Set of all contraction maps

Question

Consider $(X, d)$ be a compact metric space, and let $Con(X)$ denote the set of all contraction maps on $X$. We shall define the “distance” between two maps $f, g ∈ Con(X)$ as follows,

$$d_{Con(X)}(f, g) = \sup_{x\in X}{d(f (x), g(x)).}$$

How can we show that:

$$d(y_f , y_g) ≤\frac{1}{1 − \min(c_f , c_g)} d_{Con(X)}(f, g)$$

where $y_f$ and $y_g$ are the fixed points of $f$ and $g$, respectively, and $c_f$ and $c_g$ are their respective contraction factors.

I myself think that it would be shown only by implementing triangle inequality and the fact that both $f$ and $g$ are contraction mapping.

Thanks for your help in advance

Answer 1

WLOG suppose that $c_g=\min(c_f,c_g)$. Let $x$ be any point in $X$, we build two sequences

1. $f_0=x$, $f_{n+1}=f(f_n)$, and
2. $g_0=x$, $g_{n+1}=g(g_n)$.

It follows that $f_n\to y_f$ and $g_n\to y_g$. Now $$\begin{aligned} d(f_{n+1},g_{n+1})&=d(f(f_n),g(g_n))\\ &\le d(f(f_n),g(f_n))+d(g(f_n),g(g_n))\\ &\le d_{Con}(f,g)+c_gd(f_n,g_n). \end{aligned}$$ Let $n\to \infty$, we have $$d(y_f,y_g) = \lim d(f_{n+1},g_{n+1})\le d_{Con}(f,g)+c_g\lim d(f_n, g_n)=d_{Con}(f,g)+c_gd(y_f,y_g).$$ Since $c_g=\min(c_f,c_g)$, we are done.