Applied Mathematical Sciences, Vol. 10, 2016, no. 46, 2289 - 2294
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ams.2016.66190
A Direct Proof of Caristi’s Fixed Point Theorem
Wei-Shih Du
Department of Mathematics
National Kaohsiung Normal University
Kaohsiung 82444, Taiwan
c 2016 Wei-Shih Du. This article is distributed under the Creative Commons
Copyright Attribution License, which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
Abstract
The main aim of this paper is to give a new and simple proof of
Caristi’s fixed point theorem which is different from the known proofs
in the literature.
Mathematics Subject Classification: 47H09, 47H10
Keywords: Caristi’s fixed point theorem, transfinite induction, BrézisBrowder order principle, Zorn’s lemma
1. Introduction and preliminaries
It is well known that Banach contraction principle plays an important role
in various fields of nonlinear functional analysis and applied mathematical
analysis. Among these generalizations of the Banach contraction principle,
Caristi’s fixed point theorem [4] is undoubtedly one of the most valuable one
in nonlinear analysis. In fact, Caristi [4] proved his famous fixed point theorem
by using transfinite induction. Several elegant proofs of original Caristi’s fixed
point theorem were given; see, for example, [1-3, 5-20] and references therein.
In this paper, we give a new, simple and direct proof of Caristi’s fixed point
theorem which is different from the known proofs in the literature.
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Wei-Shih Du
2. A new and direct proof of Caristi’s fixed point theorem
In this section, we present a new, simple and direct proof of Caristi’s fixed
point theorem as follows.
Theorem 2.1 (Caristi [4]). Let (X, d) be a complete metric space and
f : X → R be a lower semicontinuous and bounded below function. Suppose
that T is a Caristi type mapping on X dominated by f ; that is, T satisfies
d(x, T x) ≤ f (x) − f (T x) for each x ∈ X.
(2.1)
Then T has a fixed point in X.
Proof. For any x ∈ X, define a set-valued mapping S : X → 2X (the power
set of X) by
S(x) = {y ∈ X : d(x, y) ≤ f (x) − f (y)}.
Clearly, x ∈ S(x) and hence S(x) 6= ∅ for all x ∈ X. We claim that for each
y ∈ S(x), we have f (y) ≤ f (x) and S(y) ⊆ S(x). Let y ∈ S(x) be given. Then
d(x, y) ≤ f (x) − f (y). So we have f (y) ≤ f (x). Since S(y) 6= ∅, let z ∈ S(y).
Thus d(y, z) ≤ f (y) − f (z). It follows that
f (z) ≤ f (y) ≤ f (x)
and hence
d(x, z) ≤ d(x, y) + d(y, z) ≤ f (x) − f (z).
So z ∈ Γ(x). Therefore we prove S(y) ⊆ S(x). We shall construct a sequence
{xn } in X by induction, starting with any point x1 ∈ X. Suppose that xn ∈ X
is known. Then choose xn+1 ∈ S(xn ) such that
f (xn+1 ) ≤
inf f (z) +
z∈S(xn )
1
, n ∈ N.
n
(2.2)
For any n ∈ N, since xn+1 ∈ S(xn ), we have
d(xn , xn+1 ) ≤ f (xn ) − f (xn+1 ).
(2.3)
So f (xn+1 ) ≤ f (xn ) for each n ∈ N. Since f is bounded below,
γ := lim f (xn ) = inf f (xn ) exists.
n→∞
n∈N
For m > n with m, n ∈ N, by (2.3) and (2.4), we obtain
d(xn , xm ) ≤
m−1
X
j=n
d(xj , xj+1 ) ≤ f (xn ) − γ.
(2.4)
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A direct proof of Caristi’s fixed point theorem
Since lim f (xn ) = γ, we obtain
n→∞
lim sup{d(xn , xm ) : m > n} = 0.
n→∞
Hence {xn } is a Cauchy sequence in X. By the completeness of X, there exists
v ∈ X such that xn → v as n → ∞. Since f is lower semicontinuous, by (2.4),
we get
f (v) ≤ lim inf f (xn ) = inf f (xn ) ≤ f (xj ) for all j ∈ N.
(2.5)
n→∞
n∈N
T
Next, we prove that ∞
n=1 S(xn ) = {v}. For m > n with m, n ∈ N, by (2.3)
and (2.5), we obtain
d(xn , xm ) ≤
m−1
X
d(xj , xj+1 ) ≤ f (xn ) − f (v).
(2.6)
j=n
Since xm → v as m → ∞, the inequality (2.6) implies
d(xn , v) ≤ f (xn ) − f (v) for all n ∈ N.
By (2.7), we know v ∈
T∞
n=1
T∞
S(xn ). Hence
∞
\
S(v) ⊆
n=1
(2.7)
S(xn ) 6= ∅ and
S(xn ).
n=1
For any w ∈
T∞
n=1
S(xn ), by (2.2), we have
d(xn , w) ≤ f (xn ) − f (w)
≤ f (xn ) − inf f (z)
z∈S(xn )
≤ f (xn ) − f (xn+1 ) +
1
n
for all n ∈ N. Hence lim d(xn , w) = 0 or, equivalently, xn → w as n →
n→∞
∞. By the uniqueness of limit of a sequence, we have w = v. So we show
T∞
n=1 S(xn ) = {v}. Since S(v) 6= ∅ and
S(v) ⊆
∞
\
S(xn ) = {v}
n=1
we get S(v) = {v}. On the other hand, by (2.1), we know T v ∈ S(v). Hence it
must be T v = v. Therefore T has a fixed point v in X. The proof is completed.
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Wei-Shih Du
Remark 2.2.
(a) Classic proofs of Caristi’s theorem involve assigning a partial order - on
X by setting
x - y ⇐⇒ d(x, y) ≤ f (x) − f (y),
and then either using Zorn’s lemma or the Brézis-Browder order principle
with the set
{y ∈ X : x - y} = {y ∈ X : d(x, y) ≤ f (x) − f (y)} for x ∈ X.
However, in our proof, we do not define any partial order on X, and then
Zorn’s lemma or the Brézis-Browder order principle are not applicable
here.
(b) In [7], we proved the Caristi’s fixed point theorem with a proof by contradiction by using the set
Γ(x) = {y ∈ X : y 6= x, d(x, y) ≤ f (x) − f (y)} for x ∈ X.
So our proof in this paper is different from [7].
Acknowledgements. This research was supported by grant no. MOST
104-2115-M-017-002 of the Ministry of Science and Technology of the Republic
of China.
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Received: June 29, 2016; Published: July 11, 2016