Locally nonconical unit balls in Orlicz spaces

Ryszard Grząślewicz , Witold Seredyński

Locally nonconical unit balls in Orlicz spaces

Abstract. The aim of this paper¹ is to investigate the local nonconicality of unit ball in Orlicz spaces, endowed with the Luxemburg norm. A closed convex set Q in a locally convex topological Hausdorff space X is called locally nonconical (LNC), if for every x, y ∈ Q there exists an open neighbourhood U of x such that (U ∩ Q) + (y− x)/2 ⊂ Q. The following theorem is established: An Orlicz space L^ϕ(µ)has an LN C unit ball if and only if either L^ϕ(µ)is finite dimensional or the measure µ is atomic with a positive greatest lower bound and ϕ satisfies the condition ∆⁰_r(µ)and is strictly convex on the interval [0, b], or c(ϕ) = +∞ and ϕ satisfies the condition

∆2(µ)and is strictly convex onR. A similar result is obtained for the space E^ϕ(µ).

2000 Mathematics Subject Classification: 52Axx, 46Axx,46Cxx.

Key words and phrases: stable convex set.

1. Introduction. A convex set Q of a real Hausdorff topological vector space X is called locally nonconical (LN C), if for every x, y ∈ Q there exists an open neighbourhood U of x, such that (U ∩ Q) + (y − x)/2 ⊂ Q, cf. [1], [3], [19], [20].

LN C sets are new class of convex sets first considered by N. Weaver, who showed that the range of finite, nonatomic vector measure taking values in R

ⁿ

has the LNC property (cf. Theorem 2.7, [19]). It is proved, that intersection and product LNC sets are LNC sets in [1]. Convex sets on the plane, stricly convex or open convex sets, polytypes, cylinders, zonoides (images of vector measures),the unit ball in c

0

, (but not the unit ball in l

^∞

) belong to the class of LNC sets. The classical cone is an example of a convex closed set which is not LNC. It is proved in [20], that any LN C set is stable, (i. e. the midpoint map Φ : Q × Q, Φ(x, y) = (x + y)/2 is open with respect to the inherited topology in Q).

1Supported by a grant from the KBN.

The aim of this paper is to investigate the local nonconicality of the unit ball B(L

^ϕ

(µ)) (and B(E

^ϕ

(µ))) of Orlicz spaces L

^ϕ

(µ) (and E

^ϕ

(µ)) of functions defined on an arbitrary σ–finite measure space, endowed with the Luxemburg norm.

2. Basic definitions and auxiliary results. Let (Ω, Σ, µ) be a measure space with a nonnegative, σ–finite and complete measure µ (µ(Ω) > 0) and ϕ: R → [0, +∞] be a convex, even function which is non-identically equal to 0 and left- continuous for t > 0, such that ϕ(0) = 0, c(ϕ) := sup{ t > 0 : ϕ(t) < ∞ } > 0.

Such functions are called Young functions. This definition is somewhat stronger than the one used in [17]. We use the notation a(ϕ) := sup{ t : ϕ(t) = 0 }. By an Orlicz space L

^ϕ

(µ) ( [14], [15], [17]), we mean the set of all measurable functions x : Ω → R, such that I

^ϕ

(λx) < ∞ for some λ > 0, where the modular I

ϕ

is defined by

I

ϕ

(x) := Z

Ω

ϕ (x(ω)) dµ.

L

^ϕ

(µ) is equipped with the Luxemburg norm [13]

kxk

^ϕ

:= inf { λ > 0 : I

^ϕ

(x/λ) ≤ 1 } .

(Note that kxk

^ϕ

≤ 1 iff I

^ϕ

(x) ≤ 1; I

ϕ

(x) = 1 implies kxk

ϕ

= 1; I

ϕ

(x) < 1 ⇒ (kxk

ϕ

= 1 iff I

ϕ

(λx) = +∞ for every λ > 1); kx

n

− xk

^ϕ

→ 0 iff I

^ϕ

(λ(x

n

− x)) → 0 for every λ > 0). If atomless part of µ is positive, then the subspace

E

^ϕ

(µ) := { x ∈ M : ∀λ > 0 I

ϕ

(λx) < +∞ } .

is called the space of finite elements, where M is the set of all measurable functions x: Ω → R. If µ is purely atomic then definition of the finite elements is

E

^ϕ

(µ) :=

(

(x

n

) : ∀λ > 0∃n

λ

∈ N X

∞ n=nλ

ϕ(λx

n

) < +∞

) .

Note, that for c(ϕ) = ∞ both definitions are equivalents, but for L

^ϕ

(µ) = l

^∞

we have E

^ϕ

(µ) = c

0

(cf.[22], p. 489).

Let r > 1. The function ϕ is said to satisfy the condition ∆

r

(µ), cf. [21], [23]

(denoted ϕ ∈ ∆

^r

(µ)) if one of the following three conditions is satisfied:

(a) µ is atomless and there exist constants c > 1 and a

0

≥ 0 such that ϕ(a

⁰

) < +∞, (or in the case µ(Ω) = +∞ then a

⁰

= 0), such that for every t ≥ a

0

, we have ϕ(rt) ≤ cϕ(t);

b) when µ is purely atomic measure with { e

ⁿ

: n ∈ N }, N ⊂ N, being the set of all atoms of Ω and there exist b > 0, c > 1 and a nonnegative sequence (d

n

) such that P

n

d

n

< + ∞, and ϕ(rt)µ(e

ⁿ

) ≤ cϕ(t)µ(e

n

) + d

n

for every t with ϕ(t)µ(e

n

) ≤ b and every n ∈ N;

c) a combination of a) and b) when Ω has both an atomless part Ω

1

and purely

atomic part Ω

2

of positive measure, such that ϕ ∈ ∆

^r

(µ|Ω

1

) and ϕ ∈ ∆

r

(µ|Ω

2

).

If c(ϕ) = ∞, then ϕ ∈ ∆

^r

(µ) for some r > 1 ⇐⇒ ϕ ∈ ∆

r

(µ) for every r > 1 ⇐⇒ ϕ ∈ ∆

²

(µ).

These equivalences remain true if µ is atomless (in this case ϕ ∈ ∆

^r

(µ) for some r > 1 implies that c(ϕ) = ∞). If µ is purely atomic measure with P

n

µ(e

n

) = ∞ and ϕ ∈ ∆

^r

(µ) for some r > 1, then ϕ(x) = 0 ⇔ x = 0 (indeed, d

n

≥ ϕ(ra(ϕ))µ(e

ⁿ

) for every n ∈ N). Thus these equivalences hold in the case of a purely atomic measure µ with an infinite number of atoms provided 0 < inf

n

µ(e

n

) ≤ sup µ(e

ⁿ

) < ∞ — no matter whether ϕ takes only finite values or not (if ϕ ∈ ∆

^r0

(µ), then evidently ϕ ∈ ∆

^r

(µ) for every 1 < r ≤ r

0

; for r > r

0

, consider b

r

= ϕ(a

⁰

r

0

/r) · inf

ⁿ

µ(e

n

) > 0, where a

⁰

= sup{ a > 0 : ϕ(a) ≤ b

r0

/ sup

_n

µ(e

n

) } > 0). If dim L

^ϕ

(µ) < ∞ (i. e. , Ω consists of a finite number of atoms), then ϕ ∈ ∆

^r

(µ) for some r > 1 if and only if L

^ϕ

(µ) is not isometric to L

^∞

(µ) (take any a

0

∈ (a(ϕ), c(ϕ)), 1 < r < c(ϕ)/a

⁰

and set b = ϕ(a

0

)·inf

n

µ(e

n

) > 0, d

n

= ϕ(ra

0

)·sup

n

µ(e

n

) < ∞). However, if 0 < a(ϕ) ≤ c(ϕ) < ∞, then ϕ does not satisfy the condition ∆

^r

(µ) for any r > c(ϕ)/a(ϕ).

Note that if c(ϕ) = ∞ and L

^ϕ

(µ) is finite dimensional, then L

^ϕ

(µ) = E

^ϕ

(µ).

If c(ϕ) = ∞ and dim L

^ϕ

(µ) = ∞, the equality L

^ϕ

(µ) = E

^ϕ

(µ) holds if and only if ϕ ∈ ∆

²

(µ) (cf. [14, Theorem 8. 13, p. 52]). Thus, applying the Lebesgue dominated convergence theorem, we obtain

(I

ϕ

(x) = 1 ⇐⇒ kxk

ϕ

= 1) if and only if ϕ ∈ ∆

²

(µ).

In fact, we can replace the condition ∆

2

(µ) by ∆

r

(µ) for some r > 1 in the last equivalence. The assumption c(ϕ) = ∞ is used only in the ”if” part of the proof.

