Midpoint Locally Uniform Rotundity of Musielak-Orlicz

Journal of Convex Analysis Volume 21 (2014), No. 2, 477–494

Midpoint Locally Uniform Rotundity of Musielak-Orlicz-Bochner Function Spaces Endowed with the Orlicz Norm∗ Shaoqiang Shang† Department of Mathematics, Northeast Forestry University, Harbin 150040, China [email protected]

Yunan Cui Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, China [email protected]

Yongqiang Fu Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China [email protected] Received: May 17, 2012 Revised manuscript received: July 31, 2012 Accepted: April 29, 2013 The criteria for midpoint locally uniform rotundity in the classical Orlicz function spaces have been given already. However, because of the complicated structure of Musielak-Orlicz-Bochner function spaces, up to now the criteria for midpoint locally uniform rotundity have not been discussed yet. In this paper, criteria for midpoint locally uniform rotundity of Musielak-OrliczBochner function spaces equipped with Orlicz norm are given. Keywords: Strongly extreme point, midpoint locally uniformly rotund, Musielak-OrliczBochner function spaces, Orlicz norm 2010 Mathematics Subject Classification: 46E30, 46B20

1.

Introduction

A lot of rotundity concepts in Banach spaces are known. Among them rotundity (R for short) and midpoint locally uniform rotundity are important notions. One of reasons is that (see [11]) a Banach X space is midpoint locally uniformly rotund if and only if every closed ball in X is approximately compact Chebyshev set. In 2007, S. T. Chen, H. Hudzik, W. Kowalewski and Y. W. Wang (see [2]) proved ∗

This research is supported by "Fundamental Research Funds for the Central Universities", DL12BB36 and by the National Natural Science Foundation of China (Grant No 11371110). † Corresponding author. ISSN 0944-6532 / $ 2.50

c Heldermann Verlag

478

S. Shang, Y. Cui, Y. Fu / Midpoint Locally Uniform Rotundity of ...

that a nonempty closed convex C of a midpoint locally uniformly rotund space is approximately compact if and only if C is Chebyshev set and the metric projection operator PC is continuous. The criteria for midpoint locally uniform rotundity in the classical Orlicz function spaces have been given in [3, 6] already. However, because of the complicated structure of Musielak-Orlicz-Bochner function spaces, up to now the criteria for midpoint locally uniform rotundity have not been discussed yet. In this paper, we will discuss criteria for midpoint locally uniform rotundity in Musielak-Orlicz-Bochner function spaces endowed with Orlicz norm. The topic of this paper is related to the topic of [1], [3]–[10] and [12]–[18]. Let (X, k · k) be a real Banach space. S(X) and B(X) denote the unit sphere and the unit ball, respectively. By X ∗ we denote the dual space of X. Let N, R and R+ denote the set of natural numbers, reals and of nonnegative reals, respectively. Let us recall some geometrical notions concerning rotundity. A Banach space X is said to be rotund if for any x, y ∈ S(X) with kx + yk = 2 we have x = y. A point ∞ x ∈ S(X) is said to be strongly extreme point if for any {xn }∞ n=1 , {yn }n=1 ∈ X with kxn k → 1, kyn k → 1 and x = 21 (xn + yn ), there holds kxn − yn k → 0 as n → ∞. If the set of all strongly extreme points of B(X) is equal to S(X), then X is said to be midpoint locally uniformly rotund. Let (T, Σ, µ) be a nonatomic finite measure space. Suppose that that a function M : T × [0, ∞) → [0, ∞] statisfies the following conditions: ′

′

(1) limu→∞ M (t, u) = ∞ and M (t, u ) < ∞ for some u > 0, for µ-a.e., t ∈ T , M (t, 0) = 0. (2) M (t, u) is convex on [0, ∞) with respect to u for µ-a.e., t ∈ T . (3) M (t, u) is a Σ-measurable function of t on T for each u ∈ [0, ∞). Let p(t, u) denote the right derivative of M (t, ·) at u ∈ R+ (where if M (t, u) = ∞, let p(t, u) = ∞) and q(t, ·) be the generalized inverse function of p(t, ·) defined on R+ by q(t, v) = sup{u ≥ 0 : p(t, u) ≤ v}. u≥0

Rv Then N (t, v) = 0 q(t, s)ds for any v ∈ R and µ-a.e. t ∈ T . It is well known that the Young inequality uv ≤ M (t, u) + N (t, v) holds for µ-a.e. t ∈ T . Moreover, uv = M (t, u) + N (t, v) ⇔ u = q(t, v) or v = q(t, u). Definition 1.1 (see [1]). We say that M (t, u) satisfies condition ∆2 (M ∈ ∆2 ) if there R exist K ≥ 1 and a measureable nonnegative function δ(t) on T such that T M (t, δ(t))dt < ∞ and M (t, 2u) ≤ KM (t, u) for almost all t ∈ T and all u ≥ δ(t). Definition 1.2 (see [1]). A Musielak-Orlicz function M (t, u) is said to be strictly convex with respect u for t ∈ T , a.e., if for almost every t ∈ T , a.e., and any u, v ∈ R, u 6= v, we have u + v M (t, u) + M (t, v) < . M t, 2 2


479

Moreover, for a given Banach space (X, k·k), we denote by XT the set of all strongly Σ-measurable functions from T to X, and for each u ∈ XT , we define the modular of u by Z ρM (u) = M (t, ku(t)k)dt. T

Let us define the Musielak-Orlicz-Bochner function space LM (X) = {u ∈ XT : ρM (λu) < ∞ for some λ > 0}. It is known that the space LM (X) is Banach spaces equipped with the Luxemburg norm o u n ≤1 kuk = inf λ > 0 : ρM λ or the Orlicz norm 1 kuk0 = inf [1 + ρM (ku)]. k>0 k If X = R, then L0M (R) is said to be the Musielak-Orlicz function space. Let us set 1 0 K(u) = k > 0 : (1 + ρM (ku)) = kuk . k In particular, the set K(u) can be nonempty. However, we have the following. Proposition 1.3 (see [13]). If limu→∞ ∅ for any u ∈ L0M (X). 2.

