# Neumann problem with zero average sobolev space

Let us considère the following Laplace-Neumann problem $$-\Delta u=0$$ with homogenuous boundary condition of type neumann, i:e $$\frac{{\partial u}}{{\partial n}} = 0$$. The variational formulation is given as follows $$a(u,v)=l(v)$$, where $$a(u,v) = \int_\Omega {\nabla u\nabla v} dx = 0$$ and $$l(v)=0$$. Let us cinsidère the space
$$\left\{ V={u \in H^1(\Omega), \int_\Omega v(x)ds=0} \right\}$$ $$l$$ is continuous ( is zero), and $$a$$ is continuous and corecive. By lax-Milgram theoremn there exists a unique solution of the variational problem $$a(u,v)=l(v)$$. We can notice that the constant are alos solutions of the problem, but the constants don't satisfy the zero average condition, I don't understand that.

In your case the problem is $$-\Delta u = 0$$ subject to $$\frac{\partial u}{\partial n} = 0,$$ which only has trivial solutions. Indeed if $$u$$ is such a solution with zero average $$\int_{\Omega} u \,\mathrm{d}x = 0,$$ since $$a(u,u) = \int_{\Omega} |\nabla u|^2\,\mathrm{d}x = 0$$ we conclude that $$u$$ is a.e. constant and hence zero as it has zero average.
So the solution $$u \equiv 0$$ is the unique solution subject to the zero average condition. The other constant solutions $$u \equiv k \neq 0$$ are also solutions, but they don't satisfy the zero average condition. I assume this is what is meant here.
For a non-homogenous problem $$-\Delta u = f$$, we get $$\ell(v) = \int_{\Omega} fv \,\mathrm{d}x$$ is non-trivial and hence we get non-trivial solutions. However if $$u$$ is a solution, then so is $$u+k$$ for any constant $$k.$$ The zero average condition removes this extra degree of freedom.