# 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? Commented Sep 30, 2020 at 20:28
• Question seems right to me, I don't know why it has downvotes without an comment for the downvote-reason. Uptoving
– L F
Commented Sep 30, 2020 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
– L F
Commented Sep 30, 2020 at 21:17
• @GiuseppeNegro I couldn't find English translation. Do you know any other books? Commented Oct 1, 2020 at 13:18

A function $$\phi \in L^1_{\mathrm{loc}}(\Bbb R)$$ is Laplace-transformable if its support lies within $$[0,+\infty)$$ in the sense of $$L^p$$ functions and there exists $$\lambda \in \Bbb R$$ such that $$t\mapsto e^{-\lambda t}\phi(t) \in L^1(\Bbb R)$$. This is a sufficiently general request to ensure the absolute integrability of the Laplace transform: $$|(\mathfrak L\phi)(s)|\leq \int_0^\infty | e^{-(\operatorname{Re}(s)+i\operatorname{Im}(s))t} \phi(t)| dt \leq \| e_{{-\!}\operatorname{Re}(s)}\phi\|_{L^1(\Bbb R)} < +\infty,$$ for all $$s\in\Bbb C$$ such that $$\operatorname{Re}(s)>\lambda_*(\phi):=\inf\{\lambda \in \Bbb R\ |\ e_{-\lambda}\phi\in L^1(\Bbb R)\}.$$ (I use the notation $$e_z$$ to indicate the function $$t\mapsto e_{z}(t) := e^{zt}$$ for $$z\in\Bbb C$$.) The above condition identifies a half-plane in $$\Bbb C$$ where $$\mathfrak L\phi$$ is guaranteed to be absolutely convergent; there, it defines a holomorphic function that might be analytically extended to a larger domain. This is the case of the Heaviside function $$\Theta$$, for example, whose transform is $$(\mathfrak L\Theta)(s) = 1/s$$.
Extending this idea to distributions, it is natural then to say that $$u\in \mathcal D'(\Bbb R)$$ is Laplace-transformable if it is supported within $$[0,+\infty)$$ in the sense of distributions and there exists $$\lambda\in\Bbb R$$ such that $$e_{-\lambda}\cdot u$$ is tempered. A sensible guess for the definition of $$\mathfrak Lu$$ is the following: $$(\mathfrak Lu)(s) = \langle u, e_{-s}\rangle.$$ Here, angled brackets indicate evaluation of the distribution on the left upon the test function on the right. However, this formula is only sensible for distributions whose support is a compact subset of $$[0,+\infty)$$, like $$\delta$$ – which is the situation addressed by user @md2perpe in their answer. In this restricted setting (which certainly falls under the purview of our definition of Laplace-transformability), the Laplace transform is even an entire (i.e. everywhere holomorphic) function due to the powerful Paley–Wiener–Schwartz theorem, whereas we know from applications that in many practical cases the Laplace transform has poles somewhere in the complex plane. But our definition of Laplace-transformability comes to our aid. To see why, observe that formally speaking $$\langle u, e_{-s}\rangle = \langle e_{-\lambda} u, e_{-(s-\lambda)}\rangle.$$ If we fix $$s,\lambda$$ such that $$\operatorname{Re}(s)>\lambda > \lambda(u) := \{\lambda \in \Bbb R\ |\ e_{-\lambda}u\in\mathcal S'(\Bbb R)\}$$, then $$e_{-\lambda} u$$ is tempered and supported in $$[0,+\infty)$$, and $$e_{-(s-\lambda)}$$ is rapidly decreasing as its argument goes to $$+\infty$$. To retrieve a bona fide rapidly decreasing function, we can multiply it by a smooth cut-off $$\vartheta$$ that $$\equiv 1$$ on $$[0,+\infty)$$ and $$\equiv 0$$ on a neighbourhood of $$-\infty$$, which kills the exponential growth in that region. Then we are allowed to test $$e_{-\lambda} u$$ on $$\vartheta\cdot e_{-(s-\lambda)}$$, and we define $$(\mathfrak L u)(s) := \langle e_{-\lambda} u, \vartheta \cdot e_{-(s-\lambda)}\rangle.$$ One may check that this definition is actually independent of the value $$\lambda$$ and the cut-off $$\vartheta$$, so it makes sense for the whole half-plane $$\{\operatorname{Re}(s)>\lambda(u)\}$$.
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 $\langle T(t), e^{-st} \rangle$? Commented Oct 1, 2020 at 9:44
• $\langle T(t), e^{-st} \rangle$ denotes $T$ applied to the function $t \mapsto e^{-st}$. For example, $\langle \delta(t), e^{-st} \rangle = e^{-s\cdot 0} = 1.$ Commented Oct 1, 2020 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 $+\infty$? Commented Oct 1, 2020 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. Commented Oct 1, 2020 at 13:25