

Z 11 - 1

= '22iqnw·-t + 1

Since {ct11 } is decreasing to zero, choose h such that Ph+ 1

-

Ph > k

+ 'qn

and

145

SYMMETRIC BASIC SEQUENCES

Vol. 15, 1973

rtf<:: 1 f2 2 k for all i such that

= aph+i'

qn+ 1 =Ph+ 1 -

Ph+ qn,

qn; and z11 = Liqn;n +1 f3ixi. Notice that the
coefficients of zn are the same _as the coefficients of yh; hence, zn = 1. Now
Pqn+i

i

= 1,2, ···,

1 ~ i ~Ph+ 1 • Define

Ph+

1

qn+1-

Ph+ 1 - Ph+q
f3 fa.
~
~

L

i =qn+ 1

i

a;h-qn+i ai

=qn+ 1
Ph+ 1 - Ph+qn

Ph+ 1- Ph
L

a;h+i ai

L
i =qn+ 1

i=1

a;h-qn+i(ai-q_-ai)

qn+k

"L

1

a;h-qn+i(ai-qn- aJ

i=qn+1
Ph+ 1- Ph+qn

L:

a;h-qn+i (ai-qn-aJ·

i =qn +k+ 1

But
qn+k

qn+k

L

~

a;h-qn+i (ai-qn- ai)

"L

i =qn+ 1

i =qn + 1

1

1
k

+ ... +

1
k

1
--·
22'
}

k tirri"es

and
Ph+1- Ph+qn

Ph+ 1-ph+qn

L

a~h-qn+i (ai-qn-

~

ai)

i =qn+k+ 1

"L

(ai-qn- aJ

i =qn+k+ 1
qn

qn

-

"L

ak+i -

L

i=1

i=1

a Ph+ 1 - Ph+i

Hence 'L["::+q~.+ 1 f3fai~ ~.By induction, we construct {zn} satisfying (1). Since {zn}
is merely a translation of a subsequence of the block basic sequence {Yn}, it follows
that {zn} is equivalent to a subsequence of {Yn}.
Finally, we claim that {zn} is equivalent to the unit vector basis of ZP. By Proposition 5, {zn} is a p-Hilbertian basic sequence. For any nonnegative scalars

b1, b2, .. ·, bn, we have
1 1/p n
-

2

"L bPz.

i=1

n

1/p

~

"L bf

L

~

i=1

j

=qi+ 1

n

lfp

qi + 1

f3~a.
J

J

-~

L~l

b.z. •
~ '

· ······················

Z. ALTSHULER ET AL.

146

- -

Israel J. Math.,

Hence {zn} is a p- Besselian basic sequence. It follows that there is a subsequence
of {Yn} equivalent to the unit vector basis of the space [P.

Q.E.b



CoROLLARY 2. Let {xn} be the unit vector basis of the Banach space d(a, P).

For every bounded block basic sequence {y 11 } of {x11 }, either there is a sub . .
sequence of {y 11 } which is equivalent to the unit vector basis of [P or {y 11 } dominates
{xn}. In particular, every symmetric basic sequence in d(a,p) dominates {xn}.
CoROLLARY 3. Let {xn} be the unit vector basis of d(a,p). If {Yn} is a bounded

block basic sequence of {xn}, then there is a block basic sequence of {Yn} which
is equivalent to the unit vector basis of [P.
PROOF. Let Yn

= Lf~+q~+l cxixi, n = 1,2,-··.

Notice that infn II L{n;~+l cxixi Jl > 0

implies that L~ 1 cxixi does not converge in d(a,p). Since {xn} is a boundedly
complete basis (see, e.q., [1]), it follows that supk:::n jl 2:7=kYi = + oo. Therefore
there exists a sequence p 1 < p2 < ···of integers such that supnll Lf~~~+ 1 Yi

II=+ oo.

Let
Pn+1

Pn+1

Zn =

L

i = Pn+ 1

2:

Yi
i

=Pn+ 1

Yi .

