We define the norm in $\mathbb{R^2}:$ by: $$\forall (x,y)\in\mathbb{R^2}, ||(x,y)||=|x|+|y|$$ Let $f$ be defined in $\mathbb{R^2}$ and $$f(x,y)=\left(\frac{1}{4}\sin(x+y),1+\frac{2}{3}\arctan(x-y)\right)$$ Prove $f(x,y)$ is a contraction in $(\mathbb{R^2},||.||)$
I try to use the mean value theorem in this problem but I don't know how since it's a multivariable fucntion.
From $$\left \| f(x_1,y_1) - f(x_2,y_2) \right \|=\\ \left \| \left(\frac{1}{4}\sin{(x_1+y_1)},1+\frac{2}{3}(\arctan{(x_1-y_1)}) \right)-\left( \frac{1}{4}\sin{(x_2+y_2)},1+\frac{2}{3}(\arctan{(x_2-y_2)}) \right) \right \|=\\ \left \| \left( \frac{1}{4}\left(\sin{(x_1+y_1)}-\sin{(x_2+y_2)}\right), \frac{2}{3}\left(\arctan{(x_1-y_1)}-\arctan{(x_2-y_2)}\right)\right) \right \|=\\ \left|\frac{1}{4}\left(\sin{(x_1+y_1)}-\sin{(x_2+y_2)}\right)\right|+\left|\frac{2}{3}\left(\arctan{(x_1-y_1)}-\arctan{(x_2-y_2)}\right)\right|=\\ \frac{1}{4}\left|\sin{(x_1+y_1)}-\sin{(x_2+y_2)}\right|+\frac{2}{3}\left|\arctan{(x_1-y_1)}-\arctan{(x_2-y_2)}\right|=...$$ Using mean value theorem and the facts that $|\cos(x)|\leq1$ and $\left|\frac{1}{1+x^2}\right|\leq1$: $$...=\frac{1}{4}\left|\cos(\varepsilon_1)\left((x_1+y_1)-(x_2+y_2)\right)\right|+\frac{2}{3}\left|\frac{1}{1+\varepsilon _2^2}\left((x_1-y_1)-(x_2-y_2)\right)\right| \leq \\ \frac{1}{4}\left|(x_1+y_1)-(x_2+y_2)\right|+\frac{2}{3}\left|(x_1-y_1)-(x_2-y_2)\right|\leq \\ \frac{1}{4}\left(\left|(x_1-x_2\right|+\left|y_1-y_2\right|\right)+\frac{2}{3}\left(\left|(x_1-x_2\right|+\left|y_2-y_1\right|\right)=\\ \left(\left|(x_1-x_2\right|+\left|y_1-y_2\right|\right)\left(\frac{1}{4}+\frac{2}{3}\right)=\frac{11}{12}\left \| (x_1-x_2,y_1-y_2) \right \|=\\ \frac{11}{12}\left \| (x_1,y_1)-(x_2,y_2) \right \|$$ As a result: $$\left \| f(x_1,y_1) - f(x_2,y_2) \right \| \leq \frac{11}{12}\left \| (x_1,y_1)-(x_2,y_2) \right \|$$