= Solution
The <contraction mapping theorem> states that a map $f$ of a nonempty <complete metric space> into itself, satisfying $d(fx,fy)\le qd(x,y)$ for some $0\le q<1$, has a unique <fixed point>. Iteration from any point converges to it; for example $d(f^nx,\tau(f))\le q^n d(x,fx)/(1-q)$.
Assume $X$ is nonempty. Its boundedness makes $\delta$ finite. Symmetry, positivity and separation follow pointwise from $d$, and taking the supremum of $d(fx,hx)\le d(fx,gx)+d(gx,hx)$ proves the triangle inequality. Hence $\delta$ is a <uniform metric>.
If $(f_n)$ is a <Cauchy sequence> in this <uniform metric>, each $(f_n(x))$ converges by completeness of $X$; denote its limit by $f(x)$. Given $\epsilon>0$, choose $N$ so that $\delta(f_n,f_m)<\epsilon$ for $n,m\ge N$. Letting $m\to\infty$ gives $d(f_n(x),f(x))\le\epsilon$ for all $x$, so $f_n\to f$ uniformly. To prove continuity, fix $x$ and use
$$
d(f(y),f(x))\le d(f(y),f_N(y))+d(f_N(y),f_N(x))+d(f_N(x),f(x)).
$$
Choose $N$ so the outside terms are small, then use continuity of $f_N$ for the middle term. Thus $f\in\operatorname{Maps}(X,X)$, proving \b[completeness] of the continuous-map space in the <uniform metric>.
The contraction subspace is \b[not necessarily complete]. On $X=[0,1]$, $f_n(x)=(1-1/n)x$, $n\ge2$, are <contraction mappings> and converge to the identity by <uniform convergence>. Their only possible uniform limit is not a contraction, so this <Cauchy sequence> has no limit in $\mathcal C$.
Nevertheless the <fixed point> assignment is continuous. Fix $f\in\mathcal C$ with contraction constant $q_f<1$, and put $u=\tau(f)$, $v=\tau(g)$. Then
$$
d(u,v)=d(fu,gv)\le d(fu,fv)+d(fv,gv)\le q_fd(u,v)+\delta(f,g).
$$
Consequently \b[$d(\tau(f),\tau(g))\le\delta(f,g)/(1-q_f)$]. This proves <continuity of the fixed-point assignment> at $f$ without requiring one contraction constant valid for every $g$.
Back to article page
