SEPARATED SEQUENCES IN NONREFLEXIVE BANACH SPACES

Transkrypt

PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 129, Number 1, Pages 155–163
S 0002-9939(00)05495-2
Article electronically published on June 21, 2000
SEPARATED SEQUENCES
IN NONREFLEXIVE BANACH SPACES
ANDRZEJ KRYCZKA AND STANISLAW PRUS
(Communicated by Dale Alspach)
Abstract. We prove that there is c > 1 such that the unit ball of any nonreflexive Banach space contains a c-separated
√sequence. The supremum of these
constants c is estimated from below by 5 4 and from above approximately
by 1.71. Given any p > 1, we also construct a nonreflexive space so that if
the convex hull of a sequence is sufficiently close to the unit sphere, then its
separation constant does not exceed 21/p .
1. Introduction
Sequences of separated elements appear quite often in the literature on Banach
spaces. They are strongly related to the problem of packing balls (see [15]), average
distances (see [2]) and infinite dimensional convexity (see [9]). The last notion in
turn plays an important role in metric fixed point theory (see [8] and [1]).
A sequence is said to be r-separated if distances between its elements are bounded
from below by r. One of the basic general results on separated sequences is due
to Elton and Odell [7]. They proved that the unit ball of an arbitrary infinite
dimensional Banach space contains an r-separated sequence for some r > 1. In this
paper we show that there exists a constant c > 1 such that in the unit ball of any
nonreflexive Banach space one can find a c-separated sequence. On the other hand
for each 1 < p < ∞ we construct a nonreflexive Banach space Yp whose unit ball
does not contain a sequence with the separation constant greater than 21/p such
that its convex hull is arbitrarily close to the unit sphere. These results provide
answers to two of the questions posed in [2]. Let us however mention that our
notation does not coincide with that of [2].
2. Preliminaries
Let X be a Banach space. By the separation constant of a sequence (xn ) in X
we mean the number
sep(xn ) = inf{kxm − xn k : m 6= n}.
Received by the editors October 29, 1998 and, in revised form, March 14, 1999.
1991 Mathematics Subject Classification. Primary 46B20.
Key words and phrases. Nonreflexive spaces, separation measure of noncompactness, James’
space.
c
2000
American Mathematical Society
155
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
156
A. KRYCZKA AND S. PRUS
Having a bounded set A ⊂ X, we put
β(A) = sup{sep(xn ) : {xn } ⊂ A}.
This quantity is called the separation measure of noncompactness of A (see [1]).
By BX we denote the closed unit ball of the space X. Kottman [14] introduced
the packing constant P (ℵ0 , X) as the supremum of all numbers r > 0 such that
there exists a sequence of pairwise disjoint balls of radius r contained in BX . We
have
P (ℵ0 , X) =
β(BX )
2 + β(BX )
(see [14]).
Let A be a subset of X. The closed convex hull of A will be denoted by co A.
Assuming that the space X is infinite dimensional and 0 ≤ < β(BX ), we set
∆() = inf{1 − inf{kxk : x ∈ co{xn }}}
where the first infimum is taken over all sequences (xn ) in BX with sep(xn ) ≥
. The function ∆ is called the modulus of noncompact convexity of X. It is
nonnegative and ∆(0) = 0. ∆ can be treated as an infinite dimensional counterpart
of the classical modulus of convexity (see [5]). We will also consider the following
coefficient:
S0 (X) = sup{ ∈ [0, β(BX )) : ∆() = 0}.
We have the formula β(Blp ) = 21/p for 1 ≤ p < ∞ (see [4] or [15], p. 91). This
coefficient may be therefore arbitrarily close to 1. On the other hand β(Bc0 ) = 2
and moreover S0 (l1 ) = S0 (c0 ) = 2. It follows that β(BX ) = S0 (X) = 2 for any
Banach space X containing an isomorphic copy of l1 or c0 (see [13]).
3. Results
Theorem 1. There exists a constant c > 1 such that if X is a nonreflexive Banach
space, then β(BX ) ≥ c.
