I have trouble understanding the following equality, which we used in a proof:
$$ c_0 \sum_{j=1}^n || \partial_{x_j} v ||^2_{L^2(\Omega)} - \frac{c_0}{2} \sum_{j=1}^n || \partial_{x_j} v ||^2_{L^2(\Omega)} - C ||v||^2_{L^2(\Omega)} = \frac{c_0}{2} ||v||^2_{H^1(\Omega)} -(C + \frac{c_0}{2})||v||^2_{L^2(\Omega)}$$
where $\Omega \subseteq \mathbb{R}^n$ open, $c_0, C >0$ and $v \in H^1_0(\Omega)=W^1_{2,0}(\Omega)$ the closure of $C_0^{\infty}(\Omega)$ in the Sobolev-space $W_2^1(\Omega)$ and $H^1(\Omega)=W_2^1(\Omega)$ the Sobolev-Space.
I'm not sure whether I made a mistake by pulling the sum into the norm, but I can't get the factor $\frac{c_0}{2} $ in the RHS of the equality before the $H^1(\Omega)$-norm.
Thank you for your help.
If you take into account the definition $\| v \|_{H^1(\Omega)}^2 = \sum_{j=1}^n || \partial_{x_j} v ||^2_{L^2(\Omega)} + ||v||^2_{L^2(\Omega)}$, everything falls into place.
Note that $\sum_{j=1}^n || \partial_{x_j} v ||^2_{L^2(\Omega)}$ appears twice on the left-hand side.