2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Estimate a Lipschitz constant for $f$. $\endgroup$ – copper.hat Dec 27 '16 at 23:17
  • 1
    $\begingroup$ 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., $\lambda \in [0,1)$). $\endgroup$ – parsiad Dec 27 '16 at 23:31
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Numbermind Dec 28 '16 at 12:14
  • $\begingroup$ 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". $\endgroup$ – copper.hat Dec 28 '16 at 19:50
1
$\begingroup$

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 $A$?

$\endgroup$
  • $\begingroup$ How do you calculate ||DF|| of infinite? $\endgroup$ – Numbermind Dec 27 '16 at 23:48
  • 1
    $\begingroup$ @MathScientist Are you familiar with this notation? You can google "sup norm" or "infinity norm." $\endgroup$ – 3-in-441 Dec 28 '16 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.