Notes on approximation in the Musielak-Orlicz spaces of vector multifunctions

Notes on approximation in the Musielak-Orlicz spaces of vector multifunctions

Andrzej Kasperski

Abstract. We introduce the spacesM_Y,ϕ¹ ,M_Y,ϕ^o,n, ˜M_Y,ϕ^o and M_Y,^o_d,ϕof multifunctions.

We prove that the spacesM_Y,ϕ¹ andM_Y,^o_d,ϕare complete. Also, we get some convergence theorems.

Keywords: Musielak-Orlicz space, multifunction, modular space of multifunctions, inte- gral operator, modular approximation

Classification: 46E99, 28B20

1. Introduction

In this paper we extend the results of [2] and [3] to the case of the spacesM_Y,ϕ¹ , M˜_Y,ϕ^o andM_Y,d,ϕ^o of multifunctions. All definitions and theorems connected with the idea of Musielak-Orlicz space can be found in [4] and [5].

Let I be a bounded interval. Let (I,Σ, µ) be the Lebesgue measure space.

Let X be a real separable Hilbert space with the normk ◦ k_X. We denote by L^ϕ(I, X) the Musielak-Orlicz space of all strongly measurable functionsx:I→X generated by a modular

̺(x) =R

Iϕ(t,kx(t)k_X)dµ,

whereϕis aϕ-function with a parameter such thatϕ:I×R→R₊, ϕ(t,◦) is an even continuous function, nondecreasing foru≥0,ϕ(t, u) = 0 iff u= 0 for every t∈I, ϕ(◦, u) is measurable for everyu∈R and lim

u→∞ϕ(t, u) =∞for a.e.t∈I.

The spaceL^ϕ(I, X) isN-complete (see [5, Corollaries 3.3]).

LetNbe the set of all positive integers.

2. Completeness

LetY be a real separable Hilbert space. Let o denote the zero element inY. Let

dist (A, B) = max(sup

x∈A

y∈Binf kx−yk_Y,sup

y∈B

x∈Ainf kx−yk_Y), for all nonempty boundedA, B⊂Y. Let

MY(I) ={F :I→2^Y :F(s) is nonempty for everys∈I, closed and bounded for a.e. s∈I}.

ForF, G∈M_Y(I) we introduce the functiond(F, G) by the formula:

d(F, G)(t) =







0, ifF(t) =G(t)

dist (F(t), G(t)), ifF(t), G(t) are bounded

∞, ifF(t)6=G(t) andF(t) orG(t) is unbounded

for everyt∈I.

Remark 1. IfX is a Banach space, then the space of all nonempty closed and bounded subsets ofX with dist is a complete metric space.

Lemma 1. LetFn∈MY(I)for everyn∈N. If:

(a) there isno>0 such thatd(Fn, Fm)are measurable form, n > no, (b) for everyε >0and everyδ >0 there existsK > no such that µ({t∈I:

d(Fn, Fm)(t)≥δ})< ε, for allm, n > K,

then there exist a subsequence{Fn_k}of the sequence{Fn}andF ∈M_Y(I)such thatd(Fnk, F)→0a.e. andd(Fn, F)are measurable forn > no.

Proof: LetFn∈M_Y(I) for everyn∈N. We have from the assumptions that there existsN(k) such thatµ({t∈I:d(Fn, Fm)(t)≥2^−k})<2^−kfor allm, n >

N(k). Letn₁=N(1),n₂= max{N(2), N(1) + 1}, . . .,nm= max{N(m), N(m− 1) + 1}. Letε >0 be arbitrary. So there isi₀such that 2ⁱ⁰⁻¹< ε. Leti₀< i < j.

LetA_i ={t∈I :d(Fni+1, Fni)(t)≥2⁻ⁱ}. It is easy to see thatµ(S∞

i=i0A_i)< ε and fort∈I\S∞

i=i0Ai we have

d(Fnj, Fni)(t)≤

j−1

X

k=i

d(Fnk+1, Fnk)(t)≤

∞

X

k=i

d(Fnk+1, Fnk)(t)< ε.

So for the subsequence {Fn_k} we have that for a.e. t ∈ I and for every ε > 0 there isK >0 such thatd(Fn_k, Fn_l)(t)< εfor allk, l > K. Hence by Remark 1 there is F ∈ M_Y(I) such that d(Fn_k, F) →0 as k → ∞ a.e. and d(Fn, F) are measurable forn > n₀ becaused(Fn, F) = lim_k→∞d(Fn_k, Fn) a.e.

