# Compactly supported function.

Can anyone explain me the features of a compactly supported functions behave when they are compactly supported. I am learning PDE and I come across it very often.

For example : When we define weak derivatives i.e
$\int_U uD^\alpha f=(-1)^{|\alpha|}\int_U vf$ , why do we take a function $f$ to be compactly supported ?

I wonder if it is has to do with the non-differentiability of $u$ at some point in the domain ?

Looking forward. Thanks

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A function $f$, not a set is compactly supported. This means that the closure of the set $\{x:f(x)\neq 0\}$ is a compact set. – Alex Becker May 9 '12 at 5:06
@ Alex , as i have come across behaviour of compactly supported functions is very useful and notable. Can you explain a bit with some example or give a an idea where i can learn about it . :) – Theorem May 9 '12 at 5:10
Please read the title and body of the question again, it could use some proofreading :) Also, could you please include your comment in the body of your question? As it stands the only answer to your question is: the only feature all compactly supported functions share is that they are, well, compactly supported... – t.b. May 9 '12 at 5:11
I downvoted. "This question does not show any research effort." I encourage you to edit your question, give it some background and direction, and to be a bit more specific. I'm confident that if you put some more time into it, more users here would spend time on writing an answer. – mixedmath May 9 '12 at 5:27
"Can anyone explain me all the features...?" In my opinion, no. – user16299 May 9 '12 at 6:05

I cannot explain all the features, but one nice property of a compactly supported function $f$ defined on an open set $\Omega$ is that there exists a compact set $K$ in $\Omega$ such that $f(x)=0$ if $x\not\in K$. This is useful since for instance when you do integration by parts on $f$, the boundary terms vanish. If $g$ is any function, you can get a compactly supported function $fg$ by multiplying it by $f$. This is called the cut-off process.