= Solution
Let $S_1u$ be the sum of the four axial neighbors and $S_2u$ the sum of the four diagonal neighbors. The two PDF stencils define
$$
\Gamma_9u=\frac23S_1u+\frac16S_2u-\frac{10}3u,
\qquad M_hu=\frac23u+\frac1{12}S_1u.
$$
The reaction stencil has four axial weights $1/12$ and center weight $2/3$, so its weights sum to one. <Taylor expansion> gives
$$
h^{-2}\Gamma_9u=\Delta u+\frac{h^2}{12}\Delta^2u+O(h^4),\qquad
M_hu=u+\frac{h^2}{12}\Delta u+O(h^4).
$$
For example, the diagonal neighbors supply the $u_{xxyy}$ term needed to turn the fourth derivatives into the full squared <Laplacian>. Adding the reaction contribution produces
$$
h^{-2}\Gamma_9u+\lambda M_hu
=(\Delta+\lambda)u+\frac{h^2}{12}\Delta(\Delta+\lambda)u+O(h^4).
$$
On a smooth solution of the constant-coefficient equation, both displayed leading terms vanish. Thus the <compact fourth-order Helmholtz stencil> has
$$
\boxed{\text{fourth-order normalized local accuracy,}\qquad
\Gamma_9u+\lambda h^2M_hu=O(h^6).}
$$
The cancellation uses the differential equation and constant $\lambda$; the <nine-point finite-difference stencil> alone still has an $O(h^2)$ operator error on an arbitrary smooth function. It also does not establish a uniform global fourth-order error without an appropriate inverse bound and compatible treatment of boundaries and resonances.
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