Laplace transform for distributions

Question

The Fourier transform for tempered distributions is well-known. It's defined by $$\langle \mathcal{F}T , \phi\rangle = \langle T,\mathcal{F}\phi\rangle$$For any Schwartz function $\phi$. For ordinary functions, it's defined by $$\mathcal{F}f(s) = \int_{-\infty}^{+\infty}e^{-2\pi ist}f(t)dt$$On the other hand, the unilateral Laplace transform for the ordinary functions is $$\mathcal{L}f(s) = \int_{0^{-}}^{+\infty}e^{-st}f(t)dt$$Where $s \in \mathbb{C}$. Is it possible to take Laplace transform of distributions? How is it defined, then? It's known that $\mathcal{L}\delta(t) = 1$ but I don't know if it's rigorous since $\delta(t)$ is not an ordinary function.

Answer 1

If $T$ is a distribution with compact support then $\langle T(t), e^{-st} \rangle$ is well-defined. One can take $\rho \in C_c^\infty$ such that $\rho \equiv 1$ on a neighborhood of the support of $T$ and define $\langle T(t), e^{-st} \rangle = \langle T(t), \rho(t) e^{-st} \rangle$. The result doesn't depend on the choice of $\rho$.