A formula used in Quantum Field Theory

Question

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?

Answer 1

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

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