estimating $L^p$ norm of $\frac{x}{|x|^3} \ast (-\text{div}_x \ldots)$

Question

I've been having problems with following an argument in a paper I'm reading, I hope someone can help me understand the point. Suppose $f(t,\cdot,\cdot)$ is a $C_0^1 (\mathbb{R}^6)$ function for all $t$ and $f_0(x,v) := f(0,x,v)$. Let $$ E(t,\cdot) := \frac{x}{|x|^3} \ast \rho(t,x)$$ where $\rho(t,x) := \int f(t,x,v) dv$. I've managed to show that $$ \rho(t,x) = - \text{div}_x \int_0^t s \int_{\mathbb{R}^3} [Ef(t-s,x-vs,v)] dv ds + \int_{\mathbb{R}^3} f_0(x-vt,v) dv$$ Let's denote $h(t,x) := \int_{\mathbb{R}^3} f_0(x-vt,v)$ We wish to show that $$||E(t,\cdot)||_p \leq C || \int_0^t s \int_{\mathbb{R}^3} [Ef(t-s,x-vs,v)] dv ds||_{L^p_x} + ||h||_r$$ with $\frac{1}{p} = \frac{1}{r} - \frac{1}{3}$ and $\frac{3}{2} < p < \infty$.

My problem is how to get rid of the divergence in formula for $\rho$. I guess what should be done is writing $$E = \frac{x}{|x|^3} \ast (-\text{div}_x \ldots) + \frac{x}{|x|^3} \ast h$$ and now I'd like to estimate the two terms separately and make use of triangle inequality. The second one should follow from Young's inequality, but the first one is problematic. I don't think I can just move the divergence to $\frac{x}{|x|^3}$ since it's too singular to be in any sobolev space so I don't think I can do that. Then again when I wrote the formula for convolution explicitly and tried to estimate this function pointwise it was even messier and didn't seem to lead to anything since I still was unable to get rid of the divergence.

I'd be grateful for any pointers regarding that!

Answer 1

Of course, you can move the divergence to $\frac{x}{|x|^3}$ by the standard procedure of integrating by parts, first cutting off the ball $|x|<\varepsilon$ and then passing to the limit $\varepsilon\to 0$. What you get this way will be singular integral operators with Zygmund–Calderon kernels. Such operators are known to be bounded from $L^p(\mathbb{R}^3)$ to $L^p(\mathbb{R}^3)$ in case $1<p<\infty$ according to the so-called Zygmund–Calderon theorems. The textbook "Singular Integral Operators" by S.Mikhlin and S.Prössdorf contains comprehensible complete proofs, and seems to stay the best introduction into the $L^p$-theory of singular integrals.

Answer 2

I don't think I can just move the divergence to $\frac{x}{|x|^3}$ since it's too singular to be in any Sobolev space

Just do it. Assuming $E,f$ smooth, we can deal with $\operatorname{div}_x\frac{x}{|x|^3}$ as a distribution. And it turns out to be a familiar distribution: $4\pi \delta_0$. Convolution with $\delta_0$ does not change the function $Ef$. The resulting $L^p$ estimates extends by density to more general $E,f$.

In a way, convolution with $\frac{x}{|x|^3}$ inverts the divergence operator.