Let:

M(I, Y) ={x:I→Y :xis strongly measurable}, M(I, R) ={q:I→R:q is measurable}.

We denote for all a ∈ Y, R, r ≥ 0, B(a, r) = {x ∈ Y : kx−ak_Y ≤ r},

R(o, r,R) ={x∈Y :r≤ kxk_Y ≤R}. Let:

M_Y^o,n(I) ={F ∈M_Y(I) :F(s) =

n

[

i=1

R(o, rⁱ_F(s), Rⁱ_F(s)) for everys∈I, rⁱ_F(◦), R_Fⁱ (◦)∈M(I, R) fori= 1, . . . , n, Rⁱ_F(t)≤r_Fⁱ⁺¹(t) fort∈I, i= 1, . . . , n−1, ifn >1},

M˜_Y^o(I) =

∞

[

i=1

M_Y^o,i(I),

M_Y^o(I) ={F ∈M_Y(I) :F(s) =B(o, R_F(s)) for everys∈I, R_F(◦)∈M(I, R)}, M_Y¹(I) ={F ∈M_Y(I) :F(s) =B(a_F(s), r_F(s)) for everys∈I, a_F(◦)∈

M(I, Y), r_F(◦)∈M(I, R)}.

IfF, G∈M_Y¹(I) andF(t) =G(t) for a.e.t∈I, thenF =Gin M_Y¹(I). IfF, G∈ M˜_Y^o(I) andF(t) =G(t) for a.e.t∈I, thenF =Gin ˜M_Y^o(I). In the setM_Y¹(I) we introduce the operations⊙:R×M_Y¹(I)→M_Y¹(I),⊕:M_Y¹(I)×M_Y¹(I)→M_Y¹(I) as follows: let F₁, F₂ ∈ M_Y¹(I), a ∈ R, F₁(s) = B(a_F1(s), r_F1(s)), F₂(s) = B(a_F2(s), r_F2(s)) for everys∈I; ifF =F₁⊕F₂ then

F(s) =B(a_F1(s) +a_F2(s), r_F1(s) +r_F2(s)) for every s∈I, if G=a⊙F1, then G(s) =B(aaF1(s), arF1(s)) for every s∈I.

It is easy to see thatF, G∈M_Y¹(I). In the set ˜M_Y^o(I) we introduce the operations

⊙:R×M˜_Y^o(I)→M˜_Y^o(I),⊕: ˜M_Y^o(I)×M˜_Y^o(I)→M˜_Y^o(I) as follows: letF1, F2∈ M˜_Y^o(I),a∈R,

F₁(s) =

n

[

i=1

R(o, rⁱ_F1(s), R_Fⁱ₁(s)), F₂(s) =

m

[

i=1

R(o, rⁱ_F2(s), Rⁱ_F2(s)) for alls∈I, ifF =F1⊕F2, thenF(s) = [

1≤i≤n 1≤j≤m

R(o, rⁱ_F1(s) +r^j_F2(s), Rⁱ_F1(s) +R_F^j₂(s))

for everys∈I, if

G=a⊙F₁, thenG(s) =

n

[

i=1

R(o, arⁱ_F1(s), aRⁱ_F1(s))

for everys∈I. It is easy to see thatF, G∈M˜_Y^o(I).

Let now

M_Y,ϕ^o (I) ={F ∈M_Y^o(I) :r_F(◦)∈L^ϕ(I, R)},

M_Y,ϕ¹ (I) ={F ∈M_Y¹(I) :aF(◦)∈L^ϕ(I, Y), rF(◦)∈L^ϕ(I, R)}, M˜_Y,ϕ^o (I) ={F ∈M˜_Y^o(I) :rⁱ_F(◦), Rⁱ_F(◦)∈L^ϕ(I, R) fori= 1, . . . , n,

ifF ∈M_Y^o,n(I)}.

Remark 2. IfF, G∈M_Y,ϕ¹ (I), thend(F, G) is measurable.

Proof: It is easy to see that

d(F, G)(s) =ka_F(s)−a_G(s)k_Y+|r_F(s)−r_G(s)| for a.e. s∈I,

sod(F, G) is measurable.