Hence, in any case, it follows that if ϕ / ∈ ∆

^r

(µ) for any r > 1, then there exists x ∈ L

^ϕ

(µ) such that kxk = 1, but I

ϕ

< 1 and this is what we need in the sequel.

Let { e

ⁿ

: n ∈ N }, N ⊂ N, be the set of all atoms of Ω and let r > 1. We say that a function ϕ satisfies the condition ∆

⁰_r

(µ)(on Ω) — denoted ϕ ∈ ∆

⁰r

(µ) — if one of the following conditions is satisfied

(a) when the atomless part of Ω is of positive measure there exist a

0

> 0 and c > 1 such that 0 < ϕ(a

0

) < ∞ and ϕ(rt) ≤ cϕ(t) for every |t| ≤ a

0

:

(b) when µ is purely atomic there exist a

0

> 0, b > 0, c > 1 and a nonnegative sequence (d

n

) such that P

_n

d

n

< + ∞, 0 < ϕ(a

⁰

) < ∞ and

ϕ(rt)µ(e

n

) ≤ cϕ(t)µ(e

n

) + d

n

for every |t| ≤ a

⁰

with ϕ(t)µ(e

n

) ≤ b and every n ∈ N.

If ϕ ∈ ∆

⁰r

(µ) for some r > 1 on the atomless part of Ω which is of positive measure, then evidently, ϕ ∈ ∆

⁰r

(µ) on the whole set Ω. Furthermore, if the measure of the atomless part of Ω is either infinite or equal to zero and ϕ ∈ ∆

^r

(µ) for some r > 1, then ϕ ∈ ∆

⁰r

(µ). Thus ϕ ∈ ∆

⁰r

(µ) for some r > 1 when dim L

^ϕ

(µ) < ∞ and L

^ϕ

(µ) is not isometric to L

^∞

(µ).

If ϕ ∈ ∆

⁰r

(µ) for some r > 1 and kxk

∞

< c(ϕ), then I

ϕ

(x) = 1 ⇐⇒ kxk

ϕ

= 1.

(see [23, p. 509]). Note that when ϕ takes only finite values, ϕ ∈ ∆

⁰r

(µ) for some r > 1 iff ϕ ∈ ∆

⁰2

(µ). Also, analogously to ∆

r

(µ), if µ is purely atomic measure with P

n

µ(e

n

) = ∞ and ϕ ∈ ∆

r

(µ) for some r > 1, then a(ϕ) = 0.

We will use the following characterization of stability in Orlicz spaces with the Luxemburg norm.

Theorem 2.1 [23, Theorem 5, p. 511] B(L

^ϕ

(µ)) is stable if and only if at least one of the following conditions is satisfied:

(i) dim L

^ϕ

(µ) < +∞;

(ii) L

^ϕ

(µ) is isometric to L

^∞

(µ);

(iii) ϕ satisfies the condition ∆

r

(µ) for some r > 1;

(iv) ϕ satisfies the condition ∆

⁰r

(µ) for some r > 1 provided c(ϕ) < +∞ and ϕ(c(ϕ)) < + ∞;

(v) ϕ satisfies ∆

⁰_r

(µ) for some r > 1 on the purely atomic part of Ω whenever c(ϕ) < + ∞, ϕ(c(ϕ)) < +∞ and the measure of the atomless part of Ω is finite;

(vi) c(ϕ) < +∞, ϕ(c(ϕ)) < +∞ and µ(Ω) < +∞.

We need the following six technical Lemmas.

Lemma 2.2 Let ϕ: R → [0, +∞] be any Young function. Set N := N or N :=

{ 1, 2, . . . , N }. Let (x

ⁱ

)

i∈N

, (y

i

)

i∈N

, (ε

i

)

i∈N

, (c

i

)

i∈N

be sequences of real numbers satisfying the following conditions:

1. c

i

> 0 for i ∈ N, 2. P

i∈N

ϕ(λx

i

+ (1 − λ)y

i

)c

i

= 1 for every 0 ≤ λ ≤ 1,

3. the set J := { i ∈ N : x

ⁱ

6= y

ⁱ

∧ (|x

ⁱ

| > a(ϕ)/2 ∨ |y

ⁱ

| > a(ϕ)/2) } is finite and for each i ∈ J the following conditions are satisfied:

(a) if x

i

6= 0, then |ε

ⁱ

| ≤ |x

ⁱ

| and if y

ⁱ

6= 0, then |ε

ⁱ

| ≤ |y

ⁱ

| (b) |ε

ⁱ

| < |y

ⁱ

− x

ⁱ

|/2

(c) |ε

ⁱ

| < min{ |(x

ⁱ

+ y

i

)/2|, a(ϕ) − |(x

ⁱ

+ y

i

)/2| }, as long as the right-hand side is positive

(d) |ε

ⁱ

| < a(ϕ)/2, as long as a(ϕ) > 0.

Let

F (t) := X

i∈N

ϕ (x

i

+ t(y

i

− x

ⁱ

)/2 + ε

i

) c

i

. Then the function F is nonincreasing on the interval [0, 1].

Proof Note that for λ = 0 and λ = 1 from Condition 2 we obtain: P

i∈N

ϕ(x

i

)c

i

= P

i∈N

ϕ(y

i

)c

i

= 1. Furthermore, for 0 < λ < 1, 1 = X

i∈N

ϕ (λx

i

+ (1 − λ)y

i

) c

i

≤ λ X

i∈N

ϕ(x

i

)c

i

+ (1 − λ) X

i∈N

ϕ(y

i

)c

i

= 1,

which means that for x

i

6= y

ⁱ

the function ϕ is affine on [x

i

, y

i

]. The numbers x

i

, y

i

may be of opposite signs only if ϕ(x

i

) = ϕ(y

i

) = 0, i.e. when |x

i

|, |y

ⁱ

| ≤ a(ϕ).

Hence, we can assume the following

either x

i

, y

i

≥ a(ϕ) ≥ 0 or x

ⁱ

, y

i

∈ [−a(ϕ), a(ϕ)], or x

ⁱ

, y

i

≤ −a(ϕ) ≤ 0 for all i ∈ N. Let k

ⁱ

be the gradient of ϕ on the segment [x

i

, y

i

] for i ∈ J. Then

ϕ ((x

i

+ y

i

)/2 + ε

i

) = ϕ ((x

i

+ y

i

)/2) + k

i

ε

i

for i ∈ J . Let k

i

(t) ≡ 0 if ε

i

= 0, k

i

(t) ≡ +∞ if ϕ (x

i

+ t(y

i

− x

ⁱ

)/2 + ε

i

) = +∞ and

k

i

(t) = ϕ (x

i

+ t(y

i

− x

ⁱ

)/2 + ε

i

) − ϕ (x

i

+ t(y

i

− x

ⁱ

)/2) ε

i

. in all other cases for i ∈ J, 0 ≤ t ≤ 1 . So k

ⁱ

(t) satisfy

ϕ (x

i

+ t(y

i

− x

ⁱ

)/2 + ε

i

) = ϕ (x

i

+ t(y

i

− x

ⁱ

)/2) + k

i

(t)ε

i

.

Note that if |x

ⁱ

|, |y

ⁱ

| ≤ a(ϕ)/2, then k

ⁱ

(t) = 0 and both sides of the above equality are equal to zero. Consider two cases A) x

i

< y

i

, B) y

i

< x

i

:

A) If ε

_i

> 0, then x

i

≤ x

ⁱ

+ t(y

i

− x

ⁱ

)/2 + ε

i

≤ (x

ⁱ

+ y

i

)/2 + ε

i

≤ (x

ⁱ

+ y

i

)/2 + (y

i

− x

ⁱ

)/2 = y

i

, so k

i

(t) ≡ k

ⁱ

.

If ε

i

< 0, then

k

i

(t) = ϕ (x

i

+ t(y

i

− x

ⁱ

)/2) − ϕ (x

i

+ t(y

i

− x

ⁱ

)/2 − (−ε

i

))

−ε

ⁱ

.

It is easy to see, that for any convex function ψ and positive constant c the function ψ(t + c) − ψ(t) is nondecreasing. Hence, the function k

ⁱ

(t) is nondecreasing

B) If ε

i

< 0, then y

i

= (x

i

+ y

i

)/2 − (x

i

− y

ⁱ

)/2 ≤ (x

i

+ y

i

)/2 + ε

i

≤ x

ⁱ

+ t(y

i

− x

i

)/2 + ε

i

< x

i

+ t(y

i

− x

ⁱ

)/2 ≤ x

i

, so k

i

(t) ≡ k

i

.

