I'm given the following problem:
Let $f: \mathbb{R}^N \rightarrow \mathbb{R}^N$ be a contraction with contraction-constant $\lambda \in [0, 1)$, i.e. we have for all $x, y \in \mathbb{R}^N$ that
$$||f(x) - f(y)|| \leq \lambda||x-y||.$$
Let $x_0 \in \mathbb{R}^N$ and $x_n = f(x_{n-1})$ for $n \in \mathbb{N}$. Let further $\bar x$ be a fixed point of $f$, i.e. $f(\bar x) = \bar x$. Show that we have
$$||x_n - \bar x|| \leq \frac{\lambda^n}{1-\lambda}||x_0 - \color{red}{n_1}||$$
The red $n_1$ is obviously a typo and I'm wondering what is actually to prove here.
- Should the $n_1$ be $\bar x$? In this case I'm wondering how to prove the statement. I'm able to prove $||x_n - \bar x|| \leq \lambda^n||x_0 - \bar x||$, but where does the denominator $1 - \lambda$ come from?
- Should the $n_1$ be $x_1$? In this case I'm wondering how to eliminate the $\bar x$ from the equation?
Hint: $$\|x_{n+m}-x_n\|\leq \|x_{n+m}-x_{n+m-1}\|+...+\|x_{n+1}-x_n\|$$ $$\leq (\lambda ^{m-n-1}+ \lambda ^{m-1}+...+1) \|x_{n+1}-x_n\| \leq (\lambda ^{n+m-1}+ \lambda ^{m-1}+...+\lambda ^{n}) \|x_1-x_0\|$$. Compute the geometric sum and use the fact that $x_{n+m} \to \overset {-} x$ as $ m\to \infty$ by Contraction Mapping Theorem.