Contractive mappings are Kannan mappings, and Kannan mappings are contractive mappings in

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Seria I: PRACE MATEMATYCZNE XLV (1) (2005), 43-56

Tomonari Suzuki^∗

Contractive mappings are Kannan mappings, and Kannan mappings are contractive mappings in

some sense

Abstract. In this paper, we prove that a mapping T on a metric space is contractive with respect to a τ -distance if and only if it is Kannan with respect to a τ -distance.

2000 Mathematics Subject Classification: Primary 54E40, Secondary 54H25.

Key words and phrases: Contractive mapping, Kannan mapping, τ -distance, fixed point.

1. Introduction. In 1922, Polish mathematician Stefan Banach proved the following famous fixed point theorem [1]: Let X be a complete metric space with metric d. Let T be a contractive mapping on X, i.e., there exists r ∈ [0, 1) satisfying

d(T x, T y) ≤ r d(x, y)

for all x, y ∈ X. Then there exists a unique fixed point x₀ ∈ X of T . This theorem called the Banach contraction principle is a forceful tool in nonlinear analysis.

This principle has many applications and many useful generalizations. For example, Caristi [2], Downing and Kirk [4], Edelstein [5], Ekeland [6, 7], Meir and Keeler [10], Nadler [11], Tataru [21], Zhong [22] and others. On the other hand, in 1969, Kannan proved the following fixed point theorem [9]: Let X be a complete metric space with metric d. Let T be a Kannan mapping on X, i.e., there exists α ∈ [0, 1/2) such that

d(T x, T y) ≤ α d(T x, x) + d(T y, y)

for all x, y ∈ X. Then there exists a unique fixed point x₀ ∈ X of T . We note that Kannan’s fixed point theorem is not an extension of the Banach contraction

∗The author is supported in part by Grants-in-Aid for Scientific Research, the Japanese Ministry of Education, Culture, Sports, Science and Technology.

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principle. We also know that a metric space X is complete if and only if every Kannan mapping has a fixed point, while there exists a metric space X such that X is not complete and every contractive mapping on X has a fixed point; see [3, 13].

In [16], the author introduced the notion of τ -distance. Let X be a metric space with metric d. Then a function p from X × X into [0, ∞) is called a τ -distance on X if there exists a function η from X × [0, ∞) into [0, ∞) and the following are satisfied:

(τ 1) p(x, z) ≤ p(x, y) + p(y, z) for all x, y, z ∈ X;

(τ 2) η(x, 0) = 0 and η(x, t) ≥ t for all x ∈ X and t ∈ [0, ∞), and η is concave and continuous in its second variable;

(τ 3) lim_nx_n = x and lim_nsup{η(z_n, p(z_n, x_m)) : m ≥ n} = 0 imply p(w, x) ≤ lim inf_np(w, x_n) for all w ∈ X;

(τ 4) lim_nsup{p(x_n, y_m) : m ≥ n} = 0 and lim_nη(x_n, t_n) = 0 imply lim_nη(y_n, t_n) = 0;

(τ 5) lim_nη(z_n, p(z_n, x_n)) = 0 and lim_nη(z_n, p(z_n, y_n)) = 0 imply lim_nd(x_n, y_n) = 0.

The metric d is a τ -distance on X. Many useful examples are stated in [8, 12, 14, 15, 16, 17, 18, 19, 20]. The author proved in [16] the following fixed point theorem, which is a generalization of the Banach contraction principle.

Theorem 1.1 ([16]) Let X be a complete metric space, let p be a τ -distance on X and let T be a contractive mapping on X with respect to p, i.e., there exists r ∈ [0, 1) such that

p(T x, T y) ≤ r p(x, y) for all x, y ∈ X. Then T has a unique fixed point x₀∈ X.

In [18], the author also proved the following fixed point theorem which is a generalization of Kannan’s fixed point theorem.

Theorem 1.2 ([18]) Let X be a complete metric space, let p be a τ -distance on X and let T be a Kannan mapping on X with respect to p, i.e., there exists α ∈ [0, 1/2) such that either (a) or (b) holds:

(a) p(T x, T y) ≤ α p(T x, x) + α p(T y, y) for all x, y ∈ X;

(b) p(T x, T y) ≤ α p(T x, x) + α p(y, T y) for all x, y ∈ X.

