Let $C=\Omega \times (0,\infty)$. We want to find a solution $v \in H^1(C)$ such that given $u \in H^{\frac 12}(\Omega)$, $$\int_0^\infty\int_\Omega \nabla v \nabla \varphi + v_y\varphi_y = 0\quad\forall \eta \in H^1(C), \quad\eta(x,0) \equiv 0$$ $$v(x,0) = u(x)$$ where eg. the $n(x,0) = 0$ means in the sense of trace, so $T\eta = 0$ for $T:H^1(C) \to H^{\frac 12}(\Omega)$.

Now, if $u \equiv 1$, then $v \equiv 1$ would be solution however $1 \notin L^2(C)$. If instead we ask for $v \in X(C)$, where $X$ is a space that involves only the derivatives and the trace onto $\Omega$ at $y=0$, then $1$ is a solution in that space, and is unique. The standard way to get solutions in $H^1(C)$ is to impose a mean value zero condition on the data $u$.

My problem is I seem to have a proof that there is a solution $v \in H^1(C)$ with $Tv=u$ for $u\equiv 1$. Here is my "proof":

Let $U \in H^1(C)$ be such that $TU = 1$. Then the difference $d=v-U$ solves $$\int_0^\infty\int_\Omega \nabla d \nabla \varphi + d_y\varphi_y = -\int_0^\infty\int_\Omega \nabla U \nabla \varphi + U_y\varphi_y \quad\forall \eta \in H^1(C), \quad\eta(x,0) \equiv 0\tag{1}$$ $$d(x,0) = 0.$$ There is a unique $d \in H^1(C)$ solving this because we can define a functional $J:\{d \in H^1(C) \mid Td = 0 \text{ and } \int_\Omega d(x,y)\;\mathrm{d}x = 0 \text{ a.e. $y$}\} \to \mathbb{R}$ such that $$J(d) = \frac{1}{2}\int\int|\nabla d|^2 + d_y^2 + \int\int \nabla U \nabla d + u_y d_y.$$ This is convex, coercive (by Poincare, due to the mean value zero part of the domain), proper etc. It is then easy to show (1) holds (the minimiser solves the related variational problem for all test functions in the domain of $J$, and then we can remove the mean value condition required on the test function). Now it remains to set $v= d+ U \in H^1(C)$, and this solves the original problem.

So where is the fault in my argument??

Note that the domain of $J$ does not require a $d$ with $\int_\Omega d(x,0) = 0$; which would rule out the initial data.