ANNALES
POLONICI MATHEMATICI
LXV.2 (1997)
Banach–Saks property in some Banach sequence spaces
by Yunan Cui (Harbin), Henryk Hudzik (Poznań)
and Ryszard Pluciennik (Zielona Góra)
Abstract. It is proved that for any Banach space X property (β) defined by Rolewicz
in [22] implies that both X and X ∗ have the Banach–Saks property. Moreover, in
Musielak–Orlicz sequence spaces, criteria for the Banach–Saks property, the near uniform
convexity, the uniform Kadec–Klee property and property (H) are given.
1. Introduction. Let N, R, and R+ stand for the set of natural numbers,
the set of reals and the set of nonnegative reals, respectively. Let (X, k·k) be
a real Banach space, and X ∗ be the dual space of X. By B(X) and S(X) we
denote the closed unit ball and the unit sphere of X, respectively. For any
subset A of X by conv(A) (conv(A)) we denote the convex hull (the closed
convex hull) of A. In [2], Clarkson has introduced the concept of uniform
convexity.
A norm k · k is called uniformly convex (written UC) if for each ε > 0
there is δ > 0 such that for x, y ∈ S(X) the inequality kx − yk > ε implies
1
(1)
2 (x + y) < 1 − δ.
A Banach space X is said to have the Banach–Saks property if every
bounded sequence (xn ) in X admits a subsequence(zn ) such that the sequence of its arithmetic means n1 (z1 + z2 + . . . + zn ) is convergent in norm
(see [1]).
It is well known that every Banach space X with the Banach–Saks property is reflexive and the converse is not true (see [7]). Kakutani [12] has
proved that any uniformly convex Banach space X has the Banach–Saks
property. Moreover, he has also proved that if X is a reflexive Banach space
1991 Mathematics Subject Classification: 46E30, 46E40, 46B20.
Key words and phrases: Banach–Saks property, property (β), nearly uniform convexity, uniform Kadec–Klee property, property (H), Musielak–Orlicz sequence space.
The first author supported by Chinese National Science Foundation Grant.
The second and third authors supported by KBN grant 2 P03A 031 10.
[193]
194
Y. A. Cui et al.
such that there is Θ ∈ (0, 2) such that for every sequence (xn ) in S(X)
weakly convergent to zero there are n1 , n2 ∈ N satisfying kxn1 + xn2 k < Θ,
then X has the Banach–Saks property.
A Banach space X is said to have property (H) (or the Kadec–Klee
property) if every weakly convergent sequence on the unit sphere S(X) is
convergent in norm (see [11]).
Recall that a sequence (xn ) is said to be an ε-separated sequence if for
some ε > 0,
sep(xn ) = inf{kxn − xm k : n 6= m} > ε.
A Banach space X is said to have the uniform Kadec–Klee property (written
UKK) if for every ε > 0 there exists δ > 0 such that if x is the weak limit
of a norm one ε-separated sequence, then kxk < 1 − δ. Every UKK Banach
space has property (H) (see [10]).
A Banach space is said to be nearly uniformly convex (written NUC) if
for every ε > 0 there exists δ ∈ (0, 1) such that for every sequence (xn ) ⊂
B(X) with sep(xn ) > ε, we have
conv({xn }) ∩ (1 − δ)B(X) 6= ∅.
Huff [10] has proved that X is NUC if and only if X is reflexive and UKK.
A Banach space X is said to be nearly uniformly smooth (NUS for short)
if for any ε > 0 there exists δ > 0 such that for each basic sequence (xn ) in
B(X) there is k > 1 such that
kx1 + txk k ≤ 1 + tε
for each t ∈ [0, δ]. Prus [20] has shown that a Banach space X is NUC if
and only if X ∗ is NUS.
For any x 6∈ B(X), the drop determined by x is the set
D(x, B(X)) = conv({x} ∪ B(X))
(see [5]). A Banach space X has the drop property (written (D)) if for every
closed set C disjoint from B(X) there exists an element x ∈ C such that
D(x, B(X)) ∩ C = {x}.
In [22], Rolewicz has proved that if the Banach space X has the drop
property, then X is reflexive. Montesinos [18] has extended this result showing that X has the drop property if and only if X is a reflexive Banach space
with property (H).
