# A Remark on the Stability of Approximative Compactness

Authors: Zhenghua Luo; Longfa Sun; Wen Zhang
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

We study the stability of approximativeτ-compactness, where τ is the norm or the weak topology. Let Λ be an index set and for every λ
∈
Λ
, let Y
λ be a subspace of a Banach space X
λ. For 1
≤
p
<
∞, let X
=
⊕
l
p
X
λ and Y
=
⊕
l
p
Y
λ. We prove that Y (resp., B
Y) is approximatively τ-compact in X if and only if, for every λ
∈
Λ, Y
λ (resp., B
Y
λ) is approximatively τ-compact in X
λ.


---

## Body

## 1. Introduction

LetX be a real Banach space and let K be a subset of X. We denote by τ either the norm or the weak topology on X. The metric projection of X onto K is the set valued map defined by P
K
(
x
)
=
{
y
∈
K
:
x
-
y
=
d
(
x
,
K
)
} for x
∈
X, where d
(
x
,
K
) denotes the distance from x to K. If, for every x
∈
X
,
P
K
(
x
)
≠
∅, we say that K is a proximinal subset of X. A sequence {
y
n
}
⊂
K is calledminimizing for x
∈
X, if x
-
y
n
→
d
(
x
,
K
).The notion of approximative compactness was introduced by Efimov and Stechkin [1] in connection with the study of Chebyshev sets in Banach spaces and plays an important role in approximation theory (see, e.g., [2, 3]). Deutsch [4] extended this notion to define approximative τ-compactness.Definition 1.
LetK be a τ-closed subset of X and x
0
∈
X. We say that K is approximatively τ-compact for x
0 if every minimizing sequence {
y
n
}
⊂
K for x
0 has a τ-convergent subsequence. If K is approximatively τ-compact for every x
∈
X, we say that K is approximatively τ-compact in X.It is easy to verify that approximativeτ-compactness implies proximinality. Clearly, compact sets or finite-dimensional subspaces of a Banach space are approximatively compact; weakly compact sets or reflexive subspaces of Banach spaces are approximatively weakly compact. Approximative τ-compactness has been studied in detail in [1, 3–7].When it comes to the stability of approximativeτ-compactness, we suppose that Λ is an index set and for every λ
∈
Λ
,
Y
λ is a subspace of a Banach space X
λ. And let X
=
⊕
l
p
X
λ
,
Y
=
⊕
l
p
Y
λ, where 1
≤
p
<
∞. Bandyopadhyay et al. [5] proved that if Y is approximatively τ-compact in X, then Y
λ is approximatively τ-compact in X
λ for every λ
∈
Λ. In this paper, we prove that the converse is also true. On the other hand, the proximinality of the unit ball of subspaces has been the subject in many recent papers (see, e.g., [8–11]). In this paper, under the above assumption, we also prove that the unit ball of Y is approximatively τ-compact in X if and only if, for every λ
∈
Λ, the unit ball of Y
λ is approximatively τ-compact in X
λ.For a real Banach spaceX, we denote by B
X the unit ball of X and denote by X
∗ the dual space of X. Before we prove the main conclusions we first show a simple property on approximative τ-compactness of the unit ball of subspaces.Proposition 2.
LetY be a subspace of a Banach space X. If B
Y is approximatively τ-compact in X, then so is Y. But the converse is not true.Proof.
Suppose thatx
∈
X and {
y
n
}
⊂
Y is a minimizing sequence of x in Y; that is, x
-
y
n
→
d
(
x
,
Y
). Then {
y
n
}
⊂
λ
B
Y for sufficiently large λ
>
0. This means that d
(
x
,
Y
)
=
d
(
x
,
λ
B
Y
) and {
y
n
} is also a minimizing sequence of x in λ
B
Y. By approximative τ-compactness of λ
B
Y (which is equivalent to the one of B
Y), {
y
n
} has a τ-convergent subsequence.
