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Suppose $\Vert s_{\ast }{\Vert }_{\infty }=1$ and $s_{\ast }(x_j)=(-1{)}^j$ at increasing sites satisfying the Schoenberg–Whitney theorem conditions. For the B-spline collocation matrix $A$, the coefficient linear functional in the dual basis is ${\mu }_i(s)={\sum }_j(A^{-1}{)}_{ij}s(x_j)$. Therefore $\Vert {\mu }_i\Vert \le {\sum }_j|(A^{-1}{)}_{ij}|$. The checkerboard inverse of a totally nonnegative matrix makes ${\mu }_i(s_{\ast })=(-1{)}^i{\sum }_j|(A^{-1}{)}_{ij}|$, attaining that bound. Since ${\mu }_i(s_{\ast })=a_i^{\ast }$,The conclusion is an extremal property of an alternating unit-norm spline, not a claim that every bounded spline has these extremal coefficients.

The equality holds when an alternating unit-norm spline $s_{\ast }$ has as many ordered extremal sites as the dimension, and those sites satisfy the Schoenberg–Whitney theorem. For every admissible set, sampling has operator norm at most one, giving $\kappa (\mathcal{S})\le \Vert A_{\mathbf{x}}^{-1}{\Vert }_{\infty }$. At the alternating sites the checkerboard inverse of a totally nonnegative matrix identifies each inverse absolute row sum with the magnitude of the corresponding coefficient of $s_{\ast }$. The Chebyshev spline coefficient and dual norm equality therefore gives equality. This minimizes amplification into B-spline coefficients; it does not by itself prove optimality for the distinct Lebesgue constant of interpolation.

Write the spline interpolation operator as $P_{\mathbf{x}}=T{A}_{\mathbf{x}}^{-1}R$, where $R$ samples at the interpolation sites and $T$ synthesizes the normalized B-spline expansion. Sampling has operator norm at most one, and nonnegative B-splines with sum at most one give $\Vert T\Vert \le 1$. HenceHere the matrix norm is the maximum absolute row sum. Existence of the inverse matrix requires the usual ordered distinct interpolation sites and the Schoenberg–Whitney theorem support conditions, with endpoint values interpreted by their one-sided limits. If the displayed diagonal condition is read without the implicit ordering and distinctness, it is insufficient: all three sites equal to $1/2$ in the quadratic example give positive diagonal entries but three identical rows. The estimate applies whenever the interpolating linear map in the question is defined uniquely.

For the quadratic Bernstein basis, $N_1=(1-x{)}^2$, $N_2=2x(1-x)$ and $N_3=x^2$. Sampling and inverting giveThe three cardinal Lagrange interpolation polynomials areThe Lebesgue constant of interpolation gives the exact operator norm: the upper bound follows from $|Pg(x)|\le \Vert g{\Vert }_{\infty }{\sum }_i|{\ell }_i(x)|$. To attain it at a maximizing point, prescribe at the three sites the signs of the corresponding cardinal polynomials and extend those values by a continuous piecewise linear function of norm one.

On $0\le x\le 1/2$, the first two cardinal polynomials are nonnegative and the last is nonpositive. Since their sum is one, their absolute sum is $1-2{\ell }_2(x)=1+2x-4x^2$. Its maximum is $5/4$ at $x=1/4$. Reflection gives the same maximum at $3/4$ on the other half. In particular the data $(1,1,-1)$ produce $1+2x-4x^2$, attaining $5/4$. ThusThe coefficient estimate loses cancellation among the B-splines, which explains why it is not sharp for the resulting function norm.

Fix a clamped nondecreasing spline knot sequence $t_1,\dots ,t_{n+k}$ on $[a,b]$, with $t_1=\cdots =t_k=a$ and $t_{n+1}=\cdots =t_{n+k}=b$. For continuous splines assume interior multiplicities at most $k-1$, and assume $t_i<t_{i+k}$. An order-$k$ spline has degree at most $k-1$ on each nonempty knot interval; at a knot of multiplicity $r$ it has $k-1-r$ continuous derivatives. Its dimension is $n$, and a normalized B-spline basis $N_1,\dots ,N_n$ has local support, nonnegative values and sum one on $[a,b]$.

