Collocation polynomial derivatives scale with the step size

For each real component polynomial $p$ of degree at most $M-1$, $p$ and $I_{M}p$ have the same $M$ nodal data and degree bound, so uniqueness in [F2] gives $I_{M}p=p$; all reproduction uses below are componentwise. The stage equations in [F1] and boundedness of $f$ on the tube first give $U_i-u=O(h)$. Since $D$ is open and the finitely many partial derivatives used in [F3] are continuous at $(t,u)$, choose a convex open ball about $(t,u)$ contained in $D$ and shrink it so those derivatives are bounded there. For sufficiently small $h$, every stage point $(t+c_{j}h,U_j)$ and its segment to $(t,u)$ lie in this ball, so [F3] applies with uniform Taylor remainders. The exact path satisfies $y^{\mathrm{\prime }}=f(t,y)$ by [F5]. We prove by induction on $s=1,\dots ,M$ that there are vector polynomials $P_s$ of degree at most $s$ such that $U_i=u+{\sum }_{r=1}^s{h}^r{P}_r(c_i)+O(h^{s+1})$ uniformly over the finite node set. For the induction step, substitute the expansion through order $s$ into each scalar coordinate of $f(t+h{c}_j,U_j)$ and apply the scalar Taylor supplier [F3] at $(t,u)$. Each coefficient of $h^m$ is a polynomial in $c_j$ of degree at most $m$, because every time increment contributes one factor $c_{j}h$ and each state coefficient at order $r$ has degree at most $r$. For $m\le M-1$, [F2] makes interpolation exact on that coefficient; the row identity $Q_{ij}={\int }_0^{c_i}{\mathrm{\ell }}_j$ then integrates it to a polynomial in $c_i$ of degree at most $m+1$. Multiplication by the outer $h$ gives the next $P_{m+1}$. Taylor remainders are uniform on the chosen ball for sufficiently small $h$, and the stated local branch supplies the stage solution.

The coefficient construction in step 1.1 and its Taylor expansion give vector polynomials $R_m$ of degree at most $m$ with $f(t+h{c}_j,U_j)={\sum }_{m=0}^{M-1}{h}^m{R}_m(c_j)+{\delta }_j(h)$, where ${\max }_j\mathrm{\parallel }{\delta }_j(h)\mathrm{\parallel }=O(h^M)$. Apply componentwise reproduction from [F2] and define $e_h(\tau )={\sum }_j{\mathrm{\ell }}_j(\tau ){\delta }_j(h)$. Then $I_M{g}_h(\tau )={\sum }_{m=0}^{M-1}{h}^m{R}_m(\tau )+e_h(\tau )$ and every derivative of $e_h$ through order $M-1$ is $O(h^M)$, because its basis is fixed. Substituting into the defining integral in [F1], rather than identifying a degree-$M$ polynomial from only its $M$ stage values, gives $u_h(\tau )=u+{\sum }_{s=1}^M{h}^s{P}_s(\tau )+R_h(\tau )$ with $P_s(\tau )={\int }_0^{\tau }{R}_{s-1}(a)da$ and $R_h(\tau )=h{\int }_0^{\tau }{e}_h(a)da$. These are the same integrated coefficients used in step 1.1 through the row integrals $Q_{ij}$. Thus $\mathrm{deg}{P}_s\le s$, $P_s(0)=0$, $\mathrm{\parallel }{R}_h{\mathrm{\parallel }}_{\mathrm{\infty }}=O(h^{M+1})$, and for $1\le j\le M$, $R_h^{(j)}=h{e}_h^{(j-1)}=O(h^{M+1})$. Differentiating the expansion annihilates every term with $s<j$ and bounds the remaining terms by $O(h^j)$. Hence $\mathrm{\parallel }{u}_h^{(j)}{\mathrm{\parallel }}_{\mathrm{\infty }}=O(h^j)$ for $1\le j\le M$; derivatives above $M$ vanish by the defining integrated cardinal sum. This argument also covers $M=1$ and a node at zero.

Repeated chain and product rules from [F3] express each coordinate of $g_h^{(j)}$ as a finite sum of derivatives of that coordinate of $f$ applied to derivatives of $(t+h\tau ,u_h(\tau ))$ whose orders add to $j$. Each argument derivative of order $s\le M$ is $O(h^s)$ by step 2.1; higher derivatives vanish for the polynomial state and affine time coordinate. The bounded derivatives in [F4] therefore give $\mathrm{\parallel }{g}_h^{(j)}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_j{h}^j$ through $j=2M$.

For $j<M$, subtract from each coordinate of $g_h$ its Taylor polynomial of degree $M-1$ at $0$. Its remainder and its first $j$ derivatives are $O(h^M)$ by step 3.1 and the one-variable Taylor remainder in [F3]. Interpolation reproduces that Taylor polynomial by [F2]; derivatives of the interpolated remainder are bounded by the fixed constants ${\max }_{\tau }{\sum }_i\mathrm{\mid }{\mathrm{\ell }}_i^{(j)}(\tau )\mathrm{\mid }$, so $(I_M{g}_h-g_h{)}^{(j)}=O(h^M)$ coordinatewise. Multiplying by $h$ gives $r_h^{(j)}=O(h^{M+1})\le O(h^{j+1})$ for $j<M$ and $h\le 1$. For $M\le j\le 2M$, $I_M{g}_h$ has degree at most $M-1$, so its $j$th derivative vanishes and $r_h^{(j)}=-h{g}_h^{(j)}=O(h^{j+1})$ by step 3.1.