Let $f=p(x)e^{-x^2}$, where $p \in \mathbb{C}[x]$. I am trying to find out if the Fourier transform (FT) of $$\sum_{j=1}^n i^{j+k}\frac{d^k}{dx^k}(x^{j}\cdot f) \quad (\text{for some fixed } k \text{ such that } 0\le k\le n)$$ is equal to $$\left(\sum_j \omega^k\frac{d^j}{d\omega^j}\right) \left(\hat{f}\right)\text{?}$$ (Here $\hat{f}=\mathscr{F}(f)$ denotes the FT of $f$.)
I thought about using the property that the Fourier transform transforms differentiation into multiplication and vice-verse (i.e., $\mathscr{F}(\frac{d}{dx}f)=i\omega \mathscr{F}(f)$ and $\mathscr{F}(x\cdot f)=i\frac{d}{d\omega}\mathscr{F}(f)$) and obtained:
$$\mathscr{F}\left(\sum_j i^{j+k}\frac{d^k}{dx^k}(x^j \cdot f) \right) = \sum_j i^{2(j+k)}\omega^k \frac{d^j}{d\omega^j}\hat{f}.$$
There is now a small problem of getting rid of the $i$'s. I was wondering if there is something about the FT of $f$ that would perhaps "absorb" them? Sorry if that sounds ridiculous; indeed I am desperate to find any reason that allows to get rid of them. This area of mathematics isn't exactly my speciality and I have stumbled upon a problem that requires this result to be true. I would appreciate any help at all with this problem. Thanks.
Edit. I think the proof can be fixed if I define $\Phi$ differently, but unfortunately I can't think of how. The problem, for anyone interested, came from here: https://www.math.ru.nl/~tcrisp/teaching/crisp_497B.pdf (see pg. 80). See this image https://i.stack.imgur.com/R60p2.png for the incomplete 'proof' that I have written so far.
Edit. I have found that the answer to my initial question is negative. The mapping $\Phi$ as defined in the linked image is not an $A_1$-module homomorphism. First note that because $A_1$ is an algebra, it suffices to prove that $\Phi$ is linear and $\Phi(sf)=s\Phi(f)$ for any $s$ in the generating set of $A_1$ (so only $x$ and $\frac{d}{dx})$ and any $f$ in $M_\sigma$. For $\Phi : f \mapsto \hat{f}$, we have $\Phi(x \cdot f)=-x \hat{f}$ and $\Phi(\frac{d}{dx}\cdot f)=-\frac{d}{dx}\hat{f}$. Fortunately, this does suggest a remedy; define $\Phi$ as $f\mapsto -\hat{f}$.