28
$\begingroup$

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?

  • 0
    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 :)2012-01-23
  • 0
    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.2012-01-23
  • 5
    Six years later I return to this question in search of other bump functions for reasons that have nothing to do with smooth manifolds and lo and behold, I find my own comment. Perhaps I should retract it...2018-02-23
  • 3
    Nah, irony is great!2018-02-23

5 Answers 5

23

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 $$ \frac{\left( {x}^{2}+1 \right)\ {{\rm e}^{{\frac {4x}{{x}^{2}-1}}}}}{\left( \left( {x}^{2}-1 \right) \left( 1+{{\rm e}^{{\frac {4x}{{x}^{ 2}-1}}}} \right)\right)^2} $$

  • 0
    +1 Nice antiderivative. How did you come up with it?2012-01-23
  • 1
    I worked backwards: start with a nice "ramp" function and take its derivative.2012-01-23
  • 2
    Funny, I wanted a ramp function so my first idea was to integrate a bump function. Therefore this question. ;-)2012-01-23
16

There is even simpler example from this book of Loring Tu.

You simply start with $$f(t)=\left\{\begin{array}{lr} 0&t\leqslant 0\\ e^{-1/t}&t>0 \end{array}\right..$$ Then you define $$g(t)=\frac{f(t)}{f(t)+f(1-t)}$$ and shift it to the right by creating $$h(t)=g(t-1).$$ To make it symmetric you put $$k(t)=h(t^2)$$ and finally to make it look like bump function you put $$\rho(t)=1-k(t).$$ It will look like this

$\hspace{4,5cm}$ enter image description here

As you can tell from the picture, Loring Tu covered more general case (he even generalised this construction to smooth manifolds of arbitrary dimension).

For more details and picutres go to Tus book.


BONUS I created plots of functions $f,g,h,k,\rho$ in this geogebra sheet. Enjoy.

13

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 $\mathbb C.$

  • 0
    Can we construct $\rho$ such that up to its $p$th derivatives, where $p\in\mathbb{N},$ are controlled by prescribed constants?2017-12-28
-1

Seems to me that if you take the second derivative of a bump function you get another bump function, which should be not so hard to show. In fact all even derivatives are bump functions. Then you linearly transform them (shift in the x or y axis and/or rescale the axes) to fit your criteria.