I have been give the following problem.
Let X be a solution to the SDE $X(t) = X(0) + \int_0^1 b(X(s))ds + \int_0^1 \sigma(X(s)) dW(s)$ where $W$ is a standard one-dimensional Brownian motion, and the functions $b : \mathbb{R} \rightarrow \mathbb{R}$ and $\sigma : \mathbb{R} \rightarrow \mathbb{R}$ have at most linear growth; that is, there is a constant $K$ such that $|b(x)| \leq K(1 + |x|), |\sigma(x)| \leq K(1 + |x|)$. Suppose that $E|X(0)|^2 < \infty$. Show that there is a constant $0 < C_2(T) < \infty$ such that
$$E [\sup_{0≤t≤T} |X(t)|^2]< C_2(T).$$
The way I am trying to prove this statement is by first finding $d|X(t)|^2$ by using Ito's Lemma. Then take the expectation of $d|X(t)|^2$ and bound this expectation in a way so that I can use Gronwall's Inequality. From here, I think I would be able to arrive at the conclusion of the proof. Below is a sketch of the work I have done so far.
Sketch of my work: First, we find that
$$d|X(t)|^2 = [2X(t) b(X(t)) +\sigma^2(X(t))]dt + 2X(t)\sigma(X(t)) dW(t).$$
Taking the expectation, we have
$$dE|X(t)|^2 = E[2X(t) b(X(t)) +\sigma^2(X(t))]dt.$$
We can bound this expectation by using the growth conditions on $b$ and $\sigma$ to get
$$E|X(t)|^2 \leq E|X(0)|^2 + \int_0^t E[2X(s)K(1+|X(s)|) +K^2(1+|X(s)|)^2]ds.$$
This is the point were I am getting stuck. I am unsure how I can manipulate the right side of the inequality to achieve something that will resemble Gronwall's Inequality. So far, does my work seem correct? Is this a valid way of trying to prove this statement? Any advice on how to proceed?
Yes, this looks correct so far.
It looks like the parts that don't immediately fit Gronwall's are the $E[|X(s)|]$ term and the $E[(1+|X(s)|)^2]$ term. For the first, you can just use that $|a| \le 1+a^2$ for all $a \in \mathbb{R}$. For the second, you can use that $(a+b)^2 \le 2(a^2+b^2)$ for all $a,b \in \mathbb{R}$. This should be the exact form you need to apply Gronwall's inequality to the function $E[|X(t)|^2]$.