- #1
AxiomOfChoice
- 533
- 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.
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.
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