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HIGHER RING DERIVATION AND INTUITIONISTIC FUZZY STABILITY

ICK-SOON CHANG

Abstract. In this paper, we take account of the stability of higher ring derivation in intuitionisticfuzzy Banach algebra associated to the Jensen type functional equation. In addition, we deal with thesuperstability of higher ring derivation in intuitionistic fuzzy Banach algebra with unit.

1. Introduction and preliminaries

The stability problem of functional equations has originally been formulated by Ulam [21] : under whatcondition does there exists a homomorphism near an approximate homomorphism? Hyers [6] answeredthe problem of Ulam under the assumption that the groups are Banach spaces. A generalized version ofthe theorem of Hyers for approximately additive mappings was given by Aoki [1], and for approximately

linear mappings was presented by Rassias [17] by considering an unbounded Cauchy difference. The paperwork of Rassias [17] has had a lot of influence in the development of what call the generalized Hyers-Ulamstabilityof functional equations. Since then, more generalizations and applications of the generalized Hyers-Ulam stability to a number of functional equations and mappings have been investigated (for example,[7, 18, 20]) In particular, Badora [2] gave a generalization of the Bourgins result [4] and he also dealt withthe stability and the Bourgin-type superstability of derivations in [3]. Recently, fuzzy version is discussedin [8, 9]. Quite recently, the intuitionistic fuzzy stability problem for Jensen functional equation and cubicfunctional equation is considered in [10, 11, 14] respectively; while the idea of intuitionistic fuzzy normedspace was introduced in [19] and there are some recent and important results which are directly related tothe central theme of this paper, that is, intuitionistic fuzziness (see for example [12, 13, 15, 16]).

In this paper, we establish the stability of higher ring derivation in intuitionistic fuzzy Banach algebraassociated to the Jensen type functional equation lf(x + y/l) = f(x) + f(y). Moreover, we consider thesuperstability of higher ring derivation in intuitionistic fuzzy Banach algebra with unit.

We now recall some notations and basic definitions used in this paper.

Definition 1.1. [7] Let A and B be algebras over the real or complex field F. Let N be the set of thenatural numbers. From m N {0}, a sequence H = {h0, h1, , hm} (resp., H = {h0, h1, , hk, })of additive operators from A into B is called a higher ring derivation of rank m (resp., infinite rank) if the

functional equation hk(xy) =

k

i=0 hi(x)hki(y) holds for each k = 0, 1, , m (resp., k = 0, 1, ) andfor all x, y A. A higher ring derivation H of additive operators on A, particularly, is called strong if h0is an identity operator.

Of course, a higher ring derivation of rank 0 from A into B (resp., a strong higher ring derivation ofrank 1 on A) is a ring homomorphism (resp., a ring derivation). Note that a higher ring derivation is ageneralization of both a ring homomorphism and a ring derivation.

Definition 1.2. A binary operation : [0, 1] [0, 1] [0, 1] is said to be a continuous t-normif it satisfiesthe following conditions :

(1) is associative and commutative, (2) is continuous, (3) a 1 = a for all a [0, 1],(4) a b c d whenever a c and b d for each a,b,c,d [0, 1].

Definition 1.3. A binary operation : [0, 1] [0, 1] [0, 1] is said to be a continuous t-conorm if itsatisfies the following conditions :

2000 Mathematics Subject C lassification: 39B52, 39B72, 39B82, 43A22,46S40, 47B47.Keywords and phrases : Higher ring derivation ; Intuitionistic fuzzy stability ; Intuitionistic fuzzy Banach algebra. This

research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF)funded by the Ministry of Education, Science and Technology (No. 2012-0002410).

1

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2 I.-S. CHANG

(1) is associative and commutative, (2) is continuous, (3) a 0 = a for all a [0, 1],(4) a b c d whenever a c and b d for each a,b,c,d [0, 1].

