= Solution
{c}
Put $e=u-u_h\in V$. The <Aubin–Nitsche duality argument> uses the dual <weak formulation>
$$
z\in V,\qquad a(v,z)=(e,v)_0\quad(v\in V).
$$
By symmetry of $a$ and the <Lax-Milgram theorem>, this has a unique solution. Taking $v=z$ and using the <Cauchy-Schwarz inequality> shows
$$
\|z\|_1^2=(e,z)_0\leq\|e\|_0\|z\|_0\leq\|e\|_0\|z\|_1,
\qquad \|z\|_1\leq\|e\|_0.
$$
We 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 $\Gamma_N$, and
$$
\|w\|_{H^2(\Omega)}\leq C_3\|z-e\|_0\leq2C_3\|e\|_0.
$$
By <Green's first identity>, for every $v\in V$,
$$
(\nabla w,\nabla v)_0=-(g,v)_0=(e-z,v)_0=(\nabla z,\nabla v)_0.
$$
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 $z=w$, proving
$$
\boxed{z\in H^2(\Omega)\cap V,\qquad\|z\|_{H^2(\Omega)}\leq2C_3\|e\|_0.}
$$
This 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> $L_hz$, 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 estimate>
$$
\|z-L_hz\|_1\leq C_Lh\|z\|_{H^2(\Omega)}.
$$
Thus no $H^3$ assumption on the dual solution is necessary. Taking $v=e$ in the dual <weak formulation> and using <Galerkin orthogonality> gives
$$
\begin{aligned}
\|e\|_0^2
&=a(e,z)=a(e,z-L_hz)\\
&\leq\|e\|_1\|z-L_hz\|_1\\
&\leq2C_3C_Lh\|e\|_1\|e\|_0.
\end{aligned}
$$
If $\|e\|_0=0$, the desired bound holds. Otherwise divide by $\|e\|_0$ and use part (d):
$$
\boxed{\|u-u_h\|_{L^2(\Omega)}\leq2C_3C_Lh\|u-u_h\|_{H^1(\Omega)}\leq C_2h^3\|u\|_{H^3(\Omega)},\qquad C_2=2C_3C_LC_1.}
$$
All constants depend only on the fixed <triangle> and <shape regularity>. The acute-angle assumption enters through the supplied mixed-boundary <elliptic regularity> result.
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