<s>[INST] There exists a unique $f \in A$ such that $Tf(x) = f(x)$ Let $A$ be a subset of $C([0,1])$ such that $A = \{ f \in C([0,1]) \ | \ f(0)=0, \ f(1)=1\}$. \
Let
\begin{align*}
Tf(x)=
\begin{cases}
\frac{3}{4}f(3x), \ 0 \leq x \leq \frac{1}{3} \\
\frac{1}{2}f(2-3x) + \frac{1}{4}, \ \frac{1}{3} < x \leq \frac{2}{3} \\
\frac{3}{4}f(3x-2) + \frac{1}{4}, \ \frac{2}{3} < x \leq 1 \\
\end{cases}
\end{align*}
I want to find an $f \in A$ such that $Tf(x)=f(x)$ and show that its nowhere differentiable, but I cant seem to find a function $f$ that agrees with $Tf(x)$..
Thanks in advance!
 [/INST] Here's a hint to get you started.
For $Tf=f$ to hold, we need to satisfy three functional equations. On the first interval we need to satisfy $\frac{3}{4}f(3x)=f(x)$, on the second interval we need to satisfy $\frac{1}{2}f(2-3x) + \frac{1}{4}$ and on the third we need to satisfy $f(x)=\frac{3}{4}f(3x-2) + \frac{1}{4}$. You can then compose these equations over and over again to get a recursive form that lets you evaluate the function at a wide variety of points. Have you dealt with functions that are nowhere differentiable in the past? This is similar to the construction of other continuous, nowhere-differentiable, fractal-like functions.
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