 Solution

Let the three grid-edge vectors be $v_1=a$, $v_2=b$, $v_3=b-a$, and define the constant-coefficient differential operator $D={\sum }_{r=1}^3(v_r\cdot \nabla {)}^2$. For a stencil at grid spacing $h$, its displacement vectors $\delta$ about the new vertex have vanishing first and third moments by central symmetry. All four stencil types have the same second momentFor the old-vertex stencil, each pair $\pm h{v}_r$ contributes $h^2{v}_r{v}_r^T/8$. For an edge parallel to $a$, the endpoint displacements are $\pm ha/2$ with weights $3/8$, and the opposite displacements are $\pm h(b-a/2)$ with weights $1/8$. Their moment is $h^2[a{a}^T/4+b{b}^T/4-(a{b}^T+b{a}^T)/8]$, precisely the same expression; the other edge orientations follow by symmetry.

If $f$ is a multivariate polynomial of total degree at most three, its Taylor expansion is exact through the third-order term. Thus every refined vertex samples the single polynomialThe polynomial $Df$ has degree at most one, so subsequent refinement reproduces it exactly. Starting at unit spacing and successively halving $h$, the limit therefore isThe correction lowers degree by two, leaving the leading homogeneous term unchanged. Hence every sampled polynomial of degree at most three generates a polynomial of that same degree, even though quadratics and cubics need not be reproduced identically. For the extruded cases this gives $y^2\mapsto y^2+1/3$ and $y^3\mapsto y^3+y$.

Degree four is not guaranteed. Extruded data $g_j=j^4$ reduce to the Cardinal cubic B-spline $G(y)={\sum }_j{j}^4{M}_3(y-j)$, which is piecewise cubic and hence cannot be a single polynomial of degree four. It is not a lower-degree polynomial either: at integer rows, $G(k)=[(k-1{)}^4+4k^4+(k+1{)}^4]/6=k^4+2k^2+1/3$, which cannot agree with a cubic at all integers. This counterexample excludes every guaranteed degree $d\ge 4$. ConsequentlyThis distinguishes polynomial generation from exact reproduction and again refers to all polynomials through the stated degree.