Using the notions of continuous t-norm and t-conorm, Saadati and Park [19] have recently introducedthe concept of intuitionistic fuzzy normed space as follows :

Definition 1.4. The five-tuple (X,,, , ) is said to be an intuitionistic fuzzy normed space if X is avector space, is a continuous t-norm, is a continuous t-conorm, and , are fuzzy sets on X (0, )satisfying the following conditions : For every x, y X and s,t > 0,

(1) (x, t) + (x, t) 1, (2) (x, t) > 0, (3) (x, t) = 1 if and only if x = 0,(4) (x,t) = (x, t|| ) for each = 0, (5) (x, t) (y, s) (x + y, t + s),

(6) (x, ) : (0, ) [0, 1] is continuous, (7) limt (x, t) = 1 and limt0 (x, t) = 0,(8) (x, t) < 1, (9) (x, t) = 0 if and only if x = 0, (10) (x,t) = (x, t|| ) for each = 0,

(11) (x, t) (y, s) (x + y, t + s), (12) (x, ) : (0, ) [0, 1] is continuous,(13) limt (x, t) = 0 and limt0 (x, t) = 1.In this case, (, ) is called an intuitionistic fuzzy norm.

Example 1.5. Let (X, ) be a normed space, a b = ab and a b = min{a + b, 1} for all a, b [0, 1].For all x X and every t > 0, consider

(x, t) ={ 1 if t > x

0 if t x and (x, t) ={ 0 if t > x

1 if t x

Then (X,,, , ) is an intuitionistic fuzzy normed space.

Example 1.6. Let (X, ) be a normed space, a b = ab and a b = min{a + b, 1} for all a, b [0, 1].For all x X and every t > 0 and k = 1, 2, consider

(x, t) =

{t

t+xif t > 0

0 if t 0and (x, t) =

kx

t+kxif t > 0

0 if t 0

Then (X,,, , ) is an intuitionistic fuzzy normed space.

Definition 1.7. [5] The five-tuple (X,,, , ) is said to be an intuitionistic fuzzy normed algebra if Xis an algebra, is a continuous t-norm, is a continuous t-conorm, and , are fuzzy sets on X (0, )satisfying the conditions (1)(13) of the definition 1.4. Furthermore, for every x, y X and s,t > 0,

(14) max{(x, t), (y, s)} (xy,t + s), (15) min{(x, t), (y, s)} (xy,t + s).

For an intuitionistic fuzzy normed algebra (X,,, , ), we further assume that (16) a a = a anda a = a for all a [0, 1].

The concepts of convergence and Cauchy sequences in an intuitionistic fuzzy normed space are studiedin [19]. Let (X,,, , ) be an intuitionistic fuzzy normed space or intuitionistic fuzzy normed algebra.A sequence x = {xk} is said to be intuitionistic fuzzy convergent to L X if limk (xk L, t) = 1

and limk (xk L, t) = 0 for all t > 0. In this case, we write (, ) limk xk = L or xkIF

Las k . A sequence x = {xk} in (X,,, , ) is said to be intuitionistic fuzzy Cauchy sequence iflimk (xk+p xk, t) = 1 and limk (xk+p xk, t) = 0 for all t > 0 and p = 1, 2, . An intuitionisticfuzzy normed space (resp., intuitionistic fuzzy normed algebra) (X,,, , ) is said to be complete if everyintuitionistic fuzzy Cauchy sequence in (X,,, , ) is intuitionistic fuzzy convergent in (X,,, , ). Acomplete intuitionistic fuzzy normed space (resp., intuitionistic fuzzy normed algebra) is also called an

intuitionistic fuzzy Banach space (resp., intuitionistic fuzzy Banach algebra).

2. Stability of higher ring derivation in intuitionistic fuzzy Banach algebra

As a matter of convenience in this paper, we use the following abbreviation :

nj=0

aj := a1 a2 an,j=0

aj := a1 a2 .

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INTUITIONISTIC FUZZY STABILITY OF HIGHER RING DERIVATION 3

In addition,

nj=0

aj := a1 a2 an,j=0

aj := a1 a2 .

We begin with a generalized Hyers-Ulam theorem in intuitionistic fuzzy Banach space for the Jensentype functional equation. The following result is also the generalization of the theorem introduced in [10].

Theorem 2.1. Let A be a vector space, and f be a mapping from A to an intuitionistic fuzzy Banachspace (B,,, , ) with f(0) = 0. Suppose that is a function from A to an intuitionistic fuzzy normedspace (C, , , , ) such that

(

lf(x + y

l

) f(x) f(y), t + s

) ((x), t) ((y), s), (2.1)

(

lf(x + y

l

) f(x) f(y), t + s

) ((x), t) ((y), s) (2.2)

for all x, y A\{0}, t > 0 and s > 0. If l > 1 is a fixed integer, and ((l + 1)x) = (x) for some

real number with 0 < || < l + 1, then there exists a unique additive mapping L : A B such thatL(x) := (, ) limn

f((l+1)nx)(l+1)n

,

(L(x) f(x), t) j=0

M(

x,((l + 1) )t

2(l + 1)

), (L(x) f(x), t)

j=0

N(

x,((l + 1) )t

2(l + 1)

)(2.3)

for all x A and t > 0, where

M(x, t) := (

(x),l + 1

4t)

(

(x),l + 1

4t)

(

(x),l + 1

4t)

(

((l + 1)x),l + 1

4t)

,

N(x, t) := (

(x),l + 1

4t)

(

(x),l + 1

4t)

(

(x),l + 1

4t)

(

((l + 1)x),l + 1

4t)

.

