I am trying to undestand a proof that says that based on standard estimates, applying Ito formula and Gronwall lemma we obtain
$$E|X_{t}^{2}|\leq Ce^{Ct}(1+|x|^{2})$$
I was trying to prove this estimate but I could not finish:
We have the SDE:
$$dX_{t}=b(X_{t})dt + \sigma(X_{t})dW_{t}$$
Applying Ito's lemma to $F(x)=x^{2}$ I obtain
$$X_{t}^{2}=x^{2} + \int_{0}^{t}(2b(X_{t})X_{t} + \sigma^{2}(X_{t}))dt + \int_{0}^{t}2\sigma(X_{t})X_{t}dW_{t}$$
Taking absulute value and using the fact that $b$ and $\sigma$ have linear growth, i.e $|b(X_{t})|\leq C_{1}(1+|X_{t}|)$ and $|\sigma(X_{t})|\leq C_{2}(1+|X_{t}|)$ I get
$$|X_{t}^{2}|\leq |x|^{2} + \int_{0}^{t}(2C_{1}(|X_{t})| + |X_{t}|^{2}) + C_{2}^{2}(1+|X_{t}|)^{2})dt + \int_{0}^{t}2C_{2}(|X_{t}| + |X_{t}|^{2})dW_{t}$$
I dont know how to continue from here.
I think using Itô creates unnecessary difficulties. However you could proceed as follow:
$$|X_t|^2 \le 3\left(x^2 + \left(\int_{0}^{t} b(X_s) ds\ \right)^2 + \left(\int_{0}^{t} \sigma(X_s) dW_s\ \right)^2 \right)$$
Where I used the classical inequality : $(a+b+c)^2 \le 3(a^2+b^2+c^2)$. Then by Cauchy-Schwartz inequality, Fubini and linear growth assumption:
$$\mathbb{E}\left( \left(\int_{0}^{t} b(X_s) ds\ \right)^2 \right) \le t \int_{0}^{t} \mathbb{E}(b(X_s)^2) ds \le C_t^1 \int_{0}^{t} \mathbb{E}(|X_s|^2) ds$$
Besides by using BDG's inequality, Fubini and linear growth assumption:
\begin{align*} \mathbb{E}\left( \left(\int_{0}^{t} \sigma(X_s) dW_s\ \right)^2 \right) &\le \mathbb{E}\left(\underset{u \le t}{\sup}\left(\int_{0}^{u} \sigma(X_s) dW_s \right)^2 \right)\\ &\le \int_{0}^{t} \mathbb{E}(\sigma(X_s)^2) ds\\ &\le C_t^2 \int_{0}^{t} \mathbb{E}(|X_s|^2) ds \end{align*}
Therefore
$$\mathbb{E}(|X_t|^2) \le 3\left(x^2 + C_t^1 \int_{0}^{t} \mathbb{E}(|X_s|^2) + C_t^2 \int_{0}^{t} \mathbb{E}(|X_s|^2) \right)$$
So that there exist $K > 0$ such that:
$$\mathbb{E}(|X_t|^2) \le K\left(1 + x^2 + \int_{0}^{t} \mathbb{E}(|X_s|^2)\right)$$
And you conclude by using Gronwall lemma.