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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?

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4 Answers 4

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  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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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.)

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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. –  Deniz Jun 9 '12 at 13:08
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@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...) –  Micah Jun 9 '12 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.

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

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