Combinatorial optimisation in radiotherapy treatment planning

1.   Introduction and preliminaries

Fixed point theory is one of the most powerful and fundamental tools of modern mathematics and may be considered a core subject of nonlinear analysis. The theory has developed rapidly since Banach's contraction principle ^[1] was introduced. There are many theorems that have the same conclusion as the contraction principle but with different sufficient conditions. For example, Kannan ^[2], Chatterjea ^[3], Geraghty ^[4], and Ćirić ^[5]. Next, we recall the concept of Kannan mapping.

Let $ (X, d) $ be a metric space, $ T:X\to X $ is said to be a Kannan mapping if there exists a constant $ \lambda\in [0, \frac{1}{2}) $ such that

for all $ x, y\in X $. Kannan proved that every Kannan mapping in a complete metric space has a unique fixed point ^[2]. In our view, Kannan's fixed point theorem is very important because Subrahmanyam ^[6] proved that a metric space $ X $ is complete if and only if every Kannan mapping has a fixed point. Thereafter, Suzuki ^[8,9,10] further generalized this conclusion. In recent years, Lu ^[11] introduced the best area of Kannan system with degree $ s $ in $ b $-metric spaces with constant $ s $. Futhermore, Berinde and Pacurar ^[12] presented the concept of enriched Kannan mappings. Mohapatra et al. ^[13] defined the new concepts of mutual Kannan contractivity and mutual contractivity that generalized the Kannan mapping and contraction. In ^[14], Debnath generalized Kannan's fixed point Theorem and used it to solve a particular type of integral equation. For more conclusions on Geraghty type contractions, see ^{[4,16,18,19,25]}. About multi-valued mappings, see ^{[15,26,27,28,29,30]}.

On the other hand, in 2018, Górnicki ^[7] proved some extensions of Kannan's fixed point theorem in the framework of metric space. In 2021, Doan ^[17] extended a result of ^[7] and proved some generalizations of Kannan-type fixed point theorems for singlevalued and multivalued mappings defined on a complete strong $ b $-metric space. On this basis, Doan raised two open questions. Our main purpose of this paper is to give positive answers to those two questions and establish a new type of Riech's fixed point theorem to improve results of Doan.

Kirk and Shahzad ^[20] introduced the notion of strong $ b $-metric space. Some deep results about strong $ b $-metric spaces are obtained in ^{[21,22,23,24]}.

Definition 1.1. ^[20] Let $ X $ be a nonempty set, $ K\geq1 $, $ D:X\times X\to [0, \infty) $ be a mapping. If for all $ x, y, z\in X $,

(1) $ D(x, y) = 0 \Leftrightarrow x = y $;

(2) $ D(x, y) = D(y, x) $;

(3) $ D(x, y)\leq KD(x, z)+D(z, y) $.

Then $ D $ is called a strong $ b $-metric on $ X $ and $ (X, D, K) $ is called a strong $ b $-metric space.

Remark 1.2. Let $ (X, D, K) $ be a strong $ b $-metric space. From Definition 1.1, we can derive the inequality,

In fact, for all $ x, y, z\in X $, we have

Therefore, for every strong $ b $-metric $ D $ with constant $ K $, it implies that

refer to ^[21].

It is obvious that if $ (X, D) $ is a metric space, then it is a strong $ b $-metric space.

Definition 1.3. ^[20] Let $ (X, D, K) $ be a strong $ b $-metric space, $ \{x_n\} $ be a sequence in $ X $ and $ x\in X $. Then

(1) $ \{x_n\} $ is said to converge to $ x $ if $ \lim\limits_{n\to\infty}D(x_n, x) = 0 $;

(2) $ \{x_n\} $ is called Cauchy if $ \lim\limits_{n, m\to\infty}D(x_n, x_m) = 0 $;

(3) $ (X, D, K) $ is said to be complete if every Cauchy sequence converges.

