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Math Help - characteristic function

  1. #1
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    Oct 2008
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    characteristic function

    A = {(x,y) in R^2: x > 0 and 0< y < x^2}

    Define f:R^2 -->R by f(x,y) = 0 if (x,y) not in A
    = 1 if (x,y) in A

    Fix h in R^2. Define g: R -->R with g_h(t) = f(t.h)
    Show that g is continuous at 0
    I dont understand how the funtion g_h(t) = f(t.h) work?

    I just use the defintnition of continuity,
    This is what I done so far:

    given e > 0, there exists d > 0 such that |f(t.h) - 0|< e
    But I dont know how to get from |f(t.h) - 0| to |f(t.h) - 0|< e
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  2. #2
    Member
    Joined
    Jun 2009
    Posts
    113
    Fix (h_1,h_2)\in\mathbb{R}^2.

    Since f(0,0)=g_h(0)=0, you need to show that when t is small enogh (th_1,th_2)\in A^c. If
    h_1 is 0, it is clear that (th_1,th_2)\ is never in A. If h_1\neq 0, you have only to care of the limit direction in t (+ or -) which makes
    th_1 positive. The other direction gives the constant function 0. In this case you have to show

     <br />
th_1\geq t^2h_2^2<br />

    for t small enogh. If h_1 is positive it is almost immediate you can do it (when t\to 0^+). If h_1 is negative and t\to 0^- the inequality

     <br />
h_1\leq th_2^2<br />

    is also fulfilled for t small enough.
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