= Solution
The assumed inequality is equivalent to
$$
J(u):=\lVert Du\rVert_2^2-\gamma\lVert u\rVert_2^2\geq0,
$$
and equality holds at $w$. For $v\in H$ and $t\in\mathbb R$, expand $J(w+tv)\geq0$ and use $J(w)=0$:
$$
0\leq2t\big((Dw,Dv)_2-\gamma(w,v)_2\big)
+t^2\big(\lVert Dv\rVert_2^2-\gamma\lVert v\rVert_2^2\big).
$$
Since this holds for both signs of arbitrarily small $t$, the linear coefficient vanishes:
$$
\int_UDw\cdot Dv=\gamma\int_Uwv
\qquad(v\in H).
$$
Every $v\in H^1(U)$ is a mean-zero function plus a constant. The same identity holds for constants because $\int_Uw=0$, so it holds for all $v\in H^1(U)$. This is precisely the weak formulation of
$$
-\Delta w=\gamma w,
\qquad
\partial_\nu w=0,
\qquad
\int_Uw=0,
$$
where the <Neumann boundary condition> is the natural boundary condition encoded by the weak formulation.
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