I am stuck on this computation: let $\Omega$ be a domain in $\mathbb R^2$ and let $\Gamma_0$ be a relatively open proper subset of $\Gamma:=\partial\Omega$. Define $$ V=\{v \in H^2(\Omega); v=\partial_\nu v =0 \mbox{ on }\Gamma_0\}, $$ where $\nu$ denotes the outer unit normal vector field of $\Gamma$. Let $$ |v|_{2,\Omega}:=\left(\sum_{|\alpha|=2} \|\partial^\alpha v\|_{L^2(\Omega)}^2\right)^{1/2}. $$
The first part of the exercise asked to prove that $V$ is closed in $H^2(\Omega)$ (and I did it).
Then, the exercise asks to prove that $|\cdot|_{2,\Omega}$ is a norm over $V$. The positive homogeneity and the triangular property are trivial to verify.
The only thing to prove is that $|v|_{2,\Omega}=0$ implies $v=0$ and I am stuck at this point.
The book suggests proving that if $|v|_{2,\Omega}=0$ then $v(x)=c_0 + c_1 x_1 +c_2x_2$, where $c_0, c_1, c_2$ are real constants.
I verified this property using the fact that $$ \int_\Omega v \partial_i \varphi dx =0,\quad \mbox{ for all }\varphi \in \mathcal D(\Omega), \mbox{ and all } i=1,2 $$ implies that $v$ is constant.
Afterwards using the boundaries conditions, one can prove that $v \equiv 0$ as an element of $V$: first, from the fact that $v=0$ on $\Gamma_0$ we infer $c_0=0$ immediately; finally, the weak gradient of $v$ is $$ \nabla v = (c_1,c_2), $$ and I computed $$ 0=\partial_\nu v := \sum_{i=1}^2 c_i \nu_i(x),\mbox{ for all } x \in \Gamma_0. $$
How do I prove that the constants $c_1=c_2=0$?
Thanks in advance.
Notice that we can write $$c_1^2 + c_2^2 = |\nabla v|^2 = (\nabla v \cdot \tau)^2 + (\nabla v \cdot \nu)^2,$$ where $\tau$ is the direction of the tangent line. By assumption, we know that the second term on the RHS is $0$ on $\Gamma_0$. What is left to show is that also the tangential part of the gradient is $0$. Do you see why this is true?
EDIT: take $\{\tau,\nu\}$ as a basis for $\mathbb{R^2}$. Then $\nabla v = a\tau + b\nu$, where $a = \nabla v\cdot \tau$ and $b = \nabla v\cdot \nu$. Thus, $$|\nabla v|^2 = \nabla v \cdot \nabla v = (\nabla v \cdot \tau)^2 + (\nabla v \cdot \nu)^2.$$