Unique weak solution of Poisson's equation Let $\Omega$ be an open set in $\mathbb{R}^n$ and now consider the weak formulation of Poisson's equation $$\int_{\Omega} \langle Du,Dv \rangle = \int_{\Omega}{fv}$$ for $v \in H_0^1$ and $u \in H_0^1$, $f \in L^2(\Omega)$
Does this problem have a unique weak solution $u$ such that the preceding weak formulation holds for all $v$?
It is quite some time ago that I covered this, but I think I can remember that one can actually show this by considering the convex energy functional. Unfortuantely, I think one needs that $\Omega$ is bounded or something like that. So could anybody help me turning my question in a correct statement(i.e. regarding the definition of $\Omega$, existence and uniqueness?)? 
 A: Short answer. If $\Omega$ is unbounded, you still have uniqueness but you might lose existence in $H^1_0(\Omega)$.
Long answer. You need boundedness of $\Omega$ to use the typical argument based on Riesz's representation of linear functionals (that's the argument that Jose mentions in his comment). This is because Poincaré's inequality $c\int_{\Omega}u^2\, dx \le \int_\Omega \lvert \nabla u\rvert^2\, dx$ is not available on unbounded domains. 
However, you still have uniqueness in $H^1_0(\Omega)$, regardless of whether $\Omega$ is bounded or not. If $u_1$ and $u_2$ are two $H^1_0(\Omega)$ weak solutions to 
$$\tag{$\star$}
\begin{cases}-\Delta u = f, & \Omega \\ u=0, & \partial \Omega\end{cases}
$$
then the difference $v=u_1-u_2$ satisfies 
$$
\int_\Omega \lvert \nabla v\rvert^2\, dx =0$$
and so it is a constant. (I am here assuming that domain means connected open set). The only constant function in $H^1_0(\Omega)$ is the null function. Therefore $v=0$. 
Remark 1. If you do not impose the condition $u\in H^1_0(\Omega)$, but rather look for classical solutions to $(\star)$, then you do lose uniqueness in unbounded domains. For example, in $\Omega=\mathbb{R}^n$ you can add any entire harmonic function to a solution and obtain another solution. The point is that $(\star)$ is underdetermined in unbounded domains, because no boundary condition is imposed at infinity. By insisting that the solution be in $H^1_0(\Omega)$ one forces some decay at infinity and thus recovers uniqueness.
Remark 2. The existence of solutions to $(\star)$ in unbounded domains is another story. Under the sole assumption that $f\in L^2(\Omega)$ it might happen that no $H^1_0(\Omega)$ weak solution exist. For a concrete example, fix $n= 3$ and $\Omega=\mathbb{R}^3$. Let 
$$
u(x)=\left(\frac{1}{1+\lvert x \rvert^2}\right)^\frac{1}{2}$$
and 
$$
f(x)=3\left(\frac{1}{1+\lvert x \rvert^2}\right)^\frac{5}{2}.$$
(The function $u$ appears in the elliptic theory of PDEs. It is called standard bubble). You can check that 
$$-\Delta u = f$$
but $u\notin L^2(\mathbb{R}^3)$ (hence $u\notin H^1(\mathbb{R}^3)$) even if $f\in L^2(\mathbb{R}^3)$. And $u$ is the unique solution to $(\star)$ with the given source term $f$, at least in the class of tempered distributions. In particular, no $H^1(\mathbb{R}^3)$ solutions to $(\star)$ with the given source term $f$ exist.
To see things clearly in $\Omega=\mathbb{R}^n$ you can solve $(\star)$ by means of the Fourier transform (or by convolution against the fundamental solution of the Laplace operator). Then the solution is formally written at Fourier side as
$$\hat{u}=\frac{\hat{f}}{\lvert\xi\rvert^2}, $$
and this is the unique solution to $(\star)$ in the class of tempered distributions, as announced previously. Now it happens that $u\in H^1_0(\mathbb{R}^n)=H^1(\mathbb{R}^n)$ if and only if 
$$
\frac{(1+\lvert\xi\rvert^2)^\frac{1}{2}}{\lvert\xi\rvert^2} \hat{f}\in L^2(\mathbb{R}^n).$$ 
This is not guaranteed by the condition $f\in L^2(\mathbb{R}^n)$ alone: to have this, $\hat{f}$ must vanish at $\xi=0$ to second order at least. 
