# Banach Fixed Point

I'm facing this problem about Banach fixed-point theorem. The theorem says that if a function is a contraction (Lipschitz) then has only one fixed point. $$**Q:** \ f(x,y)=(e^{-1-y}+\frac{x}{3},\ e^{-1-y}+\frac{y}{3}) \ \ \ , \ A=\{(x,y) \in \mathbb{R}^2:y\geqslant 0\} \ \ \ \ \ \\ Show \ that \ the \ restriction \ in \ A \ has \ one \ and \ only \ one \ fixed \ point$$ So, for this exercise I want to show that $f$ is a contraction, that is, showing $$d(f(x),f(y))\leqslant \lambda d(x,y),\ \lambda \in [0,1]$$ But my question starts here. How can I prove this?? A hint/tip would be great.

• Estimate a Lipschitz constant for $f$. – copper.hat Dec 27 '16 at 23:17
• What you wrote, by the way, is not the definition of a contraction, it is the definition of a nonexpansion. You need $\lambda$ to be strictly less than one for a contraction (i.e., $\lambda \in [0,1)$). – parsiad Dec 27 '16 at 23:31

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 $\|f(y)-f(x) \| \le \lambda \|y-x\|$.

• That is the solution it had to be, its correct. Thanks! Can you tell me some book or document that refers this method? I want to learn it better and train. – Numbermind Dec 28 '16 at 12:14
• It is a fairly standard result, most real analysis texts will discuss the fixed point/contraction mapping principle. For example, Marsden's, "Elementary classical analysis", Kolmogorov & Fomin, "Introductory Real Analysis". – copper.hat Dec 28 '16 at 19:50

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 $A$?

• How do you calculate ||DF|| of infinite? – Numbermind Dec 27 '16 at 23:48
• @MathScientist Are you familiar with this notation? You can google "sup norm" or "infinity norm." – 3-in-441 Dec 28 '16 at 3:52