# Bump function

1. Aug 23, 2007

### noospace

This is not a homework question, just trying to understand the material in my differential geomety lecture.

Consider the function

$f : \mathbb{R} \to [0,1]$

given by $f(x) = \frac{1}{A}\int_{-\infty}^{\infty} a(t) a(1-t)dt$

where a(x) is zero for x less than or equal to 0 and $a(x) = e^{-x^{-1}}$ for x > 0 and $A = \int_{0}^{1}a(t)a(1-t)dt$.

The claim is that f(x) = 0 for x less than or equal to 0 and f(x) = 1 for x greater than or equal to 1.

The first part is obvious. The second part I'm having trouble with.

$f(x > 1) = \frac{1}{A}\int_{-\infty}^{\infty} a(t) a(1-t)dt$
$f(x > 1) = \frac{1}{A}\left( \int_{-\infty}^0 + \int_{0}^{1} + \int_{1}^x \right) a(t) a(1-t)dt$
$f(x > 1) = 1 + \frac{1}{A} \int_{1}^x a(t) a(1-t)dt$.

So the question becomes, why does the second term vanish? My lecturer claims this is obvious but I just don't see it.

2. Aug 25, 2007

### noospace

There is a typo in the way I've defined f(x). It should be

$f(x) = \frac{1}{A}\int_{-\infty}^{x} a(t) a(1-t)dt$.