These properties are visible in the Cox-de Boor recurrence. Starting with interval indicators at order one, setwith a term of zero denominator defined to be zero. On each contributing support the weights are nonnegative, and summing the recurrence makes the coefficients of each lower-order B-spline add to one. This gives partition of unity and $\mathrm{supp}{N}_{i,k}=[t_i,t_{i+k}]$, with endpoint values defined by limits.

At distinct ordered sites $x_1<\cdots <x_n$, interpolation is the linear systemIt has a unique solution for every data vector exactly when the B-spline collocation matrix is invertible. The Schoenberg–Whitney theorem characterizes this by $N_i(x_i)>0$ for every $i$; away from clamped endpoints this is $t_i<x_i<t_{i+k}$. At the clamped endpoints the first and last B-splines have value one, so the endpoint version uses this positive-value formulation rather than the strict inequalities. Both ordering and the support conditions matter. A spline space with interior knots is generally not a Chebyshev system on the entire interval: a locally supported B-spline vanishes on an interval. Thus arbitrary distinct nodes need not permit interpolation. Merely specifying all data sites as knots also need not fix the remaining freedom; for example, cubic interpolation with simple interior knots at $m-2$ data sites has dimension $m+2$, and two extra boundary conditions are needed to interpolate $m$ values uniquely.

The positivity governing this existence theorem is total nonnegativity of B-spline collocation matrices: every square minor with increasing row and column indices is nonnegative. In spline terminology this is often called total positivity, but it permits zero minors caused by disjoint supports. It is not the stronger assertion that every minor is strictly positive. A useful proof uses knot insertion. Each refinement step changes coefficients bywith the unchanged endpoint pieces. The refinement matrix is nonnegative bidiagonal, and its minors are nonnegative: any nonzero determinant has its allowed diagonal matching and is a product of nonnegative entries. Products preserve this property by the Cauchy–Binet formula. Insert each sampling site to full multiplicity, so evaluation at that site becomes selection of an ordered refined coefficient. The collocation matrix is then an ordered row submatrix of the refinement product, proving total nonnegativity. The strict support criterion is the strict part of this argument: an ordered path through the consecutive support intervals exists exactly when every diagonal B-spline value is positive, giving the Schoenberg–Whitney theorem determinant criterion.

If $A$ is invertible, its determinant is positive and the adjugate identity yieldsThis checkerboard inverse of a totally nonnegative matrix is the central stability fact. With alternating data $e_i=(-1{)}^i$, the absolute row sums satisfyFor general data, the cardinal splines are ${\ell }_i(x)={\sum }_j{N}_j(x)(A^{-1}{)}_{ji}$. The spline interpolation operator and its exact Lebesgue constant of interpolation areThe equality for the operator norm follows by extending the signs at distinct sites to a continuous unit-norm data function, as in Solution 2. That solution also shows the last inequality can be very strict. Stability concerns the range function as well as its coefficients.

An optimal interpolation set must be specified relative to an objective. For the normalized B-spline coefficients, every admissible sampling set satisfiessince $c(s)=A_{\mathbf{x}}^{-1}(s(x_i))$ and sampling cannot increase the supremum norm. If a unit-norm spline $s_{\ast }$ has $n$ ordered alternating extrema $s_{\ast }(x_i)=(-1{)}^i$ at admissible sites, the checkerboard identity givesTaking the largest row shows $\Vert A_{\mathbf{x}}^{-1}{\Vert }_{\infty }\le \kappa (\mathcal{S})$, so equality holds. This proves the optimal spline coefficient interpolation sites property and the Chebyshev spline coefficient and dual norm equality. In particular, for the cubic space of Solution 1 the alternating extrema $-1,-2/3,0,2/3,1$ satisfy the support conditions. Its coefficients have maximum magnitude $11/2$, so these sites are optimal for coefficient recovery andThe argument proves optimality whenever this admissible equioscillating spline is available; it does not identify extrema of an arbitrary spline as optimal sites, or transfer coefficient optimality automatically to the Lebesgue constant of interpolation.

For the intrinsic interpolation objective, minimize ${\Lambda }_{\mathbf{x}}$ itself. There is a minimizer for a fixed finite-dimensional continuous spline space. Indeed the sites lie in a compact ordered simplex. On admissible sets, inverse entries and hence ${\Lambda }_{\mathbf{x}}$ vary continuously. As a collocation matrix tends to a singular one, its inverse norm diverges. The uniform-norm stability of a B-spline basis gives the complementary boundchoose sample signs attaining a largest inverse row, extend them to a continuous unit-norm function, and use $\Vert Tc{\Vert }_{\infty }\ge \Vert c{\Vert }_{\infty }/\kappa (\mathcal{S})$. Thus a minimizing sequence with bounded Lebesgue constants of interpolation cannot approach a singular site set. A convergent subsequence has an admissible limit attaining the minimum.

