Let $f(x)=|x|^r$ on $B_1(0)$ real valued function.Where $B_1(0)$ is the unit ball in $\mathbb{R}^n$. I am trying to show that if $r>1-n$ f has a weak derivative.
ATTEMPT: I know from the definition of the weak derivative that I need to check whether there is a function $g:B_1(0)\rightarrow \mathbb{R}$ such that $$\int_{B_1(0)}f\phi'dx=-\int_{B_1(0)}g\phi dx,$$ for all test functions $\phi\in C_{c}^{\infty}(B_1(0))$. But since $x\in \mathbb{R}^n$, I dont know how to proceed. Please help.
First of all, $f(x)=|x|^r$ is a very good smooth function and hence you should expect that classical derivative equal to weak derivative a.e., if there is any.
Now, the only thing you worry about is the singularity at $0$. However, as weak derivative, we never care a value at a single point. That is, we could always define $\nabla f(x)=\nabla |x|^r$ if $x\neq 0$ and $\nabla f=0$ if $x=0$.
Next, we only need to show that $|\nabla f(x)|\in L^1(B_1(0))$. This is nothing but some calculation:
$$\int_{B_1(0)} |\nabla f(x)|\,dx\leq C\int_0^1 s^{r-1}s^{n-1}\,ds =C\int_0^1 s^{r-1+n-1}\,ds $$ The integration is finite iff $r-1+n-1>-1$, i.e., $r-1+n>0$ as you expected.
By the way, you need to test against $C_c^1(B_1(0))$ but not $C^1(B_1(x))$. Those two are VERY different.