Is this Fourier Transform relation correct? Is the Fourier Transform of $$\nabla f\cdot \nabla g$$ from $\vec{x}$ to $\vec{k}$ space a convolution? 
I know that, for a certain definition of the FT, $$\nabla f\to \vec{k} F $$
where $F(\vec{k})$ is $f(\vec{x})$ after FT.
I suspect  the FT of $\nabla f\cdot \nabla g$ involves a convolution though? Something like
$$\int d^3 k' \,\,\vec{k}\, F(\vec{k})\cdot (\vec{k}-\vec{k}')G(\vec{k}-\vec{k}') \,\, ?$$
Sorry for the probably simple question. Also, if anyone has good comprehensive notes on FTs that'd be much appreciated too. Thanks!
 A: Lets try to figure it out. Let $\phi(\vec{k})$ be the Fourier transform of $\nabla f \cdot \nabla g $. 
$$  \nabla f \cdot \nabla g = \int \phi(\vec{k}) \ e^{i\vec{k}\cdot\vec{x}} \ \mathrm d^3\vec{k} $$
Now we write $f$ and $g$ in terms of their Fourier transforms. 
$$  \int \int \nabla \left(\hat{f}(\vec{k})e^{i\vec{k}\cdot\vec{x}}\right) \cdot \left(\hat{g}(\vec{q})e^{i\vec{q}\cdot\vec{x}}\right)\ \mathrm d^3\vec{k}\ \mathrm d^3\vec{q}  = \int \phi(\vec{k}) \ e^{i\vec{k}\cdot\vec{x}} \ \mathrm d^3\vec{k} $$
$$  \int \int \hat{f}(\vec{k})\hat{g}(\vec{q}) \left( \vec{k} \cdot \vec{q}\right) e^{i(\vec{q}+\vec{k})\cdot\vec{x}} \ \mathrm d^3\vec{k}\ \mathrm d^3\vec{q}  = \int \phi(\vec{k}) \ e^{i\vec{k}\cdot\vec{x}} \ \mathrm d^3\vec{k} $$
To kill some of the integrals I am going to apply the fourier transform to both sides, i.e., multiply both sides by $\exp(-ip\cdot x)/ (2 \pi)^3$ and integrating with respect to $x$. On the right we will get $\phi(p)$ on the left we will get a delta function which collapses one of the integrals. 
$$  \frac{1}{(2\pi)^3} \int\int \int \hat{f}(\vec{k})\hat{g}(\vec{q}) \left( \vec{k} \cdot \vec{q}\right) e^{i(\vec{q}+\vec{k}-\vec{p})\cdot\vec{x}} \ \ \mathrm d^3\vec{x} \mathrm d^3\vec{k}\ \mathrm d^3\vec{q}  = \phi(\vec{p}) $$
$$  \frac{1}{(2\pi)^3} \int\int  \hat{f}(\vec{k})\hat{g}(\vec{q}) \left( \vec{k} \cdot \vec{q}\right)(2\pi)^3 \delta^{(3)}(\vec{q}+\vec{k}-\vec{p})\  \mathrm d^3\vec{k}\ \mathrm d^3\vec{q}  = \phi(\vec{p}) $$
$$   \int  \hat{f}(\vec{k})\hat{g}(\vec{p}-\vec{k}) \left( \vec{k} \cdot (\vec{p}-\vec{k})\right) \  \mathrm d^3\vec{k}  = \phi(\vec{p}) $$
So it seems that the identity you propose it true, or at least consistent with my result. 
