Proof for moment estimates of Ito Stochastic Differential Equation Given a Ito-SDE
$$dX_t=a(t,X_t)dt+b(t,X_t)dW_t$$
fulfilling the Lipshitz condition and linear growth condition. I want to proof 
$$E\left[\sup_{t_0\le s \le T} |X_s|^{2n}\right]\le D \cdot (E[|X_{t_0}|^{2n}]+(1+E[|X_{t_0}|^{2n}]) \cdot  (T-t_0)^n \exp(C(T-t_0)))$$
with some constant D.
My suggestion is using Doob Inequality. However the Doob inequality is only valid for the time range $[0,T]$ and here we have $[t_0,T]$
Maybe following Theorem should be helpful:
$$E[|X_t|^{2n}] \le (1+E[|X_{t_0}|^{2n}]) \exp(C(t-t_0))$$
and
$$E[|X_t-X_{t_0}|^{2n}]\le D \cdot  (1+E[|X_{t_0}|^{2n}]) \cdot (t-t_0)^n \cdot  \exp(C(t-t_0))$$
 A: Let us first assume that $X_{t_0}=0$. By Itô's formula, we have
$$\begin{align*} &|X_t|^{2n} \\ &= 2n \int_{t_0}^t |X_s|^{2n-1} b(s,X_s) \, dW_s +n\int_{t_0}^t |X_s|^{2n-2} \left( (n-1) b(s,X_s)^2 + 2 a(s,X_s) |X_s| \right) \, ds \\ &=: M_t + N_t.\end{align*}$$
Clearly,
$$\mathbb{E} \left( \sup_{t_0 \leq t \leq T} |X_t|^{2n} \right) \leq \mathbb{E} \left( \sup_{t_0 \leq t \leq T} |M_t| \right) + \mathbb{E} \left( \sup_{t_0 \leq t \leq T} |N_t| \right). \tag{1}$$
We estimate both terms separately. For the first term we can use $(|M_t|)_{t \geq 0}$ is a submartingale and Itô's isometry to conclude
$$\mathbb{E} \left( \sup_{t_0 \leq t \leq T} |M_t| \right) \leq \mathbb{E}(|M_T|) \leq \sqrt{\mathbb{E}(M_T^2)} = 2n \sqrt{\int_{t_0}^T \mathbb{E}(|X_s|^{4n-2} b(s,X_s)^2) \, ds}.$$
Since $b$ is of at most linear growth, we find that there exists a constant $C_1>0$ such that
$$\mathbb{E} \left( \sup_{t_0 \leq t \leq T} |M_t| \right) \leq C_1 \sqrt{\int_{t_0}^T \mathbb{E}(1+|X_s|^{4n}) \, ds}.$$
Now you can use the upper bound for $\mathbb{E}(|X_s-X_{t_0}|^{4n})= \mathbb{E}(|X_s|^{4n})$ which you mentioned at the very end of your question to obtain a suitable estimate for the first term on the right-hand side of $(1)$. For the second one, we note that by the linear growth condition
$$\mathbb{E} \left( \sup_{t_0 \leq t \leq T} |N_t| \right) \leq C_2 \mathbb{E} \left[ \int_{t_0}^T (1+|X_s|^{2n}) \, ds \right].$$
Again the upper bound for $\mathbb{E}(|X_s-X_{t_0}|^{2n})$ provides a suitable estimate for this term, and this proves the desired inequality.
For the general case, i.e. if $X_{t_0} \neq 0$, just note that, by Hölder's inequality,
$$|X_t|^{2n} = |X_t-X_{t_0} + X_{t_0}|^{2n} \leq c |X_t-X_{t_0}|^{2n} + c |X_{t_0}|^{2n}$$
for a suitable constant $c>0$. Consequently,
$$\mathbb{E} \left( \sup_{t_0 \leq t \leq T}|X_t|^{2n} \right) \leq c \mathbb{E}(|X_{t_0}|^{2n}) + c \mathbb{E} \left( \sup_{t_0 \leq t \leq T} |X_t-X_{t_0}|^{2n} \right)$$
and now the claim follows from the first part of the proof.