Proof. Without loss of generality, we assume that 0 < < l + 1. From (2.1) and (2.2), we get

(f(x) + f(x), lt) (

(x),l

2t)

(

(x),l

2t)

,

(f(x) + f(x), lt) (

(x),l

2t)

(

(x),l

2t)

(2.4)

for all x A and t > 0. Again, by (2.1) and (2.2), we obtain

(lf(x) f(x) f((l + 1)x), lt) (

(x),l

2t)

(

((l + 1)x),l

2t)

,

(lf(x) f(x) f((l + 1)x), lt) (

(x),l

2t)

(

((l + 1)x),l

2t)

(2.5)

for all x A and t > 0. Combining (2.4) and (2.5), we arrive at

((l + 1)f(x) f((l + 1)x), 2lt) (

(x), l2

t)

(

(x), l2

t)

(

(x), l2

t)

(

((l + 1)x),l

2t)

((l + 1)f(x) f((l + 1)x), 2lt) (

(x),l

2t)

(

(x),l

2t)

(

(x),l

2t)

(

((l + 1)x),l

2t)

(2.6)

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4 I.-S. CHANG

for all x A and t > 0. This implies that

(

f(x) f((l + 1)x)

(l + 1), t)

(

(x),l + 1

4t)

(

(x),l + 1

4t)

(

(x),l + 1

4t)

(((l + 1)x),l + 1

4t)

(

f(x) f((l + 1)x)(l + 1)

, t)

(

(x), l + 14

t)

(

(x), l + 14

t)

(

(x), l + 14

t)

(

((l + 1)x),l + 1

4t)

(2.7)

for all x A and t > 0. Now we define

M(x, t) := (

(x),l + 1

4t)

(

(x),l + 1

4t)

(

(x),l + 1

4t)

(

((l + 1)x),l + 1

4t)

,

N(x, t) := (

(x),l + 1

4t)

(

(x),l + 1

4t)

(

(x),l + 1

4t)

(

((l + 1)x),l + 1

4t)

.

for all x A and t > 0. Then we have by assumption

M((l + 1)x, t) = M(x, t), N((l + 1)x, t) = N(x, t

). (2.8)

for all x A and t > 0. Using (2.7) and (2.8), we get

(f((l + 1)nx)

(l + 1)n

f((l + 1)n+1x)

(l + 1)n+1,

nt

(l + 1)n

)= (

f((l + 1)nx) f((l + 1)n+1x)

l + 1, nt

) M((l + 1)nx, nt) = M(x, t)

(f((l + 1)nx)

(l + 1)n

f((l + 1)n+1x)

(l + 1)n+1,

nt

(l + 1)n

)= (

f((l + 1)nx) f((l + 1)n+1x)

l + 1, nt

) N((l + 1)nx, nt) = N(x, t)

for all x A and t > 0. Therefore, for all n > m, we have

(f((l + 1)mx)

(l + 1)m

f((l + 1)nx)

(l + 1)n

,n1

j=m

jt

(l + 1)j )

= ( n1j=m

[f((l + 1)jx)(l + 1)j

f((l + 1)j+1x)

(l + 1)j+1

],n1j=m

jt

(l + 1)j

)

n1j=m

(f((l + 1)jx)

(l + 1)j

f((l + 1)j+1x)

(l + 1)j+1,

jt

(l + 1)j

)

n1j=m

M(x, t),

(f((l + 1)mx)

(l + 1)m

f((l + 1)nx)

(l + 1)n,n1j=m

jt

(l + 1)j

)

= ( n1j=m

[f((l + 1)jx)(l + 1)j

f((l + 1)j+1x)

(l + 1)j+1

],n1j=m

jt

(l + 1)j

)

n1j=m

(

f((l + 1)j

x)(l + 1)j

f((l + 1)j+1

x)(l + 1)j+1

, j

t(l + 1)j

)

n1j=m

N(x, t)

(2.9)

for all x A and t > 0. Let > 0 and > 0 be given. Since limtn1

j=m M(x, t) = 1 a n d

limt

n1j=m N(x, t) = 0, there exists some t0 such that

n1j=m M(x, t0) > 1 ,

n1j=m N(x, t0) < .

