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Fix an explicit stationary-grid convention. For subdivision arity $m>1$, write the univariate rule asA column of the subdivision matrix is one translated subdivision mask; the next column is translated by $m$ fine-grid rows. Thus the column displacement is the arity. If $a_k$ is active from $k=\alpha$ to $k=\beta$, the subdivision mask width is $w=\beta -\alpha +1$ fine-grid positions, after removing zero padding. Equivalently the mask span is $\beta -\alpha$ fine intervals. This is different from counting old control points in a row stencil: that count measures how many controls influence one newly created point. Stating the convention avoids confusing these two widths.

For a convergent subdivision curve scheme, its support is the influence region of one original control point, equivalently the support of a function of the basic limit $\varphi$ obtained from data $p_j={\delta }_{j0}$. Translation gives the representation $P(t)={\sum }_j{p}_j\varphi (t-j)$. Descendant indices of index zero have parameter positionsThus the limiting support enclosure is $[\alpha /(m-1),\beta /(m-1)]$, as in the support of a stationary subdivision scheme. Endpoint activity gives equality for the ordinary positive convergent masks under discussion.

The functional precision set of a subdivision scheme consists of functions exactly recovered from their uniformly sampled data using the scheme's consistent parameter placement. Polynomial precision records the exactly reproduced polynomial space. Sampling-grid spacing and origin must be specified; the usual arbitrary-grid precision requirement uses every spacing and shift, rather than merely membership in the fixed-grid reconstruction space.

Center the given mask at indices $-1,0,1$. Its binary linear interpolatory subdivision rule isEach new control point is either an old point or an edge midpoint. Hence every refined control polygon is the same curve with extra vertices. Starting from scalar impulse data, the limit is explicitlyThe mask width is three positions, its mask span is two, its largest row stencil uses two old control points, and its subdivision arity is two. Translating the mask indexing to $0,1,2$ translates the basic support to $[0,2]$; the support width remains two original grid intervals.

On $[j,j+1]$, reconstruction is $(j+1-t)p_j+(t-j)p_{j+1}$. Constant and affine samples are therefore recovered exactly. A non-affine polynomial cannot agree with a linear expression on a whole open interval. Thus its polynomial functional precision set of a subdivision scheme isFor continuous functions reproduced at every translated sampling grid, the same conclusion holds: agreement with all chord interpolants forces the affine interpolation identity, hence affinity. At one fixed grid, the larger exactly representable class is the space of linear splines with those spline knots; this is not a higher polynomial precision claim.

For arbitrary data, the one-sided derivatives at $j$ are $p_j-p_{j-1}$ and $p_{j+1}-p_j$. They need not coincide. Therefore the generic limit is $C^0$, not $C^1$. Its first derivative is piecewise constant; exceptional affine data, or individual matching neighboring slopes, can be smoother. This conclusion follows directly from the limit formula, not just from a heuristic mask-width test.

Let the original abscissae be $x_i=ih$ and vary only the ordinate of vertex $i$. Under Chaikin subdivision, the two child values on an edge are $\frac{3}{4}{y}_j+\frac{1}{4}{y}_{j+1}$ and $\frac{1}{4}{y}_j+\frac{3}{4}{y}_{j+1}$. All influence weights are nonnegative, so no cancellation removes an active descendant.

At level $r$ the spacing is $h_r=h/2^r$. If the leftmost influenced vertex is at $a_r$, the next leftmost is the quarter point on the preceding edge, $a_{r+1}=a_r-3h_r/4$; the rightmost advances by the same amount. Starting at $x_i$, the influenced vertex range is thereforeThe neighboring zero vertices bound any extra affected polygon segments by one refined spacing, which tends to zero. Passing to the limit givesIt spans three original grid intervals, with its endpoints at original edge centers. Positive descendant weights occur arbitrarily close to each endpoint, so this is the exact support closure. The support of a stationary subdivision scheme concerns the limit curve, not only the first pair of adjacent original edges. This describes interior vertices or an infinite uniform sequence; a finite open polygon needs an endpoint convention.

Fix the indexing conventionwhere the subdivision arity $a>1$ is an integer. A change of the initial control $P_0^0$ can influence only indices $i=j_1$ after one step, where $M_{j_1}\ne 0$. After $\ell$ steps the possible descendant indices areLet $j_{\min },j_{\max }$ be the smallest and largest active subdivision mask indices. Dividing by $a^{\ell }$ gives the enclosureTherefore the support of a stationary subdivision scheme, in initial control-grid units, is enclosed byFor the stated symmetric subdivision mask with active extremes $-w,w$, the width is $2w/(a-1)$. For a nonzero compactly supported basic limit function $\varphi$, this is an exact endpoint calculation: its refinement equation is $\varphi (x)={\sum }_j{M}_j\varphi (ax-j)$. If $\beta$ is its right support endpoint, only the largest active subdivision mask index can contribute sufficiently close to the rightmost point of this sum; all smaller indices have argument beyond $\beta$. Hence $\beta =(\beta +j_{\max })/a$, giving $\beta =j_{\max }/(a-1)$. The left endpoint follows in the same way. This argument also excludes cancellation at an extreme for signed subdivision masks. The support for initial control $P_k$ is translated by $k$. If the nominal subdivision mask was padded by zero coefficients, use its actual extremes; and a limiting subdivision curve has to exist before its support is meaningful. This calculation describes the entire possible influence region, rather than the width of a single refinement stencil on its finer grid.