**2. Preliminaries**

Let (<sup>Ω</sup>, *ω*) (Ω, for short) and *CL*(Ω) be the complete metric space Ω with a metric *ω* and the set of all non-empty closed subsets of Ω, respectively.

To be more precise and to be easier for readers to see the novelty of the results in this paper, we will initially give some that are known in the literature definitions.

In 2012, Samet et al. ([3]) defined *α*-admissibility of mapping in the following way:

**Definition 1.** *([3]) Let the function α* : Ω × Ω → [0, + <sup>∞</sup>)*. The mapping* J *:* Ω → Ω *is α-admissible if:*

> *<sup>α</sup>*(<sup>l</sup>, *κ*) ≥ 1 *implies α*(J (l), J (*κ*)) ≥ 1 *for* l, *κ* ∈ Ω.

Later, Salimi et al. ([4]) defined twisted (*<sup>α</sup>*,*β*)-admissible mappings in the following way:

**Definition 2.** *([4]). Let the functions α*, *β* : Ω × Ω → [0, + <sup>∞</sup>)*. The mapping* J *:* Ω → Ω *is twisted (<sup>α</sup>,β*) *-admissible if:*

$$\left\{ \begin{array}{ll} \mathfrak{a}(\mathsf{l},\mathsf{x}) \geq 1 \\ \beta(\mathsf{l},\mathsf{x}) \geq 1 \end{array} \implies \left\{ \begin{array}{ll} \mathfrak{a}(\mathcal{J}(\mathsf{l}),\mathcal{J}(\mathsf{x})) \geq 1 \\ \beta(\mathcal{J}(\mathsf{l}),\mathcal{J}(\mathsf{x})) \geq 1 \end{array} \right. \\ \left. \begin{array}{ll} \mathfrak{a}(\mathcal{J}(\mathsf{l}),\mathcal{J}(\mathsf{x})) \geq 1 \\ \end{array} \right. \\ \left. \begin{array}{ll} \mathfrak{a}(\mathsf{l},\mathsf{x}) \in \Omega. \end{array} \right. \end{array} \right.$$

Wardowski ([10]) presented a new family of mappings named Wardowski-contractions.

**Definition 3.** *([10]) The mapping* J : Ω → Ω *is ϑ-contraction if there exists a number π* > 0 *such that:*

$$\omega\left(\mathcal{I}\left(\mathfrak{l}\right), \mathcal{I}\left(\mathfrak{k}\right)\right) > 0 \implies \pi + \mathfrak{d}\left(\omega\left(\mathcal{I}\left(\mathfrak{l}\right), \mathcal{I}\left(\mathfrak{k}\right)\right)\right) \le \mathfrak{d}\left(\omega\left(\mathfrak{l}, \mathfrak{k}\right)\right), \text{ } \mathfrak{l}, \mathfrak{k} \in \Omega \tag{1}$$

.

*where ϑ* : (0, + ∞) → R *is a function satisfying the assertions:*

*(F*1*) for all* 0 < *x* < *y the inequality <sup>ϑ</sup>*(*x*) < *<sup>ϑ</sup>*(*y*) *holds; (F*2*) for* {*xj*}<sup>∞</sup> *j*=1 ⊆ (0, + ∞) *the equality* lim*j*→ ∞ *xj* = 0 *holds if* lim*j*→ ∞ *<sup>ϑ</sup>*(*xj*) = − ∞; *(F*3*)* ∃ 0 < *k* < 1 *such that* lim*x*→0<sup>+</sup> *xkϑ*(*x*) = 0.

Let Δ be the set of all mappings *ϑ* : (0, + ∞) → R satisfying the assertions (*F*1)–(*F*3).

**Theorem 1.** *([10]) Let ϑ* ∈ Δ *and* J *:* Ω → Ω *is ϑ-contraction, then the mapping* J *has a fixed point in* Ω*, i.e., there exists a point* l ∗ ∈ Ω *such that* J (l ∗) = l ∗*.*

We will give some examples of functions from the set Δ which will be used later.

**Example 1.** *([10]) Let the function ϑ*(l) = ln(l), l > 0*. Then ϑ satisfies conditions* (*<sup>F</sup>*1)*-*(*<sup>F</sup>*3)*, i.e., ϑ* ∈ Δ*. Any function* J *:* Ω → Ω *satisfying (1) is a ϑ-contraction because:*

$$
\omega(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{x})) \le e^{-\pi} \omega(\mathfrak{l}, \mathfrak{x})
$$

∀ l, *κ* ∈ Ω *with ω*(J (l), J (*κ*)) > 0 *and π* > 0. *Note that e*<sup>−</sup>*<sup>π</sup>* ∈ (0, <sup>1</sup>), *and therefore, the above condition is also the contractive condition of Banach ([1]).*

**Example 2.** *Let the function ϑ*(l) = l − √1l, l > 0*. Then ϑ satisfies conditions* (*<sup>F</sup>*1)*-*(*<sup>F</sup>*3) *with k* ∈ ( 12 , <sup>1</sup>), *i.e., ϑ* ∈ Δ*.*

*Any function* J *:* Ω → Ω *satisfying (1) is a ϑ-contraction because:*

$$+\frac{1}{\sqrt{\omega(\mathcal{J}(\mathbb{I}), \mathcal{J}(\kappa))}} + \omega(\mathcal{J}(\mathbb{I}), \mathcal{J}(\kappa)) \le -\frac{1}{\sqrt{\omega(\mathbb{I}, \kappa)}} + \omega(\mathbb{I}, \kappa)$$

∀ l, *κ* ∈ Ω *with ω*(J (l), J (*κ*)) > 0 *and π* > 0*.*

#### **3. Fixed Point Results**

We will introduce a new type of contraction mapping.