Proof. Let X be a nonreflexive Banach space. By a result of James [12] for every
θ ∈ (0, 1) there are two sequences: (xn ) in BX and (x∗n ) in BX ∗ such that x∗k (xi ) = θ
if k ≤ i and x∗k (xi ) = 0 if i < k.
Let us put 11 = 1, 21 = −1, ji = (−1)i if 1 ≤ j < i or j = 2i and ji = (−1)i+1
if i ≤ j < 2i where i = 2, 3, 4. Passing to a subsequence, we can assume that there
exist positive constants a1 , a2 , a3 , a4 for which
X
2i j
x
θai ≤ i nj ≤ ai
j=1
SEPARATED SEQUENCES
157
for any n1 < n2 < · · · < n2i where i = 1, 2, 3, 4 (see [3]). We consider the following
elements:
x0n = xn ,
1
x1n = (x2n−1 − x2n ),
a1
1
2
xn = (x1 − x3n−1 − x3n + x3n+1 ),
a2
1
x3n = (−x1 − x2 + x4n−1 + x4n + x4n+1 − x4n+2 ),
a3
1
x4n = (x1 + x2 + x3 − x5n−1 − x5n − x5n+1 − x5n+2 + x5n+3 )
a4
for n ∈ N. They are contained in BX . Moreover, sep(x0n ) ≥ θa1 and sep(xin ) ≥
if i = 1, 2, 3. We also have the estimate
kx4m − x4n k ≥ x∗5m+3 (x4m − x4n ) =
θai+1
ai
4θ
a4
whenever m < n. It follows that
√
θa2 θa3 θa4 4θ
5
,
,
,
β(BX ) ≥ max θa1 ,
≥ θ 4.
a1 a2 a3 a4
√
Since θ ∈ (0, 1) is arbitrary, we obtain the inequality β(BX ) ≥ 5 4.
Corollary 1. There exists a constant C >
Banach space, then P (ℵ0 , X) ≥ C.
1
3
such that if X is a nonreflexive
We do not know what is the greatest possible value of the constant c in Theorem
1. From a remark made by Kottman [14] it follows that if X is James’ space (see
[11]), then β(BX ) < 2. We will calculate the exact value of this coefficient for a
generalized James’ space.
Let us first establish some notation. By an interval of natural numbers we mean
a set of the form [m, n] = {k ∈ N : m ≤ k ≤ n} where m ≤ n are natural numbers.
Having such intervals I1 , I2 , we write I1 ≤ I2 if max I1 ≤ min I2 . Let now I = [m, n]
and x = (x(k)) be a sequence of real numbers. We put
δ(I, x) = x(n) − x(m),
δ+ (I, x) =
1
(|δ(I, x)| + δ(I, x)),
2
1
(|δ(I, x)| − δ(I, x)).
2
We fix 1 < p < ∞. Following James’ construction [11], we introduce the space Jp
of all real sequences x = (x(k)) converging to 0 and such that
1/p
X
n
|δ(Ik , x)|p
: I1 ≤ I2 ≤ · · · ≤ In
kxk = sup
δ− (I, x) =
k=1
is finite. This formula defines a norm on Jp . It is easy to see that the vectors
en = (0, . . . , 0, 1, 0, . . . ), where 1 occupies nth place, form a basis of Jp . The basis
(en ) is not boundedly complete (see [10]) and consequently Jp is not reflexive.
Having an arbitrary sequence x = (x(k)) of real numbers and n ∈ N, we set
Pn x = (x(1), . . . , x(n), 0, 0, . . . )
158
and Rn x = x − Pn x. Pn restricted to Jp is a norm-one projection. Moreover,
limn→∞ kRn xk = 0 for every x ∈ Jp . In the sequel we will identify a number α ∈ R
with a constant sequence (α, α, . . . ).
Lemma 1. Let (xn ) be a bounded sequence in Jp . For every γ > 0 there exist a
subsequence (xnk ), an increasing sequence (mk ) of natural numbers and a constant
α ∈ R such that
kxnk − (Pm1 xn1 + Pm2k Rm1 α + Pm2k+1 Rm2k xnk )k ≤ γ
for every k ∈ N.