Since

j=0jt

(l+1)j< , there exists a positive integer n0 such that

n1j=m

jt(l+1)j

< for all n > m n0.

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INTUITIONISTIC FUZZY STABILITY OF HIGHER RING DERIVATION 5

Then

(f((l + 1)mx)

(l + 1)m

f((l + 1)nx)

(l + 1)n, )

(f((l + 1)mx)

(l + 1)m

f((l + 1)nx)

(l + 1)n,n1j=m

jt0(l + 1)j

)

n1

j=m

M(x, t0) > 1 ,

(f((l + 1)mx)

(l + 1)m

f((l + 1)nx)

(l + 1)n, )

(f((l + 1)mx)

(l + 1)m

f((l + 1)nx)

(l + 1)n,n1j=m

jt0(l + 1)j

)

n1j=m

N(x, t0) < .

This show that { f((l+1)nx)

(l+1)n} is a Cauchy sequence in (B, , , , ). Since B is complete, we can define a

mapping L by L(x) := (, ) limnf((l+1)nx)

(l+1)nfor all x A. Moreover, if we let m = 0 in (2.9), then

we get

(f((l + 1)nx)

(l + 1)n f(x),

n1

j=0

jt

(l + 1)j ) n1

j=0

M(x, t), (f((l + 1)nx)

(l + 1)n f(x),

n1

j=0

jt

(l + 1)j ) n1

j=0

N(x, t).

for all x A and t > 0. Therefore, we find that

(f((l + 1)nx)

(l + 1)n f(x), t

)

n1j=0

M(

x,tn1

j=0j

(l+1)j

),

(f((l + 1)nx)

(l + 1)n f(x), t

)

n1j=0

N(

x,tn1

j=0j

(l+1)j

). (2.10)

Next, we will show that L is additive mapping: Note that

(

lL(x + y

l

) L(x) L(y), t

) (

lL(x + y

l

)

lf( (l+1)n(x+y)l

)

(l + 1)n,

t

4

)

(f((l + 1)nx)

(l + 1)n L(x),

t

4) (f((l + 1)

ny)

(l + 1)n L(y),

t

4)

( lf( (l+1)n(x+y)

l)

(l + 1)n

f((l + 1)nx)

(l + 1)n

f((l + 1)ny)

(l + 1)n,

t

4

),

(

lL(x + y

l

) L(x) L(y), t

) (

lL(x + y

l

)

lf( (l+1)n(x+y)l

)

(l + 1)n,

t

4

)

(f((l + 1)nx)

(l + 1)n L(x),

t

4

) (f((l + 1)ny)

(l + 1)n L(y),

t

4

)

( lf( (l+1)n(x+y)

l)

(l + 1)n

f((l + 1)nx)

(l + 1)n

f((l + 1)ny)

(l + 1)n,

t

4

). (2.11)

On the other hand, (2.1) and (2.2) give the following.

( lf( (l+1)n(x+y)l )

(l + 1)n

f((l + 1)nx)

(l + 1)n

f((l + 1)ny)

(l + 1)n ,

t

4)

(

(x),( l + 1

)n t8

) (

(y),( l + 1

)n t8

)

( lf( (l+1)n(x+y)

l)

(l + 1)n

f((l + 1)nx)

(l + 1)n

f((l + 1)ny)

(l + 1)n,

t

4

)

(

(x),( l + 1

)n t8

) (

(y),( l + 1

)n t8

). (2.12)

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6 I.-S. CHANG

Letting n in (2.11) and (2.12), we yield

(

lL(x + y

l

) L(x) L(y), t

)= 1,

(lL(x + y

l

) L(x) L(y), t

)= 0.

