= Reaction-diffusion regularity from Laplacian regularity
{title2=$\|z\|_{H^2}\leq2C\|e\|_{L^2}$}
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 $\|w\|_{H^2}\leq C\|g\|_{L^2}$. 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 $\|z\|_{H^2}\leq2C\|e\|_{L^2}$.
Indeed, the <Lax-Milgram theorem> first constructs $z$ in the <Sobolev space with a partial Dirichlet condition>; testing with $z$ shows $\|z\|_{H^1}\leq\|e\|_{L^2}$. 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 $\|z-e\|_{L^2}\leq2\|e\|_{L^2}$. This argument supplies the dual regularity needed by the <Aubin–Nitsche duality argument> for the equation with a positive reaction term.
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