Considering equation $$-\Delta u=f\ \mathrm{in}\ \Omega$$ $$u=g\ \mathrm{on}\ \partial\Omega$$ where $f\in L^2(\Omega)$ and there exists $G\in H^1(\Omega)$ such that $G=g$ on $\Omega$.
I now want to prove the ellipticity of $\Delta$, which I suppose I need to do using the Friedrichs inequality. But that only holds for functions from $H^1_0(\Omega)$. So I suppose that the solution $u$ has the form $u=G+w$, where $w\in H^1_0$. The equation than takes the form $$-\Delta w=f+\Delta G=F$$ But I don't know if $F\in L^2(\Omega)$. So how do I prove the existence and uniqueness of solution?
It is actually clear when I write down the weak form: \begin{align} -\int_\Omega\Delta w\ v&=\int_\Omega fv+\int_\Omega\Delta G\ v\\ \int_\Omega\nabla w\nabla v&=\int_\Omega fv-\int_\Omega\nabla G\nabla v\\ \int_\Omega\nabla w\nabla v&=\int_\Omega(fv-\nabla G\nabla v)\ \ \ \forall v\in H^1_0 \end{align} So I have equation in the form \begin{equation} a(w,v)=F(v)\ \ \ \forall v\in H^1_0, \end{equation} where \begin{align} a(w,v)&=\int_\Omega\nabla w\nabla v\\ F(v)&=\int_\Omega(fv-\nabla G\nabla v) \end{align} The ellipticity (coercivity) of $a(\cdot,\cdot)$ follows from the Friedrichs inequality, which I can now use, because $w\in H^1_0(\Omega)$. Boundedness is trivial in this case.
As for the functional $F$, it is also bounded, because $f,\ \nabla G\in L^2(\Omega)$. My mistake was assuming, that I need $G\in L^2(\Omega)$, but as it turns out, I only need $\nabla G\in L^2(\Omega)$, which is satisfied, because $G\in H^1(\Omega)=W^{1,2}(\Omega)$.
Now I can use the Lax-Milgram theorem to show, that there is a unique $w\in H^1_0$ such that \begin{equation} a(w,v)=F(v)\ \ \ \forall v\in H^1_0, \end{equation} and the solution to the equation is then $u=w+G$.