1. Introduction and Preliminaries

W.A. Kirk in[4] introduced asymptotic nonlinear contractions in metric spaces and proved the existence and uniqueness of a fixed point for such mappings by using ultrafilter methods. Y.Z. Chen in[2] showed the existence and uniqueness of a fixed point for asymptotic nonlinear contractions with weaker assumptions than Kirk and without using ultrafilter methods. The following theorem was proved by Chen in[2]:

THEOREM 1.1 Let (M, d) be a complete metric space, and let T: M → M satisfy:

d(Tⁿx, Tⁿy) ≤ h_n(d(x, y)) for all x, y in M,

where h_n:[0, ∞] →[0, ∞] and lim_{n → ∞} h_n = h uniformly on any bounded interval[0, b]. Assume is upper semicontinuous, h(t) < t for t > 0, and assume there is a positive integer m such that h_m is upper semicontinuous and h_m (0) = 0. If there exists some x₀ in M with a bounded orbit, then T has a unique fixed point x* in M and lim_{n → ∞} Tⁿx = x* for all x in M.

In the above theorem, the assumption on the boundedness of the orbit of some x₀ is crucial to the proof of the existence of a fixed point in X.

In this note, we remove the assumption of the boundedness of the orbit and instead introduce the condition that liminf_{n → ∞} (d(x,Tⁿx)) = 0 for some x in M, and we are able to get the same conclusion as in Theorem 1.1.

For further details about asymptotic contractions and related topics, we refer the reader to[1, 3, 5-7] and the references therein.

2. Main Results

We shall begin with a lemma introduced in[1] which will be used later for the proof of our theorem. We will reproduce the proof for the sake of completeness.

LEMMA 2.1 Let h: R+ → R+ be upper semicontinuous and h(t) < t for t > 0. Suppose there exists two sequences of nonnegative real numbers {u_n}, {ε_n}, such that:

u_2n ≤ h(u_n) + ε_n

with ε_n → 0.

Then either sup_n>0 u_n = ∞ or liminf _{n → ∞} u_n = 0.

Proof. Assume b := sup_n>0 u_n < ∞ and liminf_{n → ∞} u_n ≠ 0. Then there exists a real number m >0 and an integer N₁ > 0 such that u_n > m for all n > N₁.

Since h is upper semicontinuous, then h(t)/t is upper semicontinuous on[m, b]. Therefore h(t)/t achieves its maximum in the compact interval[m,b]. Let

L_m = max{h(t)/t, t in[m,b]} < 1

because h(t) < t for all t > 0. Let ε > 0. Then, using our assumption on {u_n} and ε_n, there exists N₂ > N₁ such that

u_2n ≤ h(u_n) + ε ≤ L_mu_n + ε for all n > N₂

Note that L_mu_n < u_n, since L_m < 1 and u_n > 0.

We define ƒ:[0, ∞) →[0, ∞) by:

ƒ(x)= L_mx + ε

Then ƒ(x) is a contraction, since L_m < 1 and

|ƒ(x) – ƒ(y)| = |(L_mx + ε) - (L_my + ε)|

= |L_mx - L_my|

=L_m|x - y|

By Banach's fixed point theorem, ƒ(x) has a unique fixed point ε / (1 - L_m), and lim _{n → ∞} ƒⁿ(x) = ε / 1 – L_m for every x in[0, ∞).

So for any n > N₂

u₂²_n ≤ h(u_2n) + ε ≤ L_m u_2n + ε

≤ L_m(L_mu_n + ε) + ε = L_mƒ(u_n) + ε = ƒ²(u_n)

We claim that u₂^k_n ≤ ƒ^k(u_n) for all k and all n > N₂. We know it holds for k =2. By induction, let’s assume it is true for some h ≥ 2. Then we have

u₂^h+1_n ≤ h(u₂^h_n) + ε ≤ L_mu₂^h_n + ε

= ƒ (u₂^h_n) ≤ ƒ(ƒ^h(u_n)) = ƒ^h+1(u_n) (1)

Since ƒ is increasing. Therefore (1) holds for all k.

Now let r = inf_{n >} N₂ u_n ≤ ƒ^k(u_n) which converges to

ε / (1- L_m) as k → ∞. Since ε was arbitrary we must have r = 0, which is a contradiction.

Now we proceed to the main theorem.

THEOREM 2.1 Let (M, d) be a complete metric space. Let T: M → M satisfy:

d(Tⁿx, Tⁿy) ≤ h_n(d(x, y)) for all x, y in M

where h_n:[0, ∞) →[0, ∞) and lim_{n → ∞} h_n = h uniformly on any bounded interval[0, b]. Assume h is upper semicontinuous, h(t) < t for t > 0, and assume there is a positive integer n* such that h_n* is upper semicontinuous and h_n*(0) = 0. If liminf_{n → ∞} (d(x,Tⁿx)) = 0 then T has a unique fixed point x in M, and lim_{n → ∞} Tⁿy = x for all y in M.

Proof. Let us first establish the uniqueness of the fixed point. Suppose there exist two fixed points z₁ and z₂ for T with z₁ ≠ z₂.