**Definition 4.** *Let the functions ϑ* ∈ Δ *and α*, *β* : Ω × Ω → {−∞} ∪ (0, <sup>∞</sup>)*. The mapping* J *:* Ω → Ω *is (<sup>α</sup>,β*)*-type ϑ-contraction if for all* l, *κ* ∈ Ω : *ω*(J (l), J (*κ*)) > 0 *the inequality:*

$$\left(\pi + a(1,\kappa)\beta(1,\kappa)\vartheta\left(\omega(\mathcal{J}(1),\mathcal{J}(\kappa))\right) \le \vartheta\left(\omega(1,\kappa)\right)\right) \tag{2}$$

*holds where π* > 0 *is a real number.*

**Definition 5.** *Let the functions ϑ* ∈ Δ *and α*, *β* : Ω × Ω → {−∞} ∪ (0, <sup>∞</sup>)*. The mapping* J *:* Ω → Ω *is (<sup>α</sup>,β*)*-type rational ϑ-contraction if for all* l, *κ* ∈ Ω : *ω*(J (l), J (*κ*)) > 0 *the inequality:*

$$a\pi + a(1,\mathfrak{x})\beta(1,\mathfrak{x})\vartheta\left(\omega\left(\mathcal{J}(\mathfrak{l}),\mathcal{J}(\mathfrak{x})\right)\right) \le \vartheta\left(\mathcal{R}(\mathfrak{l},\mathfrak{x})\right) \tag{3}$$

*holds, where π* > 0 *is a real number and*

$$\mathcal{R}(\mathfrak{l},\mathfrak{x}) = \max\left\{\omega(\mathfrak{l},\mathfrak{x}), \frac{\omega(\mathfrak{l},\mathcal{J}(\mathfrak{l})) \omega(\mathfrak{x},\mathcal{J}(\mathfrak{x}))}{1 + \omega(\mathfrak{l},\mathfrak{x})}\right\}.\tag{4}$$

**Remark 1.** *Note that the (<sup>α</sup>,β*)*-type ϑ-contraction defined in Definition 4 is a generalization of ϑ-contraction given in [10] with <sup>α</sup>*(<sup>l</sup>, *κ*) = *β*(<sup>l</sup>, *κ*) = 1 *(see Definition 3).*

We will obtain some new fixed point results applying the introduced above types of mappings.

**Theorem 2.** *Let the functions ϑ* ∈ Δ *and α*, *β* : Ω × Ω → {−∞} ∪ (0, ∞) *and* J *:* Ω → Ω *be (<sup>α</sup>,β*)*-type ϑ-contraction and the following conditions be satisfied:*


*Then the mapping* J *has a fixed point in* Ω*, i.e., there exists a point* l∗ ∈ Ω *such that* J (l∗) = l<sup>∗</sup>*.*

**Proof.** Let l0 ∈ Ω be the element from condition (b). Define the sequence {<sup>l</sup>*j*}<sup>∞</sup>*j*=<sup>0</sup> in Ω by <sup>l</sup>*j*+<sup>1</sup> = J (<sup>l</sup>*j*) for *j* = 0, 1, 2, ... . If <sup>l</sup>*j*+<sup>1</sup> = l*j* for some *j* = 0, 1, 2, ... , then l∗ = l*j* is the fixed point of the mapping J . Assume <sup>l</sup>*j*+<sup>1</sup> = l*j* for all *j* = 0, 1, 2, ... . Then from condition (a) and the choice of l0 it follows that *<sup>α</sup>*(<sup>l</sup>1, l2) = *α*(J (<sup>l</sup>0), J (<sup>l</sup>1)) ≥ 1 and *β*(<sup>l</sup>1, l2) = *β*(J (<sup>l</sup>0), J (<sup>l</sup>1)) ≥ 1. By induction we ge<sup>t</sup> *<sup>α</sup>*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1 and *β*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1 for *j* ∈ N. Now by inequality (2) with l = <sup>l</sup>*j*−<sup>1</sup> and *κ* = l*j*, we have:

$$\begin{split} \pi + \theta\left(\omega(\mathfrak{l}\_{j}, \mathfrak{l}\_{j+1})\right) &= \pi + \theta\left(\omega(\mathcal{I}(\mathfrak{l}\_{j-1}), \mathcal{I}(\mathfrak{l}\_{j}))\right) \\ &\leq \pi + \mathfrak{a}(\mathfrak{l}\_{j-1}, \mathfrak{l}\_{j}) \beta(\mathfrak{l}\_{j-1}, \mathfrak{l}\_{j}) \theta\left(\omega(\mathcal{I}(\mathfrak{l}\_{j-1}), \mathcal{I}(\mathfrak{l}\_{j}))\right) \\ &\leq \theta(\omega(\mathfrak{l}\_{j-1}, \mathfrak{l}\_{j})) .\end{split} \tag{5}$$

From inequality (5) it follows that:

$$
\vartheta\left(\omega(\mathfrak{l}\_{\mathbf{\hat{i}}},\mathfrak{l}\_{\mathbf{\hat{j}}+1})\right) \le \vartheta(\omega(\mathfrak{l}\_{\mathbf{\hat{j}}-1},\mathfrak{l}\_{\mathbf{\hat{j}}})) - \pi. \tag{6}
$$

Therefore, applying inequality (6) step by step we obtain:

$$\begin{split} \theta\left(\omega\left(\mathbb{I}\_{\left[\cdot,\mathbb{I}\_{j}\right.}\right.\right)\right) &\leq \theta\left(\omega\left(\mathbb{I}\_{j-1},\mathbb{I}\_{j}\right)\right) - \pi \leq \theta\left(\omega\left(\mathbb{I}\_{j-2},\mathbb{I}\_{j-1}\right)\right) - 2\pi \\ &\leq \dots \leq \theta\left(\omega\left(\mathbb{I}\_{0},\mathbb{I}\_{1}\right)\right) - j\pi. \end{split} \tag{7}$$

Since *ϑ* ∈ Δ, so letting *j* → ∞ in (7), we get:

$$\lim\_{j \to \infty} \vartheta \left( \omega(\mathfrak{l}\_{\mathfrak{j}}, \mathfrak{l}\_{\mathfrak{j}+1}) \right) = -\infty \Longleftrightarrow \lim\_{j \to \infty} \omega(\mathfrak{l}\_{\mathfrak{j}}, \mathfrak{l}\_{\mathfrak{j}+1}) = 0. \tag{8}$$

From condition (*<sup>F</sup>*3), ∃ 0 < *k* < 1 such that:

$$\lim\_{j \to \infty} \omega(\mathfrak{l}\_{\mathfrak{j}}, \mathfrak{l}\_{\mathfrak{j}+1})^k \theta\left(\omega(\mathfrak{l}\_{\mathfrak{j}}, \mathfrak{l}\_{\mathfrak{j}+1})\right) = 0. \tag{9}$$

