I'm really having trouble understanding how to apply Banach's fixed-point theorem in this exercise.
Let $b_i$ and $c_{ik}$ be real numbers with $1 \leq i,k \leq n$ such that the following equation holds $$ \sum_{i,k=1}^n c_{ik}^2 < 1 $$
Now I have to show the following nonlinear system has exactly one solution $(x_1^*, x_2^*, \dots, x_n^*)$ using the fixed-point theorem.
$$ x_i = \sum_{k=1}^n \sin{(c_{ik} x_k)} + b_i, \qquad 1 \leq i \leq n $$
So let's define a function $$ f: \mathbb{R}^n \to \mathbb{R}^n, x = (x_1, x_2, \dots, x_n) \mapsto f(x) = \begin{pmatrix}f_1(x_1, x_2, \dots, x_n) \\ f_2(x_1, x_2, \dots, x_n) \\ \vdots \\ f_n(x_1, x_2, \dots, x_n) \end{pmatrix} $$
$(X, d) = (\mathbb{R}^n, |\;|)$ is complete and $f$ is a self-map on $X$. Now I only need to show the contraction property holds, i.e. $$ d(f(x), f(y)) \leq qd(x,y), \quad q \in (0,1) $$
I think I can use the usual euclidean norm here but I'm not even sure how to write this out.
Right now I got this $$ \sum_{i=1}^n \left(f_i(x) - f_i(y)\right)^2 \leq q^2 \left(\sum_{i=1}^n \left(x_i-y_i\right)^2\right) $$
with $f_i(x) = \sum_{k=1}^n \sin{(c_{ik} x_k)} + b_i$
First, note that $\sin$ is a non-expansive mapping, due to its derivative being bounded by $1$. It therefore follows that, using Cauchy-Schwarz inequality, \begin{align*} |f_i(\vec{x}) - f_i(\vec{y})| &= \left|\sum_{k=1}^n(\sin(c_{ik}x_k) - \sin(c_{ik}y_k))\right| \\ &\le \sum_{k=1}^n|c_{ik}|\cdot |x_k - y_k| \\ &\le \sqrt{\sum_{k=1}^n c_{ik}^2}\cdot \sqrt{\sum_{k=1}^n |x_k - y_k|^2} \\ &= \|\vec{x} - \vec{y}\| \cdot \sqrt{\sum_{k=1}^n c_{ik}^2}. \end{align*} Therefore, $$\|f(\vec{x}) - f(\vec{y})\|^2 = \sum_{i=1}^n (f_i(\vec{x}) - f_i(\vec{y}))^2 \le \|\vec{x} - \vec{y}\|^2 \cdot \underbrace{\sum_{i=1}^n\sum_{k=1}^n c^2_{ik}}_{< 1},$$ proving the map is a Banach contraction.