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Use the standard $L_{\infty }$-oriented partition normalization of B-splines, in contrast to the unit-integral normalization. Thus $N_i\ge 0$ and ${\sum }_i{N}_i(t)\le 1$ on $[0,1]$, with equality for a complete basis on its basic knot interval. This terminology does not mean that each basis function's maximum is exactly one; for example an order-three cardinal spline has maximum $3/4$. Assume the basis has positive support on the integration interval and is continuous, as is needed for a projection taking values in $C[0,1]$.

LetThe ordinary Gram matrix $H$ is positive definite by linear independence. The appropriate row-normalized, or mixed-normalization spline Gram matrix, isHere $M_i$ is a unit-integral nonnegative spline. Unlike $H$, $G$ need not be symmetric. If the full support lies inside the interval, $d_i=(t_{i+k}-t_i)/k$; otherwise use the actual interval integral rather than that full-support formula.

The normal equations for the $L_2$ orthogonal projection $P_{\mathcal{S}}f={\sum }_i{a}_i{N}_i$ say $Ha=c$, with $c_i=\int f{N}_i$. Divide each row by $d_i$ to obtainNonnegativity and unit integral give $|b_i|\le \Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines then givesTaking the supremum over $t$ and then over nonzero $f$ proves the uniform norm bound for B-spline orthogonal projection:The row normalization is essential; replacing $G$ with the ordinary Gram matrix would not be the same asserted estimate. The operator is defined on $C[0,1]$ and maps into its finite-dimensional spline subspace, onto $\mathcal{S}$, not onto all of $C[0,1]$ as one phrase in the PDF says. Its range follows both from the formula and from reproducing every $s\in \mathcal{S}$.

If “$L_{\infty }$-normalized” were instead read literally as individually peak-one splines, their sum can exceed one. The basis-dependent factor $K=\Vert {\sum }_i{N}_i{\Vert }_{\infty }$ must then be retained: $\Vert P_{\mathcal{S}}{\Vert }_{\infty }\le K\Vert (D^{-1}H{)}^{-1}{\Vert }_{{\ell }^{\infty }}$. Equivalently the displayed bound without an outside factor uses $G=K^{-1}{D}^{-1}H$. The standard partition-normalized convention has $K\le 1$ and yields the sharper bound proved above.

Write $N_{i,k}$ for the order-$k$ B-spline. At order one, $N_{i,1}=1_{[t_i,t_{i+1})}$, and the finite sum is one on its basic knot interval. For larger order, the Cox-de Boor recurrence iswith a zero-denominator term defined as zero. Sum over $i=1,\dots ,n$. For each interior lower-order index $j=2,\dots ,n$, the combined coefficient isThe two unmatched lower-order terms have supports $[t_1,t_k]$ and $[t_{n+1},t_{n+k}]$, respectively, and vanish on the open basic interval $(t_k,t_{n+1})$. Induction therefore givesFor simple knots and $k\ge 2$, continuity extends the identity to both endpoints. In the order-one case and at admissible repeated endpoint knots, use the one-sided value from the basic interval. Extending the knot sequence in both directions also proves the subpartition of unity for B-splines, $0\le {\sum }_i{N}_i\le 1$ everywhere: the finite sum is a subset of a nonnegative full partition. That weaker global inequality will be needed in Question 7(a).

The orthogonal projection condition is $(f-s^{\ast },N_i)=0$ for all $i$. Since $M_i=k{N}_i/(t_{i+k}-t_i)$, it is equivalentlyThe mixed-normalization spline Gram matrix is invertible: $G=D^{-1}H$, where $D_{ii}=(t_{i+k}-t_i)/k>0$ and $H_{ij}=(N_i,N_j)$ is the positive-definite ordinary Gram matrix of the basis. Nonnegativity and the unit-integral normalization of a B-spline give $|b_i|\le \Vert f{\Vert }_{\infty }{\int }_0^1{M}_i=\Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines gives, on all of $[0,1]$,ConsequentlysoThis is the maximum-norm bound for spline projection. Using a subpartition matters when the basic knot interval is a proper subinterval of $[0,1]$. As in the paper's assertion that the image lies in $C[0,1]$, the spline space here must consist of continuous functions; order-one step splines or full-multiplicity interior knots do not have that property.

