Prove that a function is contractive

Question

I'm stuck with the following. I need to prove that in $D:=[0,1]\times[0,1]$ the function $F$ is contractive, where $F:\mathbb{R}^2\rightarrow\mathbb{R}^2$ is defined as:

\begin{align} F(x,y):=(\frac{1}{2} e^{-x}+\frac{y}{2},\frac{1}{3} e^{-y}+\frac{x}{3}) \end{align}

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Actually the problem is to prove: $\exists !_{(x^*,y^*)\in D}: F(x^*,y^*)=(x^*,y^*)$ using Banach fixed point theorem. First of all $D$ must be closed and $F(D)\subset D$, and both are true in this case. But I have to prove $F$ is contractive to use Banach fixed point theorem. I tried Mean Value theorem in both components $F_1(x,y)$ and $F_2(x,y)$ but that didn't help.

I would love to see methods for proving contraction in general.

Thanks!

Answer 1

For all $(x,y)\in D$ and all $(h,k)\in\mathbb{R}^2$, one has: $$\mathrm{d}_{(x,y)}F\cdot(h,k)=\left(\frac{1}{2}(k-e^{-x}h),\frac{1}{3}(h-e^{-y}k)\right).$$ Let $\|\cdot\|$ be the operator norm associated with $\|\cdot\|_1$ on $\mathbb{R}^2$, where $\|(x,y)\|_1:=|x|+|y|$. Let me remind you, that by definition, one has: $$\forall u\in\mathcal{L}(\mathbb{R}^2),\|u\|=\sup_{x\in\mathbb{R}^2\setminus\{0\}}\frac{\|u(x)\|_1}{\|x\|_1}.$$ Therefore, if $\|(h,k)\|_1=1$, one gets: $$\begin{align}\|\mathrm{d}_{(x,y)}F\cdot(h,k)\|_1&\leqslant\frac{1}{2}|k-e^{-x}h|+\frac{1}{3}|h-e^{-y}k|\\&\leqslant\frac{1}{2}(|k|+e^{-x}|h|)+\frac{1}{3}(|h|+e^{-y}|k|)\\&\leqslant\left(\frac{1}{2}+\frac{1}{3}\right)(|h|+|k|)\\&=\frac{5}{6}.\end{align}$$ Hence, one has: $$\|\mathrm{d}_{(x,y)}F\|\leqslant\frac{5}{6}<1.$$ Finally, $F$ is strictly contractive using mean value inequality.

Answer 2

The Jacobian matrix of $F$ is given by $$\begin{pmatrix}-\frac{1}{2}e^{-x} & \frac{1}{2}\\ \frac{1}{3} & -\frac{1}{3}e^{-y} \end{pmatrix}$$ where the absolute value of the largest eigenvalue in absolute value (sorry for the pun) is $\leq \frac{1+e^{-x}}{2}$ by the Gershgorin circle theorem. It follows that $F$ is a contraction and has a unique fixed point over the compact set $[0,1]\times [0,1]$ by the Banach(-Caccioppoli) theorem.

As pointed by C.Falcon, it is probably easier to show that $F$ is a strong contraction with respect to the $\|\cdot\|_1$ norm by exploiting the Lipschitz-continuity of $g(t)=e^{-t}$ over $[0,1]$.

Numerically, the fixed point is close to $(0.499336,0.391739)$.