Then T has a unique fixed point x₀∈ X.

In this paper, we prove that a mapping T on a metric space is contractive with respect to a τ -distance if and only if it is Kannan with respect to a τ -distance.

2. Preliminaries. Throughout this paper we denote by N the set of positive integers. In this section, we give some preliminaries.

Let X be a metric space with metric d. We denote by τ (X) the set of all τ - distances on X. A τ -distance p on X is called symmetric if p(x, y) = p(y, x) for all

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x, y ∈ X. We also denote by τ₀(X) the set of all symmetric τ -distances on X. It is obvious that d ∈ τ₀(X) ⊂ τ (X). We denote by T C(X) the set of all mappings T on X such that there exist p ∈ τ (X) and r ∈ [0, 1) satisfying

p(T x, T y) ≤ r p(x, y)

for all x, y ∈ X. We define the sets T K(X), T C₀(X), T K₀(X), T C₂(X) and T C₃(X) of mappings on X as follows: T ∈ T K(X) if and only if there exist p ∈ τ (X) and α ∈ [0, 1/2) satisfying either of the following holds:

p(T x, T y) ≤ α p(T x, x) + p(T y, y) for all x, y ∈ X, or

p(T x, T y) ≤ α p(T x, x) + p(y, T y)

for all x, y ∈ X. T ∈ T C₀(X) if and only if there exist p ∈ τ₀(X) and r ∈ [0, 1) satisfying

p(T x, T y) ≤ r p(x, y)

for all x, y ∈ X. T ∈ T K₀(X) if and only if there exist p ∈ τ₀(X) and α ∈ [0, 1/2) satisfying

p(T x, T y) ≤ α p(T x, x) + p(T y, y)

for all x, y ∈ X. T ∈ T C₂(X) if and only if there exist p ∈ τ (X) and r ∈ [0, 1) satisfying

p(T x, T y) ≤ r p(y, x)

for all x, y ∈ X. T ∈ T C₃(X) if and only if there exist ` ∈ N, p ∈ τ (X) and r ∈ [0, 1) satisfying

p(T^`x, T^`y) ≤ r p(x, y)

for all x, y ∈ X. We recall that a mapping T on X belongs to T C(X) if and only if T is contractive with respect to some τ -distance p on X, and a mapping T on X belongs to T K(X) if and only if T is Kannan with respect to some τ -distance p on X.

Let X be a metric space with metric d and let p be a τ -distance on X. Then a sequence {x_n} in X is called p-Cauchy [16] if there exist a function η from X × [0, ∞) into [0, ∞) satisfying (τ 2)–(τ 5) and a sequence {z_n} in X such that lim_nsup{η(z_n, p(z_n, x_m)) : m ≥ n} = 0. We know the following.

Lemma 2.1 ([16]) Let X be a metric space with metric d and let p be a τ -distance on X. If {x_n} is a p-Cauchy sequence, then {x_n} is a Cauchy sequence. Moreover, if {y_n} is a sequence satisfying lim_nsup{p(x_n, y_m) : m ≥ n} = 0, then {y_n} is also a p-Cauchy sequence and lim_nd(x_n, y_n) = 0.

Lemma 2.2 ([16]) Let X be a metric space with metric d and let p be a τ -distance on X. If a sequence {x_n} in X satisfies lim_np(z, x_n) = 0 for some z ∈ X, then {x_n} is a p-Cauchy sequence. Moreover, if a sequence {y_n} in X also satisfies lim_np(z, y_n) = 0, then lim_nd(x_n, y_n) = 0. In particular for x, y, z ∈ X, p(z, x) = 0 and p(z, y) = 0 imply x = y.

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Lemma 2.3 ([16]) Let X be a metric space with metric d and let p be a τ -distance on X. If a sequence {x_n} in X satisfies lim_nsup{p(x_n, x_m) : m > n} = 0, then {x_n} is a p-Cauchy sequence. Moreover if a sequence {y_n} in X satisfies lim_np(x_n, y_n) = 0, then {y_n} is also a p-Cauchy sequence and lim_nd(x_n, y_n) = 0.