Throughout this paper, we denote $ \mathbb{N^{*}} $ as the set of all positive integers. Let $ (X, D) $ be a metric space. We denote by $ CB(X) $ the collection of all nonempty bounded closed subsets of $ (X, D) $. Let $ T:X\to CB(X) $ be a multi-valued mapping, we say that $ x $ is a fixed point of $ T $ if $ x\in Tx $. Let $ H:CB(X)\times CB(X)\to [0, \infty) $ be the Hausdorff metric on $ CB(X) $ defined by

where $ A, B\in CB(X) $ and $ d(x, A): = \inf\limits_{y\in A}D(x, y) $.

In order to characterize the open problems posed by Doan ^[17]. We will use the following class of functions

where $ q\in(0, \frac{1}{2}) $. We call $ \Psi_{q} $ the class of Geraghty functions. We next introduce the two questions raised by Doan.

Theorem 1.4. [17, Theorem 2.4] Let $ (X, D, K) $ be a complete strong $ b $-metric space, $ T:X\to X $ be a mapping, $ q\in(0, \frac{1}{2}) $. If there exists $ \psi\in\Psi_{q} $ satisfying for all $ x, y\in X $ with $ x\neq y $,

implies

Then, $ T $ has a unique fixed point $ x^{*}\in X $.

Question 1.5. Does there exist $ q = \frac{1}{2} $ such that the above theorem holds?

For brevity, we denote $ \Psi_{\frac{1}{2}}: = \{\psi:(0, \infty)\to [0, \frac{1}{2})\; |\; \psi(t_n)\to \frac{1}{2} \; \text{implies} \; t_n\to 0\} $.

Theorem 1.6. [17, Theorem 3.3] Let $ (X, D, K) $ be a complete strong $ b $-metric space and $ T:X\to CB(X) $ be a multi-valued mapping. Suppose there exists $ s\in(0, k) $ with $ 0 < k < \frac{1}{2} $ satisfying

for each $ x, y\in X $. Then $ T $ has a fixed point.

Question 1.7. Does there exist $ k = \frac{1}{2} $ such that mapping $ T $ in Theorem 1.6 has a fixed point free?

2.   Answer to questions

2.1.   Answer to question 1

In this section, we answer question 1, and first we give the following lemma.

Lemma 2.1. Let $ (X, D, K) $ be a strong $ b $-metric space, $ T:X\to X $ be a mapping. If there exists $ q\in (0, \frac{1}{2}] $ and $ \psi\in\Psi_{q} $ satisfying for all $ x, y\in X $ with $ x\neq y $,

implies

Then,

(1) $ D(Tx, T^2x)\leq D(x, Tx) $, for each $ x\in X $;

(2) for all $ x, y\in X $, either $ \frac{1}{K+1}D(x, Tx)\leq D(x, y) $ or $ \frac{1}{K+1}D(Tx, T^2x)\leq D(Tx, y) $.

Proof. (1) Let $ x\in X $ be an arbitrary point. Without loss of generality, we can suppose that $ x\neq Tx $. From $ \frac{1}{K+1}D(x, Tx)\leq D(x, Tx) $, we have

which implies that

(2) By contradiction, assume that there exists $ x', y'\in X $ such that $ D(x', y') < \frac{1}{K+1}D(x', Tx') $ and $ D(Tx', y') < \frac{1}{K+1}D(Tx', T^2x') $. Using the triangle inequality and (2.1), we have

which contradicts the fact that $ D(x', Tx') > 0 $ (because $ D(x', Tx') > (K+1)D(x', y')\geq0 $). Thus, we proved (2). □

Theorem 2.2. Let $ (X, D, K) $ be a complete strong $ b $-metric space, $ T:X\to X $ be a mapping. If there exists $ \psi\in\Psi_{\frac{1}{2}} $ satisfying for all $ x, y\in X $ with $ x\neq y $,

implies

Then, $ T $ has a unique fixed point $ x^{*}\in X $.