If ε

i

> 0, then

k

i

(t) = ϕ (x

i

− t(x

ⁱ

− y

ⁱ

)/2 + ε

i

) − ϕ (x

i

− t(x

ⁱ

− y

ⁱ

)/2) ε

i

. As above, the function k

i

(t) is nonincreasing.

In all cases the function:

g

i

(t) := k

i

(t)ε

i

c

i

is nonincreasing. Let:

I := { i : x

ⁱ

= y

i

}, K := { i : x

ⁱ

6= y

ⁱ

∧ |x

ⁱ

|, |y

ⁱ

| ≤ a(ϕ)/2 }.

Obviously, N = I ˙∪J ˙∪K. Let λ := λ(t) = 1 − t/2. Note that X

i∈K

ϕ (x

i

+ t(y

i

− x

ⁱ

)/2 + ε

i

) c

i

= X

i∈K

ϕ (x

i

+ t(y

i

− x

ⁱ

)/2) c

i

= 0,

since the arguments of the function ϕ are contained in [−a(ϕ), a(ϕ)]. Hence, F (t) = X

i∈I

ϕ(x

i

+ ε

i

)c

i

+ X

i∈J

ϕ (x

i

+ t(y

i

− x

ⁱ

)/2 + ε

i

) c

i

=

= X

i∈I

ϕ(x

i

+ ε

i

)c

i

+ X

i∈J

ϕ (x

i

+ t(y

i

− x

ⁱ

)/2) c

i

+ X

i∈J

k

i

(t)ε

i

c

i

=

= X

i∈I

ϕ(x

i

+ ε

i

)c

i

+ X

i∈N

ϕ (λx

i

+ (1 − λ)y

i

) c

i

+

− X

i∈I

ϕ (λx

i

+ (1 − λ)y

i

) c

i

+ X

i∈K

ϕ (x

i

+ t(y

i

− x

ⁱ

)/2) c

i

+

+ X

i∈J

g

i

(t)

= X

i∈I

(ϕ(x

i

+ ε

i

) − ϕ(x

i

)) c

i

+ 1 + X

i∈J

g

i

(t).

Thus F (t) is nonincreasing on [0, 1].

_

Lemma 2.3 Let a(ϕ) = 0. If the Young function ϕ is not strictly convex in any neighbourhood of 0, and (c

n

)

_n∈N

is a sequence of real numbers satisfying inf{ c

ⁿ

: n ∈ N } > 0, then there exist two sequences of real numbers (x

ⁿ

), (y

n

) such that y

k

< x

k

for infinitely many k ∈ N, lim

_n

x

n

= lim

n

y

n

= 0, ϕ  [x

n

∧ y

ⁿ

, x

n

∨ y

ⁿ

] is affine for every n ∈ N and P

n∈N

ϕ(x

n

)c

n

= P

n∈N

ϕ(y

n

)c

n

= 1.

Additionally, if we assume c(ϕ) = +∞ or ϕ(c(ϕ)) · P

∞

n=1

c

n

> 1 instead of inf{ c

n

: n ∈ N } > 0 we obtain

(∀λ > 0)(∃n

λ

∈ N) X

∞ n=nλ

ϕ(λx

n

)c

n

< + ∞ and X

∞ n=nλ

ϕ(λy

n

)c

n

< + ∞

! .

Proof First we prove the Lemma with an additional assumption. Fix N ∈ N such that ϕ(c(ϕ)) · P

2N−1

n=1

c

n

> 1 if c(ϕ) < + ∞ and N = 1 if c(ϕ) = +∞. Let U = S

.

U

i

be the sum of disjoint open intervals U

i

of (0, c(ϕ)) such that the restricted function ϕ  U

i

is affine. U 6= ∅ and inf U = 0 by assumption. Now we define two sequences (x

n

),(y

n

), such that for n ≥ 2N, x

n

, y

n

∈ U, ϕ  [x

ⁿ

, y

n

] (respectively ϕ  [y

n

, x

n

] ) is affine, ϕ(nx

n

) < 1/(c

n

2

ⁿ

), x

n

< 1/n, ϕ(ny

n

) < 1/(c

n

2

ⁿ

), y

n

< 1/n and

X

∞ n=2N

ϕ(x

n

)c

n

= X

^∞

n=2N

ϕ(y

n

)c

n

.

The construction will be inductive. Suppose that we have constructed sequences (x

i

)

2N≤i≤2k−1

, (y

i

)

_{2N≤i≤2k−1}

for some k ∈ N satisfying the assumptions, y

²ⁱ

< x

2i

for i < k and:

2k

X

−1 n=2N

ϕ(x

n

)c

n

=

2k

X

−1 n=2N

ϕ(y

n

)c

n

.

For i ∈ { 2k, 2k+1 } we choose x

ⁱ

∈ U satisfying ϕ(ix

ⁱ

) < 1/(c

i

2

ⁱ

), x

i

< 1/i and take y

2k

< x

2k

sufficiently close to x

2k

that the appropriate assumptions hold. Moreover, x

2k

and x

2k+1

may be such small, that

((ϕ(x

2k

) − ϕ(y

2k

)) c

2k

c

2k+1

+ ϕ(x

2k+1

) < ϕ(1/2k + 1) and

(2k + 1)ϕ

⁻¹



(ϕ(x

2k

) − ϕ(y

2k

)) c

2k

c

2k+1

+ ϕ(x

2k+1

) 

< ϕ

⁻¹

1/(c

2k+1

2

^2k+1

).

Then y

2k+1

defined by the formula:

y

2k+1

:= ϕ

⁻¹



(ϕ(x

2k

) − ϕ(y

2k

)) c

2k

c

2k+1

+ ϕ(x

2k+1

) 

satisfies the appropriate assumptions, too. We may assume that ϕ is a continuous and increasing function on [0, c(ϕ)) and U is an open set. The following formula

ϕ(x

2k

)c

2k

+ ϕ(x

2k+1

)c

2k+1

= ϕ(y

2k

)c

2k

+ ϕ(y

2k+1

)c

2k+1

, holds by construction. Hence,

^2k+1

P

n=2N

ϕ(x

n

)c

n

=

^2k+1

P

n=2N

ϕ(y

n

)c

n

and the inductive step is complete. The following inequality

X

∞ n=2N

ϕ(x

n

)c

n

= X

∞ n=2N

ϕ(y

n

)c

n

≤ X

∞ n=2N

ϕ(ny

n

)c

n

≤ X

∞ n=2N

1 2

ⁿ

≤ 1

2 < 1 holds by construction. By the condition put on N at the begining of the proof, there are x

n

for 1 ≤ n ≤ 2N − 1 such that:

2N

X

−1 n=1

ϕ(x

n

)c

n

= 1 − X

^∞

n=2N

ϕ(x

n

)c

n

. Let y

n

= x

n

for n ≤ 2N − 1 Hence,

X

n∈N

ϕ(x

n

)c

n

= X

n∈N

ϕ(y

n

)c

n

= 1

Now consider λ > 0. Fix n

λ

∈ N, n ≥ 2N such that λ < n

^λ

. Thus X

∞

i=nλ

ϕ(λx

i

)c

i

≤ X

∞ i=nλ

ϕ(ix

i

)c

i

≤ X

∞ i=nλ

1

2

ⁱ

< + ∞, what ends the proof.

It remains to prove the Lemma without this additional assumption. We construct sequences (x

_2k−1

)

_k∈N

, (y

_2k−1

)

_k∈N

contained in U satisfying

0 < 1 − c := X

k∈N

ϕ(x

2k−1

)c

2k−1

= X

k∈N

ϕ(y

2k−1

)c

2k−1

< 1

and y

_2k−1

< x

_2k−1

for infinitely many k ∈ N using the methods from the previous case. To complete the proof it suffices to construct a sequence (x

2k

) converging to zero such that P

k∈N

ϕ(x

2k

)c

2k

= c and set y

2k

:= x

2k

. This sequence is constructed by induction in the following way:

Let x

2

∈ U be any real number satisfying ϕ(x

²

)c

2

< c and x

2(k+1)

= x

2k

, as long as

^k+1

P

i=1

ϕ(x

2i

)c

2i

< c. If

k+1

P

i=1

ϕ(x

2i

)c

2i

≥ c we set x

^2(k+1)

to be any x ∈ U, such that P

k

i=1

ϕ(x

2i

)c

2i

+ ϕ(x)c

2(k+1)

< c. Observe

0 < X

i∈N

ϕ(x

2i

) ≤ 1 inf{ c

ⁱ

: i ∈ N }

X

i∈N

ϕ(x

2i

)c

2i

< + ∞.