Let $y=(f(x_1),\dots ,f(x_n){)}^T$ and write the interpolating spline as $P_{x}f={\sum }_{j=1}^n{c}_j{N}_j$. The interpolation equations are $A_{x}c=y$, 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 $N_j\ge 0$ and ${\sum }_{j=1}^n{N}_j\le 1$. 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. HenceTaking the supremum over $t$ and then over unit-norm $f$ gives the B-spline interpolation operator norm estimateHere 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.

Use the standard knot-domain convention in which the spline space contains every polynomial of degree at most $k-1$ on $[a,b]$, with completed or clamped endpoint knots. The De Boor–Fix functional extracts a local spline coefficient, and its Hahn-Banach theorem extension is linear on continuous local data with the stated bound. Only that bound and spline reproduction are needed here.

For $x$ in a knot cell $[t_j,t_{j+1}]$, support and Gram bandwidth of B-splines implies that only indicescan contribute, restricted also to $1\le i\le n$. Every corresponding local support $[t_i,t_{i+k}]$ lies insideNonnegativity and subpartition of unity for B-splines therefore giveTaking the supremum over the cell proves the local stability estimate for spline quasi-interpolation:Endpoint limits follow by continuity for $k\ge 2$; the order-one version uses the usual one-sided interval convention.

Let $h={\max }_i|t_{i+1}-t_i|$ and $f\in C^k[a,b]$. Choose any point $y_j$ in the stencil and form its degree-$k-1$ Taylor polynomial $p_j$. The Taylor theorem remainder givesThe polynomial is itself a spline, so $Q{p}_j=p_j$. By linearity and the local stability estimate,Taking the maximum over cells yieldsThe constant depends only on $k$ and the supplied functional bound, not on mesh ratios or the number of knots.

The endpoint convention is essential to the global assertion. The domain can be the basic knot interval, with any needed exterior data extended smoothly, or the full approximation interval can use clamped endpoint knots. If one instead takes a finite, unclamped collection on the entire outer support interval $[t_1,t_{n+k}]$, every order-$k\ge 2$ basis function vanishes at $t_1$, so $Q1(t_1)=0$ and uniform approximation to the constant one there is impossible. This does not affect the local estimate; it explains the usual boundary completion implicit in the global spline approximation statement.

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.

We first derive the recurrence needed for the partition of unity. Denote an order-$r$ partition-normalized spline by $N_{i,r}$. For $r\ge 2$, set $h_t(u)=(u-t{)}_{+}^{r-2}$, so $(u-t)h_t(u)=(u-t{)}_{+}^{r-1}$. The elementary product identities for divided differences areThey follow from the Leibniz rule for divided differences because the divided differences of the linear factor vanish above first order. Use the first identity on $t_i,\dots ,t_{i+r-1}$ and the second on $t_{i+1},\dots ,t_{i+r}$. Subtracting cancels the common middle-block term. The defining divided-difference recurrence for $N_{i,r}$ then givesThis is the Cox-de Boor recurrence, derived here from the specified normalization rather than assumed as a partition theorem.

Extend the distinct knot sequence in both directions to a strictly increasing sequence $(t_i{)}_{i\in \mathbb{Z}}$ tending to $\pm \infty$. For order one, $N_{i,1}=1_{[t_i,t_{i+1})}$, so the full sum equals one. All higher-order splines are supported on $[t_i,t_{i+r}]$, making the following sums locally finite. Reindexing the second term of the recurrence yieldsInduction proves that the full sum is one for every order. The same recurrence gives nonnegativity: each coefficient is nonnegative wherever its associated lower-order spline is nonzero. Removing splines therefore gives the subpartition of unity for B-splines on the whole line.

For $t_k<t<t_{n+1}$, a spline with index $i\le 0$ ends no later than $t_k$, and one with $i\ge n+1$ starts no earlier than $t_{n+1}$. They all vanish, leaving exactly the indices $1,\dots ,n$. For $k\ge 2$ the splines are continuous and vanish at their support endpoints, so the equality extends to the closed basic knot interval:If order one is allowed, the standard half-open indicator convention gives equality on $[t_1,t_{n+1})$; to include its final endpoint literally, assign the left-limit value there. This harmless endpoint convention does not change integrals or $L^{\infty }$ norms. For the continuous splines of order at least two no such convention is necessary.