Proof. Let $ x $ be an arbitrary point in $ X $. Let $ x_n = T^{n}x $, $ n\in\mathbb{N}^{*} $. If for some $ n_0\in\mathbb{N}^{*} $, $ x_{n_0} = x_{n_0+1} $, then $ x_{n_0} $ will be a fixed point of $ T $. So, we can suppose that $ x_n\neq x_{n+1} $ for all $ n\in\mathbb{N}^{*} $. From Lemma 2.1, for all $ n\in\mathbb{N}^{*} $, we have

Therefore, $ \{D(x_{n}, x_{n+1})\}_{n = 1}^{\infty} $ is a decreasing sequence of nonnegative real numbers, which implies that it has a limit. Let $ \lim\limits_{n\to\infty}D(x_{n}, x_{n+1}) = t\geq0 $. In order to prove that $ t = 0 $, suppose that $ t > 0 $. In such a case, since $ 0 < \frac{1}{K+1}D(x_{n}, x_{n+1})\leq D(x_{n}, x_{n+1}) $, for all $ n\in\mathbb{N}^{*} $, we have

Then

Passing to the limit as $ n\to\infty $, we get $ \lim\limits_{n\to\infty}\psi(D(x_{n}, x_{n+1})) = \frac{1}{2} $, which implies that $ \lim\limits_{n\to\infty}D(x_{n}, x_{n+1}) = 0 $, which is a contradiction. Therefore, $ t = 0 $ and $ \lim\limits_{n\to\infty}D(x_{n}, x_{n+1}) = 0 $.

Given $ \varepsilon > 0 $, there exists $ N\in\mathbb{N}^{*} $ such that

From Lemma 2.1, for all $ n, m\in \mathbb{N}^{*} $ with $ m > n > N $, either $ \frac{1}{K+1}D(x_{n-1}, Tx_{n-1})\leq D(x_{n-1}, x_{m-1}) $ or $ \frac{1}{K+1}D(Tx_{n-1}, T^2x_{n-1})\leq D(Tx_{n-1}, x_{m-1}) $. We consider two cases.

Case 1. If $ \frac{1}{K+1}D(x_{n-1}, Tx_{n-1})\leq D(x_{n-1}, x_{m-1}) $. In this case, notice that $ D(x_{n-1}, Tx_{n-1}) = D(x_{n-1}, x_{n}) > 0 $, we have

Case 2. If $ \frac{1}{K+1}D(Tx_{n-1}, T^2x_{n-1})\leq D(Tx_{n-1}, x_{m-1}) $. In this case, notice that $ D(Tx_{n-1}, T^2x_{n-1}) = D(x_{n}, x_{n+1}) > 0 $, we have

Thus, combining all the cases we have

Therefore, $ \{x_n\} $ is a Cauchy sequence in $ (X, D, K) $. As it is complete, there exists $ x^*\in X $ such that $ \lim\limits_{n\to\infty}x_n = x^{*} $.

Since $ \lim\limits_{n\to\infty}x_n = x^{*} $ and $ \lim\limits_{n\to\infty}D(x_{n}, x_{n+1}) = 0 $, for all $ \varepsilon' > 0 $, there exists $ N'\in\mathbb{N}^{*} $ such that

Obviously, the sequence $ \{x_n\} $ has an infinite number of terms not equal to $ x^* $. By Lemma 2.1, for all $ x_{n} $, where $ x_{n}\neq x^* $ and $ n > N' $, either $ \frac{1}{K+1}D(x_{n}, Tx_{n})\leq D(x_{n}, x^*) $ or $ \frac{1}{K+1}D(Tx_{n}, T^2x_{n})\leq D(Tx_{n}, x^*) $. Clearly, there exists $ x_{n_0} $, where $ x_{n_0}\neq x^* $ and $ n_0 > N' $, such that $ \frac{1}{K+1}D(x_{n_0}, Tx_{n_0})\leq D(x_{n_0}, x^*) $. Then

From (2.2), we have

Then, $ D(x^*, Tx^*) = 0 $, $ x^* $ is a fixed point of $ T $.

