Entire Bump Functions? Where?

October 24, 2010

Here’s the cute proof of the week.  Liouville’s Theorem (in complex analysis) is coming up (because this theorem is so important to me, I’m trying to scrounge up a lot of applications for it) and so I wanted to just give a cute little corollary that comes directly from Liouville’s Theorem.  First, let me just state Liouville really quickly:


Theorem (Liouville).  A function from f:{\mathbb C}\rightarrow {\mathbb C} which is bounded (in the sense that |f(z)| < M for some real number M) and entire (complex differentiable) everywhere is constant.


Now, let’s talk about Bump Functions for a second.  A bump function is a function f:{\mathbb R}^{n} \rightarrow {\mathbb R} which is bounded, smooth, and has compact support (it is zero everywhere but on compact set), but we can generalize this to complex functions by simply changing the domain and range to the complex numbers g:{\mathbb C}^{n}\rightarrow {\mathbb C}.  So why don’t we ever hear about complex bump functions? 

A short story before tossing the corollary at you: bump functions are nice, because sometimes we need to partition up spaces and nicely distribute functions about.  For example, there are things call partitions of unity which allow us to talk about functions on manifolds nicely.  While working through one of my books for diffi-manifolds today, I noticed that while most of the time the book worked in an arbitrary field, one particular theorem only was only stated for the reals.  I wasn’t sure if it was a typo or not, so I attempted to adapt the proof for complex numbers (at least!) but I got to a point where I needed to use the complex equivalent of a bump function.  No matter where I looked, I couldn’t find any mention of them.  And then it hit me.


Corollary.  Suppose that g:{\mathbb C}\rightarrow {\mathbb C} is a bump function as we’ve defined above.  Then g is the zero function.


Proof.  By Liouville’s theorem, since g is bounded, entire, and smooth (which, in this case, means holomorphic as complex functions differentiable once are infinitely differentiable) it is constant.  Because it is zero at least on some non-compact set, it follows that it must be zero everywhere.


So there’s not much bump in bump functions on the complex plane.  That’s kind of sad!  Nonetheless, it’s a good (or at least cute) application of Liouville’s theorem.

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