Order of a corrector-predictor method Given an explicit method:
$$ x_{i+1} = x_i+ h \Phi(t_i,x_i,h) $$
as predictor method and an implicit method:
$$ x_{i+1} = x_i + h \Psi(t_i,x_i,x_{i+1},h) $$
as corrector method, it follows that
$$
x_{i+1} = x_i + h \Psi(t_i,x_i,x_{i+1}^*, h), \quad
(x^*_{i+1} = x_i+\Phi(t_i,x_i,h))
$$
is an explicit predictor-corrector method. If $\tau_p(h)$, $\tau_c(h)$ and $\tau_{pc}(h)$ denote respectively the local truncation error of predictor, corrector and predictor-corrector method, then
$$ \tau_{pc}(h) = \tau_c(h) + \mathcal O(h \tau_p(h)) $$
(which is what I haven't proved).
From the definition of local truncation error, we have:
$$
\begin{array}{rlrl}
x^{(*)}(t_{i+1}) &= x(t_i) + h \Phi(t_i, x(t_i), h) + h \tau_p(t_i,h),& \tau_p(h) &= \max_i \tau_p(t_i,h)\\
x(t_{i+1}) &= x(t_i) + h \Psi(t_i, x(t_i), x(t_{i}+h), h) + h \tau_c(t_i,h),& \tau_c(h) &= \max_i \tau_c(t_i,h)\\
x(t_{i+1}) &= x(t_i) + h \Psi(t_i, x(t_i), x^*(t_i+h), h) + h \tau_{pc}(t_i,h),& \tau_{pc}(h) &= \max_i \tau_{pc}(t_i,h)\\
\end{array}
$$
So, substracting from the third equation the second ecuation, and since $h > 0$:
$$ \tau_{pc}(t_i,h) = \tau_c(t_i,h) - [ \Psi(t_i,x(t_i),x(t_{i+1}),h)-\Psi(t_i,x(t_i),x^*(t_{i+1}),h)] $$
And only is needed to prove that the last substraction is $\mathcal O(h \tau_p(h))$. So,
$$
\|\Psi(t_i,x(t_i),x(t_{i+1}),h)-\Psi(t_i,x(t_i),x^*(t_{i+1}),h)\|\leq
L\|x(t_{i+1})-x^*(t_{i+1})\|
$$
$$
\leq L\|x(t_{i+1})-x(t_i)-h \Phi(t_i,x(t_i),h)-h\tau_p(t_i,h)\|\leq L\|...\| + Lh\|\tau_p(h)\|
$$
where $L$ is a Lipschitz constant.
How could I justify this? Any hint? Thanks in advance.

Pure predictor and pure corrector schemes
$$
\begin{array}{rlrlc}
x(t_{i+1}) &= x(t_i) + h \Phi(t_i, x(t_i), h) + h \tau_p(t_i,h),& \tau_p(h) &= \max_i |\tau_p(t_i,h)| & (1)\\
x(t_{i+1}) &= x(t_i) + h \Psi(t_i, x(t_i), x(t_{i+1}), h) + h \tau_c(t_i,h),& \tau_c(h) &= \max_i |\tau_c(t_i,h)| & (2)
\end{array}
$$
And the predictor-corrector scheme
$$
\begin{array}{rlrlc}
x^* &= x(t_i) + h \Phi(t_i, x(t_i), h) & & & (3)\\
x(t_{i+1}) &= x(t_i) + h \Psi(t_i, x(t_i), x^*, h) + h \tau_{pc}(t_i,h),& \tau_{pc}(h) &= \max_i |\tau_{pc}(t_i,h)| & (4)
\end{array}
$$
Subtracting $(2)$ from $(4)$:
$$
\tau_{pc}(t_i, h) = \tau_{c}(t_i, h) + \Psi(t_i, x(t_i), x(t_{i+1}), h) - \Psi(t_i, x(t_i), x^*, h)
$$
Thus
$$
\tau_{pc}(h) \leq \tau_{c}(h) + L ||x(t_{i+1}) - x^*||
$$
Now subtract $(1)$ from $(3)$ and get the desired result.