Proof. Since the sequence (xn ) is bounded, passing to a subsequence, we can assume
that for each i ∈ N there exists x(i) = limn→∞ xn (i). It is easy to see that (x(n))
is a Cauchy sequence, so it has a limit α. Clearly x − α ∈ Jp .
Therefore, having γ > 0, we can find m1 such that kRm1 (x− α)k ≤ 16 γ. Next, we
put m2 = m1 + 1 and choose n1 so that kPm2 (xn1 − x)k ≤ 16 γ. There is m3 > m2
for which kRm3 xn1 k ≤ 16 γ.
Proceeding in this way, we obtain sequences (mk ), (nk ) such that
kPm2k (xnk − x)k ≤
1
γ
6
and kRm2k+1 xnk k ≤
1
γ
6
for every k ∈ N. Then
kxnk − Pm1 xn1 −Pm2k Rm1 α − Pm2k+1 Rm2k xnk k
≤ kPm1 (xnk − xn1 )k + kPm2k Rm1 (xnk − α)k + kRm2k+1 xnk k
4
≤ kPm1 (xn1 − x)k + kPm2k Rm1 (x − α)k + γ ≤ γ.
6
Theorem 2. Let 1 < p < ∞. Then β(BJp ) = (1 + 2p−1 )1/p .
Proof. Clearly k2−1/p en k = 1 and
k2−1/p em − 2−1/p en k = (1 + 2p−1 )1/p
for any m 6= n. Consequently β(BJp ) ≥ (1 + 2p−1 )1/p .
In order to prove the opposite inequality, we take a sequence (xn ) in BJp and
γ ∈ (0, 1). By Lemma 1 we obtain corresponding sequences (nk ), (mk ) and a
constant α. Let us put
u1 = Pm1 xn1 + Pm2 Rm1 α + Pm3 Rm2 xn1 ,
u2 = Pm1 xn1 + Pm4 Rm1 α + Pm5 Rm4 xn2
and w = u2 − u1 .
There is a sequence I1 ≤ I2 ≤ · · · ≤ In of intervals of natural numbers such that
((1 − γ)kwk)p ≤
n
X
|δ(Ik , w)|p .
k=1
Clearly w(i) = 0 if i ≤ m2 . Therefore we can assume that m2 ≤ min I1 . We can
also assume that there is an interval Is = [j1 , j2 ] for which j1 ≤ m3 < j2 .
SEPARATED SEQUENCES
159
If k < s, then δ(Ik , w) = −δ(Ik , u1 ) and if k > s, then δ(Ik , w) = δ(Ik , u2 ). It
follows that
X
|δ(Ik , w)|p + |u1 (j1 )|p ≤ ku1 kp ≤ (1 + γ)p
k<s
and
|α − u2 (j2 )|p +
X
|δ(Ik , w)|p ≤ (1 + γ)p .
k>s
Here we treat a sum over the empty set as 0. Consequently
((1 − γ)kwk)p ≤ 2(1 + γ)p − |u1 (j1 )|p + |u2 (j2 ) − (α − u1 (j1 ))|p − |α − u2 (j2 )|p
≤ 2(1 + γ)p + (1 − 21−p )|u1 (j1 ) + u2 (j2 ) − α|p .
But
21−p |2u1 (j1 ) − α|p ≤ |u1 (j1 )|p + |α − u1 (j1 )|p ≤ ku1 kp ≤ (1 + γ)p
and similarly
21−p |2u2 (j2 ) − α|p ≤ (1 + γ)p .
Hence
|u1 (j1 ) + u2 (j2 ) − α| ≤ 21−1/p (1 + γ)
which gives us the inequality ((1 − γ)kwk)p ≤ (1 + γ)p (1 + 2p−1 ). On the other
hand kxn1 − xn2 k ≤ kwk + 2γ. Therefore
sep(xn ) ≤ (1 − γ)−1 (1 + γ)(1 + 2p−1 )1/p + 2γ.