M (t,u) u

= ∞ µ−a.e. t ∈ T , then K(u) 6=

Main results

Theorem 2.1. The space L0M (X) is midpoint locally uniformly rotund if and only if: (a) (b) (c) (d)

For any u ∈ L0M (X)\{0}, the set K(u) consists of one element from (0,+∞); M ∈ ∆2 ; M (t, u) is strictly convex with respect u for almost all t ∈ T ; X is midpoint locally uniformly rotund.

In order to prove the theorem, we give some lemmas. Lemma 2.2. Let z/kzk be a strongly extreme point. Then for any ε > 0 and α ∈ (0, 1) we have (a)

δ(α, z) := inf {max(kxk − kzk , kyk − kzk) : kx − yk ≥ ε, z = αx + (1 − α)y} > 0; x∈X

(b)

if zn → z, αn → α, then δ(αn , zn ) → δ(α, z).

480


∞ Proof. (a) Suppose that δ(α, z) = 0. Then there exists {xn }∞ n=1 ⊂ X, {yn }n=1 ⊂ X such that kxn − yn k ≥ ε, z = αxn + (1 − α)yn , but kxn k − kzk < n1 and kyn k − kzk < n1 , hence

xn 1

kzk ≤ 1 + n kzk ,

yn 1

kzk ≤ 1 + n kzk ,

xn 1

kzk ≤ 1 + n kzk ,

zn 1

kzk ≤ 1 + n kzk ,

α

xn yn z + (1 − α) = . kzk kzk kzk

We may assume without loss of generality that α ≥ 21 . Then there exists zn ∈ [xn , yn ] such that z = 21 xn + 12 zn . It is easy to see that 1 xn 1 zn z · + · = . 2 kzk 2 kzk kzk

Since z/kzk is a strongly extreme point, we have

xn

xn z z n

kzk − kzk → 0 ⇒ kzk − kzk → 0 ⇒ kxn − zk → 0(n → ∞),

a contradiction!

(b) Suppose that lim supn→∞ δ(αn , zn ) > δ(α, z), where zn → z, αn → α. Then there exist r > 0 and a subsequence of {n} denoted again by {n}, such that δ(αn , zn ) − δ(α, z) ≥ r. By the definition of δ(α, z), there exist x0 , y0 ∈ X such that r δ(α, z) + > max(kx0 k − kzk , ky0 k − kzk), 8 kx0 − y0 k ≥ ε,

z = αx0 + (1 − α)y0 .

Since zn → z, αn → α, there exists n1 ∈ N such that kzn1 − zk < k(αn1 − α)(x0 − y0 )k < 81 r. Let hn = (αn − α)(x0 − y0 ). Hence we have zn1 = = = = = = =

(1) 1 r 8

and

zn1 zn1 + αn1 x0 + (1 − αn1 )y0 − (αn1 − α)(x0 − y0 ) − [αx0 + (1 − α)y0 ] zn1 + αn1 x0 + (1 − αn1 )y0 − hn1 − z zn1 + αn1 x0 + (1 − αn1 )y0 − [αn1 hn1 + (1 − αn1 )hn1 ] − [αn1 z + (1 − αn1 )z] zn1 + αn1 (x0 − hn1 − z) + (1 − αn1 )(y0 − hn1 − z) αn1 zn1 + (1 − αn1 )zn1 + αn1 (x0 − hn1 − z) + (1 − αn1 )(y0 − hn1 − z) αn1 (x0 − hn1 − z + zn1 ) + (1 − αn1 )(y0 − hn1 − z + zn1 )

and k(x0 − hn1 − z + zn1 ) − (y0 − hn1 − z + zn1 )k = kx0 − y0 k ≥ ε.


481

By (1), we get the following inequality kx0 − hn1 − z + zn1 k − kzn1 k ≤ kx0 k + khn1 k + kzn1 − zk − kzn1 k 1 ≤ kx0 k + r + 8 1 ≤ kx0 k + r + 8

1 r − (kzk − kzn1 − zk) 8 1 1 r − kzk + r 8 8 3 = kx0 k − kzk + r 8 1 3 < δ(α, z) + r + r 8 8 1 = δ(α, z) + r. 2

Similarly, we have ky0 − hn1 − z + zn1 k − kzn1 k ≤ δ(α, z) + 21 r. Hence, δ(αn1 , zn1 ) ≤ δ(α, z) + 21 r, a contradiction! Similarly, lim infn→∞ δ(αn , zn ) < δ(α, z) is impossible. Thus, limn→∞ δ(αn , zn ) = δ(α, z). This completes the proof. Lemma 2.3 (see [13]). Let u ∈ L0M (X) \ {0}. If K(u) = ∅, then kuk0 = R A(t) · ku(t)k dt, where A(t) = limu→∞ M (t,u) . u T Lemma 2.4. For any u ∈ L0M (X), we have the inequalities kuk ≤ kuk0 ≤ 2 kuk. Proof. For any k > 0, we have ρM

u 1 [ρ (ku) + 1] k M

= ρM

ku ρM (ku) + 1

≤

ρM (ku) ≤ 1. ρM (ku) + 1

Therefore kuk ≤

1 1 [ρM (ku) + 1] whence kuk ≤ inf [ρM (ku) + 1] = kuk0 . k>0 k k

On the other hand,

u 0 u 0

kuk ≤ 1 + ρM kuk ≤ 2 whence kuk ≤ 2 kuk .

