Fixed point theorems for enriched Kannan-type mappings and application

1.   Introduction and preliminaries

The Banach contraction principle ^[1], proposed by Banach in 1922, was a fundamental consequence of fixed-point theory. It states that any self-mapping $S$ on a complete metric space $(H, d)$ satisfying

for all $o, z \in H$, then $S$ has a unique fixed point in $H$. After that, many authors have generalized, improved, and extended this celebrated result by changing either the conditions of the mappings or the construction of the space; see ^[2,3,4,5,6].

The well-known Nemytzki-Edelstein's results for contractive mappings on compact metric spaces are as follows:

Theorem 1. ^[7] Let the self-mapping $S$ on a compact metric space $(H, d)$ satisfy

for any $o, z \in H$ with $o \neq z$. Then $S$ has a unique fixed point in $H$.

Suzuki ^[8] established the following version of Edelstein's fixed-point theorem:

Theorem 2. ^[8] Let the self-mapping $S$ on a compact metric space $(H, d)$ satisfy

for all $o, z \in H$ with $o \neq z$. Then $S$ has a unique fixed point in $H$.

Especially Kannan ^[9,10] established the following results, which differ from the Banach contraction principle:

Theorem 3. ^[9,10] Let the self-mapping $S$ on a complete metric space $(H, d)$ satisfying

for all $o, z \in H$ and $\alpha \in [0, \frac{1}{2})$. Then $S$ has a unique fixed point in $H$.

It is not difficult to see that contractions are always continuous, while Kannan maps are not necessarily continuous. Another beauty of Kannan mappings is that Kannan's theorem characterizes metric completeness. In 1975, Subrahmanyam ^[11] presented the following result:

Theorem 4. ^[11] A metric space is complete if and only if every Kannan mapping $S$ has a fixed point.

In addition, Fisher ^[12] proved the following variant of Theorem 3 for a compact metric space:

Theorem 5. ^[12] Let the continuous self-mapping $S$ on a compact metric space $(H, d)$ satisfy

for all $o, z \in H$ with $o \neq z$. Then $S$ has a unique fixed point in $H$.

Rencently, Berinde, and Păcurar ^[13] introduced the notion of enriched Kannan mapping, which is a generalization of that Kannan mapping. A mapping $S:H \to H$ is called an enriched Kannan mapping or a $(a, k)$-enriched Kannan mapping if there exist $k \in [0, \frac{1}{2})$ and $a \in [0, +\infty)$ such that

We will denote the set of all fixed points of $S$ by $F(S)$. They proved the following:

Theorem 6. ^[13] Let $(H, \parallel\cdot \parallel)$ be a Banach space and $S:H\rightarrow H$ be a $(a, k)-$enriched Kannan mapping. Then the following holds:

(i) $F(S) = \left\{ o \right\};$

(ii) There exists $\lambda \in [0, 1)$, the sequence $\left\{ {{o_n}} \right \}_{n = 0}^{+\infty}$ defined by

converges to $o$ in $H$;

(iii) set $\mu = \frac{k}{{1 - k}}$, for any $n \in \mathbb{N}$, then

And then, a question was raised following the above theorem in ^[13].

Question 1. Does the enriched Kannan mapping fixed point theorem still characterize the metric completeness?

On the other hand, Takahashi ^[14] introduces the notion of the convex structure in metric space as follows:

Definition 1. ^[14] Let $(H, d)$ be a metric space. Define a function $W:H\times H\times [0, 1] \to H$ is said to be a convex structure in $H$ if

holds for each $z, o, z \in H$ and $\lambda \in [0, 1]$. $(H, d, W)$ is called convex metric space.

Remark 1. It is worth mentioning that a linear normed space embedded with the natural convex structure

is a convex metric space, but it is not valid for some metric spaces, see ^[15,16].

Lemma 1. ^[14,15,16] Let $(H, d, W)$ be a convex metric space, and $\lambda, {\lambda _1}, {\lambda _2} \in [0, 1]$. For any $o, z \in H$, the following holds:

(i) $W(o, o;\lambda) = o$; $W(o, z;0) = z$ and $W(o, z;1) = o$;

(ii) $d(o, z) = d(o, W(o, z;\lambda)) + d(z, W(o, z;\lambda))$;

(iii) $d(o, W(o, z;\lambda)) = (1 - \lambda)d(o, z)$ and $d(z, W(o, z;\lambda)) = \lambda d(o, z)$;

(iv) $\left| {{\lambda _1} - {\lambda _2}} \right|d(o, z) \le d(W(o, z;{\lambda _1}), W(o, z;{\lambda _2}))$.

