I'm trying to solve this question.
Let $0\leq \lambda \lt 1$ and $f:X\subset\mathbb R\to \mathbb R$ a $\lambda$-contraction, i.e., $|f(x)-f(y)|\leq \lambda |x-y|$ for any $x,y \in X$. Prove that if $X$ contains a closed interval $[a-r,a+r]$ and $|f(a)-a|\leq (1-\lambda)r$, then $f$ has a fixed point in $[a-r,a+r]$.
In order to begin to solve this question, I'm trying to prove that this function is derivable, I need help, particularly in this part.
thanks a lot
If the contraction $f$ is defined all over $\mathbb{R}$, then by Banach fixed-point theorem $f$ will have a fixed point, which can be obtained as the limit point of $x_{0} = a$, $x_{i+1} = f(x_{i})$. (That is, $x_{n} = f^{n}(a)$.)
Now $f$ is only defined over $X$. But the second condition guarantees that the sequence $(x_i)$ is all contained in $[a-r,a+r]$, so you are done.
To see this, first prove by induction that $$ \lvert f^{n+1}(a) - f^{n}(a)\rvert \le \lambda^{n} (1-\lambda) r, $$ so that $$ \begin{align} \lvert f^{n+1}(a) - a\rvert &\le \lvert f^{n+1}(a) - f^{n}(a)\rvert + \lvert f^{n}(a) - f^{n-1}(a)\rvert + \dots + \lvert f(a) - a\rvert \\&\le (\lambda^{n} + \lambda^{n-1} + \dots + \lambda + 1) (1 - \lambda) r = (1 - \lambda^{n+1}) r < r. \end{align}$$