We prove the generalized Hyers-Ulam stability of the following quadratic functional equations 2f((x+y)/2)+2f((x−y)/2)=f(x)+f(y) and f(ax+ay)+(ax−ay)=2a2f(x)+2a2f(y) in fuzzy Banach spaces for a nonzero real number a with a≠±1/2.

1. Introduction and Preliminaries

The stability problem of functional equations originated from a question of Ulam [1] concerning the stability of group homomorphisms. Hyers [2] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers' theorem was generalized by Aoki [3] for additive mappings and by Th. M. Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The work of Th. M. Rassias [4] has provided a lot of influence in the development of what we call generalized Hyers-Ulam stability of functional equations. A generalization of the Th. M. Rassias theorem was obtained by Găvruţa [5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Th. M. Rassias' approach.

J. M. Rassias [6] proved a similar stability theorem in which he replaced the factor ∥x∥p+∥y∥p by ∥x∥p·∥y∥q for p,q∈ℝ with p+q≠1 (see also [7, 8] for a number of other new results). The papers of J. M. Rassias [6–8] introduced the Ulam- Găvruţa-Rassias stability of functional equations. See also [9–11].

The functional equation
f(x+y)+f(x-y)=2f(x)+2f(y)
is called a quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. A generalized Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [12] for mappings f:X→Y, where X is a normed space and Y is a Banach space. Cholewa [13] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. In [14], Czerwik proved the generalized Hyers-Ulam stability of the quadratic functional equation.

J. M. Rassias [15] introduced and solved the stability problem of Ulam for the Euler-Lagrange-type quadratic functional equation

f(rx+sy)+f(sx-ry)=(r2+s2)[f(x)+f(y)],
motivated from the following pertinent algebraic equation

|ax+by|2+|bx-ay|2=(a2+b2)(|x|2+|y|2).
The solution of the functional equation (1.2) is called a Euler-Lagrange-type quadratic mapping. J. M. Rassias [16, 17] introduced and investigated the relative functional equations. In addition, J. M. Rassias [18] generalized the algebraic equation (1.3) to the following equation
mn|ax+by|2+|nbx-may|2=(ma2+nb2)(n|x|2+m|y|2),
and introduced and investigated the general pertinent Euler-Lagrange quadratic mappings. Analogous quadratic mappings were introduced and investigated in [19, 20].

These Euler-Lagrange mappings are named Euler-Lagrange-Rassias mappings and the corresponding Euler-Lagrange equations are called Euler-Lagrange-Rassias equations. Before 1992, these mappings and equations were not known at all in functional equations and inequalities. However, a completely different kind of Euler-Lagrange partial differential equations are known in calculus of variations. Therefore, we think that J. M. Rassias' introduction of Euler-Lagrange mappings and equations in functional equations and inequalities provides an interesting cornerstone in analysis. Already some mathematicians have employed these Euler-Lagrange mappings.

Recently, Jun and Kim [21] solved the stability problem of Ulam for another Euler-Lagrange-Rassias-type quadratic functional equation. Jun and Kim [22] introduced and investigated the following quadratic functional equation of Euler-Lagrange-Rassias type:

∑i=1nriQ(∑j=1nrj(xi-xj))+(∑i=1nri)Q(∑i=1nrixi)=(∑i=1nri)2∑i=1nriQ(xi),
whose solution is said to be a generalized quadratic mapping of Euler-Lagrange-Rassias type.

During the last two decades a number of papers and research monographs have been published on various generalizations and applications of the generalized Hyers-Ulam stability to a number of functional equations and mappings (see [9, 23–26]).

Katsaras [27] defined a fuzzy norm on a vector space to construct a fuzzy vector topological structure on the space. Some mathematicians have defined fuzzy norms on a vector space from various points of view [28–30]. In particular, Bag and Samanta [31], following Cheng and Mordeson [32], gave an idea of fuzzy norm in such a manner that the corresponding fuzzy metric is of Kramosil and Michálek type [33]. They established a decomposition theorem of a fuzzy norm into a family of crisp norms and investigated some properties of fuzzy normed spaces [34].

We use the definition of fuzzy normed spaces given in [31] and [35–38] to investigate a fuzzy version of the generalized Hyers-Ulam stability for the quadratic functional equations

2f(x+y2)+2f(x-y2)=f(x)+f(y),f(ax+ay)+f(ax-ay)=2a2f(x)+2a2f(y)
in the fuzzy normed vector space setting.