For any subset C of X we denote by α(C) its Kuratowski measure of
noncompactness, i.e. the infimum of those ε > 0 for which there is a covering
of C by a finite number of sets of diameter less than ε.
Goebel and Sȩkowski [8] have extended the definition of uniform convexity replacing condition (1) by a condition involving the Kuratowski measure
of noncompactness. Namely, they called a norm k · k in a Banach space X
Banach–Saks property
195
∆-uniformly convex (written ∆UC) if for any ε > 0 there is δ > 0 such
that for each convex set E contained in the closed unit ball B(X) such that
α(E) > ε, we have
inf{kxk : x ∈ E} < 1 − δ.
It is well known that ∆UC coincides with NUC.
Rolewicz [22], studying the relationships between NUC and the drop
property, has defined property (β). A Banach space X is said to have property (β) if for any ε > 0 there exists δ > 0 such that
α(D(x, B(X)) \ B(X)) < ε
whenever 1 < kxk < 1 + δ. It is well known that if a Banach space X has
property (β), then its dual space X ∗ has the normal structure (see [16]).
The following result will be very helpful for our considerations (see [15]):
A Banach space X has property (β) if and only if for every ε > 0 there
exists δ > 0 such that for each element x ∈ B(X) and each sequence (xn )
in B(X) with sep(xn ) ≥ ε there is an index k such that
x + xk 2 ≤ 1 − δ.
A map Φ : R → R+ is said to be an Orlicz function if Φ vanishes only at 0,
and Φ is even, convex, and continuous on the whole R+ (see [17], [19], [21]).
A sequence Φ = (Φn ) of Orlicz functions is called a Musielak–Orlicz
function. By Ψ = (Ψn ) we denote the complementary function of Φ in the
sense of Young, i.e.
Ψn (v) = sup{|v|u − Φn (u) : u ≥ 0},
n = 1, 2, . . .
Denote by l0 the space of all real sequences x = (x(i)). For a given Musielak–
Orlicz function Φ, we define a convex modular IΦ : l0 → [0, ∞] by the
formula
∞
X
IΦ (x) =
Φi (x(i)).
i=1
The Musielak–Orlicz sequence space lΦ is
lΦ = {x ∈ l0 : IΦ (cx) < ∞ for some c > 0}.
We consider lΦ equipped with the so-called Luxemburg norm
kxk = inf{ε > 0 : IΦ (x/ε) ≤ 1},
under which it is a Banach space (see [3], [19]).
The subspace hΦ defined by
hΦ = {x ∈ lΦ : IΦ (cx) < ∞ for every c > 0}
is called the subspace of finite (or order continuous) elements.
196
Y. A. Cui et al.
We say an Orlicz function Φ satisfies the δ2 -condition (Φ ∈ δ2 for short)
if there exist constants
P∞ k ≥ 2, u0 > 0 and a sequence (ci ) of nonnegative
numbers such that i=1 ci < ∞ and the inequality
Φi (2u) ≤ kΦi (u) + ci
holds for every i ∈ N and u ∈ R satisfying |u| ≤ u0 .
It is well known that hΦ = lΦ if and only if Φ ∈ δ2 (see [13]).
We say a Musielak–Orlicz function Φ satisfies condition (∗) if for any
ε ∈ (0, 1) there exists δ > 0 such that Φi ((1+δ)u) ≤ 1 whenever Φi (u) ≤ 1−ε
for u ∈ R and all i ∈ N (see [14]).
For more details on Musielak–Orlicz spaces we refer to [3] or [19].
2. Auxiliary facts. In order to obtain some new results, we will use
the following well-known facts.
Lemma 1 (see [4]). If a Musielak–Orlicz function Φ = (Φi ) with all Φi
finitely valued satisfies condition (∗) and Φ ∈ δ2 , then for each ε > 0 there
is δ > 0 such that kxk < 1 − δ whenever IΦ (x) < 1 − ε.
Lemma 2 (see [14]). If a Musielak–Orlicz function Φ = (Φi ) with all Φi
finitely valued satisfies condition (∗) and Φ ∈ δ2 , then for every ε > 0 and
c > 0 there exists δ > 0 such that
|IΦ (x + y) − IΦ (x)| < ε
whenever IΦ (x) ≤ c and IΦ (y) < δ.
