J. Math. Anal. Appl. 421 (2015) 1377–1395

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Journal of Mathematical Analysis and Applications

www.elsevier.com/locate/jmaa

Approximative compactness and Asplund property in Banach

function spaces and in Orlicz–Bochner spaces in particular with

application ✩

Shaoqiang Shang a,∗, Yunan Cui b

a Department of Mathematics, Northeast Forestry University, Harbin 150040, PR Chinab Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, PR China

a r t i c l e i n f o a b s t r a c t

Article history:Received 2 January 2014Available online 29 July 2014Submitted by G. Corach

Keywords:Approximative compactnessAsplund spaceRadon–Nikodym propertyBanach spaceOrlicz–Bochner function space

In this paper it is shown that: (1) If every weak∗ hyperplane of X∗ is approximatively compact, then (a) X is an Asplund space; (b) X∗ has the Radon–Nikodym property. (2) Criteria for approximative compactness of every weakly∗ hyperplane of Orlicz–Bochner function spaces equipped with the Orlicz norm are given. (3) If X has a Fréchet differentiable norm, then (a) Orlicz–Bochner function spaces L0

M (X∗) have the Radon–Nikodym property if and only if M ∈ Δ2; (b) Orlicz–Bochner function spaces EN (X) are Asplund spaces if and only if M ∈ Δ2. (4) We give an important application of approximative compactness to the theory of generalized inverses for operators between Banach spaces and Orlicz–Bochner function spaces.

© 2014 Elsevier Inc. All rights reserved.

1. Introduction

Let X be a Banach space and X∗ the dual space of X. Denote by B(X) and S(X) the closed unit ball and the unit sphere of X. Let C ⊂ X be a nonempty subset of X. Then the set-valued mapping PC : X → C

PC(x) ={z ∈ C : ‖x− z‖ = dist(x,C) = inf

y∈C‖x− y‖

}

is called the metric projection operator from X onto C.A subset C of X is said to be proximinal if PC(x) �= ∅ for all x ∈ X (see [2]). It is well known that X is

reflexive if and only if each closed convex subset of X is proximinal (see [2]).

Definition 1. A nonempty subset C of X is said to be approximatively compact if for any {yn}∞n=1 ⊂ C

and any x ∈ X satisfying ‖x− yn‖ → infy∈C ‖x− y‖ as n → ∞, there exists a subsequence of {yn}∞n=1

✩ Supported by Heilongjiang Provincial Department of Education Funds 12521070.* Corresponding author.

E-mail addresses: sqshang@163.com (S. Shang), cuiya@hrbust.edu.cn (Y. Cui).

http://dx.doi.org/10.1016/j.jmaa.2014.07.0540022-247X/© 2014 Elsevier Inc. All rights reserved.

1378 S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395

converging to an element in C. X is called approximatively compact if every nonempty closed convex subset of X is approximatively compact.

Let us present the history of the approximative compactness and related notions. This notion has been introduced by Jefimow and Stechkin in [13] as a property of Banach spaces, which guarantees the existence of the best approximation element in a nonempty closed convex set C for any x ∈ X. In [16] it was established that the approximative compactness of C in a rotund Banach space X gives also the continuity of the projector operator PC . Fully rotund Banach spaces were defined by Fan and Glicksberg in [6]. In [9] it has been proved that fully rotund Banach spaces are approximatively compact. In [2] it has been proved that approximative compactness of a Banach space X means that X is reflexive and it has the H property.

Some criteria for approximative compactness of every weak∗ hyperplane of Orlicz function spaces equipped with the Luxemburg norm are given (see [22]). Moreover, it is well known that Orlicz func-tion spaces has the Radon–Nikodym property if and only if M ∈ Δ2. However, because of the complicated structure of Orlicz–Bochner function spaces, up to now the criteria for approximative compactness of every weak∗ hyperplane and Radon–Nikodym have not been discussed yet. The paper is organized as follows. In Section 1 some necessary definitions and notations are collected including the Orlicz–Bochner function spaces. In Section 2 we prove that if every weak∗ hyperplane of X∗ is approximatively compact, then (a) X is an Asplund space. (b) X∗ has the Radon–Nikodym property. In Section 3 criteria for approximative compactness of every weakly∗ hyperplane of Orlicz–Bochner function spaces equipped with the Orlicz norm are given. Moreover, we also prove that if X has a Fréchet differentiable norm, then (a) Orlicz–Bochner function spaces L0

M (X∗) have the Radon–Nikodym property if and only if M ∈ Δ2; (b) Orlicz–Bochner function spaces EN (X) are Asplund spaces if and only if M ∈ Δ2. In Section 4 we give an important ap-plication of approximative compactness to the theory of generalized inverses for operators between Banach spaces and Orlicz–Bochner function spaces. The topic of this paper is related to the topic of [6,13,16] and [1–5,7,8,10,11,17–22].

Let us recall some geometrical notions which will be used in the further part of the paper.

Definition 2. Let D be a nonempty open convex subset of X and f a real-valued continuous convex function on D. The function f is said to be Gateaux differentiable at the point x in D if the limit

f ′(x)(y) = limt→0

f(x + ty) − f(x)t

(∗)

exists for all y ∈ X. When this is the case, the limit is a continuous linear function of y, denoted by f ′(x). If the difference quotient in (∗) converges to f ′(x)(y) uniformly with respect to y in the unit ball, then f is said to be Fréchet differentiable at x. X is called a weak Asplund space [Asplund space] or said to have the weak Asplund property if for every f and D as above, f is “generically” Gateaux [Fréchet] differentiable, that is, there exists a dense Gδ subset G of D such that f is Gateaux [Fréchet] differentiable at each point of G.

Definition 3. A Banach space X is said to have a Fréchet differentiable norm if the norm ‖ · ‖ is Fréchet differentiable at every point of S(X).

It is well known (see [14]) that a norm ‖ · ‖ in a Banach space X is Fréchet differentiable at every point of S(X) if and only if for any x ∈ S(X), if x∗

n ∈ S(X∗) and x∗n(x) → 1, then there exists an x∗ ∈ S(X∗)

such that ‖x∗n − x∗‖ → 0 as n → ∞.

Definition 4. A Banach space X is said to have the Radon–Nikodym property whenever if (T, Σ, μ) is a nonatomic measure space and v is a vector measure on Σ with values in X which is absolutely continuous with respect to μ and has a bounded variation, then there exists f ∈ L1(X) such that for any A ∈ Σ,

S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395 1379

v(A) =∫A

f(t)dt.

It is well known that if X has a Fréchet differentiable norm, then X is an Asplund space.

Definition 5. A point x ∈ A is said to be an extreme point of A if 2x = y + z and y, z ∈ A imply y = z. The set of all extreme points of A is denoted by ExtA.

Definition 6. A functional fx ∈ S(X∗) is called a supporting functional of x ∈ S(X) if fx(x) = 1. A point x ∈ S(X) is called a smooth point if it has a unique supporting functional fx. If every x ∈ S(X) is a smooth point, then X is called a smooth space.

Definition 7. A function M : R → R+ is called an N -function if it has the following properties:

(1) M is even, continuous convex and M(0) = 0;(2) M(u) > 0 for any u �= 0;(3) limu→0 M(u)u = 0 and limu→∞ M(u)u = ∞.