Considering {zn} as a bounded block basic sequence of {xn}, it is easily seen that
{zn} satisfies the hypotheses of Lemma 1. Hence, there is a subsequence {zni} of

{zn} which is equivalent to the unit vector basis of fP.
REMARK 1.

Q.E.D.

If {Yn} is a symmetric block basic sequence then it is known

[e.g., 8] that there is a projection from [ {Yn}J onto [ {znj}].
Let {xn} be the unit vector basis of d(a,p). For any infinite-dimensional subspace
X of d(a, p), by a result of B. Bessaga and A. Pelczynski (see, e.g., [12, p. 442]),

X contains a bounded basic sequence {Yn} which is equivalent to a block basic
sequence {zn} of {xn}. By Corollary 3, the subspace [ {zn}] contains a subspace
which is isomorphic to lP. Thus X contains a subspace Y which is isomorphic to
[P.

In view of the previous remark, if X has a symmetric basis, then Y is

comple~

mented in X. Hence we obtain the following result.

Every infinite dimensional subspace X of d(a, p) contains a
subspace Y which is isomorphic to [P. If X has a symmetric basis then Y can be
chosen to be comple1nented in X.
THEOREM l.

REMARK 2.

In [7, Proposition 4], it is proved that d(a,p) has a complemented

subspace isomorphic to fP •.

________

............ .........

SYMMETRIC BASIC SEQUENCES

Vol. 15, 1973

147

3. Uniqueness of symmetric basis in d(a, p)
Let {xn} be a symmetric basis of a Banach space X. If {xn} is not equivalent to

zv,

the unit vector basis of c0 or

1~p<

+ oo, then we know that there are

bounded block basic sequences of {xu} which are not symmetric. On the other
hand, if {Yn} is a symmetric basic sequence in X, then either {Yn} is equivalent to
1

the unit vector basis of the space 1 or {Yn} is weakly convergent to zero. In the
latter case, {Yn} is equivalent to a bounded block basic sequence of {xn}. In this
section, we shall construct some special symmetric basic sequences in X and in
the d(a,p) spaces we will determine all the bounded block basic sequences of the
unit vector basis which are symmetric. A new type of block basic sequence is
introduced which seems to play an important role in determining symmetric
basic sequences in Banach spaces with symmetric bases.
PROPOSITION

{a11 }
p1

E 11 ,

6. Let {xn} be a symmetric basis in a Banach space X. For any

a 1 =I= 0, and any monotone increasing sequence of natural numbers,

< p 2 < ··· < Pn < ···, let Yn = 'Lf";!+ 1 ai-p .. xifor n = 1,2, ···. Then {Yn} is a

basic sequence in X which is equivalent to the basis {xn}·
PROOF.

By Proposition 4, it is clear that the basic sequence {Yn} dominates the

basis {x11}. Conversely, let

I:;:'= 1 anX EX. Then
11

n

oo

L rtiYi ~

L

i=1

J

i=1

p .• + 1

ad

L aixi ,

n

i=1

= 1, 2, · · ·.

Thus I:~= 1 aiyi converges in X. Hence {y n} is equivalent to {xn}.

Q.E.D.

Let {xu} be a symmetric basis of a Banach space X. For any
element 'L;:'_ 1 r:xnxn EX, a 1 =I= 0, and for any natural numbers p 1 < p 2 < .. ·, let
THEOREM

2.

Yn ='Lt.~;~+ 1 ai- P"x, for n = 1, 2, ·· ·. Then there is a subsequence of {Yn} which
is a symmetric basic sequence in X.
If sup 1 ~u< +oo (Pn+ 1 - Pn) < + oo, then {Yn} is equivalent to the basis
{xn} and we are done. Assume that sup 1 ~n< + 00 (Pn+ 1 - Pn) = + oo. By switching
PROOF.

to a subsequence, if necessary, we may assume that Pn+ 2 - Pn+ 1 > Pn+ 1 - Pn for

n = 1,2, · · ·.
Let

{Ni}i= 1 , 2 , ...