Thus lim

n

x

2n

= 0. Hence, there exists infinitely many k ∈ N such that x

2(k+1)

< x

2k

it follows that:

c ≥ X

k i=1

ϕ(x

2i

)c

2i

> c − ϕ(x

^2k

)c

2k

Letting k tend to infinity, it follows that P

i∈N

ϕ(x

2i

)c

2i

= c, what completes the

proof.

_

Lemma 2.4 If 0 < c(ϕ) < +∞, ϕ(c(ϕ)) < +∞ and if there exists an infinite sequence of disjoint sets of finite positive measure (A

n

) such that

ϕ(c(ϕ)) · X

∞ n=1

µ(A

n

) ≤ 1,

then B(L

^ϕ

(µ)) is not LNC. In particular, B(L

^∞

(µ)) is not LNC when dim L

^∞

(µ) =

∞.

Proof Let n

0

∈ N, n

⁰

> 1/c(ϕ). Set

x :=

X

∞ n=n0

(c(ϕ) − 1/n) χ

An

and y := X

^∞

n=n0

c(ϕ)χ

An

By assumption I

ϕ

(y) = R

Ω

ϕ(y) dµ = P

^∞

n=n0

ϕ(c(ϕ))µ(A

n

) ≤ 1. It follows that I

ϕ

(x) ≤ 1. Moreover, for λ > 1 and m > max{ n

0

, λ/((λ − 1)c(ϕ)) } the following holds:

I

ϕ

(λx) = X

^∞

n=n0

ϕ (λ (c(ϕ) − 1/n)) µ(A

ⁿ

) ≥ ϕ ((λ − λ/(mc(ϕ))) c(ϕ)) µ(A

m

) = +∞

and I

ϕ

(λy) = +∞. Thus kxk

ϕ

= kyk

ϕ

= 1. Set x

n

:= x + (1/n)χ

An

for n ≥ n

⁰

.

Then I

ϕ

(x

n

) ≤ I

ϕ

(y) ≤ 1, so kx

n

k

^ϕ

≤ 1. Moreover, the following holds for λ > 0:

lim

_n→∞

I

ϕ

(λ(x

n

− x)) = lim

n→∞

ϕ(λ/n)µ(A

n

) = 0. Hence, lim

_n→∞

kx

ⁿ

− xk

^ϕ

= 0.

However,

I

ϕ

(x

n

+ (y − x)/2) = I

^ϕ

(x

n

− x + (x + y)/2) =

= Z

Ω

ϕ (1/n)χ

An

+ X

^∞

k=n0

(c(ϕ) − 1/2k) χ

Ak

! dµ ≥

≥ Z

Ω

ϕ ((1/n)χ

An

+ (c(ϕ) − 1/2n) χ

An

) dµ =

= Z

An

ϕ (c(ϕ) + 1/2n) dµ = ϕ (c(ϕ) + 1/2n) µ(A

n

) = +∞.

Thus x

n

+

¹₂

(y − x) 6∈ B(L

^ϕ

(µ)), which means B(L

^ϕ

(µ)) is not LNC.

_

Note that the assumptions of the above Lemma are satisfied provided c(ϕ) <

+∞, ϕ(c(ϕ)) < +∞ and either the atomless part of Ω has positive measure or inf{ µ(e

n

) : n ∈ N } = 0.

In the next Lemma we assume that µ has infinitely many disjoint atoms { e

ⁿ

: n ∈ N } and set

c

n

:= µ(e

n

)

Lemma 2.5 Let (x

n

), (y

n

) be sequences of nonnegative real numbers, such that either x

k

= y

k

or ϕ is affine and increasing on [x

k

∧ y

^k

, x

k

∨ y

^k

] for k ∈ N.

Assume P

i∈N

ϕ(x

i

)c

i

= P

_i∈N

ϕ(y

i

)c

i

= 1. Then I

ϕ

(u

k

) = 1 and I

ϕ

(u

n

+

¹₂

(y − x)) > 1 for any n ∈ N satisfying y

ⁿ

< x

n

, where u

k

:= x − 2x

k

χ

ek

, x := P

i∈N

x

i

χ

ei

, and y := P

i∈N

y

i

χ

ei

.

Proof Note I

ϕ

(u

n

) = P

_i∈N\{n}

ϕ(x

i

)c

i

+ ϕ(−x

n

)c

n

= P

_i∈N

ϕ(x

i

)c

i

= 1 what completes the proof of the first part. Set δ

i

:= (y

i

− x

ⁱ

)/2. Let n ∈ N be such that y

n

< x

n

. Thus ϕ(x

n

+ δ

n

) < ϕ(x

n

− δ

ⁿ

). Hence,

I

ϕ

 u

n

+ 1

2 (y − x) 

= I

ϕ



x − 2x

ⁿ

χ

en

+ 1

2 (y − x) 

=

= X

i∈N\{n}

ϕ(x

i

+ δ

i

)c

i

+ ϕ(−x

n

+ δ

n

)c

n

=

= X

i∈N

ϕ(x

i

+ δ

i

)c

i

− ϕ(x

ⁿ

+ δ

n

)c

n

+ ϕ(x

n

− δ

ⁿ

)c

n

>

> X

i∈N

ϕ(x

i

+ δ

i

)c

i

= X

i∈N

ϕ

 x

i

+ y

i

2

 c

i

=

= X

i∈N

 1

2 ϕ(x

i

)c

i

+ 1 2 ϕ(y

i

)c

i



= 1,

what ends the proof of the Lemma.

_

Lemma 2.6 Let (A

n

)

_n∈N

be a sequence of disjoint sets with positive measure such that lim

n→∞

µ(A

n

) = 0 and let a, b be real numbers such that 0 < a < b and the function ϕ is affine and increasing on [a, b]. Assume that µ(A) < (ϕ((a + b)/2))

⁻¹

for A := S

n∈N

A

n

. Let B ⊂ Ω \ A be a set of finite positive measure. Fix d > 0 satisfing ϕ(d)µ(B) = 1 − µ(A)ϕ((a + b)/2). Let

0 < δ < min  b − a

2 , µ(A \ A

¹

) µ(A

1

) ·

b − a 2



and let η := δµ(A

1

) µ(A \ A

¹

)

Put x := a + b

2 χ

A

+ dχ

B

, y :=

 a + b 2 − η



χ

_A\A1

+  a + b 2 + δ 

χ

A1

+ dχ

B

,

x

n

:= x (1 − 2χ

An

) for n ∈ N.

Then I

ϕ

(x) = I

ϕ

(y) = I

ϕ

(x

n

) = 1 and I

ϕ



x

n

+ y − x 2



> 1 for n ≥ 2.

Proof Observe that 0 < δ, η < (b − a)/2. Thus x(A) ∪ y(A) ⊂ [a, b]. Let k > 0 be the slope of the function ϕ on [a, b]. We have I

ϕ

(x) = ϕ ((a + b)/2)) µ(A) + ϕ(d)µ(B) = 1 and

I

ϕ

(y) = ϕ ((a + b)/2 − η) µ(A \ A

1

) + ϕ ((a + b)/2 + δ) µ(A

1

) + ϕ(d)µ(B) =

= (ϕ ((a + b)/2) − kη) µ(A \ A

1

) + (ϕ ((a + b)/2) + kδ) µ(A

1

) + ϕ(d)µ(B) =

= ϕ ((a + b)/2) (µ(A \ A

1

) + µ(A

1

)) + k (δµ(A

1

) − ηµ(A \ A

1

)) + ϕ(d)µ(B) =

= ϕ ((a + b)/2) µ(A) + 0 + ϕ(d)µ(B) = 1.

Hence,

I

ϕ

(x

n

) = Z

Ω\An

ϕ(x) dµ + Z

An

ϕ( −x) dµ = Z

Ω

ϕ(x) dµ = 1.

Moreover, for all n ≥ 2, the following equalities hold:

I

ϕ

“ x

n

+ y − x 2

” = Z

Ω

ϕ “

x

n

− x + x + y 2

” dµ = Z

Ω

ϕ “

−(a + b)χ

^An

+ x + y 2

” dµ =

= Z

An

ϕ “

−(a + b) + x + y 2

” dµ +

Z

A\An

ϕ “ x + y 2

” dµ +

Z

Ω\A

ϕ(x) dµ =

= Z

An

ϕ “

a + b − x + y 2

” dµ + Z

A

ϕ “ x + y 2

” dµ − Z

An

ϕ “ x + y 2

” dµ +

+ 1 2

Z

Ω\A

ϕ(x) dµ + 1 2

Z

Ω\A

ϕ(y) dµ =

= Z

An

ϕ “

a + b − x + y 2

” dµ −

Z

An

ϕ “ x + y 2

” dµ +

Z

A

ϕ(x) + ϕ(y)

2 dµ +

+ Z

Ω\A

ϕ(x) + ϕ(y)

2 dµ =

Z

An

“ ϕ “

a + b − x + y 2

” − ϕ “ x + y 2

”” dµ +

+ 1 2 Z

Ω

ϕ(x) dµ + 1 2 Z

Ω

ϕ(y) dµ =

= Z

An

„ ϕ

„

a + b − a + b − η 2

«

− ϕ „ a + b − η 2

««

dµ + 1 =

= 1 +

„

ϕ „ a + b 2 + η

2

«

− ϕ „ a + b 2 − η

2

««

µ(A

n

) > 1,

what ends the proof of the Lemma.

