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Math Help - Calculus Contraction Mapping

  1. #1
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    Calculus Contraction Mapping

    T(x) = X+ 1/x X=[1,INFINITY]

    |T(X) - T(Y)| < |X-Y|

    |(x+ 1/x) - (y+ 1/y)| < |x-y|

    Can i solve the left side further?
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  2. #2
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    Quote Originally Posted by littlefire View Post
    T(x) = X+ 1/x X=[1,INFINITY]

    |T(X) - T(Y)| < |X-Y|

    |(x+ 1/x) - (y+ 1/y)| < |x-y|

    Can i solve the left side further?
    I am not sure what you want.
    \left| x + \frac{1}{x} - y - \frac{1}{y} \right| = \left| x -  y\right| + \left| \frac{1}{x} - \frac{1}{y} \right|
    This is bounded by, |x - y| + \frac{|x-y}{|xy|} \leq |x-y|+|x-y| = 2|x-y|.
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  3. #3
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    Thanks a million. Still now too sure on your answer after the bounded part.

    This is my full question that I have to answer -

    Let X=[1,infinity) with the metric d(x,y) = |x-y|. Define T:X->X by T(x) = x+1/x. Prove T satisfies

    d(T(x),T(y)) < d(x,y) for all x,y E X

    but T has no fixed points in X. Does this contradict Banach's fixed point theorem.
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  4. #4
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    Quote Originally Posted by littlefire View Post
    Let X=[1,infinity) with the metric d(x,y) = |x-y|. Define T:X->X by T(x) = x+1/x. Prove T satisfies
    d(T(x),T(y)) < d(x,y) for all x,y E X
    By simple differentiation we see the function is increasing on its domain.
    T(x) = x + \frac{1}{x} \Rightarrow \quad T'(x) = 1 - \frac{1}{{x^2 }} \geqslant 0.
    So, x > y \Rightarrow \quad T(x) - T(y) > 0\,\& \,{ \color{blue}\frac{1}{x} - \frac{1}<br />
{y} < 0}.
    \begin{array}{rcl}   {\left| {T(x) - T(y)} \right|} &  =  & {T(x) - T(y)}  \\<br />
   {} &  =  & {x - y +{\color{blue} \frac{1}{x} - \frac{1}{y}}}  \\   {} &  <  & {x - y}  \\<br />
   {} &  =  & {\left| {x - y} \right|}  \\ \end{array} .

    By symmetry this holds if y>x.
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