Fourier transform of a functional derivative Suppose that $x(q)$ is the Fourier transform of the function $x(r)$, where $r$ is the real-space variable and $q$ is the Fourier-space variable. Then, suppose that $E$ is an energy functional which can be differentiated either in real-space (with respect to $x(r)$) or Fourier-space (with respect to $x(q)$).
Representing the Fourier-transform operator as FT[], is the following relation logically valid?
$$\mathrm{FT}\left[\frac{\delta E}{\delta x(r)}\right] = \frac{\delta E}{\delta x(q)}$$
If not, what would be the correct mapping between these functional derivatives?
 A: *

*Note that if a functional $S[\phi]$ is local in position space with variable $x$, it is typically not local in wavevector space with Fourier transformed variable $k$, where$^1$
$$\widetilde{\phi}(k)~=~ \int \!d^dx~e^{-ik\cdot x}\phi(k)\tag{1}$$
is the Fourier transform.


*In physical applications, it is often convenient to define functional derivatives in $x$- and $k$-space with slightly different conventions for an infinitesimal variation:
$$\begin{align}\delta S~=~&\int \!d^dx~\frac{\delta S}{\delta \phi(x)}\delta\phi(x)\cr
~=~&\int \!\frac{d^dk}{\color{red}{(2\pi)^d}}~\frac{\delta S}{\delta \widetilde{\phi}(\color{red}{-}k)}\delta\widetilde{\phi}(k),\end{align}\tag{2}$$
which is inspired by the Plancherel theorem/convolution theorem.


*Then it is easy to see that OP's conjecture is correct:
$$ FT\left[x\mapsto \frac{\delta S}{\delta \phi(x)}\right](k)~=~ \frac{\delta S}{\delta \widetilde{\phi}(k)}.\tag{3} $$


*If one doesn't like to use a different convention (2) for the functional derivative in $k$-space marked in $\rm \color{red}{red}$ color, then the formula (3) will be modified accordingly.
--
$^1$ Here we assume for simplicity a real field $\phi$, and leave to the reader to generalize to a complex field.
A: Without paying much attention to mathematical hypotheses (however they can be fixed), your idea is correct in view of the following "proof".
$$E[\hat{x}(q)]= E\left[ \frac{1}{(2\pi)^{n/2}}\int e^{irq} \hat{x}(q) dr\right]$$
Hence
$$\int \frac{\delta E}{\delta \hat{x}(q)} \hat{h}(q) dq = \frac{d}{d\alpha}|_{\alpha=0} E\left[ \frac{1}{(2\pi)^{n/2}}\int e^{irq} (\hat{x}(q) + \alpha \hat{h}(q)) dr \right] $$
$$=\frac{d}{d\alpha}|_{\alpha=0} E\left[ \frac{1}{(2\pi)^{n/2}}\int e^{irq} \hat{x}(q) dr+ \alpha (2\pi)^{-n/2}\int e^{irq} \hat{h}(q) dr \right] $$
$$ =\frac{d}{d\alpha}|_{\alpha=0} E\left[x(r)+ \alpha (2\pi)^{-n/2}\int e^{irq} \hat{h}(q) dr \right] = \int  \frac{\delta E}{ \delta x(r)} 
 \frac{1}{(2\pi)^{n/2}}\int e^{irq} \hat{h}(q) dr dq$$
Using the Plancherel theorem, the found integral can be recast to
$$\int \left( \frac{1}{(2\pi)^{n/2}}\int e^{-iqr} \frac{\delta E}{ \delta x(r)} dr\right) \:
 \hat{h}(q)  \:dq$$
This identity says that
$$  \frac{\delta E}{\delta \hat{x}(q)}  = \frac{1}{(2\pi)^{n/2}}\int e^{-iqr} \frac{\delta E}{ \delta x(r)} dr$$
which is your thesis.
