Trouble deriving DE for fourier transform from DE of function I am trying to derive an equation which is a standard result in physics (the momentum space Schrödinger equation).
(Background: The wavefunction is a complex valued function of position coordinates and time $\psi:\mathbb{R^3}\times\mathbb{R^+}\rightarrow\mathbb{C}$. It is square integrable ($L^2$) and is generally assumed to behave such that as  ${r\rightarrow\pm\infty}$ then $\psi$ and $\nabla \psi$ fall off to zero sufficiently fast. Starting from the fact the position-space ($\psi$) and momentum space ($\phi$) wave functions are fourier transforms of each other):-
$$\phi(\mathbf{p},t) = \int \psi(\mathbf{r'},t) e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}d^3r'$$
(Integral runs over all space)
Taking the partial time derivative $$\partial_t\phi(\mathbf{p},t) =\partial_t\left(\int \psi(\mathbf{r'},t) e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}d^3r' \right) = \int \partial_t\psi(\mathbf{r'},t) e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}d^3r'$$
Because $t$ is a parameter while $p$ and $r$ are dynamical variables, i.e have no explicit $t$ dependence. Now using a postulated differential equation of $\psi$ $$i\hbar \partial_t\psi(\mathbf{r},t)=-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r},t) + V(\mathbf{r})\psi(\mathbf{r},t)  $$ I get $$i\hbar \partial_t\phi(\mathbf{p},t)= \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\left[ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r'},t) + V(\mathbf{r'})\psi(\mathbf{r'},t)\right]d^3r' $$
First Question: I moved my exponential term to the left, so that the operator $\hat{H} =-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbb{r}) $ acts only on $\psi(\mathbb{r},t)$ and not on $\psi(\mathbb{r},t)e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}$. What is the justification for this exchange mathematically?
Evaluating the first integral by using integration by parts and identifying the resulting integral as the fourier transform I get $$ -\frac{\hbar^2}{2m}\int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\nabla^2\psi(\mathbf{r'},t)d^3r' = \frac{p^2}{2m}\phi(\mathbf{p},t)$$
Second Question: I am having trouble understanding the second integral perhaps for the same reason as my first question: after fourier transforming $\psi$ $$ \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}V(\mathbf{r'})\psi(\mathbf{r'},t)d^3r'= \int \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}V(\mathbf{r'})e^{\frac{i}{\hbar}\mathbf{p'}\cdot\mathbf{r'}}\phi(\mathbf{p'},t)d^3r'd^3p' $$  I thought I had done it but when I looked it up, there was a caveat that $V(\mathbb{r})$ must be analytic and the following must be true:$$\color{blue}{e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}V(\mathbf{r'}) = V(i\hbar\nabla_p)e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}}$$
Under this assumption I have the correct equation in momentum representation. $$i\hbar \partial_t \phi(\mathbf{p},t) = \frac{p^2}{2m}\phi(\mathbf{p},t)+V(i\hbar\nabla_p)\phi(\mathbf{p},t)$$
I am not getting where the blue thing came from. Can someone explain what conditions hold for $\hat{f}(x)\hat{g}(y)=\hat{g}(y)\hat{f}(x)$ particularly in this context. 
Third Question: Any other less clumsy  way to derive this?
 A: First question: You didn't move anything on the right-hand side, you just Fourier-transformed both sides of the previous equation and then commuted the time derivative with the Fourier transform on the left-hand side.
Second question: You seem to be missing $\phi(\mathbf p,t)$ in the double integral, I believe the sign in the exponential with $\mathbf p'\cdot \mathbf r'$ is wrong, there's a supernumerary closing parenthesis in the equation after that, and you didn't introduce $\hat{f}(x)$ and $\hat{g}(y)$; the question may either resolve itself or become easier to answer if you address all these issues.
[Edit in response to the comment] To make the answer to the first question more explicit: Starting from the time-dependent Schrödinger equation,
$$i\hbar \partial_t\psi(\mathbf{r},t)=-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r},t) + V(\mathbf{r})\psi(\mathbf{r},t)\;,$$
you Fourier-transform both sides:
$$\int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}i\hbar \partial_t\psi(\mathbf{r'},t)d^3r'= \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\left[ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r'},t) + V(\mathbf{r'})\psi(\mathbf{r'},t)\right]d^3r'\;.$$
Then you exchange the time derivative and the integral:
$$\partial_t\int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}i\hbar \psi(\mathbf{r'},t)d^3r'= \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\left[ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r'},t) + V(\mathbf{r'})\psi(\mathbf{r'},t)\right]d^3r'\;.$$
Mathematically one might be interested in the conditions under which this is valid, but physicists tend not to worry about that sort of thing; in any case, this is certainly much less problematic than exchanging the Laplacian and the phase factor would be. Now you just have to substitute the definition of $\phi(\mathbf{p},t)$ to obtain
$$i\hbar\partial_t \phi(\mathbf{p},t)= \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\left[ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r'},t) + V(\mathbf{r'})\psi(\mathbf{r'},t)\right]d^3r'\;.$$
