 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 $\frac{3}{4}{y}_j+\frac{1}{4}{y}_{j+1}$ 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.