Let (T, Σ, μ) be a nonatomic finite measurable space. Moreover, given any Banach space (X, ‖ · ‖), we denote by XT the set of all strongly μ-measurable function from T to X, and given any N -function M for each u ∈ XT , we define the modular of u by

ρM (u) =∫T

M(∥∥u(t)

∥∥)dt.Put

LM (X) ={u(t) ∈ XT : ρM (λu) < ∞ for some λ > 0

},

EM (X) ={u(t) ∈ XT : ρM (λu) < ∞ for all λ > 0

}.

It is well known that the Orlicz–Bochner function space LM (X) is a Banach space, when it is equipped with the Luxemburg norm

‖u‖ = inf{λ > 0 : ρM

(u

λ

)≤ 1

}

or with the Orlicz norm

‖u‖0 = infk>0

1k

[1 + ρM (ku)

].

If X = R, then L0M is said to be Orlicz function space. Let p(u) denote the right derivative of M(u) at

u ∈ R+, q(v) be the generalized inverse function of p(u) defined on R+ by

q(v) = supu≥0

{u ≥ 0 : p(u) ≤ v

}

Then we call N(v) =∫ |v|0 q(s)ds the complementary function of M . It is well known that there holds the

Young inequality uv ≤ M(u) + N(v) and uv = M(u) + N(v) ⇔ u = |q(v)| sign v or v = |p(u)| sign u. Moreover, it is well known that M and N are complementary to each other. It is well known that K(u) �= ∅, where

1380 S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395

K(u) ={k > 0 : 1

k

[1 + ρM (ku)

]= ‖u‖0

}.

For every N -function M we define the complementary function N : R → [0, ∞) by the formula

N(v) = supu>0

{u|v| −M(u)

}

for every v ∈ R. We say that an N -function M satisfies condition Δ2 if there exist K > 2 and u0 ≥ 0 such that

M(2u) ≤ KM(u) (u ≥ u0)

In this case, we write M ∈ Δ2 or N ∈ ∇2. Moreover, it is well known that M ∈ Δ2 if and only if for any l > 1, there exist v0 > 0 and δ > 0 such that N(lv) ≥ (l + δ)N(v) whenever v ≥ v0.

Let M be an N -function. An interval [a, b] is called a structural affine interval of M , or simply, SAI of M , provided that M is affine on [a, b] and it is not affine on either in [a − ε, b] or in [a, b + ε] for any ε > 0. Let {[ai, bi]}i be all the SAIs of M . We call SM = R\[

⋃i(ai, bi)] the set of strictly convex points of M .

A continuous function M : R → R is called strictly convex if

M

(u + v

2

)<

12M(u) + 1

2M(v)

for all u �= v.Let us recall some lemmas which will be used in the further part of the paper.

Lemma 1. (See [22].) The following statements are equivalent:

(1) Every weak∗ hyperplane of X∗ is approximatively compact.(2) If x∗

n ∈ S(X∗), x ∈ S(X) and x∗n(x) → 1 as n → ∞, then {x∗

n}∞n=1 is relatively compact.

Lemma 2. (See [22].) X is a smooth space if and only if any x∗ ∈ S(X∗) is an extreme point of B(X∗)provided that x∗ is norm attainable on S(X).

Lemma 3. (See [17].) Suppose that X is a Banach space. Then the following statements are equivalent:

(1) X is an Asplund space.(2) If E is a separable closed subspace of X, then E∗ is a separable space.(3) X∗ has the Radon–Nikodym property.

Lemma 4. (See [7].) Suppose that X∗ has the Radon–Nikodym property. Then (EM (X))∗ = L0N (X∗) and

(E0M (X))∗ = LN (X∗).

Lemma 5. (See [4].) Let A(x) = {x∗ ∈ S(X∗) : x∗(x) = ‖x‖ = 1}, B ⊂ S(X∗) and let B ∩ A(x) be a nonempty set for all x ∈ S(X). Assume that

B ⊂ co(F + αB

(X∗))

for some countable set F ⊂ X∗ and some α < 1. Then X∗ is the closed linear span of F .

S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395 1381

2. Approximative compactness and Asplund property in Banach spaces

Theorem 1. Suppose that every weak∗ hyperplane of X∗ is approximatively compact. Then (1) X is an Asplund space. (2) X∗ has the Radon–Nikodym property.

Proof. Let E be a separable closed subspace of X. We will prove that if x∗n ∈ S(E∗), x ∈ S(E) and

x∗n(x) → 1 as n → ∞, then sequence {x∗

n}∞n=1 is relatively compact. By the Hahn–Banach theorem, there exists y∗n ∈ S(X∗) such that y∗n ∈ S(X∗) and y∗n(z) = x∗

n(z) whenever z ∈ E. Since y∗n ∈ S(X∗), x ∈ S(X)and y∗n(x) → 1 as n → ∞, by Lemma 2, we obtain that the sequence {y∗n}∞n=1 is relatively compact. Hence we obtain that the sequence {x∗

n}∞n=1 is relatively compact. Let A(x) = {x∗ ∈ S(E∗) : ‖x‖ = x∗(x) = 1, x ∈S(E)} and {xn}∞n=1 be a dense subset of S(E). Then for any x ∈ S(E), we have

inf{dist

(A(xn), A(x)

): n ∈ N

}= 0. (1)

In fact, since {xn}∞n=1 is a dense subset of S(E), there exists a subsequence {xnk}∞k=1 of {xn}∞n=1 such that

xnk→ x as k → ∞. Pick x∗

nk∈ A(xnk

). Then

∣∣x∗nk

(x) − 1∣∣ =

∣∣x∗nk

(x) − x∗nk

(xnk)∣∣ ≤ ∥∥x∗

nk

∥∥‖xnk− x‖ → 0 as k → ∞.

From the previous proof, we obtain that {x∗nk}∞k=1 is relatively compact. Hence there exists a subsequence

{x∗nl}∞l=1 of {x∗

nk}∞k=1 such that {x∗

nl}∞l=1 is a Cauchy sequence. Let x∗

nl→ x∗

0 as l → ∞. Noticing that x∗

nk(x) → 1, we have x∗

nl(x) → x∗

0(x) = 1 as l → ∞. This implies that x∗0 ∈ A(x). Hence

inf{dist(A(xn), A(x)) : n ∈ N} = 0.Let {z∗k}∞k=1 ⊂ A(x) for any x ∈ S(E). Then z∗k(x) = 1. From the previous proof, we obtain that {z∗k}∞k=1

has a Cauchy subsequence. Moreover, it is easy to see that A(x) is a closed set. Hence we obtain that A(x)is compact. Then A(x) is separable. Let Dn be a countable dense subset of A(xn) for each n ∈ N . Then D =

⋃∞n=1 Dn is a countable set. Pick α ∈ (0, 1). Therefore, by (1), we obtain that there exist an α(x) ∈ A(x)

and n(x) ∈ N such that dist(A(xn(x)), α(x)) < α for any x ∈ S(E). Hence, for every x ∈ S(E), we obtain that B = {α(x) : x ∈ S(E)} ⊂ S(E∗) and B ∩ A(x) �= ∅ for any x ∈ S(E). Since dist(A(xn(x)), α(x)) < α

and Dn(x) is a countable dense subset of A(xn(x)), we obtain that

dist(α(x), D

)≤ dist

(α(x), Dn(x)

)= dist

(α(x), A(xn(x))

)< α

for any x ∈ S(E). This implies that

α(x) ∈ D + αB(E∗) ⊂ co

(D + αB

(E∗))

for any x ∈ S(E). Noticing that B = {α(x) : x ∈ S(E)}, we have

B ⊂ D + αB(E∗) ⊂ co

(D + αB

(E∗)).