Lemma 2.5. If M ∈ ∆2 , then for any u ∈ L0M (X) is norm absolotely continuous. Proof. Given any ε > 0, there exists k > 0 such that k1 < 21 ε. Since M ∈ ∆2 , we have ρM (ku) < ∞. Since integral absolutely continuity, we can find δ > 0 such that µE < δ implies that ρM (ku · χE ) < 1. Hence, if µE < δ, then ku · χE k0 ≤

1 2 1 [1 + ρM (ku · χE )] < [1 + 1] = < ε. k k k

482


Lemma 2.6 (see [1]). Suppose M ∈ ∆2 and e(t) = 0 µ-a.e. on T . Then ρM (un ) → 0 ⇔ kun k → 0

and ρM (un ) → 1 ⇔ kun k → 1(n → ∞),

where e(t) = sup{u ≥ 0 : M (t, u) = 0}. Lemma 2.7 (see [1]). The following are equivalent: (a) (b)

M∈ / ∆2 For each ε ∈ (0, 1), there exists u ∈ LM (X) such that ρM (u) = ε, kuk = 1 and ku(t)k < E(t) µ−a.e. on T , where E(t) = sup{u > 0 : M (t, u) < ∞}.

Lemma 2.8 (see [13]). The space L0M (X) is rotund if and only if: (a) (b) (c)

For any u ∈ S(L0M (X)), the set K(u) consists of one element from (0, +∞); X is rotund; M (t, u) is strictly convex with respect to u for almost all t ∈ T .

Proof. Necessity. By Lemma 2.8, (c) is obvious. For any u ∈ L0M (X)\{0}, we

u 0 1

= [1+ρM (k u 0 )] by Lemma 2.8, where k ∈ K( u 0 ). Hence kuk0 = have kuk 0 k kuk kuk

u 0 kuk0 1 k u [1+ρM ( kuk0 ·u)]. If k1 , k2 ∈ K(u), then kuk0 = k kuk0 [1+ρM (k1 kuk0 · kuk 0 )]. k 1

0

This implies that k1 kuk ∈

u K( kuk 0 ).

u Similarly, we have that k2 kuk0 ∈ K( kuk 0 ).

By Lemma 2.8, we get k1 = k2 . (b) Suppose that M ∈ / ∆2 . By Lemma 2.7, there exists u ∈ L0M (X) such that ρM (u) < 12 , kuk = 1 and ku(t)k < E(t) µ−a.e. on T . Then for any L > 1, we have ρM (Lu) = ∞. Indeed, suppose that there exists L1 > 1 such that ρM (L1 u) < ∞. R We know that the function F (k) = T M (t, k ku(t)k)dt is continuous on [1, L1 ]. Then there exists L2 > 1 such that ρM (L2 u) = 1. This implies that kuk ≤ L12 , contradicting kuk = 1. Decompose T into R that µE1 = µG1 . Then for any R E1 and G1 in such a way L > 1, we have E1 M (t, L ku(t)k)dt = ∞ or G1 M (t, L ku(t)k)dt = ∞. Without R loss of generality, we may assume that E1 M (t, L ku(t)k)dt = ∞. Decompose E1 into E2 and G2 in such a way R that µE2 = µG2 . Then for any L > 1, we have R M (t, L ku(t)k)dt = ∞ or G2 M (t, L ku(t)k)dt = ∞. Without loss of generalE2 R ity, we may assume that E2 M (t, L ku(t)k)dt = ∞. In generical, decompose En into ERn+1 and Gn+1 in such a way that R µEn+1 = µGn+1 . Then for any L > 1, we have En+1 M (t, L ku(t)k)dt = ∞ or Gn+1 M (t, L ku(t)k)dt = ∞. Without loss of R generality, we may assume that En+1 M (t, L ku(t)k)dt = ∞. Hence we have 1 E1 ⊃ E2 ⊃ E3 ⊃ · · ·, µEi = µEi+1 2

and

ku(t) · χEi (t)k = 1, i = 1, 2, ....

Let k ∈ K(u). It is easy to see that k ≤ 1. We will derive a contradiction for each the following two cases.


