A b-metric on a nonempty set X is a function such that for a constant b ≥ 1 and all the following conditions hold:

(M1) if and only if x = y;

(M2)

(M3)

Then is called a .-metric space.

One may note that each metric space is a .-metric space considering b = 1. However, the converse does not hold. Furthermore, the topology on a b-metric space, the concept of Cauchy sequences, convergent sequences and the completeness of the setting are analo- gous to that of standard metric spaces. However, in general, a b-metric is not a continuous mapping in both variables (see, e.g., [13]).

On the other hand, Cosentino et al. in [5] introduced the set of functions TF. in the line of Wardowski’s [19] approach to b-metric space as follows:

Let b ≥ 1 be a real number. TF. denotes the family of all functions . : R+ → R with the following properties:

(F1) F is strictly increasing;

(F2) For each sequence of positive numbers, if and only if

(F3) For each sequence of positive numbers with, there existsk such that

(F4) There exists such that for each sequence of positive numbers,if for all forall.

Let b ≥ 1 be a real number. denotes the family of all functions F : R+ → R having the properties (F1)–(F4) and the following property:

(F5) F(inf A) = inf F(A) for all with inf A > 0.

It is easy to see that the function F (X) = ln x or .F(x) = x + ln x satisfies properties (F1)–(F5) for x > 0.

We recall the following result from [5]. In this result, denotes the .-Hausdorff– Pompeiu metric.

(See [5, Thm. 3.4].) Let .X, db. be a complete b-metric space, and let T : X → P_cb(X). Assume that there exists a continuous from the right function F ∈ TF. and τ ∈ R+ such that

for all Then has a fixed piont.

Moreover, Feng and Liu obtained the subsequent result (recall that a function f : X → R is said to be lower semicontinuous if for all sequences in X with it satisfies

Let (X, d) be a metric space, T : X > CL(X), and let the function f : X → R, f (x) = d(x, Tx) be lower semicontinuous. If there exist b, c ∈ (0, 1) with b < c such that for any x _E X, there is y _E Tx satisfying

then T has a fixed point.

Against this background, we obtain fixed point results for multivalued mappings satis- fying Wardowski–Feng–Liu-type conditions for orbitally lower semicontinuous functions in orbitally complete .-metric spaces. These results generalize, complement and unify the findings proposed in [2, 12]. Besides, we illustrate a couple of examples to validate our obtained results, and also, we consider the ordered version of the attained results. Finally, we dish out an interesting application concerning our derived theorem and employing the notion of fractals to a certain type of fractal integral equations. An illustrative example is presented to show the applicability of the obtained result in reducing the energy of an antenna.

Let T : X → 2^xbe a multivalued mapping on a .-metric space (X, d_b). An orbit for . at a point x₀ ∈ X is denoted by O(x₀) and is defined as a sequence

Let be a multivalued mapping on a .-metric space (X, d_b). A b-metric space X is said to be T -orbitally complete if every Cauchy sequence of the form converges in X.

It is obvious that an orbitally complete .-metric space may not be complete. Now we present one of our main result.

Let (X, d_b) be an orbitally complete b-metric space with b ≥ 1, T : X . CL(X) and F ∈. Assume that the following conditions hold:

Then T has a fixed point.

Proof. Suppose that . has no fixed points. Then for all x ∈ X, we have d_b(x, Tx) > 0. Since for every x ∈ X and F ∈ then it is easy to prove that the setis nonempty for every x ∈ X (proof will follow in the line of [12]). If x₀ ∈ X is any initial point, then there exists such that

and for x1 ∈ X, there exists satisfying

Continuing this process, we get an iterative sequence {x_n}, where and

for all n ∈ N ∪ {0}. It follows by (1) and property (F4) that

for all n ∈ N. We verify that {x_n} is a Cauchy sequence. Since, then by the definition ofwe have

which implies that

From (2) and (3) we have

i.e.,

Let for n ∈ N; then and from (5) is decreasing. Therefore, there exists δ > 0 such that Now let δ > 0. Let B(t) = Then using (5), the following holds:

Let p_nbe the greatest number in {0, 1, . . . , n − 1} such that

for all n ∈ N. In this case, is a nondecreasing sequence. From (6) we get

In a similar way, from (4) we can obtain

Now consider the sequence. We distinguish two cases.