Lemma 2. ^[17] Let the self-mapping $S$ on a convex metric space $(H, d, W)$ and ${S_\lambda }:H \to H$ defined by

Then, we have $F(S) = F({S_\lambda })$ for any $\lambda \in [0, 1)$.

Berinde and Păcurar ^[17] gave the concept of enriched Kannan mapping on a convex metric space as below:

Definition 2. ^[17] A self-mapping $S$ on a convex metric space $(H, d, W)$ is said to be an enriched Kannan mapping if there exist $k \in [0, \frac{1}{2})$ and $\lambda \in [0, 1)$ satisfying

Notice that the continuous Kannan contractive mapping $S:H\rightarrow H$ is such that

for all $o, z \in H$ with $o\neq z$, in a complete but nocompact metric space may be fixed-point free (^[18,19,20]). Hence, a question about enriched Kannan contractive mapping may arise:

Question 2. Does there exists a complete but noncompact convex metric space $(H, d)$ and continuous enriched Kannan contractive mapping $S:H\rightarrow H$ satisfying

and $S$ is fixed-point-free?

In this work, we first give an affirmative answer to Question 1 by proving that every enriched Kannan contractive mapping has a fixed point and characterizes the completeness of the underlying normed space. Furthermore, we provide an example answer to Question 2. Moreover, we present some new fixed point results for Suzuki-enriched Kannan-type mappings in the setting of convex metric spaces. Finally, we apply the fixed point result to approximating the solution of nonlinear Volterra integral equations.

2.   Main results

In what follows, the symbol $F$ represents the set of all functions $f:[0, +\infty) \to [0, \frac{1}{2})$ such that

the symbol $\Psi$ represents the set of all strictly monotonic, increasing, and continuous functions $\psi :[0, + \infty) \to [0, + \infty)$ such that

We start with the following theorem, which is an affirmative answer to Question 1.

Theorem 7. Let $(H, \|\cdot\|)$ be a normed space and $S:H \to H$ be a mapping satisfying

for any $o, z \in H$ with $o \ne z$. If $S$ has a fixed point, then $(H, \|\cdot\|)$ is a Banach space.

Proof. If $a = 0$, the result follows from Theorem 4. Suppose that $a > 0$, we observe that (2.1) can be rewritten as follows:

Let $\lambda = \frac{a}{{a + 1}}$, clearly $a = \frac{\lambda }{{1 - \lambda }}$, then (2.1) becomes

Set ${S_\lambda }o = \lambda o + (1 - \lambda)So$, we deduce that

For the purpose of contradiction, suppose that $\left\{ {{o_n}} \right\} \in H$ is a Cauchy sequence but does not converge. Set $D(o, M) = \inf \left\{ {\left\| {o - z} \right\|:z \in M} \right\}$ and $M = \left\{ {{o_n}:n \in \mathbb{N}} \right\}$ be a set of divergent sequences of distinct elements in $H$. Let $o \in H$, and we have the subsequent cases:

Case 1. If $o \notin H$, as $\left\{ {{o_n}} \right\}$ is a Cauchy sequence, there exists an integer ${N_0}(o)$ such that

for any $l \in \mathbb{N}$. In particular

for any $l \in \mathbb{N}$, $m \ge {N_0}(o)$.

Case 2. If $o \in H$, then $o = {o_{n(o)}}$ for some $n(o) \in \mathbb{N}$, and there exists an integer ${n_0}(o) \in N$ such that

for any $m \ge {n_0}(o) > n(o)$. Define $S:H \to H$ by

Let $o, z \in H$, and we will show that $S$ is an enriched Kannan contractive mapping. Indeed, we need to consider the following four cases:

Case 1. If $o, z \notin M$, then $So = \frac{{{o_{{N_0}(o)}} - \lambda o}}{{1 - \lambda }}$ and $Sz = \frac{{{o_{{N_0}(z)}} - \lambda z}}{{1 - \lambda }}$, which imply that ${S_\lambda }o = {o_{{N_0}(o)}}$ and ${S_\lambda }z = {o_{{N_0}(z)}}$. We can suppose that ${o_{{N_0}(z)}} > {o_{{N_0}(o)}}$. It follows that

which shows that $\left\| {{S_\lambda }o - {S_\lambda }z} \right\| < \frac{1}{2}\left[{\left\| {o - {S_\lambda }o} \right\| + \left\| {z - {S_\lambda }z} \right\|} \right]$.

