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Math Help - Contraction Mapping Principle

  1. #1
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    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?
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    Quote Originally Posted by mylestone View Post
    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)
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  3. #3
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    Quote Originally Posted by aliceinwonderland View Post
    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.)
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  4. #4
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    Quote Originally Posted by mylestone View Post
    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)))<d(x_0,\Phi(x_0)) = \delta. That contradiction shows that \delta=0 and x_0 is a fixed point of \Phi.
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