Fekete interpolation sites for a continuous function space offer another useful, constructive criterion: maximize the absolute evaluation determinant. A nonzero maximum exists by compactness and linear independence. Replacing one row by evaluation at $x$ multiplies the determinant by the corresponding cardinal function, so maximality gives $|{\ell }_i(x)|\le 1$, and therefore ${\Lambda }_{\mathbf{x}}\le n$. These sites are determinant-optimal, with a proven interpolation bound; one should not claim without further argument that they minimize ${\Lambda }_{\mathbf{x}}$. Greville abscissae provide inexpensive admissible sites for the usual continuous clamped spline spaces of order at least two. Admissibility alone does not give the best stability bound.

Finally, the coefficient condition number of a normalized B-spline basis is bounded by a constant depending only on order. To see the mechanism, for each coefficient choose a longest nonempty knot cell in that B-spline's support. If its length is $h$ and the support width is $W$, then $W/h\le k$. On that cell the spline is a degree-at-most-$k-1$ polynomial; repeated Markov polynomial inequalities bound its $r$th derivative by $C_k{h}^{-r}\Vert s{\Vert }_{\infty }$. In the De Boor–Fix spline coefficient functional, the accompanying derivative of the knot polynomial is bounded by $C_k{W}^r$. Each term is consequently at most $C_k(W/h{)}^r\Vert s{\Vert }_{\infty }$, yielding $\Vert c(s){\Vert }_{\infty }\le C_k\Vert s{\Vert }_{\infty }$ independently of the number and spacing of knots. This is basis stability; it does not bound $\Vert A_{\mathbf{x}}^{-1}\Vert$ at arbitrarily poor sampling sites.

For approximation, reproduction gives the Lebesgue inequality for approximation projectorsAs the maximum knot gap tends to zero with fixed order, spline quasi-interpolation gives an approximant with error at most $\omega (f,kh)$, so the infimum tends to zero. A sequence of uniformly bounded interpolation linear operators therefore converges to every continuous target. Local support and a stable choice of sites make spline interpolation effective; mesh refinement without control of the interpolation operator norm is not by itself a convergence proof.

First take $k\ge 2$, so the simple-knot B-splines are continuous, vanish at both support endpoints, and are strictly positive inside their supports. Thus the diagonal condition is $t_i<x_i<t_{i+k}$. We prove both directions of the Schoenberg–Whitney theorem, deriving the weighted zero-count fact needed for sufficiency rather than relying on an ambiguous coefficient-free hint.

For necessity, suppose $N_i(x_i)=0$. If $x_i\le t_i$, then for every row $r\le i$ and column $j\ge i$ we have $x_r\le x_i\le t_i\le t_j$, hence $N_j(x_r)=0$. The first $i$ rows have nonzero entries in at most $i-1$ columns and are linearly dependent. If instead $x_i\ge t_{i+k}$, then for $r\ge i$ and $j\le i$ we have $x_r\ge x_i\ge t_{i+k}\ge t_{j+k}$, so those entries vanish. The last $n-i+1$ rows have nonzero entries in at most $n-i$ columns and are again dependent. In either case the B-spline collocation matrix is singular. Thus invertibility implies the diagonal positivity condition.

We establish two elementary spline facts for the converse. For local linear independence of B-splines, extend the strictly increasing knots in both directions. On a cell $(t_j,t_{j+1})$ the $k$ active basis functions have indices $j-k+1,\dots ,j$. Their knot polynomials are ${\psi }_i(y)={\prod }_{\ell =1}^{k-1}(y-t_{i+\ell })$. The Cox-de Boor recurrence proves Marsden's identityHere is the induction behind that identity. For order one the indicator functions sum to one. On passing from order $k-1$ to $k$, collect the coefficient of each lower-order basis function. With $u=t_i$, $v=t_{i+k-1}$, its extra factor isThus the identity at one order multiplies by $y-x$ at the next.

