= Solution
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 values
$$
N_{j,3}(j)=0,\quad N_{j,3}(j+1)=\frac12,\quad
N_{j,3}(j+2)=\frac12,\quad N_{j,3}(j+3)=0.
$$
At 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> is
$$
A=\frac12(I+S),\qquad
S_{ij}=\begin{cases}1,&j=i+1,\\0,&\text{otherwise}.\end{cases}
$$
The shift satisfies $S^n=0$. A finite geometric expansion gives the exact inverse
$$
\boxed{A^{-1}=2\sum_{r=0}^{n-1}(-S)^r,\qquad
(A^{-1})_{ij}=
\begin{cases}
2(-1)^{j-i},&j\ge i,\\
0,&j<i.
\end{cases}}
$$
Its $i$th absolute row sum is $2(n-i+1)$, so $\|A^{-1}\|_{\ell^\infty}=2n$. Use the preceding bounds and the supplied stability constant $d_3=3$:
$$
\boxed{\frac{2n}{3}\le\|P_{\mathbf x}\|_{L^\infty}\le2n}.
$$
Thus the <linear growth of shifted quadratic spline interpolation> is \b[$\Theta(n)$], 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.
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