Since γ > 0 is arbitrary, we obtain sep(xn ) ≤ (1 + 2p−1 )1/p .
Let us notice that the minimal value of (1 + 2p−1 )1/p is approximately equal to
1.71. It is an√upper bound for the constant c in Theorem 1. Our proof of Theorem
1 gives c = 5 4 which is about 1.32.
In [6] it was shown that if S0 (X) < 1, then the space X is reflexive. In contrast
to Theorem 1 we will prove that 1 is the greatest lower bound of the values of S0 (X)
for nonreflexive spaces X.
Having x ∈ Jp , we put
X
1/p
n
(δ+ (Ik , x))p
: I1 ≤ I2 ≤ · · · ≤ In ,
|x|+ = sup
k=1
|x|− = sup
X
n
1/p
(δ− (Ik , x))p
: I1 ≤ I2 ≤ · · · ≤ In
k=1
and
kxk± = (|x|p+ + |x|p− )1/p .
This formula gives an equivalent norm on Jp . Namely
kxk ≤ kxk± ≤ 21/p kxk
for every x ∈ Jp . The space Jp with the norm k · k± will be denoted by Yp .
Considering the sequence (2−1/p en ), we see that β(BYp ) ≥ (1 + 2p−1 )1/p .
Theorem 3. Let 1 < p < ∞. Then S0 (Yp ) = 21/p .
160
Pn
Proof. Let us consider the sequence of vectors yn = k=1 ek . Clearly kyn k± = 1
and kym − yn k± = 21/p for any m 6= n. It is also easy to see that kxk± = 1 for
every x ∈ co{yn }. This shows that S0 (Yp ) ≥ 21/p .
We will show the opposite inequality. To this end we take ∈ (21/p , β(BYp )) and
choose γ > 0 so that
1 −p p
(8 ( − 2))p .
3
Now let (xn ) be a sequence in BYp with sep(xn ) ≥ . From Lemma 1 we obtain
sequences (nk ), (mk ) and a constant α such that
γ
kxnk − uk k± ≤ p
p2
γ≤
for every k ∈ N, where uk = Pm1 xn1 + Pm2k Rm1 α + Pm2k+1 Rm2k xnk . We set
PN
u = Pm1 xn1 + Pm2 Rm1 α and zN = N1 k=1 uk where N ∈ N. Clearly it suffices to
consider the case when α ≥ 0.
We will show that
X
1 N
(1)
x
nk ≤ kuk± + 2γ
N
±
k=1
for some N ∈ N. It is easy to see that
p
X
N
p−1 X
1 N
22p−1
p
≤2
P
R
x
kP
R
x
k
≤
m
m
n
m
m
n
2k+1
2k
k
2k+1
2k
k
N
Np
N p−1
k=1
k=1
for every N ∈ N. Therefore we can find N such that
N
1 X
Pm2k+1 Rm2k xnk ≤ γ.
N
±
k=1
But one can easily check that
N
1 X
= kuk± .
(P
x
+
P
R
α)
m
n
m
m
1
1
1
2k
N
±
k=1
Hence
N
1 X
x
nk ≤ kzN k± + γ ≤ kuk± + 2γ.
N
k=1
±
Our next aim is to prove that
(2)
1
kukp± < 1 − (8−p (p − 2))p .
3
There are sequences T11 ≤ T21 ≤ · · · ≤ Tr1 and T12 ≤ T22 ≤ · · · ≤ Ts2 of intervals for
which
r
s
X
X
(δ+ (Tk1 , u))p , (1 − γ)|u|p− ≤
(δ− (Tk2 , u))p .