Case 2. If $o, z \in M$, there exist $n(o), n(z) \in \mathbb{N}$ such that $o = {o_{n(o)}}, z = {o_{n(z)}}$. Then $So = \frac{{{o_{{n_0}(o)}} - \lambda o}}{{1 - \lambda }}$ and $Sz = \frac{{{o_{{n_0}(z)}} - \lambda z}}{{1 - \lambda }}$, which implies that ${S_\lambda }o = {o_{{n_0}(o)}}$ and ${S_\lambda }z = {o_{{n_0}(z)}}$. Suppose that ${n_0}(z) > {n_0}(o)$. We obtain

which shows that $\left\| {{S_\lambda }o - {S_\lambda }z} \right\| < \frac{1}{2}\left[{\left\| {o - {S_\lambda }o} \right\| + \left\| {z - {S_\lambda }z} \right\|} \right]$.

Case 3. If $o \in M$, $z \notin M$, there exists $n(o) \in N$ satisfying $o = {o_{n(o)}}$, then $So = \frac{{{o_{{n_0}(o)}} - \lambda o}}{{1 - \lambda }}$ and $Sz = \frac{{{o_{{N_0}(z)}} - \lambda z}}{{1 - \lambda }}$ which imply that ${S_\lambda }o = {o_{{n_0}(o)}}$ and ${S_\lambda }z = {o_{{N_0}(z)}}$.

Subcase 1. If ${n_0}(o) \ge {N_0}(z)$, we obtain

which shows that $\left\| {{S_\lambda }o - {S_\lambda }z} \right\| < \frac{1}{2}\left[{\left\| {o - {S_\lambda }o} \right\| + \left\| {z - {S_\lambda }z} \right\|} \right]$.

Subcase 2. If ${n_0}(o) < {N_0}(z)$, we obtain

which shows that $\left\| {{S_\lambda }o - {S_\lambda }z} \right\| < \frac{1}{2}\left[{\left\| {o - {S_\lambda }o} \right\| + \left\| {z - {S_\lambda }z} \right\|} \right]$.

Case 4. If $o \notin M$ and $z \in M$, in this case, ${S_\lambda }o = {o_{{N_0}(o)}}$ and ${S_\lambda }z = {o_{{n_0}(z)}}$, similarly to Case 3, we can also get that $\left\| {{S_\lambda }o - {S_\lambda }z} \right\| < \frac{1}{2}\left[{\left\| {o - {S_\lambda }o} \right\| + \left\| {z - {S_\lambda }z} \right\|} \right]$.

Therefore, by summarizing all cases, for any $o, z \in H$ with $o\neq z$, we have

which shows that $S$ is an enriched Kannan contractive mapping. Notice that $S$ is fixed point-free, which is a contradiction. Therefore, $(H, \|\cdot\|)$ is a Banach space. □

Now, we give the following example to answer Question 2.

Example 1. Let $H = \mathbb{N}$ and $W(o, z;\lambda) = \lambda o + (1 - \lambda)z$, $\lambda \in [0, 1)$. We define the metric $d:H \times H \to [0, +\infty)$ by

Then $(H, d, W)$ is a complete convex metric space, but $H$ is not compact since the sequence (n) has no convergent subsequence in $H$. A mapping $S:H \to H$ is defined by $So = 7o$ for any $o \in H$. Clearly, $S$ is continuous. Moreover, for $\lambda = \frac{1}{2}$, we have $W(o, z;\frac{1}{2}) = 4o$. For all $o, z \in H$ with $o < z$, we conclude that

Notice that

Thus

for all $o, z \in H$ with $o < z$. Similarly, one can prove it for the case $o, z \in H$ with $o > z$. Therefore, $S$ is an enriched Kannan contractive mapping, but $S$ has no fixed point.

