How to prove $F$ is a contraction mapping?

Question

Given a $n\times n$ matrix $\mathbf{D}$ in which each entry $\mathbf{D}_{ij} \in [0,1]$. Let $i$-th row vector of $\mathbf{D}$ denote by $\mathbf{D}_{i\ast}$. The mapping $F:[0,1]^n \mapsto [0,1]^n$, defined as $$F(\mathbf{x})=[f_1(\mathbf{x}),f_2(\mathbf{x}),\cdots,f_n(\mathbf{x})]^T$$ where $$f_i(\mathbf{x})=\frac{1}{1+\mathbf{D}_{i\ast}\cdot \mathbf{x}}$$

Now, how can I prove $F$ is a contraction mapping? Please help me!

Answer 1

Hints:

Let's use the max norm: $|| x||=\max |x_i|$.

The $(i,j)$ entry of the Jacobian: $Jf(x)_{ij} =-\dfrac{d_{ij}}{(1+D_{i*}x)^2}$.

Now $\forall y$ such that $||y||=1$, we have $|| Jf(x) \cdot y ||\le \max_{i,j} d_{ij}<1$. Why? The are gaps here for you to fill. You need to use the fact $x\in [0,1]^n$.

So the norm of the Jacobian $||Jf(x)||:=\sup\limits_{y\,:\,||y||=1} || Jf(x) \cdot y ||<1$.

Remark: $x$ is an arbitrary point in the domain (not necessarily a fixed point).

Hint: To finish the proof, apply the mean value theorem...