How to prove that a given function vanish on the complement of the Cantor set? Let $E=\{f\in C([0,1],\Bbb R)\mid f(0)=0,~f(1)=1\}$ be the complete metric space   endowed  with the sup distance $\|\cdot\|_\infty$
Consider the operator, T defined for $f\in E$  as follows.
$$ Tf(x) = \begin{cases}
\frac{1}{2}f(3x)& x\in [0,\frac{1}{3}]\\
\frac{1}{2}& x\in [\frac{1}{3},\frac{2}{3}]\\
\frac{1}{2} (1+f(3x-2))& x\in [\frac{2}{3},1]
\end{cases}$$

  
*
  
*Prove that, T as a unique fix point. 
  
*Let $\phi\in E$ be fix point of $T$. Shows that, $\phi$ is differentiable and  $\phi'$ on the complement of the Cantor set

I was able to prove that, for all $f,g\in E$
$$\|T(f-g)\|_\infty\le \frac{1}{2}\|f-g\|_\infty$$
Then $T$ has a unique fix point. 

But don't how to prove that $\phi$ is differentiable and    $\phi'$ vanish on the complement of the Cantor set.

 A: The easiest way to answer the second question is to understand what the functional $T$ actually do on a function $f$. Let us analyze the three subintervals given by the definition of $T$. In the interval $[0,1/3]$ the argument $3x$ is monotonically increasing from $0$ to $1$, hence $\frac{1}{2}f(3x)$ is precisely a copy of $f$, but rescaled by a factor $\frac{1}{2}$ and sqeezed in the subinterval $[0,1/3]$. Then we have a constant $1/2$ on $[1/3,2/3]$. And finally again an argument $3x-2$ monotonically increasing from $0$ to $1$ on $[2/3,1]$, so that $\frac{1}{2}(1+f(3x-2))=\frac{1}{2}+\frac{1}{2}f(3x-2)$ is again a copy of $f$ rescaled by $1/2$, squeezed in the subinterval $[2/3,1]$ and lifted of $1/2$ (in this way $Tf\in E$).
So we said that we have: small copy of $f$, then $1/2$, then another small copy of $f$ lifted of $1/2$. Now take $\phi$ the fixed point of $T$, that is $T\phi=\phi$ (but also $T^n\phi=\phi$ for all $n\ge1$). From $T\phi=\phi$ we have that $\phi|_{[1/3,2/3]}=T\phi|_{[1/3,2/3]}=\frac{1}{2}$. Now let us study the interval $[0,1/3]$ first. For what we said above, if we take $T^2\phi=T(T\phi)$, the operator $T$ makes a small copy of $T\phi$ in $[0,1/3]$, hence $\phi|_{[1/9,2/9]}=\frac{1}{4}$. Taking $T^3\phi$, by the same argument we obtain that $ \phi|_{[1/27,2/27]}=\frac{1}{8}$. By continuing iterating in this way and by the fact that in $[2/3,1]$ the situation is simmetric, we obtain that $\phi$ is constant on the complement of the Cantor set. Then $\phi'=0$ on such complement.
However, the classical definition of Cantor function (https://en.wikipedia.org/wiki/Cantor_function) satisfies the fact of being a fixed point for $T$, then by uniqueness $\phi$ is exactly the Cantor function. So actually it is not differentiable, but almost everywhere differentiable, with derivative $\phi'$ equal to zero where it exists (that is the complement of the Cantor set).