Remark 2’. IfF, G∈M˜_Y,ϕ^o (I), thend(F, G) is measurable.

Proof: Let F(s) =

n

[

i=1

R(o, rⁱ_F(s), Rⁱ_F(s)), G(s) =

m

[

j=1

R(o, r^j_G(s), R^j_G(s))

fors∈I. It is easy to see that d(F, G)(s) = dist (

n

[

i=1

[rⁱ_F(s), R_Fⁱ (s)],

m

[

j=1

[r_G^j(s), R_G^j(s)]) for a.e. s∈I,

sod(F, G) is measurable (see [1, Remark 1, p. 120]).

Definition 1. LetF, Fn∈M_Y(I) for everyn∈N. We writeFn d,ϕ

−→F, if there existsno>0 such thatd(Fn, F) are measurable forn > no and

Z

I

ϕ(t, ad(Fn, F)(t))dt→0 asn→ ∞ for everya >0.

Definition 2. Let Fn ∈ M_Y(I) for every n ∈ N. We say that the sequence {Fn}fulfils the (C,d, ϕ)-condition, if there existsno>0 such thatd(Fn, Fm) are measurable forn, m > no and for everyε >0 and every a >0 there is K > no

such thatR

Iϕ(t, ad(Fn, Fm)(t))dt < εfor allm, n > K.

Definition 3. LetA⊂M_Y(I). We say thatAis (C,d, ϕ)-complete, if for every sequence{Fn} such thatFn ⊂A for everyn∈N and the sequence{Fn} fulfils the (C,d, ϕ)-condition, there isF ∈Asuch thatFn

−→d,ϕ F.

Theorem 1. M_Y,ϕ¹ (I)is(C,d, ϕ)-complete.

Proof: Let Fn ∈ M_Y,ϕ¹ (I) for every n ∈ N and let the sequence {Fn} fulfil the (C,d, ϕ)-condition. LetFn(s) =B(a_Fn(s), r_Fn(s)) for every s∈Iand every n∈N. Then{aFn} is a Cauchy sequence in the Musielak-Orlicz spaceL^ϕ(I, Y) and{r_Fn}is a Cauchy sequence in the Musielak-Orlicz spaceL^ϕ(I, R). So there area∈L^ϕ(I, Y) andr∈L^ϕ(I, R) such that

̺(a(a−aFn))→0, ̺(a(r−rFn))→0 asn→ ∞for everya >0.

LetF(s) =B(a(s),r(s)) for everys∈I. It is easy to see thatF∈M_Y,ϕ¹ (I) and Fn

−→d,ϕ F.

Remark 3. M˜_Y,ϕ^o (I) is not (C,d, ϕ)-complete.

Now, let us denote

M_Y,d^o (I) ={F ∈M_Y(I) :d(Fn, F)→0 a.e. for some Fn∈M˜_Y,ϕ^o (I), n∈N}, M_Y,d,ϕ^o (I) ={F ∈M_Y,d^o (I) :Fn d,ϕ

−→F for someFn∈M˜_Y,ϕ^o (I), n∈N}.

Remark 4. IfF, G∈M_Y,d^o (I), thend(F, G) is measurable.

Proof: Let F, G∈M_Y,d^o (I). So there areFn, Gn ∈M˜_Y,ϕ^o (I),n∈N such that d(Fn, F) → 0 and d(Gn, G) → 0 as n → ∞ a.e. So d(Fn, Gn) → d(F, G) as n→ ∞a.e. Henced(F, G) is measurable because from Remark 2’d(Fn, Gn) are

measurable forn∈N.

Theorem 2. M_Y,d,ϕ^o (I)is(C,d, ϕ)-complete.

Proof: LetFn ∈ M_Y,d,ϕ^o (I) for every n ∈N, and let the sequence {Fn} fulfil the (C,d, ϕ)-condition. It is easy to prove that the sequence {Fn} fulfils the assumptions of Lemma 1, so there exist a subsequence {Fn_k} of the sequence {Fn}andF∈M_Y(I) such thatd(Fnk, F)→0 a.e. andd(Fn, F) are measurable.