Use the standard nonnegative partition-normalized B-splines $N_j$; their sum is at most one, and is exactly one for the complete clamped basis on $[0,1]$. The printed question introduces $M_i$ only in its last matrix formula without defining it. The intended normalization is the unit-integral normalization of a B-spline:For supports wholly inside the interval, $c_i=(t_{i+k}-t_i)/k$. The positivity and partition convention, sometimes called $L_{\infty }$ normalization, should not be confused with individually rescaling every spline to have maximum exactly one.

The characterization in part 1(a) gives the normal equations for the orthogonal projection:Dividing each equation by $c_i$ yieldsThis is the mixed-normalization spline Gram matrix. It need not be symmetric, but $G=D^{-1}H$ with $D=\mathrm{diag}(c_i)$ and $H_{ij}=(N_i,N_j{)}_{L_2}$. The ordinary Gram matrix $H$ is positive definite because the splines are linearly independent, so $G$ is invertible and $a=G^{-1}d$.

Since each $M_i$ is nonnegative and has integral one,The induced matrix operator norm is the maximum absolute row sum, soFinally, nonnegativity and the subpartition of unity for B-splines giveTaking the supremum over $t$ and then over $\Vert f{\Vert }_{\infty }\le 1$ proves the maximum-norm bound for spline projection:The formulas also establish linearity and boundedness of the projection on continuous inputs. Its range is exactly the finite-dimensional spline space $\mathcal{S}$, since it fixes every spline in that space. Thus, when the splines are continuous, it is an operator $C[0,1]\to C[0,1]$ with range $\mathcal{S}$; it is not onto all of $C[0,1]$ as the printed wording says. For example, $e^t$ cannot belong to a fixed finite-degree spline space because it is not a polynomial on any nonempty knot interval. The missing $M_i$ normalization and this range correction are needed for a precise reading of the result.

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.

Combine the B-spline interpolation operator norm estimate from part 1 with the inverse operator norm computed in part 2(b):so in particular $\Vert P_x{\Vert }_{L^{\infty }}\le 2$, uniformly in $n$. The subpartition of unity for B-splines is important at the outer boundary: the finite basis may sum to less than one there, but that only strengthens the estimate.

Use $s_j=t_{i+j}$ again and take $k\ge 2$. Set $g(s)=(s-t{)}_{+}^{k-2}$, so $(s-t)g(s)=(s-t{)}_{+}^{k-1}$. LetThe Leibniz rule for divided differences, applied to the linear factor $s-t$, readsMultiply by $s_k-s_0$ and substitute $G$ to get $(t-s_0)H+(s_k-t)K$. Converting $H$ and $K$ to the lower-order normalized B-splines proves the Cox-de Boor recurrence:It starts with $N_{i,1}=1_{[t_i,t_{i+1})}$. The coefficients are nonnegative wherever the corresponding lower-order B-spline is nonzero, so induction proves nonnegativity.

For later use, extend the finite knot sequence to a strictly increasing sequence in both directions. The full order-one sum is one. When the Cox-de Boor recurrence is summed, the two coefficients of each fixed lower-order B-spline add to one. Induction therefore gives a full partition of unity. Removing terms proves the subpartition of unity for B-splines, ${\sum }_{i=1}^n{N}_{i,k}(t)\le 1$. For repeated knots, the same statement follows by a limit of strictly increasing knot sequences, with the usual convention that a recurrence term with zero denominator is zero; the bound holds on each open knot interval and for the standard one-sided endpoint values. The terminology “$L_{\infty }$-normalized” refers to this standard basis normalization; it does not claim that each individual B-spline has supremum norm exactly one.

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.

The divided difference is a finite linear combination of spline knot values, so it commutes with the finite integral. For every spline knot $t_j$ in the defining interval,It follows thatbecause the order-$k$ divided difference of a monic degree-$k$ polynomial is its leading coefficient. This proves the unit-integral normalization of a B-spline.

For the partition of unity, let $N_{i,r}$ denote the partition-normalized order-$r$ B-spline. The Cox-de Boor recurrence isIt follows by substituting the explicit divided difference formula into both sides and collecting each truncated power function. The starting case is $N_{i,1}=1_{[t_i,t_{i+1})}$, with endpoint conventions irrelevant when $r\ge 2$. The two weights are nonnegative wherever their respective lower-order B-splines are nonzero. Induction therefore gives $N_{i,r}\ge 0$ and strict positivity in its open support.