The $k$ polynomials ${\psi }_i$ are independent: evaluate successively at $y=t_j,t_{j-1},\dots ,t_{j-k+1}$. The first evaluation isolates ${\psi }_j$; after its coefficient has been eliminated the next isolates ${\psi }_{j-1}$, and so on, with nonzero diagonal factors because the knots are distinct. Comparing powers of $y$ in the identity then shows that the active $N_i$ span all polynomials of degree at most $k-1$ on the cell. There are $k$ of them, so they are independent. In particular, a spline vanishes on a whole open cell exactly when all its active coefficients vanish. Padding the finite coefficient vector with zeros makes this fact apply also near the outer knots.

Next let $s={\sum }_{i=p}^q{c}_i{N}_i$ be a real spline on $I=(t_p,t_{q+k})$, with no identically zero knot cell there. The simple-knot B-spline regularity gives $s\in C^{k-2}$, and its derivatives through order $k-2$ vanish at both endpoints because it is supported on the closed interval. Suppose it has $Z$ distinct interior zeros. Its zero set initially has $Z+2$ connected components, counting the endpoints; the zeros are isolated because no cell vanishes identically.

A useful form of Rolle's theorem counts zero components, including any zero intervals that arise after differentiation. Between two successive zero components a real differentiable function is nonzero and has an interior extremum, supplying a zero of its derivative. These derivative-zero components are distinct: a zero interval connecting two such extrema would make the original function constant across an intervening zero and nonzero gap, which is impossible. If its derivative also vanishes at both endpoints, those supply two further distinct components. Thus differentiation increases the zero-component count by at least one. Apply this successively through derivative order $k-2$; every required endpoint derivative vanishes. No derivative in this process can vanish identically: that would make $s$ a global polynomial of degree below $k-1$, with all its endpoint derivatives through that degree zero, forcing $s=0$. The final derivative has at least $Z+k$ zero components.

That final derivative is continuous and piecewise affine on $q-p+k$ knot cells. Each closed affine cell can meet at most one zero component: two different zeros would force the whole cell to vanish, joining them into one component. Every zero component meets a cell, so there are at most $q-p+k$ of them. For $k=2$ this argument applies directly without differentiation. We have proved the compact-support spline zero countThis proof also accommodates zero intervals of the derivatives; counting them as infinitely many distinct zeros would invalidate a naive repeated-Rolle argument.

Now assume $N_i(x_i)>0$ for all $i$ and suppose $A_{x}c=0$ for a nonzero coefficient vector. Since $A_x$ is real, a nonzero real null vector can be chosen. Set $s={\sum }_i{c}_i{N}_i$, so $s(x_i)=0$ for all $i$. Consider one connected component of the union of the open supports of the terms with $c_i\ne 0$. If its first and last indices are $p,q$, the component is exactly $I=(t_p,t_{q+k})$. On this interval $s$ equals $s_0={\sum }_{i=p}^q{c}_i{N}_i$. Each knot cell in $I$ has at least one active nonzero coefficient, so local independence ensures $s_0$ does not vanish on any whole cell.

For every $p\le i\le q$, diagonal positivity gives $t_i<x_i<t_{i+k}$, and this places $x_i$ inside $I$. Thus $s_0$ has the $q-p+1$ distinct zeros $x_p,\dots ,x_q$, contradicting the upper bound $q-p$. There is no nonzero null vector, proving sufficiency. Together the two directions giveFor completeness, order $k=1$ consists of disjoint interval indicators, with the usual left-closed, right-open convention. An invertible ordered collocation requires exactly one site in each interval; ordering makes this equivalent to $N_i(x_i)>0$, and the matrix is then the identity. This handles possible sites at left knots without incorrectly replacing positivity by a strict interior inequality for order one.

There are $n$ basis splines for the stated $n+k$ knots; the upper limit $k$ in the printed basis list is an indexing slip. LetThe strictly increasing sites and $t_i<x_i^{\ast }<t_{i+k}$ give invertibility by the Schoenberg–Whitney theorem. The needed additional matrix property is total nonnegativity of B-spline collocation matrices: every ordered minor of $A$ is nonnegative.

