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All issues Volume 28 / No 1 (February 2023) Wuhan Univ. J. Nat. Sci., 28 1 (2023) 11-14 Full HTML
Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 28, Number 1, February 2023
Page(s) 11 - 14
DOI https://doi.org/10.1051/wujns/2023281011
Published online 17 March 2023

© Wuhan University 2023

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction and Preliminaries

Since Banach[1] established the famous Banach Contraction Principle (Theorem 1) in 1922, fixed point theory has become a hot reasearch field in functional analysis. Then, Ciric[2,3], Rhoades[4,5] and many other authors extended the Banach Contraction Principle into various forms. And the most remarkable form is Caristi[6,7] type fixed point theorem based on Banach Contraction Principle. Since Caristi type fixed point theorem equals to Ekeland's variational principle[8,9], it has various applications in nonlinear analysis and variational inequalities[10-13].

In recent years, the combination of two contraction type has become more and more popular. In 2013, Du and Karapinar[14] firstly merged Banach Contraction into Caristi Theorem to get Theorem 2. Then in 2019, Erdal Karapiner[15] merged Ciric-Type contraction into Caristi Theorem to get Theorem 3.

We give some important definitions and theorems mentioned above.

Definition 1   Given T : X → X, we will say that a point x∈ X is a fixed point of T if Tx = x.

Theorem 1   (Banach[1]) Let(X, d) be a complete metric space, and T :X → X be a self-mapping. Suppose that there exists q∈ (0,1) such that d(Tx, Ty) ≤ qd(x, y)for every x, y ∈ X. Then T has a unique fixed point in X.

Theorem 2   (Theorem 1 in Ref.[14]) Let(X, d) be a complete metric space, and T : X → X be a map. Suppose that there exists a function φ : X → R with φ is bounded from below (inf φ(x) > -∞)

d(x, Tx) > 0 implies d(Tx, Ty) ≤ (φ(x) - φ(Tx))d(x, y), for each x, y ∈ X. Then T has a fixed point in X.

Theorem 3   (Theorem 4 inRef.[15]) Suppose that T is a self-mapping on complete metric space (X, d). If there is a φ : X → [0,∞) with d (x, Tx) > 0 , we get

d(Tx, Ty) ≤ (φ(x) - φ(Tx))N (x, y)

where

N (x, y)

= max{d(x, y), d(x, Tx), d(y, Ty), d(y, Tx), d(x, Ty)}

for all x, y ∈ X. Then T has a fixed point.

We can easily see that Theorem 3 is a generalization of Theorem 2.

And our main work of this paper is to generalize Theorem 3.

In this paper, we firstly generalize the renowned Theorem 3 into Theorem 4. Then we show Example 1 which satisfies Theorem 4 while not satisfies Theorem 3.

1 Main Result

Theorem 4   Suppose that T is a self-mapping on complete metric space (X, d). If there is a φ : X → [0,∞) with d(x, Tx) > 0, then we get:

(1)

where

for all x, y ∈ X. Then T has a fixed point.

Proof   If there exists x0 ∈ X such that , the proof is completed.

Now we assume for x0 ∈ X , and let xn+1 = Txn. If there exists a k such that d(xk, Txk) = 0, then xk is a fixed point of T and we proved the result. So we suppose for , d(xn, Txn) > 0,then we get:

d (xn, xn+1) = d(xn, Txn) > 0.

Suppose that cn= d (xn-1, xn) > 0, from equation (1), we derive that

(2)

Step 1   We prove {d(xn, xn+1)} is non-increasing and bounded below.

To prove {d(xn, xn+1)} is non-increasing, we only need to find that there exists ε ∈ (0,1) such that

d(xn, xn+1) ≤ εd(xn-1, xn),∀n ∈ ∗.

We confirm it into two cases:

Case 1:

If

max{[d(xn-1, xn) + d(xn, xn+1)], d(xn-1, xn+1)}

= d(xn-1, xn) + d(xn, xn+1).

Then equation (2) implies that

(3)

And we have cn= d(xn-1, xn), so the above equation (3) can be also written as

cn+1 ≤ (φ (xn-1) - φ (xn))(cn+ cn+1).

Therefore, we have

(4)

From the above equation (4), we can get {φ(xn)} is positive and non-increasing.

So we can assume that it converges to some r ≥ 0. For each n ∈ , we have

= (φ (x0) - φ (x1)) + (φ (x1) - φ (x2)) + …

+ (φ (xn-1) - φ (xn))

= φ (x0) - φ (xn) → φ (x0) - r < ∞,as n → ∞.

