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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 }.