Decomposition of $C_0^{\infty}(\mathbb{R}^n)$ -function I got the following question as part of a fourier-analysis course..
Consider $\phi\in C_0^{\infty}(\mathbb{R}^n)$ with $\phi(0)=0$. Apparantly then we can write $$\phi =\sum_{j=1}^nx_j\psi_j $$ for functions $\psi_j$ in the same space, and I would like to prove this.
However I'm little stuck on this. The steps would involve that we start integrating $$\int_0^{x_1}D_1\phi(t,x_2,\cdots, x_n)dt+\phi(0,x_2,\cdots, x_n) $$
and continue doing so w.r.t. to the other variables, and change the interval of integration to [0,1]. Then get everything in $C_0^{\infty}(\mathbb{R}^n)$ by writing $$\phi = \sum \frac{x_i^2\phi(x)}{\left\|x\right\|^2
} $$ 
and patch everything together with a partition of unity. I just dont quite see what they mean...A partition of unity is a sequence of functions that sums up to 1 for all $x\in \mathbb{R}^n$. 

The result can be shown by induction on $n$. 
For $n=1$, just write $\phi(x)=\int_0^x\phi'(t)dt=x\int_0^1\phi'(sx)ds$, and the map $x\mapsto \int_0^1\phi'(sx)ds$ is smooth, and has a compact support. 
Assume the result is true for $n-1\geqslant 1$. We have 
$$\phi(x)=x_n\int_0^1\partial_n\phi(x_1,\dots,x_{n-1},tx_n)dt+\varphi(x_1,\dots,x_{n-1},0).$$
As similar argument as in the case $n=1$ shows that $(x_1,\dots,x_n)\mapsto \int_0^1\partial_n\phi(x_1,\dots,x_{n-1},tx_n)dt$ is smooth with compact support. By the induction, as $(x_1,\dots,x_{n-1})\to \phi(x_1,\dots,x_{n-1},tx_n)dt+\varphi(x_1,\dots,x_{n-1},0)$ is smooth with compact support, we can write it as $\sum_{j=1}^{n-1}x_j\psi_j(x_1,\dots,x_{n-1})$. 
But it's not finished yet, as $(x_1,\dots,x_n)\mapsto \psi_j(x_1,\dots,x_{n-1})$ doesn't have a compact in $\Bbb R^n$ unless $\psi_j\equiv 0$. However, we can find a smooth function with compact support $\chi$ such that $\chi(x)=1$ whenever $x\in\operatorname{supp}(\phi)$. As $\phi(x)=\phi(x)\chi(x)$, we are done.