_

Lemma 2.7 If X ⊂ L

^ϕ

(µ) is a linear subspace of L

^ϕ

(µ) with the norm inherited from L

^ϕ

(µ), such that the equivalence

kxk

^ϕ

= 1 ⇔ I

ϕ

(x) = 1,

holds for any x ∈ X and ϕ is strictly convex, then B(X) is strictly convex.

Proof Suppose there exist x, y ∈ X, x 6= y, kxk

^ϕ

, kyk

^ϕ

≤ 1, 0 < α < 1 such that kαx + (1 − α)yk = 1. Since ϕ is strictly convex, we have

ϕ(αx(ω) + (1 − α)y(ω)) < αϕ(x(ω)) + (1 − α)ϕ(y(ω)) on the set {ω : x(ω) 6= y(ω)} wich is of positive measure. Thus

1 = kαx + (1 − α)yk

^ϕ

= I

ϕ

(αx + (1 − α)y) = Z

Ω

ϕ(αx + (1 − α)y) dµ <

<

Z

Ω

(αϕ(x) + (1 − α)ϕ(y)) dµ = αI

ϕ

(x) + (1 − α)I

ϕ

(y) ≤ α + (1 − α) = 1

and we get a contradiction.

_

3. Main results. Now we characterize the LNC properties of unit balls in Orlicz spaces L

^ϕ

(µ) and the space of finite elements E

^ϕ

(µ) respectively equipped with the Luxemburg norm assuming that ϕ is a Young function and µ is a σ–finite measure. In the cases of a purely atomic measure µ, we fix a partition of Ω on disjoint atoms { e

n

: n ∈ N }, where N denotes N if the set of atoms is infinite (i. e.

if dim L

^ϕ

(µ) = ∞) or is equal to the number of atoms if the cardinality of this set is finite (i. e. if dim L

^ϕ

(µ) < ∞).

Theorem 3.1 The unit ball B(L

^ϕ

(µ)) is LNC if and only if at least one of the following conditions is satisfied:

(i) dim L

^ϕ

(µ) < ∞

(ii) µ is purely atomic measure with inf{ µ(e

n

) : n ∈ N } > 0, ϕ ∈ ∆

⁰r

(µ) for some r > 1 and ϕ is strictly convex on the interval [0, b] for some b > 0.

(iii) c(ϕ) = +∞, ϕ ∈ ∆

²

(µ) and ϕ is strictly convex on R.

Proof (⇒) Assume that B(L

^ϕ

(µ)) is LNC and L

^ϕ

(µ) is infinite dimensional. It is necessary to prove that (ii) or (iii) are satisfied. We consider two cases:

A) Suppose µ is purely atomic and inf{ µ(e

ⁿ

) : n ∈ N } > 0. Because B(L

^ϕ

(µ)) is LNC, it follows from [20, Theorem 3.1, p. 196] that it is stable. From Wisła’s Theorem one of the six conditions from Theorem 2.1 hold, ( [23], Theorem 5).

Obviously, (i) is excluded by asumption and (ii) is excluded by Lemma 2.4, (vi) is excluded by assumption A). Thus one of conditions (iii)—(v) is satisfied. Either ϕ ∈ ∆

⁰r

(µ) for some r > 1 when c(ϕ) < +∞ and ϕ(c(ϕ)) < +∞, or ϕ ∈ ∆

^r

(µ).

Hence, ϕ ∈ ∆

⁰r

(µ) for some r > 1 in both cases. Thus a(ϕ) = 0.

Suppose that (ii) is not satisfied. Thus the assumptions of Lemma 2.3 are satis- fied. Let sequences (x

n

) and (y

n

) satisfy the conditions given in Lemma 2. Let x, y and u

n

(n ∈ N) satisfy the conditions given in Lemma 2.5. In order to show that B(L

^ϕ

(µ)) is not LNC, it suffices to prove that lim

_n→∞

ku

ⁿ

− xk = 0.

Because ϕ(x

n

)c

n

→ 0 we have lim

ⁿ

I

ϕ

(

¹₂

(x − u

n

)) = lim

n

I

ϕ

(x

n

χ

en

) = 0. By condition ∆

⁰_r

(µ) we have kx − u

n

k → 0. This completes the proof for case A).

B) It remains to consider the case when either the measure µ is not purely atomic or inf{ µ(e

ⁿ

) : n ∈ N } = 0. In other words, we may assume that there exists a sequence of mutually disjoint sets of positive finite measures (A

n

), such that lim

n

µ(A

n

) = 0. Fix such a sequence. We will show that condition (iii) of Theorem 1 is satisfied.

Because B(L

^ϕ

(µ)) is LNC and thus stable, at least one of the conditions (i)—

(vi) of Theorem 2.1 is satisfied. Conditions (i) and (ii) are excluded, as in case A).

We claim that the case where both c(ϕ) < +∞ and ϕ(c(ϕ)) < +∞ is excluded, i.e.

conditions (iv)—(vi). are excluded, too. Suppose c(ϕ) < +∞ and ϕ(c(ϕ)) < +∞.

Then there exists an increasing sequence of positive integers (n

k

) such that ϕ(c(ϕ)) · X

k∈N

µ (A

nk

) ≤ 1.

Hence, by Lemma 2.4, the unit ball B(L

^ϕ

(µ)) is not LNC — a contradiction. Thus condition (iii) of Theorem 2.1 is satisfied, so ϕ ∈ ∆

^r

(µ) for some r > 1.

We claim c(ϕ) = +∞. If µ has positive atomless part Ω

¹

, then ϕ ∈ ∆

^r

(µ  Ω

1

).

Hence, c(ϕ) = +∞.

If µ is purely atomic and inf{ µ(e

n

) : n ∈ N } = 0, then there exists n ∈ N such that

µ(e

n

) ≤ b  ϕ

 c(ϕ) r

⁰



for all r

⁰

∈ (1, r) (the constant b is choosen according to the definition of ∆

^r

(µ)).

Thus ϕ c(ϕ)/r

⁰

µ(e

n

) ≤ b. Hence, +∞ = ϕ 

r c(ϕ) r

⁰



µ(e

n

) ≤ cϕ  c(ϕ)

r

⁰



µ(e

n

) + d

n

< + ∞

and we obtain a contradiction, which proves our claim. Hence, ϕ ∈ ∆

^r

(µ) for any r > 1, in particular ϕ ∈ ∆

²

(µ).

Now we show a(ϕ) = 0. Assume a(ϕ) > 0. Let x

1

be a positive real number such that ϕ(x

1

)µ(A

1

) = 1. Set

x := x

1

χ

A1

, y := x

1

χ

A1

+ a(ϕ) ·

+∞

X

i=2

χ

Ai

. and

u

n

:= x + a(ϕ)χ

An

for n = 2, 3, . . . . Then

I

ϕ

(x) = I

ϕ

(y) = ϕ(x

1

)µ(A

1

) = 1, I

ϕ

(u

n

) = 1.

Thus x, y, u

n

∈ B(L

^ϕ

(µ)). Let λ > 0. Then:

I

ϕ

(λ(u

n

− x)) = I

^ϕ

(λa(ϕ)χ

An

) = ϕ(λa(ϕ))µ(A

n

).

Hence, lim

n

I

ϕ

(λ(u

n

− x)) = 0 and we have lim

_n

ku

ⁿ

− xk = 0. Moreover,

I

ϕ

 u

n

+ 1

2 (y − x) 

= I

ϕ



u

n

− x + 1

2 (x + y) 

= I

ϕ

a(ϕ)χ

An

+ x

1

χ

A1

+ X

^∞

i=2

a(ϕ) 2 χ

Ai

!

=

= I

ϕ



x

1

χ

A1

+ 3

2 a(ϕ)χ

An

+ X

i∈N\{ 1,n }

a(ϕ) 2 χ

Ai



 =

= ϕ(x

1

)µ(A

1

) + ϕ  3 2 a(ϕ)



µ(A

n

) = 1 + ϕ  3 2 a(ϕ)

 µ(A

n

)

> 1.