It means that

So we can get:

For , there exists , for all , we have

It equals to

Since that , we have . By taking , then we have:

Case 2

If

max{[d(xn-1, xn) + d(xn, xn+1)], d(xn-1, xn+1)}

= d(xn-1, xn+1).

Revisiting equation (2), we can have:

Then we transpose it into:

Then it is similar to the process in Case 1.

Step 2   We prove {xn} converges to some .

Note that Step 1 shows that {d(xn, xn+1)} is nonincreasing and has inferior. So we can assume d(xn, xn+1) converges to some p ≥ 0. From

d(xn, xn+1) ≤ εd(xn-1, xn), ε ∈ (0,1),∀n ∈ ∗,

we can easily get p = 0. For each m, n ∈ with m > n, we can get:

That means that m > n. So from definition we can get is a Cauchy sequence. And since X is complete, there exists , such that converges to some .

Step 3   We prove is a fixed point of T.

From equation (1), we can get:

Since that {φ(xn)} converges to some r ≥ 0 and {xn} converges to some u ∈ X, let n → +∞, n ∈ , we have

d(u,Tu) ≤ lim{d(u, xn+1) + (φ(xn) -φ(xn+1)) · (*)} → 0.

Consequently, we prove that d(u, Tu) = 0. That means Tu =u. We completed the proof.

Then, we also get some corollaries based on Theorem 4.

Firstly, we combine Theorem 4 with Theorem 3 to get Corollary 1 and Corollary 2.

Then, we respectively combine contraction fixed point with Caristi type fixed point theory of Refs.[16,17] to get Corollary 3-6 on the basis of Theorem 4.

Corollary 1   Suppose that T is a self-mapping on complete metric space (X, d). If there is a φ : X → [0,∞) with d(x, Tx) > 0, then we get:

where

for all x, y ∈ X. Then T has a fixed point.

Corollary2 Suppose that T is a self-mapping on complete metric space (X, d). If there is a φ : X → [0,∞) with d(x, Tx) > 0, then we get:

where

for all x, y ∈ X. Then T has a fixed point.

Corollary 3   Suppose that T is a self-mapping on complete metric space (X, d). If there is a φ : X → [0, ∞) with d(x, Tx) > 0, then we get:

where

for all x, y ∈ X. Then T has a fixed point.

Corollary 4   Suppose that T is a self-mapping on complete metric space (X, d). If there is a φ : X → [0,∞) with d(x, Tx) > 0, then we get:

where

for all x, y ∈ X and1,2,3,4. Then T has a fixed point.

Corollary 5   Suppose that T is a self-mapping on complete metric space (X, d). If there is a φ : X → [0,∞) with d(x, Tx) > 0, then we get:

where

for all x, y ∈ X and . Then T has a fixed point.

Corollary 6   Suppose that T is a self-mapping on complete metric space (X, d). If there is a φ : X → [0,∞) with d(x, Tx) > 0, then we get:

where

for all x, y ∈ X and there exists monotonically decreasing function . Then T has a fixed point.

2 Example

In this section, we give an example (Example 1) to show that our conclusion contains fixed point.

Example 1 Let{0,1,2,…,n}(n), endowed with the following metric:

Define T : X → X by T 0 = 0, T 1 = 2, T 2 = 3,…, Tk = k +1,…, T (n-1) = n, Tn = 0 and φ : X → [0,+∞) by

If x ∈ X and d(x, Tx) > 0, then x 0. We have

d(Tn, T 0) ≤ (φ(n) - φ(0))K(n,0).

d(Tk, T 0) ≤ (φ(k) - φ(k + 1))K(k,0),0 < k ≤ n - 1.

d(Tp, Tq) ≤ (φ(p) - φ(p + 1))K(p, q),0 < p < q ≤ n - 1.

d(Tp, Tq) ≤ (φ(p) - φ(p + 1))K(p, q),0 < q < p ≤ n - 1.

We can easily confirm that this condition satisfies Theorem 4, and T really has a fixed point.

However, if we apply this example into Theorem 3, we find that:

d(Tk, T 0) > (φ(k) - φ(k + 1))N (k,0),0 < k ≤ n - 1.

d(Tp, Tq) > (φ(p) - φ(p + 1))N (p, q), 0 < p < q ≤ n - 1.

d(Tp, Tq) > (φ(p) - φ(p + 1))N (p, q),0 < q < p ≤ n - 1.

It shows that Example 1 does not satisfy the condition of Theorem 3.

But T really has a fixed point T 0 = 0.

Remark 1   We can easily see that N (x, y) ≤ K(x, y). So, Theorem 4 is a real generalization of Theorem 3.

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