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Math Help - Contractive Map

  1. #1
    Junior Member
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    Contractive Map

    Can anyone help me with the following proof please:

    Let A>reals and Φ: A>reals Є C1.
    Then the smallest k with
    | Φ(x) Φ(y)| ≤ |x-y| for all x,y in A is given by
    k = max |Φ(ε)| with ε ЄA.

    Show that the contraction constant k is given by k = max |Φ(ε)| with ε ЄA.

    Thanks to anyone who is able to help.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by smiler View Post
    Can anyone help me with the following proof please:

    Let A>reals and Φ: A>reals Є C1.
    Then the smallest k with
    | Φ(x) – Φ(y)| ≤ |x-y| for all x,y in A is given by
    k = max |Φ’(ε)| with ε ЄA.

    Show that the contraction constant k is given by k = max |Φ’(ε)| with ε ЄA.

    Thanks to anyone who is able to help.
    The inequality |\phi(x)-\phi(y)|\leq k|x-y| can be written as  \left|\frac{\phi(x)-\phi(y)}{x-y}\right|\leq k for all x,y\in A,\, x\neq y. Letting x\to y you can derive that k \geq \max_{\varepsilon\in A}|\phi'(\varepsilon)|, can't you ? Then, to get k=\max_{\varepsilon\in A}|\phi'(\varepsilon)|, it remains to be shown that k>\max_{\varepsilon\in  A}|\phi'(\varepsilon)| is impossible (hint : use a proof by contradiction and the mean value theorem).
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  3. #3
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    Quote Originally Posted by flyingsquirrel View Post
    Hi,

    The inequality |\phi(x)-\phi(y)|\leq k|x-y| can be written as  \left|\frac{\phi(x)-\phi(y)}{x-y}\right|\leq k for all x,y\in A,\, x\neq y. Letting x\to y you can derive that k \geq \max_{\varepsilon\in A}|\phi'(\varepsilon)|, can't you ? Then, to get k=\max_{\varepsilon\in A}|\phi'(\varepsilon)|, it remains to be shown that k>\max_{\varepsilon\in A}|\phi'(\varepsilon)| is impossible (hint : use a proof by contradiction and the mean value theorem).
    thanks for your assistance, but no i dont know how to derive that because we havent been shown, nor do i know what the mean value theorem is. brief help on this would be great please
    thanks, smiler
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by smiler View Post
    thanks for your assistance, but no i dont know how to derive that
    Did you try ? That's less difficult than it looks like. Starting from

     \left|\frac{\phi(x)-\phi(y)}{x-y}\right|\leq k for all x,y\in A,\, x\neq y,
    if we let x tend to y we get that for all y \in A,
    \lim_{x\to y}\left|\frac{\phi(x)-\phi(y)}{x-y}\right| \leq \lim_{x\to y} k

    What is the LHS (think about the definition of the derivative) ? What is the RHS ?

    because we havent been shown
    That sounds like "I don't have the solution so I can't do it"...
    nor do i know what the mean value theorem is.
    I can't think of a proof that doesn't use this theorem.
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