The optimal modified Hamiltonian differs from the original by order two

Summable correction. Put $b=C_{0}M\mathrm{/}\rho$ and $u_j=(bhj{)}^{j-1}$ for $j\ge 3$. Direct division gives $u_{j+1}\mathrm{/}{u}_j=bh(j+1)(1+1\mathrm{/}j{)}^{j-1}\le ebh(j+1)$. For $j<N$, $h(j+1)\le hN\le a$, so this ratio is at most $e{C}_0\mathrm{/}{C}_2<1\mathrm{/}2$. Therefore ${\sum }_{j=3}^N{u}_j\le 2u_3=18b^2{h}^2$, with the empty sum understood as zero.

Hamiltonian comparison. The bounded set $D(\rho \mathrm{/}8)$ has $R={\sup }_{Z\in D(\rho \mathrm{/}8)}\mathrm{\parallel }Z\mathrm{\parallel }<\mathrm{\infty }$, since $U$ is bounded. By [F1], [F2] and step 1.1, $\mathrm{\mid }{H}_h-\tilde{H}\mathrm{\mid }\le R\sqrt{2d}(8M\mathrm{/}3)18b^2{h}^2$, and the same bound without $R$ controls $\mathrm{\parallel }\mathrm{\nabla }(H_h-\tilde{H})\mathrm{\parallel }$. As $\mathrm{\parallel }{X}_{\tilde{H}}\mathrm{\parallel }\le 2M$ and $J$ is an isometry, $\mathrm{\parallel }\mathrm{\nabla }\tilde{H}\mathrm{\parallel }\le 2M$. The gradient is therefore bounded independently of $h$ after fixing $h\le a\mathrm{/}3$. Reality, analyticity and normalization follow term by term in the finite sum.