$L^2$ regularity of a convolution with Newtonian potential I am reading Vorticity and incompressible flow (Bertozzi, Majda) and on page 71-72, we are concerned with recovering the velocity field of a flow from its vorticity.
At some point we need to have the $L^2$ regularity of the second derivatives of some vector field which is defined trough a convolution, more explicitely take:
$\psi (x) := \int_{\mathbb{R}^3}\frac{w(y)}{|x-y|}dy$, $h:= \operatorname{curl}\operatorname{curl} \psi$ and $k:= \nabla \operatorname{div} \psi$.
Then in the book it is said ' Because $w \in L^2$ is smooth and vanish sufficiently rapidly as $|x| \nearrow \infty$, we have $h(x)= O(|x|^{-3})$ and $k(x)=O(|x|^{-3})$  for $|x| \gg 1$, so $h$, $k$ $\in L^2$.' with not any more details.
So what I don't understand is why h and k should be $L^2$. I mean even if we consider that we can differentiate two times under the integral, we should get something that should be like (roughly speaking) $\int_{\mathbb{R}^3}\frac{w(y)}{|x-y|^3}dy$ which is $\frac{1}{|x|^3}\ast w = 1_{B(0;1)}\frac{1}{|x|^3}\ast w + 1_{B(0;1)^c}\frac{1}{|x|^3}\ast w$ and assuming that w is in $L^1$ (which I don't know if it can come out from: $w$ is vanishing sufficiently rapidly as $|x| \nearrow \infty $) the second term is in $L^2$ this is ok, but the second term is a convolution with a kernel that belongs to no $L^p$,  $p\geq 1$ so I am a bit confused. But maybe I am not attacking this with the right angle, so any help will be much appreciated, because for now this is a little messy. I don't even know the exact assumptions we need for the result to hold.
 A: In "because $w\in L^2$ is smooth"  the emphasis is on smooth.  The convolution of $w$ with an $L^1$ function is as smooth as $w$ itself (assuming nothing bad comes from the tail at infinity). This is because we can put the derivative on $w$: 
$$
\nabla_x \int w(y)\phi(x-y)\,dy  = \nabla_x \int w(x-y)\phi(y)\,dy  = \int \nabla_x w(x-y)\phi(y)\,dy 
$$
and if the last integral converges absolutely, the formal differentiation is justified. 
In your case $\phi(x)=1/|x|$ is not globally integrable, but it is locally integrable. So, let's introduce a smooth cutoff function $\chi$ such that 


*

*$\chi(x)=1$ when $|x|\le 1$

*$\chi(x)=0$ when $|x|\ge 2$


Then split the kernel as $\phi_1(x) = \chi(x)/|x|$ and $\phi_2(x) = (1-\chi(x))/|x|$. 
Observe that $\phi_2$ has bounded derivatives of all orders, with appropriate decay at infinity: $D^k \phi_2 = O(1/|x|^{k+1})$. Thus,   you can write
$$
D^k_x \int w(y)\phi_2(x-y)\,dy =\int w(y) D^k_x \phi_2(x-y)\,dy \tag{1}
$$
and estimate the integral accordingly.
With $\phi_1$, differentiate $w$ itself, 
$$
D^k_x \int w(y)\phi_1(x-y)\,dy  = \int D^k_x w(x-y)\phi_1(y)\,dy  \tag{2}
$$
The integral converges because $\phi_1$ is compactly supported and $w$ is smooth. However, the decay of $(2)$ as $x\to\infty$ is unclear without an assumption on the derivatives of $w$. I think the authors may have meant that $w$ decays at infinity together with its first and second derivatives. 
