= Solution
A <strict diagonal dominance> argument proves the estimate even on a nonuniform strictly increasing <spline knot sequence>. Put $h_i=t_{i+1}-t_i>0$. Then $N_i$ has a rising piece of length $h_i$ and a falling piece of length $h_{i+1}$, while $M_i=2N_i/(h_i+h_{i+1})$. The same integrations as in part (b) give the <linear-spline mixed Gram matrix>
$$
g_{ii}=\frac23,\qquad g_{i,i-1}=\frac{h_i}{3(h_i+h_{i+1})},\qquad
g_{i,i+1}=\frac{h_{i+1}}{3(h_i+h_{i+1})}.
$$
Only entries with indices in $1,\ldots,n$ are present. Thus $\sum_{j\ne i}|g_{ij}|\le1/3$ in every row.
For $Ga=b$, choose an index $i$ with $|a_i|=\|a\|_{\ell^\infty}$. The <reverse triangle inequality> then yields
$$
|b_i|\ge g_{ii}|a_i|-\sum_{j\ne i}|g_{ij}||a_j|
\ge\left(\frac23-\frac13\right)\|a\|_{\ell^\infty}.
$$
Hence $\|a\|_{\ell^\infty}\le3\|b\|_{\ell^\infty}$. Applying this to every $b$ proves
$$
\boxed{\|G^{-1}\|_{\ell^\infty}\le3,\qquad\|P_{\mathcal S}\|_\infty\le3\quad(k=2).}
$$
This is the <inverse infinity-norm bound from diagonal dominance>; no total positivity theorem is needed. The proof also gives invertibility directly by taking $b=0$.
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