Let $$|\mu(x,t)|+|\sigma (x,t)|\leq K(1+|x|),$$ for all $x\in\mathbb R$ and some $K>0$. Let $$\begin{cases}d X_t=\mu(X_t,t)\,\mathrm d t+\sigma (X_t,t)d W_t\\X_0=x_0\in\mathbb R\end{cases}.$$
I would like to prove that $\sup_{0\leq s\leq t}\mathbb E(|X_s|^2)<\infty $
Using Itô formulae gives $$X_t^2=x_0^2+2\int_0^tX_s\mu(X_s,s) d s+\underbrace{\int_0^t \sigma (X_s,s)d W_s}_{=:M_t}+\int_0^t\sigma (X_s,s)^2 d s.$$ Let $\tau_n\nearrow \infty $ s.t. $(M_{\tau_n\wedge t})$ is a martingale. Then $$\mathbb E[X_t^2]\leq x_0^2+2\mathbb E\int_0^{t\wedge \tau_n} X_s\mu(X_s,s) d s+2\mathbb E\int_0^{t\wedge \tau_n}(1+X_s^2)\,\mathrm d s.$$
Q1) How can I conclude ?
Moreover, I have to deduce that $\mathbb E(|X_t|^2)\leq Ce^{Dt}$ for some $C,D>0$. So, I could easily prove that $$\mathbb E(|X_t|^2)\leq C+D\int_0^t\mathbb E(|X_s|^2)ds,$$ for some $C,D>0$, and thus the result follows from Gronwall's inequality.
Q2) Nevertheless, I wonder why $\sup_{0\leq s\leq t}\mathbb E(|X_s|^2)<\infty $ is required here ?
Q3) Moreover, in my Gronwall inequality, continuity of $s\mapsto \mathbb E(|X_s|^2)$ is continuous, and I failed to prove it. Does the continuity of $s\mapsto \mathbb E(|X_s|^2)$ comes from $\sup_{0\leq s\leq t}\mathbb E(|X_s|^2)<\infty $ ?