Lemma 3 (see [6]). If a Musielak–Orlicz function Ψ = (Ψi ) ∈ δ2 , then
there
P∞ exists θ ∈ (0, 1) and a sequence (hi ) of nonnegative numbers such that
i=1 Φi (hi ) < ∞ and the inequality
1−θ
u
≤
Φi (u)
Φi
2
2
holds for every i ∈ N and u ∈ R with Φi (hi ) ≤ Φi (u) ≤ 1.
3. Results. We start with the following general result.
Theorem 1. If a Banach space X has property (β), then both X and X ∗
have the Banach–Saks property.
P r o o f. Assume X has property (β). First, we will prove that X has the
Banach–Saks property. Since property (β) implies reflexivity, it is enough to
prove that there exists Θ ∈ (0, 2) such that for each sequence (xn ) in S(X)
weakly convergent to zero, there are n1 , n2 ∈ N such that kxn1 + xn2 k < Θ.
Since (xn ) is weakly convergent to zero, the set of its elements cannot
be compact in S(X). So, there are ε0 > 0 and a subsequence (zn ) of (xn )
Banach–Saks property
197
with sep(zn ) ≥ ε0. By property (β) for X, there exists δ > 0 depending on
ε0 only such that for every z ∈ S(X) there exists k ∈ N for which
kz + zk k < 2 − δ
(cf. Proposition 1 in [15]). In particular, setting z = z1 , a natural number
k(1) 6= 1 can be found such that
kz1 + zk(1) k < Θ,
where Θ = 2 − δ. This means that X has the Banach–Saks property.
Next, we will prove that X ∗ has the Banach–Saks property. For each
sequence (xn ) in S(X) weakly convergent to zero, by the Bessaga–Pelczyński
selection principle, there exists a basic subsequence (zn ) of (xn ) (see [7]).
Property (β) for X implies that X ∗ is NUS (see [20]), i.e. for any ε > 0
there is δ ∈ (0, 1) such that there is k ∈ N, k > 1, such that
kz1 + tzk k < 1 + tε
for any t ∈ [0, δ]. In particular, taking ε = 1/2, numbers δ0 ∈ (0, 1) and
k > 1, k ∈ N, can be found such that
kz1 + δ0 zk k < 1 +
δ0
.
2
Hence
kz1 + zk k = kz1 + δ0 zk + (1 − δ0 )zk k < 1 +
δ0
δ0
+ (1 − δ0 ) = 2 − ,
2
2
i.e. X ∗ has the Banach–Saks property.
Theorem 1 cannot be reversed in general. Indeed, note that c0 as well
as its dual l1 have the Banach–Saks property, but they fail property (β).
However, both c0 and l1 are not reflexive. It is natural to ask the following
Question. Assume that X is a reflexive Banach space. Does the
Banach–Saks property for X and X ∗ imply property (β) for X?
Now, we will describe some geometric properties in Musielak–Orlicz sequence spaces.
Theorem 2. If a Musielak–Orlicz function Φ = (Φi ) with all Φi finitely
valued satisfies condition (∗), then the following statements are equivalent:
(a) lΦ is UKK;
(b) lΦ has property (H);
(c) Φ ∈ δ2 .
198
Y. A. Cui et al.
P r o o f. (a)⇒(b). This holds true for any Banach space (see [10]).
(b)⇒(c). If Φ 6∈ δ2 , we can find an element x = (x(1), x(2), . . .) ∈ S(lΦ )
such that IΦ (x) ≤ 1 and IΦ (λx) = ∞ for any λ > 1 (see [13]). Consequently,
there is an increasing sequence (ni ) of natural numbers such that
1
k(0, . . . , 0, x(ni + 1), . . . , x(ni+1 ), 0, . . .)k ≥ .
2
Putting
xi = (x(1), . . . , x(ni ), 0, . . . , 0, x(ni+1 + 1), . . .),
i = 1, 2, . . . ,
we get
(1) kxi k = 1, i = 1, 2, . . . ;
(2) xi → x weakly.