3. Main Result. In this section, we prove our main result.

Theorem 3.1 Let X be a metric space with metric d. Then

T C₀(X) = T K₀(X) = T C(X) = T K(X) = T C₂(X) = T C₃(X) holds.

In order to prove it, we need some lemmas. In the following lemmas and the proof of Theorem 3.1, we define sets T K₁(X) and T K₂(X) of mappings on X as follows: T ∈ T K₁(X) if and only if there exist p ∈ τ (X) and α ∈ [0, 1/2) satisfying

p(T x, T y) ≤ α p(T x, x) + p(T y, y)

for all x, y ∈ X. T ∈ T K₂(X) if and only if there exist p ∈ τ (X) and α ∈ [0, 1/2) satisfying

p(T x, T y) ≤ α p(T x, x) + p(y, T y) for all x, y ∈ X. We note that T K(X) = T K₁(X) ∪ T K₂(X).

Lemma 3.2 Let X be a metric space with metric d, let p be a τ -distance on X, and let % be a function from X × X into [0, ∞). Suppose that %(x, z) ≤ %(x, y) + %(y, z) for all x, y, z ∈ X, and that there exists c > 0 satisfying

min{p(x, y), c} = min{%(x, y), c}

for all x, y ∈ X. Let T be a mapping on X and let u be a point of X such that

m,n→∞lim %(T^mu, Tⁿu) = 0.

Then for every x ∈ X, lim_k%(T^ku, x) and lim_k%(x, T^ku) exist. Moreover, define functions β and γ from X into [0, ∞) by

β(x) = lim

k→∞%(T^ku, x) and γ(x) = lim

k→∞%(x, T^ku).

Then the following hold:

(i) If β(x) = 0, then {T^ku} converges to x.

(ii) If a sequence {x_n} in X satisfies lim_nβ(x_n) = 0, then lim_nd(Tⁿu, x_n) = 0.

(iii) If a sequence {x_n} in X converges to x and limnβ(x_n) = 0, then β(x) = 0.

(iv) A function q₁ from X × X into [0, ∞) defined by q₁(x, y) = β(x) + β(y) is a symmetric τ -distance on X.

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(v) A function q₂ from X × X into [0, ∞) defined by q₂(x, y) = γ(x) + β(y) is a τ -distance on X.

Proof Since limm,n%(T^mu, Tⁿu) = 0, we have lim_m,np(T^mu, Tⁿu) = 0. So, {T^ku}

is a p-Cauchy sequence by Lemma 2.3, and hence {T^ku} is a Cauchy sequence in X by Lemma 2.1. Let x ∈ X. For m, n ∈ N, since

%(T^mu, x) ≤ %(T^mu, Tⁿu) + %(Tⁿu, x) and

%(Tⁿu, x) ≤ %(Tⁿu, T^mu) + %(T^mu, x), we have

| %(T^mu, x) − %(Tⁿu, x) | ≤ max{ %(T^mu, Tⁿu), %(Tⁿu, T^mu) }.

Similarly we also have

| %(x, T^mu) − %(x, Tⁿu) | ≤ max{ %(T^mu, Tⁿu), %(Tⁿu, T^mu) }

for m, n ∈ N. Therefore {%(T^ku, x)} and {%(x, T^ku)} are Cauchy sequences. So, β and γ are well-defined. If β(x) = 0, then lim_k%(T^ku, x) = 0 and hence lim_kp(T^ku, x)

= 0. So by Lemma 2.3, (i) holds. We next show (ii). Assume a sequence {x_n} in X satisfies lim_nβ(x_n) = 0. Then there exists a mapping f on N satisfying

f (n) < f (n + 1) and %(T^{f (n)}u, x_n) < β(x_n) +1 n

for n ∈ N. We have limn%(T^{f (n)}u, x_n) = 0 and hence lim_np(T^{f (n)}u, x_n) = 0.

