Poisson problem on unbounded domain I am looking at the weak Poisson problem. For $f \in L^{2}(\Omega)$ find $u \in H^{1}_{0}(\Omega)$ so that for all $v \in H^{1}_{0}(\Omega)$:
$$\int_{\Omega}\nabla{u}\nabla{v} = \int_{\Omega}fv$$
In order to apply the Lax-Milgram-Lemma, one needs the Poincaré-inequality:
$$||u||_{H^{1}(\Omega)} \le C ||\nabla{u}||_{L^{2}(\Omega)}$$
I read online that this inequality however does not require $\Omega$ to be bounded - finite width is sufficient. 
My question: Does this mean that I can also find unique solutions for the weak Poisson Problem on every domain of finite width - for example $(-\pi,\pi) \times (0,\infty)$? I do not see what could go wrong... 
And furthermore, how does this link to the fact that the lapalcian does not have a discrete spectrum on domains of finite width like $\Omega$ as above.
 A: First, a proof of the Poincaré inequality for domains of bounded width is given here: poincaré inequality direct proof
You are absolutely right that this implies that the Poisson problem has a unique solution for all $f\in L^2(\Omega)$ if the domain $\Omega$ has bounded width. The usual proof using the Lax-Milgram lemma goes through without change.
Here is a different way to think of it using spectral theory: One can easily verify that the $L^2$ Poincaré inequality is equivalent to a spectral gap of the Dirichlet Laplacian, i.e., the existence of $\lambda_0>0$ such that $\sigma(\Delta^{(D)})\subset [\lambda_0,\infty)$. Of course this means that $\Delta^{(D)}$ is invertible, i.e., the Poisson problem has a unique solution.
This spectral gap is not directly related to the discreteness of the spectrum, it may be continuous spectrum that just starts at $\lambda_0$. However, if you already know that $\Delta$ has purely discrete spectrum, then it's easy to check whether it has a spectral gap - you just have to check if $0$ is an eigenvalue, and since an eigenfunction for $0$ satisfies $\int |\nabla f|^2=0$, this boils down to checking whether the constant functions are in the domain of your particular realization of the Laplacian.