Here is a structural justification of that sign property. Knot insertion refines spline coefficients by rules of the form $b_j={\alpha }_j{a}_j+(1-{\alpha }_j)a_{j-1}$, $0\le {\alpha }_j\le 1$, with the usual unchanged coefficients before the insertion and shifted coefficients after it. Each coefficient map is a nonnegative rectangular bidiagonal matrix; all its ordered minors are nonnegative, as its nonzero terms preserve row/column order. Products preserve this property by the Cauchy–Binet formula. Insert each ordered sampling site until it has full knot multiplicity. Evaluation at such a break is one of the refined coefficients, with a consistent one-sided convention. Consequently the B-spline collocation matrix is an ordered row submatrix of the product of refinement matrices, proving its nonnegative minors. This also applies to repeated admissible original knots.

Since $A$ is invertible and totally nonnegative, $\det A>0$. The adjugate identity gives the checkerboard inverse of a totally nonnegative matrix:Now for any spline $s={\sum }_j{a}_j{N}_j$, its sampled-value vector is $y=Aa$. Hence the coefficient linear functional in the dual basis is exactlyIn particular ${\mu }_i(N_{\ell })={\delta }_{i\ell }$, verifying the hinted formula rather than assuming it. The operator norm of this linear functional is taken for the uniform norm on the spline's interval, containing all sampling sites. Evaluation at a sampling site has norm at most one, soFor the given unit-norm alternating spline, $s_{\ast }(x_j^{\ast })=(-1{)}^j$, and the checkerboard signs make all terms in a fixed inverse row align:Since $\Vert s_{\ast }{\Vert }_{\infty }=1$, this supplies the reverse operator norm inequality and proves the Chebyshev spline coefficient and dual norm equalityThis spline attains the operator norm of each coefficient linear functional.

This addresses the general interpolation bound in part (I); the subsequent (a)–(c) address the cubic case in part (II). Write the interpolant in the B-spline basis asThe stipulated uniquely defined spline interpolation operator presupposes invertibility of the B-spline collocation matrix. For the standard increasing distinct sites, the conditions $N_i(x_i)>0$ give it by the Schoenberg–Whitney theorem. Thus $c=A_{\mathbf{x}}^{-1}Rf$.

Use the standard partition-normalized B-splines of the question: they are nonnegative and satisfy the subpartition of unity for B-splines, ${\sum }_j{N}_j(t)\le 1$. This also holds outside the basic knot interval; one may extend the knots and regard the finite family as a subfamily of a full partition of unity. The recurrence proof of the full partition is given in Question 6(c). Repeated knots retain the same nonnegativity and subpartition properties, with zero-denominator recurrence terms interpreted as zero or by knot limits. ConsequentlySampling has norm at most one, $\Vert Rf{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$, and soHere the induced matrix norm is the maximum absolute row sum. The label $L^{\infty }$-normalized means the usual partition normalization, not rescaling each spline to have maximum value one. Such a rescaling would invalidate the specific coefficient bound and the cardinal values used below.

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.

For the usual spline interpolation problem the sites are distinct and ordered; the Schoenberg–Whitney theorem then makes $A_x$ invertible under the stated support positivity. More generally, the following argument applies whenever the interpolation map in the question is well defined and $A_x$ is invertible.

Factor the spline interpolation operator as $P_x=B{A}_x^{-1}R$, whereSampling has operator norm at most one from the supremum norm to ${\ell }_{\infty }$. Nonnegativity and the subpartition of unity for B-splines giveConsequently $\Vert P_x\Vert \le \Vert A_x^{-1}{\Vert }_{{\ell }_{\infty }}$.

For the reverse bound, select a row $r$ of $A_x^{-1}$ with maximal absolute row sum, and choose $y_j=\mathrm{sgn}((A_x^{-1}{)}_{rj})$, assigning any value of modulus at most one when the entry is zero. Then $\Vert y{\Vert }_{{\ell }_{\infty }}=1$ andAt distinct sites there is a continuous piecewise linear function $f$ with $f(x_j)=y_j$ and $\Vert f{\Vert }_{\infty }=1$: interpolate linearly between the ordered sites and extend constantly to the endpoints. The given uniform-norm stability of a B-spline basis therefore impliesThus the B-spline interpolation operator norm satisfiesPositivity $N_i(x_i)>0$ alone, without the usual distinct-site convention, is insufficient to define an interpolation operator. For example, take order-two B-splines with knots $0,1,2,3$ and $x_1=x_2=3/2$. Both diagonal values are $1/2$, but the B-spline collocation matrix has identical rows $(1/2,1/2)$ and is singular. The displayed proof uses the well-defined interpolation map posited in the question; ordered distinct sites supply its standard existence guarantee.