By Lemma 5, we obtain that E∗ = span(D), and so E∗ is separable.This means that if E is a separable closed subspace of X, then E∗ is a separable space. By Lemma 3,

we obtain that X is an Asplund space and X∗ has the Radon–Nikodym property, which completes the proof. �Corollary 1. Suppose that X has a Fréchet differentiable norm. Then (1) X is an Asplund space. (2) X∗

has the Radon–Nikodym property.

1382 S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395

3. Approximative compactness and Asplund property in Orlicz–Bochner function spaces

Theorem 2. The following statements are equivalent:

(1) EN (X) has a Fréchet differentiable norm.(2) Every weak∗ hyperplane of L0

M (X∗) is approximatively compact.(3) (3a) M ∈ Δ2;

(3b) M(u) is strictly convex;(3c) X has a Fréchet differentiable norm.

In order to prove the theorem, we will give some auxiliary lemmas.

Lemma 6. (See [1].) L0M has the Radon–Nikodym property if and only if M ∈ Δ2.

Lemma 7. (See [1].) u ∈ L0M is an extreme point of B(L0

M ) if and only if μ{t ∈ T : ku(t) ∈ R\SM} = 0, where k ∈ K(u).

Lemma 8. The following statements are equivalent:

(a) u ∈ L0M is an extreme point of B(L0

M );(b) If u =

∑∞i=1 tiui where ui ∈ B(L0

M ), ti ∈ (0, 1) and ∑∞

i=1 ti = 1 belongs to S(L0M ), then the sequence

{ui}∞i=1 is relatively compact.

Proof. (b) ⇒ (a). Suppose that u ∈ L0M is not an extreme point of B(L0

M ). Then, by Lemma 7, we have μ{t ∈ T : ku(t) ∈ R\SM} > 0, where k ∈ K(u). Hence there exist an interval (a, b) and ε > 0 such that

μ{t ∈ T : ku(t) ∈ (a + ε, b− ε)

}> 0

and M is affine on [a, b]. Hence there exist p ∈ R and β ∈ R such that M(h) = ph + β for h ∈ [a, b]. Let G = {t ∈ T : ku(t) ∈ (a + ε, b − ε)}. Then it is easy to see that there exist δ > 0 and n0 ∈ N such that

μ

{t ∈ G :

(1 − 1

n0

)ku(t) ∈ (a + δ, b− δ)

}> 0.

Let H = {t ∈ G : (1 − 1/n0)ku(t) ∈ (a + δ, b − δ)}. Decompose H into E11 and E1

2 such that E11 ∪ E1

2 = H, E1

1 ∩E12 = ∅ and

∫E1

1p( k

n0u(t))dt =

∫E1

2p( k

n0u(t))dt. Decompose E1

1 into E21 and E2

2 such that E21 ∩E2

2 = ∅, E2

1 ∪E22 = E1

1 and ∫E2

1p( k

n0u(t))dt =

∫E2

2p( k

n0u(t))dt. Decompose E1

2 into E23 and E2

4 such that E23∩E2

4 = ∅, E2

3 ∪E24 = E1

2 and ∫E2

3p( k

n0u(t))dt =

∫E2

4p( k

n0u(t))dt. Generally, decompose En−1

i into En2i−1 and En

2i such that

En2i−1 ∩En

2i = ∅, En2i−1 ∪ En

2i = En−1i ,

∫E2

2i−1

p

(k

n0u(t)

)dt =

∫E2

2i

p

(k

n0u(t)

)dt,

and

(n = 1, 2, ..., i = 1, 2, ..., 2n−1).

Define

S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395 1383

un(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

u(t), t ∈ T\H(1 − 1

n0)u(t), t ∈ En

1

(1 + 1n0

)u(t), t ∈ En2

... ...

(1 − 1n0

)u(t), t ∈ En2n−1

(1 + 1n0

)u(t), t ∈ En2n ,

u′n(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

u(t), t ∈ T\H(1 + 1

n0)u(t), t ∈ En

1

(1 − 1n0

)u(t), t ∈ En2

... ...

(1 + 1n0

)u(t), t ∈ En2n−1

(1 − 1n0

)u(t), t ∈ En2n ,

and

(yn(t)

)∞n=1 =

(u1(t), u′

1(t), u2(t), u′2(t), ..., un(t), u′

n(t), ...).

Then

‖un‖0 ≤ 1k

[1 + ρM (kun)

]= 1

k

[1 +

∫T\H

M(kun(t)

)dt +

∫En

1

[p(kun(t)

)+ β

]dt +

∫En

2

[p(kun(t)

)+ β

]dt

+ · · · +∫

En2n−1

[p(kun(t)

)+ β

]dt +

∫En

2n

[p(kun(t)

)+ β

]dt

]

= 1k

[1 +

∫T\H

M(ku(t)

)dt +

∫En

1

[p

((1 − 1

n0

)ku(t)

)+ β

]dt +

∫En

2

[p

((1 + 1

n0

)ku(t)

)+ β

]dt

+ · · · +∫

En2n−1

[p

((1 − 1

n0

)ku(t)

)+ β

]dt +

∫En

2n

[p

((1 + 1

n0

)ku(t)

)+ β

]dt

]

= 1k

[1 +

∫T\H

M(ku(t)

)dt +

∫En

1

[p(ku(t)

)+ β

]dt +

∫En

2

[p(ku(t)

)+ β

]dt

+ · · · +∫

En2n−1

[p(ku(t)

)+ β

]dt +

∫En

2n

[p(ku(t)

)+ β

]dt

]

= 1k

[1 +

∫T\H

M(ku(t)

)dt +

∫En

1

M(ku(t)

)dt +

∫En

2

M(ku(t)

)dt

+ · · · +∫

En2n−1

M(ku(t)

)dt +

∫En

2n

M(ku(t)

)dt

]

≤ 1k

[1 + ρM (ku)

]≤ 1.

Similarly, ‖u′n‖0 ≤ 1. Hence ‖yn‖0 ≤ 1. This implies that yn ∈ B(L0

M ). On the other hand, we have

∞∑n=1

(12 · 1

2nun(t) + 12 · 1

2nu′n(t)

)=

∞∑n=1

12n+1

(un(t) + u′

n(t))

=∞∑

n=1

22n+1u(t) = u(t),

and

1384 S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395

∞∑n=1

(12 · 1

2n + 12 · 1

2n

)=

∞∑n=1

(12n

)= 1.

But {yn}∞n=1 is not relatively compact. In fact, since M is a convex function, there exists l > 0 such that M(v) > lv whenever v ≥ a/(kn0). Let km,n ∈ K(um − un). Then [1 + ρM (km,n(um − un))]km,n =‖um − un‖0 ≤ 2. Hence 2km,n ≥ 1. This implies that if ku(t) ∈ G, then

km,n2n0

u(t) ≥ 12

2kkn0

u(t) = 1n0k

ku(t) ≥ a

n0k.