To illustrate that the converse is not true, first, we show thatB
c
0 is not approximatively weakly compact in c
0. Take x
=
(
2,0
,
0
,
…
) and for every n
∈
N, y
n
=
(
1,1
,
1
,
…
,
1,0
,
0
,
…
), where 1 appears n times. Then d
(
x
,
B
c
0
)
=
1 and {
y
n
} is a minimizing sequence of x in B
c
0. But {
y
n
} has no weakly convergent subsequence. Hence B
c
0 is not approximatively weakly compact in c
0.
Next, letX
=
c
0
⊕
l
1
R and Y
=
c
0
⊕
l
1
{
0
}. For any α
=
(
z
,
r
)
∈
X, it is easy to see that d
(
α
,
Y
)
=
r and P
Y
(
α
)
=
{
(
z
,
0
)
}. Now, suppose {
β
n
}
=
{
(
z
n
,
0
)
}
⊂
Y is a minimizing sequence of α in Y; that is, (1)α-βn=z-zn+r⟶dα,Y=r.This implies that z
n
→
z. Hence β
n
=
(
z
n
,
0
)
→
(
z
,
0
). Therefore Y is approximatively compact in X. But, by the above discussion, B
Y is not approximatively weakly compact in Y, and not in X either.In order to prove our conclusions, we need the following lemmas.Lemma 3.
Let{
X
i
:
i
∈
N
} be a sequence of Banach spaces and let Y
i be a subspace of X
i, respectively, for i
∈
N. Consider X
=
⊕
l
p
X
i and Y
=
⊕
l
p
Y
i, where 1
≤
p
<
∞. Let x
=
(
x
i
)
∈
X and {
y
n
=
(
y
n
,
i
)
}
⊂
B
Y be a minimizing sequence of x in B
Y. Then, for every ε
>
0, there exists some j
∈
N such that, for all n, ∑
i
>
j
‍
y
n
,
i
p
<
ε
p.Proof.
If the conclusion does not hold, then, for everyj, there exists infinitely many n such that ∑
i
>
j
‍
y
n
,
i
p
≥
ε
p. We can choose some j
0 such that ∑
i
>
j
0
‍
x
i
p
<
(
ε
/
3
)
p and infinite subset {
n
k
} of N such that ∑
i
>
j
0
‍
y
n
k
,
i
p
≥
ε
p for every k. Therefore for every k,(2)x-ynkp=∑i≤j0xi-ynk,ip+∑i>j0xi-ynk,ip=∑i≤j0xi-ynk,ip+∑i>j0xip+∑i>j0xi-ynk,ip-∑i>j0xip≥dx,BYp+∑i>j0ynk,ip1/p-∑i>j0xip1/pp-∑i>j0xip≥dx,BYp+2ε3p-ε3p≥dx,BYp+ε3p.
Butx
-
y
n
→
d
(
x
,
B
Y
)
(
n
→
∞
); then x
-
y
n
p
<
(
d
(
x
,
B
Y
)
)
p
+
(
ε
/
3
)
p for sufficiently large n. This is a contradiction.Remark 4.
In Lemma3, if we replace B
Y by Y, that is, {
y
n
=
(
y
n
,
i
)
}
⊂
Y is a minimizing sequence of x in Y, then the conclusion still holds.Lemma 5.
Under the assumption in Lemma3, if, moreover, l
i
m
n
→
∞
y
n
=
r and for every i
∈
N, l
i
m
n
→
∞
y
n
,
i
=
r
i, then(1)
r
=
(
∑
i
∈
N
‍
r
i
p
)
1
/
p;
(2)
d
(
x
,
B
Y
)
=
d
(
x
,
⊕
l
p
r
i
B
Y
i
)
=
(
∑
i
∈
N
‍
[
d
(
x
i
,
r
i
B
Y
i
)
]
p
)
1
/
p.Proof.
