Let $$ f:\:[0,1]\rightarrow\mathbb{R};\\ \Phi: \mathcal{F}([0,1], \mathbb{R}) \rightarrow \mathcal{F}([0,1], \mathbb{R}),\\ f\mapsto\Phi(f):=Φf := \left(x\mapsto\left\{ \begin{array}{l@{\quad:\quad}l} \frac{1}{2}f(3x) & 0 \le x \le \frac{1}{3} \\ \frac{1}{2}f(1) & \frac{1}{3} \le x \le \frac{2}{3} \\ \frac{1}{2}f(3(x-\frac{2}{3}))+\frac{1}{2}f(1) & \frac{2}{3} \le x \le 1 \end{array}\right.\right) $$ Where $\mathcal{F}(A, B)$ is the set of all functions from $A$ to $B$.
Since the actual assignment is to show that $(\Phi^n(\mathrm{id}))_n$ converges uniformly to a continuous function $g$, I have to find a metric $d$ such that $Φ$ becomes a contraction, or equivalently: $$∃k\in[0,1):\: ∀f,g\in\mathcal{F}([0,1],\mathbb{R}): d(Φf,Φg)≤k\:d(f,g)$$.
I assumed that this should be the case for the uniform norm, so i looked at $[0,1]=[0,\frac{1}{3}]\cup[\frac{1}{3},\frac{2}{3}]\cup[\frac{2}{3},1]:=A\cup B\cup C$ separately:
$$ {∥Φf-Φg∥}_A = \sup\left\{\frac{1}{2}|f(3x)-g(3x)|,\:x\in A\right\} = \frac{1}{2}\sup\left\{|f(y)-g(y)|,\:y\in [0,1]\right\} = \frac{1}{2}{∥f-g∥} $$
$$ {∥Φf-Φg∥}_B = \sup\left\{\frac{1}{2}|f(1)-g(1)|,\:x\in B\right\} = \frac{1}{2}{∥f-g∥_{\{1\}}} ≤ \frac{1}{2}{∥f-g∥} $$
$$ {∥Φf-Φg∥}_C = \sup\left\{\frac{1}{2}|f(3(x-\frac{2}{3}))-g(3(x-\frac{2}{3}))+f(1)-g(1)|,\:x\in C\right\} = \frac{1}{2}\sup\left\{|f(y)-g(y)+f(1)-g(1)|,\:y\in [0,1]\right\} ≤ \frac{1}{2}\sup\left\{|f(y)-g(y)|+|f(1)-g(1)|,\:y\in [0,1]\right\} = \frac{1}{2}(∥f-g∥ + ∥f-g∥_{\{1\}}) ≤ {∥f-g∥} $$
but the last part does not fit into what I'm trying to show, since $k$ is supposed to be $≤1$.
as far as I can see, there are two possibilities:
- The upper estimation of the limit was too big, there is a smaller one leading to a valid $k$.
- The uniform norm does not make $Φ$ a contraction, I have to find another metric.
Which one of these possibilities holds? If it is the latter, what other metric should I seek for?
Let $X$ be the set of $f:[0,1]\to[0,1]$ such that $f(1)=1$. Then $X$ is complete in the uniform norm, the identity function is in $X$, and it's easy to show that $\Phi$ is a strict contraction on $X$.