Why is $\Phi$ contracting on $D$ here? I was wondering why the function $\phi(x,y)=(\cos(\frac{1}{8}xy),\sin(\frac{1}{6}(x+y)))$ is contracting (which means that $d(\Phi(x,y),\Phi(u,v))\leq L\cdot d((x,y),(u,v))$ with $L\in[0,1)$) on $D:=[-1,1]\times[-1,1]$, where $(D,d)$ is a complete metrical space. I don't know which value I should take for $L$ and why. Can someone help me?
 A: $\newcommand{\R}{\mathbb R}$Let $f\colon\R^2\to\R^2$ be any continuously differentiable function. Then for any $z$ and $w$ in $\R^2$
\begin{equation*}
    f(w)-f(z)=\int_0^1 da\,f'(z+a(w-z))(w-z)
\end{equation*}
and hence
\begin{equation*}
    |f(w)-f(z)|\le\int_0^1 da\,\|f'(z+a(w-z))\|\,|w-z| \\ 
    \le|w-z|\max_{a\in[0,1]}\|f'(z+a(w-z))\|, \tag{1}\label{1}
\end{equation*}
where $|\cdot|$ is any norm on $\R^2$ and $\|\cdot\|$ is the corresponding operator norm.
Let now $|z|:=\max(|z_1|,|z_2|)$ for $z=(z_1,z_2)\in\R^2$. If the matrix of a linear operator $T\colon\R^2\to\R^2$ in the standard basis of $\R^2$ is $(c_{ij}\colon i,j=1,2)$, then $\|T\|=\max_{i=1,2}\sum_{j=1}^2|c_{ij}|$.
Note that $\phi=g\circ f$, where $f(x,y):=(\frac{xy}8,\frac{x+y}6)$ for $(x,y)\in D=[-1,1]^2$ and $g(s,t):=(\cos s,\sin t)$. For $(x,y)\in D=[-1,1]^2$, the matrix of the linear operator $f'(x,y)$ is $\begin{pmatrix}y/8&x/8\\ 
1/6&1/6\end{pmatrix}$ and hence
\begin{equation*}
    \|f'(x,y)\|\le\max(\tfrac{|y|}8+\tfrac{|x|}8,\tfrac16+\tfrac16)=\tfrac13. 
\end{equation*}
So, by \eqref{1}, the function $f$ is $\tfrac13$-Lipschitz, that is, Lipschitz with Lipschitz constant $\tfrac13$.
Similarly, for all $(s,t)\in\R^2$,
\begin{equation*}
    \|g'(s,t)\|\le\max(|-\sin s|,|\cos t|)\le1, 
\end{equation*}
so that the function $g$ is $1$-Lipschitz.
The composition $\phi$ of Lipschitz functions $f$ and $g$ is a Lipschitz function with the Lipschitz constant equal to the product of the Lipschitz constants of $f$ and $g$.
Thus, the function $\phi$ is $\tfrac13$-Lipschitz.