Let (X,N) be a fuzzy normed vector space. A sequence {xn} in X is said to be convergent or converge if there exists an x∈X such that limn→∞N(xn-x,t)=1 for all t>0. In this case, x is called the limit of the sequence {xn} and we denote it by N-limn→∞xn=x.

Let (X,N) be a fuzzy normed vector space. A sequence {xn} in X is called Cauchy if for each ε>0 and each t>0 there exists an n0∈ℕ such that for all n≥n0 and all p>0, we have N(xn+p-xn,t)>1-ε.

It is well known that every convergent sequence in a fuzzy normed vector space is Cauchy. If each Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy normed vector space is called a fuzzy Banach space.

We say that a mapping f:X→Y between fuzzy normed vector spaces X and Y is continuous at a point x0∈X if for each sequence {xn} converging to x0 in X, then the sequence {f(xn)} converges to f(x0). If f:X→Y is continuous at each x∈X, then f:X→Y is said to be continuous on X (see [34]).

In this paper, we prove the generalized Hyers-Ulam stability of the quadratic functional equations (1.6) and (1.7) in fuzzy Banach spaces.

Throughout this paper, assume that X is a vector space and that (Y,N) is a fuzzy Banach space. Let a be a nonzero real number with a≠(±1/2).

2. Fuzzy Stability of Quadratic Functional Equations

We prove the fuzzy stability of the quadratic functional equation (1.6).

Theorem 2.1.

Letf:X→Ybe an even mapping withf(0)=0. Suppose thatφ is a mapping from Xto a fuzzy normed space(Z,N')such thatN(2f(x+y2)+2f(x-y2)-f(x)-f(y),t+s)≥min{N'(φ(x),t),N'(φ(y),s)}for allx,y∈X∖{0}and all positive real numberst,s. Ifφ(3x)=αφ(x)for some positive real numberαwithα<9, then there is a unique quadratic mappingQ:X→Ysuch thatQ(x)=N-limn→∞f(3nx)/9n and
N(Q(x)-f(x),t)≥M(x,(9-α)t18),
where
M(x,t):=min{N'(φ(x),32t),N'(φ(2x),32t),N'(φ(3x),32t),N'(φ(0),32t)}.

Proof.

Putting y=3x and s=t in (2.1), we get
N(2f(2x)+2f(-x)-f(x)-f(3x),2t)≥min{N'(φ(x),t),N'(φ(3x),t)}
for all x∈X and all t>0. Replacing x by 2x,y by 0, and s by t in (2.1), we obtain
N(4f(x)-f(2x),2t)≥min{N'(φ(2x),t),N'(φ(0),t)}.
Thus
N(9f(x)-f(3x),6t)≥min{N'(φ(x),t),N'(φ(2x),t),N'(φ(3x),t),N'(φ(0),t)},
and so
N(f(x)-f(3x)9,t)≥min{N'(φ(x),32t),N'(φ(2x),32t)N'(φ(3x),32t),N'(φ(0),32t)}.
Then by the assumption,
M(3x,t)=M(x,tα).
Replacing x by 3nx in (2.7) and applying (2.8), we get
N(f(3nx)9n-f(3n+1x)9n+1,αnt9n)=N(f(3nx)-f(3n+1x)9,αnt)≥M(3nx,αnt)=M(x,t).
Thus for each n>m we have
N(f(3mx)9m-f(3nx)9n,∑k=mn-1αkt9k)=N(∑k=mn-1(f(3kx)9k-f(3k+1x)9k+1),∑k=mn-1αkt9k)≥min{⋃k=mn-1{N(f(3kx)9k-f(3k+1x)9k+1,αkt9k)}}≥M(x,t).
Let ε>0 and δ>0 be given. Since limt→∞M(x,t)=1, there is some t0>0 such that M(x,t0)>1-ε. Since ∑k=0∞αkt0/9k<∞, there is some n0∈ℕ such that ∑k=mn-1αkt0/9k<δ for n>m≥n0. It follows that
N(f(3mx)9m-f(3nx)9n,δ)≥N(f(3mx)9m-f(3nx)9n,∑k=mn-1αkt9k)≥M(x,t0)≥1-ε
for all t≥t0. This shows that the sequence {f(3nx)/9n} is Cauchy in (Y,N). Since (Y,N) is complete, {f(3nx)/9n} converges to some Q(x)∈Y. Thus we can define a mapping Q:X→Y by Q(x):=N-limt→∞f(3nx)/9n. Moreover, if we put m=0 in (2.10), then we observe that
N(f(3nx)9n-f(x),∑k=0n-1αkt9k)≥M(x,t).
Thus
N(f(3nx)9n-f(x),t)≥M(x,t∑k=0n-1(α/9)k).
Next we show that Q is quadratic. Let x,y∈X. Then we have
N(2Q(x+y2)+2Q(x-y2)-Q(x)-Q(y),t)≥min{N(2Q(x+y2)-2f(3n(x+y)/2)9n,t5),N(2Q(x-y2)-2f(3n(x-y)/2)9n,t5),N(f(3nx)9n-Q(x),t5),N(f(3ny)9n-Q(y),t5),N(2f(3n(x+y)/2)9n+2f(3n(x-y)/2)9n-f(3nx)9n-f(3ny)9n,t5)}.
The first four terms on the right-hand side of the above inequality tend to 1 as n→∞ and the fifth term, by (2.1), is greater than or equal to
min{N'(φ(3nx),9nt10),N'(φ(3ny),9nt10)}=min{N'(φ(x),(9α)nt10),(φ(y),(9α)nt10)},
which tends to 1 as n→∞. Hence
N(2Q(x+y2)+2Q(x-y2)-Q(x)-Q(y),t)=1
for all x,y∈X and all t>0. This means that Q satisfies the Jensen quadratic functional equation and so it is quadratic.

