Question on the existence/uniqueness of the fixed point

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?

PS : If it helps, we may use the norm $$\|x\|=\max_{0\le t\le T}|x(t)|$$ for $$x\in E$$.
• What is the norm on $E\times E$? Also, what do you mean by "contraction constant"? May 20, 2022 at 16:39
• @IosifPinelis Thanks for the interest and quick interest. You may take whatever suits you better, e.g. $\|(x,y)\|:=\max(\|x\|,\|y\|)$ or $\|(x,y)\|:=a\|x\|+b\|y\|$ with $a,b>0$ May 20, 2022 at 16:47
• @IosifPinelis $\epsilon>0$ is a contraction constant if and only if $\|F(x,y)-F(x',y')\|\le \epsilon\|(x-x',y-y')\|$ everywhere, the same for $G$ and $(F,G_b)$ May 20, 2022 at 16:48
$$\newcommand\ep\epsilon$$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. $$\quad\Box$$