How can we show that $(x_n)_n$ is a convergent sequence?

Question

Let $f:\Bbb R^2\to\Bbb R^2$ be a contraction. Fix some $x_1$ in $\Bbb R^n$ and define the sequence $(x_n)_n$ by $x_{n+1} = f(x_n)$ for all $n\in\Bbb N$. Show that $(x_n)_n$ is a convergent sequence.

Now, this is what I understand so far. Through the Contraction Mapping Theorem, f has a unique fixed point. Let that be $x^*$. We have that $d(f(x_n), x^*) = d(x_{n+1}, x^*)\leqslant\lambda d(x_n, x^*)\leqslant\lambda^2 d(x_{n-1},x^*)\leqslant\ldots\leqslant\lambda^nd(x_1,x^*)\,.$

We can fix $N$ after which $\lambda^Nd(x_{n-(N-1)},x^*)<\varepsilon$.
I am not sure on this last part. Is it right and how do we proceed after this?

Answer 1

So far you have proved that $$ d(x_n,x^*)\leq \lambda^n d(x_1,x^*), $$ with $0<\lambda<1$.

To proof the convergence you need to show that for $\varepsilon >0$ there exists $n_0$ such that $n\geq n_0$ implies $d(x_n,x^*)<\varepsilon$.

Consider any $\varepsilon>0$, since $0<\lambda<1$ there exists $n_0$ such that $$ \lambda^{n_0}<\frac{\epsilon}{d(x_1,x^*)} $$ (since $\{\lambda^n\}$ converges to 0).

So, therefore, if $n>n_0$ then $\lambda^n<\lambda^{n_0}$ and $$ d(x_n,x^*)\leq \lambda^n d(x_1,x^*)<\lambda^{n_0}d(x_1,x^*)<\varepsilon, $$ as desired.