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.
Laplace transform for distributions
distribution-theoryfourier transformlaplace transform
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Instead of doing that trick, we will use a clever property of the Fourier Transform involving homogeneity.
$\mathbf{\text{Def.}}$ A function $f\in\mathcal{S}(\Bbb R^n)$ is homogeneous of degree $s$ if $\forall a \neq 0$
$$f(ax) = a^sf(x)$$
$\mathbf{\text{Lemma.}}$ Let $f$ be homogeneous degree $s$. Then $\hat{f}$ is homogeneous degree $-n-s$.
Proof: $$ \begin{split} \hat{f}(ak) &= \frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\Bbb R^n}f(x)e^{-i(ak) \cdot x}dx \\ &= \frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\Bbb R^n}f\left(\frac{u}{a}\right)e^{-ik \cdot u}\frac{du}{a^n}\\ &=\frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\Bbb R^n}\left(\frac{1}{a}\right)^sf(u)e^{-ik \cdot u}\frac{du}{a^n} = a^{-n-s}\hat{f}(k) \end{split} $$
Even though we proved this in the Schwartz function case, the same property applies to tempered distributions, albeit with a lot more work.
Notice that $|x|^{-1}$ is radial and homogeneous degree $-1$. Using the above lemma, we have that its Fourier transform should be homogeneous with degree $-3+1 = -2$. The only possible radial homogeneous distribution with that degree is $C|k|^{-2}$ for some constant $C$.
There's a nice proof of this in Hörmander's book "The Analysis of Linear Partial Differential Operators I" (Theorem 7.1.14). The main tool needed is a distributional version of Fubini's theorem (Theorem 5.1.1 in Hörmander's book). I'll summarize the main points:
Preliminary step 1: Fubini's theorem
If $u$ and $v$ are distributions on ${\mathbb R}^m$ and ${\mathbb R}^n$ respectively, then there is a unique product distribution $u \otimes v$ on ${\mathbb R}^{m+n}$ characterized by the condition $$ (u \otimes v)(\varphi \otimes \psi) = u(\varphi) v(\psi) $$ for all $\varphi \in {\mathcal D}({\mathbb R}^m)$ and $\psi \in {\mathcal D}({\mathbb R}^n)$, where we write $(\varphi \otimes \psi)(x,y) := \varphi(x) \psi(y)$. Moreover, $u \otimes v$ can be evaluated on an arbitrary test function $\varphi \in {\mathcal D}({\mathbb R}^{m+n})$ by $$ (u \otimes v)(\varphi) = u(x \mapsto v(\varphi(x,\cdot))) = v(y \mapsto u(\varphi(\cdot,y))). $$ (Note that there is a slightly nontrivial exercise required to show that both expressions on the right make sense, e.g. that $x \mapsto v(\varphi(x,\cdot))$ defines a smooth compactly supported function on ${\mathbb R}^m$. This rests mainly on the fact that by uniform continuity, $x \mapsto \varphi(x,\cdot)$ is a continuous map to the space of test functions, and other arguments of this sort.) If you write down this formula in the case where $u$ and $v$ are given by locally integrable functions, you'll find that it follows easily from the classical Fubini's theorem.
If you know that the product distribution is unique, then the formula follows by verifying directly that both expressions on the right hand side define distributions that satisfy the defining property of a product distribution. Uniqueness can be proved via mollification: if $u \otimes v$ were not unique, then there would exist a nontrivial distribution $w$ on ${\mathbb R}^{m+n}$ such that $w(\varphi \otimes \psi) = 0$ for all $\varphi \in {\mathcal D}({\mathbb R}^m)$ and $\psi \in {\mathcal D}({\mathbb R}^n)$. Choose approximate identities, i.e. sequences of smooth functions $\rho_j : {\mathbb R}^m \to [0,\infty)$ and $\sigma_j : {\mathbb R}^n \to [0,\infty)$ with shrinking compact supports near $\{0\}$ that converge in the space of distributions to $\delta$-functions. Then the classical Fubini theorem implies that the sequence $\rho_j \otimes \sigma_j : {\mathbb R}^{m+n} \to [0,\infty)$ also defines an approximate identity in the same sense, and it follows that the sequence of smooth functions $(\rho_j \otimes \sigma_j) * w$ converges to $w$ in the space of distributions. But those functions are all $0$ due to the defining property of $w$, thus $w=0$.
Preliminary step 2: polynomial growth
Before we can view the function $g(\xi) := u(\chi(x) e^{-i x \xi})$ as a plausible candidate for the Fourier transform of $\chi u$, we need to know that it behaves reasonably enough at infinity to define a tempered distribution. As I indicated in my comments on the previous answer, $g$ is definitely not a Schwartz function in general, but one can show that it has polynomial growth. Perhaps the quickest way is to use standard properties of the Fourier transform and rewrite $g$ as $$ g(\xi) = \left( ( {\mathcal F}\chi)^- * {\mathcal F}^*u\right)(-\xi), $$ where I'm using the notation $f^-(x) := f(-x)$. As the convolution of a Schwartz function with a tempered distribution, it follows from standard results about the convolution that this function has polynomial growth.
The main argument
As stated in the question, we need to prove that the relation $$ \int_{{\mathbb R}^n} u(\chi(x) e^{-i x \xi}) \phi(\xi) \, d\xi = u\left( \int_{{\mathbb R}^n} \chi(x) e^{- i x \xi} \phi(\xi)\, d\xi \right) $$ holds for every $u \in {\mathcal D}'({\mathbb R}^n)$, $\chi \in {\mathcal D}({\mathbb R}^n)$ and $\phi \in {\mathcal S}({\mathbb R}^n)$. By step 2, we already know that both sides give well-defined tempered distributions when regarded as functionals of $\phi$, so by density, it will suffice to assume $\phi \in {\mathcal D}({\mathbb R}^n)$. The key observation now is that by the theorem in step 1, both sides can be identified with $(u \otimes 1)(f)$, where $1 \in {\mathcal D}'({\mathbb R}^n)$ is the distribution $1(\varphi) := \int_{{\mathbb R}^n} \varphi(x)\, dx$ and $f \in {\mathcal D}({\mathbb R}^{m+n})$ is given by $$ f(x,\xi) := \chi(x) \phi(\xi) e^{-i x \xi}. $$
Best Answer
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 $\{\operatorname{Re}(s)>\lambda(u)\}$.