Banach fixed-point theorem for a recursive functional equation

Question

I was asked to prove that the functional equation

$$ x(t) = \begin{cases} \frac{1}{2} x(3t) + \frac{1}{2},& 0 \leq t \leq \frac{1}{3}\\\ f(t), & \frac{1}{3} < t\leq \frac{2}{3}\\\ \frac{1}{2}x(3t - 2) - \frac{1}{2},&\frac{2}{3} < t \leq 1 \end{cases} $$

(where $f(t) $ is the equation of the straight line that passes through $(\frac{1}{3}, \frac{1}{2}x(1) + \frac{1}{2})$ and $(\frac{2}{3}, \frac{1}{2} x(0) - \frac{1}{2})$) has exactly one solution in $C([0, 1])$ using the Banach Fixed Point Theorem. As you can see, $x(t)$ is recursive for some $t$, e.g. $\frac{1}{2} x(0) - \frac{1}{2}$ is $\frac{1}{2} (\frac{1}{2} (\frac{1}{2} [...] + \frac{1}{2}) + \frac{1}{2}) - \frac{1}{2}$.

I know that the metric in $C([0, 1])$ may be defined as

$$ d_{1}(x, y) = \int_{0}^{1} \left| x(t) - y(t) \right| dt, $$

and that I'll need to use it as per the statement of the theorem. However, I can't find a way to do that.

Any pointers would be greatly appreciated!

Answer 1

Idea:

Consider $C[0,1]$ with the supremum metric $d(x, y) = \sup_{t\in [0,1]} |x(t) - y(t)|$ and $F : C[0,1]\to C[0,1]$ be given by

$$(Fx) (t) = \begin{cases} \frac{1}{2} x(3t) + \frac{1}{2},& 0 \leq t \leq \frac{1}{3}\\\ f(t), & \frac{1}{3} < t\leq \frac{2}{3}\\\ \frac{1}{2}x(3t - 2) - \frac{1}{2},&\frac{2}{3} < t \leq 1 \end{cases} $$

If you can show that $F$ is a contraction, since $d$ is a complete metric, there is a unique $x\in C[0,1]$ so that $Fx = x$, which is what you want.