# Contraction Mapping Principle

• January 29th 2009, 05:44 PM
mylestone
Contraction Mapping Principle
Let M be a compact metric space. Let $\Phi : M \rightarrow M$ be such that $d(\Phi (x), \Phi (y)) < d(x, y), \forall x, y \in M, x \neq y$. Show that $\Phi$ has a unique fixed point.

I'd like to use the Contraction Mapping Principle. I can see that M is complete (as it is a compact metric space), but am not sure where to find a constant $c \in [0,1)$ such that $d(\Phi (x), \Phi (y)) \leq c \cdot d(x, y)$. Any advice?
• January 29th 2009, 07:22 PM
aliceinwonderland
Quote:

Originally Posted by mylestone
Let M be a compact metric space. Let $\Phi : M \rightarrow M$ be such that $d(\Phi (x), \Phi (y)) < d(x, y), \forall x, y \in M, x \neq y$. Show that $\Phi$ has a unique fixed point.

I'd like to use the Contraction Mapping Principle. I can see that M is complete (as it is a compact metric space), but am not sure where to find a constant $c \in [0,1)$ such that $d(\Phi (x), \Phi (y)) \leq c \cdot d(x, y)$. Any advice?

It seems like you don't need to find a "c" to show the existence of a fixed point of x such that $\Phi (x) = x$.

First, choose a point $x_{1}$ in M and define
$x_{2} = \Phi (x_{1}), x_{3} = \Phi (x_{2}),...,x_{n}= \Phi (x_{n-1})$ for $n \ge 2$.

You might need to mark some points in M (to figure out the points indeed converge) and make sure each $d(x_{k-1}, x_{k}) > d(\Phi (x_{k-1}), \Phi (x_{k}))$ where $k=2,3,...,n$.

Since $\Phi$ is a contractive function, $\{x_{n}\}_{n=1}^{\infty}$is a Cauchy sequence. Since M is a complete metric space, the sequence has a limit in M and we call it x. A contractive function in a metric space is a continuous, so $\{\Phi (x_{n})\}_{n=1}^{\infty}$ converges. We know that $\{\Phi (x_{n})\}_{n=1}^{\infty}$ is simply $\{x_{n}\}_{n=2}^{\infty}$ whose limit is x. Thus $\Phi (x) = x$.

To show the uniqueness, suppose on the contrary that you have another point $\Phi (y) = y$. Now you can draw a contradiction if you check your distance function formula $d(\Phi (x), \Phi (y)) < d(x, y)$
• January 30th 2009, 11:47 AM
Opalg
Quote:

Originally Posted by aliceinwonderland
It seems like you don't need to find a "c" to show the existence of a fixed point of x such that $\Phi (x) = x$.

First, choose a point $x_{1}$ in M and define
$x_{2} = \Phi (x_{1}), x_{3} = \Phi (x_{2}),...,x_{n}= \Phi (x_{n-1})$ for $n \ge 2$.

You might need to mark some points in M (to figure out the points indeed converge) and make sure each $d(x_{k-1}, x_{k}) > d(\Phi (x_{k-1}), \Phi (x_{k}))$ where $k=2,3,...,n$.

Since $\color{blue}\Phi$ is a contractive function, $\color{blue}\{x_{n}\}_{n=1}^{\infty}$ is a Cauchy sequence. Since M is a complete metric space, the sequence has a limit in M and we call it x. A contractive function in a metric space is a continuous, so $\{\Phi (x_{n})\}_{n=1}^{\infty}$ converges. We know that $\{\Phi (x_{n})\}_{n=1}^{\infty}$ is simply $\{x_{n}\}_{n=2}^{\infty}$ whose limit is x. Thus $\Phi (x) = x$.

To show the uniqueness, suppose on the contrary that you have another point $\Phi (y) = y$. Now you can draw a contradiction if you check your distance function formula $d(\Phi (x), \Phi (y)) < d(x, y)$

Unless I'm missing something, the sentence in blue needs some justification. The usual proof that the sequence is Cauchy relies on estimating $d(x_m,x_n)$ (where n>m) by using the triangle inequality to get $d(x_m,x_n) \leqslant d(x_m,x_{m+1}) + d(x_{m+1},x_{m+2}) + \ldots + d(x_{n-1},x_n)$. If the stronger condition $d(\Phi (x), \Phi (y)) \leqslant c\cdot d(x, y)$ holds, for some c<1, then you can estimate the sum of those distances by using a geometric series. But with the weaker condition $d(\Phi (x), \Phi (y)) < d(x, y)\ (\forall x \neq y)$, that method will not work.

A mapping satisfying the stronger condition is called a contraction map. A mapping satisfying the weaker condition is sometimes called a strictly metric map. A strictly metric map on a compact space need not be a contraction map. For example, the map $\Phi(x) = \tfrac12x^2$ is strictly metric by not contractive on the closed unit interval [0,1]. It is not clear to me that a strictly metric map on a compact space needs to have a fixed point. (If it has, then the fixed point is certainly unique.)
• January 31st 2009, 01:30 AM
Opalg
Quote:

Originally Posted by mylestone
Let M be a compact metric space. Let $\Phi : M \rightarrow M$ be such that $d(\Phi (x), \Phi (y)) < d(x, y), \forall x, y \in M, x \neq y$. Show that $\Phi$ has a unique fixed point.

I can now see how to do this problem. Let $\delta = \inf\{d(x,\Phi(x)):x\in M\}$. It follows from the compactness of M that the infimum is attained, so there exists $x_0\in M$ with $d(x_0,\Phi(x_0)) = \delta$. If $\delta\ne0$ then $\delta \leqslant d(\Phi(x_0),\Phi(\Phi(x_0))). That contradiction shows that $\delta=0$ and $x_0$ is a fixed point of $\Phi$.