TITLE: A formula used in Quantum Field Theory
QUESTION [7 upvotes]: There is a formula:
$$ \left. F \left( -i \frac{\partial}{\partial \mathbf{x}} \right) G(\mathbf{x}) = G \left( -i \frac{\partial}{\partial \mathbf{y}} \right) F(\mathbf{y}) e^{i\mathbf{x} \cdot \mathbf{y}} \right|_{\mathbf{y}=0} $$
In perturbation theory this formula is used in the form: 
$$ \left. \exp \left[ -\int d^4z\, V\left( \frac{-\delta}{\delta J(z)} \right)\right] \exp\left[{\frac{1}{2}\int d^4x d^4y\,J(x)\Delta(x-y) J(y)}\right] \\ = \exp\left[\frac{1}{2}\int d^4x d^4y\,\frac{\delta}{\delta \phi(x)}\Delta(x-y) \frac{\delta}{\delta \phi(y)}\right] \exp\left[{-\int d^4z\, V(\phi(z))-\int d^4x\,\phi(x) J(x)}\right] \right|_{\phi=0} $$
I know one way of proving it, which way is shown by Cosmas Zachos.
And I hear this formula can be proved by Fourier analysis too.
How can we prove?

REPLY [8 votes]: $$ \left.  G \left( -i \frac{\partial}{\partial \mathbf{y}} \right) F(\mathbf{y}) e^{i\mathbf{x} \cdot \mathbf{y}} \right|_{\mathbf{y}=0}  =\\
=\left.  G \left( -i \frac{\partial}{\partial \mathbf{y}} \right) F(  -i \frac{\partial}{\partial \mathbf{x}}) e^{i\mathbf{x} \cdot \mathbf{y}} \right|_{\mathbf{y}=0}  =\\ 
= \left.  F\left( -i \frac{\partial}{\partial \mathbf{x}} \right) 
 G \left( -i \frac{\partial}{\partial \mathbf{y}} \right)
 e^{i\mathbf{x} \cdot \mathbf{y}} \right|_{\mathbf{y}=0} =\\ 
= \left.  F\left( -i \frac{\partial}{\partial \mathbf{x}} \right) 
 G \left(\mathbf{x}\right)
 e^{i\mathbf{x} \cdot \mathbf{y}} \right|_{\mathbf{y}=0} =\\ 
=F\left( -i \frac{\partial}{\partial \mathbf{x}} \right) G(\mathbf{x}) ~. $$
The separation of conjugate variables is the same as in working in Fourier space.

Edit in response to comment.  Detail of equivalent integration by parts in the end, 
$$
F(-i\partial_x)G(x)=\int dy F(y) \tilde{G}(y) e^{ixy}/\sqrt{2\pi}= \\
=\int dy \left( \frac{1}{2\pi}\int dz  e^{-izy} G(z)\right) F(y) e^{ixy}=
$$
$$
=\int dy \left( \frac{1}{2\pi}\int dz  e^{-izy} G(i\stackrel{\leftarrow}{\partial} _y) \right) F(y) e^{ixy}=\\
=\int dy \left( \frac{1}{2\pi}\int dz  e^{-izy}  \right)  G(-i {\partial} _y)F(y) e^{ixy}=\\
=\int dy~  \delta(y) ~ G(-i {\partial} _y)F(y) e^{ixy} ~.
$$