So we see that L is additive mapping.Now, we approximate the difference between f and L in an intuitionistic fuzzy sense. By (2.10), we get

(L(x) f(x), t) (

L(x) f((l + 1)nx)

(l + 1)n, t

2

) (

f((l + 1)nx)(l + 1)n

f(x), t2

)

j=0

M(

x,((l + 1) )t

2(l + 1)

),

(L(x) f(x), t) (

L(x) f((l + 1)nx)

(l + 1)n,

t

2

) (f((l + 1)nx)

(l + 1)n f(x),

t

2

)

j=0

N(

x,((l + 1) )t

2(l + 1)

)

for all x A and t > 0 and sufficiently large n.In order to prove the uniqueness of L, we assume that T be another additive mapping from A to B,

which satisfies the inequality (2.3). Then

(L(x) T(x), t) (

L(x) f(x), t2

) (

T(x) f(x), t2

)

j=0

M(

x,((l + 1) )t

4(l + 1)

),

(L(x) T(x), t) (

L(x) f(x),t

2

) (

T(x) f(x),t

2

)

j=0

N(

x,((l + 1) )t

4(l + 1)

)

for all x A and t > 0. Therefore, due to the additivity of L and T, we obtain that

(L(x) T(x), t) = (L((l + 1)nx) T((l + 1)nx), (l + 1)nt)

j=0M

(x,

(l + 1

)n ((l + 1) )t

4(l + 1) ),

(L(x) T(x), t) = (L((l + 1)nx) T((l + 1)nx), (l + 1)nt) j=0

M(

x,( l + 1

)n ((l + 1) )t4(l + 1)

),

Since 0 < < l + 1, limn(l+1

)n = and we get

limn

M(

x,( l + 1

)n ((l + 1) )t4(l + 1)

)= 1, lim

nN(

x,( l + 1

)n ((l + 1) )t4(l + 1)

)= 0,

that is, (L(x) T(x), t) = 1 and (L(x) T(x), t) = 0 for all x A, t > 0. So L = T, which completesthe proof.

In particular, we can prove the preceding result for the case when > l + 1. In this case, the mappingL(x) := (, ) limn(l + 1)

nf((l + 1)nx). We now establish a generalized Hyers-Ulam stability inintuitionistic fuzzy Banach algebra for the higher ring derivation.

Theorem 2.2. Let A be an algebra, and F = {f0, f1, , fk, } be a sequence of mappings from A toan intuitionistic fuzzy Banach algebra (B, ,, , ) with fk(0) = 0 for each k = 0, 1 . Suppose that isa function from A to an intuitionistic fuzzy normed algebra (C, , , , ) such that for each k = 0, 1 ,

(

lfk

(x + yl

) fk(x) fk(y), t + s

) ((x), t) ((y), s), (2.13)

(

lfk

(x + yl

) fk(x) fk(y), t + s

) ((x), t) ((y), s) (2.14)

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INTUITIONISTIC FUZZY STABILITY OF HIGHER RING DERIVATION 7

for all x, y A\{0}, t > 0 and s > 0, and that is a function from A to an intuitionistic fuzzy normedspace (D, , , , ) such that for each k = 0, 1 ,

(fk(xy) k

i=0fi(x)fki(y), t + s) max{

((x), t), ((y), s)}, (2.15)

(fk(xy) k

i=0

fi(x)fki(y), t + s) min{((x), t), ((y), s)} (2.16)

for all x, y A, t > 0, and s > 0. If l > 1 is a fixed integer, ((l + 1)x) = (x), and ((l + 1)x) = (x)for some real numbers and with 0 < || < l + 1 and 0 < || < l + 1, then there exists a unique higherring derivation H = {L0, L1, , Lk, } of any rank such that for each k = 0, 1 ,

(Lk(x) fk(x), t) M(

x,((l + 1) )t

2(l + 1)

), (Lk(x) fk(x), t) N

(x,

((l + 1) )t

2(l + 1)

)(2.17)

for all x A and t > 0. In this case,

M(x, t) := (

(x),l + 1

4t)

(

(x),l + 1

4t)

(

((l + 1)x),l + 1

4t)

,

N(x, t) := (

(x),l + 1

4t)

(

(x),l + 1

4t)

(

((l + 1)x),l + 1

4t)

.

Moreover, the identity

ki=0

Li(y){Lki(y) fki(y)} = 0 (2.18)

holds for each k = 0, 1 and all x, y A.