483

Case 1. Let k < 1. Define un (t) =

n−1 X

u(t) · χFi (t),

vn (t) =

i=1

n−1 X i=1

∞

1X u(t) · χFi (t) + u(t) · χFi (t), k i=n

where Fi = Ei \E k)un + kvn = u, kun k0 ≤ kuk0 and kuRn − vn k0 ≥

(1 −

1 i+1 . Then P kun − vn k = k u · χEn = k1 . For any ε > 0, we have k1 ∞ i=n T M (t, kk· 1 u(t)χFi (t)k)dt < ε, when n is large enough. Therefore, k kuk0 ≤ kvn k0 n−1

1 1X ≤ + k k i=1

Z

M (t, kku(t) · χFi (t)k)dt

T

∞

1X + k i=n ≤ kuk0 + ε

Z

T

1

M t,

k · k u(t)χFi (t) dt

This implies that kvn k0 → kuk0 as n → ∞, a contradiction! Case 2. Let k = 1. By the proof of Case 1, without loss of generality, we may assume that µ{t ∈ T : ku(t)k = 0} > 0. Since M (t, u) is strictly convex with respect u for almost all t ∈ T , then p(t, u) < ∞ for almost all t ∈ T . Hence, there 6 0} ⊂ {t ∈ T : ku(t)k = 0}, exists v ∈ L0M (X) \ {0} such that {t ∈ T : kv(t)k = ρN (p( 21 v)) > 1 and ρM (2v) < ∞. Put u0 = u + v. Pick 0 < k1 < 1, 0 < k2 < 1, we have Z ρM (k1 u0 χTn ) ≥ k1 ku0 (t)χTn (t)k · p(t, kk2 u0 (t)χTn (t)k)dt − ρN (p(k2 u0 χTn )) T

ρM (k2 u0 χTn ) =

Z

k2 ku0 (t)χTn (t)k · p(t, kk2 u0 (t)χTn (t)k)dt − ρN (p(k2 u0 χTn )),

T

where Tn = {t ∈ T : k2 ku0 (t)k ≤ n, p(t, kk2 u0 (t)k) ≤ n}. It is follows that 1 1 [1 + ρM (k1 u0 χTn )] − [1 + ρM (k2 u0 χTn )] k1 k2 k1 − k2 k2 = −1 + [ρM (k1 u0 χTn ) − ρM (k2 u0 χTn )] − ρM (k2 u0 χTn ) k1 k2 k1 − k2 Z k1 − k2 k2 ≥ (k1 − k2 ) ku0 (t)χTn (t)k · p(t, kk2 u0 (t)χTn (t)k)dt −1+ k1 k2 k1 − k2 T − ρM (k2 u0 χTn ) =

k1 − k2 (ρN (p(k2 u0 χTn )) − 1). k1 k2

484


Taking n → ∞, we obtain k1 − k2 1 1 [1 + ρM (k1 u0 )] ≥ (ρN (p(k2 u0 )) − 1) + [1 + ρM (k2 u0 )]. k1 k1 k2 k2

(2)

By (2), it is easy to see that h < 1, where h ∈ K(u0 ). Define n−1 X

∞

n−1 X

1X u0 (t) · χFi (t) + u0 (t) · χFi (t), vn (t) = u0 (t) · χFi (t), un (t) = h i=1 i=1 i=n

′ 0

′

0 ′ ′ ′ where Fi = Ei \Ei+1 . Then (1−h)un +hvn = u0 , un ≤ ku0 k0 and un − vn ≥

′

un − vn′ = 1 u0 · χEn ≥ 1 . Using the method of Case 1, we can easily deduce k h

0

′ that un − u0 → 0, a contradiction! ′

′

∞ Suppose that (d) is not true. Then there exist x ∈ S(X), {xn }∞ n=1 ⊂ X, {yn }n=1 ⊂ X with x = 21 (xn + yn ), we have kxn k → 1, kyn k → 1 and xn − yn 9 0. Pick h(t) ∈ S(L0M (X)), then there exists d > 0 such that µH > 0, where H = {t ∈ ′ T : kh(t)k ≥ 2d}. Since h(t) ∈ S(L0M (X)), then there exists k > 0 such that Z Z ′ ′ M (t, k kh(t)k)dt ≤ M (t, k kh(t)k)dt < ∞. H

T

Setting u(t) = d · x · χH (t), we have Z

un (t) = d · xn · χH (t),

′

M (t, k ku(t)k)dt =

T

Z

Z

′

M (t, k d)dt ≤

H

′

M (t, k kun (t)k)dt ≤

T

Z

vn (t) = d · yn · χH (t).

Z

′

M (t, k kh(t)k)dt < ∞,

H

′

M (t, 2k d)dt ≤

H

Z

M (t, k kh(t)k)dt < ∞

Z

M (t, k kh(t)k)dt < ∞.

′

H

and Z

T

′

M (t, k kvn (t)k)dt ≤

Z

′

M (t, 2k d)dt ≤

H

′

H

0 and {vn }∞ ⊂ This implies that Ru ∈ n=1 ⊂ LM (X). Since M ∈ ∆2 , we have T M (t, k k2d · x · χH (t)k)dt < ∞, where k ∈ K(u). Notice that M (t, k kd · xn · χH (t)k) ≤ M (t, k k2d · x · χH (t)k). By the dominated convergence theorem, we have Z 1 0 1 + M (t, k kd · xn · χH (t)k)dt lim kun k = lim n→∞ n→∞ k T Z 1 1+ = lim M (t, k kd · xn · χH (t)k)dt k T n→∞ Z 1 1 + M (t, lim k kd · xn · χH (t)k)dt = n→∞ k T Z 1 1 + M (t, k kd · x · χH (t)k)dt = kuk0 . = k T

