Is there a reason that test functions are chosen to have a compact support?

Question

If $f$ is a distribution, we choose test functions from $C^\infty_c(\mathbb R^n)$. My question is,

1. Is there a reason that we want test function to be smooth?
2. Why do we want the function to have a compact support? Can't we have replace the condition to be $f \in L^1(\mathbb R^n)$ (or $L^p$)?

Thank you in advance.

Answer 1

1. By having smooth test functions we can define derivatives of distributions of all orders: $$\langle u^{(k)}, \phi \rangle = (-1)^k \langle u, \phi^{(k)} \rangle.$$

2. By having test functions with compact support we can

   1. give distributions a local identity,
   2. make extremely rapidly growing functions like $e^{e^{x^2}}$ be distributions.