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For unit-spaced knots, the quadratic cardinal B-spline with support $[i,i+3]$ is $N_i(t)=B(t-i)$, where the Cox-de Boor recurrence givesIn particular $B(1)=B(2)=1/2$ and $B(0)=B(3)=0$. At $x_i=i+2$, the only nonzero entries in row $i$ are $A_{ii}=1/2$ and, if $i<n$, $A_{i,i+1}=1/2$. Write $U$ for the upper shift matrix, with ones immediately above its diagonal. Thensince $U^n=0$. Thus $(A_x^{-1}{)}_{ij}=2(-1{)}^{j-i}$ for $j\ge i$, and zero otherwise. The largest absolute row sum is the first, giving $\Vert A_x^{-1}{\Vert }_{{\ell }_{\infty }}=2n$.

Using (1) and the permitted $d_3=3$ proves the stronger linear growth of shifted quadratic spline interpolation estimateTherefore $\Vert P_x\Vert =\Theta (n)$ and is not uniformly bounded. An $O(n)$ upper bound alone would not imply unboundedness; the lower bound is essential.

For the uniform spline knots, translation reduces every degree-two basis function to a quadratic cardinal B-spline. From the Cox-de Boor recurrence, an order-three function takes the valuesAt the prescribed site $x_i=i+2$, only $N_i$ and, when $i<n$, $N_{i+1}$ have nonzero values. Hence the B-spline collocation matrix isThe shift satisfies $S^n=0$. A finite geometric expansion gives the exact inverseIts $i$th absolute row sum is $2(n-i+1)$, so $\Vert A^{-1}{\Vert }_{{\ell }^{\infty }}=2n$. Use the preceding bounds and the supplied stability constant $d_3=3$:Thus the linear growth of shifted quadratic spline interpolation is $\Theta (n)$, which in particular is $O(n)$. The lower bound, rather than the $O(n)$ upper bound alone, proves that these spline interpolation operators are not uniformly bounded.