From Equation (7) we get:

$$\begin{split} & \left(\omega\left(\mathbb{I}\_{\boldsymbol{\mathsf{t}}^{\boldsymbol{\mathsf{t}}}}\mathbb{I}\_{\boldsymbol{\mathsf{t}}+1}\right)\right)^{k}\theta\left(\omega\left(\mathbb{I}\_{\boldsymbol{\mathsf{t}}^{\boldsymbol{\mathsf{t}}}}\mathbb{I}\_{\boldsymbol{\mathsf{t}}+1}\right)\right) - \left(\omega\left(\mathbb{I}\_{\boldsymbol{\mathsf{t}}^{\boldsymbol{\mathsf{t}}}}\mathbb{I}\_{\boldsymbol{\mathsf{t}}+1}\right)\right)^{k}\theta\left(\omega\left(\mathbb{I}\_{\boldsymbol{\mathsf{t}}}\mathbb{I}\_{\boldsymbol{\mathsf{t}}}\right)\right) \\ & \leq \quad \left(\omega\left(\mathbb{I}\_{\boldsymbol{\mathsf{t}}^{\boldsymbol{\mathsf{t}}}}\mathbb{I}\_{\boldsymbol{\mathsf{t}}+1}\right)\right)^{k}\left(\theta\left(\omega\left(\mathbb{I}\_{\boldsymbol{\mathsf{t}}}\mathbb{I}\_{\boldsymbol{\mathsf{t}}}\right)\right) - j\pi\right) - \left(\omega\left(\mathbb{I}\_{\boldsymbol{\mathsf{t}}^{\boldsymbol{\mathsf{t}}}}\mathbb{I}\_{\boldsymbol{\mathsf{t}}+1}\right)\right)^{k}\theta\left(\omega\left(\mathbb{I}\_{\boldsymbol{\mathsf{t}}}\mathbb{I}\_{\boldsymbol{\mathsf{t}}}\right)\right) \\ & \leq \quad - \left(\omega\left(\mathbb{I}\_{\boldsymbol{\mathsf{t}}}\mathbb{I}\_{\boldsymbol{\mathsf{t}}+1}\right)\right)^{k}j\pi \leq 0, \quad j \in \mathbb{N}. \tag{10} \end{split}$$

From inequality (10) for *j* → ∞ and (8), (9) we obtain:

$$\lim\_{j \to \infty} \left( j \left( \omega(\mathfrak{l}\_{\backslash}, \mathfrak{l}\_{\backslash^{+}}) \right)^{k} \right) = 0. \tag{11}$$

Thus there exists *j*1 ∈ N such that *j <sup>ω</sup>*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>)*<sup>k</sup>* ≤ 1 for *j* ≥ *j*1, or:

$$
\omega(\mathfrak{l}\_{\mathfrak{h}}\mathfrak{l}\_{\mathfrak{j}+1}) \le \frac{1}{j^{\mathfrak{k}}}, \quad j \ge j\_1. \tag{12}
$$

Then for *m*, *j* ∈ N with *m* > *j* ≥ *j*1, we have:

$$\begin{split} \omega\left(\mathbb{I}\_{\left[i\right.}\mathbb{I}\_{m}\right) \\ \leq & \omega\left(\mathbb{I}\_{\left[i\right.}\mathbb{I}\_{\left[i+1\right.}\right]\right) + \omega\left(\mathbb{I}\_{\left[i+1\right.}\mathbb{I}\_{\left[i+2\right.}\right) + \omega\left(\mathbb{I}\_{\left[i+2\right.}\mathbb{I}\_{\left[i+3\right.}\right) + \dots + \omega\left(\mathbb{I}\_{m-1}\mathbb{I}\_{m}\right) \\ = & \sum\_{i=j}^{m-1} \omega\left(\mathbb{I}\_{i\left.\mathbb{I}\_{i+1}\right.}\right) \leq \sum\_{i=j}^{\infty} \omega\left(\mathbb{I}\_{i\left.\mathbb{I}\_{i+1}\right.}\right) \leq \sum\_{i=j}^{\infty} \frac{1}{i^{\frac{1}{k}}} < \infty. \end{split} \tag{13}$$

Hence {<sup>l</sup>*j*} is a Cauchy sequence in Ω. From completeness of Ω there exists an element l∗ ∈ Ω*L* lim *j*<sup>→</sup>∞ <sup>l</sup>*j*+<sup>1</sup> = l<sup>∗</sup>. As J is continuous, we have J (l∗) = lim*j*<sup>→</sup>∞ J (<sup>l</sup>*j*) = lim*j*<sup>→</sup>∞ <sup>l</sup>*j*+<sup>1</sup> = l<sup>∗</sup>. It proves the claim.

In the partial case of *α*-admissible mapping we ge<sup>t</sup> the following result:

**Corollary 1.** *Let the assumptions be satisfied:*

*1. The functions ϑ* ∈ Δ *and α* : Ω × Ω → {−∞} ∪ (0, <sup>∞</sup>)*, the mapping* J : Ω → Ω *is α-admissible mapping and for* l, *κ* ∈ Ω *and ω*(J (l), J (*κ*)) > 0 *the inequality:*

$$
\pi + \mathfrak{a}(\mathfrak{l}, \mathfrak{k}) \mathfrak{b}\left(\omega(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{k}))\right) \le \mathfrak{b}\left(\omega(\mathfrak{l}, \mathfrak{k})\right),
$$

*holds.*


*Then the mapping* J *has a fixed point in* Ω*.*

**Proof.** The claim follows from Theorem 2 with *β*(<sup>l</sup>, *κ*) ≡ 1 for l, *κ* ∈Ω.

In the case when the mapping J is not continuous we ge<sup>t</sup> the following result:

**Theorem 3.** *Let* J : Ω → Ω *be an (<sup>α</sup>,β*)*-type rational ϑ-contraction and the following condition be satisfied:*


*Then the point* l∗ *from condition (c) is a fixed point of the mapping* J *.*

**Proof.** As in the proof of Theorem 2 we construct the sequence {<sup>l</sup>*j*}<sup>∞</sup>*j*=<sup>0</sup> and obtain the inequalities *<sup>α</sup>*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1, *β*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1. The sequence {<sup>l</sup>*j*}<sup>∞</sup>*j*=<sup>0</sup> is a Cauchy sequence in Ω and lim*j*<sup>→</sup>∞ *<sup>ω</sup>*(<sup>l</sup>*j*, l∗) = 0 with l∗ ∈ Ω.