Let Tm,n = {t ∈ H : um(t) �= un(t)}. Then

‖un − um‖0 = 1kn,m

[1 + ρM

(kn,m(un − um)

)]≥

∫Tn,m

M(kn,m 2n0

u(t))kn,m

dt

≥∫

Tm,n

M(km,n2n0

u(t))km,n

2n0

u(t)·(

2n0

u(t))dt ≥

∫Tm,n

l2n0

u(t)dt =∫

Tm,n

2lk

k

n0u(t)dt

= 2lkp

∫Tm,n

p

(k

n0u(t)

)dt = l

kp

∫H

p

(k

n0u(t)

)dt > 0

for any m �= n. Hence we obtain that the sequence {yn}∞n=1 is not relatively compact, a contradiction!The implication (a) ⇒ (b) is obvious, which completes the proof. �

Proof of Theorem 2. Since every weak∗ hyperplane of L0M (X∗) is approximatively compact, by Theorem 1,

we obtain that L0M (X∗) has the Radon–Nikodym property. This implies that the subspace L0

M of L0M (X)

has the Radon–Nikodym property. By Lemma 6, we have M ∈ Δ2.Let u be norm attainable on S(EN ), i.e.,

∫Tu(t)v0(t)dt = 1 for some v0(t) ∈ S(EN ). If u =

∑∞i=1 tiui,

then

1 =∫T

u(t) · v0(t)dt =∫T

( ∞∑i=1

tiui(t))

· v0(t)dt =∞∑i=1

ti

∫T

ui(t) · v0(t)dt,

where ui ∈ B(L0M ), ti ∈ (0, 1) and

∑∞i=1 ti = 1. Hence for any i ∈ N , we have

∫Tui(t) · v0(t)dt = 1. This

implies that {ui}∞i=1 is relatively compact by Lemma 1. By virtue of Lemma 8, we obtain that u is an extreme point. By Lemma 2, we obtain that EN is smooth. Hence q(u) is continuous. This implies that M(u) is strictly convex.

We will prove that if x∗n ∈ S(X∗), x ∈ S(X) and limn→∞ x∗

n(x) = 1, then {x∗n}∞n=1 is relatively

compact. In fact, suppose that {x∗n}∞n=1 is not relatively compact. Then there exists r > 0 such that

r = infm�=n ‖x∗n − x∗

m‖ > 0. By the Hahn–Banach theorem, there exist v ∈ S(EN ) and u ∈ S(L0M ) such

that ∫Tu(t)v(t)dt = 1. Define

−→v(t) = x · v(t), −−→un(t) = x∗n · u(t), n = 1, 2, ...

Then −→v ∈ S(EN (X)), −−→un ∈ S(L0M (X∗)) and

∫T

(−−→un(t),−→v(t))dt → 1. Hence

1 ←∫ (−−→un(t),−→v(t)

)dt = 1

kn

∫ (kn

−−→un(t),−→v(t))dt

T T

S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395 1385

≤ 1kn

(∫T

M(kn

−−→un(t))dt +

∫T

N(−→v(t)

)dt

)

= 1kn

(∫T

M(kn

−−→un(t))dt + 1

)= ‖−−→un‖0 = 1,

where kn ∈ K(−−→un). However,

‖−−→un − −−→um‖0 = infk>0

[1k

(∫T

M(k(−−→un(t) − −−→um(t)

))dt + 1

)]

= infk>0

[1k

(∫T

M(kru(t)

)dt + 1

)]

= r‖u‖0 = r > 0,

a contradiction. This implies that if x∗n ∈ S(X∗), x ∈ S(X) and x∗

n(x) → 1 as n → ∞, then {x∗n}∞n=1 is

relatively compact.We next will prove that X is smooth. Suppose that X is not smooth. Then there exist x ∈ S(X),

y∗ ∈ S(X∗) and z∗ ∈ S(X∗) such that y∗ �= z∗ and y∗(x) = z∗(x) = 1. It is easy to see that there exist v ∈ S(EN ) and u ∈ S(L0

M ) such that u(t) > 0 and ∫Tu(t)v(t)dt = 1. Decompose T into T 1

1 , T 12 such that

T 11 ∩ T 1

2 = ∅, T 11 ∪ T 1

2 = T and ∫T 11M(‖y∗ − z∗‖u(t))dt =

∫T 12M(‖y∗ − z∗‖u(t))dt. Decompose T 1

1 into T 21 ,

T 22 such that T 2

1 ∩T 22 = ∅, T 2

1 ∪T 22 = T 1

1 and ∫T 21M(‖y∗ − z∗‖u(t))dt =

∫T 22M(‖y∗ − z∗‖u(t))dt. Decompose

T 12 into T 2

3 , T 24 such that T 2

3 ∩T 24 = ∅, T 2

3 ∪T 24 = T 1

2 and ∫T 23M(‖y∗ − z∗‖u(t))dt =

∫T 24M(‖y∗ − z∗‖u(t))dt.

Generally, decompose Tn−1i into Tn

2i−1 and Hn2i such that

Tn2i−1 ∩ Tn

2i = ∅, Tn2i−1 ∪ Tn

2i = Tn−1i and∫

Tn2i−1

M(∥∥y∗ − z∗

∥∥u(t))dt =

∫Tn2i

M(∥∥y∗ − z∗

∥∥u(t))dt,

where n = 1, 2, ...., i = 1, 2, ...., 2n−1. Set

−−→un(t) =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

y∗ · u(t), t ∈ Tn1

z∗ · u(t), t ∈ Tn2

... ...

y∗ · u(t), t ∈ Tn2i−1

z∗ · u(t), t ∈ Tn2i,

−→v(t) = x · v(t), n = 1, 2, ...

It is easy to see that −−→un ∈ S(LM (X∗)), −→v ∈ S(EN (X)) and ∫T(−−→un(t), −→v(t))dt = 1. Since every weak∗

hyperplane of L0M (X∗) is approximatively compact, by Lemma 1, we obtain that {−−→un}∞n=1 is relatively

compact. Let Tm,n = {t ∈ T : −−→um(t) �= −−→un(t)}. Then

ρM (um − un) =∫T

M(∥∥−−→un(t) − −−→um(t)

∥∥)dt=

∫Tm,n

M(∥∥y∗ − z∗

∥∥u(t))dt

= 12

∫M

(∥∥y∗ − z∗∥∥u(t)

)dt > 0,

T

1386 S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395

for any m �= n. Therefore, by M ∈ Δ2, we obtain that there exists l > 0 such that ‖−−→un − −−→um‖0 ≥‖−−→un − −−→um‖ ≥ l, a contradiction. This implies that X is smooth. We have proved that if x∗

n ∈ S(X∗), x ∈ S(X) and limn→∞ x∗

n(x) = 1, then {x∗n}∞n=1 is relatively compact. This implies that x∗

n ∈ S(X∗), x ∈ S(X) and x∗

n(x) → 1 as n → ∞, then {x∗n}∞n=1 is a Cauchy sequence. Hence we obtain that X has a

Fréchet differentiable norm.(3) ⇒ (1). Let un ∈ S(L0

M (X∗)), v ∈ S(EN (X)) and limn→∞∫T

(un, v)dt = 1. By the Hahn–Banach theorem and Lemma 4, there exists u ∈ S(L0

M (X∗)) such that ∫T

(u(t), v(t))dt = 1. Hence we have ∫T

(un(t) + u(t), v(t))dt → 2 as n → ∞. The proof requires the consideration of few cases separately.Case 1. Let sup{kn} < ∞, where kn = K(un). We may assume without loss of generality that kn → l as

n → ∞. We claim that ‖un(t)‖ μ−→ ‖u(t)‖ in measure. Otherwise, we may assume without loss of generality that for each n ∈ N , there exists En ⊆ T , ε0 > 0 and σ0 > 0 such that μEn ≥ ε0, where

En ={t ∈ T :

∣∣∥∥un(t)∥∥−

∥∥u(t)∥∥∣∣ ≥ σ0

}.

