The Unapologetic Mathematician

Mathematics for the interested outsider

Bump Functions, part 2

As an immediate application of our partitions of unity, let’s show that we can always get whatever bump functions we need.

Let U be an open subset of M, and V be a set whose closure \bar{V} is contained within U. I say that there is a nonnegative smooth function \phi:M\to\mathbb{R} which is identically 1 on \bar{V}, and which is supported within U.

To find this function, we start with a cover of M. Specifically, let U be one set of the cover, and let M\setminus\bar{V} be the other set. Then we know that there is a countable smooth partition of unity subordinate to this cover. That is, for every k we either have \phi_k supported in U, or \phi_k supported in M\setminus\bar{V} (or possibly both).

In fact, no matter what countable partition we come up with, we can take all the \phi_k supported within U and add them all up into one function \phi, and then take all the remaining functions and add them all up into one function \psi. Then \{\phi,\psi\} is a partition of unity subordinate to our cover, and I say that \phi is exactly the function we’re looking for.

Indeed, as a part of a partition of unity, \phi is a nonnegative smooth function, and we know it’s supported in U. The only thing we need to determine is if it’s identically 1 on \bar{V}. But for p\in\bar{V} we know that \phi(p)+\psi(p)=1, and yet we also know that \psi(p)=0, since \psi is supported in M\setminus\bar{V}. Thus we must have \phi(p)=1, and \phi is indeed our bump function.


March 16, 2011 - Posted by | Differential Topology, Topology


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