L0M (X),

{un }∞ n=1

L0M (X),


485

Similarly, we have that limn→∞ kvn k0 = kuk0 . Since xn − yn 9 0, there exist ε0 > ∞ 0 and a subsequence {xnk − ynk }∞ k=1 of {xn − yn }n=1 such that kxnk − ynk k ≥ ε0 . Then Z 1 0 kun − vn k ≥ inf 1 + M (t, k kd · ε0 · χH (t)k)dt > 0, k>0 k T which implies that un − vn 9 0. This shows that L0M (X) is not midpoint locally uniformly rotund, a contradiction! Sufficiency. Put kun k0 → 1, kvn k0 → 1, un + vn = 2u, we want to prove that kun − vn k0 → 0 as n → ∞. The proof requires of considering of few cases separately. Case 1. Let max{sup{kn }, sup{hn }} < ∞, where kn = K(un ), hn = K(vn ). Without loss of generality, we may assume that kn → k and hn → h. Since kun k0 = k1n [1 + ρM (kn un )] → 1, kvn k0 = h1n [1 + ρM (hn vn )] → 1 as n → ∞, without loss of generality, we may assume that kn ≥ in view of the Fatou Lemma, we get

3 4

and hn ≥ 34 . Let l =

2kh , k+h

1 = kuk0 1 [1 + ρM (lu)] l kn + hn 2kn hn ≤ lim inf 1 + ρM u n→∞ 2kn hn kn + hn hn kn kn + hn 1 + ρM ( kn un ) + (hn vn ) ≤ lim inf n→∞ 2kn hn kn + hn kn + hn 1 1 1 = lim inf (1 + ρM (kn un )) + (1 + ρM (hn vn )) n→∞ 2 kn hn 1 = lim inf kun k0 + kvn k0 n→∞ 2 = 1. ≤

Consequently, l = K(u). Next we will prove that for any σ > 0, ε > 0 there exists N such that µ{t ∈ T : kkn un (t) − hn vn (t)k ≥ σ} < ε whenever n > N . Otherwise, without loss of generality, we may assume that for ′ ′ each n ∈ N there exists En ⊆ T , ε0 > 0 and σ0 > 0 such that µEn ≥ ε0 , where ′

En = {t ∈ T : kkn un (t) − hn vn (t)k ≥ σ0 }. We define the function η(t) = inf max(kxk − klu(t)k , kyk − klu(t)k) : kx − yk ≥ σ0 , x∈X h k lu(t) = x+ y , k+h k+h

486


where t ∈ {t ∈ T : u(t) 6= 0}. By Lemma 2.2, we have η(t) > 0. Let hn (t) → u(t) µ-a.e. on T , where hn are simple functions. Then the functions ηn (t) = inf

x∈X

max(kxk − klhn (t)k , kyk − klhn (t)k) : kx − yk ≥ σ0 , k h x+ y lhn (t) = k+h k+h

are Σ-measurable. By Lemma 2.2, we have ηn (t) → η(t) µ-a.e. on T . Then η(t) is Σ-measurable. Using the inclusion ∞

T ⊃ ∪

n=1

1 1 < η(t) ≤ , u(t) 6= 0 , t∈T : n+1 n

we get that there exists 0 < η0 < σ0 such that µH
N then |λn (t) − η(t)| < η20 on F , where F ⊂ T ′ ′ ′′ ′′ 1 and µ(T \F ) < 16 ε0 . Let En3 = En3 \(H ∪ (T \F )), then µ(En3 − En3 ) < 81 ε0 . ′′ Hence, if t ∈ En3 , then

2kn hn

≥ λn (t) > η(t) − η0 ≥ 2η0 − η0 > η0 , kkn un (t)k − u(t)

kn + hn

2 2

(3)


487

when n is large enough. By kkn un (t)k − khn vn (t)k < 41 η0 and (3), we get the following inequality

2kn hn

kn hn

kkn un (t)k + khn vn (t)k − u(t)

kn + hn kn + hn kn + hn

hn hn

2kn hn u(t) = kkn un (t)k −


kn 2kn hn kn

+ khn vn (t)k − u(t)


2kn hn hn kn kn

≥ η0 + khn vn (t)k − u(t)

kn + hn kn + hn kn + hn kn + hn

2k h hn kn kn η0 kn n n

≥ η0 + kkn un (t)k − · − u(t)

kn + hn kn + hn kn + hn 4 kn + hn kn + hn

hn kn kn

2kn hn u(t) − kn · η0 = η0 + kkn un (t)k −

kn + hn 4 kn + hn kn + hn kn + hn kn + hn hn kn kn η0 ≥ η0 + η0 − · kn + hn kn + hn kn + hn 4 hn kn 3 ≥ η0 + · η0 kn + hn kn + hn 4 3 ≥ η0 , 4 when n is large enough. Let En = En1 ∪ En2 ∪ En3 , then µEn ≥ 87 ε0 , when n is large enough. Let us define the sets 8 , An = t ∈ T : M (t, kkn un (t)k) > ε0 8 Bn = t ∈ T : M (t, khn vn (t)k) > . ε0 ′

Then 1=

Z

M (t, kkn un (t)k)dt ≥

T

′

Z

′′

M (t, kkn un (t)k)dt ≥

An

8 µAn . ε0

Similarly, we have that µBn ≤ 81 ε0 . For µ-a.e. This implies that µAn ≤ t ∈ T , we define the bounded closed set 8 8 1 2 Ct = (u, v) ∈ R : M (t, u) ≤ , M (t, v) ≤ , |u − v| ≥ η0 ε0 ε0 4 1 ε. 8 0

in the two dimensional space R2 . Since Ct is compact, then for µ-a.e. t ∈ T there exists (ut , vt ) ∈ Ct such that 1>

h ut + M (t, ( k+h h M (t, ut ) k+h

+

k v )) k+h t

k M (t, vt ) k+h

≥

h u+ M (t, ( k+h h M (t, u) k+h

+

k v)) k+h

k M (t, v) k+h

(4)