We have by Fatou Lemma Z

I

ϕ(t, ad(Fn, F)(t))dt≤εforn > K, soFn

−→d,ϕ F. For everyn∈N,ε >0,a >0 there existsF_nⁿ∈M˜_Y,ϕ^o (I) such that R

Iϕ(t, ad(F_nⁿ, Fn)(t))dt < ε, so we have Z

I

ϕ(t,a

2d(F_nⁿ, F)(t))dt≤

≤ Z

I

ϕ(t, ad(F_nⁿ, Fn)(t))dt+ Z

I

ϕ(t, ad(Fn, F)(t))dt <2ε forn > K, henceF ∈M_Y,d,ϕ^o (I) andM_Y,d,ϕ^o (I) is (C,d, ϕ)-complete.

The spaces M_Y,ϕ¹ (I) andM_Y,d,ϕ^o (I) will be called the Musielak-Orlicz spaces of vector multifunctions.

3. On the operator H LetH:I×Y →Y and let

H(F)(t) ={H(t, x) :x∈F(t)} for everyt∈I, F ∈M_Y(I).

Lemma 2. Let the functionH fulfil the following conditions:

(a) H(s, x) is a strongly measurable function as a function of s for every x∈Y,

(b) there exists L >0 such that kH(s, x)−H(s, y)kY ≤Lkx−ykY for all s∈I,x, y∈Y,

(c) H(s, o) =ofor everys∈I,

(d) ifkxk_Y <kyk_Y, thenkH(s, x)k_Y <kH(s, y)k_Y and ifkxk_Y =kyk_Y, then kH(s, x)kY =kH(s, y)kY for everys∈I,

(e) for everyt∈I and everyy∈Y there isx∈Y such thaty=H(t, x).

ThenH:M_Y,ϕ^o (I)→M_Y,ϕ^o (I)andH: ˜M_Y,ϕ^o (I)→M˜_Y,ϕ^o (I).

Proof: We only prove that H : M_Y,ϕ^o (I) → M_Y,ϕ^o (I). The proof that H : M˜_Y,ϕ^o (I) → M˜_Y,ϕ^o (I) as analogous is omitted. Let F ∈ M_Y,ϕ^o (I). We prove that there exists r_H(F) ∈ L^ϕ(I, R), r_H(F)(t) ≥ 0 for every t ∈ I, such that H(F)(t) =B(o, r_H(F)(t)) for everyt∈I. Letx∈Y,x6=obe arbitrary. Let now ξ(t) =xr_F(t)/kxk_Y for everyt ∈I. It is easy to see thatξ∈M(I, Y)∩F and kξ(t)k_Y =r_F(t) for every t ∈ I. Let r_H(F)(t) =kH(t, ξ(t))k_Y for every t ∈I.

We have

sup

z∈H(F)(t)

kzk_Y = sup

x∈F(t)

kH(t, x)k_Y ≤ kH(t, ξ(t))k_Y,

for everyt∈I, soH(F)(t)⊂B(o, r_H(F)(t)) for everyt∈I. For everya >0 we have

Z

I

ϕ(t, ar_H(F)(t))dt= Z

I

ϕ(t, akH(t, ξ(t))k_Y)dt≤ Z

I

ϕ(t, aLkξ(t)k_Y)dt

= Z

I

ϕ(t, aLr_F(t))dt.

Sor_H(F)∈L^ϕ(I, R). Let t∈I be arbitrary, lety∈B(o, r_H(F)(t)).

From (e) we obtain that there existsx∈Y such thaty=H(t, x). SokH(t, x)k_Y ≤ kH(t, ξ(t))k_Y. Hence from (d) we obtain that kxk_Y ≤r_F(t). So x∈F(t) and y∈H(F)(t). HenceH(F)(t) =B(o, r_H(F)(t)) for everyt∈I.

Remark 5. Let C(F)(t) = H(F + (−aF))(t) for every t ∈ I, where F(t) = B(a_F(t), r_F(t)) for everyt∈I. If the assumptions of Lemma 2 hold, then

C:M_Y,ϕ¹ (I)→M_Y,ϕ^o (I).

Remark 6. Let the assumptions of Lemma 2 hold. If

(i) H(s, A) is closed for every nonempty and closedA⊂Y and for a.e.s∈I, thenH:M_Y,d,ϕ^o (I)→M_Y,d,ϕ^o (I).

