= Solution
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 $\tfrac34y_j+\tfrac14y_{j+1}$ and $\tfrac14y_j+\tfrac34y_{j+1}$. 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 therefore
$$
\left[x_i-\frac{3h}{4}\sum_{j=0}^{r-1}2^{-j},\ x_i+\frac{3h}{4}\sum_{j=0}^{r-1}2^{-j}\right].
$$
The neighboring zero vertices bound any extra affected polygon segments by one refined spacing, which tends to zero. Passing to the limit gives
$$
\boxed{\operatorname{supp}\text{ influence}=[x_i-3h/2,\ x_i+3h/2].}
$$
It 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.
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