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The question is:

Let $k\in C^{0}(\mathbb{R}^{n}-\{0\})$ be a function such that $$k(xt)=t^{-n}k(x)$$ for $0\not=x\in\mathbb{R}^{n},t>0$. Show that the principal value $$\int k(x)\phi(x)dx=\lim_{|x|>\epsilon}k(x)\phi(x)dx,\phi\in C^{\infty}_{c}(\mathbb{R}^{n})$$ exists if and only if $$\int_{S^{n-1}}k(\theta)d_{\omega}(\theta)=0$$ where $d\omega$ is the usual surface measure on the unit sphere $\mathbb{S}^{n-1}$.

Show also that the first is a distribution when the second equation holds.

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I use $dr$ as the Lebesgue measure in $\mathbb{R}^{n}$, not polar coordinates. So you can think about it as $dx_{1}dx_{2}...dx_{n}$. Thus we have $dr=ds*dt$ with ds the surface measure on $S^{n-1}$ and $t=|r|$. –  Bombyx mori Jun 15 '12 at 3:28
    
Sorry I deleted my comment, I don't think I got polar coordinates right... –  Jose27 Jun 15 '12 at 3:29
    
I would love to see how this may be solved by standard curvilinear coordinates, say $x_{1}=r\cos(\theta_{1})$, $x_{2}=r\sin(\theta_{1})\cos(\theta_{2})$, etc. But it feels too complicated for this problem(or maybe I am wrong?). –  Bombyx mori Jun 15 '12 at 3:31
    
On another note, notice that your argument as is should work for any homogeneous $k$, but, for example, $\int_{|x|>\epsilon} |x|^{-n}dx=\infty$. –  Jose27 Jun 15 '12 at 3:34
    
I am not defending my definitely wrong argument, but I think your example only implies $|x|^{-n}$ cannot be used to establish such a principal value. Note $\int ds(\theta)=Cr^{n-1}\not=0$ in this situation. –  Bombyx mori Jun 15 '12 at 3:39

1 Answer 1

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We denote $ds(\theta)$ the surface measure on $\mathbb{S}^{n-1}$ at angle $\theta$. We use $|x|=r,\theta=\frac{x}{|x|}\in \mathbb{S}^{n-1}$. Then we have $dx=r^{n-1}drds$. Thus we have: \begin{align*} \int_{|x|\ge \epsilon} k(x)\phi(x)dx &=\int_{|x|\ge \epsilon} k(\frac{x}{|x|})\cdot |x|^{-n}\phi(x)dr\\ &=\int_{r\ge \epsilon} \int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\cdot \frac{1}{r^{n}}\phi(r\theta)r^{n-1}drds(\theta)\\ &=\int_{r\ge \epsilon} \int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\cdot \frac{1}{r}\phi(r\theta)drds(\theta) \end{align*} To prove the `if' part, consider the function $\phi$ that is 1 on the unit ball $B^{A}_{0}$ and has compact support in $\mathbb{R}^{n}$ with maximum value 1. Then we have \begin{align*} \int_{r\ge \epsilon} \int_{\theta\in \mathbb{S}^{n-1}}k(\theta) \frac{1}{r}\phi(r\theta)drds(\theta) &=\int^{A}_{r\ge \epsilon}\int_{\theta\in \mathbb{S}^{n-1}} k(\theta) \frac{1}{r}drds+\int^{\infty}_{r\ge A}\int_{\theta\in \mathbb{S}^{n-1}} k(\theta) \phi(r\theta)\frac{1}{r}drds\\ &=(\log(A)-\log(\epsilon))\int_{\theta\in \mathbb{S}^{n-1}} k(\theta)ds+\int^{\infty}_{r\ge A}\int_{\theta\in \mathbb{S}^{n-1}} k(\theta) \phi(r\theta)\frac{1}{r}drds \end{align*} Since $\phi$ has compact support, we may assume its support is in a ball of radius $B$, while $k(\theta)$ as maximum $K$ in the compact set $S^{n-1}$. With $C=\int_{\mathbb{S}^{n-1}}ds$ we have \begin{align*}|\int^{\infty}_{r\ge A}\int_{\theta\in \mathbb{S}^{n-1}} k(\theta) \phi(r\theta)\frac{1}{r}drds| &\le \int^{\infty}_{r\ge A}\int_{\theta\in \mathbb{S}^{n-1}} |k(\theta)||\phi(r\theta)|\frac{1}{r}drds\\ &\le\int^{B}_{A}\int_{\mathbb{S}^{n-1}}K\frac{1}{r}drds=K(\log(B)-\log(A))C \end{align*} Thus the original integral exist or not is depended on $$(\log(A)-\log(\epsilon))\int_{\theta\in \mathbb{S}^{n-1}} k(\theta)ds$$ only. And it diverges with $\epsilon\rightarrow 0$ unless $\int_{\theta\in \mathbb{S}^{n-1}} k(\theta)ds=0$.

