Fourier transform of distribution without physicist's $\delta$-function

Say I want to find the Fourier transform of the following distribution: The solution I have uses the physicist's definition of the Dirac $\delta$-function $\delta(x) = \int_\mathbb{R} e^{-i x y} dy$ which is mathematically ill-defined: I cannot figure out how to solve this problem in a mathematically rigorous way, using only the allowed operations on distributions, though it should be possible to do...

Any help is appreciated.

• First, $\langle U, \phi\rangle = \langle 1\otimes \delta\rangle$; second, you can apply FT to each part of the tensor product separately. – TZakrevskiy Jan 31 '17 at 9:13

First - I will apply Foubini theorem a lot of times, since test functions are very integrable=) Second - I will systematically omit all normalisation constants in Fourier transform, as well as integration limits. Third - I will denote by $\phi_y(x) = \phi(x,y)$. $\forall y\quad \phi_y\in D(\Bbb R^m)$. Fourth - the corresponding Fourier variables to $(x,y)$ will be $(\xi,\eta)$.
$$\langle \hat U, \phi(x,y)\rangle = \left\langle U, \int\phi(x,y) \exp(ix\xi + i y\eta) dx\,dy\right\rangle$$ $$=\int \left(\int\phi(x,y) \exp(ix\xi + i y\eta) dx\,dy\right)\big|_{\eta=0}d\xi$$ Since everything is smooth, we can pass the value of $\eta=0$ inside the integral: $$=\int \left(\int\phi(x,y) \exp(ix\xi) dx\,dy\right) d\xi$$ $$=\int dy\int d\xi \left(\int\phi_y(x) \exp(ix\xi) dx \right)$$ $$=\int dy \left\langle 1, \hat \phi_y(\xi)\right\rangle$$ (in the last integral $\hat \phi_y(\xi)$ is a FT with respect to $x$, and $1$ is a distribution acting on the variable $\xi$) $$=\int dy \left\langle \hat 1, \phi_y(x)\right\rangle$$ $$=\int dy \left\langle \delta_0, \phi_y(x)\right\rangle$$ $$=\int \phi_y(0) dy = \int \phi(0,y)dy.$$
In other words, you initial distriubtions is, in fact, a tensor product $1\otimes \delta_0$; you apply FT to this distribution, and you will get $$F[1\otimes \delta_0] = F\otimes F[\delta_0] = \delta_0\otimes 1,$$ which coreponds to out previous result.