**3. Application**

In this section, we shall consider an application for our main result. Let *X* = *C*[0, 1] be the space of all continuous functions defined on interval [0, 1] with the metric

$$d(\mathfrak{x}, \mathfrak{y}) = \sup\_{t \in [0, 1]} |\mathfrak{x}(t) - \mathfrak{y}(t)| \,.$$

In what follows we shall use Theorem 5 to show that there is a solution to the following integral equation:

$$\mathbf{x}(t) = \mathbf{y}(t) + \int\_{0}^{1} k(t, \mathbf{s}, \mathbf{x}(\mathbf{s})) d\mathbf{s}, t \in [0, 1] \tag{4}$$

Assume that *k*(*<sup>t</sup>*,*s*, *x*) is continuous. Let *y* ∈ *C*[0, 1]. We consider the following conditions:

(a) *k* : [0, 1] × [0, 1] × R × R → R is continuous;

(b) there exists a continuous function *γ* : [0, ∞] × R → R such that

$$\sup\_{t \in [0,1]} \int\_0^1 \gamma(t,s) \le 1;$$

(c) there exists *ε* ∈ [0, 1] such that

$$\left| \frac{1}{2} \left| \mathfrak{x}(s) - \mathfrak{y}(s) - \int\_{0}^{1} k(t, s, \mathfrak{x}(s)) ds \right| \right| \le \left| \mathfrak{x}(s) - \mathfrak{y}(s) \right|.$$

implies

$$|k(t, \mathbf{s}, \mathbf{x}(\mathbf{s})) - k(t, \mathbf{s}, \mathbf{y}(\mathbf{s}))| \le (1 - \varepsilon) \left| \mathbf{x}(\mathbf{s}) - \mathbf{y}(\mathbf{s}) \right|\_{\mathbf{s}^\*}$$

for all *x*, *y* ∈ *X*;

(d) there exists *x*0 ∈ *C*([0, 1]) such that for all *t* ∈ [0, 1], we have

$$\zeta(\mathbf{x}\_0(t), \int\_0^1 k(t, \mathbf{s}, \mathbf{x}(s)) ds) \ge 0\_\gamma$$

where *ζ* : *X* × *X* → [0, ∞);

(e) For all *t* ∈ [0, 1], *x*, *y* ∈ *C*[0, 1],

$$\mathcal{J}(\mathbf{x}(t), \mathbf{y}(t)) \ge 0 \Rightarrow \mathcal{J}(\int\_0^1 k(t, \mathbf{s}, \mathbf{x}(s)) ds, \int\_0^1 k(t, \mathbf{s}, \mathbf{y}(s)) ds) \ge 0\varphi$$

(f) If *xn* is a sequence in C[0,1] such that *xn* → *x* ∈ *C*[0, 1] and *ζ*(*xn*, *xn*+<sup>1</sup>) ≥ 0 for all *n*, then *ζ*(*xn*, *x*) ≥ 0 for all *n*.

**Theorem 6.** *Suppose that the conditions* (*a*)*–*(*f*) *are satisfied. Then, the integral Equation (4) has solution in C*[0, 1]*.*

**Proof.** Since *k* and the function *y* are continuous, now define an operator

$$\mathcal{T}: \mathbb{C}[0,1] \to \mathbb{C}[0,1]$$

write the integral Equation (4) in the form *x* = T *x*, where

$$\mathcal{T}\mathbf{x}(t) = \mathbf{y}(t) + \int\_0^1 k(t, \mathbf{s}, \mathbf{x}(\mathbf{s})) d\mathbf{s}.\tag{5}$$

It follows that

$$\frac{1}{2}|\mathbf{x}(\mathbf{s}) - \mathbf{y}(\mathbf{s}) - \int\_0^1 k(t, \mathbf{s}, \mathbf{x}(\mathbf{s}))d\mathbf{s}| \le (1 - \varepsilon)|\mathbf{x}(\mathbf{s}) - \mathbf{y}(\mathbf{s})|$$

implies

$$\begin{array}{ll}d(\boldsymbol{T}\boldsymbol{x},\mathcal{T}\boldsymbol{y}) &= \sup\_{\boldsymbol{t}\in[0,1]} |\mathcal{T}\boldsymbol{x}(\boldsymbol{t}) - \mathcal{T}\boldsymbol{y}(\boldsymbol{t})| \\ &\leq \sup\_{\boldsymbol{t}\in[0,1]} \int\_{0} |\boldsymbol{k}(\boldsymbol{t},\boldsymbol{s},\boldsymbol{x}(\boldsymbol{s})) - \boldsymbol{k}(\boldsymbol{t},\boldsymbol{s},\boldsymbol{y}(\boldsymbol{s}))|d\boldsymbol{s} \\ &\leq \sup\_{\boldsymbol{t}\in[0,1]} \int\_{0} \gamma(\boldsymbol{t},\boldsymbol{s}) d\boldsymbol{s} |\boldsymbol{k}(\boldsymbol{t},\boldsymbol{s},\boldsymbol{x}) - \boldsymbol{k}(\boldsymbol{t},\boldsymbol{s},\boldsymbol{y})| \\ &\leq (1-\varepsilon)|\boldsymbol{x}(\boldsymbol{s}) - \boldsymbol{y}(\boldsymbol{s})| \\ &\leq (1-\varepsilon)\max\left\{d(\boldsymbol{x},\boldsymbol{y}), d(\boldsymbol{x},\mathcal{T}\boldsymbol{x}), d(\boldsymbol{y},\mathcal{T}\boldsymbol{y}), \frac{1}{2}\left[d(\boldsymbol{x},\mathcal{T}\boldsymbol{y}) + d(\boldsymbol{y},\mathcal{T}\boldsymbol{x})\right]\right\} \\ &+ \Lambda(\varepsilon)^{a}\boldsymbol{\varphi}(\varepsilon)\left[1 + ||\boldsymbol{x}|| + ||\boldsymbol{y}|| + ||\boldsymbol{\mathcal{T}}\boldsymbol{x}|| + ||\mathcal{T}\boldsymbol{y}||\right]^{\beta} \quad \lambda \geq 0 \quad a \geq 1 \; and \; \beta \in [0, a]. \end{array}$$

Define the function *α* : *C*[0, 1] × *C*[0, 1] → [0, <sup>+</sup>∞) by

$$\mathfrak{a}(\mathfrak{x}, \mathfrak{y}) = \begin{cases} 1 & \text{if } \lhd(\mathfrak{x}(t), \mathfrak{y}(t)) \ge 0, t \in [0, 1],\\ 0 & \text{otherwise.} \end{cases}$$

For all *x*, *y* ∈ *C*[0, 1], we have

Therefore, all the conditions of Theorem 5 are satisfied. Consequently, the mapping T has a unique fixed point in *X*, which is a solution of integral equation.
