Using the shift property of the Fourier transform, $e^{-ix_0k}G(k)$ is the Fourier transform of $g(x-x_0)$.
In my case, $g$ is a Gaussian function $g(x-x_0)=e^{-\sigma (x-x_0)^2}$ , so $G(k)$ is also a rescaled Gaussian.
I have reason to believe that for arbitrary (differentiable) $F$
$F(k)*(e^{-ix_0k}G(k))=e^{ix_0k}(F(k)*G(k))$
where $*$ represents convolution.
But my proof of this is long-winded. Likely unnecessarily so. I want to confirm by some other means that this is the case. It looks like a simple result, and I feel there must be a more obvious way to demonstrate it.
So my question is simply:Is it true that $F(k)*(e^{-ix_0k}G(k))=e^{ix_0k}(F(k)*G(k))$ ? And if so, how can it be proven?
I think the result is not true. Since there is a one-one mapping between a function and it's Fourier transform, (In the proceeding transformations, trailing constants like factors of $2 \pi$ have been omitted and the --> symbol means taking a Fourier transform)
If, $$ g(x) --> G(k) \quad \text{and}$$ $$ f(x) --> F(k)$$ then, $$f(x)g(x - x_0) --> F(k)*(e^{-ix_0k}G(k)) \tag{1}$$ However, the other function has inverse FT $$\delta(x+x_0)*(f(x).g(x)) --> e^{ix_ok}(F(k)*G(k)) \tag{2}$$ Where $\delta$ is the delta function/Impulse function
If we define $h(x) = g(x)f(x)$, the LHS reduces to $$h(x+x_0) = f(x+x_0)g(x+x_0)$$ Which is not the same as the LHS of (1)