Laplace transform of distributions

Remarks

For $\operatorname{Re} s > \sigma_0$, the function $\chi(t) e^{-(s - \sigma_0) t}$ lies in the Schwartz space, so the pairing makes sense. The value does not depend on the choice of $\chi$ or of $\sigma_0$, because $T$ is supported in $[0, \infty)$; formally it is $\langle T, e^{-st} \rangle$. $\mathcal{L} T$ is holomorphic on $\operatorname{Re} s > \sigma_0$. For a locally integrable function $f$ of exponential order vanishing on $(-\infty, 0)$, it agrees with the ordinary Laplace transform $\int_0^\infty e^{-st} f(t) \, dt$. For all such $T$, $\mathcal{L}(T') = s \, \mathcal{L} T$, and the transform of a convolution is the product of the transforms.

Example: the Dirac delta $\delta$ has support $\set{0}$ and is tempered, so $\sigma_0 = 0$ works, and $\mathcal{L}\delta(s) = \chi(0) e^{0} = 1$. Likewise $\mathcal{L}\delta'(s) = -\frac{d}{dt}\left(\chi(t) e^{-st}\right)\big|_{t=0} = s$. With the Heaviside step function $H$, $\mathcal{L} H = 1/s$ and $H' = \delta$, consistent with $\mathcal{L}(H') = s \cdot \frac{1}{s} = 1$.

Non-example: $T = H(t) e^{t^2}$ is a distribution supported in $[0, \infty)$, but the positive function $e^{-\sigma_0 t} e^{t^2}$ grows faster than any polynomial. Hence $e^{-\sigma_0 t} T$ is tempered for no $\sigma_0$, and $T$ has no Laplace transform.