What is wrong with my `proof'?(solved) 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. 
 A: 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. 
