115
Fixed-Point Theory for Generalized Metric Spaces
Proof: We check the axioms for a d-metric. (M2) If �(x, y) = 0, then
d(x, y) + T (u(x), u(y)) = 0. Hence, d(x, y) = 0, and so x = y. (M3) Obvious by symmetry of d and T . (M4) Obvious since d and T satisfy the triangle
inequality.
•
Completeness also carries over if some continuity conditions are imposed.
4.8.3 Proposition Using the notation of Proposition 4.8.2, let u be continuous as a function from (X, d) to R
+
0 (where X is endowed with the topology
determined by d, and R
+
0 is endowed with its usual topology), and let T be
continuous as a function from the topological product space (R
+
0 )
2 to R
+
0 ,
satisfying the additional property T (x, x) = x for all x. If (X, d) is a complete
metric space, then (X, �) is a complete d-metric space.
Proof: Let (x n ) be a Cauchy sequence in (X, �). Thus, for each ε > 0, there
exists n 0 ∈ N such that for all m, n ≥ n 0 we have d(x m , x n ) ≤ d(x m , x n ) +
T (u(x m ), u(x n )) = �(x m , x n ) < ε. So (x n ) is also a Cauchy sequence in (X, d)
and therefore has a unique limit x in (X, d). In particular, we have x n → x
in (X, d), and also u(x n ) → u(x) and T (u(x n ), u(x)) → T (u(x), u(x)) = u(x).
We have to show that �(x n , x) converges to 0 as n → ∞. For all n ∈ N, we
obtain �(x n , x) = d(x n , x) + T (u(x n ), u(x)) → u(x) = u 1 (x), and it remains
to show that �(x, x) = 0. But this follows from the fact that (x n ) is a Cauchy
sequence, since it implies that u(x n ) = u 1 (x n ) = �(x n , x n ) → 0 as n → ∞,
and hence by continuity of u we obtain u(x) = 0.
•
An example of a natural function T which satisfies the requirements of
Propositions 4.8.2 and 4.8.3 is
T : R
+
0 × R
+
0 → R
+
0
1
: (x, y) � → (x + y).
2
We discuss a few more examples of d-metrics; they are partly taken from
[Matthews, 1992].
4.8.4 Example Let d be the metric d(x, y) =
1
2
+
0 , let u : R
+
0 → R
+
0
be the identity function, and define T (x, y) =
1
|x−y| on R
(x + y). Then � as defined in
2
Proposition 4.8.2 is a d-metric, and �(x, y) =
1
2
1
2 (x + y) = max{x, y}
|x − y| +
for all x, y ∈ R
+
0 .
4.8.5 Example Let I be the set of all closed intervals in R. Then d : I ×I →
R
+
0 defined by
1
d([a, b], [c, d]) = (|a − c| + |b − d|)
2
is a metric on I. Let u : I → R
+
0 be defined by
u([a, b]) = b − a
Fixed-Point Theory for Generalized Metric Spaces
Proof: We check the axioms for a d-metric. (M2) If �(x, y) = 0, then
d(x, y) + T (u(x), u(y)) = 0. Hence, d(x, y) = 0, and so x = y. (M3) Obvious by symmetry of d and T . (M4) Obvious since d and T satisfy the triangle
inequality.
•
Completeness also carries over if some continuity conditions are imposed.
4.8.3 Proposition Using the notation of Proposition 4.8.2, let u be continuous as a function from (X, d) to R
+
0 (where X is endowed with the topology
determined by d, and R
+
0 is endowed with its usual topology), and let T be
continuous as a function from the topological product space (R
+
0 )
2 to R
+
0 ,
satisfying the additional property T (x, x) = x for all x. If (X, d) is a complete
metric space, then (X, �) is a complete d-metric space.
Proof: Let (x n ) be a Cauchy sequence in (X, �). Thus, for each ε > 0, there
exists n 0 ∈ N such that for all m, n ≥ n 0 we have d(x m , x n ) ≤ d(x m , x n ) +
T (u(x m ), u(x n )) = �(x m , x n ) < ε. So (x n ) is also a Cauchy sequence in (X, d)
and therefore has a unique limit x in (X, d). In particular, we have x n → x
in (X, d), and also u(x n ) → u(x) and T (u(x n ), u(x)) → T (u(x), u(x)) = u(x).
We have to show that �(x n , x) converges to 0 as n → ∞. For all n ∈ N, we
obtain �(x n , x) = d(x n , x) + T (u(x n ), u(x)) → u(x) = u 1 (x), and it remains
to show that �(x, x) = 0. But this follows from the fact that (x n ) is a Cauchy
sequence, since it implies that u(x n ) = u 1 (x n ) = �(x n , x n ) → 0 as n → ∞,
and hence by continuity of u we obtain u(x) = 0.
•
An example of a natural function T which satisfies the requirements of
Propositions 4.8.2 and 4.8.3 is
T : R
+
0 × R
+
0 → R
+
0
1
: (x, y) � → (x + y).
2
We discuss a few more examples of d-metrics; they are partly taken from
[Matthews, 1992].
4.8.4 Example Let d be the metric d(x, y) =
1
2
+
0 , let u : R
+
0 → R
+
0
be the identity function, and define T (x, y) =
1
|x−y| on R
(x + y). Then � as defined in
2
Proposition 4.8.2 is a d-metric, and �(x, y) =
1
2
1
2 (x + y) = max{x, y}
|x − y| +
for all x, y ∈ R
+
0 .
4.8.5 Example Let I be the set of all closed intervals in R. Then d : I ×I →
R
+
0 defined by
1
d([a, b], [c, d]) = (|a − c| + |b − d|)
2
is a metric on I. Let u : I → R
+
0 be defined by
u([a, b]) = b − a