How do you define if is an arbitrary set ?
Given a distribution v and a smooth f (both on a set X) - how do we find the distribution u such that fu = v? If f is never zero, then u = v/f is the solution.
I am trying to do the case where X is the real line and where f isn't always non zero.
The book I am reading tells me that if the points x in X such that f(x) = 0 are isolated zeros of finite order, then we can reduce to the case f(x) = x^m for some positive integer m.
Can anyone explain why this is true?
Given the hint, I think I see what is wanted here. Let be the zeros of f, and let be a smooth partition of unity such that is the only zero of f in the support of (for each n).
Let . If you can find such that , it will follow that where . That means that it is sufficient to consider the case where f has only one zero, at . Also, by making the translation , you can assume that .
That reduces the problem to the case where f has just one zero, of finite order, at the origin. Thus , where g has no zeros at all. But you know how to solve the problem for a function with no zeros. So all that remains is to solve the problem for the function .
subordinate to an open cover (which means that each function in the partition has its support in one of the sets in the open cover). In this case, for each isolated zero of , let be the complement of the union of all the other isolated zeros. Then is an open cover of the space, and the support of each function in the associated partition of unity will have only one of the zeros in its support.