Banach Fixed Point

Question

I'm facing this problem with Banach's fixed-point theorem. This theorem says that if a self-map is a contraction on a complete metric space, then it has exactly one fixed point.

I have a problem with the following exercise. Given $$ f(x,y) = (e^{-1-y}+\frac{x}{3},\ e^{-1-y}+\frac{y}{3}),\\ A = \{(x,y) \in \mathbb{R}^2:y\geqslant 0\} $$ Show that the restriction of $f$ in A has one and only one fixed point.

So, for this exercise, I want to show that $f|_A$ is a contraction. That is, there exists $\lambda \in [0,1)$ such that $$ \displaystyle\mathop{\forall}_{x,y \in A} d(f(x),f(y))\leqslant \lambda d(x,y). $$ But my question starts here. How can I prove this?? A hint/tip would be great.

Answer 1

Note that $f(x) \in A$ for all $x \in A$.

Note that $Df(x) = \begin{bmatrix} {1 \over 3} & - e^{-(1+x_2)} \\ 0 & {1 \over 3} - e^{-(1+y)} \end{bmatrix} = {1 \over 3} I - e^{-(1+x_2)} e e_2^T$, where $e= (1,1)^T$.

If we use the induced 2 norm, we have $\|Df(x)\| \le {1 \over3} + {1 \over e} \|e e_2^T \| = {1 \over3} + {\sqrt{2} \over e} <1$. Let $\lambda = {1 \over3} + {\sqrt{2} \over e}$.

To show that $f$ is a contraction, we use the following result: If $a^T b \le M \|a\|$ for all $z$, then $\|b\| \le M$.

Pick $a$, then for some $t$, the mean value theorem gives $a^T(f(y)-f(x)) = a^T Df(x+t(y-x))(y-x) \le \lambda a^T \|y-x\|$, and so we have $\|f(y)-f(x) \| \le \lambda \|y-x\|$.

Answer 2

Hint: Let $w\equiv(x,y)$ for brevity. We know that, by a generalized form of the mean value theorem, $$ \left\Vert f(w)-f(w^{\prime})\right\Vert _{\infty}\leq\left\Vert w-w^{\prime}\right\Vert _{\infty}\sup_{t\in(0,1)}\left\Vert Df(\left(1-t\right)w+tw^{\prime})\right\Vert _{\infty} $$ where $Df$ is the Jacobian of $f$.

Can you show that $\left\Vert Df\right\Vert _{\infty}$ is strictly less than one on the region $A$?