Since lim_m,np(T^{f (m)}u, T^{f (n)}u) = 0, {x_n} is p-Cauchy and lim_nd(T^{f (n)}u, x_n) = 0 by Lemma 2.3. Since {Tⁿu} is a Cauchy sequence, we have lim_nd(Tⁿu, x_n) = 0.

Let us prove (iii). Assume a sequence {x_n} in X converges to x and limnβ(x_n) = 0.

We have already shown that such {x_n} is p-Cauchy. Fix ε > 0 with 2ε < c. Then there exists n₀∈ N such that

p(T^mu, Tⁿu) = %(T^mu, Tⁿu) < ε and β(x_n) < ε

for m, n ∈ N with m, n ≥ n0. There exists a mapping g on {n₀, n₀+ 1, n₀+ 2, · · · } such that

p(T^g(n)u, x_n) = %(T^g(n)u, x_n) < β(x_n) +ε n < c for n ∈ N with n ≥ n0. Then for each k ∈ N with k ≥ n0. we have

p(T^ku, x) ≤ lim inf

n→∞ p(T^ku, x_n)

≤ lim inf

n→∞ p(T^ku, T^g(n)u) + p(T^g(n)u, x_n)

≤ lim inf

n→∞ ε + β(x_n) + ε/n = ε.

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Hence lim_kp(T^ku, x) = 0. Therefore β(x) = lim_k%(T^ku, x) = 0. Let us prove (iv).

It is clear that q₁(x, z) ≤ q₁(x, y) + q₁(y, z) for x, y, z ∈ X. Define a function η from X × [0, ∞) into [0, ∞) by η(x, t) = t for x ∈ X and t ∈ [0, ∞). We assume that {x_n} converges to x and lim_nsup{η(z_n, q₁(z_n, x_m)) : m ≥ n} = 0. Then we have lim_nq₁(z_n, x_n) = 0 and hence lim_nβ(x_n) = 0. So, by (iii), β(x) = 0 holds. Hence

q₁(w, x) = β(w) ≤ β(w) + β(x_n) = q₁(w, x_n)

for w ∈ X and n ∈ N. This implies (τ 3)q1,η. (τ 4)_q₁_,η is obvious. We assume that lim_nη(z_n, q₁(z_n, x_n)) = 0 and lim_nη(z_n, q₁(z_n, y_n)) = 0. Then since lim_nq₁(z_n, x_n)

= 0, lim_nβ(x_n) = 0. So, by (ii), we have lim_nd(Tⁿu, x_n) = 0. Similarly we have lim_nd(Tⁿu, y_n) = 0. Therefore lim_nd(x_n, y_n) = 0. This implies (τ 5)_q₁_,η. Hence q₁ is a τ -distance on X. We can prove (v) similarly. This completes the proof.

As a direct consequence, we obtain the following.

Lemma 3.3 Let X be a metric space with metric d, let p be a τ -distance on X. Let T be a mapping on X and let u be a point of X such that

m,n→∞lim p(T^mu, Tⁿu) = 0.

Then for every x ∈ X, lim_kp(T^ku, x) and lim_kp(x, T^ku) exist. Moreover, define functions β and γ from X into [0, ∞) by

β(x) = lim

k→∞p(T^ku, x) and γ(x) = lim

k→∞p(x, T^ku).

Then the following hold:

(i) If β(x) = 0, then {T^ku} converges to x.

(ii) If a sequence {x_n} in X satisfies lim_nβ(x_n) = 0, then lim_nd(Tⁿu, x_n) = 0.

(iii) If a sequence {x_n} in X converges to x and limnβ(x_n) = 0, then β(x) = 0.

(iv) A function q₁ from X × X into [0, ∞) defined by q₁(x, y) = β(x) + β(y) is a symmetric τ -distance on X.

(v) A function q₂ from X × X into [0, ∞) defined by q₂(x, y) = γ(x) + β(y) is a τ -distance on X.

Lemma 3.4 For every metric space X,

T C(X) ⊂ T K₀(X) holds.

