# Why are bump functions compactly supported?

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?

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

In general the support of a function is defined to be the closure of the set of points where it's nonzero. (If this weren't true, there would be no such thing as a smooth, compactly supported function on $\mathbb{R^n}$: the set of points where a continuous function is nonzero is certainly open, and $\mathbb{R^n}$ contains no nonempty bounded clopen sets.)

• In $\mathbb{R}^n$, a bounded support is a compact support. I am interested in the case where it is important to say that the support is compact, not just bounded. Commented Jun 9, 2012 at 13:08
• @Deniz: Fair enough. In that case, you might want to edit your question to ask about functions that have a support that is also compact, not just also closed. (And I don't know the answer to that question as I've never had any reason to want to think about bump functions on spaces where the Heine-Borel theorem doesn't hold...) Commented Jun 9, 2012 at 13:22

In metric spaces a real-valued function whose domain is compact is uniformly continuous and it attains a minimum and maximum value.

These properties are not to be taking lightly, for example the function $x\mapsto\frac1{x(1-x)}$ on $(0,1)$ and zero else where would be with a bounded support, but the function itself is not bounded.

Generally speaking, compact sets are very well-behaved because everything that can be characterized by open sets has, in some sense, a finite character. Continuous functions are such object, as the continuous preimage of an open set is open, so continuous functions from a compact domain have a very well-behaved nature.

The above is compatible with the following statement: Mathematicians like well-behaved objects. I have to admit that until recently I always tried to explore the naughty terrains of the mathematics and slowly I understand more and more why well-behaved objects are good. Especially when they are enough to describe a whole lot of mathematics to go around.

For one, compactly supported continuous functions are bounded on their range.