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Thread: characteristic function

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    83

    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 $\displaystyle (h_1,h_2)\in\mathbb{R}^2$.

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

    $\displaystyle
    th_1\geq t^2h_2^2
    $

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

    $\displaystyle
    h_1\leq th_2^2
    $

    is also fulfilled for $\displaystyle t$ small enough.
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