488


for any (u, v) ∈ Ct . If we define the function 1 − δ(t) =

h M (t, ( k+h ut + h M (t, ut ) k+h

+

k v )) k+h t

k M (t, vt ) k+h

(5)

then δ(t) is Σ-measurable. Indeed, if we pick a dense set {ri }∞ i=1 in [0, ∞) and define the function  k h ri + k+h rj )) M (t, ( k+h 8 8    h , M (t, ri ) ≤ and M (t, rj ) ≤ , k ε0 ε0 1 − δri ,rj (t) = k+h M (t, ri ) + k+h M (t, rj )  8 8  0, M(t, ri ) > or M (t, rj ) > , ε0 ε0

then by the definition of M (t, u), it is easy to see that 1 − δri ,rj (t) is Σ-measurable and 1 1 − δ(t) ≥ sup 1 − δri ,rj (t) : |ri − rj | ≥ η0 . 4 ∞ On the other hand, since {ri }∞ i=1 is dense in [0, ∞) then {(ri , rj )}i=1,j=1 is dense in [0, ∞) × [0, ∞). By definition of function 1 − δ(t), for µ-a.e. t ∈ T, ε > 0, there exists (ri , rj ) ∈ {(u, v) ∈ R2 : M (t, u) ≤ ε80 , M (t, v) ≤ ε80 , |u − v| ≥ 41 η0 } such that 1 1 − δ(t) − ε < 1 − δri ,rj (t) ≤ sup 1 − δri ,rj (t) : |ri − rj | ≥ η0 4

µ-a.e. on T . Since ε > 0 was arbitrary, we have 1 1 − δ(t) ≤ sup 1 − δri ,rj (t) : |ri − rj | ≥ η0 4 µ-a.e. on T . Therefore, 1 − δ(t) = sup{1 − δri ,rj (t) : |ri − rj | ≥ 41 η0 } µ-a.e. in T . This implies that δ(t) is Σ-measurable. By (4) and (5), we have h k M t, u+ v k+h k+h k h M (t, u) + M (t, v) u, v ∈ Ct ≤ (1 − δ(t)) k+h k+h for µ-a.e. t ∈ T . Moreover, we have ∞ T ⊃ ∪ t∈T : n=1

1 1 < δ(t) ≤ n+1 n

.

Since M (t, u) is strictly convex with respect u for almost all t ∈ T , then there 1 exists 2δ0 ∈ (0, 1) such that µG < 16 ε0 , where G = {t ∈ T : δ(t) ≤ 2δ0 }.


489

We have Wn (t) − Qn (t) → 0(n → ∞) µ−a.e. on T , where Wn (t) =

n M (t, knh+h kkn un (t)k + n

hn M (t, kkn un (t)k) kn +hn

Qn (t) =

+

kn khn vn (t)k) kn +hn kn M (t, khn vn (t)k) kn +hn

h M (t, k+h kkn un (t)k + h M (t, kkn un (t)k) k+h

+

k k+h

khn vn (t)k)

k M (t, khn vn (t)k) k+h

· χE ′

n1 \(An ∪Bn )

· χE ′

n1 \(An ∪Bn )

(t)

(t).

By the Egorov theorem, there exists N such that if n > N , then |Wn (t) − Qn (t)| ′ 1 < 14 δ0 on T \E, where E ⊂ T and µE < 16 ε0 . Hence, if t ∈ En1 \(E ∪G∪An ∪Bn ), then 1 3 δ0 = 2δ0 − δ0 2 2 k h kkn un (t)k + k+h khn vn (t)k) M (t, k+h 1 − δ0 ≤ 1− h k M (t, kkn un (t)k) + k+h M (t, khn vn (t)k) 2 k+h ≤ 1−

kn khn vn (t)k) kn +hn , kn M (t, khn vn (t)k) kn +hn

n kkn un (t)k + M (t, knh+h n

hn M (t, kkn un (t)k) kn +hn

+

when n is large enough. This implies that kn hn kkn un (t)k + khn vn (t)k M t, kn + hn kn + hn hn kn ≤ (1 − δ0 ) M (t, kkn un (t)k) + M (t, khn vn (t)k) kn + hn kn + hn on En1 \(E ∪ G ∪ An ∪ Bn ). Let K = min{ 21 σ0 , 43 η0 }. Then we have M (t, K) > 0 1 1 µ-a.e. on T . By T ⊃ ∪∞ i=1 {t ∈ T : i+1 < M (t, K) ≤ i }, there exists a > 0 such that µC < 18 ε0 , where C = {t ∈ T : M (t, K) ≤ a}. ′

′

Let Hn = En \ (E ∪ G ∪ C ∪ An ∪ Bn ), Hn1 = En1 \(E ∪ G ∪ C ∪ An ∪ Bn ), Hn2 = ′′ ′ En2 \(E ∪ G ∪ C ∪ An ∪ Bn ), Hn3 = En3 \(E ∪ G ∪ C ∪ An ∪ Bn ). Then µHn ≥ 41 ε0 . Let k = sup{{kn }, {hn }}. Then kun k0 + kvn k0 − 2 kuk0 1 2kn hn 1 kn + hn ≥ 1 + ρM (kn un ) + (1 + ρM (hn vn )) − 1 + ρM u kn hn kn hn kn + hn Z kn + hn hn kn = M (t, kkn un (t)k) + M (t, khn vn (t)k) kn hn T kn + hn kn + hn

