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There is a normalization error in the PDF. With the printed unnormalized integral, testing a constant function equal to one would give a left side $|U|(1-|U|{)}^2$ and a right side zero. Thus that formulation fails whenever $|U|\ne 1$.

Use instead the average $v_U=|U{|}^{-1}{\int }_{U}v$. The Poincare-Wirtinger inequality, also called the Neumann-Poincare inequality, isIf no $C_P$ exists, subtract the average and normalize a violating sequence to obtain $w_j\in H^1(U)$ with ${\int }_U{w}_j=0$, $\Vert w_j{\Vert }_2=1$, and $\Vert \nabla w_j{\Vert }_2\to 0$. This sequence is bounded in the Sobolev space $H^1(U)$. The Rellich-Kondrachov compactness theorem supplies a subsequence converging strongly in $L^2(U)$ to $w$.

For every compactly supported test function $\phi$, integration by parts and these convergences give $\int w{\partial }_i\phi =0$. Thus $w$ has zero weak gradient. The Sobolev function with zero weak gradient result and connectedness make $w$ constant. Strong $L^2$ convergence preserves its zero integral, so $w=0$. It also preserves its norm one, a contradiction. This proves the correctly normalized Neumann-Poincare inequality.

The constant function $v=1$ is an admissible $H^1(U)$ test for the Neumann Poisson problem. Its weak gradient is zero, so the weak identity immediately givesThis is necessary, and the preceding Hilbert-space construction proves sufficiency for $f\in L^2(U)$. It is the balance condition corresponding to zero total boundary flux.

On a connected open set, a Sobolev space function with zero weak gradient is constant almost everywhere. Local mollification gives constant smooth representatives on interior balls; overlaps and connectedness identify their constants. Without connectedness, the constant may differ between components. This explains the constant ambiguity in a Neumann Poisson problem.