Write $Ba={\sum }_{j=1}^n{a}_j{N}_j$, and let $Rf=(f(x_1),\dots ,f(x_n){)}^T$. We use the standard interpolation convention of distinct ordered sites $x_1<\cdots <x_n$. Together with $N_i(x_i)>0$, the Schoenberg–Whitney theorem makes the B-spline collocation matrix $A$ invertible. Equivalently, its invertibility is implicit in the existence of the interpolation operator for every data vector.

If distinctness is not understood, the printed positivity condition alone is insufficient. For the order-two basis on spline knots $1,2,3,4$, take $x_1=x_2=5/2$. Both diagonal B-spline values are $1/2>0$, but the two rows of $A$ coincide. Arbitrary data cannot then be interpolated uniquely. The norm statement below concerns the intended well-defined interpolation operator.

Spline interpolation determines the coefficient vector by $Aa=Rf$, and henceNonnegativity and the subpartition of unity for B-splines giveFor completeness, extend the finite spline knot sequence beyond both ends. For the full sequence the order-one interval indicators sum to one. Summing the Cox-de Boor recurrence and shifting the index in its second term combines the two coefficients of each lower-order B-spline to one, so induction gives partition of unity at every order. Our finite collection is a subset of that nonnegative collection, and thus has sum at most one. Repeated spline knots are handled by the standard zero-term convention or a knot limit. This argument controls the entire interval, not only the basic knot interval.

Since $\Vert Rf{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$, the operator norm upper bound isHere the matrix operator norm is the maximum absolute row sum. For the lower bound, choose a row of $A^{-1}$ with maximum absolute row sum and a data vector $y$ whose components are the signs of that row's entries. Then $\Vert y{\Vert }_{{\ell }^{\infty }}=1$ andA continuous piecewise-linear function taking these values at the distinct ordered sites, and constant outside their range, has supremum norm one. Apply the given uniform-norm stability of a B-spline basis to its coefficient vector:ThereforeThe construction of the bounded continuous data extension is what permits the matrix norm to give a lower bound for a function-space operator norm.

A B-spline $N_j$ is positive exactly in the interior of its support $(t_j,t_{j+k})$. If the B-spline collocation matrix $A_{\mathbf{x}}$ is invertible, its determinant contains a nonzero permutation term. Hence there is a permutation $\pi$ such that

If $x_i\le t_i$, then every one of the first $i$ points satisfies $x_r<t_i$. Any support containing such a point must have left endpoint $t_j<x_r$, hence $j<i$. Only $i-1$ B-splines are available to match these $i$ rows, contradicting that $\pi$ is a permutation. Thus $x_i>t_i$. Similarly, if $x_i\ge t_{i+k}$, each of the last $n-i+1$ points can only be matched to an index $j>i$, but only $n-i$ such indices exist. Therefore $x_i<t_{i+k}$. We have proved

The Schoenberg–Whitney theorem states, for strictly increasing knots and interpolation sites, thatIndeed its determinant is positive under these inequalities.

Spline interpolation matches prescribed function values at distinct sites by an element of a finite-dimensional spline space. In a B-spline basis its existence and uniqueness for all data are equivalent to invertibility of the B-spline collocation matrix. Distinct ordered sites satisfying the Schoenberg–Whitney theorem give this condition.

For increasing sites and an ordered nonnegative normalized B-spline basis, the matrix $A_{ij}=N_j(x_i)$ is a totally nonnegative matrix. One way to see the minors' signs is knot insertion. Each inserted knot changes coefficients by $b_j={\alpha }_j{a}_j+(1-{\alpha }_j)a_{j-1}$ with $0\le {\alpha }_j\le 1$, using the standard constant endpoint pieces. This is a nonnegative rectangular bidiagonal map, whose minors are nonnegative. Products preserve that property by the Cauchy–Binet formula. Insert each site to full knot multiplicity; a suitable refined coefficient is then the value $s(x_i)$. Thus collocation is an ordered row submatrix of the refinement map, and has nonnegative minors. The strict support conditions $t_i<x_i<t_{i+k}$ give invertibility by the Schoenberg–Whitney theorem; its inverse has the checkerboard inverse of a totally nonnegative matrix sign pattern.