Therefore by condition (c) of Theorem 3, we have *<sup>α</sup>*(<sup>l</sup>*j*, l∗) ≥ 1 and *β*(<sup>l</sup>*j*, l∗) ≥ 1 for all *j* ∈ N. We will prove that J (l∗) = l<sup>∗</sup>. Assuming the contrary that J (l∗) = l<sup>∗</sup>. Then there exists a number *j*0 ∈ N such that <sup>l</sup>*j*+<sup>1</sup> = J (l<sup>∗</sup>), for all *j* ≥ *j*0. Therefore, *ω*(J (<sup>l</sup>*j*), J (l∗)) > 0, for *j* ≥ *j*0. By (2), we have:

$$\begin{split} \pi + \theta(\omega(\mathfrak{l}\_{\mathfrak{j}+1}, \mathcal{J}(\mathfrak{l}^{\mathfrak{\*}}))) &= \quad \pi + \theta(\omega(\mathcal{J}(\mathfrak{l}\_{\mathfrak{j}}), \mathcal{J}(\mathfrak{l}^{\mathfrak{\*}}))) \\ &\leq \quad \pi + a(\mathfrak{l}\_{\mathfrak{j}}, \mathfrak{l}^{\mathfrak{\*}}) \beta(\mathfrak{l}\_{\mathfrak{j}}, \mathfrak{l}^{\mathfrak{\*}}) \theta(\omega(\mathcal{J}(\mathfrak{l}\_{\mathfrak{j}}), \mathcal{J}(\mathfrak{l}^{\mathfrak{\*}}))) \\ &\leq \quad \theta(\omega(\mathfrak{l}\_{\mathfrak{j}}, \mathfrak{l}^{\mathfrak{\*}})). \end{split} \tag{14}$$

This implies that:

$$\begin{aligned} \left(\omega(\mathfrak{u}(\mathfrak{l}\_{j+1}, \mathcal{I}(\mathfrak{l}^\*)))\right) &\leq& \left.\theta\left(\omega(\mathfrak{l}\_{j'}\mathfrak{l}^\*)\right)-\pi\right\|\\ &< \left.\theta\left(\omega(\mathfrak{l}\_{j'}\mathfrak{l}^\*)\right). \end{aligned}$$

By (*F*1), we have:

$$
\omega(\mathfrak{l}\_{\dot{\mathfrak{j}}+1\prime}\mathcal{I}(\mathfrak{l}^\*)) < \omega(\mathfrak{l}\_{\dot{\mathfrak{j}}\prime}\mathfrak{l}^\*).
$$

Letting *j* → ∞ and using the fact that lim*j*<sup>→</sup>∞ *<sup>ω</sup>*(<sup>l</sup>*j*, l∗) = 0 and lim*j*<sup>→</sup>∞ *<sup>ω</sup>*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) = 0 we ge<sup>t</sup> *<sup>ω</sup>*(l<sup>∗</sup>, J (l∗)) ≤ 0 which is a contradiction. Therefore *<sup>ω</sup>*(l<sup>∗</sup>, J (l∗)) = 0, i.e., J (l∗) = l<sup>∗</sup>.

In the partial case of *α*-admissible mapping we obtain the result:

#### **Corollary 2.** *Let the assumptions be fulfilled:*

*1. The functions ϑ* ∈ Δ *and α* : Ω × Ω → {−∞} ∪ (0, ∞) *and the mapping* J : Ω → Ω *is α-admissible mapping such that for* l, *κ* ∈ Ω *and ω*(J (l), J (*κ*)) > 0 *the inequality:*

$$\left(\pi + \mathfrak{a}(\mathfrak{l}, \mathfrak{k})\mathfrak{\theta}\left(\omega(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{k}))\right) \right) \le \mathfrak{\theta}\left(\omega(\mathfrak{l}, \mathfrak{k})\right),$$

*holds.*

*2. The conditions (b) and (c) of Theorem 3 are fulfilled.* *Then the point* l∗ *from condition (c) is a fixed point of the mapping* J *.*

**Proof.** The claim follows from Theorem 3 with *β*(<sup>l</sup>, *κ*) ≡ 1 for l, *κ* ∈Ω.

We state the following property. (P) *<sup>α</sup>*(<sup>l</sup>, *κ*) ≥ 1 and *β*(<sup>l</sup>, *κ*) ≥ 1 for all fixed points l, *κ* ∈Ω.

**Theorem 4.** *Suppose that the assertions of Theorem 2 are satisfied and the property (P) holds, then the fixed point of the mapping* J *is unique.*

**Proof.** Let l∗, / l ∈ Ω be such that J (l∗) = l∗ and J ( / l) = / l but l∗ = / l. Then by (P), *<sup>α</sup>*(l<sup>∗</sup>, / l) ≥ 1 and *β*(l<sup>∗</sup>, / l) ≥ 1. Thus by (2), we have:

$$\begin{aligned} \left(\pi + \theta\left(\omega(\mathbb{I}^\*, \widehat{\mathbb{I}})\right)\right) &= \pi + \theta\left(\omega\left(\mathcal{J}(\mathbb{I}^\*), \mathcal{J}(\widehat{\mathbb{I}})\right)\right) \\ &\leq \quad \pi + \theta\left(a(\mathbb{I}^\*, \widehat{\mathbb{I}})\beta(\mathbb{I}^\*, \widehat{\mathbb{I}})\omega\left(\mathcal{J}(\mathbb{I}^\*), \mathcal{J}(\widehat{\mathbb{I}})\right)\right) \\ &\leq \quad \theta\left(\omega(\mathbb{I}^\*, \widehat{\mathbb{I}})\right). \end{aligned}$$

The above inequality is a contradiction because *π* > 0. Hence, l∗ is unique.

The fixed point result in Theorem 4 generalize the known in the literature result.

**Corollary 3.** *([10]). Let* J : Ω → Ω *be ϑ -contraction . Then the mapping* J *has a fixed point in* Ω*.*

**Proof.** The claim follows from the proof of Theorem 4 with *<sup>α</sup>*(<sup>l</sup>, *<sup>κ</sup>*)=*β*(<sup>l</sup>, *κ*) ≡ 1 for all l, *κ* ∈Ω.

