$L^2$ 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.

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.

• The author says that $w$ "is smooth", not just any $L^2$ function. I don't have access to the book, but usually in the PDE context "smooth" means $C^\infty$. If so, there are no issues with local behaviour of the derivatives of $w$. – user147263 Jan 2 '15 at 23:21
• Hello Dear @Behavior, First of all thank you for taking time to think with me on my problem. Lets write as you said $\int w(x-y)\phi(y) dy = \int_{|y|<1} w(x-y)\phi(y)dy + \int_{|y|>1}w(x-y)\phi(y)dy =: A_1(x) + A_2(x)$ If we differentiate $A_1$ by taking the derivative of $w$, (assuming that we are able to differentiate under the integral) we obtain the convolution between $\nabla w$ and $1_{B(0,1)}\phi$.The second function is indeed $L1(\mathbb{R}_3)$ but for the convolution to make sense we need some regularity about $\nabla w$ in particular if it is $L_2$ then $\nabla A_1(x)$ is $L_2$. – incas Jan 2 '15 at 23:26
• Yes smooth does mean $C^{\infty}$ but when you cut $\phi$ on the ball in the convolution $\int_{|y|<1} w(x-y)\phi(y)dy$ and you differentiate with respect to x under the integral sign and then study the $L_2$ regularity I don't see how just "smooth" is enough. Take $w=exp^{-1}(|x|)sin(exp^2(|x|))$ outside the unit ball and make it smooth inside. Then $\nabla w$ would be something like $exp(|x|)cos(exp(|x|)) + ...$ and this will hardly be $L^2$ when you make convolution with $\phi$. – incas Jan 2 '15 at 23:44
• As I suggested in the answer, in the integral over $|y|>1$ it is better to put the derivative onto $\phi$ instead of $w$. Then it's like $w$ convolved with $1/|y|^2$ where $|y|>1$. – user147263 Jan 2 '15 at 23:48
• Ok i think I don't understand your point. To prove that the derivative of $\int w(x-y)\phi(y)dy$ is $L^2$, are you proceeding by proving $\nabla A_1(x)$ and $\nabla A_2(x)$ are $L^2$ ? if so then focusing only on $\nabla A_1(x)$ and taking w for which the derivative explode at infinity as I suggested above then we see that $\int [\int_{B(0,1)} \nabla w(x-y) \phi(y)dy]^2 dx$ is not finite am I wrong ? – incas Jan 2 '15 at 23:59