Next, we approximate the difference between f and Q in a fuzzy sense. For every x∈X and t>0, by (2.13), for large enough n, we have
N(Q(x)-f(x),t)≥min{N(Q(x)-f(3ny)9n,t2),N(f(3ny)9n-f(x),t2)}≥M(x,t2∑k=0∞(α/9)k)=M(x,(9-α)t18).
The uniqueness assertion can be proved by a standard fashion; cf. [36]: Let Q' be another quadratic mapping from X into Y, which satisfies the required inequality. Then for each x∈X and t>0,
N(Q(x)-Q'(x),t)≥min{N(Q(x)-f(x),t2),N(Q'(x)-f(x),t2)}≥M(x,(9-α)t36).
Since Q and Q' are quadratic,
N(Q(x)-Q'(x),t)=N(Q(3nx)-Q'(3nx),9nt)≥M(x,(9/α)n(9-α)t36).
for all x∈X, all t>0 and all n∈ℕ.

Since 0<α<9, limn→∞(9/α)n=∞. Hence the right-hand side of the above inequality tends to 1 as n→∞. It follows that Q(x)=Q'(x) for all x∈X.

Theorem 2.2.

Letf:X→Ybe an even mapping withf(0)=0. Suppose thatφis a mapping fromXto a fuzzy normed space (Z,N')satisfying (2.1). Ifφ(3x)=αφ(x)for some real numberαwith α>9, then there is a unique quadratic mappingQ:X→Ysuch thatQ(x)=N-limn→∞9nf(x/3n) andN(Q(x)-f(x),t)≥M(x,(α-9)t2α),
where
M(x,t):=min{N'(φ(x3),t6),N'(φ(2x3),t6),N'(φ(x),t6),N'(φ(0),t6)}.

Proof.

It follows from (2.7) that
N(f(x)-f(x3),t)≥min{N'(φ(x3),t6),N'(φ(2x3),t6),N'(φ(x),t6),N'(φ(0),t6)}.
Then by the assumption,
M(x3,t)=M(x,αt).
Replacing x by x/3n in (2.22) and applying (2.23), we get
N(9nf(x3n)-9n+1f(x3n+1),9ntαn)=N(f(x3n)-9f(x3n+1),tαn)≥M(x3n,tαn)=M(x,t).
Thus for each n>m we have
N(9mf(x3m)-9nf(x3n),∑k=mn-19ktαk)=N(∑k=mn-1(9kf(x3k)-9k+1f(x3k+1)),∑k=mn-19ktαk)≥min{⋃k=mn-1{N(9kf(x3k)-9k+1f(x3k+1),9ktαk)}}≥M(x,t).