Then d(z₁, z₂) = d(Tⁿz₁, Tⁿz₂) ≤ h_n(d(z₁, z₂)).

Letting n → ∞ yields d(z₁, z₂) ≤ h(d(z₁, z₂)) < d(z₁, z₂), which is a contradiction. Thus the fixed point is unique.

Without loss of generality we set h_n(0) = 0 and

h(0) = 0.

We will first show that the sequence a_n = d(Tⁿ⁺¹x, Tⁿx) is bounded for every fixed x.

a_n = d(Tⁿ⁺¹x, Tⁿx) ≤ h_n(d(Tx, x))

hence:

limsup_{n → ∞} a_n ≤ lim_{n → ∞} h_n(d(Tx, x))

= h(d(Tx, x)) ≤ d(Tx, x) = b

Thus the sequence a_n = d(Tⁿ⁺¹x, Tⁿx) is bounded.

Now we will show that liminf_{n → ∞} a_n = 0.

We have: a_2n = d(T²ⁿ⁺¹x, T²ⁿx)

≤ h_n(d(Tⁿ⁺¹x, Tⁿx))

= h(a_n) +[h_n(a_n) - h (a_n)].

We choose: ε_n = h_n(a_n) - h(a_n)

Since h_n → h uniformly on[0, 2b], then ε_n→0. By Lemma 2.1, liminf_{n → ∞} a_n = 0.

Now we will show that lim_{n → ∞} a_n = 0. Assume by contradiction that lim_{n → ∞} a_n ≠ 0. Then limsup _{n → ∞} a_n >0.

Since liminf_{n → ∞} a_n = 0, there exists n₀ > 0 such that

0 < an₀ < limsup_{n → ∞} a_n.

Let {an_i} be a subsequence of {a_n} such that an_i > an₀ for all i and lim_{i → ∞} an_i = limsup_{n → ∞} a_n. Then:

an₀ < an_i = d(Tⁿⁱ⁺¹x, Tⁿⁱx)

≤ hn_{i -} n₀(d(Tⁿ⁰⁺¹x; Tⁿ⁰x))

= h(an₀) + hn_{i -} n₀(an₀) - h(an₀)

Letting i→ ∞ we get: an₀ ≤ h(an₀) < an₀, which is a contradiction. Thus lim_{n → ∞} a_n = 0.

Now we show that x is a fixed point for T.

Since liminf_{n → ∞} (d(Tⁿx, x)) = 0, there exists a subsequence, {m_k}, such that lim _{k → ∞} d(T^mkx, x) = 0

We will first show that T^n*x = x. We have:

limsup_{k → ∞} (d(T^mk+n*x, T^n*x)

≤ limsup_{k → ∞} h_n*(d(T^mkx, x)) ≤ h_n*(0) = 0

and

d(T^mk+n*x, T^mkx)

≤ d(T^mk+n*x, T^mk+n*-1x) + d(T^mk+1x, T^mkx)

by the triangle inequality.

= am_k+ n*-1 + am_k.

Letting k → ∞, we get:

d(T^mk+n*x, T^mkx)→0, since lim _{n → ∞} a_n = 0

Thus T^mkx→T^n*x. Hence by the uniqueness of the limit of T^mkx , T^n*x = x.

We claim that T^n* has a unique fixed point. Using the same argument as above, let z₁, z₂ be two fixed points of T^n* with z₁ ≠ z₂.

Note that z₁ = T^n*z₁, and T^n*z₁ = T^2n*z₁.

therefore z₁ = T^kn*z₁ for any positive integer k.

Then d(T^n*z₁, T^n*z₂) = d(T^kn*z₁, T^kn*z₂) ≤ h_kn*d(z₁, z₂). Letting k →∞, we get:

d(z₁, z₂)≤ lim _{k → ∞} h_kn*(d(z₁, z₂))

= h(d(z₁, z₂)) < d(z₁, z₂)

which is a contradiction. So T^n* has a unique fixed point.

Note that T^n*+1x = T(T^n*x) = Tx. Hence Tx is also a fixed point of T^n* . By the uniqueness of the fixed point of T^n*, we have:

Tx = x.

To show that lim_{n → ∞}d(Tⁿy, x) = 0, for any y in M, by Theorem 1.1 we need only show that d(Tⁿy, x) is bounded. We have:

d(Tⁿy, x) = d(Tⁿy, Tⁿx) ≤ h_n(d(y, x)).

Letting n → ∞, we get:

limsup _{n → ∞} d(Tⁿy, x) ≤limsup _{n → ∞} h_n(d(y, x))

= h(d(y, x)) ≤ d(y, x)

Thus the orbit of y is bounded, and hence Tⁿy →x for every y in M.

3. Conclusions and Future Directions for Research

We proved the existence and uniqueness of the fixed point as well as the convergence of the Picard iterates for asymptotic nonlinear contractions without assuming the boundedness of the orbits but with a weaker assumption, namely that liminf_{n → ∞} d(Tⁿx, x) = 0 for some x in M. It still remains an open problem whether the same conclusions hold by assuming only that liminf_{n → ∞} d(Tⁿx, x) <+∞.