The division problem in distribution theory.

Printable View

March 24th 2011, 05:02 PM
measureman

The division problem in distribution theory.

Hi all.

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?

Thanks.
March 25th 2011, 01:40 PM
girdav

How do you define $x^m$ if $X$ is an arbitrary set ?
March 25th 2011, 01:50 PM
measureman

I am doing the case where X = R. The hint given in the book is: by a partition of unity and a translation of the origin.
March 25th 2011, 02:29 PM
Opalg

Given the hint, I think I see what is wanted here. Let $\{x_n\}$ be the zeros of f, and let $\{\psi_n\}$ be a smooth partition of unity such that $x_n$ is the only zero of f in the support of $\psi_n$ (for each n).

Let $v_n = v\psi_n$ . If you can find $u_n$ such that $fu_n = v_n$ , it will follow that $fu=v$ where $u = \sum u_n$ . That means that it is sufficient to consider the case where f has only one zero, at $x_n$ . Also, by making the translation $x\mapsto x-x_n$ , you can assume that $x_n=0$ .

That reduces the problem to the case where f has just one zero, of finite order, at the origin. Thus $f(x) = x^mg(x)$ , 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 $f(x) = x^m$ .
March 29th 2011, 03:22 PM
measureman

Thanks for your reply, very helpful. Although I am wondering, how do we know that we can always find a partition which satisfies those conditions?
March 30th 2011, 01:12 AM
Opalg

Quote:

Originally Posted by measureman

Thanks for your reply, very helpful. Although I am wondering, how do we know that we can always find a partition which satisfies those conditions?

It is always possible to find a (smooth) partition of unity 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 $x_n$ of $f(x)$ , let $U_n$ be the complement of the union of all the other isolated zeros. Then $\{U_n\}$ 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.