(1) For everyε
>
0, by Lemma 3, there exists j
∈
N such that, for all n, ∑
i
>
j
‍
y
n
,
i
p
<
ε
p. For every fixed j
′
>
j, we can choose n
′ such that ∑
i
≤
j
′
‍
y
n
′
,
i
p
-
r
i
p
<
ε
p and y
n
′
p
-
r
p
<
ε
p. Then (3)∑i≤j′rip-rp≤∑i≤j′rip-yn′p+yn′p-rp=∑i≤j′rip-∑i∈Nyn′,ip+yn′p-rp≤∑i≤j′rip-yn′,ip+∑i>j′yn′,ip+yn′p-rp<3εp.By the arbitrariness of ε, we have r
=
(
∑
i
∈
N
‍
r
i
p
)
1
/
p.
(2) Note thatr
≤
1; hence ⊕
l
p
r
i
B
Y
i
⊂
B
Y. This implies that d
(
x
,
B
Y
)
≤
d
(
x
,
⊕
l
p
r
i
B
Y
i
). To prove that d
(
x
,
B
Y
)
≥
d
(
x
,
⊕
l
p
r
i
B
Y
i
), for every n, we define z
n
=
(
z
n
,
i
), where z
n
,
i
=
y
n
,
i for y
n
,
i
≤
r
i, and z
n
,
i
=
r
i
/
y
n
,
i
y
n
,
i for y
n
,
i
>
r
i. Then {
z
n
}
⊂
⊕
l
p
r
i
B
Y
i.
For arbitraryε
>
0, by Lemma 3, there exists j such that ∑
i
>
j
‍
r
i
p
<
ε
p, and for all n, ∑
i
>
j
‍
y
n
,
i
p
<
ε
p. Further, we can choose some n
0 such that, for all n
>
n
0, ∑
i
≤
j
‍
y
n
,
i
-
r
i
p
<
ε
p. Then for all n
>
n
0, we have(4)yn-zn=∑i∈Nyn,i-zn,ip1/p≤∑i∈Nyn,i-rip1/p≤∑i≤jyn,i-rip1/p+∑i>jyn,i-rip1/p≤∑i≤jyn,i-rip1/p+∑i>jyn,ip1/p+∑i>jrip1/p<3ε.By the arbitrariness of ε, we have y
n
-
z
n
→
0. This implies that (5)limn→∞⁡x-zn=limn→∞⁡x-yn=dx,BY.Therefore d
(
x
,
B
Y
)
≥
d
(
x
,
⊕
l
p
r
i
B
Y
i
). So we have d
(
x
,
B
Y
)
=
d
(
x
,
⊕
l
p
r
i
B
Y
i
).
For the second equality, first, it is obvious that(6)dx,⊕lpriBYi≥∑i∈Ndxi,riBYip1/p.On the other hand, let ε
>
0 be given. For every i, we can choose z
i
∈
r
i
B
Y
i such that x
i
-
z
i
p
<
[
d
(
x
i
,
r
i
B
Y
i
)
]
p
+
ε
/
2
i. Let z
=
(
z
i
)
∈
⊕
l
p
r
i
B
Y
i; then (7)x-z=∑i∈Nxi-zip1/p<∑i∈Ndxi,riBYip+ε1/p.By the arbitrariness of ε, we have d
(
x
,
⊕
l
p
r
i
B
Y
i
)
≤
(
∑
i
∈
N
‍
[
d
(
x
i
,
r
i
B
Y
i
)
]
p
)
1
/
p. Therefore the second equality holds.The following is our main result.Theorem 6.
LetΛ be an index set. For every λ
∈
Λ, let Y
λ be a subspace of a Banach space X
λ. For 1
≤
p
<
∞, let X
=
⊕
l
p
X
λ and Y
=
⊕
l
p
Y
λ. Then(1)
Y is approximatively τ-compact in X if and only if, for every λ
∈
Λ, Y
λ is approximatively τ-compact in X
λ;
(2)
B
Y is approximatively τ-compact in X if and only if, for every λ
∈
Λ, B
Y
λ is approximatively τ-compact in X
λ.Proof.
(1) Necessity has been proven in [5].
