# Laplace transform for distributions

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.

• @downvoter Why is my question inappropriate? Sep 30 '20 at 20:28
• Question seems right to me, I don't know why it has downvotes without an comment for the downvote-reason. Uptoving Sep 30 '20 at 21:16
• My friend, downvoters never waste their time with anything helpfull like leaving a comment, they are selfish and arrogant, they never will leave a comment Sep 30 '20 at 21:17
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 $$\rho$$.
• Thanks. Would you elaborate more, please? What do you mean by $\langle T(t), e^{-st} \rangle$? Oct 1 '20 at 9:44
• $\langle T(t), e^{-st} \rangle$ denotes $T$ applied to the function $t \mapsto e^{-st}$. For example, $\langle \delta(t), e^{-st} \rangle = e^{-s\cdot 0} = 1.$ Oct 1 '20 at 10:32
• So you are saying that if $T$ is a distribution with compact support the pairing is well-defined, otherwise we define $\langle T(t), \rho(t) e^{-st} \rangle$, right? Is this the way the Laplace transform is defined in the books? Also how have we taken into account the interval of integration which is from $0$ to $+\infty$ instead of $-\infty$ to $+\infty$? Oct 1 '20 at 13:11
• Not exactly. An ordinary distribution has a defined action only on compactly supported smooth functions. But if the distribution has compact support, we can extend the action to non-compactly supported smooth function by multiplying the test function with such a $\rho$. Since two such choices of $\rho$ only differ outside of the support of the distribution, this extension is well-defined. Oct 1 '20 at 13:25