Thus u

n

+

¹₂

(y − x) 6∈ B(L

^ϕ

(µ)), which contradicts the assumption that B(L

^ϕ

(µ))

is LNC. It follows that a(ϕ) = 0.

To prove (iii) suppose that ϕ is not strictly convex. Let x, y, x

n

, n ∈ N be defined as in Lemma 2.6. We have x, y, x

n

∈ B(L

^ϕ

(µ)) and x

n

+ (y − x)/2 6∈ B(L

^ϕ

(µ)) for n ≥ 2. Moreover,

I

ϕ

(λ(x − x

n

)) = Z

An

ϕ(2λx) dµ = Z

An

ϕ(λ(a + b)) dµ = ϕ(λ(a + b))µ(A

n

)

for λ > 0. It follows that ϕ(λ(a+b)) < +∞, since c(ϕ) = +∞. Also, lim

ⁿ→∞

I

ϕ

(λ(x−

x

n

)) = 0, since lim

n→∞

µ(A

n

) = 0. Therefore, lim

n→∞

kx − x

ⁿ

k = 0. Hence, we obtain a contradiction.

(⇐) We divide the proof into three parts.

a) Suppose that condition (i) holds, so µ is purely atomic and has a finte number of atoms { e

ⁱ

: 1 ≤ i ≤ N }. Set c

i

= µ(e

i

) for i = 1, 2, . . . , N. Let x, y ∈ B(L

^ϕ

(µ)), u

n

∈ B(L

^ϕ

(µ)) for n ∈ N, lim

n

ku

ⁿ

− xk

^ϕ

= 0. It is necessary to prove that for some n

0

∈ N, u

ⁿ

+

¹₂

(y − x) ∈ B(L

^ϕ

(µ)) for n ≥ n

0

. Without loss of generality we can assume that kλx + (1 − λ)yk

^ϕ

= 1 for any 0 ≤ λ ≤ 1. Set

x = X

N i=1

x

i

χ

ei

, y = X

N i=1

y

i

χ

ei

, u

n

= X

N i=1

u

ⁿ_i

χ

en

for i = 1, 2, . . . , N, n ∈ N, where x

ⁱ

, y

i

, u

ⁿ_i

are real numbers. We consider two cases:

1. There exists λ

0

∈ [0, 1] such that I

^ϕ

(λ

0

x + (1 − λ

⁰

)y) < 1. It is easy to see that the function g(λ) := I

ϕ

(y + λ(x − y)) is convex. From g(0) ≤ 1, g(1) ≤ 1 and g(λ

0

) < 1 we have P

^N

i=1

ϕ(λx

i

+ (1 − λ)y

i

)c

i

< 1 for any λ ∈ (0, 1). In particular, P

^N

i=1

ϕ((x

i

+ y

i

)/2)c

i

< 1. We have I

ϕ

(α(x + y)/2) = +∞ for α > 1, since k(x + y)/2k

^ϕ

= 1. Hence P

^N_i=1

ϕ(α(x

i

+ y

i

)/2)c

i

= +∞. This is possible only if there exists k, 1 ≤ k ≤ N, such that |(x

^k

+y

k

)/2| = c(ϕ) < +∞. If x

^k

6= y

^k

then the expression |λx

^k

+ (1 − λ)y

k

| for λ ∈ [0, 1] attains its supremum at one of the ends of the interval. Hence, either |x

^k

| > c(ϕ) or |y

^k

| > c(ϕ). But then either I

^ϕ

(x) = +∞

or I

ϕ

(y) = +∞, what contradicts to x, y ∈ B(L

^ϕ

(µ)). Hence, x

k

= y

k

= ±c(ϕ) for some fixed k, 1 ≤ k ≤ N. Thus lim

ⁿ→∞

I

ϕ

(α(x − u

n

)) = 0 holds for α > 0.

So lim

_n→∞

P

N

i=1

ϕ(α(x

i

− u

ⁿi

))c

i

= 0. Hence, lim

n

u

ⁿ_i

= x

i

for every 1 ≤ i ≤ N.

Otherwise, there exists a natural number 1 ≤ i ≤ N, increasing sequence n

^k

of natural numbers and ε > 0, such that |u

ⁿi^k

− x

ⁱ

| > ε. Setting a := max{ a(ϕ), 1 } and α :=

^2a_ε

we obtain P

N

j=1

ϕ(α(x

j

− u

ⁿj^k

))c

j

≥ ϕ(2a)c

ⁱ

> 0 for k ∈ N, which is contradiction. Set

I := { i : x

ⁱ

= y

i

= ±c(ϕ) }, J := {1, 2, . . . , N } \ I.

For any i ∈ J there exists n(i) ∈ N, such  

 ϕ



u

ⁿ_i

− x

ⁱ

+ x

i

+ y

i

2



− ϕ

 x

i

+ y

i

2

   c

ⁱ

< 1 N

 1 − I

ϕ

 x + y 2



,

for all n ≥ n(i), because |(x

ⁱ

+ y

i

)/2| < c(ϕ) and ϕ is continuous on (−c(ϕ), c(ϕ)).

Hence,

I

ϕ

 u

n

+ 1

2 (y − x) 

= X

i∈I

ϕ(u

ⁿ_i

)c

i

+ X

i∈J

ϕ



u

ⁿ_i

− x

ⁱ

+ x

i

+ y

i

2

 c

i

≤

≤ X

i∈I

ϕ

 x

i

+ y

i

2



c

i

+ X

i∈J

 ϕ



u

ⁿ_i

− x

ⁱ

+ x

i

+ y

i

2



− ϕ

 x

i

+ y

i

2



+

+ X

i∈J

ϕ

 x

i

+ y

i

2

 c

i

≤

≤ I

^ϕ

 x + y 2



+ X

i∈J

   ϕ



u

ⁿ_i

− x

ⁱ

+ x

i

+ y

i

2



− ϕ

 x

i

+ y

i

2

   c

ⁱ

≤

≤ I

^ϕ

 x + y 2



+ N · 1 N

 1 − I

ϕ

 x + y 2



= 1.

Thus, there exists an n

0

such that u

n

+

¹₂

(y − x) ∈ B(L

^ϕ

(µ)) for n ≥ n

0

. Thus, B(L

^ϕ

(µ)) is LNC.

2. We have I

ϕ

(λx + (1 − λ)y) = 1 for every 0 ≤ λ ≤ 1.

Let n

0

be large enough that for n ≥ n

⁰

and for ε

i

:= u

ⁿ_i

− x

ⁱ

the conditions (a)—(d) in item 3 of Lemma 2.2 are satisfied. It follows from this lemma that

I

ϕ

 u

n

+ 1

2 (y − x) 

= X

N i=1

ϕ



ε

i

+ x

i

+ y

i

2



c

i

= F (1) ≤ F (0) =

= X

N i=1

ϕ(x

i

+ ε

i

)c

i

= X

N i=1

ϕ(u

ⁿ_i

)c

i

= I

ϕ

(u

n

) ≤ 1,

for fixed n ≥ n

⁰

, such that u

n

+ (y − x)/2 ∈ B(L

^ϕ

(µ)) for n ≥ n

0

. Thus B(L

^ϕ

(µ)) is LNC, which finishes the proof of part a).

b) Now we assume that condition (ii) is satisfied.

Fix x, y ∈ B(L

^ϕ

(µ)), x 6= y, u

n

∈ B(L

^ϕ

(µ)) for n ∈ N, lim

n

ku

ⁿ

− xk

^ϕ

= 0.

We can assume that kλx + (1 − λ)yk

^ϕ

= 1 for every λ ∈ [0, 1]. It is necessary to prove that there exists n

₀

, such that for n ≥ n

0

, u

_n

+

¹₂

(y − x) ∈ B(L

^ϕ

(µ)). Anal- ogously to part a), set x = P

i∈N

x

i

χ

ei

, y = P

i∈N

y

i

χ

ei

, u

n

= P

_i∈N

u

ⁿ_i

χ

ei

n ∈ N.

By assumption inf{ c

ⁱ

: i ∈ N } > 0. In particular, P

i∈N

µ(e

i

) = +∞ and hence, a(ϕ) = 0. As in part a) we consider two cases:

1. There exists λ

0

∈ [0, 1] such that I

^ϕ

(λ

0

x + (1 − λ

⁰

)y) < 1.

Analogously to part a,), I

ϕ

(λx + (1 − λ)y) < 1 for every λ ∈ (0, 1). In particular, P

∞

i=1

ϕ((x

i

+ y

i

)/2)c

i

< 1. We obtain lim

n→∞

u

ⁿ_i

= x

i

for i ∈ N, because a(ϕ) = 0.

