I am working on the following problem trying to use strategy in this problem. I am trying to simplify the proof by working with $v_{i}=0,i\not=1$ case. But the result looks very different from what I expected, so I want to ask if there is something wrong in my computation. To be more precise, I do not understand why $(1-n)$ term appeared in my computation and my result may diverge.
Let $v_{1},...,v_{n}$ be functions of class $\mathbb{C}^{1}(\mathbb{R}^{n}-\{0\})$ such that $v_{i}(xt)=t^{1-n}v_{i}(x)$ for $0\not=x\in \mathbb{R}^{n}$, $t>0,i=1,...n$. Show that, for $\phi\in C^{\infty}_{c}(\mathbb{R}^{n})$ we have $\sum^{n}_{i=1}\langle \partial_{i}v_i,\phi\rangle=PV \int \phi\sum^{n}_{i=1}\partial_{i}v_{i}dx+\phi(0)\int_{\mathbb{S}^{n-1}}\sum^{n}_{i=1}\theta_{i}v_{i}(\theta)d\omega(\theta)$
This obviously follows if we can prove the one dimension claim by letting $v_{2}..v_{n}=0$: $\langle \partial_{1}v_{1},\phi\rangle=PV \int \phi[\partial_{1}v_{1}]dx+\phi(0)\int_{\mathbb{S}^{n-1}}\theta_{1}v_{1}(\theta)d\omega(\theta)$ And this follows if and only if we have $\lim_{\epsilon\rightarrow 0}\int^{|x|\le \epsilon}_{|x|\ge 0}[\partial_{1}v_{1}]\phi dx=\phi(0)\int_{\mathbb{S}^{n-1}}\theta_{1}v_{1}(\theta)d\omega(\theta)$ We know $x=r\theta$, thus we have $\frac{\partial r}{\partial x_{1}}=\frac{x_{1}}{r}=\theta_{1}$ Therefore(in the following we no longer distinguish $v_{1}$ and $v$) \begin{align*} \frac{\partial}{\partial x_{1}}v(r\theta) &=\frac{\partial}{\partial x_{1}}[r^{1-n}v(\theta)]\\ &=r^{1-n}\frac{\partial}{\partial x_{1}}v(\theta)+[\frac{\partial}{\partial x_{1}}r^{1-n}]v(\theta)\\ &=r^{1-n}\frac{\partial}{\partial x_{1}}v(\theta)+(1-n)r^{-n}\frac{\partial r}{\partial x_{1}}v(\theta)\\ &=r^{1-n}\frac{\partial v(\theta)}{\partial x_{1}}+(1-n)r^{-n}\frac{x_{1}}{r}v(\theta)\\ &=r^{1-n}(\frac{\partial v(\theta)}{\partial x_{1}}+(1-n)\frac{\theta_{1}}{r} v(\theta)) \end{align*} Substituting $x=r\theta$ we now have: \begin{align*} \lim_{\epsilon\rightarrow 0}\int^{|x|\le \epsilon}_{|x|\ge 0}[\partial_{x_{1}}v]\phi dx &=\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}r^{1-n}(\frac{\partial v(\theta)}{\partial x_{1}}+(1-n)\frac{\theta_{1}}{r} v(\theta))\phi(r\theta)drdsr^{n-1}\\ &=\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}(\frac{\partial v(\theta)}{\partial x_{1}}+(1-n)\frac{\theta_{1}}{r} v(\theta))\phi(r\theta)drds\\ &=\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}dr\int_{\theta\in \mathbb{S}^{n-1}}\frac{\partial v(\theta)}{\partial x_{1}}ds\\ &+(1-n)\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}\frac{\phi(r,\theta)}{r} v(\theta)\theta_{1}drds\\ &=(1-n)\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}\frac{\phi(r,\theta)}{r} v(\theta)\theta_{1}drds \end{align*} Let $\phi(r,\theta)$ have the Taylor expansion around 0: $\phi(r,\theta)=\phi(0,\theta)+r\psi(r,\theta)$ with $\psi=\frac{\int^{r}_{0} \phi'(t,\theta)dt}{r}$. Then the above integral become \begin{align*} (1-n)\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}[\frac{\phi(0)}{r}+\psi(r,\theta)]v(\theta)\theta_{1}drds \end{align*} Note this integral diverges if $\phi(0)\not=0$: $(1-n)\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\int_{\theta\in \mathbb{S}^{n-1}}[\psi(r,\theta)]v(\theta)\theta_{1}drds+(1-n)\lim_{\epsilon\rightarrow 0}\int^{\epsilon}_{0}\frac{\phi(0)}{r}\int_{\theta\in \mathbb{S}^{n-1}}v(\theta)\theta_{1}drds$