Suzuki enriched the Kannan-type mapping fixed point theorem as follows:

Theorem 8. Let $(H, d, W)$ be a complete convex metric space and $S:H \to H$ be a mapping. If there exists $\lambda \in [0, 1)$ such that for any $o, z \in H$,

implies

where $f \in F, \psi \in \Psi$. Then, $S$ has a unique fixed point ${o^*} \in H$, and the sequence ${o_{n + 1}} = W({o_n}, S{o_n}; \lambda)$ converges to ${o^*}$.

Proof. For any $o\in H$, set ${S_\lambda }o = W(o, So;\lambda)$. In this case, the given assumption becomes

Choose ${o_1} \in H$ and construct the Picard iteration associated with ${S_\lambda }$, this is ${o_{n + 1}} = {S_\lambda }{o_n}$. Without loss of generality, suppose that ${o_n} \ne {o_{n + 1}}$ for all $n\in \mathbb{N}$. Indeed, if ${o_n} = {o_{n + 1}}$ for some $n\in \mathbb{N}$, we have

which means that $d({o_n}, S{o_n}) = 0$. Thus, ${o_n}$ is a fixed point of $S$. Since

by applying the condition (2.3), we have

Then we see that

Hence $\psi \left({d({o_{n + 1}}, {o_{n + 2}})} \right) < \psi \left({d({o_n}, {o_{n + 1}})} \right)$. Since $\psi$ is increasing, we have $d({o_{n + 1}}, {o_{n + 2}}) < d({o_n}, {o_{n + 1}})$. Thus, $\left\{ {d({o_n}, {o_{n + 1}})} \right\}$ is a decreasing sequence of nonnegative real numbers, and hence it is convergent. Assume that $\mathop {\lim }\limits_{n \to +\infty } d({o_n}, {o_{n + 1}}) = r \ge 0$. If $r > 0$. From (2.4), we obtain that

letting $n \rightarrow +\infty$, we get $\frac{1}{2} \le \mathop {\lim }\limits_{n \to +\infty } f(d({o_n}, {o_{n + 1}}))$, a contradiction, thus $r = 0$. Now, we show that $\left\{ {{o_n}} \right\}$ is a Cauchy sequence. If not, then there exist $\varepsilon > 0$ and two sequences $\left\{ {{q_k}} \right\}, \left\{ {{p_k}} \right\}$ of positive integers such that

We obtain

Let $k\rightarrow +\infty$, we deduce that $\mathop {\lim }\limits_{k \to +\infty } d({o_{{q_k}}}, {o_{{p_k}}}) = \varepsilon$. Further, from

and

we obtain $\mathop {\lim }\limits_{k \to +\infty } d({o_{{q_k} + 1}}, {o_{{p_k} + 1}}) = \varepsilon$. Note that $\mathop {\lim }\limits_{n \to +\infty } d({o_n}, {o_{n + 1}}) = 0$, there exists ${N_0} \in \mathbb{N}$ such that

for any $n > {N_0}$. By applying the condition (2.3) again, we find

for all $n > N_{0}$. Thus

Letting $k \rightarrow +\infty$, we obtain that

which implies $\varepsilon = 0$ and leads to a contradiction. Therefore, $\left\{ {{o_n}} \right\}$ is a Cauchy sequence and ${o_n} \to {o^*}\in H$ as $n \to +\infty$. We assume that there exists $n \in \mathbb{N}$ such that

Then we obtain

which implies

a contradiction. Therefore, one of the following conditions holds:

(a) $\frac{1}{2}d({o_n}, {o_{n + 1}}) < d({o_n}, {o^*})$ for any $n$ in some infinite subset $E$ of $\mathbb{N}$;

(b) $\frac{1}{2}d({o_{n + 1}}, {o_{n + 2}}) < d({o_{n + 1}}, {o^*})$ for any $n$ in some infinite subset $U$ of $\mathbb{N}$.

