Fixed points for some non-obviously contractive operators defined in a space of continuous functions

Abstract

Let $X$ be an arbitrary (real or complex) Banach space, endowed with the norm $\left| \cdot \right| .$ Consider the space of the continuous functions $C\left( \left[ 0,T\right] ,X\right) $ $\left( T>0\right) $, endowed with the usual topology, and let $M$ be a closed subset of it. One proves that each operator $A:M\rightarrow M$ fulfilling for all $x,y\in M$ and for all $t\in \left[ 0,T\right] $ the condition \begin{eqnarray*} \left| \left( Ax\right) \left( t\right) -\left( Ay\right) \left( t\right) \right| &\leq &\beta \left| x\left( \nu \left( t\right) \right) -y\left( \nu \left( t\right) \right) \right| + \\ &&+\frac{k}{t^{\alpha }}\int_{0}^{t}\left| x\left( \sigma \left( s\right) \right) -y\left( \sigma \left( s\right) \right) \right| ds, \end{eqnarray*} (where $\alpha ,$ $\beta \in \lbrack 0,1)$, $k\geq 0$, and $\nu ,$ $\sigma :\left[ 0,T\right] \rightarrow \left[ 0,T\right] $ are continuous functions such that $\nu \left( t\right) \leq t,$ $\sigma \left( t\right)\leq t,$ $\forall t\in \left[ 0,T\right] $) has exactly one fixed point in $M $. Then the result is extended in $C\left( \mathbb{R}_{+},X\right) ,$ where $\mathbb{R}_{+}:=[0,\infty ).$