2kn hn

− M t, u(t)

dt kn + hn Z kn + hn hn kn ≥ M (t, kkn un (t)k) + M (t, khn vn (t)k) kn hn Hn1 kn + hn kn + hn kn hn − M t, kkn un (t)k + khn vn (t)k dt kn + hn kn + hn

490

≥

≥

=

≥

≥


Z kn + hn hn kn + M (t, kkn un (t)k) + M (t, khn vn (t)k) kn hn Hn2 kn + hn kn + hn

2kn hn

− M t,

kn + hn u(t) dt Z kn + hn hn kn + M (t, kkn un (t)k) + M (t, khn vn (t)k) kn hn Hn3 kn + hn kn + hn

2kn hn

u(t) − M t,

dt kn + hn Z kn + hn hn kn δ0 M (t, kkn un (t)k) + M (t, khn vn (t)k) dt kn hn Hn1 kn + hn kn + hn Z hn kn + hn kn + M (t, kkn un (t)k) + M (t, khn vn (t)k) dt kn hn Hn2 kn + hn kn + hn Z hn kn kn + hn M t, kkn un (t)k + khn vn (t)k + kn hn Hn3 kn + hn kn + hn

2kn hn

dt − M t, u(t)

kn + hn

Z kn + hn hn kn δ0 M (t, kkn un (t)k) + M (t, khn vn (t)k) dt kn hn Hn1 ∪Hn2 kn + hn kn + hn Z hn kn kn + hn M t, kkn un (t)k + khn vn (t)k + kn hn Hn3 kn + hn kn + hn

2kn hn

−

kn + hn u(t) dt Z 1 δ0 [hn M (t, kkn un (t)k) + kn M (t, khn vn (t)k)]dt kn hn Hn1 ∪Hn2 Z hn kn + hn kn M t, + kkn un (t)k + khn vn (t)k kn hn Hn3 kn + hn kn + hn

2kn hn

dt − u(t)

kn + hn

Z δ0 3 3 M (t, kkn un (t)k) + M (t, khn vn (t)k)dt kn hn Hn1 ∪Hn2 4 4 Z hn kn + hn kn M t, + kkn un (t)k + khn vn (t)k kn hn Hn3 kn + hn kn + hn

2kn hn

− u(t)

dt kn + hn Z δ0 3 3 M (t, kkn un (t)k) + M (t, khn vn (t)k)dt kn hn Hn1 ∪Hn2 4 4 Z 3 kn + hn M t, η0 dt + kn hn Hn3 4


1 1 M (t, kkn un (t)k) + M (t, khn vn (t)k)dt 2 1 ∪H 2 2 Hn n Z 3 kn + hn M t, η0 dt + kn hn Hn3 4 Z 1 3δ0 1 M t, kkn un (t)k + khn vn (t)k dt 2kn hn Hn1 ∪Hn2 2 2 Z kn + hn 3 + M t, η0 dt kn hn Hn3 4 Z Z kkn un (t) − hn vn (t)k 3 3δ0 kn + hn M t, M t, η0 dt dt + 2kn hn Hn1 ∪Hn2 2 kn hn Hn3 4 Z Z 3 3δ0 kn + hn kkn un (t) − hn vn (t)k M t, η0 dt dt + M t, 2 2 kn hn Hn3 4 2k Hn1 ∪Hn2 Z Z 3 3δ0 kn + hn σ0 M t, η0 dt dt + M t, 2 2 kn hn Hn3 4 2k Hn1 ∪Hn2 Z Z 3δ0 3 σ0 3 dt + 2 M t, M t, η0 dt 2 2 4 2k Hn1 ∪Hn2 2k Hn3 Z Z 3δ0 3δ0 M (t, K)dt + 2 M (t, K)dt 2 2k Hn1 ∪Hn2 2k Hn3 Z 3δ0 M (t, K)dt 2 2k Hn 1 3δ0 2 · a · ε0 , 4 2k

3δ0 = 2kn hn

≥

≥ ≥ ≥ ≥ ≥ = ≥

491

Z

when n is large enough. However, kun k0 + kvn k0 − 2 kuk0 → 0 as n → ∞, a contradiction. Hence, for any σ > 0 and ε > 0 there exists N such that µ{t ∈ T : kkn un (t) − hn vn (t)k ≥ σ} < ε, whenever n > N . By the Riesz theorem, there exists a subsequence of {n} denoted again by {n}, such that kn un (t) − hn vn (t) → 0 µ-a.e. on T . Moreover, hn hn 2kn hn kn un (t) + kn vn (t) = u(t) → lu(t) as n → ∞. kn + hn kn + hn kn + hn Hence we have kn un (t) → lu(t) µ-a.e. on T as n → ∞. Then k ≥ l. In fact, by the Fatou Lemma, it follows that 1 1 1 [1 + ρM (lu)] = kuk0 = lim kun k0 = lim [1 + ρM (kn un )] ≥ [1 + ρM (lu)], n→∞ n→∞ kn l k which implies the inequality k ≥ l. Similarly, we have h ≥ l. Then k = l. In fact, suppose that k − l = r1 > 0, h − l = r2 ≥ 0. Then 2(l + r1 )(l + r2 ) 2l2 + 2(r1 + r2 )l + 2r1 r2 2l + 2(r1 + r2 ) 2kh = = ≥ l > l, k+h 2l + r1 + r2 2l + r1 + r2 2l + r1 + r2

