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 $f_i(x) = \sum_{k=1}^n \sin{(c_{ik} x_k)} + b_i$
Best Answer
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.