Sufficiency: letx
∈
X and {
y
n
}
⊂
Y be a minimizing sequence for x. We will show that {
y
n
} has a τ-convergent subsequence. Without loss of generality, we can assume Λ
=
N and x
=
(
x
i
)
,
y
n
=
(
y
n
,
i
).
First, notice that ifz
i
∈
P
Y
i
(
x
i
) for every i, then z
=
(
z
i
)
∈
P
Y
(
x
)
. Hence (
d
(
x
,
Y
)
)
p
=
∑
j
∈
N
‍
(
d
(
x
j
,
Y
j
)
)
p. And for every i, (8)xi-yn,ip+∑j≠idxj,Yjp≤x-ynp=dx,Yp+x-ynp-dx,Yp=∑j∈Ndxj,Yjp+x-ynp-dx,Yp.So (9)xi-yn,ip≤dxi,Yip+x-ynp-dx,Yp⟶dxi,Yipn⟶∞.This implies that, for every i, {
y
n
,
i
} is a minimizing sequence for x
i in Y
i. Then {
y
n
,
i
} has a τ-convergent subsequence by the approximative τ-compactness of Y
i. By employing the diagonal process, we can choose a subsequence {
y
n
k
} of {
y
n
} such that, for every i, {
y
n
k
,
i
} has a τ-convergent to some y
i
∈
Y
i. Obviously, y
i
∈
P
Y
i
(
x
i
), and y
=
(
y
i
)
∈
Y
,
x
-
y
=
d
(
x
,
Y
).
We still denote the subsequence{
y
n
k
} as {
y
n
}. Next, to complete the proof, we will prove that {
y
n
} has a τ-convergent to y.
Case  1. τ is the norm topology. For every ε
>
0, by Remark 4, there exists some j such that ∑
i
>
j
‍
y
i
p
<
ε
p and for all n, ∑
i
>
j
‍
y
n
,
i
p
<
ε
p. Then we can choose n
0 such that, for n
>
n
0, ∑
i
≤
j
‍
y
n
,
i
-
y
i
p
<
ε
p. Hence for all n
>
n
0, (10)yn-y=∑i≤jyn,i-yip+∑i>jyn,i-yip1/p≤∑i≤jyn,i-yip1/p+∑i>jyn,ip1/p+∑i>jyip1/p<3ε.Therefore, by the arbitrariness of ε, we have that {
y
n
} converges to y.
Case  2. τ is the weak topology. Suppose f
=
(
f
i
)
∈
⊕
l
q
X
i
∗
=
X
∗ with f
=
1, where 1
/
p
+
1
/
q
=
1 when p
>
1 and q
=
∞ when p
=
1. For every ε
>
0, again by Remark 4, we can choose some j such that ∑
i
>
j
‍
y
i
p
<
ε
p, and for all n, ∑
i
>
j
‍
y
n
,
i
p
<
ε
p. Note that, for every 1
≤
i
≤
j, {
y
n
,
i
} weakly converges to y
i; hence there exists n
0 such that, for n
>
n
0, ∑
i
≤
j
‍
f
i
y
n
,
i
-
y
i
<
ε. Then for all n
>
n
0,(11)fyn-y=∑i∈Nfiyn,i-yi≤∑i≤jfiyn,i-yi+∑i>jfiyn,i-yi≤ε+f∑i>jyn,i-yip1/p≤ε+∑i>jyn,ip1/p+∑i>jyip1/p<3ε.
Again by the arbitrariness ofε, we have that {
y
n
} weakly converges to y.
(2) Necessity: fixλ
0
∈
Λ. Suppose that x
λ
0
∈
X
λ
0, and {
y
n
,
λ
0
}
⊂
B
Y
λ
0 is a minimizing sequence of x
λ
0 in B
Y
λ
0. Let x
=
(
x
λ
), y
n
=
(
y
n
,
λ
), where x
λ
=
0 and y
n
,
λ
=
0 for λ
≠
λ
0. Then x
∈
X
,
{
y
n
}
⊂
B
Y, and (12)x-yn=xλ0-yn,λ0⟶dxλ0,BYλ0.Notice that d
(
x
,
B
Y
)
≥
d
(
x
λ
0
,
B
Y
λ
0
)
. Hence (13)x-yn⟶dxλ0,BYλ0=dx,BY,which implies that {
y
n
} is a minimizing sequence of x in B
Y. By approximative τ-compactness of B
Y in X, {
y
n
} has a τ-convergent subsequence {
y
n
k
}. Therefore {
y
n
k
,
λ
0
} is τ-convergent.