We define the sets

An ={t ∈ T : M

(∥∥knun(t)∥∥) > 8d

ε0

}and B =

{t ∈ T : M

(∥∥ku(t)∥∥) > 8d

ε0

},

where k ∈ K(u) and d = max{k, sup{kn}}. Then

∥∥knun(t)∥∥ ≥ M−1

(8dε0

), t ∈ An and

∥∥ku(t)∥∥ ≥ M−1

(8dε0

), t ∈ B.

Noticing that ρM (knun) = kn − 1, we have

d ≥∫T

M(∥∥knun(t)

∥∥)dt ≥ ∫An

M(∥∥knun(t)

∥∥)dt ≥ 8dε0

μAn.

This implies that μAn ≤ ε0/8. Similarly, we have μB ≤ ε0/8. We define a bound closed set

C ={

(u, v) ∈ R2 : M(u) ≤ 8dε0

,M(v) ≤ 8dε0

, |u− v| ≥ 14σ0

}

in the two dimensional space R2. Since C is compact, there exists δ > 0 such that

M

(k

k + lu + l

k + lu

)≤ (1 − δ)

[k

k + lM(u) + l

k + lM(v)

]

for any (u, v) ∈ C. Then

M

(k

k + l

∥∥knun(t)∥∥ + l

k + l

∥∥ku(t)∥∥) ≤ (1 − δ)

[k

k + lM

(∥∥knun(t)∥∥) + l

k + lM

(∥∥ku(t)∥∥)]

for t ∈ En\(An ∪B). Let

Wn(t) =

⎧⎨⎩

M( kk+kn

‖knun(t)‖+ knk+kn

‖ku(t)‖)k

k+knM(‖knun(t)‖)+ kn

k+knM(‖ku(t)‖) , t ∈ En\(An ∪B)

0, t ∈ T\(En\(An ∪B))

and

S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395 1387

Qn(t) ={

M( kk+l‖knun(t)‖+ l

k+l‖ku(t)‖)k

k+lM(‖knun(t)‖)+ lk+lM(‖ku(t)‖) , t ∈ En\(An ∪B)

0, t ∈ T\(En\(An ∪B)).

It is easy to see that Wn(t) −Qn(t) → 0 for any t ∈ T . Therefore, by the Egorov theorem, there exist E ⊂ T

and a natural number N such that μE < ε0/16 and |Wn(t) −Qn(t)| < δ/4 whenever n > N and t ∈ T \E. Hence, if t ∈ (En\(An ∪B)) \ E, then

12δ = δ − 1

2δ

≤ 1 −M(t, k

k+l‖knun(t)‖ + lk+l‖ku(t)‖)

kk+lM(t, ‖knun(t)‖) + l

k+lM(t, ‖ku(t)‖)− 1

2δ

≤ 1 −M(t, k

k+kn‖knun(t)‖ + kn

k+kn‖ku(t)‖)

kk+kn

M(t, ‖knun(t)‖) + kn

k+knM(t, ‖ku(t)‖)

,

whenever n is large enough. This implies that

M

(k

k + kn

∥∥knun(t)∥∥ + kn

k + kn

∥∥ku(t)∥∥) ≤

(1 − 1

2δ)[

k

k + knM

(∥∥knun(t)∥∥) + kn

k + knM

(∥∥ku(t)∥∥)].

Let Hn = (En\(An ∪B)) \ E. Then 4μHn ≥ ε0 and

‖un‖0 + ‖u‖0 − ‖un + u‖0 ≥ 1kn

(1 + ρM (knun)

)+ 1

k

(1 + ρM (ku)

)− kn + k

knk

(1 + ρM

(knk

kn + k(un + u)

))

≥ kn + k

knk

∫Hn

[k

kn + kM

(∥∥knun(t)∥∥) + kn

kn + kM

(∥∥ku(t)∥∥)

−M

(k

kn + k

∥∥knun(t)∥∥ + kn

kn + k

∥∥ku(t)∥∥)]dt

≥ kn + k

knk

∫Hn

12δ

[k

kn + kM

(t,∥∥knun(t)

∥∥) + knkn + k

M(∥∥ku(t)

∥∥)]dt

≥ kn + k

knk

∫Hn

12δ

[M

(k

kn + k

∥∥knun(t)∥∥ + kn

kn + k

∥∥ku(t)∥∥)]dt

≥ kn + k

knk

∫Hn

12δ

[M

(kkn

kn + k

∣∣∥∥un(t)∥∥−

∥∥u(t)∥∥∣∣)]dt

≥ kn + k

knk

∫Hn

12δ

[M

(1

d + kσ0

)]dt

≥ 2dk

12δM

(1

d + kσ0

)μHn

≥ δε0

4dkM(

1d + k

σ0

)

whenever n is large enough. Since ∫T

(un(t) + u(t), v(t))dt → 2, we have ‖un + u‖0 → 2 as n → ∞. This implies that ‖un‖0 + ‖u‖0 − ‖un + u‖0 → 0 as n → ∞, a contradiction! Hence ‖un(t)‖ μ−→ ‖u(t)‖ in measure. By the Riesz theorem, there exists a subsequence {n} of {n} such that ‖un(t)‖ → ‖u(t)‖ μ-a.e. in T . Noticing that

1388 S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395

∣∣(un(t), v(t))∣∣ ≤ ∥∥un(t)

∥∥ ·∥∥v(t)∥∥, lim

n→∞

∫T

(un(t), v(t)

)dt = 1

and ∫T‖un(t)‖ · ‖v(t)‖dt ≤ ‖un‖0 · ‖v‖ ≤ 1, we obtain

limn→∞

∫T

∥∥un(t)∥∥ ·

∥∥v(t)∥∥dt = 1, limn→∞

∫T

[∥∥un(t)∥∥ ·

∥∥v(t)∥∥−(un(t), v(t)

)]dt = 0.

Moreover, it is easy to see that∫T

∣∣∥∥un(t)∥∥ ·

∥∥v(t)∥∥−(un(t), v(t)

)∣∣dt → 0 as n → ∞.

This implies that ‖un(t)‖ · ‖v(t)‖ − (un(t), v(t)) μ−→ 0 in measure. By the Riesz theorem, there exists a subsequence {n} of {n} such that ‖un(t)‖ · ‖v(t)‖ − (un(t), v(t)) → 0 μ-a.e. in T . By ‖un(t)‖ → ‖u(t)‖ μ-a.e. in T , it follows that (un(t), v(t)) → ‖u(t)‖ · ‖v(t)‖ μ-a.e. in T . We may assume without loss of generality that

limn→∞

(un(t)‖u(t)‖ ,

v(t)‖v(t)‖

)= 1

on {t ∈ T : ‖u(t)‖ · ‖v(t)‖ �= 0}. Moreover, μT1 = 0, where T1 = {t ∈ T : ‖v(t)‖ = 0} ∩ {t ∈ T : ‖u(t)‖ �= 0}. In fact, suppose that μT1 > 0. Then

‖u‖0 = 1k

[1 + ρM (ku)

]>

1k

[1 + ρM (kuχT\T1)

]≥ ‖uχT\T1‖0,

where k ∈ K(u). This implies that

1 =∫T

(u, v)dt =∫T

(uχT\T1 , v)dt ≤ ‖uχT\T1‖0 · ‖v‖ < 1,

a contradiction! Hence we may assume without loss of generality that

limn→∞

(un(t)‖u(t)‖ ,

v(t)‖v(t)‖

)= 1, t ∈

{t ∈ T :

∥∥u(t)∥∥ �= 0

}. (2)

Since X has a Fréchet differentiable norm, by (2), we obtain that sequence { un(t)‖u(t)‖}∞n=1 is convergent.

