I will very thankful if you can give a hint for this exercise.
Let $p=\frac{49}{100}$ and $f_n:[0,1]\to [0,1]$ functions defined recursively by $f_0(x)=x$ and $$f_{n+1}(x)=\begin{cases}pf_n(2x)&0\leq x\leq\frac12,\\p+(1-p)f_n(2x-1)&\frac12\leq x\leq 1.\end{cases}$$
(a) Find $C_1>0$ and $\lambda \in (0,1)$ such that
$$\forall n,m\in\mathbb N,\forall x\in[0,1], (n (b) Find $C_2>0$ and $\alpha \in (0,1)$ such that
$$\forall n\in\mathbb N,\forall x_0,x_1\in[0,1], |f(x_0)-f(x_1)|\leq C_2|x_0-x_1|^{\alpha}).$$ (c) Using (a) and (b), show that there exists a continuous function $f:[0,1]\to [0,1]$ such that $$\forall x\in [0,1],\quad \lim_{n\to\infty}f_n(x)=f(x).$$ My solution This is what I got (a)
$$|f_1(x)-f_0(x)|=(1-2p)x,\quad 0\leq x\leq 1/2$$
$$|f_1(x)-f_0(x)|=(1-2p)(1-x),\quad 1/2\leq x\leq 1$$ so: $$|f_1(x)-f_0(x)|=\dfrac{1}{2}(1-2p)\leq (1-2p)\quad \forall x\in[0,1] $$ Now: $$|f_{n+1}(x)-f_n(x)|=p|f_n(2x)-f_{n-1}(2x)|\leq (1-p)|f_n(2x)-f_{n-1}(2x)|\quad 0\leq x\leq 1/2$$
$$|f_{n+1}(x)-f_n(x)|=(1-p)|f_n(2x-1)-f_{n-1}(2x-1)|\quad 1/2\leq x\leq 1$$ Therefore: $$|f_{n+1}-f_n|\leq (1-p)|f_{n}-f_{n-1}|$$ Using this and the triangular inequality I get $$|f_{n+1}-f_n|\leq (1-2p)(1-p)^n$$ and $$|f_m-f_n|\leq (1-2p)\displaystyle\sum_{k=n}^{m-1}{}(1-p)^k=\dfrac{(1-2p)((1-p)^n-(1-p)^m)}{p}\leq \dfrac{(1-2p)(1-p)^n}{p}$$ Taking $C_1=\dfrac{1-2p}{p}$ and $\lambda=(1-p)$ i get (a). Is it ok?