**Example 3.** *Consider the set* Ω = l*j* : *j* ∈ N *where the natural numbers:*

$$A\_j = 1 \times 2 + 3 \times 4 + \dots + (2j - 1)(2j) = \frac{j(j+1)(4j-1)}{3}, \text{ for } j = 1, 2, \dots, n$$

*Let ω* (l, *κ*) = |l − *κ*| *for any* l, *κ* ∈ Ω*. Define the mapping* J : Ω → Ω *by,*

$$\mathcal{J}\left(\mathfrak{l}\_1\right) = \mathfrak{l}\_1, \quad \mathcal{J}\left(\mathfrak{l}\_j\right) = \mathfrak{l}\_{j-1}, \quad \text{for all } j \ge 2.$$

*Let the functions α* : Ω × Ω → {−∞} ∪ (0, ∞) *be defined by <sup>α</sup>*(<sup>l</sup>, *κ*) = *β*(<sup>l</sup>, *κ*) ≡ 1 *for all* l, *κ* ∈ Ω *and ϑ* : (0, <sup>+</sup>∞) → R *be defined by ϑ*(l) = l − √1l, l > 0*. According to Example 2 the function* Θ ∈ Δ*.*

*Then the mapping* J *is (<sup>α</sup>,β*)*-type ϑ-contraction, with π* = 12. *or it is ϑ-contraction (see Remark 1). Consider the following three possible cases:*

*Case 1. Let* 1 = *j* < *ι*. *Then,*

$$\left| \mathcal{J} \left( \mathbb{1} \right) - \mathcal{J} \left( \mathbb{1}\_{1} \right) \right| = \left| \mathbb{1}\_{-1} - \mathbb{1}\_{1} \right| = \mathfrak{Z} \times 4 + 5 \times 6 + \dots + (2\iota - 3)(2\iota - 2) \tag{15}$$

*and*

$$
\omega\left(\mathbb{I}, \mathbb{I}\_1\right) = \left|\mathbb{I} - \mathbb{I}\_1\right| = \mathfrak{Z} \times \mathfrak{A} + \mathfrak{Z} \times \mathfrak{G} + \dots + (2\iota - 1)(2\iota). \tag{16}
$$

*As ι* > 1, *so we get,*

$$\frac{-1}{\sqrt{3 \times 4 + \dots + (2\nu - 3)(2\nu - 2)}} < \frac{-1}{\sqrt{3 \times 4 + \dots + (2\nu - 1)(2\nu)}}.\tag{17}$$

> *From (17), we have,*

$$\begin{aligned} &12 - \frac{-1}{\sqrt{3 \times 4 + \dots + (2\iota - 3)(2\iota - 2)}} + 3 \times 4 + + 5 \times 6 + \dots + (2\iota - 3)(2\iota - 2) \\ &< 12 - \frac{-1}{\sqrt{3 \times 4 + \dots + (2\iota - 1)(2\iota)}} + [3 \times 4 + + 5 \times 6 + \dots + (2\iota - 3)(2\iota - 2)] \\ &\le -\frac{-1}{\sqrt{3 \times 4 + \dots + (2\iota - 1)(2\iota)}} + [3 \times 4 + + 5 \times 6 + \dots + (2\iota - 3)(2\iota - 2)] + (2\iota - 1)(2\iota) \end{aligned}$$

*By (15) and (16), we have,*

$$12 - \frac{1}{\sqrt{|\mathcal{J}\left(\mathbb{I}\right) / \mathcal{J}\left(\mathbb{I}\right)|}} + |\mathcal{J}\left(\mathbb{I}\_{\mathbb{I}}\right) , \mathcal{J}\left(\mathbb{I}\_{\mathbb{I}}\right)| < -\frac{1}{\sqrt{|\mathbb{I}\_{\mathbb{I}} - \mathbb{I}\_{\mathbb{I}}|}} + |\mathbb{I}\_{\mathbb{I}} - \mathbb{I}\_{\mathbb{I}}|.\tag{18}$$

*Case 2. Let* 1 = *ι* < *j This case is similar to Case 1 and therefore we omit it. Case 3. Let ι* > *j* > 1. *Then we have,*

$$|\mathcal{J}\left(\mathbb{t}\right) - \mathcal{J}\left(\mathbb{t}\right)| = (2j - 1)(2j) + (2j + 1)(2j + 2) + \dots + (2\iota - 3)(2\iota - 2) \tag{19}$$

*and*

$$|\mathfrak{l} - \mathfrak{l}\_{\bar{\jmath}}| = (2\mathfrak{j} + 1)(2\mathfrak{j} + 2) + (2\mathfrak{j} + 3)(2\mathfrak{j} + 4) + \dots + (2\mathfrak{\iota} - 1)(2\mathfrak{\iota}).\tag{20}$$

*As ι* > *j* > 1, *we get:*

$$(2\iota -1)(2\iota) \ge (2j + 2)(2j + 1) > (2j + 2)(2j + 2) = 2j(2j + 2) + 2(2j + 2) \ge 2j(2j + 2) + 12\iota$$

*We know that,*

$$\frac{-1}{\sqrt{(2j-1)(2j)+...+(2\iota-3)(2\iota-2)}} < \frac{-1}{\sqrt{(2j+1)(2j+2)+...+(2\iota-1)(2\iota)}}.\tag{21}$$

$$By \ (21), \ we \ have \ \cdot$$

$$\begin{aligned} &12 - \frac{1}{\sqrt{(2j-1)(2j)+(2j+1)(2j+2)+\ldots+(2\iota-3)(2\iota-2)}} \\ &+ (2j-1)(2j)+(2j+1)(2j+2)+\ldots+(2\iota-3)(2\iota-2) \\ &< 12 - \frac{1}{\sqrt{(2j+1)(2j+2)+(2j+3)(2j+4)+\ldots+(2\iota-1)(2\iota)}} \\ &+ (2j-1)(2j)+(2j+1)(2j+2)+\ldots+(2\iota-3)(2\iota-2) \\ &< -\frac{1}{\sqrt{(2j+1)(2j+2)+(2j+3)(2j+4)+\ldots+(2\iota-1)(2\iota)}} \\ &+ (2j-1)(2j)+(2j+1)(2j+2)+\ldots+(2\iota-3)(2\iota-2) \\ &+ (2\iota-1)(2\iota) \\ &= -\frac{1}{\sqrt{(2j+1)(2j+2)+(2j+3)(2j+4)+\ldots+(2\iota-1)(2\iota)}} \\ &+ (2j-1)(2j)+(2j+1)(2j+2)+\ldots+(2\iota-1)(2\iota) \end{aligned}$$