Equalities (1) follow by IΦ (xi ) ≤ 1 and IΦ (λxi ) = ∞ for every λ > 1
(i = 1, 2, . . .). We will now prove property (2). For every y ∗ ∈ (lΦ )∗ we
have y ∗ = y0∗ + y1∗ uniquely, where y0∗ and y1∗ are respectively the regular
and singular parts of y ∗ , i.e. y0∗ is determined by a function y0 ∈ lΨ and
y1∗ (x) = 0 for any
x ∈ hΦ (see [9]). Since y0 = (y0 (i)) ∈ lΨ , there exists
P∞
λ > 0 such that i=1 Ψi (λy0 (i)) < ∞. Since hxi − x, y1∗ i = 0, we have
ni+1
X
hxi − x, y ∗ i = hxi − x, y0∗ i =
x(j)y0 (j)
j=ni +1
ni+1
≤
1 X
(Φj (x(j)) + Ψj (λy0 (j))) → 0
λ j=n +1
as i → ∞,
i
which proves (2). We also have
(3) kxi − xk ≥ 1/2 for all i ∈ N,
which means that lΦ does not have property (H).
(c)⇒(a). Suppose lΦ is not UKK and Φ ∈ δ2 . There exists ε0 > 0 such
that for any θ > 0 there are a sequence (xn ) and an element x in S(lΦ ) with
sep(xn ) ≥ ε0 , xn → x weakly and kxk > 1 − θ. Since sep(xn ) ≥ ε0 , we can
assume without loss of generality that kxn − xk ≥ ε0 /2 for every n ∈ N.
Since Φ ∈ δ2 and Φ satisfies condition (∗) and x can be assumed to have kxk
close to 1, we may assume that there is η0 > 0 such that IΦ (xn − x) ≥ η0
and IΦ (x) > 1 − η0 /5. Using again Φ ∈ δ2 , there exists σ0 ∈ (0, η0 /5) such
that
η0
|IΦ (x + y) − IΦ (x)| <
5
whenever IΦ (y) < σ0 .
Since (xn ) ⊂ S(lΦ ) and xn → x weakly, by the lower semicontinuity
of the norm with respect to the weak topology, we conclude that there is
Banach–Saks property
199
P∞
i0 ∈ N such that i=i0 +1 Φi (x(i)) < σ0 . By virtue of xn → x weakly, which
implies that xn → x coordinatewise, there exists n0 ∈ N such that
i0
i0
i0
η
X
X
X
η0
0
Φi (xn (i)) −
Φi (x(i)) <
and
Φi (xn (i) − x(i)) <
5
5
i=1
i=1
i=1
for n ≥ n0 . So
1=
∞
X
Φi (xn (i)) =
i0
X
i=1
≥
i0
X
Φi (xn (i)) +
i=1
Φi (x(i)) −
i=1
∞
X
Φi (xn (i))
i=i0 +1
∞
X
η0
+
5
i=i
Φi (xn (i)).
0 +1
Hence
η0 ≤ IΦ (xn − x) =
∞
X
Φi (xn (i) − x(i))
i=1
=
i0
X
Φi (xn (i) − x(i)) +
i=1
∞
X
Φi (xn (i) − x(i))
i=i0 +1
i0
∞
X
X
2η0
η0
η0
η0
Φi (x(i)) +
+
≤1−
+
Φi (xn (i)) +
5
5
5
5
i=1
i=i0 +1
3η0
η0
3η0
≤ 1 − (1 − σ0 ) +
≤1− 1−
+
< η0 .
5
5
5
<
This contradiction proves the implication (c)⇒(a).
Corollary 1. If a Musielak–Orlicz function Φ = (Φi ) with all Φi finitely
valued satisfies condition (∗), then the following statements are equivalent:
(a) lΦ is NUC;
(b) lΦ has the drop property;
(c) Φ ∈ δ2 and Ψ ∈ δ2 .
P r o o f. Since NUC is equivalent to the conjunction of UKK and reflexivity, and the reflexivity of lΦ is equivalent to the fact that Φ ∈ δ2 and
Ψ ∈ δ2 , by Theorem 2, we get our corollary immediately.
Recall that a Nakano space l(pi ) is the Musielak–Orlicz space lΦ with
Φ = (Φi ), where
Φi (u) = |u|pi ,
1 ≤ pi < ∞, i = 1, 2, . . .
Corollary 2. For any Nakano space l(pi ) the following statements are
equivalent:
200
Y. A. Cui et al.
(a) l(pi ) is NUC;
(b) l(pi ) has the drop property;
(c) 1 < lim inf i pi ≤ lim supi pi < ∞.
