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The contraction mapping theorem states that a contraction mapping $f$ of a nonempty complete metric space into itself has a unique fixed point. More explicitly, if $d(f(x),f(y))\le qd(x,y)$ with $0\le q<1$, iteration from any $x_0$ converges to that fixed point.

To prove this, put $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$ the triangle inequality givesThus $(x_n)$ is a Cauchy sequence, and completeness supplies a limit $x_{\ast }$. A contraction mapping is continuous, so $f(x_{\ast })=\lim f(x_n)=\lim x_{n+1}=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing to the limit in the bound also gives the useful error estimate $d(x_n,x_{\ast })\le q^{n}d(x_1,x_0)/(1-q)$.

For the function space, take $X$ nonempty; otherwise the supremum in the printed definition is undefined without a convention. Boundedness of $X$ makes the uniform metric $\rho (f,g)$ finite. Nonnegativity and symmetry follow from $d$; $\rho (f,g)=0$ means $f(x)=g(x)$ at every $x$. Taking suprema in $d(f(x),h(x))\le d(f(x),g(x))+d(g(x),h(x))$ proves the triangle inequality. Hence $\rho$ is a metric.

Suppose $(f_n)$ is a Cauchy sequence for this uniform metric. At each $x$, $d(f_n(x),f_m(x))\le \rho (f_n,f_m)$, so completeness of $X$ supplies a value $f(x)={\lim }_n{f}_n(x)$. Given $\epsilon >0$, choose $N$ with $\rho (f_n,f_m)<\epsilon /2$ for all $n,m\ge N$. Fix $n\ge N$ and let $m\to \infty$ pointwise; then $d(f_n(x),f(x))\le \epsilon /2$ for every $x$. Thus $\rho (f_n,f)\le \epsilon /2<\epsilon$. The uniform limit theorem makes this limit continuous, so $f\in F$. This proves the completeness of the continuous-map space in the uniform metric.

Finally fix $f\in C$ with contraction constant $q_f<1$, and write $x_f=\theta (f)$ and $x_g=\theta (g)$. Using the fixed point equations,Therefore the fixed-point stability in the uniform metric estimate isChoosing $\rho (f,g)<(1-q_f)\epsilon$ proves continuity of $\theta$ at every $f$. The estimate uses the constant of the fixed map $f$ only; it does not require a common contraction constant for all nearby $g$.

The answers areWrite $M_n(f,g)={\sup }_{0\le t\le n}|f(t)-g(t)|$. The given metric describes locally uniform convergence: $d(f_j,f)\to 0$ if and only if $M_n(f_j,f)\to 0$ for every fixed $n$. Indeed,so convergence in $d$ forces $M_n\to 0$. Conversely, split the defining series at $N$: its tail is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$, and its finitely many initial terms tend to zero.

If $(f_j)$ is a Cauchy sequence in $d$, the same single-term estimate makes it a Cauchy sequence in the supremum norm on every $[0,n]$. The completeness of the continuous-map space in the uniform metric gives a continuous function $F_n$ on $[0,n]$ with $f_j\to F_n$ uniformly there. Uniqueness of the limit of a sequence gives $F_{n+1}{|}_{[0,n]}=F_n$, so these functions define a single continuous function $f$ on $[0,\infty )$. The preceding finite-head and tail argument proves $d(f_j,f)\to 0$. Thus the space is complete.

The constant functions $f_j(t)=j$ have $d(f_j,f_k)=1$ whenever $j\ne k$. They have no convergent subsequence, so the space is not sequentially compact.

For fixed $f,g$, set $\gamma (s)=(1-s)f+sg$. On $[0,n]$,as $s^{\prime }\to s$. The equivalence with locally uniform convergence proves that $\gamma$ is a continuous path for $d$. Hence the space is path-connected and connected.