Multi-valued contraction mappings in complete metric

R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI
PADOVA
K IYOSHI I SEKI
Multi-valued contraction mappings in
complete metric spaces
Rendiconti del Seminario Matematico della Università di Padova,
tome 53 (1975), p. 15-19.
<http://www.numdam.org/item?id=RSMUP_1975__53__15_0>
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Numérisation de documents anciens mathématiques
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Multi-Valued Contraction
in
Complete
Metric
KIYOSHI ISEKI
Let
of
~,
be
define
(x, d)
we
a
metric space.
Mappings
Spaces.
(*)
For any
B
nonempty subsets A,
Let CB(X) be the set of all nonempty closed and bounded subsets
of X. The space CB(X) is a metric space with respect to the above
defined distance H (see K. Kuratowski [1] p. 214). Then we have
the following theorem which is a generalization of S. Reich result [2]
(or see I. Rus [3]).
THEOREM]. Let (X, d) be a complete metric space, and let
CB(X ) be a nlulti-valued mapping with the following
every x, y
X -
for
E X,
where 0153, (3, y are
point, i. e. there is
(*)
f:
non-negative and 2a --~- 2~ -fa point x such that x E I(x).
Indirizzo dell’A.: Kobe
University,
y
1.
Then f has a
Rokko Nada Kobe,
Giappone.
16
PROOF. Let xo be
a
point
in
X,
and
xl E f (xo).
If
since H is a metric on CB(X). Therefore
then we have
0.
This contains the proof of the case a == f3 = y
we have
Next we suppose 0 2a + 2~3 + y and
=
Let h
By
=
.H( f (xo), f(x1))(p,
the definition of
Therefore there is
Hence
Hence
we
have
a
H,
we
then
we
have
have
point x, of
such that
17
Therefore
If
If
have the first case, then x2
a similar
such that
we
H( f (xl), !(X2)) &#x3E; 0, by
In general, if
all i (i
0, 1,
=
H( f (xi),
... ), H( f (x2),
=
0 for
&#x3E;
0,
which completes the proof.
there is a point X3 of
method,
i, then xi E f (xi).
there is a point Xi+2
If,
some
for
E
satisfying
Hence,
for n &#x3E; m,
is a Cauchy sequence. The completeness of X
This shows that
Let x’ be the limit of
the existence of the limit of
then
implies
Hence, by
the
assumption,
we
have
18
Let
n -
From
00, then
(1 ) implies
12013x2013~&#x3E;0y
we
the
following
relation.
have
which means
This completes the proof.
Let BN(X) be the set of all nonempty bounded subset of X. Then
we have a fixed point theorem.
THEOREM 2. Let (X, d) be a complete mctric space.
is ac
functions which satisfies
for
every x, y
then f has a
If f : X -- BN(X )
in X, where oc, f3, y are non-negative and 2~ -f- 4f3 -funique f ixed point, i. e. for some x’, f (x’ )
Theorem 2 is
is due to an idea
y
1,
-
generalization of S. Reich
by S. Reich [2].
a
result
[2]. The proof
PROOF. If oc = ~
0, then the result is trivial. We suppose
y
0 2a + 4fl + y. Now put p
(2oc + 4fl + 1 )~. Then we have p 1.
Hence there is a single-valued function g : X - X such that g(x) is a
which satisfies
point y in
=
==
=
For such
a
function g,
19
Hence
we
The
1.
For the
+ y
have
assumption 2a -f- 4~ -f- y C 1 implies
By a well known theorem, g has a fixed point x’, i. e.
point x’,
Hence
If
and
.H(z, f (z) ) &#x3E; 0,
which is impossible. Hence
To show z
consider
we
=
x’.
then
have
f(z)
=
~z~.
=
Hence we have z
x’, which shows that
The proof is complete.
f
=
has
a
unique
fixed
point.
REFERENCES
[1] K. KURATOWSKI, Topology, 1, PWN, Warszawa (1966).
[2] S. REICH, Fixed points of contractive functions, Boll. Un. Mat. Ital., (4) 5
(1972), pp. 26-42.
[3] I. A. Rus, Teoria punctului fix II, Univ. « Babes-Bolyai », Cluj (1973).
Manoscritto pervenuto in redazione il 19
marzo
1974.