Proof: The proof is analogous to that of Theorem 1’ in [2] so we give only the sketch of it. First, from the assumptions (b), (c) of Lemma 2 and from the assumption (i) H : MY(I) → MY(I). Second, from the assumption (b) of Lemma 2 we obtain that

(1) dist (H(F)(t),H(G)(t))≤Ldist (F(t), G(t))

for allF, G∈M_Y(I) and t∈I such thatF(t), G(t) are nonempty, bounded and closed. Third, from (1) and Lemma 2 we obtain that̺(ad(H(Fn),H(F)))→0 as n→ ∞for everya >0, whereF ∈M_Y,d,ϕ^o (I),Fn∈M˜_Y,ϕ^o ,n∈NandFn d,ϕ

−→F. SoH(F)∈M_Y,d,ϕ^o (I) because from Lemma 2H(Fn)∈M˜_Y,ϕ^o (I) for everyn∈N.

4. On the operators T_v^′ and T_v^′′

LetVbe an abstract set of indices and letV be a filter of subsets ofV.

Definition 4. A function g : V → R tends to zero with respect to V, written g(v)−→^V 0, if for everyε >0 there is a set V ∈ V such that |g(v)|< εfor every v∈V.

Definition 5. LetFv ∈M_Y(I) for every v∈V and let F ∈M_Y(I). We write Fv d,ϕ,V

−→ F, if there isVo∈ V such thatd(Fv, F) are measurable for everyv∈Vo

and for everyε >0, everya >0 there isV ∈ V such that Z

I

ϕ(t, ad(Fv, F)(t))dt < ε for everyv∈Vo∩V.

Definition 6. LetM(I)⊂M_Y(I). The familyT = (Tv)_v∈V of operators,Tv : M(I)→M(I) for everyv∈Vwill be called (d,V, M(I))-bounded, if there exist positive constants k₁, k₂ and a functiong : V→R+ such that g(v)−→^V 0, and for allF, G ∈M(I) such thatd(F, G) is measurable there exists a setV_F,G ∈ V such thatd(Tv(F), Tv(G)) are measurable and

Z

I

ϕ(t, ad(Tv(F), Tv(G))(t))dt≤k₁ Z

I

ϕ(t, ak₂d(F, G)(t))dt+g(v) for everya >0 and allv∈V_F,G.

Remark 7. Let the familyT be (d,V, M_Y,d,ϕ^o (I))-bounded. IfTv(F)^d,ϕ,V−→ F for everyF ∈M˜_Y,ϕ^o (I), thenTv(F)^d,ϕ,V−→ F for everyF ∈M_Y,d,ϕ^o (I).

Proof: Let a, ε > 0 be arbitrary. Let F ∈ M_Y,d,ϕ^o (I) be arbitrary. Let G ∈ M˜_Y,ϕ^o and V ∈ V be such that ̺(3ad(G, F))< ^ε₄, ̺(3ak2d(G, F)) < _4k^ε₁,

̺(3ad(Tv(G), G)) < ^ε₄, g(v) < ^ε₄ for every v ∈ V, where we may assume that k₁≥1. It is easy to see that suchG, V exist. We have for everyv∈V ∩V_F,G

̺(ad(Tv(F), F))≤

≤̺(3ad(Tv(F), Tv(G))) +̺(3ad(Tv(G), G)) +̺(3ad(G, F))< ε.

Let nowI= [0, b) and let us extendϕ b-periodically to the whole R.

Definition 7. We shall say that the functionϕisτ-bounded, if there are positive constantsk1, k2 such that

ϕ(t−v, u)≤k₁ϕ(t, k₂u) +f(t, v) for allu, v, t∈R,

where f : R×R → R₊ is measurable and b-periodic with respect to the first variable and such that writing h(v) = Rb

0f(t, v)dt for every v ∈ R, we have M = sup_v∈Rh(v)<∞andh(v)→0 asv→0 orv→b.

Let nowKv : [0, b)→R₊ for everyv∈V be integrable in [0, b) and singular, i.e.

σ(v) = Z b

0

Kv(t)dt−→^V 1, σ_δ(v) = Z b−δ

δ

Kv(t)dt−→^V 0

for every 0< δ < ₂^b,σ= sup_v∈Vσ(v)<∞. Let us extendKv b-periodically to the wholeR.