Let ε>0 and δ>0 be given. Since limt→∞M(x,t)=1, there is some t0>0 such that M(x,t0)>1-ε. Since ∑k=0∞9kt0/αk<∞, there is some n0∈ℕ such that ∑k=mn-19kt0/αk<δ for n>m≥n0. It follows that
N(9mf(x3m)-9nf(x3n),δ)≥N(9mf(x3m)-9nf(x3n),∑k=mn-19ktαk)≥M(x,t0)≥1-ε
for all t≥t0. This shows that the sequence {9nf(x/3n)} is Cauchy in (Y,N). Since (Y,N) is complete, {9nf(x/3n)} converges to some Q(x)∈Y. Thus we can define a mapping Q:X→Y by Q(x):=N-limt→∞9nf(x/3n). Moreover, if we put m=0 in (2.8), then we observe that
N(9nf(x3n)-f(x),∑k=0n-19ktαk)≥M(x,t).
Thus
N(9nf(x3n)-f(x),t)≥M(x,t∑k=0n-1(9/α)k).
The rest of the proof is similar to the proof of Theorem 2.1.

Theorem 2.3.

Letf:X→Ybe a mapping withf(0)=0. Suppose thatφis a mapping fromXto a fuzzy normed space (Z,N')satisfying (2.1). Ifφ(2x)=αφ(x)for some positive real number α withα<4, then there is a unique quadratic mappingQ:X→Ysuch thatQ(x)=N-limn→∞f(2nx)/4nandN(Q(x)-f(x),t)≥M(x,(4-α)t8)
where M(x,t):=min{N'(φ(2x),2t),N'(φ(0),2t)}.

Proof.

Letting y=0 and replacing x by 2x and s by t in (2.1), we obtain
N(4f(x)-f(2x),2t)≥min{N'(φ(2x),t),N'(φ(0),t)}.
Thus
N(f(x)-f(2x)4,t)≥min{N'(φ(2x),2t),N'(φ(0),2t)}.
Then by the assumption,
M(2x,t)=M(x,tα).
Replacing x by 2nx in (2.31) and applying (2.32), we get
N(f(2nx)4n-f(2n+1x)4n+1,αnt4n)=N(f(2nx)-f(4n+1x)4,αnt)≥M(2nx,αnt)=M(x,t).
Thus for each n>m we have
N(f(2mx)4m-f(2nx)4n,∑k=mn-1αkt4k)=N(∑k=mn-1(f(2kx)4k-f(2k+1x)4k+1),∑k=mn-1αkt4k)≥min{⋃k=mn-1{N(f(2kx)4k-f(2k+1x)4k+1,αkt4k)}}≥M(x,t).

Let ε>0 and δ>0 be given. Since limt→∞M(x,t)=1, there is some t0>0 such that M(x,t0)>1-ε. Since ∑k=0∞αkt0/4k<∞, there is some n0∈ℕ such that ∑k=mn-1αkt0/4k<δ for n>m≥n0. It follows that
N(f(2mx)4m-f(2nx)4n,δ)≥N(f(2mx)4m-f(2nx)4n,∑k=mn-1αkt4k)≥M(x,t0)≥1-ε
for all t≥t0. This shows that the sequence {f(2nx)/4n} is Cauchy in (Y,N). Since (Y,N) is complete, {f(2nx)/4n} converges to some Q(x)∈Y. Thus we can define a mapping Q:X→Y by Q(x):=N-limt→∞f(2nx)/4n. Moreover, if we put m=0 in (2.34), then we observe that
N(f(2nx)4n-f(x),∑k=0n-1αkt4k)≥M(x,t).
Thus
N(f(2nx)4n-f(x),t)≥M(x,t∑k=0n-1(α/4)k).
The rest of the proof is similar to the proof of Theorem 2.1.

Theorem 2.4.

Let f:X→Ybe a mapping withf(0)=0. Suppose thatφis a mapping fromXto a fuzzy normed space(Z,N') satisfying (2.1). Ifφ(2x)=αφ(x)for some real numberαwithα>4, then there is a unique quadratic mappingQ:X→Ysuch thatQ(x)=N-limn→∞4nf(x/2n)andN(Q(x)-f(x),t)≥M(x,(α-4)t2α),
where M(x,t):=min{N'(φ(x),t/2),N'(φ(0),t/2)}.