Proof. It follows by Theorem 2.1 that for each k = 0, 1 and all x A, there exists a unique additive

mapping Lk : A B given by

Lk(x) := (, ) limn

fk((l + 1)nx)

(l + 1)n(2.19)

satisfying (2.17) since (C, , , , ) is an intuitionistic fuzzy normed algebra.Without loss of generality, we suppose that 0 < < l + 1. Now, we need prove that the sequence

H = {L0, L1, , Lk, } satisfies the identity Lk(xy) =

k

i=0Li(x)Lki(y) for each k = 0, 1 and all

x A. Observed that for each k = 0, 1 ,

(

Lk(xy) k

i=0

Li(x)fki(y), t)

(

Lk(xy) fk((l + 1)

nxy)

(l + 1)n,

t

3

) (fk((l + 1)nxy)

(l + 1)n

ki=0

fi((l + 1)n

x)(l + 1)n

fki(y), t3

) ( k

i=0

fi((l + 1)n

x)(l + 1)n

fki(y)

ki=0

Li(x)fki(y), t3

),

(

Lk(xy) k

i=0

Li(x)fki(y), t)

(

Lk(xy) fk((l + 1)

nxy)

(l + 1)n,

t

3

) (fk((l + 1)nxy)

(l + 1)n

k

i=0

fi((l + 1)nx)

(l + 1)nfki(y),

t

3

) ( k

i=0

fi((l + 1)nx)

(l + 1)nfki(y)

ki=0

Li(x)fki(y),t

3

)(2.20)

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8 I.-S. CHANG

for all x, y A and t > 0. On account of (2.15) and (2.16), we see that for each k = 0, 1 ,

(fk((l + 1)nx y)

(l + 1)n

ki=0

fi((l + 1)nx)

(l + 1)nfki(y),

t

3

)

= (fk((l + 1)nx y) k

i=0

fi((l + 1)nx)fki(y),

(l + 1)nt

3)

max

(

(x),( l + 1

)n t6

), (

(y),(l + 1)nt

6

)},

(fk((l + 1)nx y)

(l + 1)n

ki=0

fi((l + 1)nx)

(l + 1)nfki(y),

(l + 1)nt

3

)

= (

fk((l + 1)nx y)

ki=0

fi((l + 1)nx)fki(y),

(l + 1)nt

3

)

min

(

(x),( l + 1

)n t6

), (

(y),(l + 1)nt

6

)},

(2.21)

for all x, y A and t > 0. Due to additivity of Lk, for each k = 0, 1 ,

( k

i=0

fi((l + 1)nx)

(l + 1)nfki(y)

ki=0

Li(x)fki(y),t

3

)

k

i=0

(

fi((l + 1)nx)fki(y) (l + 1)

nLi(x)fki(y),(l + 1)nt

3(k + 1)

),

( k

i=0

fi((l + 1)nx)

(l + 1)nfki(y)

ki=0

Li(x)fki(y),t

3

)

k

i=0

(

fi((l + 1)nx)fki(y) (l + 1)

nLi(x)fki(y),(l + 1)nt

3(k + 1)

)(2.22)

for all x, y A and t > 0. In addition, we feel that

(

fi((l + 1)nx)fki(y) (l + 1)

nLi(x)fki(y),(l + 1)nt

3(k + 1)

)

max

(

fi((l + 1)nx) (l + 1)nLi(x),

(l + 1)nt

6(k + 1)

), (

fki(y),(l + 1)nt

6(k + 1)

)},

(

fi((l + 1)nx)fki(y) (l + 1)

nLi(x)fki(y),(l + 1)nt

3(k + 1)

)

min

(

fi((l + 1)nx) (l + 1)nLi(x),

(l + 1)nt

6(k + 1)

), (

fki(y),(l + 1)nt

6(k + 1)

)}. (2.23)

Letting n in (2.20), (2.21), (2.22) and (2.23), we get (Lk(xy) k

i=0 Li(x)fki(y), t) = 1 and

(

Lk(xy)

k

i=0 Li(x)fki(y), t) = 0. This implies that

Lk(xy) =k

i=0

Li(x)fki(y) (2.24)

for each k = 0, 1 and all x, y A.Using additivity ofLk and (2.24), we find that

(l + 1)nk

i=0

Li(x)fki(y) = Lk((l + 1)nx y) = Lk(x (l + 1)

ny) =k

i=0

L(x)f((l + 1)ny).

