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Use the Sobolev space with a partial Dirichlet conditionwhere $T$ is the Sobolev trace operator. The restricted Sobolev trace operator is continuous into $L^2({\Gamma }_D)$; its kernel is consequently a closed vector subspace of $H^1(\Omega )$. Thus $V$, with the first-order Sobolev norm, is a Hilbert space. The endpoint convention for ${\Gamma }_D$ and ${\Gamma }_N$ does not affect this definition: finitely many boundary points have zero boundary length. It would be inappropriate to impose pointwise boundary values on arbitrary $H^1$ functions in two dimensions.

For a smooth solution, Green's first identity givesThe boundary term vanishes because $Tv=0$ on ${\Gamma }_D$ and the normal derivative is zero on ${\Gamma }_N$. This suggests the weak formulationHere $a$ is exactly the first-order Sobolev inner product. In particular the Cauchy-Schwarz inequality givesThus $a$ is a bounded bilinear form and a coercive bilinear form, both with constant one. The Lax-Milgram theorem proves existence and uniqueness, together withThe positive reaction term already controls the entire Sobolev norm, so no Poincaré inequality is needed here. The zero Neumann boundary condition is the natural part of the weak formulation; it is not an additional restriction on the test space. For this weak solution, $-\Delta u+u=f$ holds as a distributional identity, and the zero normal derivative on ${\Gamma }_N$ is understood through the corresponding Green's first identity.

Let $e=u-u_h$. Subtracting the two weak formulations gives Galerkin orthogonality, $a(e,v_h)=0$ for every $v_h\in V_h$. Since $a$ is the first-order Sobolev inner product, for every $v_h\in V_h$ we haveIf $e\ne 0$, division proves $\Vert e{\Vert }_1\le \Vert u-v_h{\Vert }_1$; if $e=0$, the same conclusion is immediate. Thus the Céa lemma has constant one in this case.

Use the quadratic nodal interpolant $I_{h}u$. The assumed $H^3(\Omega )$ regularity gives continuous nodal values by the Sobolev embedding theorem, and its zero Sobolev trace makes the nodal values on ${\Gamma }_D$ zero. The quadratic Lagrange triangle construction then gives $I_{h}u\in V_h$. The allowed quadratic finite element interpolation estimate on a shape-regular mesh iswhere $C_I$ depends on the fixed triangle and shape regularity, but not on $h$. ThereforeThis step uses the inclusion $I_{h}u\in V_h$ explicitly permitted by the question. Arbitrary smaller vector subspaces without the same approximation property would not give this rate.