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Proof Fix T ∈ T C(X). Then there exist p ∈ τ (X) and r ∈ [0, 1) satisfying p(T x, T y) ≤ r p(x, y)

for all x, y ∈ X. Fix u ∈ X. For m, n ∈ N with m > n, we have

p(T^mu, Tⁿu) ≤ p(T^mu, T^m−1u) + p(T^m−1u, T^m−2u) + · · · + p(Tⁿ⁺¹u, Tⁿu)

≤ r^m−1p(T u, u) + r^m−2p(T u, u) + · · · + rⁿp(T u, u)

≤ rⁿ

1 − rp(T u, u).

For m, n ∈ N with m < n, we also have

p(T^mu, Tⁿu) ≤ p(T^mu, T^m+1u) + p(T^m+1u, T^m+2u) + · · · + p(Tⁿ⁻¹u, Tⁿu)

≤ r^mp(u, T u) + r^m+1p(u, T u) + · · · + rⁿ⁻¹p(u, T u)

≤ r^m

1 − rp(u, T u).

Therefore we have

p(T^mu, Tⁿu) ≤ r^min{m,n}

1 − r max{p(u, u), p(T u, u), p(u, T u)}

for all m, n ∈ N. Since 0 ≤ r < 1, we have limm,np(T^mu, Tⁿu) = 0. So, by Lemma 3.3, β(x) = lim_kp(T^ku, x) is well-defined for every x ∈ X, and a function q from X × X into [0, ∞) defined by

q(x, y) = β(x) + β(y) for x, y ∈ X is a symmetric τ -distance on X. We also have

β(T x) = lim

k→∞p(T^ku, T x) ≤ lim

k→∞r p(T^k−1u, x) = r β(x) for all x ∈ X. Since

q(T x, T y) = 1 1 + r

β(T x) + β(T y)

+ r

1 + r

β(T x) + β(T y)

≤ r

1 + r

β(x) + β(y)

+ r

1 + r

β(T x) + β(T y)

= r

1 + r

q(T x, x) + q(T y, y)

for x, y ∈ X, we have T ∈ T K₀(X). Therefore T C(X) ⊂ T K₀(X).

T K₁(X) ⊂ T C₀(X) holds.

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Proof Fix T ∈ T K1(X). Then there exist p ∈ τ (X) and α ∈ [0, 1/2) satisfying p(T x, T y) ≤ α p(T x, x) + p(T y, y)

for all x, y ∈ X. Put r = α(1 − α)⁻¹∈ [0, 1). Since

p(T²x, T x) ≤ α p(T²x, T x) + p(T x, x),

we have p(T²x, T x) ≤ r p(T x, x) for x ∈ X. Fix u ∈ X. Then for m, n ∈ N, we have p(T^mu, Tⁿu) ≤ α p(T^mu, T^m−1u) + p(Tⁿu, Tⁿ⁻¹u)

≤ (αr^m−1+ αrⁿ⁻¹) p(T u, u)

and hence lim_m,np(T^mu, Tⁿu) = 0. So, by Lemma 3.3, β(x) = lim_kp(T^ku, x) is well-defined for every x ∈ X, and a function q from X × X into [0, ∞) defined by

q(x, y) = β(x) + β(y) for x, y ∈ X is a τ -distance. Since

p(T x, x) ≤ p(T x, T^ku) + p(T^ku, x)

≤ α p(T x, x) + α p(T^ku, T^k−1u) + p(T^ku, x), we have

p(T x, x) ≤ r p(T^ku, T^k−1u) + 1

1 − α p(T^ku, x) and hence

p(T^ku, T x) ≤ α p(T^ku, T^k−1u) + α p(T x, x)

≤ r p(T^ku, T^k−1u) + r p(T^ku, x)

for x ∈ X and k ∈ N. Therefore we have β(T x) ≤ r β(x) for all x ∈ X. So, we have q(T x, T y) = β(T x) + β(T y) ≤ r β(x) + β(y) = r q(x, y)

for all x, y ∈ X. This implies T ∈ T C₀(X).

T K₂(X) ⊂ T C₂(X) holds.

