Uniform-mesh linear spline approximation

Use the disjoint cells $I_j=[t_j,t_{j+1})$, $0\le j<J$, with the final cell also containing $1$, and put $J_j=I_j$. For any continuous input $g$ and $x\in [t_j,t_{j+1}]$, $I_{h}g(x)=\frac{t_{j+1}-x}{h}g(t_j)+\frac{x-t_j}{h}g(t_{j+1})$. The nonnegative weights sum to one. Continuity makes the endpoint values limits from within $I_j$ and gives $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(I_j)}={\max }_{[t_j,t_{j+1}]}\mathrm{\mid }g\mathrm{\mid }$, so $\mathrm{\parallel }{I}_{h}g{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(I_j)}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(J_j)}$. Continuous inputs agreeing almost everywhere on $J_j$ agree on its closure and hence have the same interpolant on $I_j$. The map is real-linear, reproduces every affine function, and is a continuous piecewise-linear spline by [F1]. Each neighborhood is one cell and each point belongs to exactly one neighborhood. Thus, as $J$ varies, the maps $I_h$ form a mesh-uniform family as in [F2] with $p=l=1$, $q=\mathrm{\infty }$, $M=1$, and $C_Q=1$.

Let $M_2=\mathrm{\parallel }{D}^{2}f{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(0,1)}$. On a cell $[u,v]$ of length $h$, put $a_x=x-u$ and $b_x=v-x$. By [F3] applied to $f$ and $Df$, let $f^{\ast }$ and $g^{\ast }$ be their absolutely continuous representatives; $(g^{\ast }{)}^{\mathrm{\prime }}=D^{2}f$ almost everywhere and $f^{\ast }(x)-f^{\ast }(u)={\int }_u^x{g}^{\ast }(s)ds$. The absolutely continuous function $t\mapsto (x-t)g^{\ast }(t)$ has derivative $-g^{\ast }(t)+(x-t)D^{2}f(t)$ almost everywhere, so the fundamental theorem for absolutely continuous functions gives $f^{\ast }(x)=f^{\ast }(u)+a_x{g}^{\ast }(u)+{\int }_u^x(x-t)D^{2}f(t)dt$, and the same formula at $v$. Since $I_{h}f(x)=f^{\ast }(u)+(a_x\mathrm{/}h)(f^{\ast }(v)-f^{\ast }(u))$, subtraction and simplification yield $f^{\ast }(x)-I_{h}f(x)=-{\int }_u^x\frac{b_x(t-u)}{h}{D}^{2}f(t)dt-{\int }_x^v\frac{a_x(v-t)}{h}{D}^{2}f(t)dt$. Hence $\mathrm{\mid }{f}^{\ast }(x)-I_{h}f(x)\mathrm{\mid }\le M_2\{b_x{a}_x^2\mathrm{/}(2h)+a_x{b}_x^2\mathrm{/}(2h)\}=M_2{a}_x{b}_x\mathrm{/}2\le M_2{h}^2\mathrm{/}8$, since $a_x+b_x=h$ and $a_x{b}_x\le h^2\mathrm{/}4$. This holds on every cell and at every point, so taking the supremum gives $\mathrm{\parallel }f-I_{h}f{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(0,1)}\le h^2\mathrm{\parallel }{D}^{2}f{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(0,1)}\mathrm{/}8$.