where $\beta$ is defined like this:
I'm trying to prove (2.18) but i don't know how to do, i calculated the integral but i don't find anything
EDIT1: $\beta(\phi_{\lambda,p,\rho}(y))=\displaystyle\frac{\int_{\mathbb{R}^N} x |\nabla(\phi_{\lambda,p,\rho}(y))(x)|^2 dx}{\int_{\mathbb{R}^N}|\nabla(\phi_{\lambda,p,\rho}(y))(x)|^2 dx}$
By definition of $\phi_{\lambda,p,\rho}$ we obtain:
$\beta(\phi_{\lambda,p,\rho}(y))=\displaystyle\frac{\int_{B_{\rho}(y)} x |\nabla u_{\lambda,p,\rho}(|x-y|)|^2 dx}{\int_{B_{\rho}(y)}|\nabla u_{\lambda,p,\rho}(|x-y|)|^2 dx}$
We have that $x\in B_{\rho}(y)$, $u_{\lambda,p,\rho}$ is radial so $u_{\lambda,p,\rho}(|x-y|)=u_{\lambda,p,\rho}(x-y)$
But how to obtain that $\beta(\phi_{\lambda,p,\rho}(y))=y$ ?
please help me
Thank you.
Let $\psi : \mathbb{R}^N \to [0, \infty)$ be a measurable function that is radially symmetric around some $y \in \mathbb{R}^N$ and $\int_{\mathbb{R}^N} \psi(x) dx = 1$. Then $$ \int_{\mathbb{R}^N} x \psi(x) dx = \int_{\mathbb{R}^N} (x - y) \psi(x) dx + y \int_{\mathbb{R}^N} \psi(x) dx = \int_{\mathbb{R}^N} z \psi(z + y) dz + y = y, $$ substituting $z = x - y$, and using radial symmetry of $\psi$.
Now apply this to the function $\psi := \tilde\psi / \int_{\mathbb{R}^N} \tilde\psi$, where $\tilde\psi := |\nabla [\Phi_{\lambda,p,\rho}(y)](\cdot)|^2$. The denominator is positive and finite as long as $u_{\lambda,p,\rho} \neq const$ where it is nonzero. And it is $\neq const$ because $\Phi_{\lambda,p,\rho}$ maps into $H_0^1(\Omega)$.
Hope that makes sense.