= Solution
Fix an explicit stationary-grid convention. For <subdivision arity> $m>1$, write the univariate rule as
$$
p_i^{\ell+1}=\sum_jS_{ij}p_j^\ell,\qquad S_{ij}=a_{i-mj}.
$$
A column of the <subdivision matrix> is one translated <subdivision mask>; the next column is translated by $m$ fine-grid rows. Thus \b[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 $\phi$ obtained from data $p_j=\delta_{j0}$. Translation gives the representation $P(t)=\sum_jp_j\phi(t-j)$. Descendant indices of index zero have parameter positions
$$
\frac{k_1}{m}+\frac{k_2}{m^2}+\cdots+\frac{k_\ell}{m^\ell},\qquad k_r\in[\alpha,\beta].
$$
Thus 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 is
$$
p'_{2j}=p_j,\qquad p'_{2j+1}=\tfrac12(p_j+p_{j+1}).
$$
Each 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 explicitly
$$
\phi(t)=\max(1-|t|,0),\qquad\boxed{\operatorname{supp}\phi=[-1,1].}
$$
The 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> is
$$
\boxed{\mathcal P=\operatorname{span}\{1,t\}.}
$$
For 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 \b[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.
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