Now, suppose that $ y^* $ is another fixed point of $ T $ such that $ y^*\neq x^* $. Since $ \frac{1}{K+1}D(x^*, Tx^*)\leq D(x^*, y^*) $, we have

which is a contradiction. Therefore, $ T $ has a unique fixed point $ x^* $ and $ \lim\limits_{n\to\infty}T^nx = x^* $ for all $ x\in X $. □

Remark 2.3. Theorem 1.4 is a corollary of Theorem 2.2.

Proof. Let $ (X, D, K) $ be a complete strong $ b $-metric space, $ q\in(0, \frac{1}{2}) $, $ T:X\to X $ be a mapping, which satisfying the condition of Theorem 1.4 with $ \psi\in\Psi_{q} $. It is not difficult to observe that the function $ \varphi:(0, \infty)\to[0, q) $ defined by

belongs to $ \Psi_{\frac{1}{2}} $. For all $ x, y\in X $ with $ x\neq y $, if $ \frac{1}{K+1}D(x, Tx)\leq D(x, y) $, then

According to Theorem 2.2, $ T $ has a unique fixed point. □

Corollary 2.4. [17, Theorem 2.1] Let $ (X, D, K) $ be a complete strong $ b $-metric space, $ T:X\to X $ be a mapping. If there exists $ \psi\in\Psi_{\frac{1}{2}} $ satisfying for all $ x, y\in X $,

Then, $ T $ has a unique fixed point $ x^{*}\in X $.

2.2.   Answer to question 2

In order to answer question 2, we first need a couple of lemmas.

Lemma 2.5. ^[17] Let $ (X, D, K) $ be a strong $ b $-metric space and $ A, B\in CB(X) $. If $ H(A, B) > 0 $ then for all $ h > 1 $ and $ a\in A $, there exists $ b\in B $ such that

Lemma 2.6. ^[24] Let $ (X, D, K) $ be a strong $ b $-metric space and let $ \{x_n\} $ be a sequence in $ X $. Assume that there exists $ \lambda\in[0, 1) $ satisfying

for any $ n\in\mathbb{N}^* $. Then $ \{x_n\} $ is Cauchy.

Lemma 2.7. ^[26] Let $ (X, D, K) $ be a strong $ b $-metric space, then for all $ a\in X $ and $ A, B\in CB(X) $

Proof. Let $ a\in X $, $ A, B\in CB(X) $. Using the triangular inequality, for all $ y\in B $, we have

Hence, we have

The proof is complete. □

Theorem 2.8. Let $ (X, D, K) $ be a complete strong $ b $-metric space and $ T:X\to CB(X) $ be a multi-valued mapping. Suppose there exists $ s\in(0, \frac{1}{2}) $ satisfying

for each $ x, y\in X $. Then $ T $ has at least one fixed point.

Proof. First, we construct a sequence $ \{x_n\}\subseteq X $ such that for each $ n\in\mathbb{N}^* $, $ x_{n}\in Tx_{n-1} $ and

where $ h = \frac{1}{4s}+\frac{1}{2} > 1 $. Let $ x_0\in X $ and $ x_1\in Tx_0 $. If $ H(Tx_0, Tx_1) = 0 $, which implies that $ Tx_0 = Tx_1 $, then $ x_1\in Tx_0 = Tx_1 $ and $ x_1 $ is a fixed point of $ T $. So, let us suppose that $ H(Tx_0, Tx_1) > 0 $. From Lemma 2.5, for $ h = \frac{1}{4s}+\frac{1}{2} > 1 $ and $ x_1\in Tx_0 $, there exists $ x_2\in Tx_1 $ such that

Similarly, let us suppose that $ H(Tx_1, Tx_2) > 0 $, by Lemma 2.5, there exists $ x_3\in Tx_2 $ such that

Suppose that $ H(Tx_{n-1}, Tx_n) > 0 $, for each $ n\in\mathbb{N}^*. $ Using Lemma 2.5 and proceeding inductively, we can obtain a sequence $ \{x_n\} $ such that $ x_{n}\in Tx_{n-1} $ and (2.3) holds for each $ n\in\mathbb{N}^* $.

