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Fixed-Point Theory for Generalized Metric Spaces
4.8.8 Proposition Let (X, �) be a d-metric space, and define d : X ×X → R
by setting d(x, y) = �(x, y) for x = y and by setting d(x, x) = 0 for all x ∈ X.
Then d is a metric on X.
Proof: We obviously have d(x, x) = 0 for all x ∈ X. If d(x, y) = 0, then either
x = y or �(x, y) = 0, and from the latter we also obtain x = y. Symmetry is
clear. We want to show that d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X. If
d(x, z) = �(x, z) and d(z, y) = �(z, y), then the inequality is clear. If d(x, z) =
0, then x = z, and the inequality reduces to d(x, y) ≤ d(x, y), which holds. If
d(z, y) = 0, then z = y, and the inequality reduces to d(x, y) ≤ d(x, y), which
also holds.
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4.8.9 Definition The metric d just defined from the d-metric � is called the
metric associated with �.
Considering Step (3) of the proof of Theorem 4.2.5, we easily verify that δ
is a dislocated ultrametric and note also that d is the metric associated with
δ.
The following proposition allows one to derive from completeness of d, in
general, that � itself is complete.
4.8.10 Proposition Let (X, �) be a d-metric space, and let d denote the
metric associated with �. If the metric d is complete, then so is �. If f is a
contraction relative to �, then f is a contraction relative to d with the same
contractivity factor.
Proof: Suppose that (x n ) is a Cauchy sequence in �. Then for all ε > 0,
there exists n 0 such that �(x k , x m ) < ε for all k, m ≥ n 0 . Consequently, we
also obtain d(x k , x m ) < ε for all k, m ≥ n 0 . Since d is complete, the sequence
(x n ) converges in d to some x, and d(x n , x) → 0 as n → ∞. We show that
�(x n , x) → 0 as n → ∞, and to do this we consider two cases.
Case i. Assume that the sequence (x n ) is such that there exists n 0 satisfying
the property that for all m ≥ n 0 , we have x m = x. Then �(x m , x) = d(x m , x)
for all m ≥ n 0 so that �(x m , x) → 0, and hence �(x n , x) → 0.
Case ii. Assume that there exist infinitely many n k ∈ N such that x n k = x.
Since (x n ) is a Cauchy sequence with respect to �, we obtain �(x n k , x) < ε
for all ε > 0, and so �(x, x) = 0. Hence, �(x n , x) = d(x n , x) for all n ∈ N, and
we obtain that �(x n , x) → 0 as n → ∞, as required.
Let λ ∈ [0, 1) be such that �(f (x), f (y)) ≤ λ�(x, y) for all x, y ∈ X,
and let x, y ∈ X. If f (x) = f (y), then we have d(f (x), f (y)) = 0, hence
d(f (x), f (y)) ≤ λd(x, y). If f (x) = f (y), then x = y, and so d(f (x), f (y)) =
�(f (x), f (y)) ≤ λ�(x, y) = λd(x, y), as required.
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Fixed-Point Theory for Generalized Metric Spaces
4.8.8 Proposition Let (X, �) be a d-metric space, and define d : X ×X → R
by setting d(x, y) = �(x, y) for x = y and by setting d(x, x) = 0 for all x ∈ X.
Then d is a metric on X.
Proof: We obviously have d(x, x) = 0 for all x ∈ X. If d(x, y) = 0, then either
x = y or �(x, y) = 0, and from the latter we also obtain x = y. Symmetry is
clear. We want to show that d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X. If
d(x, z) = �(x, z) and d(z, y) = �(z, y), then the inequality is clear. If d(x, z) =
0, then x = z, and the inequality reduces to d(x, y) ≤ d(x, y), which holds. If
d(z, y) = 0, then z = y, and the inequality reduces to d(x, y) ≤ d(x, y), which
also holds.
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4.8.9 Definition The metric d just defined from the d-metric � is called the
metric associated with �.
Considering Step (3) of the proof of Theorem 4.2.5, we easily verify that δ
is a dislocated ultrametric and note also that d is the metric associated with
δ.
The following proposition allows one to derive from completeness of d, in
general, that � itself is complete.
4.8.10 Proposition Let (X, �) be a d-metric space, and let d denote the
metric associated with �. If the metric d is complete, then so is �. If f is a
contraction relative to �, then f is a contraction relative to d with the same
contractivity factor.
Proof: Suppose that (x n ) is a Cauchy sequence in �. Then for all ε > 0,
there exists n 0 such that �(x k , x m ) < ε for all k, m ≥ n 0 . Consequently, we
also obtain d(x k , x m ) < ε for all k, m ≥ n 0 . Since d is complete, the sequence
(x n ) converges in d to some x, and d(x n , x) → 0 as n → ∞. We show that
�(x n , x) → 0 as n → ∞, and to do this we consider two cases.
Case i. Assume that the sequence (x n ) is such that there exists n 0 satisfying
the property that for all m ≥ n 0 , we have x m = x. Then �(x m , x) = d(x m , x)
for all m ≥ n 0 so that �(x m , x) → 0, and hence �(x n , x) → 0.
Case ii. Assume that there exist infinitely many n k ∈ N such that x n k = x.
Since (x n ) is a Cauchy sequence with respect to �, we obtain �(x n k , x) < ε
for all ε > 0, and so �(x, x) = 0. Hence, �(x n , x) = d(x n , x) for all n ∈ N, and
we obtain that �(x n , x) → 0 as n → ∞, as required.
Let λ ∈ [0, 1) be such that �(f (x), f (y)) ≤ λ�(x, y) for all x, y ∈ X,
and let x, y ∈ X. If f (x) = f (y), then we have d(f (x), f (y)) = 0, hence
d(f (x), f (y)) ≤ λd(x, y). If f (x) = f (y), then x = y, and so d(f (x), f (y)) =
�(f (x), f (y)) ≤ λ�(x, y) = λd(x, y), as required.
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