Sobolev approximation by local spline quasi-interpolants

On any neighborhood $J=[c,d]$, use the continuous representatives of $D^{j}f$, $0\le j\le l$, supplied by [F2]. Their Sobolev-class convention [F1] and the choice hypotheses for [F2] are available under [A1] by [F5]. Repeated integration by parts for absolutely continuous functions gives the Taylor identity $f(x)-T_{c,l}f(x)=\frac{1}{l!}{\int }_c^x(x-t{)}^l{D}^{l+1}f(t)dt,$ where $T_{c,l}f(x)={\sum }_{j=0}^l{D}^{j}f(c)(x-c{)}^j\mathrm{/}j!$. For completeness, the case $l=0$ is the increment formula in [F2]; passing from order $l-1$ to $l$ integrates ${\int }_c^x(x-t{)}^{l-1}{D}^{l}f(t)dt$ by parts with antiderivative $-(x-t{)}^l\mathrm{/}l$, producing the degree-$l$ endpoint term and the displayed remainder. For finite $q$, Hölder [F3] bounds the pointwise remainder by $\mathrm{\mid }J{\mathrm{\mid }}^{l+1\mathrm{/}{q}^{\mathrm{\prime }}}\mathrm{/}l!\mathrm{\parallel }{D}^{l+1}f{\mathrm{\parallel }}_{L^q(J)}$, where $1\mathrm{/}{q}^{\mathrm{\prime }}=1-1\mathrm{/}q$ and $1\mathrm{/}{q}^{\mathrm{\prime }}=0$ for $q=1$; multiplying by $\mathrm{\mid }J{\mathrm{\mid }}^{1\mathrm{/}q}$ gives $\mathrm{\parallel }f-T_{c,l}f{\mathrm{\parallel }}_{L^q(J)}\le \mathrm{\mid }J{\mathrm{\mid }}^{l+1}\mathrm{\parallel }{D}^{l+1}f{\mathrm{\parallel }}_{L^q(J)}\mathrm{/}l!$. For $q=\mathrm{\infty }$, integrating the kernel in the remainder gives the same bound.

Choose $c$ as the left endpoint of $J_m$ and write $T=T_{c,l}f$. Since $Q$ reproduces ${\mathcal{P}}_l$, $QT=T$. Because $I_m\subseteq J_m$, the triangle inequality and local stability give $\mathrm{\parallel }f-Qf{\mathrm{\parallel }}_{L^q(I_m)}\le \mathrm{\parallel }f-T{\mathrm{\parallel }}_{L^q(I_m)}+\mathrm{\parallel }Q(f-T){\mathrm{\parallel }}_{L^q(I_m)}\le (1+C_Q)\mathrm{\parallel }f-T{\mathrm{\parallel }}_{L^q(J_m)}\mathrm{.}$ The definition gives $\mathrm{\mid }{J}_m\mathrm{\mid }\le M{h}_m$; applying step 1.1 proves the stated local estimate.

Raise the local estimates to the $q$th power and sum over the disjoint mesh cells. Each point belongs to at most $M$ neighborhoods, so ${\sum }_m\mathrm{\parallel }{D}^{l+1}f{\mathrm{\parallel }}_{L^q(J_m)}^q\le M\mathrm{\parallel }{D}^{l+1}f{\mathrm{\parallel }}_{L^q([a,b])}^q$. Taking the $q$th root and using $h_m\le h$ gives the stated global bound for finite $q$. For $q=\mathrm{\infty }$, take the maximum of the local bounds; the factor $M^{1\mathrm{/}q}$ is $1$.

For a uniform mesh with interior knots repeated $p+1$ times, $S_{p,\xi }$ is the space of independent degree-$p$ polynomials on the cells by the definition and basis theorem. On a cell $I$, use the real $L^2(I)$ pairing $⟨u,v{⟩}_I={\int }_{I}uv$: it is finite by Hölder [F3], bilinear and symmetric by linearity of the integral, and positive definite since ${\int }_I{u}^2=0$ implies each set $\{\mathrm{\mid }u\mathrm{\mid }\ge 1\mathrm{/}m\}$ has measure zero, hence $u=0$ almost everywhere. Put ${\varphi }_j(x)=x^j$, $0\le j\le p$, and $G_{ij}=⟨{\varphi }_i,{\varphi }_j{⟩}_I$. For any nonzero coefficient vector $c$, $c^{T}Gc={\int }_I({\sum }_j{c}_j{x}^j{)}^2>0$, since a nonzero polynomial has only finitely many roots; hence $G$ is invertible. For $f\in L^2(I)$, put $b_i=⟨f,{\varphi }_i{⟩}_I$ and $c=G^{-1}b$, and define $Q_{h}f={\sum }_{j=0}^p{c}_j{\varphi }_j$ on $I$. The defining equations give $⟨f-Q_{h}f,{\varphi }_i{⟩}_I=0$ for every $i$. Thus $Q_h$ is linear, fixes ${\mathcal{P}}_p$, and has residual orthogonal to its range; Pythagoras [F4] gives $\mathrm{\parallel }{Q}_{h}f{\mathrm{\parallel }}_{L^2(I)}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(I)}$. Assign knot values from the cell on their right and the value at $b$ by the left limit. The cellwise polynomials then form a member of $S_{p,\xi }$. As $J$ varies, these operators form a mesh-uniform local quasi-interpolant family with one-cell neighborhoods, $M=1$, and $C_Q=1$. Applying steps 1.1–3.1 with $l=s-1$ gives $\mathrm{\parallel }f-Q_{h}f{\mathrm{\parallel }}_{L^2}\le 2h^s\mathrm{\parallel }{D}^{s}f{\mathrm{\parallel }}_{L^2}\mathrm{/}(s-1)!$ for $1\le s\le p+1$.