= Solution
The <midpoint quadratic spline collocation> matrix is
$$
\boxed{A_x=\frac18
\begin{pmatrix}
6&1&0&\cdots&0\\
1&6&1&\ddots&\vdots\\
0&1&6&\ddots&0\\
\vdots&\ddots&\ddots&\ddots&1\\
0&\cdots&0&1&6
\end{pmatrix}.}
$$
Its <strict diagonal dominance> margin is at least $3/4-2/8=1/2$. The <inverse infinity-norm bound from diagonal dominance> thus already proves $\|A_x^{-1}\|_\infty\le2$: choosing an index where a vector $c$ attains its maximum absolute entry gives $\|A_xc\|_\infty\ge\tfrac12\|c\|_\infty$, which also proves invertibility.
To evaluate the finite-dimensional norm exactly, set $D_{ii}=(-1)^i$ and $B=DA_xD$. If $T$ is the matrix with ones on its two adjacent diagonals and zeros elsewhere, then $B=(3/4)(I-T/6)$. Since $\|T/6\|_\infty\le1/3$, the <Neumann series>
$$
B^{-1}=\frac43\sum_{r=0}^\infty(T/6)^r
$$
has nonnegative entries. Thus $A_x^{-1}=DB^{-1}D$ has absolute entry values equal to those of $B^{-1}$. Its absolute row-sum vector is $v=B^{-1}\mathbf1$, characterized by
$$
6v_i-v_{i-1}-v_{i+1}=8,\qquad v_0=v_{n+1}=0.
$$
The homogeneous roots are $3\pm2\sqrt2$, and the constant particular solution is $v_i=2$. Writing $q=3-2\sqrt2\in(0,1)$ and imposing the two boundary values gives
$$
v_i=2\left[1-\frac{q^i+q^{n+1-i}}{1+q^{n+1}}\right].
$$
This is largest at the middle row or rows. Therefore, with $i_* =\lfloor(n+1)/2\rfloor$, \b[the exact norm is]
$$
\boxed{\|A_x^{-1}\|_\infty=
2\left[1-\frac{q^{i_*}+q^{n+1-i_*}}{1+q^{n+1}}\right]<2.}
$$
For example, it is $4/3$ for $n=1$ and $8/5$ for $n=2$, and it increases towards two as the system grows. Thus two is a uniform bound, rather than the exact finite-$n$ inverse norm.
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