TITLE: compact support intuition needed
QUESTION [6 upvotes]: I am studying Distribution theory. But I am curious about that why we coin compact support.  In what situation is it useful? Can any one give an intuitive way to explain this concept?

REPLY [4 votes]: Why do we require that $\varphi(x) = 0$ for $x$ large enough  ? Because it allows us to integrate by parts without fear : 
$$\int_{-\infty}^\infty T(x) \varphi'(x)dx = \lim_{x \to \infty} T(x)\varphi(x)-T(-x)\varphi(-x)-\int_{-\infty}^\infty T'(x) \varphi(x)dx$$
Here $\varphi \in C^\infty_c$ and $T$ is a distribution, so $T(x)\varphi(x)$ doesn't make sense, but if you assume that $\varphi(x) = 0$ for $|x| > M$ then clearly $\lim_{x \to \infty} T(x)\varphi(x)-T(-x)\varphi(-x) = 0$ and it makes sense to write
$$\langle T,\varphi' \rangle =\int_{-\infty}^\infty T(x) \varphi'(x)dx = -\int_{-\infty}^\infty T'(x) \varphi(x)dx=-\langle T',\varphi \rangle \tag{1}$$
which is exactly what we need for defining $\delta'$ the derivative of the Dirac delta and the derivatives of distributions in general.
Now the big question : how do you prove that $(1)$ makes sense (that it doesn't lead to some contradictions) ? Well you can take it as a definition, so nothing to prove, 
or you can show that when defining the distributions as linear operators $C^\infty_c \to \mathbb{R}$ continuous for the test function space topology,  then the differentiation operator $\langle T,.\rangle \mapsto \langle T',.\rangle$ is continuous in the sense of distributions.

See also the Schwartz space, where we replace the compact support property by a decay $o(x^{-k})$ at $\infty$, discarding the distributions with a too large grow rate and keeping the so-called tempered distributions.