Proof Fix T ∈ T K2(X). Then there exist p ∈ τ (X) and α ∈ [0, 1/2) satisfying p(T x, T y) ≤ α p(T x, x) + p(y, T y)

for all x, y ∈ X. Put r = α(1 − α)⁻¹ ∈ [0, 1). Since p(T²x, T x) ≤ α p(T²x, T x) + p(x, T x)

and p(T x, T²x) ≤ α p(T x, x) + p(T x, T²x), we have p(T²x, T x) ≤

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r p(x, T x) and p(T x, T²x) ≤ r p(T x, x) for all x ∈ X. Fix u ∈ X. For m, n ∈ N, we have

p(T^mu, Tⁿu) ≤ α p(T^mu, T^m−1u) + α p(Tⁿ⁻¹u, Tⁿu)

≤ α (r^m−1+ rⁿ⁻¹) p(T u, u) + p(u, T u)

and hence lim_m,np(T^mu, Tⁿu) = 0. So, by Lemma 3.3, β(x) = lim_kp(T^ku, x) and γ(x) = lim_kp(x, T^ku) are well-defined for all x ∈ X, and a function q from X × X into [0, ∞) defined by

q(x, y) = γ(x) + β(y) for x, y ∈ X is a τ -distance on X. Since

p(x, T x) ≤ p(x, T^ku) + p(T^ku, T x)

≤ p(x, T^ku) + α p(T^ku, T^k−1u) + α p(x, T x), we have

p(x, T x) ≤ 1

1 − α p(x, T^ku) + r p(T^ku, T^k−1u) and hence

p(T^ku, T x) ≤ α p(T^ku, T^k−1u) + α p(x, T x)

≤ r p(T^ku, T^k−1u) + r p(x, T^ku)

for x ∈ X and k ∈ N. Therefore we have β(T x) ≤ r γ(x) for all x ∈ X. Similarly, we can prove γ(T x) ≤ r β(x) for all x ∈ X. Then we have

q(T x, T y) = γ(T x) + β(T y) ≤ r β(x) + r γ(y) = r q(y, x)

for all x ∈ X. This implies T ∈ T C₂(X). This completes the proof.

T C₂(X) ⊂ T C₃(X) holds.

Proof Fix T ∈ T C2(X). Then there exist p ∈ τ (X) and r ∈ [0, 1) satisfying p(T x, T y) ≤ r p(y, x)

for all x, y ∈ X. Since

p(T²x, T²y) ≤ r p(T y, T x) ≤ r²p(x, y)

for all x, y ∈ X, we have T²∈ T C(X). This implies T ∈ T C₃(X). This completes

the proof.

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Lemma 3.8 Let X be a metric space with metric d. Then T C₃(X) ⊂ T C₀(X) holds.

Proof Fix T ∈ T C3(X). Then there exist ` ∈ N, a τ -distance p on X and r ∈ [0, 1) such that p(T^`x, T^`y) ≤ r p(x, y) for all x, y ∈ X. Define a mapping S on X by S = T^`and let u ∈ X be fixed. Then we have, for m, n ∈ N,

p(S^mu, Sⁿu) ≤ r^min{m,n}

1 − r max{p(u, u), p(Su, u), p(u, Su)}.

Since 0 ≤ r < 1, we have lim_m,np(S^mu, Sⁿu) = 0. By Lemma 3.3, β(x) = lim_kp(S^ku, x) is well-defined for every x ∈ X. We note that

β(Sx) = lim

k→∞p(S^ku, Sx) ≤ r lim

k→∞p(S^k−1u, x) = r β(x) for x ∈ X. Define a function q from X × X into [0, ∞) by

q(x, y) =

`−1

X

k=0

r^−k/`

β(T^kx) + β(T^ky)

for x, y ∈ X. Let us prove such q is a τ -distance on X. (τ 1)_q is obvious. De- fine a function η from X × [0, ∞) into [0, ∞) by η(x, t) = t for x ∈ X and t ∈ [0, ∞). Then (τ 2) and (τ 4)_q,ηare obvious. We assume lim_nη(z_n, q(z_n, x_n)) = 0 and lim_nη(z_n, q(z_n, y_n)) = 0. Then since lim_nq(z_n, x_n) = 0, we have lim_nβ(x_n) = 0. So, by Lemma 3.3, lim_nd(Sⁿu, x_n) = 0. We can similarly prove that lim_nd(Sⁿu, y_n) = 0. Hence lim_nd(x_n, y_n) = 0. This implies (τ 5)_q,η. We assume that lim_nd(x_n, x) = 0 and lim_nsup{η(z_n, q(z_n, x_m)) : m ≥ n} = 0. Then we have lim_nq(z_n, x_n) = 0 and hence lim_nβ(x_n) = 0. By Lemma 3.3, β(x) = 0 holds. From