Since $ x_{n}\in Tx_{n-1} $ for all $ n\in\mathbb{N}^* $, then $ \frac{1}{K+1}d(x_{n-1}, Tx_{n-1})\leq D(x_{n-1}, x_n) $. Hence, we have

From (2.3) and (2.4), we get

Therefore, for all $ n\in\mathbb{N}^* $, we have

where $ \lambda = \frac{hs}{1-hs} = \frac{1+2s}{3-2s}\in(\frac{1}{3}, 1) $. According to Lemma 2.6, $ \{x_n\} $ is Cauchy. Since $ (X, D, K) $ complete, there exists $ x^*\in X $ such that $ \lim\limits_{n\to\infty}x_n = x^* $.

We claim that for all $ n\in\mathbb{N}^* $, either $ \frac{1}{K+1}d(x_n, Tx_n)\leq D(x_n, x^*) $, or $ \frac{1}{K+1}d(x_{n+1}, Tx_{n+1})\leq D(x_{n+1}, x^*) $. In order to prove our claim, we argue by contradiction. If there exists $ n_0\in\mathbb{N}^* $ such that $ D(x_{n_0}, x^*) < \frac{1}{K+1}d(x_{n_0}, Tx_{n_0}) $ and $ D(x_{n_0+1}, x^*) < \frac{1}{K+1}d(x_{n_0+1}, Tx_{n_0+1}) $. By (2.5), we have

On the other hand, since $ H(Tx_{n_0}, Tx_{n_0+1}) > 0 $, then $ Tx_{n_0}\neq Tx_{n_0+1} $. Hence, $ D(x_{n_0}, x_{n_0+1}) > 0 $. This contradiction guarantees that our claim holds.

Without loss of the generality, we may assume that $ \frac{1}{K+1}d(x_n, Tx_n)\leq D(x_n, x^*) $ holds for infinity positive integers $ n $. Then, there exists $ \{x_{n_i}\}_{i = 1}^{\infty}\subseteq \{x_n\} $ such that

By Lemma 2.7, for each $ i\in\mathbb{N}^* $, we have

Then, from (2.5), we get

where $ \lambda\in(\frac{1}{3}, 1) $. Letting $ i\to\infty $ in the above inequality, we obtain $ d(x^*, Tx^*) = 0 $. Then $ x^* $ is a fixed point of $ T $. □

Remark 2.9. Notice that the Hausdorff semidistance is utilized in the fixed point theorems for multi-valued mappings, for example ^[31,32,33]. It is obvious that the Hausdorff semidistance $ e(A, B) $ and the Hausdorff distance $ H(A, B) $ are distinct. However, we can demonstrate that Lemma 2.5, Lemma 2.7, and Theorem 2.8 hold, if replacing "$ H(A, B) $" with "$ e(A, B) $", "$ e(B, A) $", and "$ e(A, B) $", respectively.

Remark 2.10. It is evident to see that Theorem 1.6 can be obtained from Theorem 2.8.

Corollary 2.11. ^[15] Let $ (X, d) $ be a complete metric space, $ 0\leq s < \frac{1}{2} $. Suppose $ T:X\to CB(X) $ is a continuous multi-valued mapping satisfying

then $ T $ has at least one fixed point.

We give an example of a multi-valued mapping $ T $ that satisfies the conditions of Theorem 2.8. It is worth noting that all points in $ X $ are fixed points of $ T $.

