Weak solution for equation $-\Delta u = f$. Suppose $f \in L^{2}$. I know that 
$$\left\{\begin{array}{c}
−\Delta u = f(x) & \text{on }\Omega \\
u(x)=0 &  \text{on } \partial\Omega
\end{array}\right.,$$
where $\Omega \subset \mathbb{R}^{N}$ is open and bounded, has a weak solution $u \in H^{1}_{0}(\Omega)$. I can obtain this solution using the Riesz Representation Theorem.
But, how obtain a weak soution for this equation with Neumann boundary conditions? And in the case where $\Omega = \mathbb{R}^{N}$?
In both cases I have been tried use the Lax Milgran Theorem, but I can't show the coercitive condition.
Another question: What the regularity of these weak solutions? I know that if $f$ is $C^{0,\alpha}$ and the problem has Dirichlet conditions, then the solution is $C^{2}$, am I right?
Thanks in advance.
 A: Let us start by  discussing the following Neumann problem:
\begin{align}
- \Delta u &= f \quad \textrm{ in  } \Omega, \\
 \frac{\partial u}{\partial \nu} &= 0 \quad   \textrm{ on } \partial \Omega.
\end{align}
Here, $\Omega \subset \mathbb{R}^n$ is a bounded Lipschitz domain.
As Michal has already mentioned, a solution $u$ of this boundary value problem is not unique; you can always add a constant. Note that 
\begin{equation}
\int_{\Omega} f = 0
\end{equation}
is a necessary condition on $f$ to ensure existence of a solution to the boundary value problem. (Why?)
The tool to prove existence of a weak solution is the Lax-Milgram theorem. We choose the following subspace 
\begin{equation}
H = \left\{ u \in H^1(\Omega) \; \colon \, \int_{\Omega} u = 0 \right\}
\end{equation}
of $H^1(\Omega)$. Note that the only constant function  $c \in H$ is $c = 0$. Moreover, for functions $u \in H$ the following Poincare inequality holds:
\begin{equation}
\int_{\Omega} \lvert u \rvert^2 \leq C_P \int_{\Omega} \lvert \nabla u \rvert^2.
\end{equation}
Now, define $b \: \colon H \times H \to \mathbb{R}$ by
\begin{equation}
b(u, v) = \int_{\Omega} \langle \nabla u , \nabla v \rangle.
\end{equation}
Similar to the Dirichlet problem, you can show that $b$ is bounded; that is, there exists a constant $C> 0$ such that for all $u, v \in H$ we find that
\begin{equation}
\lvert b(u, v)\rvert \leq C \|u \| \cdot \|v\|.
\end{equation}
Coercivity of $b$ follows from the Poincare inequality. Thus, we can apply the Lax-Milgram theorem. For every $g \in H'$ we obtain a unique $u \in H$ such that
\begin{equation}
b(u, v) = \langle g, v \rangle_{H', H}
\end{equation} 
for all $v \in H$.
Here, $\langle \cdot, \cdot \rangle_{H', H}$ denotes the dual pairing of $H'$ and $H$.
To find a solution $u \in H$ to the boundary value problem above, we have to choose the functional $g$ appropriately. Set $g_0 \: \colon H \to \mathbb{R}$,
\begin{equation}
\langle g_0, v \rangle_{H', H} = \int_{\Omega} f  v.
\end{equation} 
Note that the boundary value $\partial_{\nu} u = 0$ is encoded in $g_0$, since there is no integral over the boundary $\partial \Omega$. We claim that the function $u \in H$ that satisfies
\begin{equation}
b(u, v) = \langle g_0, v \rangle
\end{equation}
for all $v \in H$ is a solution of the boundary value problem. To see this, we choose $\varphi \in C_c^{\infty}(\Omega)$. Then 
\begin{equation}
\tilde{\varphi} = \varphi - \int_{\Omega} \varphi \in H.
\end{equation}
Moreover, we find that
\begin{equation}
\int_{\Omega} \langle \nabla u, \nabla \varphi \rangle = b(u, \tilde{\varphi}) = \langle g_0, \tilde{\varphi} \rangle = \int_{\Omega} f \varphi - \bigg( \int_{\Omega} f \bigg) \bigg( \int_{\Omega} \varphi\bigg) = \int_{\Omega} f \varphi.
\end{equation}
As $\varphi \in C_c^{\infty}(\Omega)$ was chosen arbitrarily, we have proved that $- \Delta u = f$ in $\Omega$.
