Let $B_R(x_o)$ be a closed ball with center at $x_o$ and radius $R$ on a Banach Space $X$ with $F: X \to X$ being a contraction mapping with Lipschitz constant $L<1$ on the closed ball in question.
(i) Show that if $\|x_1 - x_o\| < (1-L)R $ then $F$ has a unique fixed point in the closed ball $B_R(x_o)$
(ii) Show the iterative sequence starting at $x_o$ converges to this fixed point of $F$ on $B_R(x_o)$
This is an interesting variation on the "normal" contraction mapping / fixed point theorem, and while I understand the proof of that one, this one has me sitting staring at a blank page for a long time...
Define $x_n:=F(x_{n-1})$ for $n\geq1$. I claim that $\|x_1-x_0\|<(1-L)R$ guarantees that $x_n\in B:=B_R(x_0)$ for all $n\geq0$.
Proof. Assume that $\{x_0,x_1,\ldots,x_{n-1}\}\subset B$ for some $n\geq2$. Then $$\|x_k-x_{k-1}\|\leq L\|x_{k-1}-x_{k-2}\|\leq\ldots\leq L^{k-1}\|x_1-x_0\|$$ is true for $1\leq k\leq n$. It follows that $$\|x_n-x_0\|\leq\sum_{k=1}^n\|x_k-x_{k-1}\|\leq\sum_{k=1}^n L^{k-1}\ \|x_1-x_0\|<{1\over 1-L}(1-L)R=R\ ,$$ hence $x_n\in B$ as well.
Since all $x_n\in B$ we now can show (I leave that to you) that $(x_n)_{n\geq0}$ is in fact a Cauchy sequence, hence convergent in $X$ to a point $x_*$, which necessarily lies in $B$ since $B$ is closed.