Approximation of the characteristic function of a compact set

  • Thread starter Mystic998
  • Start date
  • #1
206
0

Homework Statement


Okay, so this is a three-part question, and I need some help with it.

1. I need to show that the function [tex]f(x) = e^{-1/x^{2}}, x > 0[/tex] and 0 otherwise is infinitely differentiable at x = 0.

2. I need to find a function from R to [0,1] that's 0 for [itex]x \leq 0[/itex] and 1 for [itex] x \geq \epsilon[/itex].

3. For K compact, U open in [itex]R^n[/itex] and [itex] K \subset U[/itex], I need to find a function from U to [0,1] that's 1 on K and and 0 outside of some closed set C containing K and contained in U.


Homework Equations




The Attempt at a Solution


1. I can do the first part up to the second derivative pretty easily, I think. And I think that any derivative of [tex]e^{-1/x^2}[/tex] is going to be a polynomial times [tex]e^{-1/x^2}/x^n[/tex] for some n, so showing that the derivative of the function above is continuous at 0 just amounts to showing that [tex]\frac{e^{-1/x^2}}{x^n}[/tex] goes to 0 as x goes to 0. I just wanted to see if I have the right idea and if there might be an easier way.

2. For this one I think you just take [tex]f(x) = e^{-{(\frac{\epsilon - x}{x})}^{2}}[/tex] on the interval [itex](0,\epsilon)[/itex], then 0 and 1 where appropriate. Just checking to see if I'm way off or not here.

3. Now on this one I have very little idea what to do. I was thinking I could take 2 closed balls with [itex]K \subset C_{1} \subset C_{2} \subset U[/itex], then use [tex]e^{-1/x^2}[/tex] to define a function that's goes smoothly from 1 to 0 on [itex]C_{2} - C_{1}[/itex], but I'm not sure that's the best way to go about it.

Also, sorry for the TeX looking so nasty. The epsilons in the functions weren't very legible with the itex tag.
 

Answers and Replies

  • #2
206
0
Just a quick bump before bed.
 

Related Threads on Approximation of the characteristic function of a compact set

  • Last Post
Replies
5
Views
986
  • Last Post
Replies
4
Views
1K
Replies
3
Views
2K
Replies
2
Views
3K
Replies
3
Views
1K
Replies
3
Views
1K
Replies
4
Views
10K
Replies
19
Views
8K
Replies
1
Views
3K
Top