Let q : [0, b) → R be measurable and let us extend q b-periodically to the wholeR. We introduce the family of operatorsA¹ = (A¹_v)_v∈V by the formula:

A¹_v(q)(t) = Z b

0

Kv(s−t)q(s)ds for everyv∈Vand everyt∈[0, b).

Letx: [0, b)→Y be strongly measurable and let us extendx b-periodically to the wholeR. We introduce the family of operatorsA² = (A²_v)v∈V by the formula:

A²_v(x)(t) = ( Rb

0Kv(s−t)x(s)ds, if Rb

0 Kv(s−t)kx(s)kY ds <∞

o, if Rb

0 Kv(s−t)kx(s)kY ds=∞ for everyv∈Vand everyt∈[0, b).

Let us extendF b-periodically to the wholeR.

LetBv(F) ={A²_v(x) :x∈M([0, b), Y)∩F}for everyF ∈M_Y([0, b)) and every v∈V.

Remark 8. IfA¹_v :L^ϕ([0, b), R)→L^ϕ([0, b), R), where R= [−∞,+∞], then Bv:M_Y,ϕ^o ([0, b))→M_Y,ϕ^o ([0, b)).

Proof: LetF ∈M_Y,ϕ^o ([0, b)),v∈V. We have forD= [0, b)

sup

x∈M(D,Y)∩F

k Z b

0

Kv(s−t)x(s)dsk_Y ≤ sup

x∈M(D,Y)∩F

{ Z b

0

Kv(s−t)kx(s)k_Y ds}

= Z b

0

Kv(s−t)r_F(s)ds.

On the other hand, forx(s) =xr_F(s)/kxk_Y for everys ∈D, where x∈Y and x6=o, we have

k Z b

0

Kv(s−t)x(s)dsk_Y =k x kxkY

Z b 0

Kv(s−t)r_F(s)dsk_Y = Z b

0

Kv(s−t)r_F(s)ds.

Let 0<Rb

0 Kv(s−t)rF(s)ds <∞and lety∈B(o,Rb

0Kv(s−t)rF(s)ds). Let xt(s) =yr_F(s)/

Z b 0

Kv(s−t)r_F(s)ds for everys∈[0, b). We have

Z b 0

Kv(s−t)xt(s)ds=y andxt∈M([0, b), Y)∩F because

kxt(s)k_Y =kyr_F(s)/

Z b 0

Kv(s−t)r_F(s)dsk_Y ≤r_F(s) for every s∈[0, b).

SoB(F)(t) =B(o, rB(F)(t)) for everyt∈[0, b), where r_B(F)(t) =

( R_b

0 Kv(s−t)r_F(s)ds, ifA¹_v(r_F)(t)<∞

0, ifA¹_v(r_F)(t) =∞

for everyt∈[0, b). It is easy to see thatrB(F)∈L^ϕ([0, b), R).

LetF ∈M_Y,ϕ¹ ([0, b)) and let F(s) = B(a_F(s), r_F(s)) for every s∈[0, b). We introduce the family of operatorsT^′= (T_v^′)_v∈V by the formula:

T_v^′(F)(s) =

B(A²_v(a_F)(s), A¹_v(r_F)(s)), ifA¹_v(r_F)(s)<∞ {A²_v(a_F)(s)}, ifA¹_v(r_F)(s) =∞

for everys∈[0, b) and every v∈V.

LetF ∈M˜_Y,ϕ^o ([0, b)) and F(s) = Sn

i=1R(o, rⁱ_F(s), Rⁱ_F(s)) for every s∈[0, b), where we receive that if there are D ⊂ [0, b), D ∈ Σ, and m < n such that F(s) = Sm

i=1R(o, rⁱ_F(s), Rⁱ_F(s)), Rⁱ_F(s) < rⁱ⁺¹_F (s) for s ∈ D, i = 1, . . . , m−1 ifm > 1, then we denote F(s) =Sn

i=1(o, r_Fⁱ (s), Rⁱ_F(s)) for everys ∈D, where r_Fⁱ (s) = rⁱ_F(s), Rⁱ_F(s) = Rⁱ_F(s) for i = 1, . . . , m, rⁱ_F(s) = Rⁱ_F(s) = Rⁱ_F(s) for i=m+ 1, . . . , nfor everys∈D.