If ($a$) holds, then we have

Hence, $\mathop {\lim }\limits_{n \in E, n \to +\infty } \psi (d({o_{n + 1}}, {S_\lambda }{o^*})) = 0$ and $\mathop {\lim }\limits_{n \to +\infty } {o_{n + 1}} = {S_\lambda }{o^*}$, that is, $\left\{ {{o_n}} \right\}$ has a subsequence converging to ${S_\lambda }{o^*}$. Similarly, if ($b$) holds, we also obtain that $\left\{ {{o_n}} \right\}$ has a subsequence converging to ${S_\lambda }{o^*}$. Since $\left\{ {{o_n}} \right\}$ is converging to $o^{*}$, ${o^*} = {S_\lambda }{o^*}$. If ${z^*}$ is another fixed point of ${S_\lambda }$, that is, ${o^*} = {S_\lambda }{o^*} \ne {S_\lambda }{z^*} = {z^*}$. Since $\frac{1}{2}d({o^*}, {S_\lambda }{o^*}) = 0 < d({o^*}, {z^*})$, then, we have

thus $d({o^*}, {z^*}) = 0$, which is a contradiction. Combining this with Lemma 2, we have that $S$ has a unique fixed point in $H$. □

Example 2. Let $H = [0, 2]$ and $d$ be a Euclidean metric on $H$. Set $W(o, z;\lambda) = \lambda o + (1 - \lambda)z$ for any $\lambda \in [0, 1)$. Then $(H, d, W)$ is a complete convex metric space. Let us define $S:H \to H$ by

We choose $\lambda = \frac{1}{8}$, we have

Clearly, $S$ satisfies the contractive (2.3) ($\psi (z) = t$ and $f(z) = - \frac{z}{{12}} + \frac{1}{2}$ for all $z\geq0$). All conditions of Theorem 8, hold and therefore $S$ has a unique fixed point 1 in $H$.

Next, we give a generalization of Theorem 8 in the setting of compact convex spaces.

Theorem 9. Let $(H, d, W)$ be a compact convex metric space with continuous convex structure and $S:H \to H$ be a continuous mapping. If there exists $\lambda \in [0, 1)$ such that for any $o, z \in H$,

implies

where $\psi \in \Psi$. Then, $S$ has a unique fixed point ${o^*} \in H$, and the sequence ${o_{n + 1}} = W({o_n}, S{o_n}; \lambda)$ converges to ${o^*}$.

Proof. For any $o \in H$, we set ${S_\lambda }o = W(o, So;\lambda)$. It is clear that ${S_\lambda }$ is continuous. In this case, the given assumption becomes

for any $o, z \in H$ and $\lambda \in [0, 1)$. Let $g(o) = d(o, {S_\lambda }o)$. It is clear that $g(o)$ is continuous. By the fact that $H$ is compact, there exists a point ${o^*} \in H$ such that $g({o^*}) = \inf \left\{ {g(o):o \in H} \right\}$, then $g({o^*}) \le g(o)$ for any $o\in H$. Suppose that ${o^*} \ne {S_\lambda }{o^*}$, thus

then we obtain

This implies that $\psi \left({d({S_\lambda }{o^*}, S_\lambda ^2{o^*})} \right) < \psi \left({d({o^*}, {S_\lambda }{o^*})} \right)$, contradicting $d({o^*}, {S_\lambda }{o^*}) \le d({S_\lambda }{o^*}, S_\lambda ^2{o^*})$. Hence ${o^*} = {S_\lambda }{o^*}$. Assume that ${z^*}$ is another fixed point, that is, ${o^*} = {S_\lambda }{o^*} \ne {S_\lambda }{z^*} = {z^*}$. Since $\frac{1}{2}d({o^*}, {S_\lambda }{o^*})$ $=$$0 < d({o^*}, {z^*})$, we have

Thus, $d({o^*}, {z^*}) = 0$, i.e. $o^{*} = {z^*}$, combining this with Lemma 2, we have that $S$ has a unique fixed point $o^{*}$ in $H$. Choose ${o_1} \in H$ and the iterative process $\left\{ {{o_n}} \right\}_{n = 0}^ {+\infty}$ defined by

Suppose that ${o_n} \ne {S_\lambda }{o_n}$ for all $n\in \mathbb{N}$. Since $\frac{1}{2}d({o_n}, {S_\lambda }{o_n}) < d({o_n}, {S_\lambda }{o_n}) = d({o_n}, {o_{n + 1}})$, we have

then $\psi \left({d({o_{n + 1}}, {o_{n + 2}})} \right) < \psi \left({d({o_n}, {o_{n + 1}})} \right)$. Assume that $\mathop {\lim }\limits_{o \to +\infty } \psi \left({d({o_n}, {o_{n + 1}})} \right) = r \ge 0$. Suppose that $r > 0$. Since $H$ is compact, there exists a convergent subsequence $\{ {o_{{n_k}}}\}$ of $\{ {o_n}\}$ such that ${o_{{n_k}}} \to o$, as $k \to +\infty$. Thus,