Ni n N.J =
let ui =

0

be subsets of the natural numbers, N, such that N

for all i i=J' and

-

-

N·' = N' i = 1' 2 ' ···.For each i'

=·Uy:_ 1 Ni,

·}.
N·={i
1
l
'
]=1

2, ...•

I:]= 1 aixi,i· It is clear that {ui} is a symmetric basic sequence in d(a, p).

Let {ni} be an increasing sequence such that / I:~
i= 1,2,···. Let zi = 'L;~i 1 -p .. iaixi _j, i = 1,2,··· .. Then

p.,i +

.

1

_

Pn. + 1

~

aixi

j

< 1/2 i,

148

Z. ALTSHULER ET AL.
00

00

L

U·Z·~
~

-

00

~

L

i=1

i=1

Israel J. Math .,

j

<L:

aJ·X· J·
l.

= Pn l· + 1 - Pn • + 1



1/2 = 1.
1

i=1

l

By a theorem of B. Bessaga and A. Pelczynski (e.g. [12, p. 93]), {ui} "' {zi}. Now
it is clear that {zi} "' {Yni}. Hence, {ui} "' {Yni} and so {YnJ is symmetric.
3. For 1 ~ p <

REMARK

'
Q.E.D.

+ oo, the symmetric basic sequences in the Lorentz

sequence space d( a, p) constructed in Theorem 2 are not equivalent to the unit
vector basis of [P. Indeed, if 0 i= Ln~ 1 anxn E d( a' p) where {xn} is the unit vector
basis of d(a,p), we may assume that 1 ~ a 1 ~ a2 ~

~an~

~

0. Since
limi~oo Ln00 1 ai+nxn j = 0, for any B > 0, there exists a positive integer Ne such
that

II L:-1 aN c+nxn II p<

•••

···

Let a = {an}· Since lim n~oo an= 0, we have that

B Pf2.

limn~ co L~~1 aifn = 0. Choose n such that
1
-- """P
j +
x .(:'or 1· -- 1' 2 ' · · · '
~i = p j + 1 v.,i- p j i 1'

L7N1aijn < sP /2. Then if yi

N

n
L YN,.+i P<
i =1

N..,
L al!£

~· ,

i=1



HJ

L

00

k=(i-1)n+ 1

ak

nN~

Hence

II L7=t YN . +i \\ < n Ps.

<n 2
11

But n P=

L7= 1 ei II

~

+·X·
l
l. P

i=l

BP

<
11

+n L aN
+n 2

where {en} is the unit vector

basis of fP. Thus {Yn} is not equivalent to {en}· Similarly, no subsequence of {Yn} is
equivalent to {en}.

DEFINITION.

Let {xn} be a symmetric basis of a Banach space X. For any

< Pz < ... < Pn < ···, let Yn= ~tn;~+ 1 ai-p ..xi
for n = 1, 2, .... Then {Yn} is a bounded block basic sequence of {xn} in X. We shall
call {Yn} a block of type I of {xn}.
L:

1 anxn

EX,

THEOREM

3.

a1

=/:. 0, and any

P1

Let {xn} be the unit vector basis of the Lorentz sequence space

d(a, p). For any bounded block basic sequence {Yn} of {xn}, {Yn} has a subsequence
equivalent either to the unit vector basis of fP or to a block basic sequence of type
'

I of {xn} .

= Lfn;~+ 1 aixi for n = 1, 2, ·· ·. We may assume that I Yn I = l
and aPn+1 ~ apn+2 ~ ... ~ aPn+1 > 0 for n = 1,2,···. If SUPl:::n<+ooCPn+1- Pn) <
+ oo, then {Yn} is equivalent to {xn} and so is equivalent to a block of type I of
{x,J. Assume now that SUP1::: n <+ooCPn+1- Pn) = + 00.
. PROOF.

Let Yn

'


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