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Consider $a(v,v)={\int }_b^c[|v^{\prime }{|}^2+V{v}^2]$ on real $H^1$ functions with $v(c)=0$, where $V_{-}=\max (-V,0)$. The pointwise estimate in the Endpoint Poincaré inequality gives ${\int }_b^c{V}_{-}{v}^2\le \theta \Vert v^{\prime }{\Vert }_2^2$ with $\theta ={\int }_b^c{V}_{-}(x)(c-x)dx$. If $\theta <1$, then $a(v,v)\ge (1-\theta )\Vert v^{\prime }{\Vert }_2^2\ge (1-\theta )[1+(c-b{)}^2/2{]}^{-1}\Vert v{\Vert }_{H^1}^2$. Thus the form is a coercive bilinear form even though the potential can change sign. For $V(x)=x$ on $(-1,1)$, $\theta ={\int }_{-1}^0(-x)(1-x)dx=5/6$, yielding the explicit constant $1/18$.

Take the affine trial space $V_1=\{v\in H^1(a,b):v(a)=1\}$ and the test space $V_0=\{v\in H^1(a,b):v(a)=0\}$. Multiplication by $v\in V_0$ and integration by parts incorporate the right Neumann boundary condition as a load:For the continuous piecewise-quadratic trial space, impose the left value exactly and use the same equation for every piecewise-quadratic test function vanishing at $a$. With a basis $({\psi }_j)$ of that test space and $u_h=1+{\sum }_j{c}_j{\psi }_j$, the finite element equations areThe Endpoint Poincaré inequality, reflected about the interval midpoint, makes the derivative norm a norm on $V_0$. Galerkin orthogonality then gives the exact best-approximation identity: for every admissible $v_h$,Therefore $u_h$ minimizes the derivative error over all admissible piecewise-quadratic functions.

To prove the requested error bound, assume $f\in H^1(a,b)$. Then $u\in H^3(a,b)$ and $u^{\prime \prime \prime }=-f^{\prime }$. We need an elementary Poincare-Wirtinger inequality on each element $I$ of length $h_I$: if $w\in H^1(I)$ has mean zero, its continuous representative has a zero somewhere, and the fundamental theorem of calculus and Cauchy-Schwarz inequality give $\Vert w{\Vert }_{L^2(I)}\le h_I\Vert w^{\prime }{\Vert }_{L^2(I)}$.

Choose an affine function $q_I$ whose slope is the average of $u^{\prime \prime }$ on $I$ and whose average equals the average of $u^{\prime }$. Then $w=u^{\prime }-q_I$ and $w^{\prime }=u^{\prime \prime }-q_I^{\prime }$ both have mean zero. Applying the just-proved bound twice yieldsDefine a quadratic interpolant on $I$ by integrating $q_I$ from the true left-endpoint value. The equal-average condition ensures that it reaches the true right-endpoint value, so these pieces join to an admissible continuous piecewise-quadratic function $I_{h}u$. Sum the squared estimates and use the best-approximation property:Only the maximum element length is used; equal spacing is unnecessary.

For the Aubin–Nitsche duality argument, let $e=u-u_h$ and solve $-z^{\prime \prime }=e$, $z(a)=0$, $z^{\prime }(b)=0$. Explicitly $z^{\prime }(x)={\int }_x^{b}e(t)dt$, so $\Vert z^{\prime \prime }{\Vert }_2=\Vert e{\Vert }_2$. Integration by parts and Galerkin orthogonality givewhere $J_{h}z$ is its nodal piecewise-linear interpolant, which belongs to the quadratic test space. On each element $(J_{h}z{)}^{\prime }$ is the average of $z^{\prime }$, so the same mean-zero inequality gives $\Vert z^{\prime }-(J_{h}z{)}^{\prime }{\Vert }_2\le h\Vert z^{\prime \prime }{\Vert }_2$. Canceling $\Vert e{\Vert }_2$ when nonzero proves the additional order:

A real bilinear form is a coercive bilinear form on $V$ when $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for all $v\in V$ and some fixed $\alpha >0$. It is a bounded bilinear form, equivalently continuous, when $|a(v,w)|\le M\Vert v{\Vert }_V\Vert w{\Vert }_V$ for one finite $M$. A continuous linear functional satisfies $|\ell (v)|\le C_{\ell }\Vert v{\Vert }_V$.

The Galerkin method seeks $u_h\in S$ with $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in S$. Subtraction from the continuous equations gives Galerkin orthogonality: $a(u-u_h,v_h)=0$ for every $v_h\in S$. Under the stated continuity and coercivity assumptions, Céa's lemma saysIndeed, with $e=u-u_h$ and any $v_h\in S$, Galerkin orthogonality eliminates $v_h-u_h$:If $e=0$ the claim is immediate; otherwise cancel $\Vert e{\Vert }_V$ and take the infimum. No symmetry of $a$ was used. These statements define an approximation when it exists; a finite-dimensional $S$ gives existence and uniqueness by the Lax-Milgram theorem, as does a closed vector subspace.

For the interval example write $L=c-b$ and $R=\max (|b|,|c|)$. The one-dimensional Sobolev representative of $v\in V$ satisfiesby the Cauchy-Schwarz inequality. The representation follows by taking the absolutely continuous function primitive of the weak derivative; its difference from $v$ has zero weak derivative, hence is constant, and $v(c)=0$ fixes that constant. Integrating the estimate gives the Endpoint Poincaré inequalityThus the nonnegative potential $x^2$ givesFor continuity, $|x|\le R$, the Cauchy-Schwarz inequality for the two integrals, and then its two-dimensional version giveSimilarly $\boxed{|\ell (v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_2\le \Vert f{\Vert }_2\Vert v{\Vert }_V}$, proving continuity of the linear functional.

For the last displayed integral take $\Omega =(-1,1)$ and retain the endpoint condition $v(1)=0$, the domain-consistent interpretation of the changed formula. The changed form is still coercive. A negative potential on part of the interval does not by itself destroy coercivity. The sharper pointwise estimate above givesDiscarding only the positive part of the potential yieldsThe proved Endpoint Poincaré inequality with $L=2$ gives $\Vert v{\Vert }_V^2\le 3\Vert v^{\prime }{\Vert }_2^2$, and thereforeThis is a concrete application of negative-potential domination by endpoint control. The constant is sufficient, without claiming to be optimal.