Weak solutions to the Neumann's problem (Evans PDE) 
Let $U$ be connected. A function $u \in H^1(U)$ is a weak solution
  of Neumann's problem
  \begin{equation}
(*)\qquad\left\{
\begin{array}{rl}
-\Delta = f & \text{in } U \\
\frac{\partial u}{\partial \nu} = 0 & \text{on } \partial U
\end{array}
\right.
\end{equation}
  if
  $$
\int_U Du \cdot Dv \; dx = \int_Ufv \; dx
$$
  for all $v \in H^1(U)$. Let $f\in L^2(U)$. Prove $(*)$ has a weak solution
  if and only if
  $$
\int_U f \; dx =0.
$$

For the only if part I set $v=1$. However I do not see where to start
the if part. I was thinking of the Lax-Milgram theorem. That's where I'm now.
 A: For the Neumann problem $\,(\ast)\,$ in a bounded domain $U\subset\mathbb{R}^n$, $n\geqslant 2$, satisfying the cone condition, to prove that assumption 
$$
f\in \{ L^2(U)\,\colon\;\int\limits_{U}f\,dx=0\}\tag{1}
$$
implies the existence of a weak solution $u\in H^1(U)$, it is convenient to introduce  the space
$$
\widetilde{H}^1(U)=\{w\in H^1(U)\colon\,\int\limits_{U}\!w\,dx=0\}.
$$
Notice that $\widetilde{H}^1(U)$ is a Hilbert space with  inner product
$$
(u,v)\overset{\rm def}{=}\int\limits_{U}\nabla u\cdot\nabla v\,dx
$$
satisfying the condition 
$$
(u,u)=0\;\;\Longrightarrow\;\;u=0
$$
by virtue of the Poincaré inequality 
$$
\|u\|^2_{L^2(U)}\leqslant C\int\limits_{U}|\nabla u|^2\,dx
\quad \forall\,u\in \widetilde{H}^1(U)\tag{2}
$$
which requires certain regularity of the boundary $\partial U$. Note that the cone condition is not
precisely the regularity of $\partial U$ for $(2)$ to be valid — it just proves to be the least complicated suitable general restriction on $\partial U$.  Denote
$$
\bar{u}\overset{\rm def}{=}\frac{1}{|U|}\int\limits_{U}u\,dx,
$$
with notation $|U|$ standing for the $n$-dimensional Lebesgue measure of domain 
$U\subset \mathbb{R}^n$.  Since $u-\bar{u}\in \widetilde{H}^1(U)$ for any 
$u\in H^1(U)$, the Poincaré inequality can be as well rewritten in the form
$$
\|u-\bar{u}\|^2_{L^2(U)}\leqslant C\int\limits_{U}|\nabla (u-\bar{u})|^2\,dx
=C\int\limits_{U}|\nabla u|^2\,dx
\quad \forall\,u\in H^1(U).
$$
        The rest of the proof is easy. Consider a linear  functional 
$$
\Lambda(v)=\int\limits_{U}fv\,dx
$$
on $\widetilde{H}^1(U)$. Due to $(2)$, the linear functional $\Lambda$ is bounded
on the Hilbert space $\widetilde{H}^1(U)$. Hence, by the Riesz representation theorem, there is a unique $u\in\widetilde{H}^1(U)$ such that
$$
\Lambda(v)=(u,v)\quad \forall\,v\in \widetilde{H}^1(U),\tag{3}
$$
which immediately implies the integral identity
$$
\int\limits_{U}\nabla u\cdot\nabla v\,dx=\int\limits_{U}fv\,dx
\quad \forall\,v\in \widetilde{H}^1(U).\tag{4}
$$
To complete the proof, notice that, in fact, $(4)$ is valid as well for all 
$u\in H^1(U)$. Indeed, due to the assumption $(1)$, for any $v\in H^1(U)$ we have
$$
\int\limits_{U}fv\,dx=\int\limits_{U}f(v-\bar{v})\,dx=
\int\limits_{U}\nabla u\cdot\nabla (v-\bar{v})\,dx=
\int\limits_{U}\nabla u\cdot\nabla v\,dx
$$
by virtue of $(3)$ since $v-\bar{v}\in \widetilde{H}^1(U)$. Thus, there is a unique
$u\in\widetilde{H}^1(U)\subset H^1(U)$ such that
$$
\int\limits_{U}\nabla u\cdot\nabla v\,dx=\int\limits_{U}fv\,dx
\quad \forall\,v\in H^1(U).
$$
Q.E.D
Remark.  Being valid for general real bilinear forms, not necessarily symmetric, the Lax-Milgram theorem looks too much advanced for this rather trivial case when all the inner product axioms are met by the symmetric bilinear form 
$\,(\cdot,\cdot)$. Generally, the Lax-Milgram theorem is to be applied in cases where
the Riesz representation theorem is inapplicable, e.g., in case of a Dirichlet problem for the equation $-\Delta u+\partial_{x_m}u=f$.