Hence there exists x(t) ∈ S(X) such that un(t)‖u(t)‖ → x(t) on {t ∈ T : ‖u(t)‖ �= 0}. Let

u0(t) ={ ‖u(t)‖x(t), t ∈ {t ∈ T : ‖u(t)‖ �= 0}

0, t ∈ {t ∈ T : ‖u(t)‖ = 0}.

It is easy to see that ‖u0‖0 = 1 and un(t) → u0(t) μ-a.e. in T . Next we will prove that l = h, where h ∈ K(u0) and l = limn→∞ kn. In fact, by the Fatou Lemma, it follows that

1h

[1 + ρM (hu0)

]= ‖u0‖0 = lim

n→∞‖un‖0 = lim

n→∞1kn

[1 + ρM (knun)

]≥ 1

l

[1 + ρM (lu0)

].

Hence ‖u0‖0 = [1 + ρM (hu0)]/h = [1 + ρM (lu0)]/l. Since M(u) is strictly convex, we have that L0M (R) is

strictly convex. Noticing that ‖u0‖0 = [1 + ρM (hu0)]/h = [1 + ρM (lu0)]/l, we have l = h. By the convexity of M , we have

S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395 1389

M(‖knun(t)‖) + M(‖hu0(t)‖)2 −M

(‖knun(t) − hu0(t)‖

2

)≥ 0

for μ-a.e. t ∈ T . Moreover, we have

ρM (knun) = kn‖un‖0 − 1 → h‖u0‖0 − 1 = ρM (hu0) as n → ∞.

Therefore, by the Fatou Lemma, we obtain the following

ρM (hu0) =∫T

limn→∞

[M(‖knun(t)‖) + M(‖hu0(t)‖)

2 −M

(‖knun(t) − hu0(t)‖

2

)]dt

≤ lim infn→∞

∫T

[M(‖knun(t)‖) + M(‖hu0(t)‖)

2 −M

(‖knun(t) − hu0(t)‖

2

)]dt

= ρM (hu0) − lim supn→∞

ρM

[12(knun − hu0)

].

This implies that ρM (12 (knun − hu0)) → 0 as n → ∞. Since M ∈ Δ2, we have ‖knun − hu0‖0 ≤

2‖knun − hu0‖ → 0 as n → ∞. Using limn→∞ kn = l = h, we obtain ‖un − u0‖0 → 0 as n → ∞.Case 2. Let sup{kn} = ∞, where kn = K(un). We consider the sequence u′

n = 12 (un + u) in place of

{un}∞n=1, because ‖un − u‖0 → 0 as n → ∞ if and only if ‖u′n − u‖0 → 0 as n → ∞. Moreover,

∥∥∥∥12(un + u)

∥∥∥∥0

≤ 12(‖un‖0 + ‖u‖0)

for every n ∈ N . Hence lim supn→∞ ‖12 (un + u)‖0 ≤ 1. Since

∫T

(12(un(t) + u(t)

), v(t)

)dt = 1

2

∫T

(un(t), v(t)

)dt + 1

2

∫T

(u(t), v(t)

)dt → 1,

we conclude that lim infn→∞ ‖12 (un + u)‖0 ≥ 1. Consequently, limn→∞ ‖1

2(un + u)‖0 = 1. Define wn =2knk(kn + k), where k ∈ K(u). Then the sequence {wn}∞n=1 is bounded. Moreover,

∥∥∥∥12(un + u)

∥∥∥∥0

≤ 1wn

[1 + ρM

(wn · un + u

2

)]

= kn + k

2knk

[1 + ρM

(knk

kn + k(un + u)

)]

≤ kn + k

2knk

[1 + ρM

(k

kn + k(knun)

)+ ρM

(kn

kn + k(ku)

)]

≤ 12

[1kn

(1 + ρM (knun)

)+ 1

k

(1 + ρM (ku)

)]

= 12[‖un‖0 + ‖u‖0] → 1 as n → ∞,

whence it follows that

kn + k[1 + ρM

(2knk · 1(un + u)

)]→ 1 as n → ∞.

2knk kn + k 2

1390 S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395

Since∫T

(u(t), v(t))dt = 1 and ∫T

(un(t), v(t))dt → 1, we have ∫T

(u′n(t), v(t))dt → 1. Therefore, we can

prove in the same way as in Case 1 that ‖u′n − u‖0 → 0. So {un}∞n=1 is relatively compact. Hence EN (X)

has a Fréchet differentiable norm.By Lemma 1, the implication (1) ⇒ (2) is obvious, which completes the proof. �

Lemma 9. (See [1].) For any N -function M and ε > 0, there exists a strictly convex N -function M1 such that

M(u) ≤ M1(u) ≤ (1 + ε)M(u), u ∈ R.

Lemma 10. (See [1].) For any N -function M and ε > 0, there exists an N -function M1 such that

M(u) ≤ M1(u) ≤ (1 + ε)M(u), u ∈ R,

and its right derivative p1 is continuous.

Lemma 11. (See [1].) (1) L0M has the Radon–Nikodym property if and only if M ∈ Δ2; (2) EN is an Asplund

space if and only if M ∈ Δ2.

Theorem 3. Let X be a Fréchet differentiable space. Then

(1) L0M (X∗) has the Radon–Nikodym property if and only if M ∈ Δ2;

(2) EN (X) is an Asplund space if and only if M ∈ Δ2.

Proof. Let M ∈ Δ2. By Lemma 9, there exists a strictly convex N -function M1 such that

M(u) ≤ M1(u) ≤ 2M(u), u ∈ R.

Since M ∈ Δ2, there exist K > 2 and u0 ≥ 0 such that M(2u) ≤ KM(u) whenever u ≥ u0. This implies that

M1(2u) ≤ 2M(2u) ≤ 2KM(u) ≤ 2KM1(u)

whenever u ≥ u0. Therefore M1 ∈ Δ2. By Theorem 2, we obtain that every weak∗ hyperplane of (L0M1

(X∗),‖ · ‖0

1) is approximatively compact. So (L0M1

(X∗), ‖ · ‖01) has the Radon–Nikodym property by Theorem 1.

Moreover, we have

‖u‖0 = infk>0

1k

[1 +

∫T

M(∥∥ku(t)

∥∥)dt] ≤ infk>0

1k

[1 +

∫T

M1(∥∥ku(t)

∥∥)dt] = ‖u‖01

and

‖u‖01 = inf

k>0

1k

[1 +

∫T

M1(∥∥ku(t)

∥∥)dt] ≤ infk>0

1k

[1 +

∫T

2M(∥∥ku(t)

∥∥)dt] ≤ 2‖u‖0

for any u ∈ L0M (X∗). This means that L0

M (X∗) has the Radon–Nikodym property.