*By (19) and (20), we have:*

$$12 - \frac{1}{\sqrt{|\mathcal{I}\left(\mathfrak{l}\_{\mathfrak{l}}\right) - \mathcal{I}\left(\mathfrak{l}\_{\mathfrak{l}}\right)|}} + |\mathcal{I}\left(\mathfrak{l}\_{\mathfrak{l}}\right) - \mathcal{I}\left(\mathfrak{l}\_{\mathfrak{l}}\right)| < -\frac{1}{\sqrt{|\mathfrak{l}\_{\mathfrak{l}} - \mathfrak{l}\_{\mathfrak{l}}|}} + |\mathfrak{l}\_{\mathfrak{l}} - \mathfrak{l}\_{\mathfrak{l}}|.$$

*Thus all the hypotheses of Theorem 3 hold and therefore, the mapping* J *has a unique fixed point* l1*.*

Now we provide some fixed point theorems for (*<sup>α</sup>*,*β*)-type rational *ϑ*-contraction.

**Theorem 5.** *Let the functions ϑ* ∈ Δ *and α* : Ω × Ω → {−∞} ∪ (0, ∞) *and* J *:* Ω → Ω *be (<sup>α</sup>,β*)*-type ϑ-contraction and:*


*Then the mapping* J *has a fixed point in* Ω*, i.e., there exists a point* l∗ ∈ Ω *such that* J (l∗) = l<sup>∗</sup>*.*

**Proof.** As in the proof of Theorem 2 we construct the sequence {<sup>l</sup>*j*}<sup>∞</sup>*j*=<sup>0</sup> in Ω. Assume that <sup>l</sup>*j*+<sup>1</sup> = l*j* for all *j* = 0, 1, 2, ... . Then from condition (a) and the choice of l0 it follows that *<sup>α</sup>*(<sup>l</sup>1, l2) = *α*(J (<sup>l</sup>0), J (<sup>l</sup>1)) ≥ 1 and *β*(<sup>l</sup>1, l2) = *β*(J (<sup>l</sup>0), J (<sup>l</sup>1)) ≥ 1. By induction we ge<sup>t</sup> *<sup>α</sup>*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1 and *β*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1 for *j* ∈ N. Now by inequality (3) with l = <sup>l</sup>*j*−<sup>1</sup> and *κ* = l*j*, we have:

$$\begin{split} \left( \pi + \theta \left( \omega(\mathsf{l}\_{\circ}, \mathsf{l}\_{\circ + 1}) \right) \right) &= \pi + \theta \left( \omega(\mathcal{I}(\mathsf{l}\_{\circ - 1}), \mathcal{I}(\mathsf{l}\_{\circ})) \right) \\ &\leq \pi + a(\mathsf{l}\_{\circ - 1}, \mathsf{l}\_{\circ}) \beta(\mathsf{l}\_{\circ - 1}, \mathsf{l}\_{\circ}) \theta \left( \omega(\mathcal{I}(\mathsf{l}\_{\circ - 1}), \mathcal{I}(\mathsf{l}\_{\circ})) \right) \\ &\leq \theta(\mathcal{R}(\mathsf{l}\_{\circ - 1}, \mathsf{l}\_{\circ})) \end{split} \tag{22}$$

where

$$\begin{split} \mathcal{R}(\mathfrak{l}\_{\{\mathfrak{l}\_{\}-1},\mathfrak{l}\_{\}}) &= \max \left\{ \omega(\mathfrak{l}\_{\{\mathfrak{l}\_{\}-1},\mathfrak{l}\_{\}}), \frac{\omega\left(\mathfrak{l}\_{\{\mathfrak{l}\_{\}-1},\mathcal{I}\left(\mathfrak{l}\_{\{\mathfrak{l}\_{\}}-1}\right)\right)\omega(\mathfrak{l}\_{\{\mathfrak{l}\_{\}},\mathcal{I}\left(\mathfrak{l}\_{\}\right))}}{1+\omega\left(\mathfrak{l}\_{\{\mathfrak{l}\_{\}-1},\mathfrak{l}\_{\}}\right)} \right\} \\ &= \max \left\{ \omega(\mathfrak{l}\_{\{\mathfrak{l}\_{\}-1},\mathfrak{l}\_{\}}), \frac{\omega\left(\mathfrak{l}\_{\{\mathfrak{l}\_{\}-1},\mathfrak{l}\_{\}}\right)\omega(\mathfrak{l}\_{\{\mathfrak{l}\_{\}},\mathfrak{l}\_{\{\mathfrak{l}\_{\}}+1})}}{1+\omega(\mathfrak{l}\_{\{\mathfrak{l}\_{\}-1},\mathfrak{l}\_{\}})} \right\}. \end{split} \tag{23}$$

If we assume max *ω*(<sup>l</sup>*j*−1, l*j*), *<sup>ω</sup>*(<sup>l</sup>*j*−1,l*j*)*ω*(<sup>l</sup>*j*,l*j*+<sup>1</sup>) <sup>1</sup>+*ω*(<sup>l</sup>*j*−1,l*j*) = *<sup>ω</sup>*(<sup>l</sup>*j*−1,l*j*)*ω*(<sup>l</sup>*j*,l*j*+<sup>1</sup>) <sup>1</sup>+*ω*(<sup>l</sup>*j*−1,l*j*) , then from (22) we obtain:

$$\left(\pi + \vartheta\left(\omega\left(\mathbb{I}\_{j}, \mathbb{I}\_{j+1}\right)\right) \right) \le \vartheta\left(\frac{\omega\left(\mathbb{I}\_{j-1}, \mathbb{I}\_{j}\right)\omega\left(\mathbb{I}\_{j}, \mathbb{I}\_{j+1}\right)}{1 + \omega\left(\mathbb{I}\_{j-1}, \mathbb{I}\_{j}\right)}\right) < \vartheta\left(\omega\left(\mathbb{I}\_{j}, \mathbb{I}\_{j+1}\right)\right).$$

