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For the distinct-knot linear B-spline hats, let $h_i=t_{i+1}-t_i>0$. The mixed-normalization spline Gram matrix has entriesOnly indices in the finite basis are retained. This matrix has a row strict diagonal dominance margin at least $1/3$, so the inverse infinity-norm bound from diagonal dominance gives $\Vert G^{-1}{\Vert }_{{\ell }^{\infty }}\le 3$. Equidistant spline knots give off-diagonal entries $1/6$. Both pieces of every hat are retained; repeated endpoint spline knots require different boundary formulas.

For the orthogonal B-spline projector $P$, the normal equations in the mixed-normalization spline Gram matrix give $Ga=r$, with $r_i=(M_i,f)$. Positivity and unit integral of $M_i$ imply $\Vert r{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$. Nonnegativity and subpartition of unity of $N_i$ give $\Vert {\sum }_i{a}_i{N}_i{\Vert }_{\infty }\le \Vert a{\Vert }_{{\ell }^{\infty }}$. Hence $\Vert P{\Vert }_{\infty }\le \Vert G^{-1}{\Vert }_{{\ell }^{\infty }}$, where the matrix norm is maximum absolute row sum. The range is the spline space, which lies in $C[0,1]$ only for continuous splines.

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.

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.

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.

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.

Use the standard partition normalization of B-splines, not rescaling each spline to have supremum exactly one. Let $h_i=t_{i+k}-t_i>0$. Then $M_i=k{N}_i/h_i$. In the ordinary real inner product, the mixed-normalization spline Gram matrix is $G=D^{-1}H$, where $D=\mathrm{diag}(h_i/k)$ and $H_{ij}=(N_i,N_j)$. Since the $N_i$ form a basis, $H$ is positive definite; $D$ is invertible, so $G$ is invertible. The undeclared upper index $h$ in the printed matrix should be $n$.

We need the standard positivity and normalization properties with their justification. The Cox-de Boor recurrence iswith zero-width terms interpreted as zero. It starts from nonnegative interval indicators. On each lower-order spline's support, its recurrence coefficients are nonnegative, so induction gives $N_i\ge 0$. Extend the finite knots outside their endpoints to a full knot sequence. In the full sum, coefficients of each lower-order spline add to one, so the recurrence preserves partition of unity. Our finite collection is a subset of nonnegative B-splines; therefore ${\sum }_{i=1}^n{N}_i(t)\le 1$ everywhere. The Marsden identity gives equality on the basic knot interval.

The divided-difference definition also gives support in $[t_i,t_{i+k}]$: below the left endpoint the knot-data function is a polynomial of degree $k-1$, whose order-$k$ divided difference vanishes; above the right endpoint the knot data vanish. Choose $a<t_i$ and $b>t_{i+k}$. Integrating in $t$ and then taking the divided difference in its knot variable givesbecause an order-$k$ divided difference of a monic degree-$k$ polynomial is one. The support lies inside $[0,1]$, so ${\int }_0^1{M}_i=1$.

Orthogonality of $f-s^{\ast }$ to the spline space gives $Ga=r$, with $r_i=(M_i,f)$. Since $M_i\ge 0$ and has unit integral, $|r_i|\le \Vert f{\Vert }_{\infty }$. The induced matrix maximum norm is the absolute row-sum norm, soTaking the supremum over nonzero $f$ proves the maximum-norm bound for spline projection:The range is $\mathcal{S}$, so the printed phrase “onto $C[0,1]$” must mean into $C[0,1]$, and onto $\mathcal{S}$. Continuity of the range requires continuous B-splines, for example $k\ge 2$ and internal multiplicities at most $k-1$. For order-one or discontinuous B-splines, the same bound is valid with codomain $L^{\infty }[0,1]$ rather than $C[0,1]$.