# The Unapologetic Mathematician

## 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.

March 12, 2011