Sectional forward Euler converges globally with first-order error

Use the finite event polynomials to extend $F$ to all real augmented states, and use the Euler map ${\mathrm{\Psi }}_h(v)=v+hF(v)$ there. This map agrees with every successful algorithm step; the error analysis may evaluate it at other states without imposing the algorithm's acceptance test. Let $B_M=\{(n,L):0\le n_i\le M\mathrm{/}{x}_i,0\le L\le M\}$, a compact box since the grid is finite and each $x_i>0$. The exact path lies in $B_M$ by [F1]; starting from $y^0=y(0)$, repeated application of [F3] puts every numerical state in the nonnegative augmented-mass simplex of mass $M$, which is contained in $B_M$. Set $R_B={\max }_{z\in B_M}\mathrm{\parallel }z{\mathrm{\parallel }}_2$ and $B_F={\max }_{z\in B_M}\mathrm{\parallel }F(z){\mathrm{\parallel }}_2$; these are finite because $F$ is polynomial. Since $h=T\mathrm{/}N\le T$, every exact-start Euler output obeys $\mathrm{\parallel }y(t_k)+hF(y(t_k)){\mathrm{\parallel }}_2\le R_B+T{B}_F$. Thus the exact path, all successful numerical states and Euler stage states, and the exact-start Euler inputs and outputs all lie in the compact tube $K=[0,T]\times \bar{B}(0,R_B+T{B}_F+1)$. This tube lies with positive margin in the bounded open cylinder $V=(-1,T+1)\times B(0,R_B+T{B}_F+2)$, whose state fibers are convex. The time-independent finite polynomial field $f(t,z)=F(z)$ and its first derivatives are bounded on $V$. Hence [F4] applies uniformly to every local defect, including the exact-start Euler output, and gives defect $C{h}^2$; its derivative bound also gives the one-step factor $1+Lh$ for inputs in $K$.

Apply [F5] with initial error zero, the fixed mesh $h=T\mathrm{/}N$, and the common compact tube from step 1.1. It gives ${\max }_{k\le N}\mathrm{\parallel }{y}^k-y(t_k){\mathrm{\parallel }}_2\le C_{T}h$, as claimed.