Open Access
 Issue Wuhan Univ. J. Nat. Sci. Volume 28, Number 1, February 2023 11 - 14 https://doi.org/10.1051/wujns/2023281011 17 March 2023 This 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 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 firstly merged Banach Contraction into Caristi Theorem to get Theorem 2. Then in 2019, Erdal Karapiner 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 xX is a fixed point of T if Tx = x.

Theorem 1   (Banach) 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.) 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.) 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 and 1,2,3,4. 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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