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I've only seen the bump function $e^\frac1{x^2-1}$ so far. Where could I find examples of functions $C^∞$ on $\mathbb{R}$ that are zero everywhere except on $(-1,1)$?

Are there others that do not involve the exponential function? Are there any with a closed form integral? Is there a preferred function?

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Perhaps you might ask, "Does it matter?" IMHO not really; once you have one bump function, it can do all the heavy lifting needed to produce the sort of results that distinguish smooth manifolds from analytic manifolds. This is not to denigrate your curiosity, just to comment that the existence of one sort is enough :) –  Neal Jan 23 '12 at 1:38
    
This question would be better if it were worded more precisely, for example by including the information in the first sentence of Robert Israel's answer below. –  Greg Martin Jan 23 '12 at 6:06
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2 Answers 2

up vote 10 down vote accepted

Presumably what you want is a function that is $C^\infty$ on $\mathbb R$, nonzero on $(-1,1)$ and zero elsewhere. It's convenient to use something involving the exponential function because it's nonzero everywhere but goes to faster than any polynomial at $-\infty$ and it's easy to differentiate. If you really want to avoid the exponential function, you might try something like $$\frac{1}{I_0(1/(1-x^2))}$$ for $-1 < x < 1$ where $I_0$ is a modified Bessel function of order $0$.

For an example that has a closed-form antiderivative, you might try $$ \frac{\left( {x}^{2}+1 \right)\ {{\rm e}^{{\frac {4x}{{x}^{2}-1}}}}}{\left( \left( {x}^{2}-1 \right) \left( 1+{{\rm e}^{{\frac {4x}{{x}^{ 2}-1}}}} \right)\right)^2} $$

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+1 Nice antiderivative. How did you come up with it? –  jnm2 Jan 23 '12 at 13:29
    
I worked backwards: start with a nice "ramp" function and take its derivative. –  Robert Israel Jan 23 '12 at 17:26
    
Funny, I wanted a ramp function so my first idea was to integrate a bump function. Therefore this question. ;-) –  jnm2 Jan 23 '12 at 22:54
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All of these involve breaking up the domain by inequalities and, whether visible or not, involve something no simpler that the exponential function. The original one-sided version is $$ f(x) = e^{-1/x} \; \mbox{for} \; x > 0 $$ but $$ f(x) = 0 \; \mbox{for} \; x \leq 0. $$

You can get a bump from this by multiplication with $$ g(x) = f( 1 + x) \cdot f(1 - x) $$

You get a smoothed step function from $$ h(x) = \int_{- \infty}^x \; g(t) dt $$

You get a plateau bump function, constant in the middle, from $$ p(x) = h(x + A) \cdot h(A -x) $$ for some $A > 1.$

We can prove some properties of this sort of thing. It has no removable singularity at the points where it is not real analytic, at best it has an essential singularity or possibly is not even defined in any neighborhood of the point in $\mathbb C.$

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