The above inequality is a contradiction because *π* > 0. Hence,

$$\max\left\{\omega\left(\mathbb{I}\_{j-1},\mathbb{I}\_{\mathfrak{j}}\right),\frac{\omega\left(\mathbb{I}\_{j-1},\mathbb{I}\_{\mathfrak{j}}\right)\omega\left(\mathbb{I}\_{\mathfrak{j}},\mathbb{I}\_{\mathfrak{j}+1}\right)}{1+\omega\left(\mathbb{I}\_{j-1},\mathbb{I}\_{\mathfrak{j}}\right)}\right\}=\omega\left(\mathbb{I}\_{j-1},\mathbb{I}\_{\mathfrak{j}}\right).$$

Therefore the inequality (22) is reduced to:

$$
\pi + \vartheta \left( \omega(\mathfrak{l}\_{\mathfrak{l}}, \mathfrak{l}\_{\mathfrak{j}+1}) \right) \le \vartheta(\omega(\mathfrak{l}\_{\mathfrak{j}-1}, \mathfrak{l}\_{\mathfrak{j}})).\tag{24}
$$

.

Following the same procedure as we did in Theorem 2, we ge<sup>t</sup> l∗ ∈ Ω such that J (l∗) <sup>=</sup>l<sup>∗</sup>. Thus l∗ is a fixed point of J

In the partial case of *α*-admissible mapping we obtain the result:

#### **Corollary 4.** *Let the following assumptions be satisfied:*

*1. The functions ϑ* ∈ Δ *and α* : Ω × Ω → {−∞} ∪ (0, ∞) *and the mapping* J : Ω → Ω *is α-admissible mapping such that for* l, *κ* ∈ Ω *and ω*(J (l), J (*κ*)) > 0 *the inequality*

$$
\pi + \mathfrak{a}(\mathfrak{l}, \mathfrak{k}) \mathfrak{b}\left(\omega(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{k}))\right) \le \mathfrak{b}\left(\mathcal{R}(\mathfrak{l}, \mathfrak{k})\right),
$$

*holds where*

$$\mathcal{R}(\mathfrak{l},\mathfrak{k}) = \max \left\{ \omega(\mathfrak{l},\mathfrak{k}), \frac{\omega(\mathfrak{l},\mathcal{I}(\mathfrak{l})) \omega(\mathfrak{k},\mathcal{I}(\mathfrak{k}))}{1 + \omega(\mathfrak{l},\mathfrak{k})} \right\};$$


*Then the mapping* J *has a fixed point in* Ω*.*

**Proof.** The claim follows from Theorem 5 with *β*(<sup>l</sup>, *κ*) ≡ 1 for l, *κ* ∈Ω.

Now we prove a result for (*<sup>α</sup>*,*β*)-type rational *ϑ*-contraction when the mapping J is not continuous.

**Theorem 6.** *Let the functions ϑ* ∈ Δ *and α*, *β* : Ω × Ω → {−∞} ∪ (0, ∞) *and* J *:* Ω → Ω *be an (<sup>α</sup>,β*)*-type rational ϑ-contraction and the following condition be satisfied:*


*Then the point* l∗ *from condition (c) is a fixed point of the mapping* J *in* Ω*.*

**Proof.** As in the proof of Theorem 2 we construct the sequence {<sup>l</sup>*j*}<sup>∞</sup>*j*=<sup>0</sup> in Ω. Similarly to the proof of Theorem 5 we obtain the inequalities *<sup>α</sup>*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1, *β*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1 and {<sup>l</sup>*j*}<sup>∞</sup>*j*=<sup>0</sup> is a Cauchy sequence in Ω which converges to l∗, i.e., lim*j*<sup>→</sup>∞*<sup>ω</sup>*(<sup>l</sup>*j*, l∗) = 0.

Therefore by condition (c) of Theorem 6, we have *<sup>α</sup>*(<sup>l</sup>*j*, l∗) ≥ 1 and *β*(<sup>l</sup>*j*, l∗) ≥ 1 for all *j* ∈ N. We will prove that J (l∗) = l<sup>∗</sup>. Assume the contrary that J (l∗) = l<sup>∗</sup>. Then there exists *j*0 ∈ N such that <sup>l</sup>*j*+<sup>1</sup> = J (l<sup>∗</sup>), for all *j* ≥ *j*0. Therefore, *ω*(J (<sup>l</sup>*j*), J (l∗)) > 0, for *j* ≥ *j*0. By (3), we have:

$$\begin{array}{rcl}\pi+\theta\left(\omega\left(\mathbb{I}\_{[\!+1\!\times},\mathcal{I}\left(\mathbb{I}^{\star}\right)\right)\right)&=&\pi+\theta\left(\omega\left(\mathcal{I}\left(\mathbb{I}\_{[\!\cdot\right\},\mathcal{I}\left(\mathbb{I}^{\star}\right)\right)\right)\\&\leq&\pi+a\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}}\mathbb{I}\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}}\right)\theta\left(\omega\left(\mathcal{I}\left(\mathbb{I}\_{[\!\cdot\right\},\mathcal{I}\left(\mathbb{I}^{\star}\right)\right)\right)\\&\leq&\vartheta\left(\max\{\omega\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right),\frac{\omega\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right)\right)\omega\left(\mathbb{I}^{\star}\left(\mathbb{I}^{\star}\right)\right)}{1+\omega\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right)}\}\}\\&=&\vartheta\left(\max\{\omega\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right),\frac{\omega\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right)\omega\left(\mathbb{I}^{\star}\right)\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right)}{1+\omega\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right)}}}{\vartheta\left(\max\{\omega\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right),\frac{\omega\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right)\omega\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right)}{1+\omega\left(\mathbb{I}\_{[\!\cdot\right\}^{\star}\right)}}\})\end{array} \tag{25}$$

which implies:

$$\begin{split} \left(\vartheta\left(\omega(\mathbb{I}\_{j+1},\mathcal{J}(\mathbb{I}^\*))\right)\right) &\leq \quad \theta\left(\max\{\omega(\mathbb{I}\_{j},\mathbb{I}^\*),\frac{\omega(\mathbb{I}\_{j},\mathbb{I}\_{j+1})\omega(\mathbb{I}^\*,\mathcal{J}(\mathbb{I}^\*))}{1+\omega(\mathbb{I}\_{j},\mathbb{I}^\*)}\}\right) - \pi \\ &< \quad \theta\left(\max\{\omega(\mathbb{I}\_{j},\mathbb{I}^\*),\frac{\omega(\mathbb{I}\_{j},\mathbb{I}\_{j+1})\omega(\mathbb{I}^\*,\mathcal{J}(\mathbb{I}^\*))}{1+\omega(\mathbb{I}\_{j},\mathbb{I}^\*)}\}\right). \end{split}$$