Since $\frac{1}{2}d(o, {S_\lambda }o) < d(o, {S_\lambda }o)$, we have

it follows that

which is a contradiction, so $r = 0$, which implies $o = o^{*}$. Since $0 = \frac{1}{2}d(o^{*}, {S_\lambda }o^{*}) < d({o_n}, {o^*})$, then we have

thus $\mathop {\lim }\limits_{n \to +\infty } \psi (d({o_{n + 1}}, {o^*})) = 0$, which implies $d({o_{n + 1}}, {o^*}) \to 0$ as $n\rightarrow +\infty$. Therefore, ${o_{n + 1}} = W({o_n}, S{o_n}; \lambda)$ converges to $o^{*}$. □

Remark 2. Let $H, d, W$ and the map $S:H \to H$ be defined as in Example 2. Then $(H, d, W)$ is a compact convex metric space, and $S$ satisfies the contractive (2.5) for any $\psi (t) = t$ for all $t\geq0$, $S$ has a unique fixed point 1 in $H$. However, it is to be noted that $S$ is not continuous, so Theorem 9 is not applicable.

Question 3. Does the conclusion of Theorem 9 still hold true if we remove the condition "S is continuous"?

The following theorem is an answer to the above question:

Theorem 10. Let $(H, d, W)$ be a compact convex metric space and $S:H \to H$ be a mapping. If there exists $\lambda \in [0, 1)$ such that for any $o, z \in H$,

implies

where $\psi \in \Psi$. Then, $S$ has a unique fixed point ${o^*} \in H$, and the sequence ${o_{n + 1}} = W({o_n}, S{o_n}; \lambda)$ converges to ${o^*}$.

Proof. For any $o\in H$, we set ${S_\lambda }o = W(o, So;\lambda)$, and the given assumption becomes

for all $o, z \in H$ and $\lambda \in [0, 1)$. We set $L = \inf \left\{ {d(o, {S_\lambda }o):o \in H} \right\}$. Then we can find e sequence $\left\{ {{o_n}} \right\} \in H$ such that

By the compactness property of $H$, we put $\mathop {\lim }\limits_{n \to +\infty } {o_n} = e$ and $\mathop {\lim }\limits_{n \to +\infty } {S_\lambda }{o_n} = c$ for some $e, c \in H$. Then we deduce

Now we check that $L = 0$. Assume the contrary, i.e., $L > 0$, there exists ${N_0} \in \mathbb{N}$ such that

for all $n \ge {N_0}$. Thus

which implies

Let $n \to +\infty$, and we obtain

which implies $\psi \left({d(c, {S_\lambda }c)} \right) \le \psi \left({d(e, c)} \right).$ Thus, we obtain $\psi \left({d(c, {S_\lambda }c)} \right) \le \psi \left(L \right)$. Moreover, since $\frac{1}{2}d(c, {S_\lambda }c) < d(c, {S_\lambda }c)$ then, we have

it follows that $\psi \left({d({S_\lambda }c, S_\lambda ^2c)} \right) < \psi \left({d(c, {S_\lambda }c)} \right) \le \psi \left(L \right)$. Hence $d({S_\lambda }c, S_\lambda ^2c) < L$, a contradiction. Therefore, $L = 0$ and $\mathop {\lim }\limits_{n \to +\infty } {o_n} = \mathop {\lim }\limits_{n \to +\infty } S{o_n} = e$. We shall show that ${S_\lambda }$ has a fixed point in $H$. Assume on the contrary ${S_\lambda }$ does not have a fixed point. Since $\frac{1}{2}d({o_n}, {S_\lambda }{o_n}) < d({o_n}, {S_\lambda }{o_n})$, then we have

which implies that

Due to $\psi$ is increasing, thus $d({S_\lambda }{o_n}, S_\lambda ^2{o_n}) < d({o_n}, {S_\lambda }{o_n})$. By using triangular inequality, we obtain

Hence $\mathop {\lim }\limits_{n \to +\infty } d(e, {{S_\lambda ^2}}{o_n}) = 0$. Assume that there exists $N \in \mathbb{N}$ such that

we deduce that

a contradiction. Thus, either

holds for any $n \in \mathbb{N}$. This yields one of the following conditions:

