How to find this integral over an annulus i have $v=\sum_{g\in G_k} g v_r\in W^{1,N}_{0,G_k}(\Omega_r)\setminus\{0\}$.
$$\Omega_r=\{x\in \mathbb{R}^N, r<|x|<r+1\}, r>0, N\geq 2, N\neq 3$$
$$O_k=\{g\in O(2): g(x)=\left(x_1 \cos\frac{2\pi l}{k}+x_2\sin\frac{2\pi l}{k},-x_1\sin\frac{2\pi l}{k}+x_2\cos\frac{2\pi l}{k} \right)$$
Where  $O(2)$ Is the group of $2\times 2$ orthogonal matrices, $x=(x_1,x_2)\in\mathbb{R}^2$  and $l\in \{0,\ldots, k-1\}$
and $G_k=O_k\times O(N-2), 1\leq k <\infty$.
$W^{1,N}_{0,G_k}(\Omega_r)=\{u\in W^{1,N}_0(\Omega_r): u(x)=u(g^{-1}x),\, \text{for all} \, g\in G_k\}$
$v_r\in W^{1,N}_0(B_{\delta,r})\setminus\{0\}$ where $B_{\delta,r}=B_\delta(((2r+1)/2,0,\ldots,0))\subset \Omega_r$ satisfies
$$
g^i B_{\delta,r}\cap g^j B_{\delta,r}=\emptyset,\, for all \, g^i\in G_k, i\neq j, i,j=0,1,...,k-1
$$
My question is why $\int_{\Omega_r}|\nabla v|^N dx=\int_{\Omega_r} |\nabla v_r|^N dx $
 A: Suppose that $f$ and $g$ have disjoint support. There are $3$ cases:

*

*$f(x)\ne0$ and $g(x)=0$

*$f(x)=0$ and $g(x)\ne0$

*$f(x)=0$ and $g(x)=0$
In each case, $|f(x)+g(x)|=|f(x)|+|g(x)|$ and $|f(x)||g(x)|=0$.
Apply the Binomial Theorem to get
$$
\begin{align}
|f(x)+g(x)|^n
&=(|f(x)|+|g(x)|)^n\\
&=\sum_{k=0}^n\binom{n}{k}|f(x)|^k|g(x)|^{n-k}\\
&=|f(x)|^n+|g(x)|^n
\end{align}
$$
since all terms except $k=0$ and $k=n$ contain a product of $|f(x)||g(x)|$, so those terms are $0$.
Note that the change of variables is due to the rotations, but the determinant of a rotation matrix is $1$, so they don't affect the value of the integral.
Therefore,
$$
\begin{align}
\int_{\Omega_r}|\nabla v(x)|^N\,\mathrm{d}x
&=\int_{\Omega_r}\left|\sum_{g\in G_k}\nabla v_r(g(x))\right|^N\,\mathrm{d}x\\
&=\int_{\Omega_r}\sum_{g\in G_k}|\nabla v_r(g(x))|^N\,\mathrm{d}x\\
&=\int_{\Omega_r}\sum_{g\in G_k}|\nabla v_r(x)|^N\,\mathrm{d}g^{-1}(x)\\
&=\int_{\Omega_r}\sum_{g\in G_k}|\nabla v_r(x)|^N\det(g)^{-1}\,\mathrm{d}x\\
&=k\int_{\Omega_r}|\nabla v_r(x)|^N\,\mathrm{d}x
\end{align}
$$
