# Does the term localized function exists?

I am looking for a term that describes function that is "localized". What I mean is a function that is non zero in a bounded range and zero else where, such as the a rectangle pulse function. But would also like that the term cover functions such as gaussian that approach zero as you get away from the peak. If you have the whole definition with a reference would be wonderful. I couldn't find anything that pleased me in a basic googling.

• If it is constant in the non-zero range, you can call it a step function. Jan 17, 2011 at 22:52
• A "bump function" has additional constraints (smoothness), but may in practice be what you're looking for. en.wikipedia.org/wiki/Bump_function Jan 17, 2011 at 22:58
• Depends on your application, in addition to "vanishing at infinity" as described by Qiaochu and Arturo below, sometimes it is also useful to specify exactly how quick it vanishes: logarithmically? at a power law? faster than any power law? This allows you to rule in functions that do look like they have bumps (the Gaussian for example) while ruling out functions that look more spread-out (something like $1/(1 +|x|^2)$. Jan 18, 2011 at 0:10
• In this question, the answerer uses the term "locally constant": math.stackexchange.com/questions/2225/… Jan 20, 2011 at 1:39

Given a function $f\colon X\to \mathbb{R}$, the support of $f$ is defined to be $$\mathrm{sup}(f) = \{x\in X\mid f(x)\neq 0\}.$$
For functions that have limit equal to $0$ at infinity you talk about functions that "vanish at $\infty$". (Thanks to Qiaochu for mentioning this, as I missed it).
Note that any function with bounded support will necessarily vanish at infinity, but $y = e^{-x^2}$ shows you can have functions that vanish at $\infty$ but are not of bounded support.