Suppose that we have a general SDE on a probability space $(\Omega,\mathcal{F},P)$ defined by: $$ dX_t = b(t,X_t) dt + \sigma(t,X_t) d W_t, $$ where $W$ is a Brownian motion and $b$ and $\sigma$ are Borel-measurable functions. Furthermore suppose that we have a strong solution $X$ to the SDE with initial condition $\xi$ if $X_0 = \xi$, $X$ has continuous paths, is $\mathbb{F}$ adapted, $$ \int_0^t (|b(s,X_s)| + \sigma(s,X_s)^2) ds < \infty \text{ for all $t \geq 0$ a.s.} $$ and it satisfies the SDE.
Furthermore suppose we have $b$ and $\sigma$ such that they are Lipschitz continuous in the second variable, i.e. there exists $K>0$ such that $$ |b(t,x)-b(t,y)| + | \sigma(t,x)-\sigma(t,y)| \leq K |x-y| \text{ for all $t \geq 0$, $x,y \in \mathbb{R}$}.$$
I now want to prove that if $X$ and $Y$ are strong solutions to the SDE then for all $T$ there exists a constant $D$ such that $$ \mathbb{E} || X- Y ||_t^2 \leq D \mathbb{E} | X_0 - Y_0 |^2, \text{ for all } t \leq T.$$
To begin proving this given $X$ and $Y$ we define $$ U_t(X) = \xi + \int_0^t b(s,X_s)ds + \int_0^t \sigma(s,X_s) d W_s, $$ and $U_t(Y)$ likewise. Note that we can write $X=U(X)$. Now since $(p+q)^2 \leq 2 (p^2 + q^2)$ for any $p,q \in \mathbb{R}$ then for $t \leq T$ we obtain that \begin{align*}&|U_t(X)-U_t(Y)|^2 \leq \\ &2\left( \int_0^t |b(s,X_s)-b(s,Y_s)|ds\right)^2 + 2\left(\int_0^t |\sigma(s,X_s)-\sigma(s,Y_s)|dW_s\right)^2 \\ &\leq 2\left( \int_0^t |b(s,X_s)-b(s,Y_s)|ds\right)^2 + 2\sup_{u \leq t}\left(\int_0^t |\sigma(s,X_s)-\sigma(s,Y_s)|dW_s\right)^2. \end{align*} Hence by Doob's $L^2$ inequality and Cauchy Schwarz: \begin{align*} &\mathbb{E} ||U(X) - U(Y) ||^2_t \leq \\ &2\mathbb{E}\left( \int_0^t |b(s,X_s)-b(s,Y_s)|ds\right)^2 + 8\mathbb{E}\left(\int_0^t |\sigma(s,X_s)-\sigma(s,Y_s)|dW_s\right)^2. \end{align*} If we then follow this line of thought and use Lipschitz continuity we obtain that $$ \mathbb{E} ||U(X) - U(Y) ||^2_t \leq 2K^2 (T+4) \int_0^t \mathbb{E} || X-Y||^2_s ds.$$
Now to obtain our result I think we have to use $(a+b+c)^2 \leq 3(a^2+b^2+c^2)$ and mimic the proof above then use Gronwall's inequality to complete the proof. But I can't seem to figure out the details, could anyone help me with those? Thanks for any help.
By definition,
$$U_t(X)-U_t(Y) = (X_0-Y_0) + \int_0^t (b(s,X_s)-b(s,Y_s)) \, ds + \int_0^t (\sigma(s,X_s)-\sigma(s,Y_s)) \, dW_s$$
and therefore
$$\begin{align*}&|U_t(X)-U_t(Y)|^2 \leq \\ & \color{red}{3 |X_0-Y_0|^2} + 3 \left( \int_0^t |b(s,X_s)-b(s,Y_s)|ds\right)^2 + 3\left(\int_0^t |\sigma(s,X_s)-\sigma(s,Y_s)|dW_s\right)^2. \end{align*}$$
The first term (colored in red) is missing in your calculation. Proceeding as in your proof, we obtain the inequality
$$\begin{align*} \mathbb{E}\|X-Y\|_t^2 &= \mathbb{E}\|U(X)-U(Y)\|_t^2 \\ &\leq 3\mathbb{E}(|X_0-Y_0|^2) + 3K^2 (T+4) \int_0^t \mathbb{E}\|X-Y\|_s^2 \, ds. \end{align*}$$
Applying Gronwall's lemma gives
$$\mathbb{E}\|X-Y\|_t^2 \leq 3\mathbb{E}(|X_0-Y_0|^2) \cdot \exp \left( 3K^2 (T+4) T \right)$$
for any $t \leq T$.