To prove the `only if' part we assume $$\int_{\theta\in \mathbb{S}^{n-1}} k(\theta)ds=0$$ Then above integral become \begin{align*} \int_{r\ge \epsilon} \int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\frac{1}{r}\phi(r\theta)drds(\theta) \end{align*} Since $\phi(r\theta)\in C^{\infty}_{c}(\mathbb{R}^{n})$, $$\frac{\partial \phi}{\partial r}=\sum \frac{\partial \phi}{\partial x_{i}}*\frac{\partial x_{i}}{\partial r}=\sum\frac{\partial \phi}{\partial x_{i}}\theta_{i}$$ Thus $\phi$ is $\infty$ differentiable in $r$ as well. We may expand $\phi(r\theta)=\phi(r,\theta)=\phi(0)+\psi(r,\theta)r$ with $$\psi(r,\theta)=\sum^{\infty}_{i=1}\frac{r^{i-1}}{i!}\frac{\partial^{i} \phi(0,\theta)}{\partial r^{i}}$$ be continuous in $r$ and $\theta$. Then $\frac{\phi}{r}=\frac{\phi(0)}{r}+\psi(r,\theta)$, substitue this into the above formula and assume the support of $\phi$ is in a ball of radius $L$, then we have: \begin{align*} \int_{r\ge \epsilon} \int_{\theta\in \mathbb{S}^{n-1}}k(\theta) \frac{1}{r}\phi(r\theta)drds(\theta) &=\int^{L}_{\epsilon}\int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\frac{\phi(0)}{r}drds(\theta)+\int^{L}_{\epsilon}\int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\psi(r,\theta)drds(\theta)\\ &=\phi(0)\int^{L}_{\epsilon}\frac{1}{r}dr\int_{\theta\in \mathbb{S}^{n-1}}k(\theta)ds(\theta)+\int^{L}_{\epsilon}\int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\psi(r,\theta)drds(\theta)\\ &=\int^{L}_{\epsilon}\int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\psi(r,\theta)drds(\theta) \end{align*} We now prove as $\epsilon\rightarrow 0$ the above integral converges to $$\int^{L}_{0}\int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\psi(r,\theta)drds(\theta)$$ Because $\psi(r,\theta)$ is continuous on $r$ and $\theta$, thus $|\psi(r,\theta)|\le M,\forall x\in B^{L}_{0}$. Thus the difference integral \begin{align*} |\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\psi(r,\theta)drds(\theta)| &\le \int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}|k(\theta)\psi(r,\theta)|drds(\theta)\\ &\le KMC\int^{\epsilon}_{0}dr\\ &\le KMC\epsilon\\ &\rightarrow 0 \end{align*} Now we show the above $k$ defines a distribution. It is suffice to show that for any $\{\phi_{n}\}\rightarrow 0,\langle k,\phi_{n}\rangle \rightarrow 0$. We have proved that for any $\phi\in C^{\infty}_{c}$, $$\langle k,\phi\rangle=\int^{L}_{0}\int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\psi(r,\theta)drds(\theta)$$ We now derive $$\psi(r,\theta)=\frac{\phi(r,\theta)-\phi(0,\theta)}{r}=\frac{\int^{r}_{0}\phi'(t,\theta)dt}{r}$$ Since $\phi_{n}\rightarrow 0$, $\partial^{\alpha}\rightarrow 0$ in the support of $\phi_{n}$ as well. Notice $$\frac{\partial \phi}{\partial r}=\sum \frac{\partial \phi}{\partial x_{i}}*\frac{\partial x_{i}}{\partial r}=\sum\frac{\partial \phi}{\partial x_{i}}\theta_{i}$$ Thus we have $$\frac{\partial \phi_{n}(r,\theta)}{\partial r}\rightarrow 0$$ in the compact region $B^{L}_{0}$. Thus we have \begin{align*} |\int^{L}_{0}\int_{\theta\in \mathbb{S}^{n-1}}k(\theta)\psi_{n}(r,\theta)drds(\theta)| &\le \int^{L}_{0}\int_{\theta\in \mathbb{S}^{n-1}}|k(\theta)|\psi_{n}(r,\theta)|drds(\theta)\\ &=\int^{L}_{0}\int_{\theta\in \mathbb{S}^{n-1}}|k(\theta)|\frac{\int^{r}_{0}\frac{\partial \phi_{n}}{\partial r}(t,\theta)dt}{r}|drds(\theta)\\ &\le\int^{L}_{0}\int_{\theta\in \mathbb{S}^{n-1}}|k(\theta)|\frac{\max(|\partial\phi_{n}|)*r}{r}|drds(\theta)\\ &=\max(|\partial\phi_{n}|)\int^{L}_{0}\int_{\theta\in \mathbb{S}^{n-1}}|k(\theta)|drds(\theta)\\ &=\max(|\partial\phi_{n}|)K\int^{L}_{0}\int_{\theta\in \mathbb{S}^{n-1}}drds(\theta)\\ &=\max(|\partial\phi_{n}|)KLC\\ &\rightarrow 0 \end{align*} And therefore $\langle k,\phi \rangle$ is a distribution.

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