 Solution

Use the standard partition normalization of the B-splines, so $N_j\ge 0$ and the subpartition of unity for B-splines gives ${\sum }_{j=1}^n{N}_j(t)\le 1$. Write the interpolating spline as $P_{x}f={\sum }_j{c}_j{N}_j$ and the sampled values as $y_i=f(x_i)$. The interpolation equations are $A_{x}c=y$, hence $c=A_x^{-1}y$ whenever the stated interpolation map is well defined. For every $t$,The sampling map also satisfies $\Vert y{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$. Taking supremum norms and then the operator norm therefore provesThis is the B-spline interpolation operator norm bound, and it applies on any interval containing the sampling sites on which these normalized B-splines are used.

There is a small qualification to the printed hypotheses: diagonal positivity alone does not ensure uniqueness unless the usual site assumptions are supplied. For example, with two quadratic B-splines on knots $1,2,3,4,5$ and $x_1=x_2=3$, both diagonal values equal $1/2$, but $A_x$ has two identical rows and is singular. The intended increasing sites $x_1<\cdots <x_n$ with $N_i(x_i)>0$ satisfy the Schoenberg–Whitney theorem, making $A_x$ invertible; alternatively invertibility is implicit in the premise that $P_x$ is the unique interpolation map. The displayed bound is proved under that premise, rather than inferring invertibility from diagonal positivity alone.