Sufficiency: suppose thatx
∈
X
∖
B
Y and {
y
n
}
⊂
B
Y is a minimizing sequence of x in B
Y. Like the proof in (1), we will prove that {
y
n
} has a τ-convergent subsequence and we can assume Λ
=
N, x
=
(
x
i
), and y
n
=
(
y
n
,
i
)
. By employing the diagonal process, we can choose a subsequence {
y
n
k
} of {
y
n
} (we still denote the subsequence as {
y
n
}) such that l
i
m
n
→
∞
y
n
=
r, and for every i
∈
N, l
i
m
n
→
∞
y
n
,
i
=
r
i. Then by Lemma 5, we have r
=
(
∑
i
∈
N
‍
r
i
p
)
1
/
p, and(14)dx,BY=dx,⊕lpriBYi=∑i∈Ndxi,riBYip1/p.
Next, for everyi
∈
N, we will show that {
y
n
,
i
} has a τ-convergent subsequence. We can assume that, for all n and i, y
n
,
i
≤
r
i. Otherwise, we can replace {
y
n
} with {
z
n
} which we define in the proof of Lemma 5(2).
Note that, for everyi
∈
N, (15)xi-yn,ip+∑j≠idxj,rjBYjp≤x-ynp=dx,BYp+x-ynp-dx,BYp=∑j∈Ndxj,rjBYjp+x-ynp-dx,BYp.Then (16)dxi,riBYip≤xi-yn,ip≤dxi,riBYip+x-ynp-dx,BYp.Hence, when n
→
∞, (17)xi-yn,i⟶dxi,riBYi.This implies that {
y
n
,
i
} is a minimizing sequence of x
i in r
i
B
Y
i. By approximative τ-compactness of r
i
B
Y
i in X
i, {
y
n
,
i
} has a τ-convergent subsequence.
Employing the diagonal process again, we can choose a subsequence{
y
n
k
} of {
y
n
} such that, for evrey i, {
y
n
k
,
i
} has a τ-convergents to some y
i
∈
r
i
B
Y
i. Let y
=
(
y
i
); then y
∈
⊕
l
p
r
i
B
Y
i. We still denote {
y
n
k
} as {
y
n
}. Finally, just like the proof in (1), we can prove that {
y
n
} has a τ-convergent to y.Remark 7.
The above theorem does not hold forp
=
∞. Indeed, suppose that Z is an infinite-dimensional proper closed subspace of l
2. By Theorem 1.4 in [12], B
Z is approximatively compact in l
2. Next, we show that Z
⊕
l
∞
Z is not approximatively compact in l
2
⊕
l
∞
l
2. Choose x
0
∈
l
2
∖
Z and y
0
∈
Z such that d
(
x
0
,
Z
)
=
1
=
x
0
-
y
0. Furthermore, we take a sequence {
z
n
}
⊂
Z with z
n
=
1 satisfying that {
z
n
} has no convergent subsequence. Note that (18)0,x0-zn,y0∞=max⁡zn,x0-y0=1;and for any (
z
,
y
)
∈
Z
⊕
l
∞
Z, (19)0,x0-z,y∞=max⁡z,x0-y≥x0-y≥1.This means that d
(
(
0
,
x
0
)
,
Z
⊕
l
∞
Z
)
=
1 and {
(
z
n
,
y
0
)
} is a minimizing sequence of (
0
,
x
0
) in Z
⊕
l
∞
Z. But {
(
z
n
,
y
0
)
} has no convergent subsequence.



---

*Source: 2734947-2016-01-21.xml*