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Constructing a smooth characteristic function

  1. Oct 31, 2011 #1
    Constructing a "smooth" characteristic function

    Suppose I'd like to construct a [itex]C^\infty[/itex] generalization of a characteristic function, [itex]f(x): \mathbb R \to \mathbb R[/itex], as follows: I want [itex]f[/itex] to be 1 for, say, [itex]x\in (a,b)[/itex], zero for [itex]x < a-\delta[/itex] and [itex]b > x + \delta[/itex], and I want it to be [itex]C^\infty[/itex] on [itex]\mathbb R[/itex]. How do I know I can do this? Namely, how do I define the function on [itex][a-\delta,a][/itex] and [itex][b,b+\delta][/itex] to make sure this will happen?

    This might be easy, but I'm not familiar enough with properties of [itex]C^\infty[/itex] functions to immediately see how to do this.
     
    Last edited: Oct 31, 2011
  2. jcsd
  3. Oct 31, 2011 #2
    Re: Constructing a "smooth" characteristic function

    What you're looking for is called a bump function. The key ingredient in the construction of bump functions is the following function:

    [tex]f(x) = \begin{cases}e^{-1/x} & \text{if } x>0 \\ 0 & \text{if } x \leq 0 \end{cases}[/tex]

    It is easy to see that f is infinitely differentiable. Then consider the following functions:

    [tex]g_1(x) = f(x-(a-\delta))f(a-x)[/tex]
    [tex]g_2(x) = f(x-b)f((b+\delta)-x)[/tex]

    g_1 is a smooth function which is positive on (a-δ, a) and 0 elsewhere, and g_2 is a smooth function which is positive on (b, b+δ) and 0 elsewhere. Then consider the following functions:

    [tex]h_1(x) = \frac{\int_{a-\delta}^{x} g_1(t)\ dt}{\int_{a-\delta}^{a} g_1(t)\ dt}[/tex]
    [tex]h_2(x) = \frac{\int_{b}^{x} g_2(t)\ dt}{\int_{b}^{b+\delta} g_2(t)\ dt}[/tex]

    h_1 and h_2 are smooth functions from R into [0, 1], h_1 is 0 for x<a-δ and 1 for x>a, and h_2 is 0 for x<b and 1 for x>b+δ. Finally, the function h=h_1 - h_2 is the bump function you're looking for.
     
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