Contraction on a metric.

Question

Let $M=\{x\in \mathbb{R}|x\geq 1\}$ with the absolute metric, being a metric space. Show that:

a) The mapping $f:M\to M$ with

$f:M\to M,~f(x)=\frac{1}{x}+\frac{x}{2}$

is a contraction and has the fixed point $x_0=\sqrt{2}$.

b)The mapping $g:M\to M$ with

$g:M\to M,~g(x)=\frac{1}{x}+x$

fulfills the inequality

$d(g(x),g(y))<d(x,y)$ for all $x,y\in M$ with $x\neq y$,

but does not have a fixed point.

We talked about the Banach fixed-point theorem in class and although I thought I understood it at first, I really didn't once I tried myself to some exercises for fixed-points.

I know that the Banach theorem says that a map is a contraction map if I can find a $q\in[0,1)$, so that $d(T(x),T(y))<qd(x,y)$. But I don't have to approach both of these tasks with that definition. Basically I don't know how to apply this to solve this problem.

Can anyone more knowledgeable with the Banach theorem help me out here?

Answer 1

No need for Banach's theorem itself, this is an exercise that illustrates this theorem.

For the first, you need to show it is a contraction, which is just a problem showing an inequality, e.g. using the mean value theorem as supinf suggests.

The mean value theorem says that for fixed $x < y$ we have some $x < z < y$ such that $f(x) - f(y) = f'(z)(x - y)$ (as $f$ is continuously differentiable on its domain). So $\left|f(x) - f(y)\right| = \left|f'(z)\right|\left|x-y \right|$, taking absolute values, and as $|f'(z)| \le \frac{1}{2}$ on all of $M$ you can take $q = \frac{1}{2}$.

In that case you know that Banach's theorem garantuees that there is a unique fixed point, and you are asked to show this in fact is $\sqrt{2}$. So substitute $x = \sqrt{2}$ in $f(x)$ and show it equals $\sqrt{2}$ again, or take the harder way and solve $f(x) = x = \frac{1}{x} + \frac{x}{2}$. The latter implies $\frac{x}{2} = \frac{1}{x}$, and cross-multiplying gives $x^2 = 2$, which has two solutions, only one of which lies in $M$.

For the second part, $g$ obeys the stated inequality, which can be shown in a similar way, and trying to solve $g(x) = x$ we get $x = x + \frac{1}{x}$ or $\frac{1}{x} = 0$. The latter is impossible so no fixed point exists.

The last serves to show that we really need some fixed $q$ in the contraction definition, or else the theorem fails ($M$ is complete but there is no fixed point for the "weak contraction" $g$, as opposed to the real contraction $f$ from before).

Answer 2

Hint:

a typical way to apply this for real functions is to use the mean value theorem: $\forall x,y \exists z \in (x,y): f(x)-f(y)=f'(z)(x-y)$. It follows that $|f(x)-f(y)| = |f'(z)| |x-y|$. Calculating the derivatives in that cases should be easy, so this helps you to proove an inequality of the type $|f(x)-f(y)| < q |x-y|$ for constant $q$.

e.g. for (a) you have $f'(x)=\frac{1}{2} - \frac{1}{x^2}$ and for $x> 1$ we have $|f'(x)| < \frac{1}{2}$. for (b) you have $f'(x)=1-\frac{1}{x^2} \Rightarrow |f'(x)| < 1$.