The Unapologetic Mathematician

Mathematics for the interested outsider

Bump Functions, part 1

Dealing with an unexpected breach of my GMail address book kept me busy yesterday. But at least I get this one up before today’s activities.

Now we come to the heart of our partitions of unity: the bump functions. These are like smooth analogues of characteristic functions. A characteristic function \chi_S is defined as 1 on a set and 0 off of it. We can use them (and have!) to “mask” off a function f; multiply f by \chi_S and suddenly f is supported on S. But doing this introduces some nasty discontinuities.

A bump function \phi fixes the problem by smoothly tailing off to zero between an inner set V and an outer open set U that contains the closure of V. Then the product f\phi will be at least as smooth as the original function f was, except in the case of analytic functions. On V, \phi is identically 1, and so f(x)\phi(x)=f(x) for points x\in V. Outside of U, f(x)\phi(x)=0.

So let C_\epsilon(0) be the open cube in \mathbb{R}^n consisting of those vectors with each of their components in the interval (-\epsilon,\epsilon). We will start by constructing a bump function between C_1(0) and C_2(0).

The real core here is the function

\displaystyle h(x)=\left\{\begin{array}{lc}e^{-\frac{1}{x}}&x>0\\{0}&x\leq0\end{array}\right.

I leave it to you to verify that this function is, in fact, smooth at x=0; show that each derivative of the function on the right is zero at this point. It’s clearly not analytic, though, since its Taylor series at this point sums to the zero function.

Now, consider the function h(2+x)h(2-x). If x\geq2 or x\leq-2, one or the other factor is zero, and so the product is supported inside [-2,2]. We can also write down h(2+x)h(2-x)+h(x-1)+h(-x-1), which is everywhere strictly greater than zero, meaning we can divide by it:

\displaystyle f(x)=\frac{h(2+x)h(2-x)}{h(2+x)h(2-x)+h(x-1)+h(-x-1)}

If x\leq1 then h(x-1)=0, while if x\geq-1 then h(-x-1)=0. So on the whole interval [-1,1], this quotient is exactly 1.

Therefore f is a bump function between the intervals (-1,1) and (-2,2) in \mathbb{R}. For \mathbb{R}^n, just define

\displaystyle\phi(a_1,\dots,a_n)=f(a_1)\dots f(a_n)

and we have a suitable bump function between the cubes C_1(0) and C_2(0) for any dimension.

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March 12, 2011 - Posted by | Differential Topology, Topology

4 Comments »

  1. Sorry about the gmail problems … have you considered adding two-factor authentication to your account — it’s the bomb! http://googleblog.blogspot.com/2011/02/advanced-sign-in-security-for-your.html

    Comment by Literalman | March 12, 2011 | Reply

  2. [...] a countable atlas ; we use for the coordinate maps since we’ll want the free. We’ve also seen that we have a smooth bump function between the two cubes and in . So let’s [...]

    Pingback by Partitions of Unity (proof) « The Unapologetic Mathematician | March 14, 2011 | Reply

  3. [...] immediate application of our partitions of unity, let’s show that we can always get whatever bump functions we [...]

    Pingback by Bump Functions, part 2 « The Unapologetic Mathematician | March 16, 2011 | Reply

  4. [...] John Armstrong: Bump Functions [...]

    Pingback by Second Xamuel.com Linkfest | March 26, 2011 | Reply


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