Fourier Transform and Dirac Delta Function A Fourier transform of function R into function Q is defined as:
$$Q(\underline{k}) = \int_{}^{}R(\underline{x}) e^{-i\underline{k}·\underline{x}} \mathrm{d}\underline{x}.$$
where I've underlined $\underline{x}$ and $\underline{k}$ to denote that they are a set of vectors.
Suppose that $R(\underline{x})$ is only a function of coordinate differences, for example, $R(\underline{x})=$($x$1-$x$2)($x$3-$x$1), where $\underline{x}$=($x$1,$x$2,$x$3).
Why must the Fourier transform $Q(\underline{k})$ contain a Dirac Delta function?
Note: The fact that the Fourier transform must contain a Dirac Delta function was mentioned on the bottom of page 25 of the following set of notes by my Physics professor, http://www-thphys.physics.ox.ac.uk/people/JohnCardy/qft/qftMT2012.pdf
 A: A somewhat physical explanation is this:
Dependence on coordinate differences only implies translation invariance, which by Noether's theorem implies conservation of momentum, which in turn is represented as $\delta(\sum_k p_k)$ in $p$-space.
A mathematical example in 1 dimension:
$$\begin{align}
\mathcal{F}\{f(x_1-x_2)\}(p_1,p_2) 
&= \iint f(x_1-x_2) \, e^{-i(p_1 x_1+p_2 x_2)} \, dx_1 \, dx_2 \\
&= \{ \text{ variable change: } s = x_1 - x_2,\ t = x_2 \} \\
&= \iint f(s) \, e^{-i(p_1(s+t)+p_2t)} \, ds \, dt \\
&= \int \left( \int f(s) \, e^{-ip_1s} \, ds \right) e^{-i(p_1+p_2)t} \, dt \\
&= \int \mathcal{F}\{f\}(p_1) e^{-i(p_1+p_2)t} \, dt \\
&= \mathcal{F}\{f\}(p_1) \int e^{-i(p_1+p_2)t} \, dt \\
&= \mathcal{F}\{f\}(p_1) \, 2\pi \, \delta(p_1+p_2)
\end{align}$$
A: This is not totally rigorous, but for Physics it should be enough :-)
In general, if $Q(\underline{k}) $ is the transform of $R(\underline{x})$, then the transform of $R(\underline{x} + a \underline{u})$ where $ \underline{u}=(1,1,1)$ and $a$ is some scalar, is
$$Q'(\underline{k})=\exp(i 2 \pi a \,  \underline{u} \cdot \underline{k}) Q(\underline{k})  = \exp(i 2 \pi a (  k_1+k_2 +k_3)) \, Q(\underline{k})  \tag{1}$$
But if $R(\underline{x})$  is only a function of coordinate differences, then $R(\underline{x} + a \underline{u})=R(x)$, and $Q'(\underline{k})=Q(\underline{k})$ for all $a$. 
Then, from $(1)$, $$Q(\underline{k})\ne 0 \implies \exp(i 2 \pi a (  k_1+k_2 +k_3))=1 \hskip{5mm} (\forall a) \implies k_1+k_2 +k_3 = 0$$ 
That is, $Q(\underline{k})$ must be zero everywere except (at most) a measure zero set (a plane in $\mathbb{R}^3$). Then either $Q(\underline{k})$ is identically zero (trivial) or it's "degenerate" (has Dirac deltas).