We introduce the family of operatorsT^′′= (T_v^′′)_v∈V by the formula:

T_v^′′(F)(s) = ( Sn

i=1R(o, A¹_v(r_Fⁱ )(s), A¹_v(Rⁱ_F)(s)), ifA¹_v(Rⁿ_F)(s)<∞

{o}, ifA¹_v(Rⁿ_F)(s) =∞

for everys∈[0, b) and every v∈V.

Remark 9. If A¹_v : L^ϕ([0, b), R) → L^ϕ([0, b), R), where R = [−∞,+∞], then T_v^′ :M_Y,ϕ¹ ([0, b))→M_Y,ϕ¹ ([0, b)).

Proof: Let F ∈M_Y,ϕ¹ ([0, b)), F(s) =B(a_F(s), r_F(s)) for every s∈[0, b). It is easy to see that

B(A²_v(a_F)(s), A¹_v(r_F)(s)) =B(A²_v(a_F)(s),0)⊕B(o, A¹_v(r_F)(s))

for everys∈[0, b) andA²_v :L^ϕ([0, b), Y)→L^ϕ([0, b), Y), soT_v^′(F)∈M_Y,ϕ¹ ([0, b)).

Corollary 1. If the assumptions of Lemma2 and Remarks5, 8hold, then

T_v^′(C) :M_Y,ϕ¹ ([0, b))→M_Y,ϕ^o ([0, b)).

Applying the proofs of Proposition 2 and Theorem 4 in [3], we obtain the following

Theorem 3. Letϕbe a convex,τ-boundedϕ-function which fulfils the∆₂ con- dition, Rb

0ϕ(t, c)dt < ∞ for every c > 0 and let (Kv)_v∈V be singular. Then

̺(a(A²_vx−x))−→^V 0 for everya >0and everyx∈L^ϕ([0, b), Y).

Corollary 2. If the assumptions of Theorem3hold, then T_v^′(F)^d,ϕ,V−→ F for everyF∈M_Y,ϕ¹ ([0, b)).

Proof:By the assumptionsT_v^′ :M_Y,ϕ¹ ([0, b))→M_Y,ϕ¹ ([0, b)). LetF ∈M_Y,ϕ¹ ([0, b)), F(s) =B(a_F(s), r_F(s)) for everys∈[0, b). We have fora >0

Z b 0

ϕ(t, ad(T_v^′(F), F)(t))dt

≤1 2

Z b

0 ϕ(t,2a|A¹_v(rF)(t)−rF(t)|)dt +1

2 Z b

0

ϕ(t,2akA²_v(a_F)(t)−a_F(t)k_Y)dt−→^V 0.

Remark 10. Let A = Sn

i=1[a_i, b_i], B = Sn

i=1[c_i, d_i], where [a_i, b_i], [c_i, d_i], i = 1, . . . , n, are nonempty and compact segments in R, then dist (A, B) ≤ Pn

i=1dist ([a_i, b_i],[c_i, d_i]).

Corollary 3. If the assumptions of Theorem3 hold, then

T_v^′′(F)^d,ϕ,V−→ F for everyF∈M˜_Y,ϕ^o ([0, b)).

Proof: Let F ∈ M˜_Y,ϕ^o ([0, b)), F(s) = Sm

i=1R(o, rⁱ_F(s), Rⁱ_F(s)), a > 0, v ∈ V.

By the assumptions and by Remark 10 (also, see the proof of Remark 2’ and [2, Remark 10]) we have

Z b 0

ϕ(t, ad(T_v^′′(F), F)(t))dt

≤ 1 2m

m

X

i=1 b

Z

0

ϕ(t,2am|A¹_v(r_Fⁱ )(t)−rⁱ_F(t)|)dt

+ 1 2m

m

X

i=1 b

Z

0

ϕ(t,2am|A¹_v(Rⁱ_F)(t)−R_Fⁱ (t)|)dt−→^V 0.

Let F ∈ M_Y,d,ϕ^o ([0, b)). Let v ∈ V be arbitrary. If there exists Gv ∈ M_Y,d,ϕ^o ([0, b)) such that limn→∞Rb

0 ϕ(t, ad(T_v^′′(Fn), Gv)(t))dt= 0 for everya >0 and every sequence{Fn}such thatFn∈M˜_Y,ϕ^o ([0, b)) for everyn∈N and limn→∞Rb

0 ϕ(t, ad(Fn, F)(t))dt= 0 for everya >0, then we defineTv(F) =Gv. Theorem 4. Let the assumptions of Theorem3hold and there are K₁, K₂>0 such that̺(ad(T_v^′′(F), T_v^′′(G))) ≤K1̺(aK2d(F, G)) for allF, G ∈ M˜_Y,ϕ^o ([0, b)), a >0and everyv∈V, thenTv(F)^d,ϕ,V−→ F for everyF ∈M_Y,d,ϕ^o ([0, b)).

