Let $X=\{f\in C[0,1]|\ f(0)=0,f(1)=1, f([0,1])\subset [0,1]\}$. I have proved $X$ is closed subset of the complete metric space $C[0,1]$. Hence, $X$ is complete.
Define $T:X\to X$ by $$\displaystyle{(Tf)(x)= \begin{cases} \frac{3}{4}f(3x)&\text{if } x\in\left[0,\frac{1}{3}\right]\\ \frac{1}{4}+\frac{1}{2}f(2-3x)&\text{if } x\in\left[\frac{1}{3},\frac{2}{3}\right]\\ \frac{1}{4}+\frac{3}{4}f(3x-2)&\text{if } x\in\left[\frac{2}{3},0\right]\\ \end{cases}}$$
I have proved that $T(f)\in X\ \forall f\in X$ and $\lVert T(f)-T(g)\rVert_\infty\le\frac{3}{4}\lVert f-g\rVert_\infty$, hence $T$ is strict contraction map.
By Banach Fixed Point Theorem, $\exists h\in X$ such that $T(h)=h$. By hypothesis, $h$ is continuous on $[0,1]$. Our target is to show that $h$ is nowhere differentiable.
Let $a\in[0,1]$, then $\forall n\in\mathbf{N}$, $a3^n\in[0,3^n]$.
We know that there is $k_n\in\{1,2,\ldots,3^n\}$ such that $k_n-1< a3^n\le k_n\implies \frac{k_n-1}{3^n}< a\le\frac{k_n}{3^n}$.
Let $\displaystyle{t_n\in\left[\frac{k_n-1}{3^n},\frac{k_n}{3^n}\right]}$ such that $$ \text{sup}\left\{|h(t)-h(a)|:\ t\in\left[\frac{k_n-1}{3^n},\frac{k_n}{3^n}\right]\right\}=|h(t_n)-h(a)| $$
Now observe that $\displaystyle{\frac{k_n-1}{3^n}< a\le\frac{k_n}{3^n}\implies -\frac{1}{3^n}< a-\frac{k_n}{3^n}\le 0}$, then by sandwich theorem we have $\displaystyle{\lim\limits_{n\to\infty}\frac{k_n}{3^n}=a}$.
Again we have $\displaystyle{\frac{k_n-1}{3^n}\le t_n\le \frac{k_n}{3^n}}$, so $\displaystyle{\lim\limits_{n\to\infty}t_n=a}$.
Now we have to show $\displaystyle{\frac{|h(t_n)-h(a)|}{|t_n-a|}\ge \frac{1}{2}\left(\frac{3}{2}\right)^n\ \forall n\in\mathbb{N}}$, so that we can conclude that $\displaystyle{\frac{|h(t_n)-h(a)|}{|t_n-a|}}$ is an unbounded sequence.
I'm stuck at this step. How to prove the above inequality? I think induction on $n$ will work, but I'm unable to prove it.
It is also okay if anyone show $\displaystyle{\frac{|h(t_n)-h(a)|}{|t_n-a|}}$ is unbounded without exactly showing the above inequality.
Can anyone help me in this regard? Thanks for help in advance.