Case 1. For each n ∈ N, there is m > n such that Then we obtain a subsequence we deduce that

Hence,

for all k. Consequently, which contradicts the fact that

Case 2. There is n₀ ∈ N such that . for all m > n₀. Then . for all m > n₀. Hence, and by (F2) which contradicts the fact that Thus,. From (F3) there exists k ∈ (0, 1) such that

By (7) the following holds for all n ∈ N:

Passing to limit as n → ∞ in (9), we obtain

Since . then there exists for all n /= n₀. Thus,

for all n ≥ n₀. Letting n → ∞ in (10), we have that is,

From (11) there exits n1 ∈ N such that .So, we have for all n ≥ n1. Now, the last limit implies that the series is convergent, and hence, {xn} is a Cauchy sequence in X. Since X is a orbitally complete b-metric space, there exists such that xn → z as n → ∞. On the other hand, from (8) and (F2) we have is orbitally lower semicontinuous, we have

Therefore, z ∈ Tz. Hence, . has a fixed point.

Next, we present an example in order to substantiate the above result.

Example 1. Let us consider the set X = [1/9, ∞) and define a mapping d_b: X × X → R by db.x, y) = |x−y|2. Then d_bis a b-metric on X with b = 2. Next, we define a mapping T : X → CL(X) by

for all x ∈ X. Let us take for all t ∈ (0, ∞). Then, clearly, for all t ≥ 0.

For any x ∈ X, we have d_b(x, Tx) = ((3x − 1)/4)². So the mapping x ›→ d_b(x, Tx) is orbitally lower semicontinuous. Again, for any x ∈ X, we have y = (x + 1)/4 ∈, and for this y, we have

Thus, we see that all the conditions of Theorem 3 hold true. So by that theorem . has a fixed point, and note that z = 1/3 is a fixed point of T .

Our second result is related to multivalued mappings T on the b-metric space X, where Tx is compact for all x X. By taking into account Case 1, we can take n ≥ 0. Therefore, the proof of the following theorem is obvious.

Let us denote k(X) as the set of all nonempty compact subsets of X.

Let (X, d_b) be an orbitally complete b-metric space, b ≥ 1, T : X → K(X)and F ∈ Assume that the following conditions hold:

Then T has a fixed point.

Now we have the ordered version of the above results. To do this, we recall the definition of ordered b-metric space. (X, d_b, “) is called an ordered b-metric space if d_bis a b-metric space on X and is a partially ordered set. Further, if is a partially ordered set, then x, y ∈ . are called comparable if holds.

For x ∈ X with d_b(x, Tx) > 0, define a set as

Letbe an ordered orbitally complete b-metric space with b ≥ 1, T : X . CL(X) and F ∈ TF.. Assume that the following conditions hold:

If the condition

holds, then T has a fixed point.

Proof. Following the line of proof of Theorem 3 and definition of, we can show thatis a Cauchy sequence in with for . N. From the orbital completeness of X there exists such that By assumption (C) for all n. Rest is followed from the proof of Theorem 3.

Let (X, d, “) be an ordered orbitally complete b-metric space with b ≥ 1, T : X → K(X) and F ∈ Assume that the following conditions hold:

Then T has a fixed point provided (C) holds.

Now we present an example to authenticate Theorem 5.

Example 2. Let us take X = NU {0} and consider a relation < on X by defining x<y if and only if y divides .. Then it is easy to verify that (X, <) is a partially ordered set. Now we define d_b : X × X → R by

Then (X, d_b) is a partially ordered b-metric space with b = 3.

Now we define a mapping T : X → CL(X) by

Let us choose F(a) = ln a for all a ∈ R+ and T (t) = 1/8, n(t) = 1.10 for all t ∈ (0, ∞). Then it is obvious that F ∈and T (t)> η(t), Further, for x ∈ X, we have

Therefore, x ›→ d_b(x, Tx) is orbitally lower semicontinuous. For x ∈ X, we have and for this y, we have

Thus, we see that all conditions of Theorem 5 hold. So by the same theorem T has a fixed point. Indeed z = 0 is a fixed point of T .

Let X = C[J, R], and let the operator defined by

whereis defined and continuous on a fractal set of fractal dimensión,is continuous function. Moreover, assume the following conditions:

Then the integral equation (12) admits a solution.

Proof. We have to prove that the operator Q achieves all the assumptions of Theorem 3 in the single-valued type. Let X ∈ X, then we obtain

Thus, we get the following inequality:

Since the natural logarithm indicates to be in, employing it on above inequality, we conclude that

Consequently, we obtain

Now let, thus, we indicate the following inequality:

Hence, in view of Theorem 3 (single-valued) with the operator admits at least one fixed point, which is corresponding to the solution of Eq. (12). This completes the proof.

Example 3. Consider the following data: From Fig. 1 we can say that the relation between . and . certainly minimize the energy of the antenna. Obviously, the√value of minimized the energy. It is clear that Hence, in view of Theorem 7, Eq. (12) has a solution, which minimize the energy.