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For $u\in H_0^2(U)$, ${\sum }_{i,j}\Vert {\partial }_{ij}u{\Vert }_2^2=\Vert \Delta u{\Vert }_2^2$. Two integrations by parts prove this for compactly supported test functions, and $H^2$ density extends it to the clamped second-order Sobolev space. Together with the Poincare-Wirtinger inequality applied to the mean-zero first derivatives and the zero-boundary Poincaré inequality, it controls the full $H^2$ norm by $\Vert \Delta u{\Vert }_2$. The boundary conditions are essential to this exact identity on bounded domains.

The mean-zero Sobolev space is the closed subspaceOn a bounded connected Lipschitz domain, the Poincare-Wirtinger inequality makes $\Vert \nabla u{\Vert }_2$ an equivalent Hilbert norm on this space.

On a smooth bounded connected domain, the homogeneous Neumann problem seeks $u\in H^1(U)$ such that $\int \nabla u\cdot \nabla v=\int fv$ for every $v\in H^1(U)$. For $f\in L^2(U)$ it is solvable exactly when $\int f=0$, and the weak solution is unique up to constants. The Poincare-Wirtinger inequality and the Riesz representation theorem give existence on the mean-zero Sobolev space. The test function one gives necessity of the compatibility condition.

Work first on $B=B_1$, and let $m=|B{|}^{-1}{\int }_{B}u$. Put $Z=\{u=0\}$. Since $|Z|\ge \theta |B|$,Also ${\int }_B{u}^2={\int }_B|u-m{|}^2+|B|m^2$, soThe Poincare-Wirtinger inequality now bounds the last integral by $C_n{\int }_B|Du{|}^2$. Here is a direct proof on the ball. For smooth $u$, the variance identity and the fundamental theorem along line segments giveFor $0\le t\le 1/2$, fix $y$ and change variables $z=(1-t)x+ty$. Convexity keeps $z$ inside $B$, and the inverse Jacobian is at most $2^n$. Thus the double integral of this gradient term is at most $2^n|B|{\int }_B|Du{|}^2$. For $1/2\le t\le 1$, fix $x$ instead and obtain the same bound. Integration over $t$ proves ${\int }_B|u-m{|}^2\le 2^{n+1}{\int }_B|Du{|}^2$. Density of smooth functions in $W^{1,2}(B)$ passes the estimate to all Sobolev functions.

For a ball centered at $a$ with radius $R$, set $v(y)=u(a+Ry)$. Its zero set has at least the same volume fraction $\theta$, while ${\int }_{B_1}{v}^2=R^{-n}{\int }_{B_R}{u}^2$ and ${\int }_{B_1}|Dv{|}^2=R^{2-n}{\int }_{B_R}|Du{|}^2$. ThereforeThis establishes the Poincare inequality with a positive-measure zero set, with an explicit admissible constant. If $\theta =1$, the function is already zero almost everywhere.

The appropriate Poincare-Wirtinger inequality, with no zero-boundary assumption, isAssume $\Omega$ is nonempty, bounded, connected and has the stated cone property. If this failed, after subtracting the mean and normalizing there would be $u_j\in W^{1,p}(\Omega )$ withThe sequence is bounded in $W^{1,p}$. By the permitted Rellich-Kondrachov compactness theorem, a subsequence converges strongly in $L^p$ to $u$. For every compactly supported test function, integration by parts and these two convergences show that each distributional derivative of $u$ is zero. A function with zero weak gradient is constant on a connected open set: mollification makes it locally constant, and connectedness identifies the local constants.

Convergence in $L^p$ on a finite-measure set also preserves the integral, so $u_{\Omega }=0$ and the constant is zero. But strong convergence preserves $\Vert u{\Vert }_p=1$, a contradiction. This proves the inequality; connectedness prevents different constants on separate components.

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.

Suppose the claimed Poincare-Wirtinger inequality were false. There would be $u_k\in H^1(U)$ such that, after settingwe have $(v_k{)}_U=0$, $\Vert v_k{\Vert }_2=1$, and $\Vert D{v}_k{\Vert }_2\to 0$. The sequence is bounded in $H^1(U)$, so part 1(b)(ii) supplies a subsequence converging strongly in $L^2(U)$ and weakly in $H^1(U)$ to some $v$.

The weak gradient of $v$ is zero. Because $U$ is connected, $v$ is a constant function; its mean is zero, so $v=0$. Strong convergence would then give $\Vert v_k{\Vert }_2\to 0$, contradicting $\Vert v_k{\Vert }_2=1$. Therefore

The Poincare-Wirtinger inequality on a ball and the assumed estimate (8) implyThis is exactly the stated Campanato space criterion with a constant $M$ independent of the ball. Therefore

Testing the weak formulation with the constant function $1$ proves the necessary compatibility conditionAssume first that $U$ is connected and this condition holds. On the mean-zero Sobolev spaceuse the norm $\Vert v{\Vert }_{†}=\Vert Dv{\Vert }_{L^2(U)}$. To prove the needed Poincare-Wirtinger inequality, suppose it failed. There would be $v_k\in H_{†}^1(U)$ with $\Vert v_k{\Vert }_2=1$ and $\Vert D{v}_k{\Vert }_2\to 0$. The Rellich-Kondrachov compactness theorem gives a subsequence converging strongly in $L^2$ and weakly in $H^1$ to a function $v$. Its weak derivative vanishes, so connectedness makes $v$ constant; its zero mean makes it zero. This contradicts $\Vert v{\Vert }_2=1$. Hence $\Vert Dv{\Vert }_2$ is an equivalent Hilbert space norm on $H_{†}^1(U)$.

DefineBoundedness of $a^{ij}$ makes $B$ a bounded bilinear form, while uniform ellipticity givesso it is a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare-Wirtinger inequality make $\ell$ a bounded linear functional. The Lax-Milgram theorem supplies a unique $u\in H_{†}^1(U)$ satisfying $B(u,v)=\ell (v)$ for every mean-zero $v$. For arbitrary $v\in H^1(U)$, subtract its mean; the omitted constant contributes zero on both sides because $D1=0$ and ${\int }_{U}f=0$. Thus $u$ solves the original problem.

If two solutions exist, their difference $w$ satisfies $B(w,w)=0$, so uniform ellipticity gives $Dw=0$. It is therefore constant on $U$. The solution is unique up to an additive constant, and its mean-zero representative is unique. If $U$ is disconnected, the precise condition is ${\int }_{U_j}f=0$ on every connected component $U_j$, and one independent additive constant remains on each component.

For the uniform distribution on $[0,1]$,The constant $1/{\pi }^2$ is sharp and is the mean-zero Poincare-Wirtinger inequality on the interval.

Suppose $u\in W^{1,2}(B_R)$ vanishes on a measurable subset of measure at least $\theta |B_R|$, with $\theta >0$. Writing $m$ for its mean, the zero set gives $\theta |B_R|m^2\le {\int }_{B_R}|u-m{|}^2$. Thus $\int u^2\le (1+{\theta }^{-1})\int |u-m{|}^2$, and the Poincare-Wirtinger inequality supplies the displayed estimate. An elementary proof of that inequality on a ball bounds differences along line segments and splits the segment parameter at $1/2$, giving the valid explicit constant $C(n,\theta )=2^{n+1}(1+{\theta }^{-1})$. The change of scale from the unit ball multiplies the constant by $R^2$.