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A nine-point finite-difference stencil on a square grid couples a central value to its four axial neighbours and four diagonal neighbours. Ordering the grid by columns produces a block tridiagonal matrix whose diagonal and off-diagonal blocks are symmetric tridiagonal Toeplitz matrices.

The centered quadratic spline interpolation matrix is the symmetric tridiagonal Toeplitz matrixFor $n=1$, this means simply $A_x=(3/4)$. To evaluate the inverse norm exactly, let $D=\mathrm{diag}((-1{)}^i{)}_{i=1}^n$ and $M=8D{A}_{x}D=\mathrm{tridiag}(-1,6,-1)$. If $J$ is the nonnegative adjacency matrix of the path, then $M=6I-J$ and $\Vert J/6{\Vert }_{\infty }\le 1/3$. The convergent Neumann seriestherefore has nonnegative entries. Since $A_x^{-1}=8D{M}^{-1}D$, its entrywise absolute values are $8M^{-1}$. Let $r_i={\sum }_j|(A_x^{-1}{)}_{ij}|$. Then $r=8M^{-1}1$, orThe homogeneous characteristic roots are $3\pm 2\sqrt{2}$. Put $\tau =3-2\sqrt{2}\in (0,1)$. The particular solution $r_i=2$ and the two boundary conditions giveThese row sums are symmetric and increase toward the centre: the exponential sum in the numerator is minimized at the middle integer or pair of integers. With $m=\lfloor (n+1)/2\rfloor$, the exact finite-size inverse norm is thusFor example it equals $4/3$ at $n=1$ and $8/5$ at $n=2$, tending to $2$ as $n\to \infty$. Stating only the bound two would not evaluate the finite matrix norm requested here.

Let ${\theta }_k=k\pi h=k\pi /(m+1)$ and define $q_j^{(k)}=\sin (j{\theta }_k)$. The artificial boundary values are $q_0^{(k)}=q_{m+1}^{(k)}=0$. For the symmetric tridiagonal Toeplitz matrix $H=[\beta ,\alpha ,\beta ]$, the sine addition formula givesThus all such matrices have the same discrete sine transform eigenvectors, with eigenvaluesThe sine vectors are mutually orthogonal vectors and nonzero, so the matrix $Q=(\sin (ij\pi h){)}_{i,j=1}^m$ is invertible. If $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$, then $HQ=QD$ and therefore