Example 2.12. Let $ X = \mathbb{N}^* $, $ D:X\times X\to[0, \infty) $ defined by $ D(x, y) = |x-y| $, for all $ x, y\in X $. It is easy to verify that $ (X, D, 1) $ is a complete strong $ b $-metric space. Let $ T:X\to CB(X) $ defined by

Then it is clear that $ d(x, Tx) = 0 $ and $ H(Tx, Ty) = 0 $ for each $ x, y\in X $. By Theorem 2.8, $ T $ has at least one fixed point. Furthermore, it is easy to see that any point in $ X $ is an fixed point of $ T $.

3.   A new type of Riech's fixed point theorem

Lemma 3.1. Let $ (X, D, K) $ be a strong $ b $-metric space, $ T:X\to X $ be a mapping. If there exists $ \varphi\in\Psi_{\frac{1}{3}} $ satisfying for all $ x, y\in X $ with $ x\neq y $,

implies

Then,

(1) $ D(Tx, T^2x)\leq D(x, Tx) $, for each $ x\in X $;

(2) for all $ x, y\in X $, either $ \frac{1}{K+1}D(x, Tx)\leq D(x, y) $ or $ \frac{1}{K+1}D(Tx, T^2x)\leq D(Tx, y) $.

Proof. For any $ x\in X $, without loss of generality, we may consider $ x\neq Tx $. By $ \frac{1}{K+1}D(x, Tx)\leq D(x, Tx) $, we have

Thus, $ D(Tx, T^2x)\leq D(x, Tx) $ for all $ x\in X $. The proof of the second part of this Lemma follows in a similar manner as Lemma 2.1 and so is omitted. □

Theorem 3.2. Let $ (X, D, K) $ be a complete strong $ b $-metric space, $ T:X\to X $ be a mapping. If there exists $ \varphi\in\Psi_{\frac{1}{3}} $ satisfying for all $ x, y\in X $ with $ x\neq y $,

implies

Then, $ T $ has a unique fixed point $ x^{*}\in X $.

Proof. Let $ x\in X $ be an arbitrary point and $ \{x_n\} $ be a sequence defined by $ x_n = T^nx $ for all $ n\in\mathbb{N}^* $, suppose that every $ D(x_n, x_{n+1}) > 0 $. By Lemma 3.1,

Then, $ \{D(x_{n}, x_{n+1})\}_{n = 1}^{\infty} $ is monotonically decreasing with a lower bound. Hence, $ \{D(x_{n}, x_{n+1})\} $ converges. For each $ n\in\mathbb{N}^* $, since $ D(x_n, x_{n+1}) > 0 $ and $ \frac{1}{K+1}D(x_n, Tx_n)\leq D(x_n, x_{n+1}) $, we get

Then

Suppose that $ \lim\limits_{n\to\infty}D(x_{n}, x_{n+1}) > 0 $. Letting $ n\to\infty $, we obtain $ \varphi(D(x_{n}, x_{n+1}))\to\frac{1}{3} $, which implies $ D(x_{n}, x_{n+1})\to0 $. This contradiction guarantees that $ \lim\limits_{n\to\infty}D(x_{n}, x_{n+1}) = 0 $.

According to Lemma 3.1, for each $ p, q\in\mathbb{N}^* $, either $ 0 < \frac{1}{K+1}D(x_p, Tx_p)\leq D(x_p, x_q) $ or $ 0 < \frac{1}{K+1}D(Tx_p, T^2x_p)\leq D(Tx_p, x_q) $. Let $ M(p, q) = (K+\frac{K+1}{3})D(x_p, x_{p+1})+\frac{1}{3}D(x_q, x_{q+1})+\frac{1}{3}D(x_p, x_{q}) $, where $ p, q\in\mathbb{N}^* $. We claim that

Now there are the following two cases.

Case 1. If $ 0 < \frac{1}{K+1}D(x_p, Tx_p)\leq D(x_p, x_q) $. In this case, we have

Case 2. If $ 0 < \frac{1}{K+1}D(Tx_p, T^2x_p)\leq D(Tx_p, x_q) $. In this case, by Lemma 3.1, we have

Therefore, we obtain (3.1).

