Why is it necessary that test functions have finite support?

Question

For example, if $\phi(x)$ is a test function, which means smooth and with finite support the following is true: $$\lim\limits_{n->\infty} \int\limits_{-\infty}^{\infty} \delta_n(x)\phi(x)\mathrm{d}x = \phi(0)$$ I don't see why it is necessary to restrict $\phi(x)$ to have finite support, wouldn't this be true for every smooth function?

Answer 1

The requirement of $\phi$ having compact support allows us to interpret every locally integrable function $f$ as a distribution, because the integral $\int f\phi$ is guaranteed to converge (and have the right kind of continuity properties) as long as $f$ is locally integrable.

Allowing $\phi$ to have unbounded support comes with a tradeoff: some assumption must be made about the behavior of $f$ at infinity. The standard example is the definition of tempered distributions, which allows unbounded support of test functions $\phi$, provided that $\phi$ decays faster than any power of $x$. Then $\int f\phi$ converges provided that $f$ is bounded by some polynomial. But for example, $f(x) = e^x$ is not a tempered distribution.