492


2kh = l. By 1l [1+ρM (lu)] = 1 and k1n [1+ρM (kn un )] → 1 contradicting the equality k+h as n → ∞, we have ρM (lu) − l + 1 = 0 and ρM (kn un ) − kn + 1 → 0 as n → ∞. Hence, ρM (kn un ) → ρM (lu) as n → ∞. By the convexity of M , we have

M (t, kkn un (t)k) + M (t, klu(t)k) −M 2

kkn un (t) − lu(t)k t, ≥0 2

for µ-a.e. t ∈ T . Therefore, by the Fatou Lemma, we obtain that ρM (ku) Z kkn un (t) − lu(t)k M (t, kkn un (t)k) + M (t, klu(t)k) − M t, dt = lim 2 2 T n→∞ Z M (t, kkn un (t)k) + M (t, klu(t)k) kkn un (t) − lu(t)k = lim inf − M t, dt n→∞ 2 2 T 1 = ρM (ku) − lim sup ρM (kn un − lu) , 2 n→∞ which implies that ρM ( 12 (kn un − lu)) → 0 as n → ∞. By Lemma 2.6, we have that kkn un − luk → 0 as n → ∞. By Lemma 2.4, we have that kkn un − luk0 → 0 as n → ∞. By limn→∞ kn = k = l, we have the convergence kun − uk0 → 0 as n → ∞. Thus, kun − vn k0 → 0 as n → ∞. Case 2. Let max{sup{kn }, sup{hn }} = ∞, where kn = K(un ), hn = K(vn ). We ′ ′ consider the sequence un = 21 (un + u) and vn = 12 (vn + u) in place of {un }∞ n=1 0 0 ∞ and {vn }n=1 , because kun − uk → 0 and kvn − uk → 0 as n → ∞ if and only if

′

un − u 0 → 0 and vn′ − u 0 → 0 as n → ∞, respectively. Moreover,

0

1

(un + u) ≤ 1 (kun k0 + kuk0 )

2

2

for every n ∈ N . Hence,

0

1 lim sup (un + u)

≤ 1.

2 n→∞

0 In the same way we can prove that lim supn→∞ 12 (vn + u) ≤ 1. Since 12 (un +

0 u)+ 12 (vn +u) = 2u, for every n ∈ N , we conclude that lim infn→∞ 12 (un + u) ≥

0

0 1 and lim infn→∞ 12 (vn + u) ≥ 1. Consequently, limn→∞ 12 (un + u) → 1 and

0 limn→∞ 1 (vn + u) → 1 as n → ∞. Define 2

wn =

2kn l kn + l

and rn =

2hn l hn + l

∞ for any n ∈ N , where l ∈ K(u). Then the sequences {wn }∞ n=1 and {rn }n=1 are


493

bounded. Moreover,

0

1

(un + u) ≤ 1 1 + ρM wn · un + u

2

wn 2 kn + l kn l = (un + u) 1 + ρM 2kn l kn + l kn + l kn l ≤ (kn un ) + ρM (lu) 1 + ρM 2kn l kn + l kn + l 1 1 1 ≤ (1 + ρM (kn un )) + (1 + ρM (lu)) 2 kn l 1 = (kun k0 + kuk0 ) → 1 as n → ∞ 2 whence it follow that kn l kn + l (un + u) → 1 as n → ∞. 1 + ρM 2kn l kn + l Analogously, rn + l 2rn l

rn l (un + u) → 1 as n → ∞. 1 + ρM rn + l

0

′ Therefore, we can prove in the same way as in Case 1 that un − u → 0 and

′

vn − u 0 → 0 as n → ∞. This completes the proof. References [1] S. T. Chen: Geometry of Orlicz spaces, Diss. Math. 356 (1996) 204 pp. [2] S. T. Chen, H. Hudzik, W. Kowalewski, Y. W. Wang, M. Wisła: Approximative compactness and continuity of metric projector in Banach spaces and applications, Sci. China, Ser. A 51 (2008) 293–303. [3] Y. Cui, H. Hudzik, R. Pluciennik: Extreme points and Strongly extreme points in Orlicz spaces equipped with the Orlicz-norm, Z. Anal. Anwend. 22 (2003) 789–817. [4] Y. Cui, H. Hudzik, J. Li, M. Wisła: Strongly extreme points in Orlicz spaces equipped with the p-Amemiya norm, Nonlinear Anal. 71 (2009) 6343–6364. [5] H. Hudzik, P. Kolwicz: A note on P -convexity of Orlicz and Musielak-Orlicz spaces of Bochner type, in: Function Spaces (Pozna´ n, 1998), H. Hudzik et al. (ed.), Lecture Notes in Pure and Appl. Math. 213, Marcel Dekker, New York (2000) 181–191. [6] H. Hudzik, M. Wisła: On extreme points of Orlicz spaces with Orlicz norm, Collect. Math. 44 (1993) 135–146. [7] H. Hudzik: Orlicz spaces without strongly extreme points and without H-points, Can. Math. Bull. 36 (1993) 173–177.

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