# Contraction on a metric.

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?

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).

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 $f'(x)=1-\frac{1}{x^2} \Rightarrow |f'(x)| < 1$.

• To be honest, I still don't see how the inequality follows from the mean value theorem. – Rafa Fafa Jun 28 '15 at 8:23
• i made an edit, maybe it helps. could you specify what exactly you don't understand? – supinf Jun 28 '15 at 8:30