From the following obvious inequality 0 ≤ ϕ(x

ⁱ

) ≤ (1/inf{ c

j

: j ∈ N }) · ϕ(x

i

)c

i

, together with P

∞

i=1

ϕ(x

i

)c

i

≤ 1 we obtain lim

ⁱ

ϕ(x

i

) = 0. Hence, lim

i

x

i

= 0.

Analogously, lim

i

y

i

= 0 and lim

i

u

ⁿ_i

= 0 for n ∈ N. Set

J := { k ∈ N : |x

^k

| = c(ϕ) or |y

^k

| = c(ϕ) }.

Obviously J is finite. Define the following two complementary subspaces of L

^ϕ

(µ):

X

1

:= span { χ

^ek

: k 6∈ J } X

2

:= span { χ

^ek

: k ∈ J } Set

x

⁰

:= Pr

X1

x, x

⁰⁰

:= Pr

X2

x, y

⁰

:= Pr

X1

y, y

⁰⁰

:= Pr

X2

y and

u

⁰_n

:= Pr

X1

u

n

, u

⁰⁰_n

:= Pr

X2

u

n

for n ∈ N,

where Pr denotes the natural projection. Obviously, x = x

⁰

+ x

⁰⁰

, y = y

⁰

+ y

⁰⁰

, u

n

= u

⁰_n

+ u

⁰⁰_n

, and kx

⁰

k

∞

< c(ϕ), ky

⁰

k

∞

< c(ϕ). Thus k(x

⁰

+ y

⁰

)/2k

∞

< c(ϕ). From this we obtain k(x

⁰

+ y

⁰

)/2k

ϕ

< 1, since I

ϕ

((x

⁰

+ y

⁰

)/2) < 1 (see the remarks after the definition of ∆

⁰_r

(µ)). Moreover,

0 ≤ I

ϕ

(λ(x

⁰

− u

⁰n

)) = X

k6∈J

ϕ (λ(x

k

− u

ⁿk

)) c

k

≤ I

^ϕ

(λ(x − u

n

)) −→ 0

for λ > 0. Therefore, lim

n→∞

kx

⁰

− u

⁰n

k

^ϕ

= 0. Since

u

⁰ⁿ

+ 1

2 (y

⁰

− x

⁰

)

_ϕ

=

u

⁰ⁿ

− x

⁰

+ x

⁰

+ y

⁰

2

_ϕ

≤ ku

⁰n

− x

⁰

k

ϕ

+ x

⁰

+ y

⁰

2

_ϕ

,

we obtain

∃ε > 0 ∃n

⁰

∈ N ∀n ≥ n

⁰

u

⁰ⁿ

+ 1

2 (y

⁰

− x

⁰

)

_ϕ

< 1 − ε and I

ϕ

(u

⁰_n

+ (y

⁰

− x

⁰

)/2) < 1 − ε.

We can find n

1

≥ n

⁰

such that I

ϕ

 u

⁰⁰_n

+ 1

2 (y

⁰⁰

− x

⁰⁰

) 

= X

k∈J

ϕ

 u

ⁿ_k

+ 1

2 (y

k

− x

^k

)  c

k

< ε

for any n ≥ n

¹

, because lim

k

u

^k_n

= x

n

for k ∈ J and J is finite. Hence, I

ϕ

 u

n

+ 1

2 (y − x)



= X

k6∈J

I

ϕ

 u

ⁿ_k

+ 1

2 (y

k

− x

^k

)



+ X

k∈J

I

ϕ

 u

ⁿ_k

+ 1

2 (y

k

− x

^k

)



=

= I

ϕ

 u

⁰_n

+ 1

2 (y

⁰

− x

⁰

)  + I

ϕ

 u

⁰⁰_n

+ 1

2 (y

⁰⁰

− x

⁰⁰

) 

< 1 − ε + ε = 1 for n ≥ n

¹

, so u

n

+ (y − x)/2 ∈ B(L

^ϕ

(µ)).

2. We have I

ϕ

(λx + (1 − λ)y) = 1 for every 0 ≤ λ ≤ 1. Obviously, there exists n

0

∈ N such that x

ⁱ

, y

i

∈ [−b, b] for i > n

⁰

. Set

I := { i ∈ N : x

ⁱ

= y

i

}, J := { i ∈ N : x

ⁱ

6= y

ⁱ

}.

Since

1 = X

i∈N

ϕ (λx

i

+ (1 − λ)y

i

) c

i

≤ λ X

i∈N

ϕ(x

i

)c

i

+ (1 − λ) X

i∈N

ϕ(y

i

)c

i

= 1,

it follows that ϕ is affine on every interval [x

i

, y

i

] (or [y

i

, x

i

]) for i ∈ J. Because ϕ is strictly convex on [−b, b], we obtain i 6∈ J for i > n

⁰

. Hence, J ⊆ { 1, 2, . . . , n

⁰

}.

In particular, x

i

= y

i

for i > n

0

. Therefore, setting ε

i

= u

ⁿ_i

− x

ⁱ

for sufficently large n, the conditions given in Lemma 2.2 for the case N = + ∞ are satisfied. Therefore, from Lemma 2.2

I

ϕ

 u

n

+ 1

2 (y − x) 

= F (1) ≤ F (0) = I

ϕ

(u

n

) ≤ 1.

c) Suppose that (iii) is satisfied. From ∆

2

(µ) and c(ϕ) = ∞ it is known that L

^ϕ

(µ) satisfies the following condition:

kxk

^ϕ

= 1 ⇔ I

ϕ

(x) = 1 x ∈ L

^ϕ

(µ).

From Lemma 2.7 B(L

^ϕ

(µ)) is strictly convex and so LNC. The proof of the Theorem

is complete.

_

Corollary 3.2 If µ is an atomless measure, then the following conditions are equivalent:

(i) B(L

^ϕ

(µ)) is LNC.

(ii) ϕ is strictly convex, c(ϕ) = +∞ and ϕ satisfies either ∆

²

globally when µ(Ω) = +∞ or for sufficently large t when µ(Ω) < +∞.

(iii) B(L

^ϕ

(µ)) is strictly convex.

Proof (i) ⇒ (ii) Since µ is atomless, conditions (i) and (ii) from Theorem 3.1 are not satisfied. Therefore, condition (iii) of this theorem must hold. The result follows from the fact that for atomless measures the condition ∆

2

(µ) is equivalent to the classic condition for ∆

2

based on µ(Ω).

(ii) ⇒ (iii) Follows from the last part of c) in the proof of Theorem 3.1.

(iii) ⇒ (i) Obvious.



Corollary 3.3 If µ is purely atomic measure with an infinite number of atoms and

0 < inf{ µ(e

ⁿ

) : n ∈ N } ≤ sup{ µ(e

ⁿ

) : n ∈ N } < +∞

then the following conditions are equivalent:

(i) B(L

^ϕ

(µ)) is LNC.

(ii) ϕ satisfies the condition δ

2

and is strictly convex on [0, b] for some b > 0.

Proof (i) ⇒ (ii) Condition (ii) holds because conditions (ii) and (iii) of Theorem 3.1 are satisfied. From the remark given after the definition of ∆

⁰_r

(µ) we see that ϕ satisfies the condition δ

2

.

(ii) ⇒ (i) follows from Theorem 3.1 and from the remark mentioned above.



From the last corollary it follows that B(l

^ϕ

) is LNC iff the condition (ii) of Corollary 3.3 is satisfied. Obviously, B(l

^p

) is LNC for 1 < p < +∞. Note that B(l

¹

) and B(l

^∞

) are stable, but not LNC (see also [1], [20]).

Example 3.4 A ball B(l

^ϕ

) satisfying LNC need not be strictly convex.

Indeed, for

ϕ(t) =

 t

²

, for|t| ≤

¹2

|t| −

¹4

, for|t| ≥

¹2

the unit ball B(l

^ϕ

) is LNC from Corollary 3.3, but

I

ϕ

(λx + (1 − λ)y) = I

ϕ

 3 4 + 1

4 λ, 3 4 −

1

4 λ, 0, 0, . . .



= 1 2 + 1

4 λ + 1 2 −

1 4 λ = 1, for x = (1, 1/2, 0, 0, . . .), y = (3/4, 3/4, 0, 0, . . .) and λ ∈ [0, 1]. Hence, kzk

^ϕ

= 1 for any z ∈ [x, y], i.e. B(l

^ϕ

) is not strictly convex.

Now we present a characterization of the LNC property for the unit ball B(E

^ϕ

(µ)).

