In this paper we study the following functional integral equation,

where $a<c<b$ are real numbers, $(\mathbb{B},\left|\cdot\right|)$ is a Banach space, $H\in C([a,b]\times[a,c]\times\mathbb{B},\mathbb{B}),\ K\in C([a,b]^{2}\times\mathbb{B},\mathbb{B}),\ g\in C([a,b],\mathbb{B})$ and $A:C([a,c],\mathbb{B})\rightarrow C([a,c],\mathbb{B})$ and $B:C([a,b],\mathbb{B})\rightarrow C([a,b],\mathbb{B})$ are given operators.

For some examples of such integral equations see [5], [6], [16], [28], [2], [4], [7].

Let $V:C([a,b],\mathbb{B})\rightarrow C([a,b],\mathbb{B})$ be defined by

In this paper we consider on the spaces of continuous functions max-norms.

Let us suppose that

• $(C_{1})$

  $\exists L_{H}>0:\left|H(t,s,\eta_{1})-H(t,s,\eta_{2})\right|\leq L_{H}\left|\eta_{1}-\eta_{2}\right|,$ for all $t\in[a,b],s\in[a,c],\ \eta_{1},\eta_{2}\in\mathbb{B}$;

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• $(C_{2})$

  $\exists L_{K}>0:\left|K(t,s,\eta_{1})-K(t,s,\eta_{2})\right|\leq L_{K}\left|\eta_{1}-\eta_{2}\right|,$ for all $t,s\in[a,b],\ \eta_{1},\eta_{2}\in\mathbb{B}$;

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• $(C_{3})$

  $\exists L_{A}>0:\ \underset{[a,c]}{\max}\left|A(y)(t)-A(z)(t)\right|\leq L_{A}\underset{[a,c]}{\max}\left|y(t)-z(t)\right|,$ $\forall y,z\in C([a,c],\mathbb{B})$;

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• $(C_{4})$

  $\exists L_{B}>0:\ \left|B(y)(t)-B(z)(t)\right|\leq L_{B}\underset{[a,t]}{\max}\left|y(s)-z(s)\right|,$ $\forall\ t\in[a,b].$

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If we apply the contraction principle, in a standard way, for equation (1.1), we have the following result:

The aim of this paper is to improve condition $(C_{5}^{\prime}),$ obtaining the same conclusions. In order to do this we shall apply instead of contraction principle, a new variant of fibre contraction principle, variant given in [13].

In a similar way we study the equation

with suitable conditions on $H,\ K,\ A$ and $B.$

Throughout this paper we shall use the notations from [28], [22] and [13].

By definition, an operator $V:C([a,b],\mathbb{B})\rightarrow C([a,b],\mathbb{B})$ is forward Volterra operator if the following implication holds:

for all $t\in[a,b].$

An operator $V:C([a,b],\mathbb{B})\rightarrow C([a,b],\mathbb{B})$ is backward Volterra operator iff:

for all $t\in[a,b].$

If $a<c<b$ then $V$ is forward Volterra operator w.r.t. the interval $[c,b]$ iff

The operator $V$ is backward Volterra operator w.r.t. the interval $[a,c]$ iff:

For more considerations on abstract Volterra operator see [9], [36]. For other examples see [2], [4], [5], [6], [7], [12], [14], [16], [17], [22], [35], …

Let us consider the equation (1.1) in the conditions $(C_{1})-(C_{4})$. For $m\in\mathbb{N},m\geq 2$ we shall use the following notations: $t_{0}:=c,\ t_{k}:=c+\dfrac{k(b-c)}{m},\ k=\overline{1,m},\ X_{0}=C([a,c],\mathbb{B}),\ X_{i}=C([t_{i-1},t_{i}],\mathbb{B}),X=\prod\limits_{i=0}^{m}X_{i}.$

We consider the spaces of continuous functions with the max-norms. In order to use the variant of fibre contraction principle given by Theorem 2.4, we need the following subsets:

For $x\in X_{0},\ U_{1x}:=\{x_{1}\in X_{1}|\ (x,x_{1})\in U_{1}\},\$for $x\in X_{i-1},\ U_{ix}:=\{x_{i}\in X_{i}|\ (x,x_{i})\in U_{i}\},\ i=\overline{2,m}.$

We remark that, $U_{i},U_{ix},\ i=\overline{1,m}$ are nonempty closed subsets.

We also need the following operators:

It is clear that, $R_{i}\left(C([a,t_{i}],\mathbb{B})\right)=U_{i}$ and $R_{i}:C([a,t_{i}],\mathbb{B})\rightarrow U_{i}$ is an increasing homeomorphism.

Since the operator, $V:C([a,b],\mathbb{B})\rightarrow C([a,b],\mathbb{B})$ defined by, $V(x)(t):=$ second part of equation (1.1), is a forward Volterra operator on $[c,b],$ it induces the following operators:

Let

If on cartesian product we consider max-norms, the operators $R_{i},\ i=\overline{1,m}$ are isometries. From the above definitions, we remark that $(T_{0},T_{1})(U_{1})\subset U_{1},\ (T_{0},T_{1},\ldots,T_{m})(U_{m})\subset U_{m.}$

In the conditions $\ (C_{1})-(C_{4})$ we have that: $T_{0}$ is $(L_{H}L_{A}+L_{K}L_{B})(c-a)$-Lipschitz.

If we suppose that

• $(C_{5})$

  $(L_{H}L_{A}+L_{K}L_{B})(c-a)<1$

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then we are in the conditions of Theorem 2.4 with $L_{i}=\max\left\{(L_{H}L_{A}+L_{K}L_{B})(c-a),\dfrac{L_{K}L_{B}(b-c)}{m}\right\}$ and $l_{i}=\dfrac{L_{K}L_{B}(b-c)}{m},\$with suitable $m\in\mathbb{N}.$

From this theorem we have that $T$ is PO.

Since $V=R_{m}^{-1}TR_{m}$ and $V^{n}=R_{m}^{-1}T^{n}R_{m},$ it follows that $V$ is PO.

So, we have:

In this section we consider the following integral equation

where $a<c<b$ are real numbers, $(\mathbb{B},\left|\cdot\right|)$ is a Banach space, $H\in C([a,b]\times[c,b]\times\mathbb{B},\mathbb{B}),\ K\in C([a,b]^{2}\times\mathbb{B},\mathbb{B}),\ g\in C([a,b],\mathbb{B})$ and $A:C([c,b],\mathbb{B})\rightarrow C([c,b],\mathbb{B})$ and $B:C([a,b],\mathbb{B})\rightarrow C([a,b],\mathbb{B})$ are operators. We suppose that: