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Put $e=u-u_h\in V$. The Aubin–Nitsche duality argument uses the dual weak formulationBy symmetry of $a$ and the Lax-Milgram theorem, this has a unique solution. Taking $v=z$ and using the Cauchy-Schwarz inequality showsWe must justify $H^2$ elliptic regularity for this dual problem from the result actually supplied. That result concerns $\Delta$, whereas the dual equation is $-\Delta z+z=e$. Put $g=z-e\in L^2(\Omega )$. The given result produces $w\in H^2(\Omega )\cap V$ with $\Delta w=g$, zero normal derivative on ${\Gamma }_N$, andBy Green's first identity, for every $v\in V$,Set $v=w-z$. Then $\nabla (w-z)=0$. On the connected triangle, a Sobolev function with zero weak gradient is constant, and its zero Sobolev trace on ${\Gamma }_D$ forces that constant to be zero. Hence $z=w$, provingThis proves the needed reaction-diffusion regularity from Laplacian regularity without assuming the dual solution already has second derivatives in the L2 space.

Only $H^2$ regularity is available for $z$. Use its continuous piecewise-linear interpolant $L_{h}z$, which belongs to the quadratic conforming finite element space $V_h$ because ${\mathcal{P}}_1(K)\subset {\mathcal{P}}_2(K)$. It has the same zero Dirichlet boundary condition and satisfies the finite element interpolation estimateThus no $H^3$ assumption on the dual solution is necessary. Taking $v=e$ in the dual weak formulation and using Galerkin orthogonality givesIf $\Vert e{\Vert }_0=0$, the desired bound holds. Otherwise divide by $\Vert e{\Vert }_0$ and use part (d):All constants depend only on the fixed triangle and shape regularity. The acute-angle assumption enters through the supplied mixed-boundary elliptic regularity result.

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.

Subtract the two weak solution identities and test with their difference $w\in H^1(U)$. ThenA Sobolev function with zero weak gradient is constant on each connected component. One justification is to mollify locally: each mollification has zero gradient and is constant on its ball, and overlaps identify the constants; taking limits gives the original assertion. Since $U$ is connected, $w$ is one constant on $U$.

Conversely, adding a constant changes neither the weak derivative nor the weak identity. The solution is unique up to an additive constant.

Suppose the mixed boundary value problem $\Delta w=g$, with zero Dirichlet boundary condition on a nonempty boundary portion and zero Neumann boundary condition on the remainder, has an $H^2$ solution with $\Vert w{\Vert }_{H^2}\le C\Vert g{\Vert }_{L^2}$. Assume the domain is connected and that zero Sobolev trace on the specified portion excludes nonzero constants. Then the same boundary conditions for $-\Delta z+z=e$ give $\Vert z{\Vert }_{H^2}\le 2C\Vert e{\Vert }_{L^2}$.

Indeed, the Lax-Milgram theorem first constructs $z$ in the Sobolev space with a partial Dirichlet condition; testing with $z$ shows $\Vert z{\Vert }_{H^1}\le \Vert e{\Vert }_{L^2}$. Apply the assumed elliptic regularity result to $g=z-e$. Its solution $w$ and $z$ satisfy the same weak Laplace equation, so testing their difference shows $\nabla (w-z)=0$. The Sobolev function with zero weak gradient property and zero Sobolev trace give $w=z$. Finally $\Vert z-e{\Vert }_{L^2}\le 2\Vert e{\Vert }_{L^2}$. This argument supplies the dual regularity needed by the Aubin–Nitsche duality argument for the equation with a positive reaction term.