[Math] Why are bump functions compactly supported

compactnessreal-analysis

Smooth and compactly supported functions are called bump functions. They play an important role in mathematics and physics.
In $\mathbb{R}^n$ and $\mathbb{C}^n$, a set is compact if and only if it is closed and bounded.
It is clear why we like to work with functions that have a bounded support. But what is the advantage of working with functions that have a support that is also closed? Why do we often work with compactly supported functions, and not just functions with bounded support?

Best Answer

  1. On spaces such as open intervals and (more generally) domains in $\mathbb R^n$, compactness of support tells us much more than its boundedness. Any function $f\colon (0,1)\to\mathbb R$ has bounded support, since the space $(0,1)$ itself is bounded. But if the support is compact, that means that $f$ vanishes near $0$ and near $1$. (Generally, near the boundary of the domain).

  2. On the other hand, when bump functions are considered on infinite-dimensional spaces (which does not happen nearly as often), the support is assumed bounded, not compact. A compact subset of an infinite-dimensional Banach space has empty interior, and so cannot support a nonzero continuous function. If you are interested in this subject (which is a subset of the geometry of Banach spaces), see Smooth Bump Functions and the Geometry of Banach Spaces: A Brief Survey by R. Fry and S. McManus, Expo. Math. 20 (2002): 143-183

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