β(T^`x) = β(Sx) ≤ r β(x) = 0, we have T^`x = x by Lemma 3.3. For k = 1, 2, · · · , ` − 1, we have

β(T^kx) = β(T^kT^`x) = β(ST^kx) ≤ r β(T^kx) and hence β(T^kx) = 0. Therefore

q(w, x) =

`−1

X

k=0

r^−k/`β(T^kw) ≤ q(w, x_n)

for all n ∈ N. This implies (τ 3)q,η. We have shown that q is a symmetric τ -distance on X. We also have

q(T x, T y) =

`−1

X

k=0

r^−k/`

β(T^k+1x) + β(T^k+1y)

=

`−1

X

k=1

r^−(k−1)/`

β(T^kx) + β(T^ky)

+ r^−(`−1)/`

β(T^`x) + β(T^`y)

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= r^1/`

`−1

X

k=1

r^−k/`

β(T^kx) + β(T^ky) + r⁻¹

β(T^`x) + β(T^`y)

!

≤ r^1/`

`−1

X

k=1

r^−k/`

β(T^kx) + β(T^ky)

+

β(x) + β(y)

!

= r^1/`q(x, y).

Therefore T ∈ T C₀(X). This completes the proof.

Proof (Proof of Theorem 3.1) It is obvious that T C0(X) ⊂ T C(X) and T K₀(X) ⊂ T K₁(X). From Lemma 3.4, we have T C(X) ⊂ T K₀(X). From Lemma 3.5, we have T K₁(X) ⊂ T C₀(X). So, we have

T C₀(X) ⊂ T C(X) ⊂ T K₀(X) ⊂ T K₁(X) ⊂ T C₀(X) and hence

T C₀(X) = T C(X) = T K₀(X) = T K₁(X).

It is obvious that T K₀(X) ⊂ T K₂(X). From Lemmas 3.6, 3.7 and 3.8, we have T C₀(X) = T K₀(X) ⊂ T K₂(X) ⊂ T C₂(X) ⊂ T C₃(X) ⊂ T C₀(X) and hence

T C₀(X) = T K₀(X) = T K₂(X) = T C₂(X) = T C₃(X).

Since T K₁(X) = T K₂(X), we have

T K(X) = T K₁(X) ∪ T K₂(X) = T K₁(X) = T K₂(X).

Therefore we obtain the desired result.

Using Theorem 3.1, we give another proof of Theorem 1.2.

Proof (Proof of Theorem 1.2) Since T ∈ T K(X), we have T ∈ T C(X) from Theorem 3.1. So, by Theorem 1.1, T has a unique fixed point. This completes the

proof.

4. Additional Result. In this section. We discuss about mappings which are connected with Edelstein’s fixed point theorem [5].

Let X be a metric space and let p be a τ -distance on X. For ε ∈ (0, ∞], X is called (p, ε)-chainable if for each (x, y) ∈ X × X, there exists a finite sequence {u₀, u₁, u₂, · · · , u_`} in X such that u₀ = x, u_` = y and p(u_i−1, u_i) < ε for i = 1, 2, · · · , `. Further such {u₀, u₁, · · · , u_`} is called a (p, ε)-chain in X linking x and y.

Theorem 4.1 Let X be a metric space. Suppose that X is (p, ε)-chainable for some ε ∈ (0, ∞] and for some τ -distance p on X. Let T be a mapping on X. Suppose that there exists r ∈ [0, 1) such that p(T x, T y) ≤ r p(x, y) for all x, y ∈ X with p(x, y) < ε. Then T ∈ T C₀(X).

