= Implicit midpoint Lipschitz estimate
{title2=$\operatorname{Lip}(\Psi_h)\leq(1+hL/2)/(1-hL/2)$}
For a uniformly <Lipschitz continuous> state field $f$ with constant $L$, an <implicit midpoint rule> step is uniquely solvable when $hL/2<1$, by the <contraction mapping theorem>. Write its step map as $\Psi_h(u)=u+h\Phi_h(u)$, where $\Phi_h(u)=f(t+h/2,u+h\Phi_h(u)/2)$. Subtracting the increment equations gives
$$
(1-hL/2)\|\Phi_h(u)-\Phi_h(v)\|\leq L\|u-v\|.
$$
Therefore $\operatorname{Lip}(\Phi_h)\leq L/(1-hL/2)$, and the <triangle inequality> gives the bound for $\Psi_h$. Combining it with the <implicit midpoint consistency bound> yields
$$
\|e_n\|\leq R_h^n\|e_0\|+\frac{Kh^2}{L}(R_h^n-1),
\qquad R_h=\frac{1+hL/2}{1-hL/2}
$$
for $L>0$. This follows by summing the geometric error recurrence with forcing $Kh^3/(1-hL/2)$. For $hL\leq1$ and $nh\leq T$, $R_h^n\leq e^{2LT}$, proving second-order <global error> with exact starting data. If $L=0$, the direct bound is $\|e_n\|\leq\|e_0\|+KTh^2$.
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