S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395 1391

Since M ∈ Δ2, then for any l > 1, there exist v0 > 0 and δ > 0 such that N(lv) ≥ (l + δ)N(v) whenever v ≥ v0. By Lemma 10, there exist δ1 = min{δ/4, δ/(4l)} > 0 and an N -function N2 such that

N(v) ≤ N2(v) ≤ (1 + δ1)N(v), v ∈ R,

and its right derivative p2 is continuous. Then M2 is a strictly convex function. Since δ1 = min{δ/4, δ/4l}, we have

δ2 = δ − lδ11 + δ1

≥ δ − l(δ/4l)1 + δ1

> 0

and

N2(lv) ≥ N(lv) ≥ (l + δ)N(v) ≥ (l + δ) 11 + δ1

N2(v) =(l + δ − lδ1

1 + δ1

)N2(v)

= (l + δ2)N2(v)

whenever v ≥ v0. This implies that M2 ∈ Δ2. Therefore, by Theorem 2, we obtain that every weak∗

hyperplane of (L0M2

(X∗), ‖ · ‖02) is approximatively compact. So (EN2(X), ‖ · ‖2) is an Asplund space by

Theorem 1. Moreover, we have ‖v‖0 ≤ ‖v‖02 and

‖v‖02 = inf

k>0

1k

[1 +

∫T

N2(∥∥kv(t)∥∥)dt] ≤ inf

k>0

1k

[1 +

∫T

(1 + δ1)N(∥∥kv(t)∥∥)dt]

≤ (1 + δ1) infk>0

1k

[1 +

∫T

N(∥∥kv(t)∥∥)dt] ≤ (1 + δ1)‖v‖0

for any v ∈ EN (X). Then

‖v‖ ≤ ‖v‖0 ≤ ‖v‖02 ≤ 2‖v‖2 ≤ 2‖v‖0

2 ≤ 2(1 + δ1)‖v‖0 ≤ 4(1 + δ1)‖v‖.

Since ‖v‖ ≤ 2‖v‖2 ≤ 4(1 + δ1)‖v‖ and (EN2(X), ‖ · ‖2) is an Asplund space, we obtain that EN (X) is an Asplund space.

If L0M (X∗) has the Radon–Nikodym property, then L0

M has the Radon–Nikodym property. Therefore, by Lemma 11, we have M ∈ Δ2. If EN (X) is an Asplund space, then EN is an Asplund space. Therefore, by Lemma 11, we have M ∈ Δ2, which completes the proof. �Corollary 2. (1) L0

M has the Radon–Nikodym property if and only if M ∈ Δ2; (2) EN is an Asplund space if and only if M ∈ Δ2.

4. Applications to the theory of generalized inverses

Let T be a linear bounded operator from X into Y . Let D(T ), R(T ) and N(T ) denote the domain, range and null space of T , respectively. If N(T ) �= {0} or R(T ) �= Y , the operator equation Tx = y is generally ill-posed. In applications, one usually looks for the best approximate solution (b.a.s.) to the equation (see [15]).

A point x0 ∈ D(T ) is called the best approximate solution to the operator equation Tx = y, if

‖Tx0 − y‖ = inf{‖Tx− y‖ : x ∈ D(T )

}

1392 S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395

and

‖x0‖ = min{‖v‖ : v ∈ D(T ), ‖Tv − y‖ = inf

x∈D(T )‖Tx− y‖

},

where y ∈ Y (see [15]).Nashed and Votruba [15] introduced the concept of the (set-valued) metric generalized inverse T .

Definition 8. Let X, Y be Banach spaces and T be a linear operator from X to Y . The set-valued mapping T ∂ : Y → X defined by

T ∂(y) ={x0 ∈ D(T ) : x0 is a best approximation solution to T (x) = y

}for any y ∈ D(T ∂), is called the (set-valued) metric generalized inverse of T , where

D(T ∂

)=

{y ∈ Y : T (x) = y has a best approximation solution in X

}.

Definition 9. A Banach space X is said to have the H property whenever if {xn}∞n=1 ⊂ S(X), x ∈ S(X)and xn

w−−→ x, then xn → x.

During the last three decades, the linear generalized inverses of linear operators in Banach spaces and their applications have been investigated by many authors (see [12,23–25]). In this section, we will investigate continuity of the metric generalized inverse of a linear operator.

Theorem 4. Let X be a smooth spaces, T : X → Y a linear bounded operator, N(T ) a one-dimensional subspace, N(T ∗) a Chebychev set and R(T ) a closed subspace. Then D((T ∗)∂) = X∗ and (T ∗)∂ is single-valued. Moreover, if every weak∗ hyperplane of X∗ is approximatively compact and Y ∗ is approximatively compact, then (T ∗)∂ is continuous.

In order to prove this theorem, we give first some lemmas.

Lemma 12. Let X be a smooth space. Then for any x ∈ S(X), if x∗n ∈ S(X∗) and x∗

n(x) → 1, then there exists a supporting functional x∗ of x such that x∗

nw∗−−→ x∗ as n → ∞.

Proof. Since X is a smooth space, we obtain that B(X∗) is w∗ sequentially compact. Hence, if x∗n(x) → 1,

then there exist x∗ ∈ B(X∗) and a subsequence {x∗nk}∞k=1 of {x∗

n}∞n=1 such that x∗nk

w∗−−→ x∗ and x∗(x) = 1. This implies that x∗ is a supporting functional of x. We claim that x∗

nw∗−−→ x∗ as n → ∞. Otherwise, there

exist ε0 > 0, y ∈ X and a subsequence {nl} of {n} such that |x∗nl

(y) − x∗(y)| > ε0. Since x∗nl

(x) → 1, there exists a subsequence {ni} of {nl} such that x∗

ni

w∗−−→ x∗ as i → ∞, a contradiction, which completes the proof. �Lemma 13. Let X be a smooth space. Then every weak∗ hyperplane of X∗ is a Chebychev set.

Proof. By [22], we know that every weakly∗ closed convex set is a proximinal set. Let H∗ = {x∗ ∈ X∗ :x∗(x) = k} be a weak∗ hyperplane. For any x∗ /∈ H∗, we may assume without loss of generality that x∗ = 0. Suppose that there exists z∗1 �= z∗2 such that ‖z∗1‖ = ‖z∗2‖ = dist(0, H∗). Let {y∗n}∞n=1 = (z∗1 , z∗2 , z∗1 , z∗2 , ...). We claim that {y∗n}∞n=1 is a weakly∗ Cauchy sequence. In fact, since H∗ is a weakly∗ closed set, we have dist(0, H∗) = r > 0. Pick y∗0 ∈ PH∗(0). Then

r = dist(0, PH∗(0)

)=

∥∥y∗0∥∥, Br(0) ∩H∗ = ∅, Br(0) ∩H∗ = PH∗(0).

S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395 1393

We may assume without loss of generality that k ≤ 0. Hence

k = sup{x(y∗)

: y∗ ∈ H∗} ≤ inf{x(y∗)

: y∗ ∈ Br(0)}

= −‖x‖ ·∥∥y∗0∥∥.

In fact, suppose that there exists y∗0 ∈ Br(0) such that x(y∗0) < k. Then there exists λ ∈ (0, 1) such that x(λy∗0) = k. It is easy to see that λy∗0 ∈ Br(0) and λy∗0 ∈ H∗, a contradiction. Since y∗0 ∈ PH∗(0) ⊂ H∗, we have

−‖x‖ ·∥∥y∗0∥∥ ≤ x

(y∗0)≤ sup

{x(y∗)

: y∗ ∈ H∗}≤ inf

{x(y∗)

: y∗ ∈ Br(0)}

= −‖x‖ ·∥∥y∗0∥∥.

This implies that

−‖x‖ ·∥∥y∗0∥∥ = x

(y∗0)

= sup{x(y∗)

: y∗ ∈ H∗}.This means that the inequality x(y∗0) ≥ x(y∗n) holds. Therefore

∥∥0 − y∗0∥∥ = x

(0 − y∗0

)≤ x

(0 − y∗n

)≤

∥∥0 − y∗n∥∥ → dist

(0, H∗) =

∥∥0 − y∗0∥∥ =

∥∥y∗0∥∥.Hence

∥∥y∗n∥∥ →∥∥y∗0∥∥ as n → ∞

and

x(0 − y∗n

)→

∥∥0 − y∗0∥∥ as n → ∞.

