I would like to verify the following computation:
Consider the subspace $V=\left\{u\in H^1(\Omega): \quad \frac{1}{|\partial\Omega} \int_{\partial \Omega}
u\,d\sigma=0 \right\}$ of $H^1(\Omega)$, which is a Hilbert space with inner
product $<,>_{1,2}$: I intend to show that the boundary value problem with weak formulation
$$\int_{\Omega} (\nabla u\cdot \nabla
v+uv)\,dx=\int_{\Omega} fv \,dx, \quad \forall v\in V$$ has a unique solution
We verify that $(V,<,>_{1,2})$ is a Hilbert space in the first
place, for which it suffices to prove that it is complete. Let
${v_k}\subset V$ be a Cauchy sequence: then it is also Cauchy in
$(H^1(\Omega),<,>_{1,2})$, which is complete, so there exists $v\in
H^1(\Omega)$ such that $\lim_{k\to \infty} ||v_k-v||_{1,2}=0$ and it
suffices to observe that actually $v\in V$, namely that
$\frac{1}{|\partial \Omega|} \int_{\partial \Omega} v\,d\sigma=0$.
Recall that by Green's identity we have that $$\int_{\Omega} v\Delta u\,dx=\int_{\partial \Omega} v\partial_{\nu}u\,d\sigma-\int_{\Omega} \nabla u\cdot \nabla v\,dx$$ so that our variational problem is equivalent to finding $u\in V$ such that $$\int_{\Omega} uv-v\Delta u\,dx+\int_{\partial \Omega} v\partial_{\nu}u\,d\sigma=\int_{\Omega} fv\,dx, \quad \forall v\in V$$
In particular this must hold for functions $v\in V_0$ (with zero trace) so that the integral equation $$\int_{\Omega} uv-v\Delta u\,dx=\int_{\Omega} fv\,dx, \quad \forall v\in V_0$$ must hold. This in turn implies that $$\int_{\partial \Omega} v\partial_{\nu}u\,d\sigma=0, \quad \forall v\in V$$ Since these functions $v\in V$ also satisfy $\frac{1}{|\partial\Omega} \int_{\partial \Omega} u\,d\sigma=0$, we conclude that $\partial_{\nu}u$ must be constant, whence the boundary value problem associated with the variational formulation above is given by
$$\left\{ \begin{array}{ll} u-\Delta u=f, & \Omega \\
\partial_{\nu}u=const, & \partial\Omega
\end{array} \right.$$
In order to establish the well posedness of the problem, we verify the conditions of the Lax-Milgram theorem for the bilinear form $B(u,v)=\int_{\Omega} (\nabla u\cdot \nabla v+uv)\,dx$ and the linear functional $Lv=\int_{\Omega} fv \,dx$.
Continuity of $B$ follows from $$|B(u,v)|=\left|\int_{\Omega} (\nabla u\cdot \nabla v+uv)\,dx\right|\stackrel{[1]}{\leq} ||\nabla u||_0||\nabla v||_0+||u||_0||v||_0 \stackrel{[2]}{\leq} ||\nabla u||_0||\nabla v||_0+C_P^2||\nabla u||_0||\nabla v||_0$$ where in [1] we use the Cauchy-Schwarz inequality and in [2] we use Poincare's inequality, which can be easily proved to hold in $V$.
With regards to coercivity, it follows from $$B(u,u)=\int_{\Omega} |\nabla u|^2\,dx+\int_{\Omega} u^2\,dx\geq ||\nabla u||_{1,2}^2$$
Continuity of $L$ follows from Cauchy-Schwarz's and Poincaré's inequalities, namely
$$|Lu|=\leq \int_{\Omega} |uv| \,dx\leq ||u||_{1,2}||v||_{1,2}\leq
C_P||u||_{1,2}||\nabla v||_{1,2}$$
Hence $L\in H^{-1}(\Omega)$ and $||L||_{H^{-1}(\Omega)}\leq C_P||u||_{1,2}$.
Thanks in advance for any insight.