Extend the finite spline knot sequence strictly in both directions, with no finite accumulation. At each point only finitely many B-splines are nonzero. For the full extended family the sum is one for order one. Summing the Cox-de Boor recurrence, the combined coefficient of $N_{j,r-1}$ isInduction gives a full partition of unity at every order. On $t_k<t<t_{n+1}$, all extended order-$k$ functions with index $i\le 0$ or $i\ge n+1$ vanish by their supports. HenceThe same argument gives the useful global subpartition of unity for B-splines, $0\le {\sum }_{i=1}^n{N}_i(t)\le 1$, including outside the basic spline knot interval.

The orthogonal projection has range $\mathcal{S}$, not all of $C[0,1]$: the latter is an infinite-dimensional vector space whereas $\mathcal{S}$ is a finite-dimensional vector space. For a concrete counterexample, $e^x$ cannot belong to any finite-degree piecewise polynomial space, since all its derivatives are nonzero on every open interval. For $k\ge 2$, the B-splines are continuous, so the orthogonal projection is a map $C[0,1]\to \mathcal{S}\subset C[0,1]$. For order one it instead has a generally discontinuous piecewise constant range. For example, a single interval indicator on $[1/4,3/4)$ is the projection of the constant function one onto its span, and is not continuous.

Write $s^{\ast }={\sum }_j{a}_j{N}_j$. Since $M_i$ is a positive scalar multiple of $N_i$, the operator normal equations $⟨f-s^{\ast },N_i⟩=0$ are equivalent toThe mixed-normalization spline Gram matrix is invertible: $G=DH$ where $D_{ii}=k/(t_{i+k}-t_i)>0$ and $H_{ij}=⟨N_i,N_j⟩$ is the ordinary positive-definite matrix of inner products of a linearly independent basis. The positivity and unit-integral normalization of a B-spline give $|b_i|\le \Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines givesHere the operator norm of a matrix on ${\ell }^{\infty }$ is its maximum absolute row sum. Consequently,This proves the maximum-norm bound for spline projection.

Put $r=k-1$. The linear-reproduction request requires $k\ge 2$, which we use for that final step. Since ${\psi }_i={\omega }_i/r!$, the useful normalized Marsden dual functional isThere is no further factor $1/r!$ outside this sum: it has already been included in ${\psi }_i$.

To derive the formula directly from Marsden's identity, Taylor's theorem for the polynomial $p$ around the arbitrary point $x$ givesOn the other hand,Differentiate Marsden's identity $r-j$ times in $x$, multiply by $(-1{)}^j{p}^{(j)}(x)/r!$, and sum over $j$. The left side becomes $p(t)$, while the right side becomesOnly finite sums and derivatives of polynomials are involved.

Independence of the auxiliary point does not require an assumption about uniqueness of an expansion. Differentiate the formula for ${\lambda }_i$ itself:The first summand at $j=r$ and the second at $j=0$ vanish because both polynomials have degree at most $r$. All remaining terms cancel after shifting the index by one. Therefore ${\lambda }_i(p)$ is a constant in $x$, and it is visibly a linear functional of $p$.

For $p(t)=a+bt$, only $j=0,1$ remain. The leading two coefficients of the monic knot polynomial giveIt follows thatSubstituting into the expansion proveson the basic knot interval. Thus the Greville abscissae are exactly the sampling coefficients that reproduce linear polynomials; the constant case also gives the subpartition of unity for B-splines with equality on this interval.

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 spline quasi-interpolant synthesizes a B-spline expansion from local bounded linear functionals, instead of solving a global interpolation system. Suppose each functional is bounded by $c_k\Vert f{\Vert }_{C[t_i,t_{i+k}]}$ and the operator reproduces every spline. On a knot cell $[t_j,t_{j+1}]$, only indices $j+1-k\le i\le j$ contribute. Their supports lie in $[t_{j+1-k},t_{j+k}]$, and subpartition of unity for B-splines proves local operator norm at most $c_k$. Reproduction of degree-$k-1$ polynomials and a local Taylor theorem remainder then give error at most $(1+c_k)(2k-1{)}^k{h}^k\Vert f^{(k)}{\Vert }_{\infty }/k!$, on the basic knot domain with the usual completed boundary basis.