Proof: The proof is analogous to that of Theorem 3’ from [2], so we give the sketch of it only. Analogously as in that proof we prove that the family (Tv)_v∈V is (d,V, M_Y,d,ϕ^o ([0, b))-bounded. So we obtain the assertion from Remark 7 and Corollary 3.

Final remarks. The results of [2] can be extended in other ways.

1. Let x, y ∈ Y. By s(x, y) we denote the closed segment joining the points x andy. Leta∈Y. Define:

Y^a={λa:λ∈R},

Y_ϕ^1,a={F ∈M_Y(I) :F(t) =s(b_F(t), e_F(t)) for everyt∈I, where b_F(·), e_F(·)∈L^ϕ(I, Y^a)},

Y_ϕ^n,a={F ∈MY(I) :F(t) =

n

[

i=1

s(bⁱ_F(t), eⁱ_F(t)) for everyt∈I, where bⁱ_F(·), eⁱ_F(·)∈L^ϕ(I, Y^a), i= 1, . . . , n,keⁱ_F(t)k_Y ≤ kbⁱ⁺¹_F (t)k_Y for every

t∈I, i= 1, . . . , n−1 ifn >1}, Y˜_ϕ^a=

∞

[

i=1

Y_ϕ^n,a,

Y_d^a={F∈M_Y(I) :d(Fn, F)→0 a.e. for someFn∈Y˜_ϕ^a, n∈N}, Y_d,ϕ^a ={F∈Y_d^a:

Z

I

ϕ(t, λd(Fn, F)(t))dt→0 asn→ ∞for everyλ >0 for someFn∈Y˜_ϕ^a, n∈N}.

The results of [2] will be in force if we replaceR byY, the space X_d,ϕ by Y_d,ϕ^a and if we introduce the other evident changes.

2. LetY =Rⁿ. By Πⁿ(a_i, b_i) we denote the Cartesian product of then closed segments [ai, bi], whereai, bi ∈R. Define

Y_ϕ^Πn ={F ∈MY(I) :F(t) = Πⁿ(a^F_i (t), b^F_i (t)) for everyt∈I, a^F_i (·), b^F_i (·)∈L^ϕ(I, Y) fori= 1, . . . , n},

D(F, G)(t) = max

1≤i≤nd([a^F_i , b^F_i ],[a^G_i , b^G_i ])(t) for allF, G∈Y^Πn, t∈I.

We easily obtain that the spacehY^Πn,Diis a complete space. For allF ∈Y^Πn, v∈V,t∈[0, b) we define:

T_vⁿ(F)(t) = Πⁿ(A¹_v(a^F_i )(t), A¹_v(b^F_i )(t)).

We easily obtain the following :

Theorem 5. If the assumptions of Theorem3hold, then T_vⁿ(F)^D,ϕ,V−→ F for everyF ∈Y_ϕ^Πn, n∈N.

References

[1] Kasperski A.,Modular approximation inX˜ϕby a filtered family of dist-sublinear operators and dist-convex operators, Mathematica Japonica38(1993), 119–125.

[2] ,Notes on approximation in the Musielak-Orlicz spaces of multifunctions, Com- mentationes Math., in print.

[3] Musielak J.,Modular approximation by a filtered family of linear operators, Functional Ana- lysis and Approximation, Proc. Conf. Oberwolfach, August 9-16, 1980; Birkh¨auser Verlag, Basel, 1981, pp. 99–110.

[4] , Orlicz spaces and Modular spaces, Lecture Notes in Mathematics Vol. 1034, Springer-Verlag, Berlin, 1983.

[5] Wis la M.,On completeness of Musielak-Orlicz spaces, Chin. Ann. of Math.10B (3)(1989), 292–300.

Institute of Mathematics, Silesian Technical University, Kaszubska 23, 44-100 Gliwice, Poland

(Received January 14, 1993,revised June 25, 1993)