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.

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 $-\Delta u+\partial_{x_m}u=f$.