= Solution
Let $y=(f(x_1),\ldots,f(x_n))^T$ and write the interpolating <spline> as $P_xf=\sum_{j=1}^nc_jN_j$. The interpolation equations are $A_xc=y$, so $c=A_x^{-1}y$. Since $P_x$ is defined as a unique interpolant for arbitrary data, its <B-spline collocation matrix> is invertible. Ordered sites satisfying diagonal positivity ensure this by question 7. Positivity alone without admissible ordering does not: two equal sites in overlapping <basis> <supports> could give two identical rows. The <norm> argument below uses the existence of the operator stipulated here.
Standard partition-normalized <B-splines> satisfy $N_j\ge0$ and $\sum_{j=1}^nN_j\le1$. The latter is a <subpartition of unity for B-splines>: extending the knot sequence gives a full partition of unity, and discarding <basis> functions cannot increase its sum. Hence
$$
|P_xf(t)|\le\sum_j|c_j|N_j(t)\le\|c\|_{\ell^\infty}
\le\|A_x^{-1}\|_{\ell^\infty}\|y\|_{\ell^\infty}
\le\|A_x^{-1}\|_{\ell^\infty}\|f\|_\infty.
$$
Taking the supremum over $t$ and then over unit-<norm> $f$ gives the <B-spline interpolation operator norm> estimate
$$
\boxed{\|P_x\|_{L^\infty}\le\|A_x^{-1}\|_{\ell^\infty}.}
$$
Here the <matrix> <norm> is the induced maximum-row-sum <norm>, not the maximum of its individual entries. The proof only needs nonnegative <basis> functions, their sum at most one, and invertible interpolation.
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