Theorem 3.5 B(E

^ϕ

(µ)) is LNC iff at least one of the following conditions is sat- isfied:

(i) µ is purely atomic measure with a finite number of atoms (equivalently dim E

^ϕ

(µ) <

∞).

(ii) c(ϕ) = +∞, µ is purely atomic measure with an infinite number of atoms, inf{ µ(e

n

) : n ∈ N } > 0 and either a(ϕ) > 0 or ϕ is strictly convex on [0, b]

for some b > 0.

(iii) c(ϕ) = +∞, ϕ is strictly convex on R and µ either has a positive atomless part or inf{ µ(e

n

) : n ∈ N } = 0.

(iv) c(ϕ) < +∞ and a(ϕ) > 0.

(v) c(ϕ) < +∞ and ϕ is strictly convex on [0, b] for some b > 0.

(vi) c(ϕ) < +∞, µ(Ω)ϕ(c(ϕ)) ≤ 1.

(vii) c(ϕ) < +∞ and µ has a positive atomless part.

Proof (⇒) Assume that B(E

^ϕ

(µ)) is LNC. Suppose dim E

^ϕ

(µ) = ∞. We con-

sider three cases:

1. c(ϕ) = +∞ and µ is purely atomic measure with an infinite number of atoms, such that inf{ µ(e

ⁿ

) : n ∈ N } > 0. We prove that in this case (ii) is satisfied.

Consider the case a(ϕ) = 0. Suppose that ϕ is not strictly convex on any interval of the form [0, b] where b > 0. Let x, y be defined as in Lemma 2.3 and u

n

as in Lemma 2.5. Then x, y, u

_n

∈ B(E

^ϕ

(µ)). Moreover, for each λ > 0 and n > 2λ we have

I

ϕ

(λ(x − u

n

)) = ϕ(2λx

n

)µ(e

n

) = ϕ(nx

n

)µ(e

n

) < 1 2

ⁿ

→ 0

holds. Hence, lim

n→∞

I

ϕ

(λ(x − u

n

)) = 0 and lim

n→∞

u

n

= x in the space E

^ϕ

(µ).

This contradicts the fact that B(E

^ϕ

(µ)) is LNC.

2. c(ϕ) = +∞, µ either has a positive atomless part or inf{ µ(e

ⁿ

) : n ∈ N } = 0.

To start, we prove that a(ϕ) = 0. By assumption there exists an infinite sequence of mutually disjoint sets with positive measures (A

n

), such that P

_n∈N

µ(A

n

) < +∞.

Suppose that a := a(ϕ) > 0. Set x := dχ

A1

+ aχ

∞S

i=2

An

, y := dχ

A1

− aχ

^∞S

i=2

An

,

where d is a positive number, such that ϕ(d)µ(A

1

) = 1.

Set u

n

:= x − 2aχ

An

. Then

I

ϕ

(λx) = ϕ(λd)µ(A

1

) + ϕ(λa)µ [

∞ i=2

A

n

!

< + ∞

for λ > 0. Hence, x ∈ E

^ϕ

(µ) and similarly y, u

n

∈ E

^ϕ

(µ). Moreover, lim

n

I

ϕ

(x − u

n

) = lim

n

ϕ(2a)µ(A

n

) = 0, Thus lim

n

ku

ⁿ

− xk = 0 Also,

I

ϕ

(x) = ϕ(d)µ(A

1

) + ϕ(a)µ [

∞ i=2

A

n

!

= 1

and analogously I

ϕ

(y) = 1 and I

ϕ

(u

n

) = 1 for n ∈ N. Hence, x, y, u

n

∈ B(E

^ϕ

(µ)).

But

u

n

+ 1

2 (y − x) = u

n

− x + x + y

2 = −2aχ

An

+ dχ

A1

and I

ϕ

 u

n

+ 1

2 (y − x) 

= ϕ(2a)µ(A

n

) + ϕ(d)µ(A

1

) = 1 + ϕ(2a)µ(A

n

) > 1.

Therefore, u

n

+

¹₂

(y − x) 6∈ B(E

^ϕ

(µ)), which means that B(E

^ϕ

(µ)) is not LNC.

Thus a(ϕ) = 0.

Now we prove that the function ϕ is strictly convex.

To obtain a contradiction, suppose that ϕ is not strictly convex. Note that x, y,

x

n

∈ E

^ϕ

(µ) for n ∈ N (defined in Lemma 2.6), since they are linear combinations

of characteristic functions of sets of finite positive measures which, by c(ϕ) = +∞,

belong to E

^ϕ

(µ). From Lemma 2.6 it follows that B(E

^ϕ

(µ)) is not LNC. Thus we obtain a contradiction.

3. Remaining cases. Then c(ϕ) < +∞. If µ has positive atomless part then (vii) is satisfied. Let µ be purely atomic measure with an infinite numbers of atoms. If a(ϕ) > 0 or ϕ is strictly convex on [0, b] for some b > 0 then (iv) or (v) is satisfied.

Let a(ϕ) = 0 and let ϕ be not strictly convex in any neighbourhood of 0. We need to prove that µ(Ω)ϕ(c(ϕ)) ≤ 1. Suppose ϕ(c(ϕ)) P

∞

n=1

µ(e

n

) > 1. Let x, y be defined as in Lemma 2.3 and u

n

as in Lemma 2.5. In an analogous way as in the step 1 we prove that B(E

^ϕ

(µ)) is LNC.

(⇐) We divide the proof into six parts.

a) Assume that condition (i) holds. But then E

^ϕ

(µ) = L

^ϕ

(µ) is of finite dimen- sion and the thesis follows from Theorem 3.1.

b) Assume that condition (ii) holds. Let x, y ∈ B(E

^ϕ

(µ)), u

n

∈ B(E

^ϕ

(µ)) for n ∈ N. Let lim

n→∞

ku

ⁿ

−xk

^ϕ

= 1 for every 0 ≤ λλ ≤ 1. It is necessary to prove that there exists n

₀

, such that for n ≥ n

0

, u

_n

+ (y − x)/2 ∈ B(E

^ϕ

(µ)). Without loss of generality we assume kλx+(1−λ)yk = 1 for 0 ≤ λ ≤ 1. Hence I

^ϕ

(λx+(1−λ)y) = 1 for 0 ≤ λ ≤ 1. If not then I

^ϕ

(α(λx+(1−λ)y)) = +∞ every α > 1, what contradicts λx + (1 − λ)y ∈ E

^ϕ

(µ) in case c(ϕ) = +∞. We set

x = X

i∈N

x

i

χ

ei

, y = X

i∈N

y

i

χ

ei

, u

n

= X

i∈N

u

ⁿ_i

χ

ei

for n ∈ N, where x

ⁱ

, y

i

, u

ⁿ_i

are real numbers. We prove that lim

n

x

n

= lim

n

y

n

= 0.

Suppose for instance that lim

n

x

n

6= 0. Then there exists ε > 0 and a subsequence (x

nk

) such that |x

nk

| > ε. Let λ > 0 satisfy ϕ(ελ) = 1. Then

I

ϕ

(λx) = X

n∈N

ϕ(λx

n

)µ(e

n

) ≥ X

k∈N

ϕ(λx

nk

)µ(e

nk

) ≥

≥ X

k∈N

ϕ(λε)µ(e

nk

) ≥ X

k∈N

inf{ µ(e

n

) : n ∈ N } = +∞,

which is impossible for x ∈ E

^ϕ

(µ). Thus lim

_n→∞

x

n

= lim

_n→∞

y

n

= 0. Hence, there exists n

0

∈ N such that for i > n

⁰

, either x

i

, y

i

∈ [−a(ϕ)/2, a(ϕ)/2] in the case a(ϕ) > 0, or x

i

, y

i

∈ [−b, b] in the case a(ϕ) = 0 and ϕ is strictly convex on [0, b]. Denote

I := { i ∈ N : x

ⁱ

= y

i

} , J :=



i ∈ N : x

ⁱ

6= y

ⁱ

and



|x

ⁱ

| > a(ϕ)

2 or |y

ⁱ

| > a(ϕ) 2

 

.

Analogously to the proof of Theorem 3.1, we observe that the function ϕ is affine

on every interval of the form [x

i

, y

i

] or [y

i

, x

i

] for i ∈ I such that x

i

6= y

ⁱ

. Moreover,

i 6∈ J for i > n

⁰

, because either i ∈ I in the case a(ϕ) = 0 (from the strict convexity

of ϕ on [−b, b]), or |x

ⁱ

|, |y

ⁱ

| ≤ a(ϕ)/2 in the case a(ϕ) > 0. Thus J is finite and