Contractions and finding Fixed Points Define a map D on $C[0,1]$ by:
$$D(f(x)) = \begin{cases}
  \frac 23 + \frac 13 f(3x), & \text{if }1 \le x \le \frac 13 , \\
  (2+f(1)) (\frac 23 - x), & \text{if }\frac 13 \le x \le \frac 23, \\
x - \frac 23, & \text {if } \frac 23 \le x \le 1
\end{cases}$$
Show that $D$ is a contraction and determine the fixed point for D
My attempt:
$$|T(f(x)−Tg(x)|=|(\frac 23 + \frac 13 f(3x)− \frac 23+ \frac 13 g(3x)|= \frac 13 |f(3x)−g(3x)| ≤ \frac 13|f−g|_\infty$$
Then, $||Tf−Tg||∞≤ \frac 13||f−g||$
$$|Tf(x)−Tg(x)|=|(2+f(1))(\frac 23−x)−(2+g(1))(\frac 23−x)|=|(\frac 23−x)−(f(1)−g(1)|=|\frac {(2−3x)}3||f(1)−g(1)≤ \frac 13||f−g|_\infty$$
But I got stuck with the last one, here is my attempt for that
$$|Tf(x)−Tg(x)|=|x− \frac 23−(x− \frac 23)|=|2x|$$
I don't know how to show the above equation is a contraction, and determine the fixed point for $D$.
EDIT $$|Tf(x)−Tg(x)|=|x− \frac 23−(x− \frac 23)|= 0 $$
 A: It looks to me like you've successfully shown that your transformation is a contraction. To find the fixed function, we can simply iterate the iteration a few times to see that the graph will look something like so:

Thus, your answer should be defined piecewise with different definitions on the intervals of the form 
$$\frac{1}{3^n} \leq x \leq \frac{2}{3^n} \: \: \text{ and } \: \:
  \frac{2}{3^{n+1}} \leq x \leq \frac{1}{3^n}.$$
It's pretty easy to see that this function is fixed by your transformation.  The last part of your transformation (over the interval $2/3\leq x \leq 1$) implies that $T(f)(x)=x-2/3$ for any function at all and the second part (over the interval $1/3\leq x \leq 2/3$) implies that $T(f)(x) = 7(2 - 3 x)/9$ for any function satisfying $f(1)=1/3$. The graph above certainly satisfies both of these. The first portion of your transformation implies that the graph of $f$ over the interval $0\leq x \leq 1/3$ should be a scaled copy of the entire graph by the factor $1/3$.  Again, the graph above appears to satisfy this:

Finally, note that application of this self-similarity to the piecewise definition of the fixed function over the intervals $1/3 \leq x \leq 2/3$ and $2/3 \leq x \leq 1$ should yield the definition of the fixed function over the general intervals of the form
$$\frac{1}{3^n} \leq x \leq \frac{2}{3^n} \: \: \text{ and } \: \:
  \frac{2}{3^{n+1}} \leq x \leq \frac{1}{3^n}.$$
