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For distinct ordered knots, put $h_i=t_{i+1}-t_i>0$. The linear B-spline $N_i$ is the triangular hat supported on $[t_i,t_{i+2}]$, equal to $(t-t_i)/h_i$ on its rising segment and $(t_{i+2}-t)/h_{i+1}$ on its falling segment. Its integral-normalized partner is $M_i=2N_i/(h_i+h_{i+1})$. ThusFor example, the left overlap is ${\int }_0^{h_i}(u/h_i)(1-u/h_i)du=h_i/6$. Nonadjacent hats have disjoint interiors of their supports, so their inner product is zero. The linear-spline mixed Gram matrix therefore has rowRetain only column indices in $1,\dots ,n$: the first or last row has no neighbor outside this range. Admissible repeated endpoint knots are obtained by a limit with zero spacings, provided $h_i+h_{i+1}>0$; a degenerate hat with zero support width is not a basis element.

For $k=2$ on an equidistant spline knot sequence, $N_i$ is the triangular hat with value one at $t_{i+1}$ and support $[t_i,t_{i+2}]$. Its slopes on the two pieces are $1/h$ and $-1/h$, and $M_i=N_i/h$. Direct integration givesNonadjacent hats have disjoint interiors of their supports, so their inner product vanishes. Dividing by $h$ therefore givesThis linear-spline mixed Gram matrix is tridiagonal. Every diagonal entry is $2/3$, including the first and last: the distinct-knot hats have their full supports inside $[0,1]$. End rows simply lack one off-diagonal neighbor; there are no repeated endpoint spline knots here.

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 matrixOnly entries with indices in $1,\dots ,n$ are present. Thus ${\sum }_{j\ne i}|g_{ij}|\le 1/3$ in every row.

For $Ga=b$, choose an index $i$ with $|a_i|=\Vert a{\Vert }_{{\ell }^{\infty }}$. The reverse triangle inequality then yieldsHence $\Vert a{\Vert }_{{\ell }^{\infty }}\le 3\Vert b{\Vert }_{{\ell }^{\infty }}$. Applying this to every $b$ provesThis is the inverse infinity-norm bound from diagonal dominance; no total positivity theorem is needed. The proof also gives invertibility directly by taking $b=0$.