Multidimensional Fourier transform of the laplacian In my course on electromagnetic field theory we use the Fourier transform to simplify Maxwell's equations, for example:
$$\frac{\partial ^2\vec E(\vec r,t)}{\partial t^2} \rightleftharpoons -\omega^2\mathcal{F}_t\vec E(\vec r,\omega)$$
I haven't officially studied multidimensional fourier transforms but I've been exposed to them through MIT's set of courses "The fourier transform and its applications" (Brad Osgood). I tried transforming things like the Laplacian of a function and this is what I got:
Start by considering the $\mathbb{R}^3$ FT of the second x-derivative of some nice function f and manupulate the integral to be in terms of the one dimensional FT with respect to x:
$$\mathcal{F}_{xyz}\{\frac{\partial ^2f(\vec r)}{\partial x^2}\}=\int\limits_{\mathbb{R}^3}e^{j\vec\xi\cdot\vec r}\frac{\partial ^2f(\vec r)}{\partial x^2}dxdydz=\int\limits_{\mathbb{R}^2}e^{j(\xi_2y+\xi_3z)}\mathcal{F}_x\{\frac{\partial ^2f(\vec r)}{\partial x^2}\}dydz$$
and so, using a well known identity for the FT of a derrivative:
$$ \mathcal{F}_{xyz}\{\frac{\partial ^2f(\vec r)}{\partial x^2}\}= \int\limits_{\mathbb{R}^2}e^{j(\xi_2y+\xi_3z)}(j\xi_1)^2\mathcal{F}_x\{f\}dydz=-\xi_1^2\mathcal{F}_{xyz}\{f\}$$
and since that is the case we can conclude that:
$$\mathcal{F}_{xyz}\{\Delta f\}= -(\xi_1^2+\xi_2^2+\xi_3^2)\mathcal{F}_{xyz}f= -\|\vec\xi \|^2\mathcal{F}_{xyz} f $$
The reason I have for doubting this is that if I use this to transform the homogeneous wave equation for the electric field:
 $$ \Delta \vec E-\mu\epsilon\frac{\partial ^2\vec E}{\partial t^2}=0$$
I get:
$$-( \|\vec\xi \|^2-\omega^2\mu\epsilon)\mathcal{F}\vec E=0 \Rightarrow \mathcal{F}\vec E=0$$
This seems to say that $\vec E$ is equal to zero for all points in space and time (which is definitely false). So either I've got my multidimensional FT wrong (maybe a $\mathcal{F}f=0$ does not imply $f=0$ like in the one dimensional case?) or I've got my electromagnetics wrong in which case this isn't the right forum. 
Anyway, I'm interested to hear anything you have to say on the subject, I've had a hard time finding anything relating the Fourier transform and vector calculus so any pointers on where to look is appreciated as well.
 A: Under the circumstances, $\mathcal{F}\vec{E}=0$  iff  $\mathcal{F}\vec{E}\,$ is locally integrable, e.g., continuous  in $\,\vec{\xi},\omega\,$.  But when you stick to the common sense, this cannot be the case, since for $\mathcal{F}\vec{E}\,$ to be continuous in $\,\vec{\xi},\omega\,$, a vector field $\vec{E}\,$ is generally required to be absolutely integrable in $\,\vec{x},t\,$ over the space-time $\mathbb{R}^4$, which for sufficiently regular $\vec{E}\,$ necessarily implies that $\vec{E}\to 0\,$ as $t\to\pm\infty$. More precisely, 
$$E_j\in L^1(\mathbb{R}^4)\;\Rightarrow\; \mathcal{F}E_j\in C_0(\mathbb{R}^4),
\quad j=1,2,3,$$
while 
$$
E_j\in L^2(\mathbb{R}^4)\;\iff\; \mathcal{F}E_j\in L^2(\mathbb{R}^4),\quad j=1,2,3.
$$
In your case, $\mathcal{F}E_j\,$ will be a generalized function, otherwise called a disribution, supported on the cone $\,\|\vec{\xi}\|^2-\mu\epsilon\omega^2=0\,$ in $\,\mathbb{R}^4$.  To avoid stumbling upon the question of What is time?, it is better to apply the Fourier transform only in spatial variables. This way, solving a Cauchy problem for the 2nd-order ODE, you will readily find 
$$
\mathcal{F}\vec{E}(\vec{\xi},t)=\mathcal{F}\vec{E_0}(\vec{\xi})\cos{(a\|\xi\|t)}+
\mathcal{F}\vec{E_1}(\vec{\xi})\frac{\sin{(a\|\xi\|t)}}{a\|\xi\|},\quad a\overset{def}{=}\frac{1}{\sqrt{\mu\epsilon}}\;,$$
with the initial data
$$
\vec{E}_0=\vec{E}|_{t=0}\;,\quad \vec{E}_1=\frac{\partial \vec{E}}{\partial t}\biggr|_{t=0}\;. 
$$ 
The inverse Fourier transform
$$\mathcal{F}^{-1}\biggl(\frac{\sin{(a\|\xi\|t)}}{a\|\xi\|}\biggr)=\delta_{at}\;,\,t>0,  $$ 
where $\,\delta_{at}\,$ denotes a $\,\delta$-function supported on the sphere $\,|x|=at\,$. Hence follows
$$\mathcal{F}^{-1}\bigl(\cos{(a\|\xi\|t)}\bigr)=\frac{\partial\,}{\partial t} \delta_{at}\;,  $$ 
whence by convolution theorem follows the classical Kirchhoff's formula for a 3D wave equation. Applying the Fourier transform to a 3D wave equation you cannot bypass using generalized functions, otherwise widely known as distributions.
For a physicist, the best mathematically rigourous textbook on the topic of Generalized Functions, otherwise Distributions, seems to be Methods of the Theory of Generalized Functions by V.S. Vladimirov (Taylor & Francis, 2002). 
A: You arrive at the equation
$$-( \|\vec\xi \|^2-\omega^2\mu\epsilon)\mathcal{F}\vec E=0$$
The solutions to this equation are $\vec E=0$ (trivial solution) or alternately:
$$ \|\vec\xi \|^2-\omega^2\mu\epsilon=0$$
If you rearrange it and rewrite it in terms $\mu\epsilon=1/c^2$ you get the solution
$$\frac{\omega}{\|\vec\xi \|}=c$$
which should be familiar (quite well known in physics). I'm not extremely knowledgeable with multidimensional fourier transforms, but it suggests you're doing it right if you've derived this equation.