P r o o f. This follows immediately by Corollary 1 and the fact that for
the Nakano function Φ = (Φi ) with Φi (u) = |u|pi we have Φ ∈ δ2 if and only
if lim supi pi < ∞, and its complementary function Ψ ∈ δ2 if and only if
lim inf i pi > 1.
Corollary 3. Let l(pi ) be a Nakano space. Then the following statements are equivalent:
(a) l(pi ) is UKK;
(b) l(pi ) has property (H);
(c) lim supi pi < ∞.
P r o o f. This follows immediately by Theorem 2 and the fact that the
Nakano function Φ = (Φi ) with Φi (u) = |u|pi satisfies the δ2 -condition if and
only if condition (c) is satisfied.
Theorem 3. If a Musielak–Orlicz function Φ = (Φi ), with all Φi finitely
valued and satisfying Φi (u)/u → 0 as u → 0, satisfies condition (∗), then lΦ
has the Banach–Saks property if and only if Φ ∈ δ2 and Ψ ∈ δ2 .
P r o o f. Since the Banach–Saks property implies reflexivity and the reflexivity of lΦ is equivalent to Φ ∈ δ2 and Ψ ∈ δ2 , we only need to prove
sufficiency. By Ψ ∈ P
δ2 , there exists Θ ∈ (0, 1) and a sequence (hi ) of positive
∞
numbers such that i=1 Φi (hi ) < ∞ and
Φi (u)
u
≤ (1 − Θ)
Φi
2
2
for all i ∈ N and u ∈ R with Φi (hi ) ≤ Φi (u) ≤ 1 (see Lemma 3).
By Φ ∈ δ2 and condition (∗) for Φ, for any ε ∈ (0, Θ/16), there exists a
δ ∈ (0, Θ) such that
ε
|IΦ (y + z) − IΦ (y)| <
2
whenever IΦ (y) ≤ 1, IΦ (z) ≤ δ (see Lemma 2).
For each sequence (xn ) of S(lΦ ) with xn → 0 weakly,
P∞ we have xn → 0
coordinatewise, so there are i0 and n0 ∈ N such that i=i0 +1 Φi (x1 (i)) < δ,
P∞
Pi0
Φi (hi ) < δ/16 and
Hence
i=i
+1
i=1 Φi (xn (i)) < δ for n > n0 .
0
P∞
i=i0 +1 Φi (xn (i)) ≥ 1/2 for n > n0 and
Banach–Saks property
IΦ
x1 + xn
2
=
i0
X
Φi
i=1
≤
i0
X
Φi (x1 (i))
2
i=1
≤
i0
X
Φi (x1 (i))
2
i=1
=
xn (i) + x1 (i)
2
1
2
+
Φi
i=i0 +1
∞
X
Φi
i=i0 +1
xn (i)
2
xn (i) + x1 (i)
2
+ε
∞
∞
X
1−Θ X
+
Φi (xn (i)) +
Φi (hi ) + ε
2 i=i +1
i=i +1
Φi (x1 (i)) +
i=1
∞
X
0
i0
nX
+
+
∞
X
201
∞
X
0
Φi (xn (i))
o
i=i0 +1
Φi (hi ) + ε −
i=i0 +1
∞
Θ X
Φi (xn (i))
2 i=i +1
0
1
Θ
Θ
Θ
Θ
≤2· +
+
−
=1− ,
2 16 16
4
8
for n > n0 .
In view of Lemma 1, there exists δ > 0 independent of x1 and xn such
that
kx1 + xn k < 2 − δ
for n > n0 .
The proof of Theorem 3 is finished.
Corollary 4. The Nakano sequence space l(pi ) has the Banach–Saks
property if and only if 1 < lim inf i pi ≤ lim supi pi < ∞.
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Department of Mathematics
Harbin University of
Science and Technology
Xuefu Road 52
150080 Harbin, China
Faculty of Mathematics
and Computer Science
Adam Mickiewicz University
Matejki 48/49
60-769 Poznań, Poland
E-mail: [email protected]
Institute of Mathematics
T. Kotarbiński Pedagogical University
Pl. Slowiański 9
65-069 Zielona Góra, Poland
E-mail: [email protected]
Reçu par la Rédaction le 15.4.1996
Révisé le 25.6.1996