Proof.

It follows from (2.31) that
N(f(x)-4f(x2),t)≥min{N'(φ(x),t2),N'(φ(0),t2)}.
Then by the assumption,
M(x2,t)=M(x,αt).
Replacing x by x/2n in (2.39) and applying (2.40), we get
N(4nf(x2n)-4n+1f(x2n+1),4ntαn)=N(f(x2n)-4f(x2n+1),tαn)≥M(x2n,tαn)=M(x,t).
Thus for each n>m we have
N(4mf(x2m)-4nf(x2n),∑k=mn-14ktαk)=N(∑k=mn-1(4kf(x2k)-4k+1f(x2k+1)),∑k=mn-14ktαk)≥min{⋃k=mn-1{N(4kf(x2k)-4k+1f(x2k+1),4ktαk)}}≥M(x,t).

Let ε>0 and δ>0 be given. Since limt→∞M(x,t)=1, there is some t0>0 such that M(x,t0)>1-ε. Since ∑k=0∞4kt0/αk<∞, there is some n0∈ℕ such that ∑k=mn-14kt0/αk<δ for n>m≥n0. It follows that
N(4mf(x2m)-4nf(x2n),δ)≥N(4mf(x2m)-4nf(x2n),∑k=mn-14ktαk)≥M(x,t0)≥1-ε
for all t≥t0. This shows that the sequence {4nf(x/2n)} is Cauchy in (Y,N). Since (Y,N) is complete, {4nf(x/2n)} converges to some Q(x)∈Y. Thus we can define a mapping Q:X→Y by Q(x):=N-limt→∞4nf(x/2n). Moreover, if we put m=0 in (2.42), then we observe that
N(4nf(x2n)-f(x),∑k=0n-14ktαk)≥M(x,t).
Thus
N(4nf(x2n)-f(x),t)≥M(x,t∑k=0n-1(4/α)k).
The rest of the proof is similar to the proof of Theorem 2.1.

Now we prove the fuzzy stability of the quadratic functional equation (1.7) for the case a≠(±1/2).

Theorem 2.5.

Let|2a|>1andf:X→Ya mapping withf(0)=0. Suppose thatφis a mapping fromXto a fuzzy normed space(Z,N')such thatN(f(ax+ay)+f(ax-ay)-2a2f(x)-2a2f(y),t+s)≥min{N'(φ(x),t),N'(φ(y),s)}for allx,y∈X∖{0}and all positive real numberst,s. Ifφ(2ax)=αφ(x)for some positive real numberαwith0<α<4a2, then there is a unique quadratic mappingQ:X→Ysuch thatQ(x)=N-limn→∞f((2a)nx)/(2a)2nandN(Q(x)-f(x),t)≥N'(φ(x),(4a2-α)t4) for all x∈Xand allt>0.

Proof.

Putting y=x and s=t in (2.46), we get
N(f(2ax)-4a2f(x),2t)≥N'(φ(x),t)
for all x∈X and all t>0. Thus
N(f(x)-f(2ax)4a2,t2a2)≥N'(φ(x),t)
and so
N(f(x)-f(2ax)4a2,t)≥N'(φ(x),2a2t).
Replacing x by (2a)nx in (2.50), we get
N(f((2a)nx)(2a)2n-f((2a)n+1x)(2a)2n+2,αnt(2a)2n)=N(f((2a)nx)-f((2a)n+1x)4a2,αnt)≥N'(φ(x),2a2t).
Thus for each n>m we have
N(f((2a)mx)(2a)2m-f((2a)nx)(2a)2n,∑k=mn-1αkt(2a)2k)=N(∑k=mn-1(f((2a)kx)(2a)2k-f((2a)k+1x)(2a)2k+2),∑k=mn-1αkt(2a)2k)≥min{⋃k=mn-1{N(f((2a)kx)(2a)2k-f((2a)k+1x)(2a)2k+2,αkt(2a)2k)}}≥N'(φ(x),2a2t).

