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Math Help - Lipschitz continuous

  1. #1
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    Question Lipschitz continuous

    Let f : R ->R be a function.

    We say that f is Lipschitz continuous if there is some
    L > 0 such that |f(x) − f(y)| < L|x − y| for all x, y in R.

    I want to show that f : R ->R defined as f(x) = x^2 / (1+ x^2)is Lipschitz continuous.

    I know I should use the Mean Value Theorem but i cant seem to apply it can some one please help thanks
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  2. #2
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    This is the absolute value of the derivative:
    \left| {\frac{{2x}}{{\left( {1 + x^2 } \right)^2 }}} \right|.
    Graph it to see that it is bounded.

    From the mean value theorem:
    x < c < y\;\& \;\left| {\frac{{2c}}{{\left( {1 + c^2 } \right)^2 }}} \right| = \left| {\frac{{f(y) - f(x)}}{{x - y}}} \right|.
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  3. #3
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    Question not Lipschitz continuous.

    Then i was wondering how do you Show that f : R -> R defined as f(x) = x^2 is not Lipschitz continuous.
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  4. #4
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    Quote Originally Posted by dopi View Post
    Then i was wondering how do you Show that f : R -> R defined as f(x) = x^2 is not Lipschitz continuous.
    Assume that there is a number A so that:
    |x^2-y^2|\leq A|x-y| for all x,y\in \mathbb{R}.
    So in particular for x\not = y this inequality should work.
    But then,
    |x-y||x+y|\leq A |x-y| \implies |x+y|\leq A.
    Which is impossible because |x+y| can be made larger than A, i.e. y=A+1,x=0.
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