Furthermore, we have

x

(− y∗n‖y∗n‖

+ y∗n‖y∗0‖

)=

(1

‖y∗0‖− 1

‖y∗n‖

)· x

(y∗n

)→ 0 as n → ∞,

which shows that

x

(− y∗n‖y∗n‖

)→ 1 as n → ∞.

By Lemma 12, we obtain that {y∗n}∞n=1 is a weakly∗ Cauchy sequence. However, it is easy to see that {y∗n}∞n=1 = (z∗1 , z∗2 , z∗1 , z∗2 , ...) is not a weakly∗ Cauchy sequence, a contradiction. This implies that every weak∗ hyperplane of X∗ is a Chebychev set, which completes the proof. �Proof of Theorem 4. Since R(T ) is closed, we obtain that R(T ∗) = N⊥ = {x∗ ∈ X∗ : x∗(x) = 0, x ∈ N(T )}by the closed range theorem. Since dimN(T ) = 1, there exists x ∈ S(X) ∩N(T ) such that R(T ∗) = {x∗ :x∗(x) = 0}. This implies that R(T ∗) is a weakly∗ hyperplane of X∗. By Lemma 13, we obtain that R(T ∗) is a Chebychev set. Let x∗ ∈ X∗. Then there exists a unique z∗ ∈ R(T ∗) such that PR(T∗)(x∗) = {z∗}. Hence there exists y∗ ∈ Y ∗ such that (T ∗)−1(z∗) = y∗ + N(T ∗). Since N(T ∗) be a Chebychev set, we obtain that D((T ∗)∂) = X∗ and (T ∗)∂ is single-valued.

Let y∗n = (T ∗)∂(x∗n), y∗0 = (T ∗)∂(x∗

0) and x∗n → x∗

0. It is easy to see that T ∗y∗n = PR(T∗)(x∗n) and

T ∗y∗0 = PR(T∗)(x∗0). Since every weak∗ hyperplane of X∗ is approximatively compact and R(T ∗) is a

1394 S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395

Chebychev set, we obtain that PR(T∗) is continuous. Then T ∗y∗n → T ∗y∗0 as n → ∞. Since T is a linearbounded operator, we obtain that N(T ∗) is a closed subspace. Put

T ∗ : Y ∗/N(T ∗) → R

(T ∗), T ∗

[y∗]

= T ∗(y∗).By the inverse operator theorem, we have

[y∗n

]= T ∗−1(PR(T∗)

(x∗n

))→ T ∗−1(PR(T∗)

(x∗

0))

=[y∗0]

as n → ∞. (3)

This implies that ‖[y∗n]‖ → ‖[y∗0 ]‖ as n → ∞. Since

y∗n =(T ∗)∂(x∗

n

),

∥∥[y∗n]∥∥ = infz∗∈N(T∗)

∥∥x∗n + z∗

∥∥, T ∗[x∗n

]= T ∗(y∗n + z∗

)= PR(T∗)

(y∗n

),

y∗0 =(T ∗)∂(x∗

0),

∥∥[x∗0]∥∥ = inf

z∗∈N(T∗)

∥∥x∗n + z∗

∥∥, T ∗[x∗

0]

= T ∗(x∗0 + z∗

)= PR(T∗)

(y∗0),

we have ‖[y∗n]‖ = ‖y∗n‖ and ‖[y∗0 ]‖ = ‖y∗0‖. Noticing that ‖[y∗n]‖ → ‖[y∗0 ]‖, we have ‖y∗n‖ → ‖y∗0‖. We will derive limn→∞ (T ∗)∂(x∗

n) = (T ∗)∂(x∗0) for each of the following two cases.

Case 1. y∗0 = 0. By ‖y∗n‖ → ‖y∗0‖, we have ‖y∗n‖ → 0 as n → ∞. This implies that (T ∗)∂(x∗n) → (T ∗)∂(x∗

0)as n → ∞.

Case 2. y∗0 �= 0. By x∗n → x∗

0, we obtain that the sequence {x∗n}∞n=1 is a bounded sequence. Since

‖x∗n − PR(T∗)(x∗

n)‖ → dist(x∗0, R(T ∗)), we obtain that {PR(T∗)(x∗

n)}∞n=1 is a bounded sequence. Since T ∗−1

is a linear bounded operator, we obtain that {[y∗n]}∞n=1 is a bounded sequence. By ‖[y∗n]‖ = ‖y∗n‖, we obtain that {y∗n}∞n=1 is a bounded sequence. Since Y ∗ is approximatively compact, Y ∗ is reflexive. Hence there exists a subsequence {y∗nk

}∞k=1 of {y∗n}∞n=1 such that y∗nk

w−−→ y∗ as k → ∞. This implies that T ∗y∗nk

w−−→ T ∗y∗ as k → ∞. By T ∗y∗n → T ∗y∗0 , we have T ∗y∗0 = T ∗y∗. By the Hahn–Banach theorem, there exists y∗∗ ∈ S(X∗∗)such that y∗∗(y∗) = ‖y∗‖. Then

∥∥y∗∥∥ = y∗∗(y∗)

= limk→∞

y∗∗(y∗nk

)≤ lim

k→∞

∥∥y∗nk

∥∥ = limk→∞

∥∥[y∗nk

]∥∥ =∥∥[y∗0]∥∥ =

∥∥y∗0∥∥.Hence ‖y∗‖ = ‖y∗0‖. Noticing that y∗0 = (T ∗)∂(x∗

0), we have y∗0 = y∗. Then y∗nk

w−−→ y∗0 as k → ∞. Since X∗

is approximatively compact, by [11], we obtain that X∗ has the H property. Therefore, by y∗nk

w−−→ y∗0 and ‖y∗nk

‖ → ‖y∗0‖, we have y∗nk→ y∗0 as k → ∞. Then y∗n → y∗0 as n → ∞. In fact, suppose that there exist

ε > 0 and a subsequence {ni} of {n} such that ‖y∗ni− y∗0‖ > ε. Then there exists a subsequence {nl} of

{ni} such that y∗nl→ y∗0 as l → ∞, a contradiction.

By Case 1 and Case 2, we have (T ∗)∂(x∗n) → (T ∗)∂(x∗

0) as n → ∞, which completes the proof. �Lemma 14. (See [7].) L0

M (X) is approximatively compact if and only if

(1) M ∈ Δ2 and N ∈ Δ2;(2) M(u) is strictly convex;(3) X is approximatively compact and rotund.

Theorem 5. Suppose that every weak∗ hyperplane of L0M1

(X∗) is approximatively compact, L0M2

(Y ∗) is approximatively compact and T is a linear bounded operator from EN1(X) to EN2(Y ). Then if N(T ) is a one-dimensional subspace and R(T ) is a closed subspace, then (T ∗)∂ is single-valued and continuous, and D((T ∗)∂) = L0

M1(X∗).

Proof. By Lemma 14 and Theorem 2, we obtain that if L0M2

(Y ∗) is approximatively compact, then EN2(Y )has a Fréchet differentiable norm and is reflexive. This implies that L0

M (Y ∗) is rotund. Hence N2(T ∗) is a

2

S. Shang, Y. Cui / J. Math. Anal. Appl. 421 (2015) 1377–1395 1395

Chebychev set. Moreover, by Theorem 2, we obtain that EN1(X) has a Fréchet differentiable norm. Then EN1(X) is smooth. Therefore, by Theorem 4, we obtain that D((T ∗)∂) = L0

M1(X∗), (T ∗)∂ is single-valued

and continuous, which completes the proof. �References

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