Let ε>0 and δ>0 be given. Since limt→∞N'(φ(x),2a2t)=1, there is some t0>0 such that N'(φ(x),2a2t0)>1-ε. Since ∑k=0∞αkt0/(2a)2k<∞, there is some n0∈ℕ such that ∑k=mn-1αkt0/(2a)2k<δ for n>m≥n0. It follows that
N(f((2a)mx)(2a)2m-f((2a)nx)(2a)2n,δ)≥N(f((2a)mx)(2a)2m-f((2a)nx)(2a)2n,∑k=mn-1αkt(2a)2k)≥N'(φ(x),2a2t0)≥1-ε
for all t≥t0. This shows that the sequence {f((2a)nx)/(2a)2n} is Cauchy in (Y,N). Since (Y,N) is complete, {f((2a)nx)/(2a)2n} converges to some Q(x)∈Y. Thus we can define a mapping Q:X→Y by Q(x):=N-limt→∞f((2a)nx)/(2a)2n. Moreover, if we put m=0 in (2.52), then we observe that
N(f((2a)nx)(2a)2n-f(x),∑k=0n-1αkt(2a)2k)≥N'(φ(x),2a2t).
Thus
N(f((2a)nx)(2a)2n-f(x),t)≥N'(φ(x),2a2t∑k=0n-1(α/(2a)2)k).
The rest of the proof is similar to the proof of Theorem 2.1.

Theorem 2.6.

Let|2a|<1andf:X→Ya mapping withf(0)=0. Suppose thatφis a mapping fromXto a fuzzy normed space(Z,N')satisfying (2.46). Ifφ(2ax)=αφ(x)for some real numberαwithα>4a2, then there is a unique quadratic mappingQ:X→Ysuch thatQ(x)=N-limn→∞(2a)2nf(x/(2a)n)andN(Q(x)-f(x),t)≥M(x,(α-4a2)t4)for allx∈Xand allt>0.

Proof.

It follows from (2.50) that
N(f(x)-(2a)2f(x2a),2t)≥N'(φ(x2a),t)
for all x∈X and all t>0. Thus
N(f(x)-4a2f(x2a),t)≥N'(φ(x2a),t2)=N'(φ(x),α2t).
Replacing x by x/(2a)n in (2.58), we get
N((2a)2nf(x(2a)n)-(2a)2n+2f(x(2a)n+1),(2a)2ntαn)=N(f(x(2a)n)-4a2f(x(2a)n+1),αnt)≥N'(φ(x),α2t).
Thus for each n>m we have
N((2a)2mf(x(2a)m)-(2a)2nf(x(2a)n),∑k=mn-1(2a)2ktαk)=N(∑k=mn-1((2a)2kf(x(2a)k)-(2a)2k+2f(x(2a)k+1)),∑k=mn-1(2a)2ktαk)≥min{⋃k=mn-1{N((2a)2kf(x(2a)k)-(2a)2k+2f(x(2a)k+1),(2a)2ktαk)}}≥N'(φ(x),α2t).
Let ε>0 and δ>0 be given. Since limt→∞N'(φ(x),(α/2)t)=1, there is some t0>0 such that N'(φ(x),(α/2)t0)>1-ε. Since ∑k=0∞(2a)2kt0/αk<∞, there is some n0∈ℕ such that ∑k=mn-1(2a)2kt0/αk<δ for n>m≥n0. It follows that
N((2a)2mf(x(2a)m)-(2a)2nf(x(2a)n),δ)≥N((2a)2mf(x(2a)m)-(2a)2nf(x(2a)n),∑k=mn-1(2a)2ktαk)≥N'(φ(x),α2t0)≥1-ε
for all t≥t0. This shows that the sequence {(2a)2nf(x/(2a)n)} is Cauchy in (Y,N). Since (Y,N) is complete, {(2a)2nf(x/(2a)n)} converges to some Q(x)∈Y. Thus we can define a mapping Q:X→Y by Q(x):=N-limt→∞(2a)2nf(x/(2a)n). Moreover, if we put m=0 in (2.60), then we observe that
N((2a)2nf(x(2a)n)-f(x),∑k=0n-1(2a)2ktαk)≥N'(φ(x),α2t).
Thus
N((2a)2nf(x(2a)n)-f(x),t)≥N'(φ(x),αt2∑k=0n-1((2a)2/α)k).
The rest of the proof is similar to the proof of Theorem 2.1.

Acknowledgment

Dr. Sun-Young Jang was supported by the Research Fund of University ofUlsan in 2008, and Dr. Choonkil Park was supported by National ResearchFoundation of Korea (NRF-2009-0070788).

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