Next, we demonstrate that $ \{x_n\} $ is a Cauchy sequence reasoning by contradiction. If not, it is easy to show that there exists $ \varepsilon_0 > 0 $ and two subsequence $ \{x_{n_k}\} $ and $ \{x_{m_k}\} $ of $ \{x_n\} $ such that for each $ k\in\mathbb{N}^* $, we have

From $ \lim\limits_{n\to\infty}D(x_n, x_{n+1}) = 0 $, there exists $ N\in\mathbb{N}^* $ such that $ D(x_n, x_{n+1}) < \frac{\varepsilon_0}{7K+2} $ for each $ n\geq N $. For all $ k > N $, since $ \min{\{n_k, m_k, m_k-1\}}\geq K-1\geq N $, then

By (3.1) and (3.2), we have

Hence, we obtain

which contradicts (3.2). This contradiction shows that $ \{x_n\} $ is Cauchy. As $ (X, D, K) $ is complete, there exists $ x^*\in X $ such that $ \lim\limits_{n\to\infty}x_n = x^* $.

According to Lemma 3.1, for each $ n\in\mathbb{N}^* $, either $ \frac{1}{K+1}D(x_{n}, Tx_{n})\leq D(x_{n}, x^*) $ or $ \frac{1}{K+1}D(Tx_{n}, T^2x_{n})\leq D(Tx_{n}, x^*) $. Similarly, let us consider two cases.

Case 1. If $ \frac{1}{K+1}D(x_{n}, Tx_{n})\leq D(x_{n}, x^*) $, since $ D(x_{n}, Tx_{n}) = D(x_{n}, x_{n+1}) > 0 $, we have

Then

Case 2. If $ \frac{1}{K+1}D(Tx_{n}, T^2x_{n})\leq D(Tx_{n}, x^*) $, by $ D(Tx_{n}, T^2x_{n}) = D(x_{n+1}, x_{n+2}) > 0 $, we get

Then

Therefore, for all $ n\in\mathbb{N}^* $, we have

Letting $ n\to\infty $ in the above inequality, we obtain $ D(x^*, Tx^*) = 0 $ and $ x^* $ is a fixed point of $ T $.

Suppose that $ y^* $ is another fixed point of $ T $ and $ D(y^*, x^*) > 0 $. Since $ D(x^*, Tx^*) = 0 $, it follows that $ \frac{1}{K+1}D(x^*, Tx^*)\leq D(x^*, y^*) $. Then

which is a contradiction with the fact that $ D(x^*, y^*) > 0 $. As a consequence, $ T $ has a unique fixed point $ x^* $ and $ \lim\limits_{n\to\infty}T^nx = x^* $ for all $ x\in X $. □

Corollary 3.3. [17, Theorem 2.7] Let $ (X, D, K) $ be a complete strong $ b $-metric space, $ T:X\to X $ be a mapping. If there exists $ \varphi\in\Psi_{\frac{1}{3}} $ satisfying for all $ x, y\in X $ with $ x\neq y $,

Then, $ T $ has a unique fixed point $ x^{*}\in X $ and for any $ x\in X $ the sequence of iterates $ \{T^nx\} $ converges to $ x^* $.

4.   Conclusions

We focus on a new type of Kannan's fixed point theorem in the setting of strong $ b $-metric spaces. Using some useful lemmas, we derive three fixed point theorems. The first two theorems give positive answers to Questions 1.5 and 1.7, respectively. The third theorem is a new type of Reich's fixed point theorem and also a generalization of Doan's result (Theorem 2.7 in ^[17]).

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors are thankful to the referees for their valuable comments and suggestions to improve this paper.

Research supported by the National Natural Science Foundation of China (12061050, 11561049) and the Natural Science Foundation of Inner Mongolia (2020MS01004).

Conflict of interest

The authors declare that there is no conflict of interest.