Let $E$ a Banach space ($E$ is the space of continuous functions on $[0,T]$ for my case). Let $F, G: E\times E\to E$ be contraction maps of contraction constant $\epsilon>0$. Given $b\in\mathbb R$, consider the map


\begin{eqnarray}
(F,G\_b) : E\times E &\to& E\times E \\
 (x,y) &\mapsto& (F(x,y),G(x,y)+bx).
\end{eqnarray}


For any $b$, can we always find $\epsilon>0$ small enough s.t.


1. $(F,G\_b)$ admits a fixed point?
2. $(F,G\_b)$ admits a unique fixed point?
3. If not, do we have some conditions to ensure the existence/uniqueness?


Any answer, comments and references are highly appreciated!


PS : If it helps, we may use the norm $\|x\|=\max\_{0\le t\le T}|x(t)|$ for $x\in E$.



Question on the existence/uniqueness of the fixed point
$\newcommand\ep\epsilon$Yes, $H:=(F,G\_b)$ has a unique fixed point, for each real $b$. Indeed, let
$$\ep:=1/2$$
and then take any
$$a\in\Big(0,\frac\ep{\ep+|b|}\Big).$$
Let
$$\|(x,y)\|:=\|x\|+a\|y\|$$
for $(x,y)\in E\times E$. Then for all $(u,v)$ and $(x,y)$ in $E\times E$ we have
$$\begin{aligned}
\|H(u,v)-H(x,y)\|&=\|F(u,v)-F(x,y)\|+a\|G\_b(u,v)-G\_b(x,y)\| \\
&\le\ep\|(u,v)-(x,y)\|+a\ep\|(u,v)-(x,y)\|+a|b|\,\|u-x\| \\ 
&=(\ep+a\ep+a|b|)\|u-x\|+(\ep+a\ep)a\|v-y\| \\ 
&\le k(\|u-x\|+a\|v-y\|)=k \|(u,v)-(x,y)\|, 
\end{aligned}$$
where
$$k:=\max[\ep+a\ep+a|b|,\ep+a\ep]=\max[\ep+a(\ep+|b|),\ep+a\ep]\in(0,1).$$
So, $H$ is a contraction map and thus has a unique fixed point. $\quad\Box$


