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Math Help - Contraction Mappings without a fixed point

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    Contraction Mappings without a fixed point

    Show that the following mapping f:X \rightarrow X do not have a fixed point and explain why the Contraction Mapping Principle is not contradicted.
    Problem: Let X=R and f(x) = x+1 for all x in X.

    My Solution:
    f do not have a fixed point because x does not equal to x+1 for any real number x.

    d(f(x),f(y)) = |(x+1) - (y+1)| = |x-y|,
    and d(x,y) = |x-y|.

    So d(f(x),f(y)) = d(x,y), now I know I have to find a constant c with 0 <= C < 1 for the Contraction Mapping Principle to hold. But I can't find that c.

    Please help.
    Last edited by tttcomrader; September 14th 2007 at 01:35 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by tttcomrader View Post
    Problem: Let X=R and f(x) = x+1 for all x in X.

    Show that the mapping doesn't have a fixed point, and the Contraction Mapping Theorem is not contradicted.

    Solution: d(f(x),f(y)) = |(x+1) - (y+1)| = |x-y|,
    and d(x,y) = |x-y|.

    So d(f(x),f(y)) = d(x,y), now I know I have to find a constant c with 0 <= C < 1 for the Contraction Mapping Principle to hold. But I can't find that c.

    Please help.
    Look at the question and answer what is asked.

    1. Does the mapping have a fixed point? Justify your answer.

    2. If there was no fixed point show that at least one of the conditions
    of the the contraction mapping theorem is violated.

    RonL
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    I think I didn't word the question correctly, let me edit.

    "Show that the following mapping f:X \rightarrow X do not have a fixed point and explain why the Contraction Mapping Principle is not contradicted."

    So if I understand this question correctly, I have to show the mapping do not have a fixed point, but yet it is still a contraction. (I guess because it is not in a complete space?)

    And yeah, I forgot to put the first part of my answer on there, it is added now.
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    Quote Originally Posted by tttcomrader View Post
    I think I didn't word the question correctly, let me edit.

    "Show that the following mapping f:X \rightarrow X do not have a fixed point and explain why the Contraction Mapping Principle is not contradicted."

    So if I understand this question correctly, I have to show the mapping do not have a fixed point, but yet it is still a contraction. (I guess because it is not in a complete space?)

    And yeah, I forgot to put the first part of my answer on there, it is added now.
    No, "the Contraction Mapping Principle is not contradicted." You need to show why the mapping does not have a fixed point and why there is no contradiction.

    -Dan
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    So what I need is to find a c with 0 <= c < 1 such that d(f(x),f(y)) <= cd(x,y) right?

    Now as how my calculation goes, d(f(x),f(y)) = d(x,y), so I"m stuck here. Or perhaps I have done the whole thing wrong.
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