1. Nov 21, 2006

chy1013m1

any thoughts to this question?
Give an example of a C^oo (C infinity) function f : R->R which is positive on the interval (-1, 1) and 0 elsewhere

2. Nov 21, 2006

Hurkyl

Staff Emeritus
For an indirect approach, you could try constructing one as a limit.

For a more direct approach, figure out what the obstacle is.

You know smooth functions that are zero outside of (-1, 1), and you know smooth functions that are positive inside (-1, 1).

So, what's the problem? You need all of the derivatives to match up. In other words, for your function positive inside of (-1, 1), you need all of its derivatives to be zero at 1 and at -1.

So that's your problem: you need to find a function (other than the zero function) whose derivatives are all zero at some point.

3. Nov 21, 2006

chy1013m1

just to check if i am on the right track..
let f(x ) = { e ^ (1 / (x ^ 2 - 1)) if -1 < x < 1
0 otherwise
then it is easy to show that
lim f( x) = 0 (split into 2 cases)
x->1 or -1

it is also true that
lim (f(x + h) - f(x)) / h = 0 for x = 1, -1 (show with limit proof)
h->0
then f is C1

so then induction on f being Ck, show it is C(k+1)..(if i did it correctly, it should work out..)
then by induction, f is Cinf.

4. Nov 21, 2006

Hurkyl

Staff Emeritus
Hrm. I find it very plausible that function will work. Incidentally, I was thinking of using the function

$$g(x) := \begin{cases} 0 & x \leq 0 \\ e^{-1/x^2} & x > 0$$

as a building block, since we already know it's smooth. Actually, I think you can build your function out of this, so I'm even more convinced.