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Proof In the case of ε = ∞, T ∈ T C(X). So, by Theorem 3.1, T ∈ T C0(X). We shall prove this theorem in the case of ε < ∞. We define a function % from X × X into [0, ∞) by

%(x, y) = inf ( _`

X

i=1

p(u_i−1, u_i) : {u₀, u₁, · · · , u_`} is a (p, ε)-chain linking x and y )

for all x, y ∈ X. From (τ 1), p(x, y) ≤ %(x, y) for all x, y ∈ X, and p(x, y) = %(x, y) for all x, y ∈ X with p(x, y) < ε. Therefore

min{p(x, y), ε} = min{%(x, y), ε}

for all x, y ∈ X. We next prove %(x, z) ≤ %(x, y) + %(y, z) for all x, y, z ∈ X. Fix x, y, z ∈ X and δ > 0. Then there exist (p, ε)-chains {u₀, u₁, · · · , u_k} linking x and y, and {v₀, v₁, · · · , v_`} linking y and z such that

k

X

i=1

p(u_i−1, u_i) ≤ %(x, y) + δ and

`

X

i=1

p(v_i−1, v_i) ≤ %(y, z) + δ.

Since {u₀, u₁, · · · , u_k, v₁, v₂, · · · , v_`} is a (p, ε)-chain linking x and z, we have

%(x, z) ≤

k

X

i=1

p(u_i−1, u_i) +

`

X

i=1

p(v_i−1, v_i)

≤ %(x, y) + %(y, z) + 2δ.

Since δ > 0 is arbitrary, we obtain %(x, z) ≤ %(x, y) + %(y, z). Let us prove

%(T x, T y) ≤ r %(x, y)

for all x, y ∈ X. Let x, y ∈ X be fixed. Then for each δ > 0, there exists a (p, ε)- chain {u₀, u₁, · · · , u_`} linking x and y such thatP`

i=1p(u_i−1, u_i) ≤ %(x, y) + δ. For i ∈ {1, 2, · · · , `}, since p(u_i−1, u_i) < ε, we have

p(T u_i−1, T u_i) ≤ r p(u_i−1, u_i) < ε.

Hence {T u₀, T u₁, T u₂, · · · , T u_`} is a (p, ε)-chain linking T x and T y. So, we have

%(T x, T y) ≤

`

X

i=1

p(T u_i−1, T u_i) ≤ r

`

X

i=1

p(u_i−1, u_i)

≤ r %(x, y) + rδ ≤ r %(x, y) + δ.

Since δ > 0 is arbitrary, we obtain %(T x, T y) ≤ r %(x, y). Fix u ∈ X. Then we have

%(T^mu, Tⁿu) ≤ r^min{m,n}

1 − r max{%(u, u), %(T u, u), %(u, T u)}

for m, n ∈ N. Since 0 ≤ r < 1, we have limm,n%(T^mu, Tⁿu) = 0. So, by Lemma 3.2, β(x) = lim_k%(T^ku, x) is well-defined for every x ∈ X, and a function q from X × X into [0, ∞) defined by

q(x, y) = β(x) + β(y)

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for x, y ∈ X is a symmetric τ -distance on X. We also have β(T x) = lim

k→∞%(T^ku, T x) ≤ lim

k→∞r %(T^k−1u, x) = r β(x) for all x ∈ X. So, we have

q(T x, T y) = β(T x) + β(T y) ≤ r β(x) + β(y) ≤ r q(x, y)

for all x, y ∈ X. This implies T ∈ T C₀(X). This completes the proof. We give another proof of a generalization of Edelstein’s fixed point theorem.

Theorem 4.2 ([18]) Let X, ε, p and T as in Theorem 4.1. If X is complete, then T has a unique fixed point.

Proof From Theorem 4.1, we have T ∈ T C0(X) = T C(X). So, by Theorem 1.1, T has a unique fixed point. This completes the proof. Acknowledgement. The author wishes to express his thanks to Professor P.

V. Subrahmanyam for giving the historical comment.

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Tomonari Suzuki

Department of Mathematics, Kyushu Institute of Technology Sensuicho, Tobata, Kitakyushu 804-8550, Japan

E-mail: suzuki-t@mns.kyutech.ac.jp

(Received: 12.12.2003)