$X_t$ is a process defined by an SDE:
$d X_t=\mu\left(t, X_t\right) d t+\sigma\left(t, X_t\right) d B_t$
Under the global Lipschitz continuity and linear growth conditions, suppose that $E\left[\|\xi\|^{2 p}\right]$ is finite for some $p \geq 1$. Show that for any finite $T>0$, there exists a constant $C$, depending only on $K, T, p$, such that $$ E\left[\sup _{0 \leq t \leq T}\left\|X_t\right\|^{2 p}\right] \leq C\left(1+E\left[\|\xi\|^{2 p}\right]\right) . $$
Does this estimate still hold if global Lipschitz continuity is replaced by local Lipschitz continuity while the linear growth condition remains unchanged?
Hint: Use the following vector inequality $\left(\left|a_1\right|+\cdots+\left|a_n\right|\right)^{2 p} \leq n^{2 p}\left(\left|a_1\right|^{2 p}+\cdots+\left|a_n\right|^{2 p}\right)$.
A related theorem and conclusions are:
Theorem (existence of strong solutions). Suppose $\mu, \sigma$ satisfy the global Lipschitz and linear growth conditions $$ \begin{aligned} & \|\mu(t, x)-\mu(t, y)\|+\|\sigma(t, x)-\sigma(t, y)\| \leq K\|x-y\|, \\ & \|\mu(t, x)\|^2+\|\sigma(t, x)\|^2 \leq K^2\left(1+\|x\|^2\right), \end{aligned} $$ for every $t \geq 0$ and $x, y \in \mathbb{R}^n$. Let $\xi$ be a $\mathbb{R}^n$ random vector, independent of $B$, and with finite second moment $E\left[\|\xi\|^2\right]<\infty$. Then there exists a continuous, adapted process $X$ which is a strong solution of equation (1) with initial condition $\xi$. Moreover, this process is squareintegrable: for any finite $T>0$, there exists a constant $C$, depending only on $K$ and $T$, such that $$ E\left[\left\|X_t\right\|^2\right] \leq C\left(1+E\left[\|\xi\|^2\right]\right) e^{C t}, 0 \leq t \leq T . $$
Remark 1. Global Lipschitz continuity implies linear growth. To see this, just set $y=0$ in the Lipschitz continuity condition and we obtain $$ \begin{aligned} \|\mu(t, x)\| & \leq\|\mu(t, 0)\|+K\|x\|, \\ \|\sigma(t, x)\| & \leq\|\sigma(t, 0)\|+K\|x\| . \end{aligned} $$
If we replace global Lipschitz continuity by local Lipschitz continuity in Theorem 2 while keeping the other conditions, the conclusions of Theorem 2 still hold (Friedman (1975), Chapter 5, Theorem 2.2). The proof is based on localizing the initial condition and $\mu(t, x)$ and $\sigma(t, x)$.
In class I learned about the proof of the global Lipschitz + linear growth version as below, and I tried to replicate that but I'm not quite sure how these conditions play a role, especially when changing to local Lipschitz:
For $0 \leq t \leq T$, $$E\left[\left\|x_t^{(k+1)}\right\|^2\right] \leqslant 9 E\left[\|\xi\|^2\right]+9 K^{\top} \int_0^t\left(1+E\left[\left\|X_s^{(k)}\right\|^2\right]\right) ds$$
Let $$C=\max (9,9 T K^2, 9 T^2 K^2 ).$$ we have
$$E\left[\left\|X_t^{(k)}\right\|^2\right] \leqslant C\left(1+E\left[\|\xi\|^2\right]\right)+C \int_0^t E\left[\left\|X_s^{(k)}\right\|^2\right] d s$$
$$\leqslant C\left(1+E\left[\|\xi\|^2\right]\right)+C \cdot C\left(1+E\left[\|\xi\|^2\right]\right) t$$
$$+C \int_0^t C \int_0^1 E\left[\left\|X_u^{(k-1)}\right\|^2\right] d u d s$$
Applying the estimate repeatedly
$$\leqslant C\left(1+E\left[\|\xi\|^2\right]\right)\left(1+ C t+\cdots+\frac{(C t)^{k+1}}{(k+1) !}\right)$$
$$\leqslant C\left(1+E\left[\|\xi\|^2\right]\right) e^{C t}$$