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INTUITIONISTIC FUZZY STABILITY OF HIGHER RING DERIVATION 9

So we obtaink

i=0 Li(x)fki(y) =k

i=0 Li(x)fki((l+1)

ny)

(l+1)n. Hence for each k = 0, 1 ,

( k

i=0

Li(x)fki(y) k

i=0

Li(x)fki((l + 1)

ny)

(l + 1)n, t)

= 1,

( k

i=0 Li(x)fki(y)

ki=0 Li(x)

fki((l + 1)ny)

(l + 1)n , t)

= 0

for all x, y A and t > 0. This relation yields that for each k = 0, 1 ,

( k

i=0

Li(x)Lki(y) k

i=0

Li(x)fki(y), t)

( k

i=0

Li(x)Lki(y) k

i=0

Li(x)fki((l + 1)

ny)

(l + 1)n,

t

2

) ( k

i=0

Li(x)fki((l + 1)

ny)

(l + 1)n

k

i=0

Li(x)fki(y),t

2

)

k

i=0

(Li(x)Lki(y) Li(x)fki((l + 1)

ny)

(l + 1)n,

t

2(k + 1)),( k

i=0

Li(x)Lki(y) k

i=0

Li(x)fki(y), t)

( k

i=0

Li(x)Lki(y) k

i=0

Li(x)fki((l + 1)

ny)

(l + 1)n,

t

2

) ( k

i=0

Li(x)fki((l + 1)

ny)

(l + 1)n

k

i=0

Li(x)fki(y),t

2

)

k

i=0

(

Li(x)Lki(y) Li(x)fki((l + 1)

ny)

(l + 1)n,

t

2(k + 1)

)(2.25)

for all x, y A and t > 0. On the other hand, we see that

(

Li(x)Lki(y) Li(x) fki((l + 1)ny)

(l + 1)n, t

2(k + 1)

)

max

(

Li(x),(l + 1)nt

4(k + 1)

), (

Lki(y) fki((l + 1)

ny)

(l + 1)n,

t

4(k + 1)

)},

(

Li(x)Lki(y) Li(x)fki((l + 1)

ny)

(l + 1)n,

t

2(k + 1)

)

min

(

Li(x),(l + 1)nt

4(k + 1)

), (

Lki(y) fki((l + 1)

ny)

(l + 1)n,

t

4(k + 1)

)}. (2.26)

Sending n in (2.25) and (2.26), we have that for each k = 0, 1 ,

( k

i=0

Li(x)Lki(y) k

i=0

Li(x)fki(y), t)

= 1,

(

ki=0

Li(x)Lki(y) k

i=0

Li(x)fki(y), t)

= 0.

for all x, y A and t > 0. Thus, we conclude that

ki=0

Li(x)Lki(y) =k

i=0

Li(x)fki(y) (2.27)

for each k = 0, 1 and all x, y A.

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10 I.-S. CHANG

Therefore, by combining (2.24) and (2.27), we get the required result, which completes the proof.

As a consequence of Theorem 2.2, we get the following superstability.

Corollary 2.3. Let (B,,, , ) be an intuitionistic fuzzy Banach algebra with unit, and let a sequenceof operators F = {f0, f1, , fk, } on A satisfy fk(0) = 0 for each k = 0, 1 , where f0 is an identityoperator. Suppose that is a function from A to an intuitionistic fuzzy normed algebra (C, , , , )

satisfying (2.13) and (2.14), and that is a function from A to an intuitionistic fuzzy normed space(D, , , , ) satisfying (2.15) and (2.16). If l > 1 is a fixed integer, ((l + 1)x) = (x), and ((l +1)x) = (x) for some real numbers and with 0 < || < l + 1 and 0 < || < l + 1, then F is a stronghigher ring derivation on A.

Proof. According to (2.19), we have L0(x) = x for all x A, and so L0(=f0) is an identity operator onA. By induction, we lead the conclusion. If k = 1, then it follows from (2.18) that f1(x) = L1(x) holdsfor all x A since A contains the unit element. Let us assume that fm(x) = Lm(x) is valid for all x Aand m < k. Then (2.18) implies that x{Lm(y) fm(y)} = 0 for all x, y A. Since A has the unit element,fk(y) = Lk(y) for all x A. Hence we conclude that fk(y) = Lk(y) for each k = 0, 1, 2 and all x A.So this tell us that F is a higher ring derivation of any rank from A and B. The proof of the corollary iscomplete.

We remark that we can prove the preceding result for the case when > l + 1 and > l + 1.

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Ick-Soon Chang, Department of Mathematics, Mokwon University, Daejeon 302-729, Republic of Korea.E-mail address: ischang@mokwon.ac.kr