(1) There exists an infinite subset I of $\mathbb{N}$ such that

(2) There exists an infinite subset J of $\mathbb{N}$ such that

For the first case, let $n\rightarrow +\infty$, we obtain

and consequently, $\psi \left({d(e, {S_\lambda }e)} \right) = 0$. This yields $d(e, {S_\lambda }e) = 0$. Thus, $e = {S_\lambda }e$. Also in the second case, let $n\rightarrow +\infty$, we obtain that

Similarly, we can conclude that $e = {S_\lambda }e$. Hence, $e$ is a fixed point of $S_{\lambda}$ in both cases, a contradiction. If we assume that $z^{*}$ is another fixed point of $S_{\lambda}$, that is, ${o^*} = {S_\lambda }{o^*} \ne {S_\lambda }{z^*} = {z^*}$. Since $\frac{1}{2}d({o^*}, {S_\lambda }{o^*}) =$$0$$< d({o^*}, {z^*})$, we have

Thus $d({o^*}, {z^*}) = 0$, i.e., $o^{*}$ is the unique fixed point of ${S_\lambda }$. Combining with Lemma 2, we have that $S$ has a unique fixed point in $H$. □

3.   Application

In this part, we establish the existences to solution of nonlinear Volterra integral equations

where $l > 0$, $G:[0, l] \times [0, l] \times \mathbb{R} \to \mathbb{R}$, and $g:[0, l] \to \mathbb{R}$.

Theorem 11. Assume that

(1) The function $G$ is continuous;

(2) For any $\lambda \in [0, 1)$, there exists $f \in F$ such that

Then (3.1) has a unique solution. Moreover, the solution is exhibited as follows:

where $z(x) = \mathop {\lim }\limits_{n \to +\infty } {o_n}(x)$, ${o_0}(x) = o_{0} \in H$, and

Proof. Let $H = C\left({[0, l], \mathbb{R}} \right)$ be the set of all continuous functions on the interval $[0, l]$. Define the metric $d:H \times H \to {\mathbb{R}^ + }$ by

and the mapping $W:H \times H \times [0, 1) \to H$ by the formula

Clearly, $(H, d, W)$ is a complete convex metric space. Consider the mapping

It is clear that $o(x)$ is a solution of Eq (3.1) if and only if $o(x)$ is a fixed point of $S$, that is, $So = o$. Obviously, $S$ is well defined. Define ${S_\lambda }o$ by

and set ${o_{n + 1}}(x) = ({S_\lambda }{o_n})(x) = \lambda {o_n}(x) + (1 - \lambda)(S{o_n})(x), n\in \mathbb{N}$. Let $o, z \in H$, we have

This implies that

Hence, $S$ is an enriched Kannan-type mapping. Meanwhile, by Theorem 8 ($\psi (o) = o$ for all $o > 0$), $S$ has a unique fixed point $z(x)$, satisfying $z(x) = (Sz)(x) = ({S_\lambda }z)(x)$, which means that $z(x)$ is the solution of (3.1). Now, we will show that $z(x) = f(x) + \int_0^x {G(x, r, z(r))dr}$. Note that

which implies that $\mathop {\lim \sup }\limits_{n \to +\infty } \left| {z(x) - {o_{n + 1}}(x)} \right| = 0$. Thus

Hence, we have

□

4.   Conclusions

In this paper, we prove three questions about enriched Kannan-type mapping, including the open question raised by Berinde and Păcurar ^[13]. We defined and studied Suzuki-enriched Kannan-type mappings in convex metric spaces. Several examples related to theorems are also provided to show the validity of our main results. The solution of an integral equation is also investigated. Suzuki enriched Kannan-type mappings are natural generalizations of enriched Kannan mappings. Our results extend fundamental findings previously established in related research.

Author contributions

Yao Yu, Chaobo Li and Dong Ji: Conceptualization, Methodology, Validation, Writing-original draft and Writing-review & editing. All authors contributed equally to the writing of this article. All authors have read and approved the final version of the manuscript for publication.

Use of AI tools declaration

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

Acknowledgments

This work was supported by Harbin Science and Technology Plan Project (Grant No. 2023ZCZJCG039) and Heilongjiang Key Research and Development Program Guide category (Grant No. GZ20220077).

Conflict of interest

The authors declare no conflict of interest.