Consider the bivariate spline quasi-interpolant $S2$ defined on a bounded rectangle $R$ with simple interior knots. It is known that, if $f \in C^k(R)$, then $||f-S_2f||_{\infty}$ behaves like $O(H^k)$ for $k=0, \dots 3$, where $H$ is the maximum of the steps of the partitions on both sides of the rectangle. If I calculate $||S_2g-S_2(S_2g)||_{\infty}$ for a function $g \in C^{\infty}(R)$ I expect a behaviour $O(H)$ because $S_2g$ is only $C^1(R)$, but I find $O(H^3)$. Can someone give me a proof of this? Thanks.
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