By (*F*1), we have:

$$
\omega(\mathfrak{l}\_{j+1}, \mathcal{J}(\mathfrak{l}^\*)) < \max \{ \omega(\mathfrak{l}\_j, \mathfrak{l}^\*), \frac{\omega(\mathfrak{l}\_{j'}\mathfrak{l}\_{j+1}) \omega(\mathfrak{l}^\*, \mathcal{J}(\mathfrak{l}^\*))}{1 + \omega(\mathfrak{l}\_{j'}\mathfrak{l}^\*)} \}.
$$

Letting *j* → ∞ and using the fact that lim*j*<sup>→</sup>∞ *<sup>ω</sup>*(<sup>l</sup>*j*, l∗) = 0 and lim*j*<sup>→</sup>∞ *<sup>ω</sup>*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) = 0 we ge<sup>t</sup> *<sup>ω</sup>*(l<sup>∗</sup>, J (l∗)) ≤ 0 which is a contradiction. Therefore *<sup>ω</sup>*(l<sup>∗</sup>, J (l∗)) = 0, i.e., J (l∗) = l<sup>∗</sup>.

**Example 4.** *Let* Ω = {0} ∪ [ 94 , 5] *and ω* (l, *κ*) = |l − *κ*| *for* l, *κ* ∈ Ω*. Clearly* (<sup>Ω</sup>, *ω*) *is a complete metric space. Consider the function ϑ* (l) = −√1l + l ∈ Δ *for* l ∈ Ω *(see Example 2) and π* ∈ 0, <sup>112</sup>−<sup>3</sup>√<sup>5</sup> 15 *. Define* J : Ω → Ω *and α*, *β* : Ω → {−∞} ∪ (0, ∞) *by:*

$$\mathcal{J}(\mathfrak{l}) = \left\{ \begin{array}{ll} \mathfrak{l}', & \mathfrak{l} \in \{0\} \cup [\mathfrak{l}', \mathfrak{k}), \\ 0, & \mathfrak{l} = \mathfrak{s}. \end{array} \right\}$$

*and*

$$
\alpha\left(\mathfrak{l},\mathfrak{x}\right) = \beta\left(\mathfrak{l},\mathfrak{x}\right) = 1
$$

*We prove that* J *is (<sup>α</sup>*, *β)-type rational ϑ-contraction. Consider these possible cases: Case I. For* l = 0 *and κ* = 5, *we have*

$$
\omega\left(\mathcal{I}(0), \mathcal{J}(5)\right) = \omega\left(\{\frac{9}{4}\}, 0\right) = \frac{9}{4} > 0
$$

*and*

$$\mathcal{R}\left(0,5\right) = \dots = \max\left\{\omega\left(0,5\right), \frac{\omega\left(0,\mathcal{I}(0)\right) \cdot \omega\left(5,\mathcal{I}(5)\right)}{1 + \omega\left(0,5\right)}\right\}.$$

.

*Since,*

$$
\omega\left(\mathcal{J}(0), \mathcal{J}(5)\right) = \frac{9}{4} < 5 = \omega\left(0, 5\right) \le \mathcal{R}\left(0, 5\right)\,.
$$

*So, we have*

$$-\frac{1}{\sqrt{\omega \left( \mathcal{J}(0), \mathcal{J}(5) \right)}} < -\frac{1}{\sqrt{\mathcal{R} \left( 0, 5 \right)}}\gamma$$

*which further implies:*

$$-\frac{1}{\sqrt{\omega\left(\mathcal{J}\!\!/\mathcal{O},\mathcal{J}\!\!/\mathcal{S}\right)}} + \omega\left(\mathcal{J}\!\!/\mathcal{O},\mathcal{J}\!\!/\mathcal{S}\right) < -\frac{1}{\sqrt{\mathcal{R}\left(\mathcal{O},\mathcal{S}\right)}} + \mathcal{R}\left(\mathcal{O},\mathcal{S}\right) \dots$$

*Thus we obtain:*

$$\begin{split} &\pi + \mathfrak{a}\left(0,5\right)\mathfrak{f}\left(0,5\right)\mathfrak{f}\left(\omega\left(\mathcal{I}\left(0\right),\mathcal{I}\left(5\right)\right)\right) = \pi + \mathfrak{d}\left(\omega\left(\mathcal{I}\left(0\right),\mathcal{I}\left(5\right)\right)\right) \\ &= \pi - \frac{1}{\sqrt{\omega\left(\mathcal{I}\left(0\right),\mathcal{I}\left(5\right)\right)}} + \omega\left(\mathcal{I}\left(0\right),\mathcal{I}\left(5\right)\right) = \pi - \sqrt{\frac{4}{9}} + \frac{9}{5} \\ &\leq \quad - \sqrt{\frac{1}{5}} + 5 \leq -\frac{1}{\sqrt{\mathcal{R}\left(0,5\right)}} + \mathcal{R}\left(0,5\right) = \mathfrak{d}\left(\mathcal{R}\left(0,5\right)\right). \end{split}$$

*Hence,*

$$\left(\pi + \mathfrak{a}\left(0, 5\right) \mathfrak{z}\left(0, 5\right) \mathfrak{z}\left(\omega\left(\mathcal{I}\left(0\right), \mathcal{I}\left(5\right)\right)\right) \le \mathfrak{z}\left(\mathcal{R}\left(5, \kappa\right)\right).$$

*Case II.*

*For* l ∈ [ 94 , <sup>5</sup>), *κ* = 0

$$
\omega\left(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{l})\right) = \omega\left(\{\frac{9}{4}\}, \{\frac{9}{4}\}\right) = 0.
$$
