Commutator identities and Fourier transform Is it possible to derive one side of the arrow below from the other by using only the Fourier transform and its reciprocal?
$$[\hat{p},f(\hat{x})]=-i\hbar f'(\hat{x}) \leftrightarrow [\hat{x},f(\hat{p})]=i\hbar f'(\hat{p})$$
 A: Under some hypotheses on $f$ the answer is positive. I consider the simplest case below.
If $U$ is a unitary operator on the Hilbert space $H$ and $A: D(A) \to H$ is a self-adjoint operator over the same Hilbert space, form spectral calculus it arises that $$Uf(A)U^{-1} = f(UAU^{-1})\tag{1}$$ for every measurable function $f : \mathbb R \to \mathbb R$.
Regarding momentum $P$ and position $X$ operators over $H= L^2(\mathbb R, dx)$, it holds $$U P U^{-1} =-X\:,\quad  U X U^{-1} =P \tag{2}$$
where $U : L^2(\mathbb R, dx) \to L^2(\mathbb R, dx)$ is the unitary operator given by Fourier(-Plancherel) transform. If $f : \mathbb R \to \mathbb R$ is for instance in the space ${\cal S}(\mathbb R)$ of Schwartz' functions (also using the fact that (1) this space is invariant under $X,P$ and functions of each such operator $g(X)$, $g(P)$ for $g \in {\cal S}(\mathbb R)$, (2) the explicit form of $P$ over ${\cal S}(\mathbb R)$, and that (3) $f \in {\cal S}(\mathbb R)$ entails $f' \in {\cal S}(\mathbb R)$),
$$[P, f(X)] =-i\hbar f'(X)\:.$$
As a consequence, since $U$ preserves ${\cal S}(\mathbb R)$,
$$[UPU^{-1}, Uf(X)U^{-1}]=  U[P, f(X)]U^{-1} =-i\hbar Uf'(X)U^{-1}\:.$$
From (1) and (2). the found result can be written
$$[X, f(P)]  =i\hbar f'(P)\:.$$
