Question on the existence/uniqueness of the fixed point

Question

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 $x\in E$.

Answer 1

$\newcommand\ep\epsilon$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. $\quad\Box$