(1 − γ)|u|p+ ≤
k=1
k=1
≤ m2 and max Ts2 = m2 + 1. As in the
Since α ≥ 0, we can assume that
proof of Theorem 2 we put w = u2 − u1 and find sequences I11 ≤ I21 ≤ · · · ≤ In1 and
max Tr1
SEPARATED SEQUENCES
161
2
I12 ≤ I22 ≤ · · · ≤ Im
of intervals such that m2 ≤ min I1j , j = 1, 2,
(1 − γ)|w|p+ ≤
n
X
(1 − γ)|w|p− ≤
(δ+ (Ik1 , w))p ,
k=1
m
X
(δ− (Ik2 , w))p .
k=1
We can assume that there are s1 , s2 for which kj = min Isjj ≤ m3 < max Isjj = lj ,
j = 1, 2. Let us put K1 = [k1 , m3 + 1], K2 = [m3 + 1, l2 ]. We have
X
X
(δ+ (Ik1 , w))p + (δ+ (K1 , w))p +
(δ− (Ik2 , w))p
k<s1
X
=
k<s2
(δ− (Ik1 , u1 ))p
k<s1
Similarly
X
(δ+ (Ik2 , u1 ))p
k<s2
ku1 kp±
≤
X
+ (δ+ (K1 , w))p +
< 1 + γ.
X
(δ+ (Ik1 , w))p + (δ− (K2 , w))p +
k>s1
(δ− (Ik2 , w))p < 1 + γ.
k>s2
Hence
(1 − γ)kwkp± ≤ 2(1 + γ) + (δ+ (Is11 , w))p − (δ+ (K1 , w))p
+ (δ− (Is22 , w))p − (δ− (K2 , w))p .
But
kwkp± ≥
−
γ
p2p−1
p
> p − γ.
It follows that
1 p
( − 2) < (δ+ (Is11 , w))p − (δ+ (K1 , w))p + (δ− (Is22 , w))p − (δ− (K2 , w))p .
2
We have two cases.
I. 14 (p − 2) < (δ+ (Is11 , w))p − (δ+ (K1 , w))p .
By the mean value theorem
1
(δ+ (Is11 , w))p − (δ+ (K1 , w))p ≤ pkwkp−1
± (δ+ (Is1 , w) − δ+ (K1 , w))
1
≤ p4p−1 (|w(l1 ) − α| + w(l1 ) − α)
2
= p4p−1 (u2 (l1 ) − α).
Therefore 8−p (p − 2) < u2 (l1 ) − α, so we obtain
ku2 kp± ≥
r
X
(δ+ (Tk1 , u))p + (u2 (l1 ) − α)p +
k=1
s
X
(δ− (Tk2 , u))p
k=1
> (1 − γ)kukp± + (8−p (p − 2))p .
Consequently
kukp± < (1 − γ)−1 (1 + γ − (8−p (p − 2))p )
which gives (2).
162
II. 14 (p − 2) < (δ− (Is22 , w))p − (δ− (K2 , w))p .
Similarly as in the previous case we see that
(δ− (Is22 , w))p − (δ− (K2 , w))p
2
≤ pkwkp−1
± (δ− (Is2 , w) − δ− (K2 , w))
≤ p4p−1 (−u1 (k2 )).
Hence 8−p (p − 2) < −u1 (k2 ). But
r
s
X
X
(δ+ (Tk1 , u))p +
(δ− (Tk2 , u))p + (−u1 (k2 ))p ≤ ku1 kp± .
k=1
k=1
Therefore
(1 − γ)kukp± + (8−p (p − 2))p < 1 + γ
which implies (2).
From (1) and (2) we see that
1/p
1 −p p
p
+ 2γ.
inf{kxk : x ∈ co{xn }} < 1 − (8 ( − 2))
3
Since γ > 0 can be arbitrarily small, we obtain
1/p
1
∆() ≥ 1 − 1 − (8−p (p − 2))p
3
where ∆ is the modulus of noncompact convexity of Yp . This shows that S0 (Yp ) ≤
21/p .
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MR 99e:47070
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294–299. MR 55:11004
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Department of Mathematics, M. Curie-Sklodowska University, 20-031 Lublin, Poland
E-mail address: [email protected]
Department of Mathematics, M. Curie-Sklodowska University, 20-031 Lublin, Poland
E-mail address: [email protected]

SEPARATED SEQUENCES IN NONREFLEXIVE BANACH SPACES

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