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Math Help - sum of two lipschitz funtions

  1. #1
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    Question sum of two lipschitz funtions

    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.


    letting f, g : R -> R be lipschitz continious funtions,
    (a) i want to show that f+g is lipschitz continious too

    and also if f,g are bounded lip continuos funtions
    (b) then i want to show that f.g is lipschitz continous too

    does anyone know how to show these?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dopi View Post
    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.


    letting f, g : R -> R be lipschitz continious funtions,
    (a) i want to show that f+g is lipschitz continious too
    Proof:

    Since f and g are Lipschitz continuous, we have that |f(x) - f(y)| < \frac {L_1}2|x - y| for L_1 >0 and |g(x) - g(y)| < \frac {L_2}2|x - y| for L_2 > 0

    Now let L = \mbox{max} \{ L_1, L_2 \}, and note that (f + g)(x) = f(x) + g(x), then we have:

    |(f + g)(x) - (f + g)(y)| = |f(x) + g(x) - f(x) - g(x)| \le |f(x) - f(y)| + |g(x) - g(y)|  < \frac {L_1}2|x - y| + \frac {L_2}2|x - y| < \frac L2|x - y| + \frac L2|x - y| = L|x - y|

    QED

    you do (b), it should be similar
    Last edited by Jhevon; November 13th 2007 at 01:48 PM.
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  3. #3
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    Quote Originally Posted by dopi View Post
    Let f : R ->R be a function.

    and also if f,g are bounded lip continuos funtions
    (b) then i want to show that f.g is lipschitz continous too
    If f is Lipschitz then |f(x)-f(y)|\leq A|x-y| similarly g so |g(x)-g(y)|\leq A|x-y|. We want to show |f(x)g(x)-f(y)g(y)|\leq C|x-y| for some C. Note that |f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)|\leq  |f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)| \leq M_1B|x-y|+M_2A|x-y| = C|x-y| where C=M_1B+M_2A. Q.E.D.
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